by Judith Curry
Last month I attended a Workshop on The Ethics of Communicating Scientific Uncertainty: Understanding How Scientists, Environmental Lawyers, and Journalists Treat Uncertainty.
The Workshop was organized by the Environmental Law Institute, their website for the Workshop is [here]. On the site is a webinar How Professional Standards Shape Scientists’, Lawyers’, and Journalists’ Approaches to Uncertainty (plus ppt slides), which the participants watched prior to the Workshop.
About 50 individuals attended the Workshop, roughly equal numbers from the 3 groups. In terms of WG1-topic climate scientists, there was only one other in attendance besides myself. The Workshop was funded by the National Science Foundation Paleoclimate Program. Huh? I was sort of wondering about that, but I quickly understood after meeting the Program Manager, Dave Verardo. Fascinating; I hope to do a Q&A with Dave Verardo in the near future.
The ELI will make all materials/transcripts from the Workshop publicly available, and will write a synthesis report. Below I provide excerpts from some of the material we received and my own reflections. I will do a Part 2 on this once the additional materials from the ELI are available.
Webinar
Here are some notes and punchlines from the Webinar:
Scientists’ perspective, summary points:
- Disagreeement arises from differences in science, interpretation of science, judgment about acceptability of risk, others?
- Risk means uncertainty: causal relationships, likelihood of occurrence, consequences
- Scientific norms encourage disclosure of sources and magnitude of uncertainty
- Communicating uncertainty is hard
Lawyers’ perspective:
- “A court proceeding, such as a trial, is not simply a search for dispassionate truth. The law must be fair.”
- Two competing mandates: i) Advocacy for client (present the case with persuasive force); ii) Candor to the tribunal (avoid false evidence)
- Burden of proof: civil cases – preponderance of evidence; criminal cases – beyond reasonable doubt
- Scientific evidence is traditionally presented by experts
Journalists’ perspective:
- Traditional values: be fair, be accurate, be engaging
- Fewer than 10% of journalists are skeptical about climate change, but more than one third of journalists believed that coverage of climate change science must reflect a balance of viewpoints or present all sides of controversy (half did not believe in the need for balance).
- When are journalists going to start treating climate skeptics like flat earthers? Not soon.
- At best, journalists handle uncertainty by: Deeply researching explanations and explaining which are probable and why some are wrong or unlikely; Reporting potential sources of bias of those advocating for particular explanations; Engaging an audience
Reading list
We were given a reading list of 12 papers, below are excerpts from the ones I found most interesting.
How Rachel Carson Spurred Chemical Concerns by Highlighting Uncertainty, by Andy Revkin. Excerpts:
But of all the fresh considerations of Carson’s work, there’s one that stands out for me at the moment — a recent paper by two researchers of rhetoric and writing who dug in on “Silent Spring” drafts, notes and revisions and found that Carson had a remarkable and rare trait for someone so committed to raising public concern about a pressing environmental issue. Rather than downplay scientific uncertainty and gaps in understanding, she progressively amplified what was unclear about the human impacts of DDT and other synthetic compounds on humans and wildlife.
The authors, Kenny Walker, a doctoral candidate at the University of Arizona, and Lynda Walsh, an assistant professor of English at the University of Nevada, Reno, describe how surprised they were by this pattern, given a large body of work, including “Merchants of Doubt,” the book by Naomi Oreskes and Erik Conway, showing that inflating uncertainty has mainly been the task of industry-backed efforts to blunt public concern and policy.
With Carson’s approach to conveying risk, they write, she appears to have created “a bridge across the is–ought divide in science-related policy making, using the uncertainty topos to invite the public to participate by supplying fears and values that would warrant proposals for limiting pesticide use.”
This stands in stark contrast to the “Be Worried” approach that some have tried (in vain) on global warming over the years. Mind you, Carson’s book has plenty of passages, from beginning to end, warning powerfully of the “grim specter” of a poisoned future.
FASCINATING article, read the whole thing. Back in the days, circa 2006/2007 when I was not skeptical of AGW, i did emphasize uncertainty in my public presentations, and I did not find that this lessened concern or left anyone thinking that this wasn’t a problem worthy of serious consideration. They left the room trusting me because I was honest about the uncertainties.
Journalism ethics and climate change reporting in a period of intense media uncertainty, by Bud Ward. Excerpts:
Consider another aspect of journalistic ethics as it applies to covering climate change. One element of the SPJ Code of Ethics urges reporters to ‘give voice to the voiceless; official and unofficial sources of information can be equally valid’. Another cautions reporters to ‘support the open exchange of views, even views they find repugnant’ (note that it is ‘repugnant’ and not ‘factually inaccurate’ here).
In reporting on climate change and the findings in the physical and earth sciences defining it, US reporters for many years practiced what critics contend is a ’false balance’, providing space disproportionate to its scientific credibility to perspectives running counter to what is now widely accepted as the ‘established’ scientific judgment. In effect, reporters may for too long have been balancing opinions about science when in fact they might better have been evaluating and reporting evidence based on the science. Accuracy can trump balance in such a case, so that one perspective gets 90% of the column inches, based on the standard of evidence, and another perhaps 5 or 10%, or maybe none at all.
Just as that approach to covering climate science based on evidence and not sheer opinion has changed throughout much of the western world, it appears in recent years to have also changed, albeit much later, among many US news reporters.
So, instead of the over-simplified notion of providing ’balance’ in reporting on news involving differing perspectives, journalists increasingly, and rightly, take their clues from the leading and acknowledged scientific experts when it comes to the facts and causes of global climate change. That means, in effect, reporting as a given—until science shows otherwise—that warming of Earth actually is occurring and that human activities have a significant role, though not the only one, in that warming.
Issues of journalism ethics in dealing with climate change go much farther than that. Assume, for the purposes of discussion, that the effects and impacts of global warming just may be as dire as a number of leading scientists suggest, threatening not only vast ecological systems and the natural resources dependent on them, but also posing great risks of accelerated extinctions, forced human migrations from low-lying lands, diminished food resources for a growing population, and so forth. What then are the ethical responsibilities facing news reporters? Is it up to them to sustain the clarion call by way of front-page headlines and repeated broadcast ‘breaking news’ alerts, despite what some observers now dismiss as ‘climate fatigue’ on the parts of their already-harried audiences? Isn’tthat too much like making and not merely reporting the news that others—acclaimed scientists or leading policy makers—have as part of their portfolios?
So what if the nature of the threatened worst-case climate change outcomes are to be most clearly manifested only for future generations, or only as they directly affect audiences geographically far removed from one’s own readers or viewers? Do the journalists’ ethical responsibilities differ if it is just ‘some other population’ (or perhaps even some other species) that is at greatest risk, and not the one closest to them in time and space?
Journalists have profound ethical responsibilities covering issues as expansive and critical (not many are, perhaps) as climate change. That they are dealing with these issues during a time of profound change in their own field, and during a time of profound global economic and financial uncertainty, compounded by ongoing threats of divisive wars and terrorist activities, only confounds their approach to these issues.
JC comment: This article emphasizes the colossal and unnecessary damage to science, public debate and policy making by the Oreskes ‘merchant of doubt’ inspired denier witch hunt. This situation is getting out of control by a new Oreskes documentary and book, I’m planning a future post on this.
When questions of science come to a courtroom, by Cornelia Dean. Excerpts:
Idealistic lawyers and idealistic scientists often describe themselves as engaging in a search for truth.
The scientists follow the scientific method. They state their hypotheses, describe the ways they test them, present their findings — and wait for another researcher to prove them wrong. Lawyers’ practice is built on the idea that the best way to shake the truth out of a complex dispute is for advocates on each side to argue it, as vigorously as they can, in front of an impartial judge or jury.
These approaches work more or less well on their own. But when a legal issue hinges on questions of science, they can clash. And the collision can resound all the way up to the Supreme Court.
Last Wednesday, the nine justices heard arguments in the first global warming case to come before the court. Massachusetts, 11 other states and several cities and environmental groups are saying that the federal Environmental Protection Agency has ignored the requirements of the Clean Air Act and otherwise shirked its responsibilities by failing to regulate emissions of heat-trapping gases, chiefly carbon dioxide.
But much of the argument hinged on scientific questions. Is the earth’s climate changing? If so, are human activities contributing to the change?
Mainstream science has answers to these questions (yes and yes). But while it is impossible to argue that earth has not warmed up a bit in the last century, there are still some scientists with bright credentials and impressive academic affiliations who argue that people don’t have much do to with it.
One issue is the standard of proof. Typically, scientists don’t accept a finding unless, statistically, the odds are less than 1 in 20 that it occurred by chance. This standard is higher than the typical standard of proof in civil trials (“preponderance of the evidence”) and lower than the standard for criminal trials (“beyond a reasonable doubt”).
The justices may also consider that when scientists confront a problem, they collect all the information they can about it and then draw conclusions.
Lawyers work in reverse. They know their desired outcome at the outset, so they gather arguments to support it. While it would be unethical for scientists reporting on their work to omit findings that don’t fit their hypotheses, lawyers are under no compunction to introduce evidence that hurts their cases; that’s the other side’s job.
Perhaps the knottiest problem, though, has been deciding what scientific evidence or testimony should be considered in the first place.
For years, the standard was “general acceptance” by scientists, which a federal appeals court enunciated in 1923 in Frye v. United States, a case involving lie detectors. The court ruled that lie detector technology had yet to win wide acceptance and barred its use.
In these rules, the test became not wide acceptance, but whether the scientific, technical or other specialized information would assist the judge or jury in reaching a decision and whether witnesses seeking to testify about it had enough knowledge or expertise to make a valuable contribution.
Critics of this standard say it flooded courtrooms with junk science, as people with good (or seemingly good) credentials but bad ideas took the stand before judges and juries unable to differentiate between credible scientific claims and those with only an aroma of scientific respectability.
But even if lawyers and judges could routinely absorb a thorough grounding in the scientific issues they confront, there would still be trouble. For one thing, the state of scientific knowledge changes rapidly. Sometimes, there are multiple scientific views of a given issue, all potentially credible. And sometimes research on an issue does not even begin until it works its way into court. To an extent, that was the case with the silicone breast implants.
But without a method of providing courts with reliable scientific information, scientific research has an uncertain role in the courts, and especially the Supreme Court, as David L. Faigman put it in “Laboratory of Justice” (Times Books, 2004).
JC comment: This article raises the really key issues here, particularly: They know their desired outcome at the outset, so they gather arguments to support it. While it would be unethical for scientists reporting on their work to omit findings that don’t fit their hypotheses, lawyers are under no compunction to introduce evidence that hurts their cases; that’s the other side’s job. While this describes lawyers, it also describes scientist-advocates; in principle this behavior is ok, but in the legal system the ‘other side’ has a fair chance to make their case; in the climate change debate, the ‘other side’ is smeared as ‘deniers.’
JC reflections
This was a fascinating Workshop; I went into this with more of a philosophy of science perspective on uncertainty and came away with a more ‘real world’ perspective.
One issue that came up in a break out group was the different codes of conducts. Lawyers and journalists have clear codes of conducts, scientists do not (beyond admonitions related to research misconduct – fabrication, falsification, plagiarism). Scientists employed by the government have some codes of conduct to adhere to; academic scientists do not. As a result there is some pretty irresponsible public behavior by academic and think tank/advocacy group scientists, and there are absolutely no repercussions (I’m sure we can all think of examples).
In any event, much food for thought, and I expect to have a follow up post on this once the Workshop summaries are written up.
You have a pretty key issue here, related to public debate and information.
Lawyers can be one-sided in courtroom because the “other side” will have the same means to present their case. Scientists are encouraged to become activist-scientists because of thermaggedon. But, most of the media think there is no need to present both sides of the controversy. So, you get a mixture of systems with the perfect strategy to avoid any critical thinking.
In a courtroom, a lawyer presenting clearly false or absurd “evidence” will be called out by the opposition, will look totally stupid, and might lose the case on that basis. So there is clear incentive to only present evidence and arguments that can be justified. That is why there is the rule that the prosecution must share evidence favorable to the defense and outrage when prosecutors make stuff up. There are no such checks and balances in the climate debate because the media have picked a side. It becomes like a kangaroo court or an old West hanging judge.
As I understand the process, the Burden of Proof falls on the Prosecution and is summarized in their opening argument. If the Prosecution falls to present a compelling opening argument, the Defense can simply call for a Directed Verdict which will require the Judge to either dismiss the suit or rule.
Pre-trial disclosure levels the playing field prior to opening argument.
Given the extensive uncertainties in Climate Science, as documented by the ARs, its highly unlikely anyone would pursue a suit based on Climate Change (warming or cooling).
Suits related to NGOs which intentionally mislead the public and policy decision-makers is another matter as their intent creates damages. A Class Action suit Public vs NGOs would be interesting. An International suit Governments vs UNFCCC would also be interesting if its allowed to occur.
“…the [News] media have picked a side.”
Ethics in News Journalism is an interesting topic. News Journalists, other than opinion pieces, are not in the business of reporting Truth. Their sole responsibility is to report facts related to a given situation or event.
Give Journalists the correct facts, or some basis to interpret a situation or event, and then there would be an ethical issue if they chose to ignore the facts in reporting.
The burden falls on the Scientific community to provide the facts. As far as I know, there isn’t any single repository of factual information the News Media can trust so they simply Report the events.
John: the news media simply report the facts? They often publish press releases from Greenpeace with hardly any editing, or simply repeat absurd statements from Al Gore about 20 foot sea level rise–a fact?
Craig: the “facts” relate to an event or a situation which just occurred aka the “news”. PR groups get paid to promote placement of press releases and its not unusual to find individuals like Gore promoting disinformation or incomplete information.
Gore example: consequence of sea level rise based on the complete melt of Greenland glaciers with missing information related to extent over time.
It isn’t particularly scary to discover it will require thousands of years for this rise to occur so Gore left that part unstated. Yet, it doesn’t change the news fact, Gore stated today that sea level rise will (fill in some foolish claim of your choice). The reporter is covering what Gore said not the accuracy of his statement.
The difficulty of analogizing climate debates at any level with legal proceedings is that in court both lawyers have equal standing. Lawyer A does not say his opponent, Lawyer B, is a dunce or denier.
The equivalent argument wrt climate change would be Lawyer A not only calling Lawyer B a dunce and denier, but saying he/she is not even a lawyer.
Judith, it is not you who need to plumb the depths of uncertainty past a given point–it is your opponents who need to be introduced to it.
Tom Fuller,
“Lawyer A does not say his opponent, Lawyer B, is a dunce or denier.”
You don’t spend much time in court do you? :-)
Insults are common, usually preceded by the phrase “with all due respect.” There is one state court judge in Chicago who has a sheet of paper taped to the front of his bench that says “The words ‘with all due respect’ shall not be used in this courtroom.”
@ John
> The reporter is covering what Gore said not the accuracy of his statement
You’ve avoided the issue (what a surprise)
WHY is Gore’s intellectual thuggery considered NEWS ? That’s the issue
Tom –
==> “Lawyer A does not say his opponent, Lawyer B, is a dunce or denier.”
Now that you’ve seen Judith quoted, I imagine that you’ll be the first in line to express your outrage over her comparing people at the Pentagon to holocaust deniers?
Even still, with your expression of outrage a certainty, I hope you won’t mind if I don’t hold my breath waiting for you to get started?
“The equivalent argument wrt climate change would be Lawyer A not only calling Lawyer B a dunce and denier, but saying he/she is not even a lawyer.”
As far as I know, we do have a right to defend ourselves in court. So if we say that, We will leave it to the scientists, it reminds me of being told I do not have a right to represent myself when the point of the proceedings might very well be my fate. Lawyers have ‘monopolized’ their profession by and large. They do have an enforceable ‘code of ethics’ I think. Perhaps the two things go together.
Joshua, since 1988 the consensus has engaged in a sustained battle to deny their opponents standing. That includes your work here over the past few years. It is contemptible. As are you, in my opinion.
Your team has lied about Richard Lindzen, a respected climate scientist with a string of publications and responsible positions to his credit, accusing him falsely of being in the pay of Big Somebody and libeling him, saying his view that the statistics linking second hand smoke to cancer were far weaker than those tying smoking to cancer were denying that smoking caused cancer. Lindzen’s quote: “I actually think it does.”
Your team says that Spencer is religious, hence cannot be granted standing. That Freeman Dyson is old, hence cannot be granted standing. That Ivar Giaevar, Nobel Prize winner in physics, hasn’t specialized in climate change and cannot be granted standing, but Al Gore, sharer of the IPCC’s Nobel Peace Prize, should be. You say that John Christy is religious, hence should not be granted standing.
Are you beginning to see a pattern here? From outside the consensus team it looks very much as though you are attacking the standing of your opponents rather than engaging on the science.
Joshua, you have been attacking Judith Curry for years, albeit using a different tactic to your brainless companion Michael and the grab-bag tossing FOMD and others. But never on the science. Always on some hypothesized slight, failure to address topics you personally want discussed and now for her turning the denier word back on the slimebags that have been using it against her for years.
Most of the regular commenters here know exactly what you have been up to. Most of us share my contempt for you. You are polite. ‘You can smile and smile and be a villain too.’
And that’s what you are, IMO. You’re one of the bad guys.
John,
RE: “The reporter is covering what Gore said not the accuracy of his statement.”
Accuracy is suppossed to be one of the cornerstones of good reporting. Now one could argue in the above example that this extends only to accurately quoting Mr Gore and leaving it the responsibility of the audience to assess credibility. I have no argument with that.
I also believe that one can make a distinction between a reporter covering a press conference and a journalist covering a wider topic. The latter tend to comment on the topic at hand, not just report quotes. In such an instance, if Mr Gore states that the seas will rise 20 ft should the Greenland icecap melt and neglect to mention that would take 1000 years to happen, then it is up to the journalist to include that information. Otherwise he is not proving balanced, accurate commentary.
Joshua,
Don’t you have an egg to suck on somewhere?
Tom,
+1000 on your reply to Joshua.
The only possible point of disagreement is calling him one of the bad guys. Guys like Josh, Michael and fan are more like annoying clowns and jackasses. They don’t have any standing to matter and can be ignored.
John, I’ve worked (long ago) as a newspaper and radio reporter. Of necessity, they can’t print everything, they have to select. And in selecting, they are concerned not with balance or truth but what makes a good story. Added to that, many studies have shown the great majority of journalists to be left-wing. The result in the alleged CAGW context, which I’ve been following since the 1980s, has been a severe bias to the scary stories and warmist line, with its left-wing policy agenda. It’s never “Just the facts, ma’am.”
Tom: “Joshua, since 1988 the consensus has engaged in a sustained battle to deny their opponents standing.” Quite so, and the media have been complicit in this. Good post @ 8.53.
Jim D @ 9.27, see my first para. Almost the only non-alarmist outlet in Australia over many years has been The Australian, and their editorial line has consistently been that AGW is occurring and must be dealt with by GHG emissions reductions. They have given space to Lomborg et al, but It’s only in recent months that their coverage has shifted – the pause is acknowledged – and sceptical letters have begun to dominate.
Tom Fuller | October 21, 2014 at 8:53 pm |
Tom, that was a long-winded evasion of Joshua’s point.
timg56,
I agree with your reply yet, with the exception of a handful of journalists, the analysis of Gore’s sea level rise claim requires scientific understanding and some math skills. Journalists aren’t scientists and typically aren’t allowed to do in-depth coverage due to the cost. This is why I commented that its up to the scientific community to dispel the spin used by NGOs and people like Al Gore by establishing a repository of factual information for the Press and for policy decision-makers.
ianl8888,
“WHY is Gore’s intellectual thuggery considered NEWS ? That’s the issue”
Gore has celebrity status. When he plays the Mad Hatter and throws a Global Warming tea party, the Press is inclined to cover the antics.
To be honest, I think he is being misled by NGO stakeholders as he clearly lacks critical skills required for scientific research and commentary.
Its worth pointing out, many of the articles released from NASA and other scientific agencies are equally misleading. Technical communications is an art and they aren’t hiring skilled communicators. This is deeply disturbing as NASA has numerous programs for the classroom.
NASA programs for the classroom should be peer reviewed and fact checked before release.
John “Technical communications is an art and they aren’t hiring skilled communicators. This is deeply disturbing as NASA has numerous programs for the classroom.” – is BANG on.
When I do a google image search for the terms “voyager location jpl.nasa”, MOST (7/8) of the NASA published illustrations that come up have the planets oriented perpendicular to reality, I have to guess their engineers know where they really are, or there would be a lot of misses with rockets. But on the communication end, their accuracy is sorely lacking in some regards.
Then again, when I think about it, their engineers believe in time dilation instead of clock calibration, so maybe they have a thing or two to learn as well. :-D
Alistair Riddoch,
LOL, they do have a tendency to poorly present information although I’m a fan of their 3D modeling and the photography is beyond amazing.
Its very disappointing to read their climate press releases and many of their climate related articles. Sadly, NOAA isn’t any better. Some jack-wagon from NOAA is currently claiming 2014 will be “the hottest year on record” which was picked up by over 300 media outlets.
Panic! 2014 hottest year ever (Not so fast, say the satellites)
http://joannenova.com.au/2014/10/panic-2014-hottest-year-ever-no-no-no-say-the-satellites/
Michael | October 22, 2014 at 8:41 am |
“Tom, that was a long-winded evasion of Joshua’s point.”
But Michael, Joshua doesn’t have a point.
He never had a point, he never will have a point.
In short, he is entirely pointless.
Nor is he the only one.
The other side can present their view, and we see it in editorials and through various journalists sympathetic to their view. Their main problem is pushing to the front among all the crazies that also occupy “the other side”. It is very hard for them to set out a sensible view for the public when the people over their shoulders are shouting out conspiracy theories and vent their general hate towards government and academia. It’s tough to be heard over the din. I don’t know what the solution to that is.
And which conspiracy theories does Dr Curry ascribe to?
Keep this type of commenting up and we’ll be able to see every pimple and mole on your hind end, so much of it will be hanging in the breeze.
Crazies like Richard Lindzen, Freeman Dyson, Judith Curry, Ivar Giaevar, Roy Spencer, Roger Pielke Sr., Roger Pielke Jr.?
I don’t agree that these are the crazies. I think the crazies are the conspiracy theorists, people suspicious of government takeovers, or anti-science, anti-academic elitism, etc. That doesn’t describe people who are within academia. They have a problem when they don’t tell the crazies to be quiet or why they are wrong, because it implies support. There’s too little denouncing going on.
Way too many crazies on both sides.
The Eco-religious fanatics and terrorists, ELF, ALF, Greenpeace, etc, are far more dangerous. IMHO
Jim D, I would welcome the opportunity to see where people such as yourself have led by example, calling to account the crazies that have over-egged the consensus arguments. Or where anyone on the consensus side has done so, for that matter.
Scientists do call out the egregious errors among their own ranks too. Lindzen and Spencer, to name two, have been subjected to some critiques and forced to make corrections. There might have been errors on the consensus side, but the loud noise about CRUTEM went away when others redid their work with independent methods. Some papers stand the criticism, some get modified, updated or corrected, and some are just too weak to even try submitting for review. That’s how science progresses.
Kind of proving my point, aren’t you JimD? When Lewandowsky made a fool of himself with his paper, where was the Consensus Team criticizing him? Prall, Andersen et al 2010 PNAS? What has the Consensus Team said about Peter Gleick or Phil Jones’ request to delete all emails?
You don’t criticize bad science. You don’t criticize bad ethics. You don’t critcize bad practice. You just criticize your policy opponents.
Only the dead are free of doubt.
AGW theory is science without doubts!
Note the urge to use uncertainty about climate prognostication to amplify fear of the future.
And journalists, hah. In your own words above the mandate for fairness of journalists is juxtaposed to the revelation that half of them won’t abide by fairness in the climate debate. And this, presumably, among the elite and most thoughtful invited to the conference.
========================
We ought to hammer them back by turning their beloved precautionary principle on its head. The cost of mitigation will be massive, especially for those who can least afford it, while any benefits look more and more uncertain.
alarmists almost never discuss the cost side of the equation in any serious way. Funny that. What they do instead is wildly exaggerate the dangers ….while telling lies about how “green” energy and the conversion thereto will pay for itself.
Rachel Carson got it. The IPCC gets it. If you ignore uncertainty, you are vulnerable. But if you say “There are many uncertainties but the message is clear” then with a compliant media you are ring-fenced.
The technique is deadly. Literally.
Most journalists are no longer fit for purpose. They are constrained by the amount of space or time they have for their story, mainly confined to sound-bites. More recently, fewer people are going into journalism programs and the expanded “eligible” pool of people for admission bring fewer qualifications to work at the craft. Today’s journalists, as they have less experience and generalized educational training including science/math/history, look to resources to fill those gaps; i.e. experts, where they tend to ask the wrong people about the wrong issues. That’s how Greenpeace and World Wildlife Federation get asked by journalists to discuss climate science topics that are much more nuanced and detailed than the journalist can comprehend. WWF and Greenpeace have ready handouts, slick eye-popping spreads sheets with permission of copy-write, easily inserted into one’s computer: voila, “your” story to beat the dead-line.
If it bleeds it leads. There is already enough bleeding in the world today, and, journalists, looking for their own Pulitzer Prize, are sure to enhance the blood, guts, and glory if they can’t find it lying around and readily picked up.
Decades ago (1960s), I was a news reporter on UK daily newspapers and Canadian commercial radio. I found a high level of dishonesty in journalism – most often, the story was the thing, not the truth. So I decided to make my career as an economist rather than a journalist. I don’t have inside knowledge of the present situation, the one journalist I know well has high standards.
Heh, selection bias. Extreme selection bias even.
===============
Speaking of selection bias, I wonder how many climate scientists out there didn’t get into it because they believed in dangerous global warming.
Ecologists and evironmental scientists often have an anti-human bias going in. I think this also turns a lot of more objective and intelligent people off of these fields.
In fact, when I was choosing my major I was interested in those fields but didn’t pursue them because I didn’t like the zealots in the fields.
Journalist > economist
Out of the frying pan etc. etc.
Part of the trouble is they,re just not very bright. I got into an argument with the public editor,s office at the nyt’s about the history of Clinton’s impeachment. They insisted over a series of 3 or 4 emails that he was impeached in the house but not the senate…which of course is not how the process works. They were so contemptuous…or whomever I was talking to was…and so sure of themselves…they could’ve even be bothered to look it up.
This is the nyt’s which I’m guessing mostly highers Ivy Leaguers. Really pretty digraceful.
Isn’t it ironic?
It is hires.
I don’t know that I’d say they aren’t bright. But they are not particularly well educated, at least in areas relating to science and engineering. (Anything related to military topics is also well outside their sphere of comprehension.)
Go on line and see the course requirements for a degree in Journalism at Columbia. You can party til the sun don’t rise and still graduate with a degree.
In the end, journalism is more and more just entertainment. It won’t sell unless there is either someone calling someone else dishonest or someone saying it is worse than anyone thought. Scientific consensus by itself doesn’t make the cut for entertainment. Newspapers only report controversies, however made up, and shifts, not the status quo and incremental advances.
Can’t say I find anything to disagree with you here.
If you want to know about the science, go to the source, and that is not the newspapers or blogs.
You could say I have, with 2 graduate science degrees.
I scraped through the physics and statistics, so there are a lot of specific discussions I don’t try to argue. But i did learn enough to know that the “peer-reviewed” argument is a smoke screen. And enough to know that there were considerable “leaps of faith” taking place between what was (and is) known science fact and what was basically conjecture. The whole extreme weather storyline is nothing but conjecture. The impact on the polar vortex from climate change is as well. The more one looks, the more they find this is standard in the field of climate science.
“Rather than downplay scientific uncertainty and gaps in understanding, she [Carson] progressively amplified what was unclear about the human impacts of DDT and other synthetic compounds on humans and wildlife.” Don’t see what’s surprising about this. When an advocate has a scientific consensus on her side, she doesn’t want to hear about uncertainty that goes against her advocacy. Whereas Carson was using the uncertainty as a tool to make people more afraid than the consensus then warranted.
Miker –
==> “Whereas Carson was using the uncertainty as a tool to make people more afraid than the consensus then warranted.”
Reminds me of when SWIRLCAREs ignore uncertainty to scare people about the economic ruin that will result from ACO2 mitigation.
Try being specific about the costs of proposed CO2 mitigation actions vs. what they will accomplish.
“Whereas Carson was using the uncertainty…Reminds me of when SWIRLCAREs ignore uncertainty” – I’m guessing you got your point a little garbled; these seem to be opposite. But the point is a good one nevertheless.
Still, I think it’s more complicated than we’re saying here. I’m guessing that most conservatives aren’t just _uncertain_ about economic ruin, they’re pretty sure about it. The fact that economists don’t have a consensus on the subject is a reflection of the fact that no one expects economists to have a consensus on anything. The conservatives are pushing for a point of view that they believe is true.
The same is true in reverse for liberals who are quite comfortable with enforcing mitigation by top-down means – they believe it’ll be fine, trust their economists, and are unconcerned with the other economists who think it’s a terrible idea.
Now what about Rachel Carson? Was she uncertain, or was she pushing uncertainty because she believed that the consensus science was wrong and she was right?
Now what about AGW skeptics? Are they pushing uncertainty, or is uncertainty a way to get consensus scientists to give at least some credence to the lower sensitivity/low damage point of view that the skeptics actually think is more correct than the consensus?
In between all these groups who aren’t really uncertain at all, there are a few others who are actually uncertain and need to pick their way through to a policy that deals with their uncertainty.
miker613:
Those are excellent points. I think that there are a few general principles that can be used to determine how to make decisions under uncertainty but the actual situation may dictate whether they can or should be applied. In large measure, many people act as if they are certain. In military and business situations, dithering or hesitation can be more deadly than the wrong decision.
The Precautionary Principle, which many environmentalists, extremists and idealists depend on, is used to defend extreme actions (including the refusal to take action) that have their own significant costs. For example, do you ban vaccines because some may die as a result of taking the vaccine? The needed type of cost benefit thinking typically is beyond the purview of scientists addressing an issue and frequently requires the balancing of accounts using the ultimate currency, peoples’ lives. The use of DDT certainly falls into this category.
==> I’m guessing you got your point a little garbled; these seem to be opposite.”
Maybe – but maybe not:
<blockquote“Whereas Carson was using the uncertainty as a tool to make people more afraid than the consensus then warranted.”
SWIRLCAREs use uncertainty to make people more afraid (of economic collapse) than the consensus (that ACO2 emissions present an uncertain risk) warrants.
I guess maybe it depends on your definition of consensus. It’s one of those terms that gets thrown around selectively so as to conveniently bolster arguments. Does the “consensus” say that ACO2 represents certain catastrophe? I don’t think so – yet many SWIRLCAREs use the uncertainty of economic outcome to make the public afraid of certain disaster should we enact policies to mitigate emissions. Because it can’t be proven beyond all certainty that benefits of mitigation will be greater than costs, we see arguments all the time that there is no doubt that costs will be greater then benefits.
==> “But the point is a good one nevertheless.”
Ok, fine. Maybe my comparison is flawed, but the point remains that uncertainty is used as a rhetorical tool to advance ideological agendas. It happens on both sides of these types of issues.
==> “The conservatives are pushing for a point of view that they believe is true.”
Sure. They absolutely believe their opinion to be true. No uncertainty. Look around these threads. You will see dooming and glooming from SWIRLCAREs that allow for no uncertainty: Mitigation = billions of starving children in Africa, and anyone who thinks that mitigation (to some degree) is a viable option is either indifferent to that certain suffering, or outright desiring that as goal. That isn’t to say that all SWIRLCAREs think that mitigation will certainly bring about the end of the world as we know it….but a certain vocal segment does..
==> “The same is true in reverse for liberals who are quite comfortable with enforcing mitigation by top-down means – they believe it’ll be fine, trust their economists, and are unconcerned with the other economists who think it’s a terrible idea.”
“Unconcerned?” Sure, there is a segment that dismisses any possibility that the costs will outweigh the benefits. There are combatants on both sides who ignore the reality that most economic analyses bracket various outcomes with confidence intervals and error margins.
==> “Now what about AGW skeptics? Are they pushing uncertainty, or is uncertainty a way to get consensus scientists to give at least some credence to the lower sensitivity/low damage point of view that the skeptics actually think is more correct than the consensus?”
As many SWIRLCAREs seem to conveniently forget, “skeptics” are not monolithic. From what I see, a lot of them push uncertainty about something like climate sensitivity by ignoring the possibility of high sensitivity. So the approach to uncertainty is inconsistent. Further, they often push uncertainty about something like sensitivity even as the ignore the uncertainty about economic outcomes. That angle is particularly beautiful, because they complain that “consensus” uncertainty is based on unvalidated and unverified models even as they rely on unvalidated and unverified modeling to base their certainty about economic disaster as an outcome of mitigation.
==> “In between all these groups who aren’t really uncertain at all, there are a few others who are actually uncertain and need to pick their way through to a policy that deals with their uncertainty.”
What I see, something for which I think there is solid empirical evidence, is that many, many folks are quite certain even though they don’t actually understand the technical issues or what scientists actually say, and pick and choose their experts on the basis of how well what those experts say aligns with their cultural or ideological orientation. There are those who aren’t uncertain at all, and there are many others who are far more certain than their technical understanding justifies. The number focused on picking their way through evaluating policies in the face of uncertainty, seems to me to be quite small.
Miker –
Related:
My guess is that if we compared these numbers to numbers on Avian flue when Bush was president, we might see the opposite effect. My guess is that evaluating risk in the face of uncertainty is largely related to who is in the executive office. Funny how that works, ain’t it?
http://www.people-press.org/2014/10/21/ebola-worries-rise-but-most-are-fairly-confident-in-government-hospitals-to-deal-with-disease/
Joshua,
Skeptics, monolithic or not, do not ignore the possibility of high climate sensitivity. They are arguing against the built in belief that climate sensitivity can only be high.
Built in by your words and description of them as ignoring that possibility.
That is only true in your perception of climate sceptics and description of what you see their beliefs to be, not what their beliefs actually are.
Matt Ridley had an interesting piece in the Times (UK) recently. It was pay-walled, so I was only able to read part. The EU ban on neo-nicotinoids has caused crop failures, hasn’t seemed to have helped bees, and he suggests crop failures may put more stress on bees.
http://www.rationaloptimist.com/blog/bees-and-pesticides.aspx
Systemic Spread and Propagation of a Plant-Pathogenic Virus in European Honeybees, Apis mellifera
Emerging and reemerging diseases that result from pathogen host shifts are a threat to the health of humans and their domesticates. RNA viruses have extremely high mutation rates and thus represent a significant source of these infectious diseases. In the present study, we showed that a plant-pathogenic RNA virus, tobacco ringspot virus (TRSV), could replicate and produce virions in honeybees, Apis mellifera, resulting in infections that were found throughout the entire body. Additionally, we showed that TRSV-infected individuals were continually present in some monitored colonies. While intracellular life cycle, species-level genetic variation, and pathogenesis of the virus in honeybee hosts remain to be determined, the increasing prevalence of TRSV in conjunction with other bee viruses from spring toward winter in infected colonies was associated with gradual decline of host populations and winter colony collapse, suggesting the negative impact of the virus on colony survival. Furthermore, we showed that TRSV was also found in ectoparasitic Varroa mites that feed on bee hemolymph, but in those instances the virus was restricted to the gastric cecum of Varroa mites, suggesting that Varroa mites may facilitate the spread of TRSV in bees but do not experience systemic invasion. Finally, our phylogenetic analysis revealed that TRSV isolates from bees, bee pollen, and Varroa mites clustered together, forming a monophyletic clade. The tree topology indicated that the TRSVs from arthropod hosts shared a common ancestor with those from plant hosts and subsequently evolved as a distinct lineage after transkingdom host alteration. This study represents a unique example of viruses with host ranges spanning both the plant and animal kingdoms.
http://mbio.asm.org/content/5/1/e00898-13
That’s interesting.
Can that be related to neo-nicotiniods in some way? Perhaps they prevent it?
“While this describes lawyers, it also describes scientist-advocates; in principle this behavior is ok, but in the legal system the ‘other side’ has a fair chance to make their case; in the climate change debate, the ‘other side’ is smeared as ‘deniers.’” -JC
Except that it’s quite obvious, ‘deniers’ have had their views covered in the media far out of proportion to the extent that those views occur within the scientific community.
Those in the climate consensus who have cried for carbon cuts now for the last 20-years and their accomplices in biosci, psychology, sociology, etc. who publish disaster after disaster press releases printed without comment, without thought without any concern for the accuracy of GCM “projections” have dominated the media blitz because disaster and panic sells. I only read about sceptic scientists on the blogs, never in the news papers.
It’s for the children, why are you against a free and independent press?
Re pundit predictions, Philip Tetlock collected a vast number
of expert forecasts of political and economic events over more
than 20 years and showed that these forecasts were not much
better than chance, or the predictions of non-experts.
http://press.princeton.edu/titles/7959.html
Of course cli sci experts or cli sci fi experts are exempt from
the above on account of the nobility of the cause.
beththeserf,
You might like a book called “The Fortune Sellers”. I think it covered differing types of predictions and actual outcomes, about various things, including finance, sport, and politics, from memory. I no longer have a copy, but the local library can no doubt dig one up.
If you like it, it was my pleasure to help. If you don’t, I deny all knowledge of said book, and any rubbish it contained.
Live well and prosper,
Mike Flynn.
Thx Mike Flynn. Proviso duly noted. )
Most hilarious cli sci prediction would have to be David Jones, head of climate at BoM, in 2009 predicting Big Crisp just before Big Wet.
I’m not predicting or planning anything for anyone’s great-grandchildren. I don’t even know those guys, and I’m pretty sure they’ll spend a lot of time giggling at our various fads, fashions and theories. They’ll mock our climate science as they ride along in their thorium-fuelled self-guided levitation machines through the Brisbane snows. I already resent those smug unborn so-and-so’s.
Howard
Don’t you think you are being a bit truculent?
Yer must remember this:
‘Children just aren’t going to know what snow is,’
said Dr David Viner, a senior research scientist
at CRU, the Climactic Research Unit, or make that
the Cassandra Reseach Unit, at the University of
East Anglia?
“Except that it’s quite obvious, ‘deniers’ have had their views covered in the media far out of proportion to the extent that those views occur within the scientific community.”
Journalism, if we believe what is written here, has an obligation to be
fair, accurate, and engaging. And very often these obligations are in conflict or apparent conflict.
You think that fair means the coverage should be proportional to the number of people who believe a certain proposition. Why? There are many cases where minority views are covered in a manner that is disproportional
to the number of people who believe it. Typically this is done because it’s engaging and interesting to people. Or it is justified on the basis of being accurate. in covering say “christian attitudes on gay marriage” It would be
perfectly normal to report on the variety of christian attitudes, from nut jobs who think “god hates fags” to more broadly held views. One would not judge the fairness of these reports by counting the number of nut jobs who hold extremist views and limiting the words expended by the proportion of the population. being accurate and engaging means that sometimes you cover the extremes.
peter gleick stole documents and faked a document. What’s the fair way to report that? Should one point out that is but one case of thousands of scientists who dont do this kind of crap. Do we allocate 10 words to the story? Do we “in fairness” write articles about scientists who dont do this so that our coverage is balanced? Or does the charge to be engaging take on more significance?
Some nut jobs, climate extremists, have argued that the arctic would disappear by 2013 or so. How does one cover this “fairly” Do we report that this guy is a minority. do we give his fever brained nonsense the 10 words of print it deserves? nope. And what of the nuts who blathered on about model results that predicted 10C of warming. Did their results get the one word it deserved ( that word being BUNK). Nope. measuring fairness is harder than estimating sensitivity
I don’t see “skeptics” publishing very many papers except in low impact journals. It’s not surprising they get very little press coverage.
What is the newspeak narrative now? Too much press, no press due to peer review gate-keeping? Please stay on message Joseph, else you confuse the proles.
There is no dearth of “skeptic” opinion in the media. Not much actual science, though..
Joseph – yes.
Howard – that’s right, even just plain old skeptic ‘opinion’ gets a very good run.
Joe,
With you being blind in one eye and unable to see out of the other, this comes as no surprise.
Jojo and Mikey
It’s amusing to see how blind you are to your own bias.
The sceptics get equal coverage in the MSM? :)
Really?
I’m a scientist. It’s not obvious to me at all. I see it as about 50:1
on alarmism versus skepticism. Think of every heavy rainstorm, drought, hurricane, tornado, or forest fire which is “linked” to “climate change” with no evidence at all and no rebuttal by skeptics. Even when both sides are represented, the broadcaster may roll his eyes at the “denier” side or undermine their arguments by making it clear that he agrees with the other side. The idea that “deniers” are invited and given equal weight to “true scientists” is one that is being pushed by “green-totalitarians” (I put it in quotes so don’t cry about it) that brook no dissent and can’t stand open discussion and debate.
‘the scientific community’
Could you define that for me please?
Michael: “Except that it is quite obvious, ‘deniers’ have had their views covered far out of proportion to the extent that those views occur within the scientific community”.
I dispute that claim. First of all Climate Science news coverage has plummeted in recent years as it is considered a ‘big yawn’ with no one listening. Second is that coverage breaks down to political affiliation of the news organizations. The BBC, NYT, LAT, WP, ABC, NBC, CBS, CNN, to name a few all refuse to have any coverage of skeptics or deniers or doubters as they are described. The NYTs and the LA times refuse to even publish them in their opinion sections of regular readers. Now the Daily Mail, WSJ and Fox news do cover the skeptics. They are widely portrayed by Mainstream news analysts as highly biased. They say Fox news is only 28% accurate. That means 28% of their coverage is dedicated to consensus views and 72% is dedicated to doubters.Their is a huge bias in the mainstream media and doubters are largely either dismissed outright or portrayed as nut cases. I believe that is largely in line with your thinking. So what’s your problem? Perhaps you should get your facts straight before you post such ridiculous nonsense.
Michael said “… it’s quite obvious, ‘deniers’ have had their views covered in the media far out of proportion to the extent that those views occur within the scientific community…”
Are you serious? Honest to God, is there a day that goes by without the media quoting some hare-brained study where “Global Warming causes X” ? – where X is anything from cow infertility to increases in crime to deaths of moose in Minnesota to increased acne. The press will joyfully report on any such claim – so long as there is a so-called “researcher” who supposedly looked into the matter. Here’s a site that documents all the published claims… http://www.numberwatch.co.uk/warmlist.htm . And another here: http://cuttingthroughthefog.blogspot.com/2014/06/ridiculous-climate-change-claims.html .
So yeah… the media are SO much more disproportionate in their coverage of skeptics over warmists. Get real.
“As a result there is some pretty irresponsible public behavior by academic and think tank/advocacy group scientists, and there are absolutely no repercussions (I’m sure we can all think of examples).” – JC
And what examples to follow we have with journos and lawyers…..a great argument against codes of conduct.
Michael:
You may be aware that the United States issues professional licenses to civil engineers and geologists. There are codes of conduct for science and business practices because much of what geologists and civil engineers do impacts the public health and safety. While not perfect (what is!), this system has resulted in safe and secure drinking water and the appropriate handling of wastes. The environmental improvements designed by engineers and geologists, constructed by blue collar ruffians and operated by private and public technicians goes on unnoticed without any recognition from “environ-mentalists” and largely without significant input from “academia nuts”.
Some of in the real world of adult sanity want to see a proper risk assessment done to professional standards. It’s for the children, so I don’t understand why you support policy based on ad hoc methods.
a ‘risk assessment done to professional standards.’
Sounds fascinating.
In response to this:
Judith says the following (emphasis mine)
==> “While this describes lawyers, it also describes scientist-advocates
Well, it doesn’t describe advocates – it describes a form of advocacy. Interesting that you jump to characterizing people….but let’s consider this further, as it is a good description of your Congressional testimony, where you, quite specifically, omitted findings that didn’t support your hypothesis. And I note that you have continued to repeat this behavior, in your advocacy, such as in your WSJ editorial.
What isn’t good for the goose is just find for the gander, eh Judith?
==> “… in the climate change debate, the ‘other side’ is smeared as ‘deniers.’”
Well now, isn’t that fascinating? Apparently you forgot what you said a few days ago, eh? Here, let me remind you:
Interesting how none of your “denizens” have stepped forward to you comparing people in the Pentagon to holocaust deniers, eh Judith?
What isn’t good for the goose is just find for the gander, eh Judith?
Dog with a bone, Joshua. For Judith to throw the term used against her back at those who have been using it–sure, exactly equivalent. In your mind only.
Oh, right, I forgot Tom –
In your book, “They did it first” counts as a justification.
Sorry for forgetting that.
Nasty ankle bite there Judy, make sure your tetanus jab is up to date and seriously consider a rabies shot as the mutt is quite evidently frothing at the mouth.
Joshua, are you adopting FOMD’s annoying “eh”? I vote against.
Well, four of them in one comment might be a smidgen over the top. A smidgen…
“Joshua, are you adopting FOMD’s annoying…”
He don’t need FOMD to be annoying. ;)
Andrew
For Joshua, being annoying has been elevated to a kind of performance art.
The man has it down
miker613 – “…adopting FOMD’s …eh…”
The “eh” is a rhetorical pointed stick in the ribs, perfect for sadists. BTW, I read an account of a Saudi beheading in which the stick in the ribs prods the condemned to raise his head straight up for the sword. Watch out for the “eh”, it’s a taunt to get you to raise your head.
I am unclear where the politicians stand in the debate. For example, Boxer (CA), Markey (MA), Waxman (CA), Sanders (VT), Brown (OH), … and I almost forgot, Al Gore. Do they line up ideologically, for constituents, behind party lines, following the money … all of the above? (A more fundamental question is what do any politicians “truly believe”? I look forward to seeing this material presented and discussed in congressional reviews. As much as anyone the politicians need to understand uncertainty. Since they all have “science advisors” on staff what you hear in their streaming vitriolic litanies on climate change have little to do with actual understanding of science.
Danley
I think you can look to Obama’s appointment of a political apparatchik to Ebola czar
a serious infectious disease requires perception management
same with climate change
“science advisors” are for standing on the podium behind the politician
same as flags
In administrative law, there is no effective means of challenging agency scientific findings in court because of extreme judicial deference to such findings. Paradoxically, the greater the uncertainty the greater the deference. This is radically distinct from the resolution of scientific questions in conventional civil or criminal cases. So when EPA cooks up a crock like the Endangerment Finding on AGW, the courts, applying the doctrines of administrative law, abdicate any searching judicial review, and defer almost entirely to whatever EPA says, no matter how nonsensical it might be. EPA’s attribution statement, a bonfire of logical fallacies lifted verbatim from AR4, emerged unscathed from this extremely deferential review, and the challengers were actually mocked in the DC Circuit panel opinion for their trouble. In my view, if the attribution statement were vigorously challenged in a normal civil jury trial context, it would not survive.
STRIKING CONTRASTS REGARDING “UNCERTAINTY”
Climate Etc readers are invited to verify for themselves that the free-as-in-freedom Hansen et al survey “Earth’s Energy Imbalance and Implications“ — an article much-admired and oft-cited by many (including FOMD) — discusses “uncertainty” on eighteen of its thirty-nine pages.
In striking contrast, Judith Curry and Marcia Wyatt’s “stadium wave” preprint does not use the word “uncertain” — or any of its derivatives — even once (ouch).
Conclusion Judith Curry and Marcia Wyatt, please take a lesson regarding “uncertainty” in scientific discourse from James Hansen and colleagues!
One of my favorite hansen paper
here is what he writes
“Fast-feedback climate sensitivity has been estimated in innumerable climate model
studies, most famously in the Charney et al. (1979) report that estimated equilibrium global
warming of 3°C ± 1.5°C for doubled CO2 (a forcing of 4 W/m2
), equivalent to 0.75°C ± 0.375°C
per W/m2 . Subsequent model studies have not much altered this estimate or greatly reduced the error estimate, because of uncertainty as to whether all significant physical processes are included in the models and accurately represented. ”
Chareney 1979 using a forcing of 4W concluded that ECS was 3C.
what do you see?
Mosher: I see what has been causing my brain to boil for the last few weeks. WGII AR5 Chapter 7 on clouds describes at lenght the uncertainty of cloud feedback, and the almost complete lack of knowledge of low cloud feedback; it could be positive or negative and the magnitude is unknown. However it then concludes that cloud feedback is likely positive, providing only murky reasoning. One reason is that there is no evidence that the models are inaccurate. Similiar to your example.
Normally I do not read your posts as they are full of stupidity like “everyone knows X”, when there is very little that fits that classification. However, in this case I did skim both Hanson’s and Curry’s papers for uncertainty. Did you even skim these papers? Look at Curry’s paper, Figure 4. The uncertainty (error) bars are very visible on the graph. One does not need to use words to describe uncertainty when a picture works. Your statement that Judith Curry and Marcia Wyatt’s “Stadium Wave” preprint does not use the word “uncertain” — or any of its derivatives — even once (ouch), is false because the text talks about the error bars of Figure 4. Maybe you did not include enough “derivatives” in your analysis, but may I suggest that you look at the pictures before you post?
AT,
Of course you’ve got him dead to rights. But it doesn’t matter. Fan’s mind is a filter which will not admit any information with the wrong shape…which is to say any information not in accord with his beliefs. The “FOMD “moniker is as amusing as R. Gates’ “aka the skeptical warmist”.
Surely you don’t think John Sidles is acting in an honourable manner in making his comments? He is trying to enrage people and stop them analyzing of the ATL and BTL posts.
Fan, Dude! Feed us trolls the actual paper next time, not just the stinkin abstract. You Harsh my Mellow with your selfish Bogarting.
http://www.atmos-chem-phys.net/11/13421/2011/acp-11-13421-2011.pdf
Uncertain is mentioned 8-times in the 29-page paper.
Aerosols gets the big uncertainty headline:
Inspite of the mortal mystery that is aerosol forcing, Jesus hisself proclaims to know the unmeasured and proselytizes:
Can I get an AMEN, Brother!
Climate Etc readers are invited to verify for themselves (by text search) that both the preprint and the (substantially identical) published text of Hansen et al “Earth’s Energy Imbalance and Implications” (2011) discuss “uncertainty” in twenty-eight (resp. twenty-four) passages.
Take a lesson in the thorough, effective communication of scientific uncertainty, Judith Curry and Marcia Wyatt!
Howard, ATAndB, and pokerguy, your collective climate-science understanding might be improved by a thoughtful reading of Hansen’s respectful, rational, responsible (and well-referenced) uncertainty-analysis.
*EVERYONE* appreciates *THAT* scientific reality, eh Climate Etc readers?
The stadium wave is sold as a hypothesis which is very very uncertain by definition. This is obvious when you read the paper and when you read Dr. Wyatt’s web page
http://www.wyattonearth.net/thestadiumwave.html
Finally, a real geologist is working the problem.
“*EVERYONE* appreciates *THAT* scientific reality” by FOMD.
After a discourse on uncertainty???
Where is the uncertainty in your concluding statement? In what universe would that even be true? Human nature is such that I would not expect that even 50% of people appreciate what you wrote. I don’t even expect that 50% of people appreciate what I write. Further, whether your statement is a “scientific reality” or not is inconclusive as well as I am not sure that Hanson is a very good example of presenting uncertainty. He puts error bars on some things on his graphs, and not on others. Are the other things known to perfect precision, or more likely, is the error so large as to be indeterminate? Take a couple of minutes and review the pictures, it is not that hard.
Only a Sith deals in absolutes.
As an engineer, I adhere to a code of ethics and conduct and I’m professionally accountable for my work. An immoral, unethical or incompetent engineer is found out rather quickly (and hopefully before somebody is hurt).
That is perhaps the main reason the unaccountable academic scientists do not impress me in the least and I usually have little regard for what they say unless they have earned my trust.
Well…
Lawyers can walk away from their mistakes, doctors get to bury their mistakes, scientists are judged by the number of articles and citations even though double checking shows most scientific papers are wrong (or at least not repeatable).
Engineers have to fix their mistakes. That process is painful (and can be painful to their career). Engineers believe in getting it right the first time.
Peer review apparently is like playing footsie compared to an engineering design review and ineffective by comparison.
Yeah, and look at how much more ethical they are than scientists. :)
NO EVIDENCE? NO UNCERTAINTY
Climate Etc readers are invited to verify for themselves that WUWT has taken to smearing Naomi Oreskes in no “uncertain” terms, without presenting *ANY* direct evidence to support the charges
Good on `yah, WUWT reader John Whitman, for asking to see the evidence.
It’s a shame that (to date) WUWT has not seen fit to answer your reasonable respectful responsible question!
Boo on `yah, WUWT editors and publisher Anthony Watts, for smearing scientists without evidence, and (especially) for allowing outrageously misogynistic comments to pass unmoderated.
————-
Judith Curry, you may disagree with Naomi Oreskes professionally, yet still speak out against gratuitously misogynistic smears on women scientists!
————-
Professor Oreskes’ chief scientific offense — it appears — has been to thoroughly document the racketeering exploitation of faux-uncertainty by the tobacco industry … racketeering that cost the lives of tens of millions of people.
Boo on `yah, faux-scientific institutes that for decades have supported for-profit yet all-too-lethal anti-scientific faux-uncertainty racketeering!
Conclusion Climate Etc readers are to listen Oreskes speak for herself, and consider one simple question …
Who’s on the side of the angels? The world wonders!
My cats tell me that Oreskes is full of fleas.
===============
“Misogynistic”.
The woman’s creepy beyond words. So I guess you’ll have to lock me up for a hate crime. Then again, I also hate Michael Mann. So maybe what I’m really guilty of is discriminating against despicable creeps. Lock me up with the rest of the deniers, fan. And don’ forget to throw away the key.
Creepy! A perfect way to assess intellectual discourse. Go have a lollypop and snuggle up in your blanky and the bad people will just go away.
“intellectual discourse.”
Spare me the piety. Naomi is in fact advocating for closed minds, which is pretty stunning for an academic. Can,t put up a link on this little tinker toy I,m typing on. Just do a quick search on “open-mindedness is the wrong approach Naomi Orestes.”
Here you go, Howard. “Creepy” doesn’t begin to cover it. But of course it goes right past you. Like all warmists, you’re afflicted with a tin ear and a blind eye
http://judithcurry.com/2012/01/23/open-mindedness-is-the-wrong-approach/
It’s just an idea to be debated or ignored. By demonizing and making fun of her sex, sexuality and looks, you give her too much power and attention. Your tells shine like a purple pimple on the end of an alabaster nose.
a fan of *MORE* discourse: Professor Oreskes’ chief scientific offense — it appears — has been to thoroughly document the racketeering exploitation of faux-uncertainty by the tobacco industry … racketeering that cost the lives of tens of millions of people.
In my view, Oreskes’s chief “scientific” offense is to argue by innuendo and association and analogy that of all the scientific debates of the last few centuries, the one most closely related to the CO2 debate is the cigarette debate; and that opposition to the extreme view of the CO2 theory (that increased CO2 will necessarily cause disaster) is as “criminal” (evil, etc) as the opposition to the claim that cigarettes cause cancer.
Oreskes also lumped together those who claimed cigarettes posed no public health risk and those who claimed that the risk of second-hand smoke had been exaggerated.
A possibly related error is being made now by people who confuse the issue of CO2 caused warming with all human impacts on climate. And there is a vilification of people who point out the possibility that the warming since the end of the LIA has not in fact been detrimental.
Boo on `yah, faux-scientific institutes that for decades have supported for-profit yet all-too-lethal anti-scientific faux-uncertainty racketeering!
Where exactly do you find racketeering in the many demonstrations that the extreme view of the CO2 case (that increased CO2 will necessarily cause disaster) is full of holes, liabilities, cavities, etc? Consider the scientific case that the climate with 400 ppm CO2 in the atmosphere is worse than the climate was with 280 ppm CO2: everything good and bad can can be well-documented occurs with approximately the same distributions of frequency and intensity as at any time in recorded history; except possibly that crop failures and other disasters related to low temperatures are a little less frequent.
Matthew,
While your comments are to the point and well reasoned, they are also wasted. fan has repeatedly proven that honest discourse is well beyond his interest, if not ability.
That the majority of WUWT comments abusively focus upon Oreskes’ gender and appearance is regrettable … yet perhaps unsurprising, given deplorable editorial leniency that WUWT moderators have shown toward smearing, doxxing, and personal abuse.
It’s dismaying to find that a considerable portion of Climate Etc commenters excuse these practices.
Consider the very first WUWT comment, for example:
Yikes.
“Let us not assassinate this scientist further. Have you no sense of decency, WUWT editors? At long last, have you left no sense of decency?”
Questions Judith Curry, do you and your female students similarly encounter vile misogyny and personal smears? In your view, should the public speak out against these practices?
The world wonders … and young climate-science students especially wonder.
Fan of His Own Discharge:
We can agree on this.
WUWT is all about a safe haven for right-wing nutters and conspiracy theorists. Dr. Lewandownsky’s hypothesis is exactly correct for that despicable blog where little boys can feel safe and find acceptance.
a fan of *MORE* discourse: That the majority of WUWT comments abusively focus upon Oreskes’ gender and appearance is regrettable …
I agree.
“That the majority of WUWT comments abusively focus upon Oreskes’ gender and appearance is regrettable …”
Fan, I’ve only skimmed the first few lines of the WUWT blog post, and have seen no comments. But I’m willing to bet that the majority of the comments do not focus on gender and appearance. If I’m wrong of course I’ll cheerfully admit it. If you’re wrong, you could perhaps cheerfully explain why you feel it necessary to exaggerate. To sweeten the pot, if I’m wrong I’ll donate 100 bucks to PETA, which is my favorite charity.
Step up to the plate, Fan. Do we have a bet?
One proviso, to count for you they must abusively focus, in your words. Those critical of such comments….if their are some…certainly shouldn’t count for your side.
pokerguy: Fan, I’ve only skimmed the first few lines of the WUWT blog post, and have seen no comments. But I’m willing to bet that the majority of the comments do not focus on gender and appearance.
At the time of writing it, FOMD’s comment was true. There have been long runs of irritating and deplorable comments on her looks, demeanor, carriage etc, and comments that are generally misogynistic. Whether they still constitute a majority, there are too many of them. Last I checked, there were a few snips, but a bunch of the comments stood without deprecation by the site administrators.
I actually looked up the WUWT post in question:
http://wattsupwiththat.com/2014/10/20/slimed-by-naomi-oreskes-in-defense-of-dr-fred-singer/
It consists mostly of an excerpt from “EcoTerror: The Violent Agenda to Save Nature”.
The post is about a Nightline hit piece “Environmental Science For Sale,” that Ted Koppel personally caused to be rewritten as a more balanced segment.
You can read the comments yourself – the worst comments are at a DailyKos level.
As far as her original consensus paper:
http://scienceandpublicpolicy.org/monckton/consensuswhatconsensusamongclimatescientiststhedebateisnotover.html
“928 abstracts, published in refereed scientific journals between 1993 and 2003, and listed in the ISI database with the keywords ‘climate change’.
> She concluded that 75% of the papers either explicitly or implicitly accepted the “consensus” view…
“Remarkably, none of the papers disagreed with the consensus position. ”
A number of people have look at this and with only 1% of the 928 (there were about 12000 and she for whatever reason took a smail sample) explicitly endorsed the consensus. She might actually be right though – given the political climate in science at the time and the focus of government funding on CAGW the likelyhood of someone being funded to critically review climate change theory and actually get the paper through peer review is pretty small. There were about 44 natural forcing papers that may have disputed part of CAGW theory.
http://quadrant.org.au/opinion/qed/2014/09/doomed-kittens-puppies/
Quote from her book: “The Collapse of Western Civilization: A View from the Future.”
“The loss of pet cats and dogs garnered particular attention among wealthy Westerners, but what was anomalous in 2023 soon became the new normal. A shadow of ignorance and denial had fallen over people who considered themselves children of the Enlightenment.”
The climategate files proved basically all the claims of the skeptics were true. Studies that didn’t support CAGW were being suppressed, Careers of skeptical scientists and editors were being deliberately harmed, Peer review panels were being used to screen out skeptical papers. You can read the climategate emails and judge for yourself.
It is what it is. She is a radical activist who is not objective and is somewhat deceptive. Her writings don’t sound particularly accurate or scientific.
However the phrase “Liar for Hire” doesn’t appear to have originated with Oreskes and is attributed to a reviewer of the movie based on her book.
FOMD, perhaps your criticism of WUWT’s commenters should actually take place at WUWT.
Just a thought.
John, I am sure you more than any other commentator here knows how painful and emotionally debilitating it is to be ugly in a society that values beauty. The trials you must go through everyday from the unaffected must really scar your soul.
“That the majority of WUWT comments abusively focus upon Oreskes’ gender and appearance is regrettable
Fan,
I find 22 comments out of 159 currently, that could be classified as misogynistic. There are likely just as many speaking out against. In short fan, not even close to a majority. However, just to show what a sport I am, I will go ahead with my donation anyway. Maybe in a show of good fellowship and peace, you could do the same …
“I am sure you more than any other commentator here knows how painful and emotionally debilitating it is to be ugly in a society that values beauty.
I found the photo of Fan unusual, but not unattractive. If anyone’s not seen “Laurence, Anyways” directed by the astonishingly young (for such a talent) Mr. Dolan, I can’t recommend it enough. Imvho a masterpiece, a word I don’t toss around lightly..
Pardon me fan if my assumption is wrong. But either way I’m grateful for having been born a straight fellow with no gender ambiguities to worry about. Life is hard enough as it is.
As to being unattractive, we all get to experience that if we live long enough.
I just skimmed the WUWT comments on O and most of the off color ones are from one person, with a couple of more commenters thrown in. Per usual, Fan of More BS lives up to his name. Also, you never know when an Alinsky liberal will go on WUWT and intentionally insert offensive commentary. Kind of like farting at the symphony.
From the article:
…
Another idea I had that almost came to fruition was directed at the Rochester Philharmonic, which was the establishment’s — and Kodak’s — cultural jewel. I suggested we pick a night when the music would be relatively quiet and buy 100 seats. The 100 blacks scheduled to attend the concert would then be treated to a preshow banquet in the community consisting of nothing but huge portions of baked beans. Can you imagine the inevitable consequences within the symphony hall? The concert would be over before the first movement — another Freudian slip — and Rochester would be immortalized as the site of the world’s first fart-in.
…
http://www.progress.org/tpr/saul-alinsky-on-the-struggle-with-eastman-kodak/
Jim,
I thought about not counting those kinds of comments from the same poster, but that wasn’t the terms of the bet I offered. Still, it’s a good point. Of course the whole gambit’s a distraction. Easier to go on the attack about misogyny than to defend Oreskes’ fantastical and nauseating ideas.
And not for nothing, but I see plenty of insulting comments about Christopher Monckton’s appearance on alarmist blogs. Doesn’t count I guess, because he’s not a woman?
Fan – just as bad, another recent thread about Oreskes over at WUWT:
9 “snips” for being “over the top” and look what kind of comments don’t merit such a distinction:
Joshua,
Just a small sampling.
So not the best bits then?
Perhaps the person is completely sane, she writes articles with Lewindowsky whose every word is approved of completely by Joshua.
There, does that make her look better?
A fan of *MORE* discourse | October 21, 2014 at 12:59 pm
“NO EVIDENCE? NO UNCERTAINTY
Climate Etc readers are invited to verify for themselves that WUWT has taken to smearing Naomi Oreskes in no “uncertain” terms, without presenting *ANY* direct evidence to support the charges”
She lambastes her policy opponents as unscrupulous ideologues and views the “science” of global warming as unassailable. Her presentations are unabashed propaganda without any balance. This probably doesn’t win her any friends among those with opposing viewpoints.
She makes a number of dubious inferences:
1. Regulation is good.
2. Environmental extremism (or what many regard as environmental extremism) is good.
3. AGW is bad (any warming or human influence is by definition bad).
4. Her opponents are motivated by ideology and money to obscure the truth.
5. Any scientific paper that doesn’t explicitly reject AGW supports AGW.
6. She makes no distinction between AGW and CAGW and treats them as the same thing (they aren’t).
7. More CO2 and AGW are without benefit and are “externalities” that sources should pay for.
Most of the tenets of CAGW don’t withstand even superficial engineering analysis. This is problematic if CAGW is used as a basis for policy.
BINGO!
http://arstechnica.com/science/2014/10/people-think-their-opponents-are-hate-filledunless-you-pay-them-money/
This is Field Marshal Montgomery’s caravan. The pictures on the wall are the German Generals, Rommel, Model and Kesselring, who were his adversaries.
http://www.movilleinishowen.com/history/moville_heritage/henry_montgomery/life_magazine/enemy_faces.jpg
People on both sides of a divide think of their opponents as hate-filled, ignorant, unscientific, etc etc etc, where, in reality, they may simply have different beliefs.
Beliefs trump reason.
Oreskes is NOT a scientist and nothing she has published is science. I agree that people focusing on the fact the she is hideously ugly are out of bounds. That has nothing to do with the important point that she is wrong on most everything.
TAKE FOMD’S “NAOMI” QUIZ !!!
Question Who is most trustworthy regarding climate-change?
—————
http://www.eurweb.com/wp-content/uploads/2011/04/naomi-campbell.jpg
Naomi C?
—————
http://stephenleahy.files.wordpress.com/2010/07/naomi-orseskes-profile-pix.jpg
Naomi O?
—————
http://blogs.tennessean.com/tunein/files/2012/01/Naomi-Judd.jpg
Naomi J?
—————
The Obvious Answer *ALL* of them are *FAR* more trustworthy than Chris Monckton!
Oh heck … let’s try again for “Naomi C” !!!
http://imstars.aufeminin.com/stars/fan/naomi-campbell/naomi-campbell-20080928-458929.jpg
or maybe
http://i.telegraph.co.uk/multimedia/archive/01583/naomi_1583812c.jpg
but not
http://www.desmogblog.com/sites/beta.desmogblog.com/files/Christopher-Monckton-.jpg
`Cuz it’s appearance that matters, eh Climate Etc readers?
AFOMD,
Maybe I’m missing something here. What have people’s faces to do with their intellectual attainments?
Is this an American perceptions? You appear to choose politicians on their appearances – is it the same for supposed scientists?
In the UK, Steven Hawkings is accepted as being quite useful in at least one field, but you might well deride his physical appearance as being less beautiful than a fashion model. Is he more or less beautiful than Naomi? Is beauty (to Western eyes) a good indicator of intellectual prowess?
I think you are drawing a long bow here. A few facts to back up your unsubstantiated – and therefore meaningless – assertions or implications, might advance your cause. Or maybe not. What say you?
Live well and prosper,
Mike Flynn.
a fan of *MORE* discourse: The Obvious Answer *ALL* of them are *FAR* more trustworthy than Chris Monckton!
You would have a lot of trouble showing that.
Mike Flynn: What have people’s faces to do with their intellectual attainments?
FOMD is mocking some posts at WUWT.
Deduction by Marler, outrageous examples by WUWT commenters.
`Cuz if beauty counts, then angry old denialists are in a world of hurt!
How about none of the above? (Though I’ll admit I have no idea who the last Naomi is.)
None is qualified to discuss the subject.
You sir are barred from the fly-over states!
FOMD’S Rule Don’t mess with *ANYONE* named Naomi!
Fan
No I had never heard of the third Naomi either.
However your general point was that there were a number of unpleasant and sexist comments about Oreskes. Why some commentators took that unpleasant and outdated approach I don’t know.
tonyb
Anthony Watts has removed and snipped misogynistic comments about Naomi Oreskes before:
http://wattsupwiththat.com/2014/09/03/climate-craziness-of-the-week-naomi-oreskes-says-climate-change-will-kill-your-pets/#comment-1726737
With a blog as popular as his, there’s no way he can personally screen every comment.
Here’s a recent item, relevant to uncertainty and how it is used, on the evolution from tobacco to fossil fuels by Margo T. Oge.
http://www.huffingtonpost.com/margo-t-oge/blowing-smoke-at-global-w_b_6004774.html
Dealing with risk and uncertainty in policy (business) decisions requires a bit of attention to definition and detail. Uncertainty has a fairly clear and relatively uniform interpretation across scientists, decision makers, and the lay public. Risk can be perceived much less precisely, but should be subject to precise definition. The components you list, causal relationships, likelihoods, and consequences are fundamental but “risk” as used in decision analysis does not mean uncertainty, it is defined as the expectation value of a consequence, usually as a combination of a probability (or likelihood) and a consequence. The combination depends on the nature of the probabilities. Often independence is assumed and the combination is taken to be multiplicative. In reality consequences are rarely independently distributed.
Risks in the climate arena seem to be essentially anecdotally brought up without any integrating view. Studies keep coming out that evaluate the likelihood of more or worse heat waves, more or worse or more variable tornadoes, worse floods, inundated shorelines, etc. all under differing sets of assumptions (such as model outputs). But the individual assessments are used more to inflame risk perception than to incorporate in a consistent integrated analysis of expected outcomes.
A well framed decision would consider what objectives are to be maximized (or minimized if they are costs not benefits), what quantities (attributes, independent variables, etc) contribute to the objectives, what decision alternatives are to be decided amongst, and what impact those alternatives would have on the attributes and hence the objectives. In many engineering or business applications it is a simple analysis that involves only engineering or monetary alternatives, attributes, and objectives.
A complicating factor that seems to not have been a topic at the workshop is the varying perceptions of risk (how averse or not averse to bad expected effects or how desirous of good expected effects are different impacted communities, or stakeholders). Most policy issues involve perception of risk as an unavoidable part of the analysis. The climate policy discussion seems not to have explicitly recognized this, at least not formally. Expected outcomes, particularly those involving future costs such as sickness, death, economic loss, etc., can be valued much differently by various stakeholders in the decision. (Attorneys in my experience tend to be very risk averse.) This valuation can be estimated via fairly rigorous preference elicitation methods (although this gets into social science and hard scientists often cringe at this point). Unfortunately the discussion on risk preferences in the climate policy realm seems more oriented toward demonization of one side and is too often not attempting a positive contribution to objective analysis.
A variety of policy issues suffer from lack of attention to risk preferences such as nuclear power, GMOs, carbon pollution, etc. A fundamental fact that needs to be accepted is that it is OK for different communities to view risk differently. Only by understanding the different preferences in the analysis can a decision find balance among the perceptions of risk of all stakeholders in a manner such that the stakeholders know how well their risk perceptions are addressed in the final decision. We are used to this implicitly in motor vehicle use, air transportation, vaccine administration, etc. There are fairly well characterized risks and stakeholders may in many cases choose how much exposure to the risk they will tolerate. Policy decisions can impact the risks by requiring safety equipment, procedures, or a regimen of progressive human trials. The treatment of risk in climate policy seems to me to lack a notion of completeness, balance, objectiveness, and appreciation for alternative views that we have applied to other policy decision areas.
Unfortunately, in my experience and perhaps in the climate arena, policy decision makers have not truly desired an objective, balanced, transparent decision process. Instead they more routinely have opted to make the decision themselves, satisfying those they wish to satisfy, and the others can pound salt. The most important part of any rational policy decision is the desire of the decision maker to make such a decision.
Good post, might need repeating in a few months.
I think if scientists would be clear about what they don’t “know” the climate change initiative would die an instant death.
To the best of my knowledge, we (everyone) don’t know:
– WHAT is inside an atom (we simulate the energy field we think is probably there):
https://www.youtube.com/watch?v=J3xLuZNKhlY
– WHAT comprises 96% of the unverse (we have quantified the existence of Dark Matter, and Dark Energy):
https://www.youtube.com/watch?v=R5orcCuprG4
– WHETHER the Higg’s Bosun that was found is THE Standard Model Higg’s Bosun, ie how gravity actually works, (see 5:40 in video):
https://www.youtube.com/watch?v=649iUqrOKuE
– knowledge of the cause of the Grand minimums and Grand maximums in the solar cycle?
At best, even IF the 97% concensus were true, what are they armed with?? A non-clear-cut, quasi-knowledge, of FOUR percent of the universe.
What are the non-alarmists armed with? The same 4%.
Please keep in mind that 50% of the population of the U.S. of A. are BELOW average intelligence.
Does the term “ship of fools” ring a bell??
I like to remind people that when listening to the advice of their doctor, there is a 50% change that said doctor graduated in the bottom half of their class.
Wonder what percentage of climate researchers graduated in the upper half of their physics and statistics classes.
timg56,
Even worse, if a student attempts 100% of the questions, and correctly answers 70% to pass, this shows the student wrongly assumed they knew the answer in 30% of cases. Not particularly reassuring, if depending on such a person for reliable advice.
As far as I know, climate scientists are self appointed. Climate is the result of calculation of an average of weather parameters over an arbitrary period. Hardly the sort of thing any first or even second rate mind would bother with – unless extremely gullible or possibly congenitally duplicitous.
I work on the assumption that anyone who claims their main scientific area is climate science is either a fool or a fraud. So far, I don’t seem to have suffered any adverse effects as a result of this approach.
Live well and prosper,
Mike Flynn.
It is not always true that a lawyer does not have to present evidence that hurts his case. For Prosecuting attorneys there is the “Brady” rule that requires that they inform the defense of any exculpatory evidence.
@Alistair
Agree totally. I think a misconception of non-scientists it that we (scientists) KNOW some things about reality, when in fact we only have momentarily successful ways of conceptualizing phenomena that are useful. In the past we have KNOWN that particles were particles, until we discovered they could be viewed as sometimes waves. The (particulate) electron is not something that exists in nature, it exists as a model in the human mind that helps us interpret measurements. Or we have KNOWN that the vacuum was empty, until we found out it could be viewed as seething with activity. The notion that we know Reality at any stage I think detracts from scientific progress. We should always be able to accept that what we KNEW as reality was just a Platonic reflection of reality on the cave wall that we need to be ready to change when more data becomes available.
@ FAH Thanks. It’s a fair sized “bee in my bonnet”.
This statement by a renowned heliophysicist brings it to the finish line for me…
Jenkins said.
“We haven’t known the solar neutrino to interact significantly with anything, but it fits with the evidence we’ve gathered as the likely source of these fluctuations,” he said. “So, what we’re suggesting is that something that can’t interact with anything is changing something that can’t be changed.”
And that statement to me is the equivalent of saying (harshly worded, but accurate):
“hahaha, OK climatologists,
“sure” the climate is changing,
“sure” your consensus is important,
“sure” you “know” what your talking about,
“sure” CO2 is VERY important, “really I believe you, really”.
“97%, my that is impressive, and so BIG a number.”
“Now why don’t you be good little boys and good little girls, and run along and play.”
“And try not to cause too much societal alarm, or unnecessary economic destruction”
“and try not to distract real scientists from continuing their adult work with your silly childish uninformed observations, opinions and claimed “knowledge”.
“And for God’s sake, learn to think before you speak with “certainty” because you are unable to properly assess your own ability to be “certain””
Climatologist who think they KNOW what they are talking about
= Puck from Shakespeare’s Midsummer Nights Dream
This phrase: “… momentarily successful ways of conceptualizing phenomena that are useful.”
Is of course another word for MODELS.
But let’s not let the term “MODELS” interfere with the fact that they are based on an inadequate understanding of about 4% of the known universe. :-)
Percentages are a deceptive way of trying to indicate the importance of or uncertainty something. A single bullet represents less than 0.0000000000000001% of the known mass of the universe but it can still kill you. As humans seem to occupy the non-dark matter, non dark energy portion of the known universe, which seems to be smaller than the dark energy/dark matter portion, best we still with things that are real, and not try to deceptively use numbers to indicate uncertainty.
“Models” is another word that can be interpreted differently. I think the meaning I had in the phrase was emphasizing the conceptual nature, that is the idea we have even before we start writing equations and generating computer code. Modern parlance in climate science seems to equate “model” with some number of lines of specific computer code. Many such codes seem to me to have similar conceptual frameworks (hesitating to use the word model here) and differ largely in how they arrange the input and output and time and space slicing.
I guess I would call the standard Model in particle physics, or the expanding universe Model, or a genetic Model of behavior, or the plum pudding model of the nucleus of Rutherford’s time as conceptual models. They can each have different realizations in equations and computer codes. Things like the CMIP models or the MODTRAN model for transmission are, in my view, models, not Models.
I am not a climate scientist, but it looks like the notion of “forcings” radiative or otherwise as an accurate and complete accounting for climate thermodynamic behavior might classify as a Model.
A pet peeve I do have is the apparent current practice of calling a set of runs of a model (small case model, usually upper case acronym) as an “experiment.” I am not ready for that yet. Perhaps when all consciousness and reality is borged into computers that might be relevant, but not yet for me.
are not “dark” matter and energy an ad hoc hypothesis created to explain the unexpected observation of accelerated universal expansion?
just as “missing heat” is a postulated explanation for predicted 21st century warming that has failed to happen
’cause the basic theories in both cases have major flaws
R Gates. that is true.
AND every second TRILLIONS of neutrinos pass through the bullet (even if it is in a human), then through the earth.
I don’t think we have a good clue as to what they are, what they interact with, or what they do, or what they can do.
I believe (100%), I will NOT get a good clear explanation of neutrinos, or the above stated areas of limited awareness from a Climatologist??
Will I??
Maybe the problem is the word “Climatologist”…. Since they can ONLY have “opinions”, and do NOT have “knowledge”, and since they are so obviously easily swayed by an even less informed public opinion, their title should be reduced to “clima-tician” or “clima-polit-ologist”. That would be fair and clear forewarning of their actual agenda and source of opinion for all to see and understand. (in keeping with the nature of this particular post).
The real John Smith….their existence has been “quantified”. I think it is accepted. I accept it. NASA “accepts” it. and many others have used it in their calculations and observations (Laniakea, Perseus findings) that lend credibility to their existence. they just don’t know what they are. Only what they do.
A celebrated science-poem, communicated by FOMD to Beth and Kim!
Mr. FAH I had little experience with hydrogeolocical models. One of the lessons I received is “garbage in”, “garbage out”. I remember how some models predicted the drying of some springs in an aquifer for decades ago, with perfectly continue with his nowadays flow. Evidently the model had introduced very bad data inside, the “conceptual” model was very bad, and the “forcings” (we didn’t call them like that, then) were not very accurated. But models when well accurated were very useful to see if some of the hypotheses were false, or to inferr the range of some variables of the conceptual model of the aquifer system.
Hey Fan,
Your literary tastes are improving. Updike a nice change from Wendell. I’ve proposed a bet to you a bit upstream. Check it out…
Gates,
While said bullet “can” kill you, try checking on the chances of any one bullet doing so.
His name is Gary Hobson. He gets tomorrow’s newspaper today. He doesn’t know how. He doesn’t know why. All he knows is when the early edition hits his doorstep, he has twenty-four hours to set things right. (IMDb, Early Edition, TV Series 1996-2000, Drama, Fantasy, Sci-Fi).
Is anybody reminded of Naomi Oreskes? After a substitution of a name and a change of Sci-Fi to Fi-Sci the fit is almost perfect.
#curious George… I think you have inspired a term that should exist if it does not….Cli-Sci-Fi
Cli-Fi would do.
.. Chi_Fi bubble in the twitterverse
True. Putting the “sci” in their may lend them false credibility if it’s stickiness to the “fi” wasn’t emphasized. :-)
• Yes, Cli Fi exists … it’s become one of the “hottest” publishing genres (LoL).
• Yes, neutrino observatories exist too.
From where we’ve been at least we know where we are –e.g.,
No fair, many nice people do not believe in the pause, try W Connelly and Tamino and the general public.
“Communicating uncertainty is hard”
Baloney. Communicating uncertainty is easy. What is hard for the warmists is trying to convince voters to accept massive policy initiatives in the face of uncertainty.
Thus you get the “uncertainty principle” and absurd “percentages” of certainty based on “expert judgment” in the ARs.
Here’s how you communicate uncertainty.
“We don’t know why there is an 18+ year ‘pause.'”
“We don’t know why there is an increasing divergence in the temperature reports and model projections.”
“We don’t know how to model El Ninos or the other oceanic oscillations with any degree of accuracy.”
“We don’t know the effect of water vapor and clouds as feed backs with any accuracy.”
“We don’t know how to model the global climate sufficient to give any realistic projection of future temperatures in the next 10, 20, 50 or 100 years, with an accuracy that would justify massive public policy changes”
See how easy that is?
+1
And not only that, Mr GaryM, we don’t have the possibility to introduce those uncertainties in the current computer codes, or to make new computer codes that include those conceptual models. Here some people talks about attorneys ethics, and one of the principles of justice is “equality of processal arms”
You are missing the most important example – 2009 L’Aquila earthquake
http://www.wired.com/2014/10/italian-scientists-appeal-laquila-quake-conviction/
What happened with the L’Aqula earthquake is a classic example of communicating uncertainty in the context of competing scientific and pseudoscientific claims. It all went horribly wrong
Interesting thread on Twitter now, between Shub and Jim Bouldin re SciAm paper, Ed Hawkins and Mannian statements regarding 2C N hemisphere danger zone etc. https://twitter.com/jim_bouldin/status/524641938839183360
Ethics anyone?
Communicating uncertainty is hard, but mainly because it requires self-awareness of what one does and does not know, whereas most people (even scientists) tend to be too sure of what they “know”.
re: risk. We tend to fear that which is more horrible, even if rare. So we are terrified of sharks and ebola whereas flu kills 3000 to 50,000/yr (CDC) in the USA or more (hard to estimate) and autos kill more than that. In the climate debate therefore the emphasis is on scary “disruption” and flooded new York city, even if NYC is already likely to flood because they built the S end in a swamp and didn’t fill it enough.
NY is overdue for a cat4/5 hurricane, yet we want to impoverish the developing world because we may make that hurricane half a percent stronger, or weaker… we don’t know the sign. Or come a day earlier, or later.
There is more than one ethics of communicating scientific uncertainty. Unfortunately, one used frequently in climatology, is to sweep the uncertainty under the carpet. Make a simplifying assumption, don’t include it in the list of approximations used, and you don’t have to bother with an estimate of its impact. Then run the model and yell .. temperatures in 2100 MAY be six degrees higher. When asked about the accuracy of the model, yell .. how accurate our model is is NOT HUGELY IMPORTANT.
Example: an approximation (neglect the heat contents of any airborne water) used in CAM5 model is undocumented, and CAM modelers don’t even know why they are forced to assume a constant latent heat of water vaporization. Fortunately, “Gavin” in my second reference could shine some light on it.
Long live false (scientific?) certainty.
http://judithcurry.com/2013/06/28/open-thread-weekend-23/#comment-338257
http://judithcurry.com/2012/08/30/activate-your-science/#comment-234131
“Fewer than 10% of journalists are skeptical about climate change, but more than one third of journalists believed that coverage of climate change science must reflect a balance of viewpoints or present all sides of controversy (half did not believe in the need for balance).”
So even allowing for the usual rigging, push-polling and slob terminology behind such “findings”, we can safely conclude that the bulk of journalists are mental adolescents. But maybe I’m wrong. Maybe they’re like “a bridge across the is–ought divide in science-related policy making, using the uncertainty topos to invite the public to participate by supplying fears and values that would warrant proposals for limiting pesticide use.” (A journalist wrote that.)
Climatologists and Cosmetologists share the same goal: making something appear better than it really is. In the case of Climatologists, making their conjectures of man’s monstrous behavior causing distant future calamity upon earth’s environment is a task of control knob manipulation until the outcome says what it is you want it to say. In cosmetology, there are all sorts of specialties: shampoo technicians, hair stylists, colorists, estheticians, etc. who dabble and dab until the outcome looks just right. Models anyone?
Both practitioners are concerned about the appearance of the outcome. The search is not for the truth of internal variability, nor for the goodness of the inner person; rather, both are concerned what the product looks like. Can the model projection, can the person leaving the establishment impress others of their value, wholesomeness, and appearing to be better than they really are.
Climatologists should be required to obtain a license from the state, just like cosmetologists are, to practice on the public.
I once interviewed a young woman whose paperwork indicated that she was an oncologist. Intrigued, because she was about 20 years old, I inquired if she was in the medical field. Her answer was ‘Yes, but I only do nails’. She was an onychologist.
==================
Is that a true story Kim?
If yes I will tell my colleagues at work.
Yes, and I thought you’d be amused.
=======
Kim
Pardon me. I can not find the word: onychologist; i.e., one who attends to nails. At times, on Sunday, I hear something about being nailed to a cross. I associate being nailed to a cross with martyrdom and suffering. Are you saying that climate scientists are martyrs? nailed to the cross of CAGW? destined to lead others through the valley of death? turn water into wine? and otherwise forgive and forget? Are you speaking of Michael Mann?, Kevin Trenberth and others as the apostles? Or, as the drama plays out, has climate science become a charade, bestowed upon millions, bejeweled with pomp and circumstance, yet, in the end, peddlers in gowns instead of rags, nevertheless, peddling pittance, vulgarity, and, with eyes wary, collecting the money of an unsuspecting public?
RiH, onychology (uncountable) (medicine) the study of fingernails and toenails.
Q&A with an online doctor: “I understand that there are onychologists, who, I’m informed, specialize in nail problems.” “ Few physicians limit their practice to nail problems, but yes, they exist, and are called onychologists.”
DocMartyn, something else for you. My son just dropped in to solicit my blood for his research project: Immunological susceptibility to pulmonary Nontuberculous mycobacterial disease. Some children seek to get their hands on their parents’ wealth; our kids are better of than us, so need to seek an alternative. (I’ve been organising asbestos removal for another one today. I hope that doesn’t compromise the PNM study.)
I very much doubt the claim that fewer than 10% of journalists are skeptical. Even democrats do not poll that low, more like 20%, with conservatives running around 80%. (There are a goodly number of conservative journalists.) When I was a journalist in the electric power industry ten years ago every journalist I knew was a skeptic. Perhaps the journalists attending the ethics workshop were all liberals. Liberals like this sort of feel good stuff.
David,
How many of those electric utility journalists had a science or engineering background?
I took a couple of journalism courses after seeing how msm covered commercial nuclear power generation. The quality was barely above appalling.
Looking for something on the ethics of scientists, this is from Nancy L. Jones:
Principles for the practice
Objectivity Honest assessment and minimization of the biases inherent in science, i.e. cultural and other influences on the experimental design, techniques and interpretation of the data
Questioning certitude Questioning current authoritative view or dogma in order to continue the process of advancing new knowledge
Research freedom Allowing ideas to flourish within the scientific community because wrong or true concepts will eventually be proven as such
Research reproducibility Quality scientific research can be re-proven and is openly available to all qualified scientists to move knowledge forward
Respect for subjects The highest ethical standards are upheld to respect all living things, with profound respect granted to human life and dignity
Scientific community The scientific community is the guardian for the integrity of science by proving the veracity of individual findings through peer review and reproducing experimental results, and by training and accrediting future scientists
Virtues
Duty Scientists are committed to serve and guard humanity and seek to advance scientific understanding and respect for the truth
Integrity Scientists strive to be objective, fair, truthful, and accurate
Accountability Scientists are accountable to their profession and society
Altruism Scientists’ primary focus is the best interests of humanity and not self interest, commercial interests, or the promotion of the industry of science
Excellence Scientists are committed to a lifestyle of learning and transmitting knowledge and skills
Respect for colleagues Scientists treat associates and trainees with respect and credit their contributions
http://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&ved=0CCkQFjAB&url=http%3A%2F%2Fwww.openu.ac.il%2Fcode_of_ethics%2Fdownload%2FA_code_of_ethics_Jones_2007.pdf&ei=mMpGVNfCGY63yASvhILgDA&usg=AFQjCNFvgxYlYB_5GSWAjpfjbtDE0ZJa7g&bvm=bv.77880786,d.aWw&cad=rja
The above points are further explained at the link. It’s easy for me to say that scientists should voluntarily adopt some form of a code of ethics. I find there are a lot of positives of CPAs having one though ours is mandatory and we can be punished for violations.
Well, that pretty well sums up how I behave, let’s not confine it to scientists. But, as always, it is the core volition which counts, with good volition and understanding/wisdom, you don’t need external codes of conduct. And you can’t guarantee compliance from those who lack the volition.
I think that it would be interesting that other options, other conceptual models, could be someway tested in the models (knowing it’s not, by far, an experiment). In someway models are a tool, a “weapon”, that is only in one side. And we are seeing there are not another opposite side, but many other sides with different hypothesis. But I suppose models are very expensive and so they are only in the hands of government or quasi government agencies, or require of government money to be developed. So the introduction of other “conceptual models” in the “computer codes”, as mr. Fah wrote.
Jeff Conlon on The Air Vent asked: “So you are stating that your idea is not new and are of the belief that Loshmidt had the right answer correct?”
To this I replied:
The fact that the temperature gradient forms autonomously at the molecular level (without any specific need for upward convection) was first explained in the 19th century, and has never been correctly rebuked. But this “gravito-thermal effect” has been overlooked by James Hansen et al. Hence 255K is not the right “starting point” and there is not “33 degrees of warming” but more like 10 to 12 degrees of cooling by radiating molecules, mostly water vapour of course, because the radiating properties of these molecules have a temperature levelling effect working against the gravitationally-induced temperature gradient that is not due to any lapsing process..
But what has not been explained prior to the 21st century is how the necessary energy transfers over the sloping thermal profile just like new rain water falling on a small section of a lake spreads out evenly over the whole lake. This is what happens (and must happen) in planetary tropospheres, and it happens because the Second Law of Thermodynamics is all about thermodynamic equilibrium evolving. Thermodynamic equilibrium has a density gradient (because there must be more kinetic energy per molecule at lower altitudes) and that density gradient thus has a temperature gradient.
Thermodynamic equilibrium is what it says – an equilibrium state just as much as is mechanical equilibrium which keeps the surface of a lake more-or-less following the curvature of the Earth. Gravity spreads new rain water over the lake, raising the level all around the lake. Likewise, new thermal energy absorbed in a planet’s upper troposphere or elsewhere (such as in and above clouds) spreads out in all accessible directions by convection, where I use the term to mean both diffusion and advection in accord with normal usage in physics. And that’s how the required energy gets down to the base of the Uranus troposphere to maintain temperatures hotter than Earth’s surface. Likewise on Venus and likewise on Earth because solar radiation directly to the surface is nowhere near sufficient and we would freeze on cloudy days if this downward convection were not a reality.
Moderator:
This comment is being posted on about six other blogs as I don’t like wasting my time on just one blog, unless a blog owner runs an article on the content of my book and agrees not to delete any of my comments replying to comments on that thread. I may do likewise with any future such questions and answers in the interests of disseminating correct physics and gradually wearing down the greatest error ever made since the flat Earth garbage.
LoL you ain’t never run a molecular simulation in a gravity field, have yah “physicist”?
http://upload.wikimedia.org/wikipedia/commons/6/6d/Translational_motion.gif
FOMD is pleased to supply simulation code that will help to amend the errors in your understanding!
That is fake.
Particles skim slowly along the base and then shoot up, moreover the particles do not follow parabola.
Thus we have the juxtaposition of the other two dimensions.
The utterly mad and the completely dishonest.
What part of “hamiltonian flows induce symplectic isomorphisms“ troubles your dynamical intuition, Rob Ellison?
The part where you link to random Wikipedia pages you clearly don’t understand for obscure but quite likely pathological reasons.
The chief o’ unintentional irony strikes again!!!!1111!!!
We get this: “Thus we have the juxtaposition of the other two dimensions.
The utterly mad and the completely dishonest.”
The part where you link to random Wikipedia pages you clearly don’t understand for obscure but quite likely pathological reasons.
Along with this:
> The utterly mad and the completely dishonest
I must admit that there are still serendipities that astonish me :)
An amazing exchange ….
Rob Ellison, it’s definitely true that at least one of us has in insecure grasp of the foundations of computational statistical mechanics … even at the Wikipedia level.
http://upload.wikimedia.org/wikipedia/commons/6/6d/Translational_motion.gif
Example As one learns more-and-more about symplectic integration (per the above dynamical simulation), statistical mechanics and thermodynamics become less-and-less “art” and more-and-more “science” … until the point is reached that the gravito-thermal fallacy dissipates like a childhood dream.
Best wishes for enjoyable computational learning are extended to *ALL* Climate Etc regulars!
What part of “hamiltonian flows induce symplectic isomorphisms“ troubles your dynamical intuition,
The part of bringing a constraint ( the temporal horizon of weather prediction) that is well described in the mathematical literature eg Arnold.
the strategy of applying geometric methods to the infinite dimensional problems is as follows. Having established certain facts in the finite dimensional situation (of geodesics for invariant metrics on finite-dimensional Lie groups), one uses the results to formulate the corresponding facts for the infinite-dimensional case of the diffeomorphism groups. These final results often can be proved directly, leaving aside the difficult questions of foundations for the
intermediate steps (such as the existence of solutions on a given time interval). The results obtained in this way have an a priori character: the derived identities or inequalities take place for any reasonable meaning of “solutions,” provided that such solutions exist. The actual existence of the solutions remains an open
question.
For example, we deduce the formulas for the Riemannian curvature of a group endowed with an invariant Riemannian metric. Applying these formulas to the case of the infinite-dimensional manifold whose geodesics are motions of the ideal fluid, we find that the curvature is negative in many directions. Negativeness
of the curvature implies instability of motion along the geodesics (which is well-known in Riemannian geometry of finite-dimensional manifolds). In the context of the (infinite-dimensional) case of the diffeomorphism group, we conclude that the ideal flow is unstable (in the sense that a small variation of the initial data implies large changes of the particle positions at a later time). Moreover,
the curvature formulas allow one to estimate the increment of the exponential deviation of fluid particles with close initial positions and hence to predict the time period when the motion of fluid masses becomes essentially unpredictable.
For instance, in the simplest and utmost idealized model of the earth’s atmosphere (regarded as two-dimensional ideal fluid on a torus surface), the deviations grow by the factor of 10^5 in 2 months. This circumstance ensures that a dynamical weather forecast for such a period is practically impossible (however powerful the
computers and however dense the grid of data used for this purpose)
The habit of dropping Wikipedia links to esoteric subjects of no possible relevance is everywhere in evidence – that he has very little competence in the subjects is obvious over time with errors compounding. The point of all this bad faith – effectively lies and fraud – is not evident on the surface. It is not about understanding and communicating science obviously but of the delusion of advancing an agenda of societal and economic transformation.
Joshua thinks it is about me and quotes the Allinsky rule I quoted. But both FOMBS and Joshua demonstrate the workings of the rule. FOMBS in using – superficially – the trappings of science to ridicule a ‘denier’. Joshua in using the tired old complaint of unintended irony. Loosely based on the idea that quoting Allinsky suggests that I see myself as purer – and that objectively criticising FOMBS for bad faith shows that I am not and just don’t realise it. So a bit of unintended irony there from Joshua.
It all relies on a chain of reasoning that is false. First that my observation is ridicule rather than a genuinely held opinion on a puzzling – and quite likely pathological behaviour. Second that quoting Allinsky’s rules for radicals means that I am professing to be beyond using ridicule. Neither is the case.
Arnold’s Corollary As dynamical flows become more-and-more unstable, thermodynamic predictions become more-and-more reliable.
That’s the microscopic basis of Onsager’s reciprocity relations!
Nowadays pretty much *EVERY* STEM student codes up Hamiltonian simulations at *SOME* point in their education. That’s why young 21st century scientists develop a *MUCH* deeper and more integrated understanding of thermodynamics and statistical mechanics than 20th century scientists!
Conclusion If you’ve never computationally integrated a high-dimension symplectomorphic flow … now’s a *GREAT* time to get started!
Rob Ellison, are you *ABSOLUTELY* certain that symplectic integration is of “no possible relevance” to computational statistical mechanics and/or thermodynamics?
http://i2.kym-cdn.com/photos/images/newsfeed/000/621/992/b85.jpg
Well there’s your problem right there!
FOMD’s Thermodynamic Conundrum
On the surface of the airless Moon sits a box of ordinary Earth-air at standard temperature and pressure. Upon the box’s top surface is a pinhole leak, from which the molecules fly upward and (eventually) fall back (without interacting with one another during flight).
Question Of those molecules that fly one kilometer high or higher, what is their average kinetic energy as they pass through a height of one kilometer? Through two kilometers? Through three kilometers?
Answer The average kinetic energy of the molecules that freely fly to any given height is independent of that height.
At larger heights there are fewer molecules … but the average per-molecule kinetic energy is unchanged.
Recommendation Don’t take FOMD’s word … calculate the molecular-energy distribution for yourself, straight from first-principles Newtonian dynamics!
So what he is saying is that at every level –
1/2(V1^2 +V2^2+ … +Vn^2) = a constant despite the fact that all are losing kinetic energy as they rise.
The whole idea is utterly irrelevant to global energy dynamics.
But a simple test – let him provide the proof.
In the absence of gravity the result is standard. In particular, the equations of motion allow for stationary states with uniform density and temperature. The effect of gravity was included by adding the gravitational potential energy to the hamiltonian; which is standard practice. The result of that modification is that, in the presence of the gravitational field there are no longer any stationary solutions with uniform temperature. Instead both density and temperature decrease with elevation. This should be welcome as being in agreement with what is observed in real atmospheres. The fact that a portion of the atmosphere of the Earth exhibits the same temperature profile is a surprise; it suggests that the temperature gradient is not a product of radiation.’
Heat and Gravitation. I. The Action Principle, Christian Fronsal, Univ. Califormia, 2011.
So adding gravity to the atmospheric Hamiltonian suggests that the gravito-thermal effect exists. But of course real science depends on empirical proof.
Context is everything in science communication. Going beyond a link to a Wikipedia page to a discussion of the relevance for hydro/thermo dynamics. So the question for FOMBS to answer is – what is it? A random link to Wikipedia or something that has a concrete application that advances understanding of atmospheric dynamics. Rational maths and science as opposed to look here’s some esoteric bit of math so this proves catastrophic global warming and we need to overthrow capitalism and democracy.
So by all means quote some science – FOMBS – that has some relevance to climate – and not just post an image of a collapsed road.
We’ll have Mathematica check the calculus!
(* start with Boltzman distribution *) NormalDistribution[0,1]// (* compute distribution at height zh *) Integrate[ PDF[#][z] z*DiracDelta[zp^2-(z^2-zh^2)], {z,zh,Infinity}, GenerateConditions->False ]&//#/Integrate[#,{zp,-Infinity,Infinity}]&// (* verify that the distribution is unchanged *) #==PDF[NormalDistribution[0,1]][zp]&// (* true iff gravito-thermal effect is zero *) Simplify//Print["Result: ",#]&; Result: TrueIt is a pleasure to assist your thermodynamical understanding, Rob Ellison!
Rob Ellison: “So what he is saying is that at every level –
1/2(V1^2 +V2^2+ … +Vn^2) = a constant despite the fact that all are losing kinetic energy as they rise.”
The point that you keep missing (or rather ignoring,) and it’s the main point FOMD made in his conundrum post (but you always snip and ignore that part) is that not *all* the molecules from one given level rise to the higher one up. That’ not just true on the Moon but within any parcel of gas in a gravitational field. That’s because they don’t all have enough kinetic energy to climb across the gravitational potential. Many fall back down before getting there. So, this drop-off effect skews the energy *distribution* up even though all individual molecules that are going up decelerate and conserve energy (KE+PE) *individually*.
There also is another subtle point that’s easy to miss; and that Doug, yourself, and all your other fellow gravito-thermalists, always miss in addition to the previous one. (To be fair I also missed it until I did some calculations yesterday.) It’s the fact that each individual free falling molecules that contributes to the distributions of two different thin horizontal layers (either on its way up or down across them) spends less time in the lower layer since it’s moving faster across it. It therefore weigh less in the energy distribution for the molecules that occupy that layer at any given time.
Once you account for those two additional effects, they cancel the effect from the variation in kinetic energy of the rising and descending particles on the temperature gradient. The three effects *together* account for the gas being isothermal at equilibrium.
“The effect of gravity was included by adding the gravitational potential energy to the hamiltonian; which is standard practice. The result of that modification is that, in the presence of the gravitational field there are no longer any stationary solutions with uniform temperature.”
Not true. That’s not merely a result of the modification of the Hamiltonian. You are misinterpreting the work of Christian Frønsdal and are drawing categorical conclusions that he presents as tentative suggestions (and that don’t mesh too well with standard statistical mechanics and the second law). He concludes his paper thus:
“The conclusion of this study is that the requirement of an isothermal equilibrium is *consistent* with thermodynamics, but at a price that *may* be too high. If the implications for the gravitational interaction are unacceptable, then work remains to be done, and the need for an experiment remains urgent.” (My emphasis)
In any case, his derivation of a temperature gradient in a gas at equilibrium in a gravity field is anything but standard (as he himself acknowledge). Ever since the debate between Maxwell and Boltzmann, on the one side, and Loschmidt on the other side, ended, the huge majority of physicists have sided with M&B, including Loschmidt himself!
‘Rob Ellison: “So what he is saying is that at every level –
1/2(V1^2 +V2^2+ … +Vn^2) = a constant despite the fact that all are losing kinetic energy as they rise.”
P-N starts to say that I have forgotten that some molecules don’t make it between 1 and 2km. I haven’t – n is of course the number of molecules at the level.
I am afraind I didn’t read further than that – extreme verbiage doesn’t constitute proof. Not is some garbled ‘code’ more than lies and fraud.
‘Not true. That’s not merely a result of the modification of the Hamiltonian. You are misinterpreting the work of Christian Frønsdal and are drawing categorical conclusions that he presents as tentative suggestions (and that don’t mesh too well with standard statistical mechanics and the second law).
So quoting the paper is misrepresenting the paper?
‘The statement that any two thermodynamic systems, each in a state of equilibrium with a well defined temperature, and in thermal equilibrium with each other, must have the same temperature, is a central tenet of thermodynamics. A natural generalization is that the temperature, in an extended but closed system in a state of equilibrium, must be uniform, and there is near universal agreement that this remains true in the presence of gravitational fields. This is important for the understanding of terrestrial and stellar atmospheres, where the gravitational forces create a non-uniform density distribution.
It must be stressed that an actual, terrestrial or stellar, atmosphere is not isolated and is not a thermodynamic state of equilibrium. It is, at best, a stationary configuration in which the effects of incoming and outgoing radiation are balanced in what is called radiative equilibrium. When such atmospheres are described by a mathematical model, then this quiescent state is mapped to the equilibrium of the mathematical model. What
is puzzling, is that the mathematical model, intended to describe an isolated gas, with the gravitational field included in standard fashion, turns out to be a model (in fact, the standard model) of an actual atmosphere that is far from being isolated. This mystery has an easy, but unpopular resolution: If one could concede that the true equilibrium state may be isentropic instead of isothermal, then the role of incoming radiation (in the case of the earthly atmosphere), or internal generation of energy (in the case of stars) would be simply to make up for the loss of energy radiated into the cosmos. So far, we have not
found an attractive alternative.’ op. cit.
I have skimmed the paper – and am going back to come to terms with the Lagrangian formulation for the Hamiltonian for the atmosphere. How’s that for substituting jargon for clarity. It quantifies energies (say kinetic and potential) in a dynamic equation in a generalised co-ordinate system (r = q,p,t).
But as I said quite explicitly – the maths suggests as isentropic atmosphere – but science is empirical Of course he suggested an experiment. It is not science without experiment.
Rob Ellison: “P-N starts to say that I have forgotten that some molecules don’t make it between 1 and 2km. I haven’t – n is of course the number of molecules at the level.”
In order to claim a vertical gravity induced temperature gradient, one must consider more than one single level. When you do so, you are considering *different* collections of molecules. I didn’t offer a proof about isothermality at equlibrium, myself. I merely countered *your* over-simplistic argument and pointed to your incorrect tacit assumptions.
Rob Ellison: “So quoting the paper is misrepresenting the paper?”
Sorry, you hadn’t put any open quote before this extract in your post and I had assumed that it was your own paraphrase. I didn’t see the orphan single closing quote at the end. My confusion… helped by your mistake.
In any case, while his form for the Hamiltonian of the dynamical system is standard, his derivation of a thermodynamical equilibrium state is quite non-stadard and, as I said, conflicts with the second law of thermodynamics. Indeed, the only tentative empirical proof to which he refers, the experimental setup of Roderich W. Graeff, allegedly behaves — and is advertised by Graeff himself — as perpetual motion machine that produces free energy indefinitely out of nothing. This is worse than cold fusion.
Rob Ellison: “P-N starts to say that I have forgotten that some molecules don’t make it between 1 and 2km. I haven’t – n is of course the number of molecules at the level.”
In order to claim a vertical gravity induced temperature gradient, one must consider more than one single level. When you do so, you are considering *different* collections of molecules. I didn’t offer a proof about isothermality at equlibrium, myself. I merely countered *your* over-simplistic argument and pointed to your incorrect tacit assumptions.
I’m not claiming a gravity induced temperature gradient. Merely reading a mathematically interesting – but difficult paper – on action principles in the atmosphere. At the very least it goes through some relevant maths and physics of the atmosphere. Including a detained and erudite history of the issue.
He seems to be talking 2 separate issues here. The first is the average kinetic energy of molecules was constant at every level. This is the context of molecular projectiles on the moon. No remote connection to the real world. With this we can consider only Newtonian laws of motion where for an individual molecule kinetic energy plus potential is a constant at any height like any other kind of projectile. This kinetic energy is not the kinetic energy of gases in the atmosphere – which according to kinetic theory – the essence of balls in the box gif from Wikipedia I posted and FOMBS has repeated 3 times now – is random in 3 dimensions.
No what he is saying is the sum of the kinetic energies of all molecules divided by the number of molecules is constant at every height for these molecules escaping a box on the moon. It is an utterly stupid idea supposed to refute something or other and I refuse to think about it any more.
Rob Ellison: “So quoting the paper is misrepresenting the paper?”
Sorry, you hadn’t put any open quote before this extract in your post and I had assumed that it was your own paraphrase. I didn’t see the orphan single closing quote at the end. My confusion… helped by your mistake.
So the citation at the end didn’t give you a clue?
In any case, while his form for the Hamiltonian of the dynamical system is standard, his derivation of a thermodynamical equilibrium state is quite non-stadard and, as I said, conflicts with the second law of thermodynamics. Indeed, the only tentative empirical proof to which he refers, the experimental setup of Roderich W. Graeff, allegedly behaves — and is advertised by Graeff himself — as perpetual motion machine that produces free energy indefinitely out of nothing. This is worse than cold fusion.
‘The effect of gravity was included by adding the gravitational potential energy to the hamiltonian; which is standard practice. The result of that modification is that, in the presence of the gravitational field there
are no longer any stationary solutions with uniform temperature. Instead both density and temperature decrease with elevation. This should be welcome as being in agreement with what is observed in real atmospheres. The fact that a portion of the atmosphere of the Earth exhibits the same temperature profile is a surprise; it suggests that the temperature gradient is not a product of radiation. The problem is that nothing could justify an application of this theory to phenomena
that are significantly influenced by radiation. Nothing that went into building up the theory suggests that the gas is not in isolation. But there is a very strong conviction among physicists that, in an isolated system, the temperature must be uniform, gravitation notwithstanding. We have built a theory that, urprisingly, seems to apply to the irradiated
and gravitating atmosphere, but we have not solved the more basic problem, to provide a dynamical theory that incorporates the isothermal atmosphere.
The problem is that nothing could justify an application of this theory to phenomena that are significantly influenced by radiation. Nothing that went into building up the theory suggests that the gas is not in isolation. But there is a very strong conviction among physicists that, in an isolated system, the temperature must be uniform, gravitation notwithstanding. We have built a theory that, surprisingly, seems to apply to the irradiated
and gravitating atmosphere, but we have not solved the more basic problem, to provide a dynamical theory that incorporates the isothermal atmosphere. Why is the prediction of a temperature gradient in an isolated atmosphere so shocking? Imagine a large heat bath located in the region z > 0 in R^3. A vertical tube, filled with an ideal gas, has its upper end in thermal contact with the bath, otherwise it is isolated. Assume that, at equilibrium, the lower part of the tube has a temperature that is higher than that of the bath. Now extract a small amount of heat from the bottom of the tube; then the restoration of equilibrium demands that heat must flow from the bath to the warmer, lower part of the tube, in violation of one of the statements of the second law, namely:
“Heat cannot pass by itself from a colder to a hotter body” (Clausius 1887).
Concerning the status of this formulation by Clausius of the second law of thermodynamics we quote I. Muller (2007):
“This statement, suggestive though it is, has often been criticised as vague. And indeed, Clausius himself did not feel entirely satisfied with it. Or else he would not have tried to make the sentence more rigorous in a page-long comment, which, however, only succeeds in removing whatever suggestiveness the original statement may have had”. And Muller continues: ”We need not go deeper into this because, after all, in the end there will be an unequivocal mathematical statement of the second law”.
We note that Maxwell (1868), in refuting Loschmidt, did not make use of the statement but argued that the arrangement could be turned into a source of energy, a second class perpetuum mobile. But the argument is incomplete.’ op. cit.
What we have is a theory for an isolated system that no one doubts – that is assumed to apply in a gravity field for lack of a standard treatment. The argument is incomplete because it does not consider the force of gravity.
We see immediately P-N’s problem – superficiality, arguments from authority, unsupported assertions, the misapplication of jargon, extreme verbosity in the pursuit of spurious abstractions, a feigned expertise, an uncritical zeal where even reading a paper in all innocence on the gravito-thermal effect without denouncing it as heresy is a sceptical sin.
Here’s Clive best’s take – http://clivebest.com/blog/?p=4101 – a simple experiment would resolve a hundred year old argument between some of the finest minds in in physics
Rob Ellison: “I’m not claiming a gravity induced temperature gradient.”
OK. You earlier wrote: “Oh – yeah. My view of thermal stratification under gravity is that it may or may not happen. The physical mechanism seems reasonable enough.”
What “physical mechanism” are you referring to? I asked you before but you ignored the question.
Rob Ellison: “Here’s Clive best’s take – http://clivebest.com/blog/?p=4101 – a simple experiment would resolve a hundred year old argument between some of the finest minds in in physics.”
The argument was resolved more than a hundred years ago, in 1896, when Loschmidt himself acknowledged that Maxwell and Bolzmann were right after all. But some people will never give up on the idea of a working perpetual motion machine and the dream of infinite free energy. No single negative experimental result will ever convince them.
In any case, while his form for the Hamiltonian of the dynamical system is standard, his derivation of a thermodynamical equilibrium state is quite non-stadard (sic)and, as I said, conflicts with the second law of thermodynamics.
He actually accused me of misrepresenting the paper by quoting it. Didn’t recognise the passage and denied that it was what the paper was about. Nor does he understand the Hamiltonian – which is a vector field defining a flow on the symplectic manifold. How’s that for jargon? The latter emerging necessarily from the Hamiltonian as a real valued differentiable function. The paper under discussion deals with an extension of thermodynamics into a flow field.
‘Hydrodynamics is a theory of continuous distributions of matter, described in the
simplest case by two fields or distributions: a density field and a velocity field, both defined
over R3 or a portion thereof. The role of temperature is often constrained, as it is
taken to be determined by the density and the pressure. Classical thermodynamics, on the
other hand, is the study of states of equilibrium, with uniform density and temperature, and relations between such states. In this context, extremum principles first formulated by Gibbs (1878) play an important role; see for example Callen (1960), but the extension of thermodynamics to systems in which the dynamical variables are fields on R3 is not immediate and in fact variational principles are seldom invoked in studies of such systems.’ op. cit.
But what does P-N actually say? That a math he doesn’t seem to understand is wrong because he says so? That for some unspecified reason it violates the second law?
The argument was resolved more than a hundred years ago, in 1896, when Loschmidt himself acknowledged that Maxwell and Bolzmann were right after all. But some people will never give up on the idea of a working perpetual motion machine and the dream of infinite free energy. No single negative experimental result will ever convince them.
Kelvin himself questioned what he called the Maxwell-Boltzmann doctrine – especially for an isolated system. But it is not about rants on perpetual motion – the perpetuum mobile anti-science stupidity of such as myself or claims of authority – no matter what the authority is. It is about the math which changes when a gravity effect is added.
P-N – as well as FOMBS – is clearly feigning an expertise he is far from possessing. Using verbiage to pretend to knowledge without having any substance. Always ready to launch into the anti-science tirades that they so depend on – and which are so misguided. As I say – it reminds me of Feynman for some reason. Only with a dimension of righteous indignation added to the overweening arrogance and ignorance.
“There were lot of fools at the conference – pompous fools – and pompous fools drive me up the wall. Ordinary fools are alright; you can talk to them and try to help them out. But pompous fools – guys who are fools and covering it all over and impressing people as to how wonderful they are with all this hocus pocus – THAT, I CANNOT STAND! An ordinary fool isn’t a faker; an honest fool is alright. But a dishonest fool is terrible!”
“But what does P-N actually say? That a math he doesn’t seem to understand is wrong because he says so?”
Did I say that his math was wrong? I didn’t check it and in fact don’t have the chops. It’s been way too long since I’ve done much classical mechanics (or any serious mathematical physics). I’m glad you can follow all the derivations. Maybe *you* can tell me where he went wrong.
I said his whole approach is non-standard as he himself acknowledges. There is more to is than simply defining the Hamiltonian for the system. Any approach that derives an equilibrium state for an ideal gas subjected to gravity that yields a non-isothermal state is non-standard.
He not only derives it for an ideal non-radiating gas but suggests that it has relevance to the actual lapse rate of the radiating atmosphere and may be exemplified in Graeff’s cranky free energy generators raises some flags. The second law precludes such an equilibrium state (with a temperature gradient any steeper than Tolman’s tiny relativistic gradient for radiation) in a gas that can radiate at all. He doesn’t seem to be realizing this. (Do you?)
Rob Ellison: “Nor does he understand the Hamiltonian – which is a vector field defining a flow on the symplectic manifold. How’s that for jargon?”
I’m very impressed. Maybe it’s my lack of knowledge of symplectic geometry that prevents me from understanding your claim that a free falling projectile has the same kinetic energy at a given level, both on the way up and down, only if this level is the mid-point of the full vertical trajectory.
Does it also have something to do with your unique understanding of the alleged fact an adiabatically expanding rising parcel of air in the atmosphere has a reduced “average kinetic energy” (whatever that is) but the same total kinetic energy?
And I had forgotten another gem: “Temperature is the result both of the average kinetic energy of the molecules and the number of molecules hitting a surface.”
Another probable consequence of Hamiltonian-Ellisonian symplectic statistical-mechanics.
Another long winded diatribe about the errors of sceptics? Seems so.
He sticks to it with huge persistence and verbosity. The energizer bunny theory of molecular moon projectiles. The theory of temperature depending only on the average KE whether there be 1 or 100 million molecules. Big claims that depend only on the authority and expertise of FOMBS and P-N. Trust them – they have Gaia and science on their side.
‘The effect of gravity was included by adding the gravitational potential energy to the hamiltonian; which is standard practice. The result of that modification is that, in the presence of the gravitational field there are no longer any stationary solutions with uniform temperature. Instead both density and temperature decrease with elevation. This should be welcome as being in agreement with what is observed in real atmospheres. The fact that a portion of the atmosphere of the Earth exhibits the same temperature profile is a surprise; it suggests that the temperature gradient is not a product of radiation.’ op. cit.
Here is a later – peer reviewed – version of the paper I’ve linked to – http://www.mdpi.com/1099-4300/16/3/1515
Now I’m an agnostic. Laboriously following the mathematical reasoning. Which tends to take me weeks – if not months to decades for certain difficult concepts. P-N doesn’t understand the math – but it’s gotta be wrong and non-standard. Says it all really.
Rob Ellison: “The theory of temperature depending only on the average KE whether there be 1 or 100 million molecules.”
I said nothing about cases where the number of particles is so small that there is no meaningful talk of statistical distributions.
Suppose there are N particles (with N very large) of an ideal gas in a container of volume V, temperature T and pressure V. Compare this with another container also of volume V where there are 2N particles (of the same gas) with the very same average kinetic energy (per particle). Is the pressure the same or different? Is the temperature the same or different?
According to my “theory”, the pressure doubles (as does the rate of collisions on the internal sides of the container) but the temperature is the same in both containers. What does your theory say? Is the temperature really dependent on both the average KE *and* the rate of collisions?
“[…] of volume V, temperature T and pressure V.”
…of volume V, temperature T and pressure *P*.
There is a very basic physical interpretation of molecules imparting momentum – more molecules in a space more momentum.
T = PV/nR
So double n and keep V constant. There is a linear relationship between P and V – but n and P is quite different relying on compressibility – so P increases faster than n as more molecules are stuffed into the space. It is progressively harder to stuff more molecules in. So T increases in a closed system – which is what we started with oh so long ago.
Rob Ellison wrote: “There is a linear relationship between P and V – but n and P is quite different relying on compressibility – so P increases faster than n as more molecules are stuffed into the space. It is progressively harder to stuff more molecules in. So T increases in a closed system – which is what we started with oh so long ago.”
You are answering a question that I didn’t ask and that doesn’t provide support to your questionable claim. I had asked about two identical containers with the same volume, same pressure, same average kinetic energy of the molecules, but twice as many molecules in the second container. You had said that temperature depends *both* on the average kinetic energy and the rate of collisions. So, it would seem that you are committed to say that the temperatures aren’t the same in the case that I just described since the pressure (and hence rate of collisions) is higher, while the average KE is the same. But you now balk at saying this and answer an unrelated question about compression instead.
You are now describing a situation where one achieved the higher pressure through injection/compression (“stuff more molecules”). This yields an increase in temperature, for sure, since there is an external energy input, but *also*, therefore, an increase in average KE of the molecules. If the container is allowed to cool to the previous temperature, the pressure and rate of collisions will *still* be twice as large as they originally were (before the compression/stuffing process) but the average kinetic energy per molecule will be restored to where it was. Will it not? This illustrates that kinetic temperature is determined *only* by the average KE.
For an ideal gas, KE_avg = (3/2)kT.
This is independent of pressure/collision rate. It relates average kinetic energy to temperature directly regardless of pressure. This is why free “Joule decompression” (in Joule’s experiment) conserves temperature for an ideal gas. As molecules from one container spread out through a pinhole in an empty adjacent container, and pressure equilibrates to a lower value, since no work is effected, average KE is conserved and temperature doesn’t change at al.
I said that the temperature depends on the kinetic energy of the molecules and the number of molecules. This is pretty obvious.
The box is a closed system as I said – the typical setup for the ideal gas law. But the original and incorrect statement form FOMBS was that more molecules were hitting the bottom than the top and yet magically top and bottom were at the same temperature.
Now he is specifying that it cools off to a local thermodynamic equilibrium – duh. .
Rob Ellison wrote: “I said that the temperature depends on the kinetic energy of the molecules and the number of molecules. This is pretty obvious.”
It is pretty obviously false. T = (2/3)*KE_avg / k
(where k is the Boltzmann’s constant)
The number of particles, or the pressure, don’t figure in this relation.
If two identical boxes are filled, one with n particles, and the other one with 2n particles, both with the *same* average KE, then the temperature in both of them is (2/3)*KE_avg / k. Right? You never answer this question.
“The box is a closed system as I said – the typical setup for the ideal gas law. But the original and incorrect statement form FOMBS was that more molecules were hitting the bottom than the top and yet magically top and bottom were at the same temperature.”
That’s not magical at all. Since the pressure varies according to the barometric formula, the density (and collision rate) diminishes on the way up. Though the molecules hit the top with the same average velocity (per molecule), the rate of the collisions is lower. Still, the temperature is the same, because the velocity distribution is the same. The Maxwell speed distribution is a function of temperature only and it determines EK_avg uniquely.
http://en.wikipedia.org/wiki/Kinetic_theory#mediaviewer/File:Translational_motion.gif
Se we have an insulated box with an ideal gas and kinetic theory says that the gases are randomly distributed. The original statement by FOMBS was that density differences between the bottom and top meant that more molecules hit the bottom surface and less the top. But that magically the surfaces were at the same temperature.
Density differences are not a factor in practically sized boxes. It is an enclosed system – the pressure is determined by the density of the gas and is uniform.
But more molecules hitting one surface would warm it more. The temperature is caused by transfer of momentum to the molecules or atoms of the box.
So we again have an artificial scenario from FOMBS that gets the simplest ideas from which kinetic theory evolves wrong. And then P-N defends the error with tendentious and fantastic – and quite mad – arguments that invent his own assumptions.
He then demands that I answer his question. Sorry.
Rob Ellison: “Se we have an insulated box with an ideal gas and kinetic theory says that the gases are randomly distributed.”
Not within a vertical gravity field, which is what we are discussing. Within a gravity field the barometric formula applies (see ref. below*), even within a closed box at equilibrium. Else the laws of Newtonian mechanics and gravitation would be violated. We would have the same pressure on the top and bottom of the box and hence the weight of the gas would not register when we weigh the box. It’s rather the energy states that are randomly distributed according to the equipartition theorem.
“The original statement by FOMBS was that density differences between the bottom and top meant that more molecules hit the bottom surface and less the top. But that magically the surfaces were at the same temperature.”
Yes, because T = 2/3*EK_avg / k. (**)
quite independently of pressure. You can’t keep ignoring laws of physics just because they aren’t convenient to your argument.
“Density differences are not a factor in practically sized boxes. It is an enclosed system – the pressure is determined by the density of the gas and is uniform.”
The density isn’t vertically uniform. If the box measures 1m*1m*1m, say, and contains air at 1atm, then the air within the box weigh about 1kg. Thus the pressure difference between top and bottom is approximately m*g/A = 10N/m^2. This pressure difference is accounted for by the density difference, not temperature. Were it not for this pressure difference, then a box filled with a gas would weigh exactly the same as an empty box.
“But more molecules hitting one surface would warm it more. The temperature is caused by transfer of momentum to the molecules or atoms of the box.”
We are considering the situation at equilibrium. You can picture the temperature being the same in the gas, the box, and the room. There is no heat flow anywhere in the vicinity of the system under consideration.
In any case, no, a higher rate of momentum transfer from individual molecules would not warm the surface more. If you have two adjacent containers at different pressures, and a conductive boundary in between, then there no heat transfer between them if both gases have the same temperature. It is irrelevant that the higher pressure gas supplies more collisions to the shared boundary than the low pressure gas.
References:
(*) http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/barfor.html#c2
(**) http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/kintem.html
At random – because I really can’t be bothered reading it all.
We would have the same pressure on the top and bottom of the box and hence the weight of the gas would not register when we weigh the box.
Yes we have the same pressure top and bottom. It is a matter of momentum transfer to do with the number and speed of the gas molecules. The movement of molecules is random. The weight of the gas is another thing entirely. The internal pressure in a rubber balloon is uniform – but if it is just air inside it will still sink.
The question was framed such that the gas was imparting energy to the walls of the box. That is the walls were cooler than the box. In that case – more molecules hitting one surface would heat that more through collision of the gas molecules with the material of the walls.
The premise is however weak – as molecules disperse randomly throughout the box.
P-N has a very poor appreciation of the simplest of physics – none at all of the broader aspects of climate – and likes to lecture on his feigned expertise. At great length.
That is the walls were cooler than the (gas)…
He is quite right though – a gas in local thermodynamic equilibrium will not heat the surroundings. Trivial but true.
Enough – I am bored with the long scroll to the end just to find more erroneous nonsense about boxes.
Rob Ellison: “Yes we have the same pressure top and bottom. It is a matter of momentum transfer to do with the number and speed of the gas molecules. The movement of molecules is random. The weight of the gas is another thing entirely. The internal pressure in a rubber balloon is uniform – but if it is just air inside it will still sink.”
So, now you are denying that a gas in a box has any vertical pressure gradient at all within the Earth surface gravitational field. This is just too bizarre a thing for an hydrological engineer to believe. You must be pulling my leg. Does this mean that if we close hermetically all the windows of a tall building then barometers within it will rapidly come to register the same pressure at all stories? Do you also believe this to be true about pressure in liquid filled container at equilibrium?
What you say about weight and momentum also makes little sense. How can the gas molecules transfer any net momentum at all to the box if they hit the top and the bottom of the box with the very same velocity distribution?
During their free falling trajectories in the box, in between the collisions with the top or bottom (side collisions have no momentum effect due to symmetry), the molecules are free falling and gaining downward momentum p = m*v = F*Δt = m*g*Δt. (from Newton’s second law and the definition of momentum p = m*v, and Δt is the time between the collisions). It is this constant gravitational momentum gain that is transferred by individual molecules to the box. There is a *net* transfer to the bottom of the box precisely because of this gain during the free falling trajectory. If there were no pressure difference — hence the same rate of collisions with the same average (vertical) velocity — between top and bottom, then the gas would be unable to transfer any amount of net momentum to the box and, as I said, its weight would not register outside of the box.
“The question was framed such that the gas was imparting energy to the walls of the box.”
No. We have been assuming equilibrium and hence an insulated container (or, equivalently, box walls at the same temperature as the gas). The molecules impart momentum to the walls but not net energy. (The box gains no net momentum due to its walls being connected together and its being held up against gravity). It’s a clear violations of the condition of equilibrium that there would be any heat transfer through any surface. Kinetic energy is exchanged back and forth at the same rate, between the gas and surface molecules, with no effect on the molecular speed distribution.
Rob Ellison,
Remember also the condition of hydrostatic equilibrium. For any macroscopic parcel of fluid, its weight is *exactly* balanced by the local pressure gradient. For the case of an isothermal gas at equilibrium, this condition is satisfied by the gradient defined by the barometric formula.
You can’t have hydrostatic equilibrium under gravity with no pressure gradient at all. Each parcel of air would be essentially in free fall together with all the parcels surrounding it — and hence it wouldn’t even encounter any inertial or viscous resistance by them). This would occur as long as there is no pressure gradient. This is of course absurd. It could last for a fraction of a second but then the fluid within the box would very rapidly come in equilibrium as the pressure gradient is established through the rapid compression of the lower layers by the (initially) free falling layers above them.
In case you like pictures better:
http://en.wikipedia.org/wiki/Hydrostatic_equilibrium#mediaviewer/File:Hydrostatic_equilibrium.svg
Dr. Curry,
You may (or may not) be right when you say that “lawyers are under no compunction to introduce evidence that hurts their cases; that’s the other side’s job.” There are in fact ethics limits on the ability of lawyers to fail to disclose to a judge information adverse to their client’s position. But, more importantly, there are significant limits on the ability of an expert witness (including a scientific expert), even though engaged by one side of the dispute, to do the same when testifying as to matters within her professional competence.
First, when an expert witness swears an oath before testifying, that commitment establishes ethical obligations for the witness (“”Do you swear to tell the truth, the whole truth and nothing but the truth”). The “whole truth” element of that oath affirms the obligation of the witness to include in her testimony all material information, even if adverse to the interests of the disputing party who engaged that expert to testify.
Second, in many court systems other than the United States courts (for example, in England), expert witnesses are barred from testifying as advocates on behalf of the party who hired them to testify. Instead, in English courts the expert witness is obligated by mandatory court rules to provide independent and impartial evidence and owes her primary duty to the court rather than the instructing party:
1. It is the duty of an expert to help the court on the matters within his expertise.
2. This duty overrides any obligation to the person from whom he has received instructions or by whom he is paid.
Also:
[E]xpert evidence should be the independent product of the expert uninfluenced by the pressures of litigation” and “an expert should assist the court by providing objective, unbiased opinion on matters within his expertise, and should not assume the role of an advocate.”
Third, many expert witnesses in the United States and elsewhere are additionally bound by binding codes of conduct imposed by the professional organizations to which they belong. Some of those organizations include in their codes specific obligations relating to testimony by members as experts before courts and other tribunals, in particular various engineering and valuation organizations whose members routinely testify in courts.
As just one example, members of the American Society of Appraisers (ASA) are often asked to serve as expert witnesses in court on damages questions. Sections 4.3 and 7.5 of the ASA’s Principles of Appraisal Practice and Code of Ethics emphatically reject the notion that members, when giving expert evidence before a tribunal, can “omit findings that don’t fit their hypotheses” or neglect to “introduce evidence that hurts their [client’s] cases.”
4.3 Appraiser’s Obligation Relative to Giving Testimony
When an appraiser is engaged by one of the parties in a controversy, it is unethical for the appraiser to suppress any facts, data, or opinions which are adverse to the case his client is trying to establish; or to overemphasize any facts, data, or opinions which are favorable to his client’s case; or in any other particulars to become an advocate. It is the appraiser’s obligation to present the data, analysis, and value without bias, regardless of the effect of such unbiased presentation on his client’s case. (Also, see Sec. 7.5)
7.5 Advocacy
If an appraiser, in the writing of a report or in giving an exposition of it before third parties or in giving testimony in a court action suppresses or minimizes any facts, data, or opinions which, if fully stated, might militate against the accomplishment of his client’s objective or, if he adds any irrelevant data or unwarranted favorable opinions or places an improper emphasis on any relevant facts for the purpose of aiding his client in accomplishing his objective, he is, in the opinion of the Society, an advocate. Advocacy, as here described, affects adversely the establishment and maintenance of trust and confidence in the results of professional appraisal practice and the Society declares that it is unethical and unprofessional. (Also, see Sec. 4.3)
Many other professional organizations include similar, if less pithy, obligations in their codes of conduct that require a member testifying as an expert witness to not surpress or minimize adverse information or otherwise engage in advocacy.
In the real world, of course, too many expert witnesses act as “hored guns” and disregard these obligations (or are not even aware of them). Those experts testify as an advocate for the disputing party who engaged that expert, despite their ethics obligations.
Interestingly, a duty to not act as an advocate in court testimony cannot be found in a code of conduct for academics. University professors in the U.S. generally belong to the American Association of University Professors (AAUP). The AAUP approved a Standard of Professional Ethics in 1987. Most of the guidance in that Standard addresses the circumstances of university life, scholarship and teaching. Article V of the Statement, though, describes the rights and obligations of professors as members of the community generally: “professors have the rights and obligations of other citizens.”
As a consequence, the AAUP Statement of Professional Ethics gives little specific guidance to university professors regarding their ethical responsibilities when appearing as an expert witness in a dispute. if a university professor takes the traditional court-administered oath before testifying, though, then like any other citizen the professor is bound in her testimony to tell the truth, the whole truth and nothing but the truth.
In my personal view, the most useful advice for expert witnesses is actually found in a report issued by the Society of Petroleum Evaluation Engineers (SPEE) on “The Ethical Considerations Involved in Expert Witnessing,” in SPEE Ethics Guidance (1998).
Are you comfortable with the reputation of the party you would be representing? An expert can be a completely ethical witness even working for someone of questionable repute. In America everyone has the right to hire the best available legal counsel and technical assistance for their “day in court.” But if you lie with dogs you’re liable to get fleas. There can be undue pressure to slant your opinions and less than full disclosure of all the facts and data. At best it is an uncomfortable situation and can become a quagmire, particularly for the inexperienced witness.”
if you or your denizens find this information about the ethics duties of expert witnesses to be useful, then Kantor, A Code of Conduct for Party-Appointed Experts in International Arbitration – Can One be Found?, 26 Arbitration International 323 (2010), may also be of interest.
Regards,
MK
Oops, “hored guns” should read “hired guns.” Apologies – I can’t type.
MK
I haven’t read the post yet and will have to come back later to read it and watch the webinar. But I’d just like to comment on what information I’d like to see and how I’d like it presented. Regarding presentation, I refer readers to Dimson (2003) Figures 3, 4 and 5 here: http://www.econ.uniurb.it/materiale/2781_triumph_of_the_optimists.pdf
I’d like to see charts like this (i.e. showing the median, top and bottom quartiles and top and bottom deciles) of the projected abatement cost and projected benefit for each policy under consideration. I’d like to see the projection out to say 2050 or 2100. Anything beyond that is meaningless, IMO, including accumulating net benefits out to 2300 as the IAMs like DICE do.
The benefits of this approach are:
1. it focuses researchers and policy advocates attention on what is relevant for policy analysis
2. the chart includes all benefits and all abatement costs
3. the probabilities will focus debate on all the inputs required to estimate the probabilities of all costs and benefits of the policies, including:
a. the probability that the policy will deliver the expected benefits for the expect6ed costs
b. the probabilities of time to the next abrupt climate change, its direction (to warmer or cooler), its rate of change, its magnitude of change
c. the impacts of abrupt cooling as well as of abrupt warming
d. The risk reduction due to the probability that human caused GHG emissions may be delaying abrupt cooling, and or reducing its magnitude, and/or reducing the rate of abrupt cooling.
4. All this combined in one or two easy to understand charts for each policy.
Dimson: ““While a country has only one past, there are many possible futures.” A good and sufficient reason not to build policy around one possible aspect of the future as if it were fundamental and all encompassing.
Hi Faustino,
Thank you. I agree with you and with the bit you quoted from Dimson. But my hope is to get researchers and policy proponents to focus on estimating the the probabilities that the policies they advocate would deliver the projected benefits and the policy be sustainable for 100 years, or whatever it takes. If we could get proponents to focus on the probability of success of the policies they advocate I suspect we could get people to understand why there is such strong resistance to those policies from rationalists.
I just received this response from a close friend (lives in Qld):
I have two responses:
1. If we can’t estimate the probabilities that a policy will deliver the projected benefits, how can it be justified?
2. My purpose is to get policy proponents to focus on the probability that their policy will deliver what they claim. If we could get people to focus on the probability of success, perhaps we could “cut the emissions of twaddle” (to quote David MacKay, UK chief science adviser to DECC and author of “Sustainable Energy – without the hot air”
Think of a jigsaw puzzle. the fact that it can only be completed if all the pieces fit together is a test of the validity of the whole result. Yet it is difficult to quantify this result as a overall measure os success. The designers of jig saws probably go to considerable lengths to ensure that only one solution is possible,whereas real life affords few examples of the possibility of such an example.
The uncertainty is actually whether it’s a fraud from the outset, with potemkin sciencey things brought in to cover it.
Somebody discovered that something brought in funds, and here we are two decades later.
The error bars are more potemkinization, a sciencey discussion designed to cover the initial fraud when the cover starts to get threadbare.
The logic behind ‘Silent Spring’ was entirely reasonable; firstly we evolved in the presence of natural environmental toxins, secondly it follows that we evolved mechanisms to deal with the major classes of environmental toxins and finally, the detoxification pathways that evolved for natural substances could deal with man-made compounds.
This was eminently reasonable, and the majority of scientists in biochemistry and pharmacology believed it was highly meritus.
Bruce Ames also believed in the paradigm and wished to prove it, and developed the Ames test for mutagenesis. He was shocked that man-made chemicals did not have attributes different from natural ones. Ames, and others, tested the paradigm and basically destroyed it.
Funny thing.
DDT and its major breakdown product, DDE, are xenoestrogens and very long lived in the biosphere. They are not the sort of lipophilic compounds we should be releasing into the environment. In the Western world at least, stopping the widespread usage of DDT was probably a good thing. In Malarial zones, perhaps less so.
You can do the right thing, stop using DDT in the US farm belt, for the wrong reason, DDT/DDE isn’t a carcinogen but is a hormone mimetic, and yet do the baby in bath water, condemning people, especially children, in the third world to die of a treatable disease as they cannot keep their homes free of disease.
The cost/benefit analysis of stopping something must include not only the wealthy ‘West’, but also all the people on the planet.
DocMartyn, how long does it take insects to develop resistance to fluoro-chloro organic pesticides?
Perhaps you don’t remember when the radio was alive with The Endrin Song and The Heptachlor Song.
Catch Phrase “The Cost Is Really Low”
But those chemical-industry claims weren’t true, were they DocMartyn?
The insects rapidly became immune; farmers had to apply higher-and-higher levels of toxins to stay in business; eventually the companies that made these toxic chemicals declared bankruptcy … leaving taxpayers with massive SuperFund clean-up costs
The cost of denailism Denial of these biochemical realities created a massive (and still ongoing) economic, environmental, and moral train-wreck … didn’t it DocMartyn?
Corollary In the long run, the”cost of carbon” ain’t so low either.
*EVERYONE* understands *THAT*, eh Climate Etc readers?
“DocMartyn, how long does it take insects to develop resistance to fluoro-chloro organic pesticides?”
I am not aware of any fluoro-chloro organic pesticides in use except for Fluvalinate which is used to treat Honey Bee hives infested with varroa mites and Tefluthrin.
“The insects rapidly became immune; farmers had to apply higher-and-higher levels of toxins to stay in business”
This assumes that farmers are stupid, and generally they are not. Farmers typically rotate through different classes of pesticide, which act on a different pathway.
“companies that made these toxic chemicals declared bankruptcy … leaving taxpayers with massive SuperFund clean-up costs”
Firstly, former pesticide, herbicide and fungicide plants are a tiny fraction of SuperFund sites, creosote plants are much more common.
Secondly, SuperFunding come from either the polluter or from a levy imposed on businesses that use chemicals, so that they, not general tax payers, fund clean ups. The levy ended in 1996, at the start of Clinton’s second term thanks to his VP. Since 2003, when he trust fund expired, general revenues have covered the shortfall.
Perhaps you could ask Al ‘casting vote’ Gore why he stopped the levy being paid?
Even DDT is still in wide use globally, DocMartyn! Not to mention the “dirty gang” that additionally includes dicofol, heptachlor, endosulfan, chlordane, aldrin, dieldrin, endrin, mirex, kepone and pentachlorophenol.
Question Why is there a persistent toxic connection between tobacco-harm denial, pesticide-harm denial, and CO2-harm denial?
The world wonders! Fortunately, healthcare researchers are well-posed to provide scientific, economic, medical, and moral answers.
Good on `yah, rational-respectful-responsible healthcare researchers!
AFOMD,
Correct me if I’m wrong, but I fear it is you, not DocMartyn, who asserts with amazing ignorance. With typical Warmist arrogance, you imply that DDT, as well as insecticides such as dieldrin etc., are flouro-chloro insecticides. They are not. To describe them as such merely indicates the amazing ability of Warmists to substitute fantasy for fact, with a straight face.
DDT is commonly described as dichlorodiphenyltrichloroethane, which unsurprisingly contains no reference to flourine at all. Maybe you confuse chlorine with flourine. They do sound similar, and do share some letters. Possibly Warmists find chemistry a little difficult.
Are you really stupid, or does your hubris prevent you from discriminating between fact and fiction?
The world has given up wondering. It is certain.
Live well and prosper,
Mike Flynn.
“DocMartun asserts [with amazing ignorance] “I am not aware of any fluoro-chloro organic pesticides in use”
Even DDT is still in wide use globally, DocMartyn! Not to mention the “dirty gang” that additionally includes dicofol, heptachlor, endosulfan, chlordane, aldrin, dieldrin, endrin, mirex, kepone and pentachlorophenol.”
Those are all organochlorines, none of them are fluoro-chloro organic pesticides; the lack of fluorine in their structures is a bit of a clue.
DDT is very useful in indoor applications, and even Malarial vectors that are DDT resistant avoid surfaces treated with DDT.
630,000 people die from Malaria annually, mostly under-5. Pesticides are the cheapest, most effective first line of defense.
Judith Curry and others:
The Clausius (“hot to cold”) statement of the Second Law is really just a corollary which has certain pre-requisites, namely that it only applies if other forms of internal energy remain constant. We know this because thermodynamic equilibrium (which the Second Law says will evolve) must take into account entropy and thus all forms of energy.
In particular, the mean gravitational potential energy of any small region in which a process takes place must be homogeneous if we are only considering mean molecular kinetic energy which determines temperature. In other words, for conduction, diffusion, advection and convection you can only claim that thermal energy transfers from hot to cold in a horizontal plane.
In a vertical plane any change in gravitational potential energy must be offset by an equivalent change in kinetic energy (temperature) if there is a state of thermodynamic equilibrium.
Unless you understand this Judith, you will never understand the thermodynamics which explains why it’s hotter than Earth at the base of the nominal troposphere of Uranus, even though it’s nearly 30 times further from the Sun than we are.
The Second Law of Thermodynamics can be applied to explain why a planet’s troposphere has a density gradient. This has nothing to do with upward convection from a surface heated by solar radiation. There is no such surface on Uranus, but the density gradient is still there. When you have such a density gradient caused by gravity acting on molecules between collisions, then you must also have a temperature gradient. Each is the state of thermodynamic equilibrium and, until you understand why, you do not understand thermodynamics. I suggest you heed what Josef Loschmidt said in the 19th century, because he was right.
When you realise that the sloping thermal profile acts like a level surface of a lake and spreads new thermal energy (rain) in all accessible directions, then, and only then, will you understand how downward convection is what supplies the missing energy that James Hansen though he could explain as being due to back radiation. But it isn’t, and thus the whole greenhouse conjecture is false.
That’s physics, Judith!
The problem with most journalists is that no matter how much “deep research” they do, they aren’t equipped to understand it. So, I suspect most of them look to authority to establish their own position.
Same for the scientists of course, because climate covers many sciences.
The position in a Court of law is more complex.
First, subject to privilege (and a few niche exceptions), a party is obliged to disclose all documents in their possession, power, custody and/or control that is relevant to the issues raised in the dispute, and this extends to documentation that derogates against the case being run, and/or supports the opponenet’s case.
Second, an Expert would be expected to be able to substantiate his position by providing full details of his workings, and matters upon which he relies in support of his opinion; the material difference between Expert evidence and evidence of fact is that an Expert is entitled to give opinion evidence on matters within his expertise.
Thus an Expert would, upon request, be expected to provide the raw data upon which he relies, and if he ‘adjusts’ the data, he would be expected to explain each and every adjustment and the opponent lawyer could test the validity of each and every such adjustment.
Reflect on that in comparison with the land based thermometer data sets, and CRU’s (and others) persistent failure to comply with FOIs and not wishing to disclose how they have produced their data sest since others wish to pick holes in it.
In a Court of Law, the starting pont would be the null hypothesis, ie., climate naturally changes and all such changes that we are preently seeing are the result of and/or are within the realms of natural variability.
It would be for those who claim that climate change is the result of AGW, to prove their case. Even on the civil burden (balance of probablilities) if properly tested, it is difficult to see how the required standard could be discharged.
. .
You are correct that people who are qualified as expert witnesses are allowed to give opinions. You are incorrect the about the level of rigor that is required to get scientific evidence in court and how it gets used. Experts don’t have to provide data in any great detail. The courts don’t want to get bogged down in minutia. Yes an opposing lawyer can question an expert and the data, but the other lawyer can block the questions. Read up on the Daubert and Kumho Tire Supreme Court decisions on scientific evidence.
From there:
“Scientific conclusions are subject to perpetual revision. Law, on the other hand, must resolve disputes finally and quickly.”
“We recognize that, in practice, a gatekeeping role for the judge, no matter how flexible, inevitably on occasion will prevent the jury from learning of authentic insights and innovations. That, nevertheless, is the balance that is struck by Rules of Evidence designed not for the exhaustive search for cosmic understanding but for the particularized resolution of legal disputes.”
An adversarial approach has much to commend.
Governments ought to adopt such an approach when considering legislation.
Thus, a government should enegage Experts and give them a copy of the IPCC report and ask the Expert to advise what is wrong with the report? What are its weaknesses? etc.
When considering legislation, the government should commision Expert evidence seeking advice on what is wrong with the proposed legislature, what are the negative effects of the proposed legislature? etc.
It is only when armed with details of both sides of the coin can effect policy be made.
In that way, a government would be better placed to avoid the unintended consequences arising from its policies.
But the over-riding problem is that there is a lack of accountability in public office. There needs to be a system whereby those who make decisions are held accountable for their decisions. I am aware that that would discourage many from seeking public office, but there will never be good decision making when there are no consequences to screwing up.
+ 10
Good post Richard +1
The government does utilize the adversarial approach in making legislation. The Congress pours over the detail of legislation, has countless hearings, sends proposed legislation through the Congressional Budget Office, has press conferences. Do you realize how many politicians are lawyers? Just because they get elected, it doesn’t mean they aren’t adversarial anymore. ;)
Regulations are subject to constant scrutiny. Regulatory decisions end up in the courts all the time. There is a whole field of law, administrative law, that deals with governmental agencies and regulations.
There is also a system of accountability for decision makers, it’s called elections. If enough people don’t like like what they are doing, the politicians get voted out. It’s called democracy.
Regarding convection, everyone needs to understand that “heat transfer by bulk fluid flow” is not a bulk flow caused by an external mechanical energy supply. All convection is driven only by a higher level of mean kinetic energy in some region where it is hotter than the state of thermodynamic equilibrium (with its associated temperature gradient) would normally be. So when direct solar radiation strikes an asphalt surface (emissivity 0.92) it can raise the temperature thereof because its intensity may be, say, 450W/m^2 which would support a temperature of about 305K and this causes air molecules at the surface interface to be hotter than what they were in the cool of the pre-dawn hours of the morning – ie near the supporting temperature. So upward diffusion and advection (that is, convection) occur. But warm air does not rise in parcels, and warmed air may even fall – convection is just a net movement of molecules during their normal free path motion between collisions. Extra kinetic energy in the surface “pushes” more molecules away so that “sheets” of warmed air appear to rise, but in fact it is mostly movement of the warmer temperature that gives this effect. A light fan will turn slowly, but the heat transfer can be faster than the apparent movement of the air which is very slow. But extra thermal energy absorbed from solar radiation in the upper troposphere of a planet can drive convection downwards.
“Extra kinetic energy in the surface “pushes” more molecules away so that “sheets” of warmed air appear to rise, but in fact it is mostly movement of the warmer temperature that gives this effect.”
So a soaring glider’s:
http://www.aerospaceweb.org/question/nature/thermals/thermals3.jpg
lower wing surface is being bombarded by these molecules thus keeping it aloft. Have we ruled out such thermals existing? To the best of my knowledge a glider descends in a rising column of air.
Raagnar wrote: “So a soaring glider’s: […] lower wing surface is being bombarded by these molecules thus keeping it aloft.”
“Physicist” (Doug C.) seems to believe one can blow off the candles on a birthday cake without disturbing any macroscopic air parcel at all in between the blower and the cake. This is up there with his earlier suggestion that cumulus clouds gather in the vicinity of mountain ranges because they are gravitationally attracted towards them. (I am unsure how he would interpret an helium filled balloon moving forward in an accelerating car).
“But extra thermal energy absorbed from solar radiation in the upper troposphere of a planet can drive convection downwards.”
A hot low-density fluid spontaneously sinking into a cooler higher-density one. That’s awesome!
Thanks for your contribution Physicist +!. Now lets hear what other pro AGW physicists have to say about your POV……? Crickets ….?
I would like to see an article on attribution bias in Climate science particularly with respect to climate sensitivity.
stay tuned!
Would that include Hubers and Knutti’s new rcp8.5, C&W with new and improved Aerosols/Solar 1.8 TRC/3.0 ECS paper?
HRT and breast cancer.
10,000 women, 5 years, 303 develop breast cancer.
Control group 300.
Cause of cancer according to patient, 303 HRT in the first group.
Cause in the second group, inevitably trauma at some stage in the 6-12 months preceding diagnosis.
HRT has the blame although it is only likely to have been the vehicle in 1% of cases. People need something to blame.
CO2 has the same problem.
If you are looking for causation you will find it . The true sensitivity is buried under the huge attribution desire and need.
[note above figures are off top of my head to give an example of attribution bias]
TonyB
Do you have any comment on tins recent paper on sea level changes.
I am particlularly interest in the conclusion that there have been very little sea level change for the past 6000 years until the last
TonyB
Do you have any comment on this recent paper on sea level changes.
I am particularly interested in the conclusion that there has been little sea level change for the past ,4000 years until the last 150 years. For the past 150 years, the rate of increase has been faster than ever before (my interpretation). The last paragraph says:
Here is the link
“Sea level and global ice volumes from the Last Glacial Maximum to the Holocene”
Kurt Lambeck, Hélène Rouby, Anthony Purcell, Yiying Sun, and Malcolm Sambridge
http://www.pnas.org/content/early/2014/10/08/1411762111.full.pdf
Peter
Thanks for this. I will have a read through during the day and will comment more after
I am unsure why there would NOT be oceanic oscillations because our climate certainly fluctuates, causing glacial melt and advance. All hugely complicated by land rise and fall and trying to determine a GLOBAL picture causes additional difficulties-especially over protracted time scales- when regional estimates probably have more merit.
Tonyb
Global parameters are
Cr*p IMHO
Excellent article citation by Peter Lang!
This is new strong scientific evidence to support the Mann-Hansen-Oreskes worldview that (1) the “hockey-stick handle” is flat and (2) the “hockey-stick handle” is lengthening, in energy-balance consequence of (3) the “human carbon volcano”.
So, is the “uncertainty monster” down for the count?
http://www.nikoojanen.com/wp-content/uploads/2014/07/motivationfordays.jpg
Most climate-science scientists say so!
So the more that is learned about the past, the more certain the hockeystick becomes. Interesting.
Peter Lang
I will be circumspect here as I don’t want to trigger moderation with too many links
If you go to my article here
http://judithcurry.com/2011/07/12/historic-variations-in-sea-levels-part-1-from-the-holocene-to-romans/
You will see in the 4th paragraph I mention it is ‘part of a ‘longer “document’ If you click on ‘document’ and scroll to the bottom of the extended version you will come to this reference and graphic
—— ——-
“A further study uses the same information and is headed; “Reconstructing sea level from paleo and projected temperatures 200 to 2100 AD”
This rather apocalyptic version-as do the other variations- starts towards the end of our studied period, but the margins of error show the high levels in earlier Roman times, then a dip, then a rise to the MWP peak, a decline, then a rise in modern times to below that of Roman and MWP before a dramatic projected increase closely linked to the future
temperature increase expected by some researchers.
http://www.glaciology.net/Home/PDFs/Announcements/gslprojection
——- ——-
It appears that sea level oscillates some 20/40cm around a mean average.
As regards glacier movements you will see my graphic here
http://wattsupwiththat.files.wordpress.com/2013/08/clip_image010.jpg
It shows glacier advances and retreats over the last 3000 years from observational evidence compiled by such as Ladurie.
Clearly there has been major advances and retreats, which in theory should correspond roughly with sea level changes. (I haven’t done the full research on Parts 2 and 3 of ‘Historic variations in sea levels’ yet, so this assumption might be incorrect)
During Roman times it is said that many of the high level Alpine glaciers disappeared. At this point dear old Max would chime in with accounts of how high level silver mines near his home in Switzerland are still buried under ice.
As I say, there are enormous complications by trying to look so far back in time as the paper you referenced and trying to attribute global changes is further complicated by land changes as it sinks or rises.
So, I don’t believe this constancy of sea levels until the last 150 years. Neither observational evidence nor borehole data supports this. There appears to have been high water stands around the 5th Century, around the 12th and 16th Century and in between the levels have been altering substantially.
The LIA was supposed to have been the coldest part of the Holocene for 8000 years or so and would have deposited vast amounts of snow and ice which has been melting for some 300 years.
I have never seen estimates as to how much water was locked up during this period but if it was unprecedented in thousands of years there is theoretically a lot of melt to come which could cause sea levels to rise above the MWP levels
Water drawn from aquifers is another wild card
tonyb
Tony makes a good point about aquifers but even without that you are looking at reconstructions that have resolutions of 15-20 cm and >200 years. I don’t see how much is going to be able to be made of this study in regards to showing the modern era is unusual.
Steven
A lot of these paleo proxy reconstructions are a very coarse sieve through which the real world ships through
tonyb
TonyB,
Got it. thank you. I’ll dig into your reference. in the meantime this seems like the takeaway message:
“It appears that sea level oscillates some 20/40cm around a mean average.”
This would seem to be consistent with the Lambeck conclusion, or at least the uncertainty range is to great to disagree with the Lambeck conclusion.
I enjoyed reading the essay. I cannot get passed more than 20 of the responses in one sitting, so I’ll just give my opinion. I see a phenomenon called information leakage. With every transfer of information, one fact gets dropped. Scientists advise Al Gore, for example. They gave him all the facts. He takes all these facts, and due to his non-scientist training, leaves out a statistic that he finds too technical. Now, we have journalists seeing his movie and understanding even less about science than Al Gore, writing a review. They report on the alarming graph that they saw, the hockey stick and GCM predictions. And then we are left with readers who read articles and have even less information to make decisions. So they accept what their favorite commentator says. And then I read it here on this blog.
Rose
I have a post responding to the first thread in moderation. I’ll re-post it here (fingers crossed).
John, I’ve worked (long ago) as a newspaper and radio reporter. Of necessity, they can’t print everything, they have to select. And in selecting, they are concerned not with balance or truth but what makes a good story. Added to that, many studies have shown the great majority of journalists to be left-wing. The result in the alleged CAGW context, which I’ve been following since the 1980s, has been a severe bias to the scary stories and warmist line, with its left-wing policy agenda. It’s never “Just the facts, ma’am.”
Tom: “Joshua, since 1988 the consensus has engaged in a sustained battle to deny their opponents standing.” Quite so, and the media have been complicit in this. Good post @ 8.53.
Jim D @ 9.27, see my first para. Almost the only non-alarmist outlet in Australia over many years has been The Australian, and their editorial line has consistently been that AGW is occurring and must be dealt with by GHG emissions reductions. They have given space to Lomborg et al, but it’s only in recent months that their coverage has shifted – the pause is acknowledged – and sceptical letters have begun to dominate.
“Skeptics” spend a lot of whining about the world not being fair to them.
Warmers spend a lot of time lying.
Andrew
Dr. Curry, how does the Bud Ward article support your take away from it? The article had nothing to do with what you claim it does. Nowhere does the article talk about denier witch hunts or loss of confidence in science.
It goes to the question of credibility – they have very little.
Dr. Curry and Bud Ward have very little?
The witch hunters and catastrophists. Journalists. Take your pick.
A Motor Trend review recently that looked at the realistic range obtainable in a Tesla. A 528i chase car allowed for somewhat of a real world but still an apples to oranges comparison. The trip was 238 miles which was about the full extent of the Tesla’s juice—i.e., 78.2 kWh.
Unfortunately, MT erroneously compared the Tesla’s performance to the gallons of fuel used by the 528i on the trip —i.e., MT compared the 528’s mpg to the amount of energy in 2.32 gallons for the Tesla. Needless to say, iMT did not bother to consider the amount of energy that is required in the real world to generate the 78.2 kWh of electricity needed to charge the Tesla’s batteries.
For example, a diesel generator can produce around 3.2 kWh /liter of fuel. So, the 78.2 kWh that the Tesla used in the MT comparo to go 238 miles would have required 24.4375L or 6.46 gallons of diesel fuel to produce. Losses to transmit electricity over the grid is 5-7% which when factored in to the calculation results in 6.78 to 6.91 gallons of diesel for the Tesla’s 238 miles… not, 2.32 gallons.
The BMW 528i, managed 30.1 mpg; however, it uses gasoline not diesel so that requires an adjustment. But, using the 6.78-6.91 gallons of gas (not diesel fuel) the 528i would have gone about 204 to 207 (not 238 so a little less); however, if the 528i had been a diesel it would have gone a lot further on the same amount of diesel fuel –e.g., instead of the 528i’s 30.1 mpg, the Mercedes E250 BlueTEC comes in at 28 city and 42 highway. The combined mpg is somewhere in-between.
In one E250 BTC review the results were as follows: “by the end of our five days with the E250 BlueTec, we’d averaged an indicated 40.6-mpg over 595 miles. At that time, the instrument display was still showing a remaining range of 235 miles—so from the 21.1-gallon tank a total driving range of more than 800 miles would be within a real-world (non-hypermiling) reality.”
Accordingly, using the 40.6 combined mpg of the E250 BTC puts the mileage at 275 – 277 miles (not just 238 miles) on 6.78 – 6.91 gallons of diesel fuel (and, the E could be driven over 500 additional miles without having to refuel).
Tesla’s biggest leverage is the arbitrage between coal at ~$2/GJ and Oil at ~$20/GJ, and then the average car’s abysmal actual efficiency vs a coal fired power plant 37%.
Where the Energy Goes: Gasoline Vehicles
Where are those ultra cheap ultra super capacitors when we need them?
Missing the R&D funding that has gone into subsidizing inefficient solutions and global climate models.
Concerning the “Costs” of “Mitigation”:
To restore economic growth and preserve liberty, the Clean Air Act should be amended. That would defang the EPA.
How? CO2 should be excluded from the scope of “Green House Gasses” (GHG) in the Clean Air Act. Technically, CO2 is a GHG. Practically, it is beneficial. Politically, it is the means for the EPA to command the economy. Political command of the economy is ruinous.
Why? Fossil Fuels provide over 80% of our energy supply [1]. Its availability and cost affect where and if one works, what one may buy and where (or if [2]) one lives. It is the ideal lever for the “Progressive” vision for our society. Politically, it is an ideal mechanism to play favorites.
There have been a number of schemes for government to run our lives. Among them are Turnover Tax, Cap And Trade, Carbon Tax, Carbon Credit Card, and Tax And Dividend [3].
In summary: Remove the political basis for a command economy run by the EPA.
Sources:
[1]
EIA. “Total Energy – Annual Energy Review (EIA).” Governmental. U.S. Energy Information Administration, September 27, 2012.
http://www.eia.gov/totalenergy/data/annual/pecss_diagram.cfm
http://www.eia.gov/totalenergy/data/annual/perspectives.cfm
[2]
Hope, Jenny. “Fuel Poverty Britain: 24,000 Will Die from Cold This Winter and 3m Worry about Heating Their Home.” Mail Online, January 19, 2014.
http://www.dailymail.co.uk/news/article-2478114/Fuel-poverty-Britain-24k-die-winter-rising-energy-prices.html
Samenow, Jason. “Cold Kills More than Heat, CDC Says; Other Researchers Caution Not Necessarily.” Washington Post, August 4, 2014. http://www.washingtonpost.com/blogs/capital-weather-gang/wp/2014/08/04/cold-kills-more-than-heat-cdc-says-but-researchers-caution-it-depends/
Vulliamy, Ed. “Cold Homes Will Kill up to 200 Older People a Day, Warns Age UK.” The Guardian, October 22, 2011, sec. Society. http://www.theguardian.com/society/2011/oct/22/older-people-cold-energy-bills
[3]
Pooh. “The Politics of ‘AGW.’” Discussion. Global Warming and Weather Discussion, October 28, 2008.
http://solarcycle24com.proboards.com/thread/192/politics-agw
Favoritism (Cronyism) already implemented:
Solyndra, A123 , Abound Solar, Beacon Power, BrightSource Energy Inc., Evergreen Solar, Eastern Energy, Fisker, GE, GM (Chevy Volt), Ener1, Range Fuels, Solar Trust of America, Spectrawatt, LightSquared, Unisolar, Bright Automotive, Olson’s Crop Service, Energy Conversion Devices, Sovello, Siag, Solon, Q-Cells, Mountain Plaza
I was a journalist for 10 years and found the column you cited interesting for what it didn’t say. Journalism is really bad at history, so it’s not surprising what was left out.
The author wrote:
“So, instead of the over-simplified notion of providing ’balance’ in reporting on news involving differing perspectives, journalists increasingly, and rightly, take their clues from the leading and acknowledged scientific experts when it comes to the facts and causes of global climate change. That means, in effect, reporting as a given—until science shows otherwise—that warming of Earth actually is occurring and that human activities have a significant role, though not the only one, in that warming.”
It seems like a minor point, but it’s important to remember what newspapers were reporting back in the heady days of “false balance.” They were “reporting” was “ice-free Arctic in five years,” forecasts of annual direct hits from super-charged hurricanes, mosquito born disease in areas that had allegedly “never seen malaria” before, switching from coal to solar panels and wind mills would be cheap, easy, fast and the only obstacle was Republicans. The people pushing back hard at these wild claims weren’t wild-eyed Tea Baggers, they were hurricane experts, epidemiologists, experts on the Arctic and more. Editors threw in “false balance” to the pretension that Al Gore’s movie was a “documentary” because they’d look like idiots otherwise.
In other words, in 2003 – 2006, climate advocates, including climate scientists, would have been cheering or silent about Peter Wadhams wild claims and somebody would have to be dragged into the TV studio for the false balance of say “well, not really.” Today editors and reporters (the ones who still have jobs) are smart enough not to get burned by trying to headline the Wadhams and it was Gavin Schmidt providing what used to be decried as “false balance.”
JeffN,
But that seems to be the problem. What motivates journalists and the newspapers they work for? It seems like only stories that can end with an exclamation point are worthy of publication. Otherwise readers turn to the internet out of boredom. The science has become secondary to the more exciting disputes among the scientists themselves. Maybe it is just easier for people to understand. They can choose a team and cheer them on, regardless of what is actually being said.
Rose
Judith,
I can’t think of anyone better than you to take on Naomi Oreskes.
I will, and I’ll take on Judith. Water vapour cools – temperature data proves it. But the IPCC wants you to be gullible enough to believe that a mean of less than 2% water vapour raises the surface temperature about 30 degrees. So 4% in a rain forest raises it 60 degrees does it? But 1% in a desert raises it 15 degrees does it?
Ragnar, Pierre 7 Peter Davies
To explain in more detail, yes of course you can get measurable advection very close to a heated solid surface on a sunny day. Adiabatic advection is just adiabatic diffusive heat transfer which is sufficiently fast to be measurable. Both are adiabatic convective heat transfer in physics and they involve transfer of kinetic energy through molecular collision. Above the relatively small proportion of the troposphere (near a heated solid surface on a clear day) where advection can be measured there is still diffusive heat transfer which will be upward in some regions (mostly above solid surfaces that are hotter than ocean surfaces) and mostly downwards above oceans where the Sun has heated the stratosphere and upper troposphere due to ozone and water vapour absorption. Late at night most adiabatic convection is only upward diffusive heat transfer.
As for your descending glider, it is primarily held aloft in the same way any winged aircraft is because of the wing design and the Bernoulli Principle. Its original gravitational potential energy is converted to kinetic energy giving it the necessary speed to stay aloft. Yes, upward thermal currents will slow the mean rate of descent a little, but they are not literally pushing the glider upwards and obviously it would fall out of the sky if it were motionless.
It is indeed a whole new paradigm, but it happens and my physics is correct. That’s why you Pierre can’t answer my questions about Uranus and Venus as to how the necessary thermal energy gets to the base of their tropospheres and into the Venus surface in order to raise its temperature during the day by the same amount (about 5 degrees) that it cooled the night before.
You only need your trusty on-line Stefan Boltzmann calculator to realise that the only regions in the Venus troposphere where the direct solar radiation can raise the temperature are those at temperatures below about 400K. Neither the temperature of the surface (varying between about 732K and 737K) nor those temperatures in the lower troposphere can be raised by direct solar radiation. Yet these temperatures do rise over the course of every four-month-long Venus day. Why is it so, Pierre-Normand?
Judith, you recently penned a “controversial” WSJ op-ed, and this past spring testified before our somewhat ill informed (and continuingly ill informed ) Congress on this issue, and above write:
“”””Fewer than 10% of journalists are skeptical about climate change, but more than one third of journalists believed that coverage of climate change science must reflect a balance of viewpoints or present all sides of controversy (half did not believe in the need for balance).”””
A balance of the “Viewpoints”?
Or a balance of the science.
The two are EXTREMELY different.
You also claim that:
“””In the legal system the ‘other side’ has a fair chance to make their case; in the climate change debate, the ‘other side’ is smeared as ‘deniers.'”””
This presents a false dichotomy, or at best a very misleading one.
There is no “other side.” There is science, which has all sorts of uncertainty, and is an ongoing process of mistake, adjustment, correction, learning, debate, re examination.
And then there is climate refutation. If you don’t believe this, fine; but it still leads to the same end: The “climate refutation” is based upon a large set of climate change myths and miscontructions.
Which is not another side, but misinformation.
Misinformation, or misunderstanding on the basic issue, is NOT a side. But, respectively, misinformation, or misunderstanding. What the media should be doing, in fact (but for the most part does not) is in fact present the story of misinformation, and the story of basic issue misunderstanding posing or being asserted as a “side,”rather than that misinformation or misunderstanding, as a “side” or “viewpoint,” devoid of relevant factual context, itself.
I ask you Judith Curry to read this humble piece, and tell me what the scientific justification, for this “other side” – that suggests that the phenomenon known casually as climate change (the present and likely future effect range of our past and current long term atmospheric alteration) is essentially non existent or insignificant – IS. Then we can scientifically examine it. Publicly. (Or privately, I am receptive to discussion for the sake of learning, and expanding awareness both on my end and for some one else.)
Additionally, your claim that unlike our “legal system,” the “other side” is “smeared as deniers,” is mischaracterizing or misleading, on multiple fronts.
First of all, this is not an issue to be decided by argument, but by science. I’m not sure that last point can be emphasized enough. Yet on this issue there has evolved a large conflation between the two; and, similarly, rhetoric is taking the place of science in this so called “debate.” (Worse, it is often misleadingly offered as, confused with, and often intermingled with, “science.” or out of context, or incomplete scientific information.)
You may be a “scientist,” in some sense, with some relevant expertise – and certainly have the academic background – but the vast majority of the so called “other side (which I imagine you are on the fringes of, but knowing we are affecting the earth in a non trivial way, not much more) IS NOT.
So if people want to claim climate scientists are wrong, by rhetoric and irrational statements – in that they are completely irrelevant so using them to “refute” climate change is irrational, such as constantly repeated if illogical claim that we aren’t much affecting the earth because “the earth has changed before,” or, inaccurate or completely issue mischaracterizing statements – such as “the earth has not been warming” (see above link or, talk to NASA or NOAA or the World Meteorological Organization) – that is not a “counter argument.” It is a fundamental misunderstanding of the issue, or flat out misinformation.
Second, the implication that the “other side” is not heard, is wildly mistaken. This “other side” practically dominates the Internet, among the three major ones dominates the leading (in terms of both viewership and “most trusted polls”) national news channel, and often comes close to dominating world discussion and much social and quasi media.
That is hardly being quashed.
In addition, the representation of the so called “other side” (look at the wide spread and incredibly disproportionate attention this scientist, because he is a rare example of an actual practicing climate scientist who leans toward the “it’s not a big deal” view, gets) is highly disproportionate – that is, much larger – in the mainstream media relative to the number of actual scientists who professional study this issue, as opposed to politically interested parties spouting forth opinions as fact. (Though a lot of the latter from this “other side” get coverage too.)
Lastly, the implication is that climate “denialism” is denigrated, while climate science is not. Yet Michael Mann (see this comment) is nevertheless called the “child molestor” of climate science, as one (albeit a particularly vile one) of countless examples. (And to the extent it is in a highly derogatory and conclusionary way, as if “everybody knows” it is bunk, I repeatedly take issue with this. This is also different from illustrating the massive degree of misinformation that such “climate change naysaying” relies upon, and correctly categorizing it as such, however, although that is often mistakenly conflated with denigration or attack.)
Yet the fact is, climate scientists repeatedly have not just their work, but their integrity attacked.
And while as you know I don’t really support the use of the term “denialism” (though a better one is still elusive, since skeptic is misleading, contrarian doesn’t really say much, and no one seems to like “naysayer”) the fact is that much climate change refutation is based on denial of the essentially widespread (among scientists who actually professionally study the issue, and instead of continually taking false issue with this, why don’t you ask them – all of them) idea or knowledge that a multi million year geologically radical increase in the long lived concentration of thermal radiation absorbing and re emitting molecules – thus adding increasingly more energy to the earth lower atmosphere system and slowly heating both the earth and increasingly changing the normally (and “relatively”) stable earth systems that drive climate – nevertheless won’t really affect the earth’s long term climate much.
So the label is not all that inaccurate. And it hardly rises to the level of smear – let alone in relation to the actual facts – that climate scientists are routinely subject to.
Again, the bottom line is that misinformation is not another side, but misinformation.
Misconstruction and misunderstanding of the basic issue is not another side, but error, or mistake.
And while viewpoints can be incorporated into science in the sense of in the sense of interpretation of objective knowledge of the physical world, viewpoints, in this context are not science.
John
I am a historical climatologist. In this respect I am fortunate to live close to the Met office which has one of the worlds finest climate related library and archives.my particular rsearch project at the moment is to research our climate back to the 11 th century by which time superstition, legend and religion start to interfe with the reliability of the records.
The trouble with paleo proxy reconstructions such as the hockey stick and it’s spaghetti derivatives is that they are a very coarse sieve through which the fine grain of natural variability in the form of annual and decadal temperatures fall through. Instrumental records are far more sensitive to these annual and decadal changes which is why the last hundred years of the hockey stick looks so dramatic compared to the static nature of paleo proxy reconstructions that preceded it.
In this respect the modern era can be seen to be nothing remarkable in its historic context of 3000 years when we can trace the advance and retreat of glaciers
You can see what I mean in this article here that I wrote last year
http://wattsupwiththat.com/2013/08/16/historic-variations-in-temperature-number-four-the-hockey-stick/
Tonyb
Tony: What’s your take on the following paper?
Historical Climate Records in China and Reconstruction Of Past Climates
http://journals.ametsoc.org/doi/abs/10.1175/1520-0442(1989)002%3C0833%3AHCRICA%3E2.0.CO%3B2
Hmmm. Colder is a bummer. Who’d a thunk it?
=======================
Rls
It looks a very comprehensive series of studies. I have seen some of them individually before but not collectively. I will work my way through them.
I note that Michael Mann and Phil jones are some of the authors.
The past often showed great variability in climate, but what I find most interesting is that some of the worst extremes, in particular drought and flooding, occurred during the little ice age episodes. This experience seems to be reflected in China and in Europe which is the opposite of what is often claimed when warmer is supposed to be wetter and more extreme.
Did you see my reply to you regarding the University of Michigan?
Tonyb
I did see your reply in which you discussed that you’ll be doing more research on borehole data but may need my help in the future. I have plenty of time on my hands; only job is to chauffeur my granddaughters to school, dance classes, the Detroit Zoo, Detroit Institute of Arts, University of Michigan events and sights, etc. The link below shows one of their favorite places: the Wave Field was designed by the woman that designed the Vietnam War Memorial.
http://m.campusinfo.umich.edu/article/wave-field-north
Tonyb,
I wonder if it is possible (he said, doing his damnedest to be ultra-diplomatic) that there is a reasonably stable quantity of surface water on the face of the Earth, whether solid or liquid.
If there is, then assuming the overall level of the crust stays more or less the same – uplift more or less compensated by sinking somewhere else, then possibly the sea level as gauged against whatever bits of crust sitting proud of the water may not change all that much overall.
Add to this isostatic rebound when ice caps, glaciers etc., melt, it might be that because the whole shebang – lithosphere, aquasphere, and so on – seems to be floating on a liquid mantle, there is a rough balance between the physical location of the various bits.
If the interaction is chaotic, then you are stuffed, because without satellites you only have relative, rather than absolute (more or less) measurements. I notice in passing that parts of the British Isles are sinking, parts are rising, and there doesn’t seem to be much rhyme or reason to the movement.
As you are aware, I’m used to people heaping ordure upon me. I’d be grateful if you could accept my thoughts as idle speculation, with some apparent observational basis.
Live well and prosper,
Mike Flynn.
Mike
The average depth of the ocean is 4000 metres. That the sea level might vary by 20 to 40 cm is a tiny variation.
Tonyb
Tonyb,
I think I am agreeing. The apparent sea levels might bob up and down even more than a metre local – as has happened many times before – without any assistance or lack thereof by supposed CO2 influence.
I guess what I’m trying to say that relating any sort of global rise or fall to a global metric which is almost impossible to establish a base line for (given that 70% of the crust is under 4000 metres of water, as you say). The displacement of liquid water due to rises and falls in the sea bed is a complete unknown, as far as I know.
Not trying to stir the pot, just curious about some peoples absolute certainty about stuff that seems a trifle uncertain, to say the least!
Thanks.
Live well
Mike
I think there are many certain uncertainties of which ‘global’ averages are the worst and become even more so when taken back thousands of years.
I much prefer regional analyses and preferably local ones.
tonyb
Rob Ellison wrote: “I said that the temperature depends on the kinetic energy of the molecules and the number of molecules. This is pretty obvious.”
It is pretty obviously false. T = (2/3)*KE_avg / k
(where k is the Boltzmann’s constant)
The number of particles, or the pressure, don’t figure in this relation.
Do we have the essence of the problem here?
Here is the box again.
http://upload.wikimedia.org/wikipedia/commons/6/6d/Translational_motion.gif
With physics – you begin by defining the boundary conditions. In this case the walls were said to be gaining heat – uniformly despite a ‘density difference’ that meant that there were more molecules transferring heat to the bottom than the top.
The transfer of heat to a surface is a physical result of collisions between the gas molecules and the walls. The statistical treatment is a view of the overall behavior of the system – statistical mechanics – in which the molecular dynamics play no part. The approach has enabled solutions of a broad range of problems without detailed consideration and summing of molecular dynamics – a task that is impossible still even for a simple box. What it can’t be used for is to deny the basis of heat transfer through the dynamics of individual molecules.
Rob Ellison: “Yes we have the same pressure top and bottom. It is a matter of momentum transfer to do with the number and speed of the gas molecules. The movement of molecules is random. The weight of the gas is another thing entirely. The internal pressure in a rubber balloon is uniform – but if it is just air inside it will still sink.”
So, now you are denying that a gas in a box has any vertical pressure gradient at all within the Earth surface gravitational field. This is just too bizarre a thing for an hydrological engineer to believe. You must be pulling my leg. Does this mean that if we close hermetically all the windows of a tall building then barometers within it will rapidly come to register the same pressure at all stories? Do you also believe this to be true about pressure in liquid filled container at equilibrium?
In the atmosphere – a pressure differential is caused by the weight of atmosphere above dropping as you rise. If we assume that the box is in an earth gravity
In the atmosphere the density gradient is caused by the weight of the atmosphere. The mass of the air above decreases with height. With the box we have whatever pressure is applied to the top – plus whatever the additional weight of the gas is at the bottom. This latter is always neglected – because the box is assumed small – the size of the box just needs to be large in relation to the distance between molecules – and what we are talking about is kinetic theory of gases – in which molecules bounce of each other and the walls and distribute randomly within the space. It is a matter of boundary conditions defined for a specific result.
‘What you say about weight and momentum also makes little sense. How can the gas molecules transfer any net momentum at all to the box if they hit the top and the bottom of the box with the very same velocity distribution?’
The momentum translates to a uniform internal pressure. Weight in a gravity field is still mass x g.
During their free falling trajectories in the box, in between the collisions with the top or bottom (side collisions have no momentum effect due to symmetry), the molecules are free falling and gaining downward momentum p = m*v = F*Δt = m*g*Δt. (from Newton’s second law and the definition of momentum p = m*v, and Δt is the time between the collisions). It is this constant gravitational momentum gain that is transferred by individual molecules to the box. There is a *net* transfer to the bottom of the box precisely because of this gain during the free falling trajectory. If there were no pressure difference — hence the same rate of collisions with the same average (vertical) velocity — between top and bottom, then the gas would be unable to transfer any amount of net momentum to the box and, as I said, its weight would not register outside of the box.
They lose kinetic energy on the way up and gain it on the way down. The molecular movements are random – so this evens out. Higher kinetic energy on the way down a higher bounce on the way back. The weight of the gas is still mass x g.
“The question was framed such that the gas was imparting energy to the walls of the box.”
No. We have been assuming equilibrium and hence an insulated container (or, equivalently, box walls at the same temperature as the gas). The molecules impart momentum to the walls but not net energy. (The box gains no net momentum due to its walls being connected together and its being held up against gravity). It’s a clear violations of the condition of equilibrium that there would be any heat transfer through any surface. Kinetic energy is exchanged back and forth at the same rate, between the gas and surface molecules, with no effect on the molecular speed distribution.
The assumption was that the gases were warming the surface.
Remember also the condition of hydrostatic equilibrium. For any macroscopic parcel of fluid, its weight is *exactly* balanced by the local pressure gradient. For the case of an isothermal gas at equilibrium, this condition is satisfied by the gradient defined by the barometric formula.
You can’t have hydrostatic equilibrium under gravity with no pressure gradient at all. Each parcel of air would be essentially in free fall together with all the parcels surrounding it — and hence it wouldn’t even encounter any inertial or viscous resistance by them). This would occur as long as there is no pressure gradient. This is of course absurd. It could last for a fraction of a second but then the fluid within the box would very rapidly come in equilibrium as the pressure gradient is established through the rapid compression of the lower layers by the (initially) free falling layers above them.
The pressure gradient in liquid is very much greater than with gases. But we are talking a box and kinetic theory of gases and not hydrostatics. In a relatively small space – say a pressure vessel – we would assume a random distribution of molecules and a uniform pressure.
https://watertechbyrie.wordpress.com/wp-admin/post.php?post=28&action=edit
The internal pressure is uniform but the gas still has a mass in a gravity field. Otherwise you would be able to lift yourself by your own bootstraps.
This is all rudimentary stuff – and the confusion stems from ill defined and changing boundary conditions, trying to fit different theories into the same box, misunderstanding basic concepts in Newtonian physics and ignoring the difference between statistical mechanics and molecular dynamics.
As I say – P-N pretends to an expertise that he is far from possessing. This was my original point in relation to FOMBS. Pretensions to science he barely understand for reasons of advancing a rhetorical agenda – rhetorically flourishing of symplectic manifolds in the service of transforming economies and societies. It is part of the meme to be scientifically superior to science deniers – no matter how little allied to reality – that is antithetical to ethical communication.
Ah… pressure vessel…
https://watertechbyrie.files.wordpress.com/2014/06/spherical.png
Rob, you are changing the topic entirely. We never were discussing heat transfer through the boundaries of a box. We were discussing a gas volume in thermodynamic *equilibrium* with a *uniform* temperature within a vertical gravitational field. There is no net heat flow anywhere under equilibrium. The gas interactions with the solid boundary (that has the same temperature as the gas) doesn’t have any effect of the speed distribution of the molecules.
FOMD and I claimed that although individual molecules lose kinetic energy on the way up (and gain it back on the way down) this does not change the local speed *distribution*, which is the same at all levels withing the box (or some isothermal bit of stable troposphere). This is because the pressure variation (according to the barometric formula) means there is a drop-off rate in the number of particles that have lower total energy (KE+PE) on the way up (since the slower particles lose *all* their KE before getting to the higher levels and fall back down) and the distribution therefore remains constant (the Maxwell distribution indeed only is a function of T, and not of pressure or n).
You argued that this was absurd, then were backed into denying that there is any vertical gradient at all within the box (which *is* absurd), and now are arguing that the pressure change is “negligible” in the context of discussing some unspecified heat transfer problem that never was under discussion.
But there *is* a non negligible (though small) pressure gradient within the box. It is the exact same gradient as the gradient outside of the box and it fully accounts for the aforementioned vertical drop-off rate needed to explain the constant speed (and KE) distribution as a function of height that so puzzles you.
I have to copy everything or I get lost in the verbiage.
Rob, you are changing the topic entirely. We never were discussing heat transfer through the boundaries of a box. We were discussing a gas volume in thermodynamic *equilibrium* with a *uniform* temperature within a vertical gravitational field. There is no net heat flow anywhere under equilibrium. The gas interactions with the solid boundary (that has the same temperature as the gas) doesn’t have any effect of the speed distribution of the molecules.
There was a box with a bottom that heated more than the top.
FOMD and I claimed that although individual molecules lose kinetic energy on the way up (and gain it back on the way down) this does not change the local speed *distribution*, which is the same at all levels withing the box (or some isothermal bit of stable troposphere). This is because the pressure variation (according to the barometric formula) means there is a drop-off rate in the number of particles that have lower total energy (KE+PE) on the way up (since the slower particles lose *all* their KE before getting to the higher levels and fall back down) and the distribution therefore remains constant (the Maxwell distribution indeed only is a function of T, and not of pressure or n).
KE is determined by temperature – molecules at the top of the box and at the bottom have the same KE because they are nominally at the same temperature. The molecules are moving randomly.
You argued that this was absurd, then were backed into denying that there is any vertical gradient at all within the box (which *is* absurd), and now are arguing that the pressure change is “negligible” in the context of discussing some unspecified heat transfer problem that never was under discussion.
But there *is* a non negligible (though small) pressure gradient within the box. It is the exact same gradient as the gradient outside of the box and it fully accounts for the aforementioned vertical drop-off rate needed to explain the constant speed (and KE) distribution as a function of height that so puzzles you.
I denied that a constant KE on the moon was proven in any way at all. By all means reference something. In a box – elastic collisions ensure molecules tend to be randomly distributed within the space. The statement of the problem was that there were more molecules hitting the bottom than the top yet the temperature transfer was equal. This was of course intended to prove that superior science means that capitalism is evil and must be abolished. It is in other words far from a serious or relevant comment. My original comment was on the bad faith routinely shown by FOMBS.
The weight of the gas in the box is the result of acceleration of molecules under gravity. Faster impact on the bottom – what I denied was that there was a density gradient. What I refuted was that more molecules with the same average kinetic energy hitting the surface would result in the same heat transfer – assuming as FOMBS did a sink – as fewer molecules hitting another surface.
It is a fundamental misunderstanding of thermal energy.
Take these cylinders of oxygen again – they are in local thermodynamic equilibrium. This means that the total thermal energy is the same in all cases. But there are more molecules in the compressed gas – which means that the average kinetic energy per molecule is less. This seems to be the fundamental problem – the confusion of average thermal energy for total thermal energy which is brought in via the Boltzmann constant. The heat of the substance is related to the total thermal energy.
The basic idea is that more molecules with the same average thermal energy per molecule hitting a surface will warm it more than fewer molecules. A straight forward outcome of molecular dynamics. Which is what I said oh so long ago.
“The statement of the problem was that there were more molecules hitting the bottom than the top yet the temperature transfer was equal.”
No. I claimed that the temperature was the same. I said nothing about any “temperature transfer(sic)”. You claimed that the temperature can’t be the same at the top and bottom since temperature is a function of both KE and collision rate. I pointed out that is isn’t; it merely depends on the average KE of the gas molecules. Now you’re trying to pretend that I was making a statement about heat transfer. I never have been.
Rob Ellison: “The internal pressure is uniform but the gas still has a mass in a gravity field. Otherwise you would be able to lift yourself by your own bootstraps.”
Look up the distinction between “mass” and “weight”. The gas in the vessel/balloon/box has weight F = m*g that presses down on the vessel. That’s because the particles are free falling and gaining downward momentum. The downward momentum delta_p gained during the free fall trajectory is given back to the vessel through collisions with it. It translates as more collisions on the bottom than on the top. And this accounts for the internal pressure gradient. It is small, sure. Just enough to hold the mass of gas up.
Rob Ellison: “The internal pressure is uniform but the gas still has a mass in a gravity field. Otherwise you would be able to lift yourself by your own bootstraps.”
tl;dr version: In fact, if the gas had a uniform pressure inside the container, as you allege, *then* it would be holding itself up by its own bootstraps, since in that case it would exert no net force at all on the vessel.
Pierre-Normand,
I must be a glutton for punishment. At least least one of you will heap scorn and derision on me. I haven’t been following the to and fro closely, but I just want to point out that density, pressure, and temperature at equilibrium are not necessarily related.
Consider three oxygen cylinders at your local gas merchant. All the same nominal size, but one contains gas at 100 bars, one at 10 bars, and one at 1 bar.
The temperature of all three is the same, both the gas and the container. I think this might have some relevance. Neither pressure nor gravity causes a rise in temperature.
Which one is going to tell me that they can determine gas pressure by measuring its temperature or vice versa?
Live well and prosper,
Mike Flynn.
Mike Flynn, you are correct. The ideal gas law relates the molar amount (how many moles of molecules), the pressure, temperature and volume of a gas parcel in one single relation. So, you can only infer one of those quantities if you know the other three.
I gave the weight as mass x g. You can be assured that an engineer understands this.
It translates as faster collisions at the bottom – but we are assuming inelastic collisions such that some of this energy is converted to kinetic energy in the walls. There was no assumption of thermal equilibrium between the gas and the walls. You need to stick to the boundary conditions of a problem that was incompetently formulated.
Consider three oxygen cylinders at your local gas merchant. All the same nominal size, but one contains gas at 100 bars, one at 10 bars, and one at 1 bar.
The temperature of all three is the same, both the gas and the container. I think this might have some relevance. Neither pressure nor gravity causes a rise in temperature.
Which one is going to tell me that they can determine gas pressure by measuring its temperature or vice versa?
Any substance will reach a local thermodynamic equilibrium. The zeroth law of thermodynamics ensures that. Ultimately this leads to the heat death of the universe. This is what’s known as a trivial result.
Rob Ellison wrote: “I gave the weight as mass x g. You can be assured that an engineer understands this.”
If you understand this then your claim that “The internal pressure is uniform but the gas still has a mass in a gravity field.” doesn’t make sense.
“It translates as faster collisions at the bottom.”
Since the temperature is uniform, the average KE of the molecules is the same across the whole box. Hence the collisions on the bottom aren’t any faster on average than the collisions on top. But there are more of them because the pressure, and hence also the molecular density, both are higher. It’s the density difference that accounts for the higher pressure in the isothermal case.
But even if we have it your way, supposing there would be more momentum transferred to the bottom due to higher average velocities, this would still contradict your claim that the weight of the gas has nothing to do with a top-to-bottom pressure difference. dp = F*dt. (Newton’s second law). F = P*A (Force is pressure over area). Hence if the gas transfers more momentum to the bottom as it does to the top, the pressure can’t be the same.
Rob Ellison wrote: “There was no assumption of thermal equilibrium between the gas and the walls.”
Right. There was no assumption that there isn’t a rabbit in the box either. We were discussing a gas in equilibrium. Why would we suddenly consider a warm gas cooling off in a cooler box? This has nothing to do with the original problem which concerns average KE distribution in a vertical column of gas where each molecules conserves TE = KE+PE. The box merely prevent the gas from expanding. That’s all.
You can either assume that the wall collisions are elastic, which generally is the case with idealized insulated walls, or, equivalently, assume that the walls have the same temperature as the gas. (And don’t forget either that there is no rabbit in the box). In either case the collisions with the wall have zero effect on the speed or KE *distributions* of the molecules. Some molecules may give up some KE to the walls, but as many gain KE from the walls.
Your trouble, I think, is that you still can’t come to grip with the fact that a lower collision rate with the top wall is consistent with the gas being at the same temperature both at the top and bottom. But as I said, kinetic temperature only is a function of average KE, and is independent of pressure. This has never been a problem about heat transfer through the walls or else we would have mentioned heat capacity and thermal conductivity. But this would be a completely irrelevant distraction from the original problem raised by FOMD about KE distribution as a function of height and TE = PE + KE being satisfied by all individual molecules.
Rob Ellison wrote: “I gave the weight as mass x g. You can be assured that an engineer understands this.”
If you understand this then your claim that “The internal pressure is uniform but the gas still has a mass in a gravity field.” doesn’t make sense.
“It translates as faster collisions at the bottom.”
Since the temperature is uniform, the average KE of the molecules is the same across the whole box. Hence the collisions on the bottom aren’t any faster on average than the collisions on top. But there are more of them because the pressure, and hence also the molecular density, both are higher. It’s the density difference that accounts for the higher pressure in the isothermal case.
You can measure the net weight of air – it is the weight less the buoyancy. Air at the density of the surrounding air is essentially weightless. It is stable – all the forces are equal.
The temperature is uniform in the box – the density is uniform because the molecules bounce elastically off each other and the walls – the downward motion of molecules is accelerated by gravity but so is the upward rebound – the molecules bounce around at hundreds of metres per second.
But even if we have it your way, supposing there would be more momentum transferred to the bottom due to higher average velocities, this would still contradict your claim that the weight of the gas has nothing to do with a top-to-bottom pressure difference. dp = F*dt. (Newton’s second law). F = P*A (Force is pressure over area). Hence if the gas transfers more momentum to the bottom as it does to the top, the pressure can’t be the same.
http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/weighgas.html
At ground level the box under the weight of the entire atmospheric column – the pressure at the bottom is the weight of the atmosphere over the area plus the weight of the gas inside the box. Let’s call the pressure uniform atmospheric. The density is uniform because the molecules randomly distribute themselves through the box through collisions. We assume with the molecular kinetic theory of gases that the gas is isotropic as a result of these random motions.
Enter your comment here…
Rob Ellison: “At ground level the box under the weight of the entire atmospheric column – the pressure at the bottom is the weight of the atmosphere over the area plus the weight of the gas inside the box.”
Yes, assuming the box isn’t pressurized, then the pressure at the bottom of the box, multiplied by the area, just is the weight of the whole atmospheric column *including* the gas in the box. And the pressure at the top of the box, multiplied by the area, just is the weight of the atmospheric column *excluding* the gas in the box. The barometric formula, and the pressure gradient that it determines, applies equally well inside or outside the box. (Assuming equal gas density).
Imagine many holes on the side of the box that allow pressure equalization over the full height of the box. Then the pressure gradient is the same in the box as it is outside. Then you can plug the holes. This will not cause the pressure gradient within the box to vanish.
“You can measure the net weight of air – it is the weight less the buoyancy. Air at the density of the surrounding air is essentially weightless. It is stable – all the forces are equal.”
The buoyant force on the box is the same irrespective of the content of the box (since it equals the weight of the outside air displaced by the box.) So, you can weigh the air-filled box, vacuum it, and weigh the box again. The difference is the weigh of the air that you have evacuated.
“The temperature is uniform in the box – the density is uniform because the molecules bounce elastically off each other and the walls – the downward motion of molecules is accelerated by gravity but so is the upward rebound – the molecules bounce around at hundreds of metres per second.”
If it were true that the density and temperature were both uniform, then the box would weight the exact same when full of empty. That’s because if the internal pressure is the same on the bottom and top walls, then there is no net force applied on the box that corresponds to the weight of the air content.
You can measure the net weight of air – it is the weight less the buoyancy. Air at the density of the surrounding air is essentially weightless. It is stable – all the forces are equal.”
The buoyant force on the box is the same irrespective of the content of the box (since it equals the weight of the outside air displaced by the box.) So, you can weigh the air-filled box, vacuum it, and weigh the box again. The difference is the weigh of the air that you have evacuated.
Well no – the buoyant force is the weight of the displaced fluid less the weight of the box and contents. You can always remove the contents to weight the box. You can then know what the contents weigh. So what?
“The temperature is uniform in the box – the density is uniform because the molecules bounce elastically off each other and the walls – the downward motion of molecules is accelerated by gravity but so is the upward rebound – the molecules bounce around at hundreds of metres per second.”
If it were true that the density and temperature were both uniform, then the box would weight the exact same when full of empty. That’s because if the internal pressure is the same on the bottom and top walls, then there is no net force applied on the box that corresponds to the weight of the air content.
The weight is provided by the gravity augmented downward molecular velocity. The molecules are randomly distributed – the gas is isotropic – therefore the density and temperature are uniform. This is assumed in the kinetic theory of gases.
Rob Ellison: “The temperature is uniform in the box – the density is uniform because the molecules bounce elastically off each other and the walls – the downward motion of molecules is accelerated by gravity but so is the upward rebound – the molecules bounce around at hundreds of metres per second.”
Not sure what you mean by gravity accelerating the rebound. Gravity accelerates the molecules downwards both after and before the rebounds with the top or bottom walls. The momentum gained by gravity is the integral of mg*dt over the whole free-fall trajectory in between rebounds. The action of gravity over the duration of the rebounds is negligible (and isn’t, in any case, directed upwards). There therefore is a constant gain in downward momentum. The total rate of this gain — for all the molecules — must be opposed by the area*pressure difference on top and bottom. Else the gas would be free falling as a whole.
Imagine a tennis ball bouncing elastically up and down inside a truck. The average weight of the truck just is the weight of the truck + the weight of the ball. The ball is constantly free-falling between the rebounds. This is why it is bouncing harder on the bottom than it is on the top. It may also be that it doesn’t even reach the top. In any case, its “pressure” (rate of momentum transfer) on the top is less than it is on the bottom. It’s the same with a very large truck and one bazillion tennis balls bouncing all around. They collectively transfer their weight to the truck merely through hitting the floor harder than the ceiling of the container.
‘Not sure what you mean by gravity accelerating the rebound. Gravity accelerates the molecules downwards both after and before the rebounds with the top or bottom walls.’
The rebound velocity is the same as the velocity it hit with. Simple?
Rob Ellison wrote: “There was no assumption of thermal equilibrium between the gas and the walls.”
Right. There was no assumption that there isn’t a rabbit in the box either. We were discussing a gas in equilibrium. Why would we suddenly consider a warm gas cooling off in a cooler box? This has nothing to do with the original problem which concerns average KE distribution in a vertical column of gas where each molecules conserves TE = KE+PE. The box merely prevent the gas from expanding. That’s all.
The statement was that more molecules hit the bottom of the container – but that warming at the top and the bottom was equal because the average kinetic energy of the particles was the same. Simply wrong.
Molecular projectiles on the moon was a different thought bubble. One where it was assumed that KE was equal at all levels – despite losing KE and gaining PE on the way up. Again simply wrong – and assertions with demonstration are utterly worthless.
‘You can either assume that the wall collisions are elastic, which generally is the case with idealized insulated walls, or, equivalently, assume that the walls have the same temperature as the gas. (And don’t forget either that there is no rabbit in the box). In either case the collisions with the wall have zero effect on the speed or KE *distributions* of the molecules. Some molecules may give up some KE to the walls, but as many gain KE from the walls.’
No the stated boundary condition was that the walls gained energy.
Your trouble, I think, is that you still can’t come to grip with the fact that a lower collision rate with the top wall is consistent with the gas being at the same temperature both at the top and bottom. But as I said, kinetic temperature only is a function of average KE, and is independent of pressure. This has never been a problem about heat transfer through the walls or else we would have mentioned heat capacity and thermal conductivity. But this would be a completely irrelevant distraction from the original problem raised by FOMD about KE distribution as a function of height and TE = PE + KE being satisfied by all individual molecules.
Temperature is a product of the total thermal energy of the mass. The problem was defined as more molecules hitting the bottom than the top – which because the gas was at a uniform temperature meant that the average kinetic energy was the same and therefore the temperature increase at the surfaces was the same. Again – simply wrong. The thermal energy of the mass is a product of the average kinetic energy per molecule times the number of molecules. The number of molecules – or mass – as I said above is brought into the equation via the Boltzmann constant.
Forgive me for insisting that the boundary conditions remain the same – otherwise we could prove with shifting goalposts that P-N’s mother was a duck.
The other problem of KE for all molecular projectiles on the moon being a constant with height is something that is described in words as the increase in potential energy at every higher level being exactly compensated for by the drop out of lower energy projectiles so that the average stayed the same precisely. It is mathematically hugely unlikely – and definitely unproven in any rigourous sense. An article of faith with P-N it seems.
It seems moreover hugely irrelevant. Total kinetic energy of molecular moon projectiles – which is what counts – drops with height. The situation in the atmosphere is much more complex.
“Molecular projectiles on the moon was a different thought bubble. One where it was assumed that KE was equal at all levels – despite losing KE and gaining PE on the way up.”
Exactly. That was the claim. Heat transfer with walls is irrelevant to it. Your formulation, of course, misses the point that the *distribution* of KE is the same at all levels despite KE being reduced on the way up for each *individual* molecules. You keep pretending to not to hear the main point, which is that there are different *populations* at different levels. Not all molecules can get to level z. Only those molecules that have TE >= mgz can get level z.
If you constantly ignore the argument and misrepresent the claim, then it’s easy for you to shoot down your own strawman and miss the point.
This is an utterly forgettable thought bubble – supply a rigourous proof of KEavg remaining constant for molecular moon projectiles or continue to be a verbose poseur. Ask your friend FOMBS.
“This is an utterly forgettable thought bubble – supply a rigourous proof of KEavg remaining constant for molecular moon projectiles or continue to be a verbose poseur.”
OK, I’ll supply a proof. It may take a couple days. Meanwhile I can be forgiven for straightening up your misunderstanding of the claim that you purport to argue against — and the rather crude logical and scientific errors that you are making in the process.
LMFAO
“Temperature is a product of the total thermal energy of the mass.”
At this point I am unsure that you grasp the difference between temperature and heat at all. Kinetic temperature is proportional to average KE, period. It has nothing to do with the gas being dense or rarefied. The terrestrial thermosphere is very rarefied, has a very low density of internal energy (KE volumetric density), hence a very low heat content, but an extremely high temperature.
Pierre-Normand,
Most people, including self anointed climate scientists, have only a hazy understanding of the differences and relationships between heat, energy, and temperature, as far as I can see.
The concept of temperature as we generally understand it becomes more or less irrelevant in the thermosphere and exosphere, for example. These thought experiments are all good fun, but generally lead nowhere. I still don’t believe the Warmist dogma, but I support your opinion on the gravito-thermal effect, at least on he basis of my admittedly brief perusal of the current discussion.
Live well and prosper,
Mike Flynn.
Mike Flynn: “These thought experiments are all good fun, but generally lead nowhere.”
The fun is the main point. If they lead to some understanding, that’s a bonus.
‘The distinguishing difference between the terms kinetic energy and thermal energy is that thermal energy is the mean energy of disordered, i.e. random, motion of the particles or the oscillations in the system. The conversion of energy of ordered motion to thermal energy results from collisions.[6]
All kinetic energy is partitioned into the degrees of freedom of the system. The average energy of a single particle with f quadratic degrees of freedom in a thermal bath of temperature T is a statistical mean energy given by the equipartition theorem as
E thermal = f.1/2. kT
where k is the Boltzmann constant. The total thermal energy of a sample of matter or a thermodynamic system is consequently the average sum of the kinetic energies of all particles in the system. Thus, for a system of N particles its thermal energy is[7]
U thermal = N.f.1/2. kT.
For gaseous systems, the factor f, the number of degrees of freedom, commonly has the value 3 in the case of the monatomic gas, 5 for many diatomic gases, and 7 for larger molecules at ambient temperatures. In general however, it is a function of the temperature of the system as internal modes of motion, vibration, or rotation become available in higher energy regimes.’ Wikipedia
P-N confuses the mean energy of the particle – which is related to temperature of course to the total energy of the system – with the total energy of the system. Two gas cylinders at different pressures at thermal equilibrium with the surroundings will have the same thermal energy. The average will be less – you get that by dividing by N.
Molecular dynamics says that more molecules at a lower average kinetic energy will deliver the same energy to the walls as in the lower density gas.
“P-N confuses the mean energy of the particle – which is related to temperature of course to the total energy of the system – with the total energy of the system.”
But I said nothing at all about the total energy of the system. I told you two dozen times that the average kinetic energy KE_avg = (3/2)kT, where T is the kinetic temperature, and hence that *since* the temperature of the gas is the same throughout the box (because I assumed thermodynamical equilibrium) then the average kinetic energy per molecule is the same at the bottom and the top of the box.
“Two gas cylinders at different pressures at thermal equilibrium with the surroundings will have the same thermal energy. The average will be less – you get that by dividing by N.”
No. They will have the same average kinetic energy per molecule KE_avg = (3/2)kT and hence *not* the same total thermal energy. It’s not the same same number of molecules in both cylinders, and hence not the same number of degrees of freedom (3 per molecules for an ideal gas).
“Forgive me for insisting that the boundary conditions remain the same ”
The boundary condition is this: insulate walls, constant volume. No heat transfer. No temperature variation.
We have more molecules hitting the bottom at the same same average kinetic energy and the bottom and top heating the same. The assumption was a heat sink. Alternatively we could have no net heat transfer and more molecules hitting the bottom and the top and bottom being the same temperature. Either is silly.
“The assumption was a heat sink.”
Do you have a quote? I’ve always said “thermodynamic equilibrium” and “isothermal” and now you bring up a heat sink. In any case a heat sink at the same temperature as the gas doesn’t sink any heat at all.
“Alternatively we could have no net heat transfer and more molecules hitting the bottom and the top and bottom being the same temperature. Either is silly.”
I don’t have a clue why you think that the top and bottom couldn’t have different pressures and yet the same temperature. Higher pressure –> higher rate of collisions. Same temperature –> same average kinetic energy per molecule. You seem to be saying that a gas could never achieve thermodynamic equilibrium in an insulated container without also maintaining an equal pressure (independent of height) in spite of gravity.
don’t have a clue why you think that the top and bottom couldn’t have different pressures and yet the same temperature. Higher pressure –> higher rate of collisions. Same temperature –> same average kinetic energy per molecule. You seem to be saying that a gas could never achieve thermodynamic equilibrium in an insulated container without also maintaining an equal pressure (independent of height) in spite of gravity.
The real question is – does P-N have a clue period?
Uniform density – uniform temperature – increased velocity of impact at the bottom. The density results from random motions of particles in the box. Gravity increases the downward velocity of molecules – but they also bounce back harder.
“Uniform density – uniform temperature – increased velocity of impact at the bottom.”
If there is increased velocity of impact, then average kinetic energy also is higher, no? This contradicts the assumption about uniform temperature. KE_avg = (3/2)kT.
“The density results from random motions of particles in the box. Gravity increases the downward velocity of molecules – but they also bounce back harder.”
Yes, they bounce back harder. So, what? They still decelerate on the way up and won’t hit the top as hard (and sometimes won’t reach the top at all). Also, if they move faster, on average, then since T = (2/3k)*KE_avg, the temperature would be higher.
Also, the pressure would be higher. delta_P/A = delta_F = dp/dt. (A is the bottom area) A higher rate of momentum transfer from molecules implies a larger average force on the bottom; and this implies a larger pressure.
The assumption of uniform pressure in violation of the condition of hydrostatic equilibrium also doesn’t make sense. That would mean that there is no buoyancy in the box. Put a helium filled balloon in the box, and it will fall down trough the denser air because there is no ambient pressure gradient to exert any buoyant force on the external shell of the balloon.
Rob Ellison: “The other problem of KE for all molecular projectiles on the moon being a constant with height is something that is described in words as the increase in potential energy at every higher level being exactly compensated for by the drop out of lower energy projectiles so that the average stayed the same precisely. It is mathematically hugely unlikely – and definitely unproven in any rigourous sense. An article of faith with P-N it seems.”
I’m glad that you at least — and at last — got a handle on the claim being made. However I think you haven’t fully come to grip yet with the fact that it also applies to any isothermal parcel of weakly interacting gas (such as air at 1atm and normal temperature) that’s subjected to gravity such that every molecules is constantly falling with an acceleration g in between collisions.
You at least see that in order to deny my claim in the terrestrial/box case, you must deny that pressure varies vertically at all in the box, or ignore the fact that kinetic temperature at any point within a gas only is a function of the average kinetic energy per molecule. But this seems to be a reductio ad absurdum of your position. My position, though as of yet unproven at least is consistent with, and seems entailed by, fairly well established laws of physics. (What is entailed is that the average KE at each height is the same, and what must be proven is that the distribution is the same Boltzmann distribution at all levels.)
I always had a handle – it is just horrendously unlikely and utterly unsupported verbiage.
The temperature of a substance is a function of its total thermal energy.
U thermal = N.f.1/2. kT.
At the same temperature – the average is proportional to the number of particles in the system.
I’m glad that you at least — and at last — got a handle on the claim being made. However I think you haven’t fully come to grip yet with the fact that it also applies to any isothermal parcel of weakly interacting gas (such as air at 1atm and normal temperature) that’s subjected to gravity such that every molecules is constantly falling with an acceleration g in between collisions.
The air is stable if it is in a local thermodynamic equilibrium. The motions are completely random. The vector addition of velocities of molecules travelling at 300m/s plus acceleration due to gravity results in molecules travelling at 300m/s. This is before molecules are convected and turbulently entrained.
‘You at least see that in order to deny my claim in the terrestrial/box case, you must deny that pressure varies vertically at all in the box, or ignore the fact that kinetic temperature at any point within a gas only is a function of the average kinetic energy per molecule. But this seems to be a reductio ad absurdum of your position. My position, though as of yet unproven at least is consistent with, and seems entailed by, fairly well established laws of physics. (What is entailed is that the average KE at each height is the same, and what must be proven is that the distribution is the same Boltzmann distribution at all levels.)’
The gas in the box under kinetic theory is isotropic. It just is. Temperature is a function of total thermal energy – obviously. The sum of the kinetic energies of all molecules in the system. It just is.
LMFAO. You need to decide on the temperature of the box on the moon – calculate the mean velocity – distribute velocities according to the Maxwell-Boltzmann distribution – calculate the trajectory of individual molecules – calculate PE at various levels and equate that to change in KE of molecules that reach that level – and then sum all KE and compare it to every other level . Over how many molecules?
I think you are gonna need an analytical proof. In the meantime just keep posing like FOMBS.
“The gas in the box under kinetic theory is isotropic. It just is.”
Not under gravity.
The theory says isentropic – but hey – you can always invent your own. All you need now is a proof.
“The theory says isentropic – but hey – you can always invent your own.”
Ah, sorry. I had misread that as ‘isotropic’. I thought you meant to suggest that kinetic theory implies that a gas in thermodynamic equilibrium under gravity would not exemplify a vertical temperature gradient. Though, it looks like you still believe that.
Did I say isentropic? Is my face red.
‘Kinetic theory is also an example of isotropy. It is assumed that the molecules move in random directions and as a consequence, there is an equal probability of a molecule moving in any direction. Thus when there are many molecules in the gas, with high probability there will be very similar numbers moving in one direction as any other hence demonstrating approximate isotropy.’ Wikipedia
It is really the very opposite of hydrostatics – which assumes isentropy.
Rob Ellison: “It is really the very opposite of hydrostatics – which assumes isentropy.”
But I agree that the speed distribution is everywhere isotropic. I assumed it is Boltzmann’s distribution for an ideal gas, and this is isotropic. You can’t infer from the fact that the speed distribution is isotropic that there is no vertical pressure gradient. At equilibrium there is no net molecular flux across any boundary. Molecules cross any boundary with the same average velocity in both direction. The pressure gradient, and density everywhere, are steady.
And you still need to account for the fact that, under your strange assumption that the hydrostatic equilibrium condition need not apply to the gas in a closed box at equilibrium, there would be no buoyancy.
As I said, put an helium filled balloon in in a thick empty (except for air at normal atmospheric pressure) hermetic safe and close the door. Let things settle down for a few hours for temperatures to equilibrate everywhere. Will the balloon fall down through the denser air? It should fall down to the bottom of the safe if there is no pressure gradient in the surrounding air. But surely that is absurd.
But I said nothing at all about the total energy of the system. I told you two dozen times that the average kinetic energy KE_avg = (3/2)kT, where T is the kinetic temperature, and hence that *since* the temperature of the gas is the same throughout the box (because I assumed thermodynamical equilibrium) then the average kinetic energy per molecule is the same at the bottom and the top of the box.
You assumed a density gradient – so if the temperature is the same at the top and the bottom then the average kinetic energy – per molecule – is not the same.
P-N said nothing about total thermal energy in the system because he assumes it makes no difference how many molecules there are. Simply not true.
It is easy to see. I have quoted Wikipedia above – it is correct – and there is added information on degrees of freedom and the average kinetic energy formula.
Rob Ellison “You assumed a density gradient – so if the temperature is the same at the top and the bottom then the average kinetic energy – per molecule – is not the same.
The average kinetic energy per molecule is (3/2)kT. The average kinetic energy per translational degree of freedom is (1/2)kT. (From your own source). There are three such degrees for particles of an ideal gas; hence (3/2)kT. The average kinetic energy per molecules does not depend on the volumetric density of the molecules. If the average Joe is 5 feet tall, it doesn’t matter if there are 1,000 or 10,000 Joes in the stadium. The average joe still is 5 feet tall.
“P-N said nothing about total thermal energy in the system because he assumes it makes no difference how many molecules there are. Simply not true.”
Sorry, forgot to say. There are 875708847507525427475325 molecules. It’s constant because they don’t move through the walls. Total thermal energy is constant because of conservation of energy and the system is in thermodynamical equilibrium. There are no heat flows. I once said the container is adiabatic but that caused you to make a huge tantrum.
“It is easy to see. I have quoted Wikipedia above – it is correct – and there is added information on degrees of freedom and the average kinetic energy formula.”
Yes, and it directly contradicts your claim that average kinetic energy per molecule varies when density varies — despite constant temperature. It doesn’t. The average kinetic energy of the molecules of an ideal gas is (3/2)kT irrespective of density. It merely depends on temperature and nothing else.
Rob Ellison “You assumed a density gradient – so if the temperature is the same at the top and the bottom then the average kinetic energy – per molecule – is not the same.
The average kinetic energy per molecule is (3/2)kT. The average kinetic energy per translational degree of freedom is (1/2)kT. (From your own source). There are three such degrees for particles of an ideal gas; hence (3/2)kT. The average kinetic energy per molecules does not depend on the volumetric density of the molecules. If the average Joe is 5 feet tall, it doesn’t matter if there are 1,000 or 10,000 Joes in the stadium. The average joe still is 5 feet tall.
If there are 5 degrees of freedom – the KE avg. = (5/3)kT
We have been through this before. If the gas cylinders are at the same temperature they have the same total thermal energy. The more compressed gas has ore molecules therefore the average kinetic energy per molecule is less. Joe turns out to be 2 foot tall.
“P-N said nothing about total thermal energy in the system because he assumes it makes no difference how many molecules there are. Simply not true.”
Sorry, forgot to say. There are 875708847507525427475325 molecules. It’s constant because they don’t move through the walls. Total thermal energy is constant because of conservation of energy and the system is in thermodynamical equilibrium. There are no heat flows. I once said the container is adiabatic but that caused you to make a huge tantrum.
I certainly do not have tantrums – I get bored little know nothings who with pretentious to pedagogical pomposity. I remember now – I said it was pretentious use of half understood jargon that subverted rather than enhanced communication and to say insulated as every rational person – including most physicists and engineers would.
Now if you add gas to the system and allow it to cool off to the surroundings – the total energy would be the same but the average kinetic energy per molecule would fall.
“It is easy to see. I have quoted Wikipedia above – it is correct – and there is added information on degrees of freedom and the average kinetic energy formula.”
Yes, and it directly contradicts your claim that average kinetic energy per molecule varies when density varies — despite constant temperature. It doesn’t. The average kinetic energy of the molecules of an ideal gas is (3/2)kT irrespective of density. It merely depends on temperature and nothing else.
We have a total thermal energy.
U thermal = N.f.1/2. kT. – now we can measure the temperature and – assuming 7 degrees of freedom say – calculate total thermal energy.
Now for this system of N molecules.
KEavg = U thermal/N
But – you say –
KEavg = f.k.T – where f is 7 for a diatomic gas.
But k = R/N
Where R is the gas constant for a specific volume and N is the number of molecules.
So if we increase N we reduce the average KE. This is the result I have discussed endlessly – and which proceeds directly from kinetic molecular theory as I originally stated. This is simply that energy is imparted by collisions and the amount is determined by the average KE and the number of particles. KE is not independent of the number of particles in the system.
“If there are 5 degrees of freedom – the KE avg. = (5/3)kT”
No. In that case the *total* energy (not just kinetic) of the molecules would be (5/2)kT. (Not 5/3).
If there are 5 degrees of freedom then the average kinetic energy still is (3/2)kT. 5 degrees of freedom would be typical of diatomic molecules where in addition to the three translational degrees of freedom, there are two rotational degrees of freedom (two axis of rotation). That give a further two 1/2kT to the total (average) energy of the molecule. But this energy does not contribute to the kinetic energy. The kinetic energy is purely translational.
“We have been through this before. If the gas cylinders are at the same temperature they have the same total thermal energy. The more compressed gas has more molecules therefore the average kinetic energy per molecule is less. Joe turns out to be 2 foot tall.”
Except there is a law that states that the average energy per degree of freedom is (1/2)kT. Each molecule brings 3 degrees of freedom (entirely translational). Hence the total thermal energy in the cylinder is N*(3/2)kT where N is the number of molecules. If the second cylinder has twice as many molecules then the total internal energy is doubled.
That should be intuitive to anyone. If you want to warm up both those cylinders 10°C more (from 15°C to 25°C), then it ought to take twice as much energy to warm up the cylinder that has twice the pressure (and density) since there are twice as many molecules to speed up to the new average KE.
I think you still are struggling with the concept of the arithmetic mean. You accept that molecules have EK_avg = (3/2)kT. You postulate that one cylinder has twice as many molecules, and both have the same temperature. And you still can’t calculate that one cylinder has twice the energy.
The formula has been given several times.
KEavg. = f.(1/2).k.T
where f is the degree of freedom.
Flynn specified that the cylinders were at an equilibrium temperature with the surroundings.
This means that the total internal energy of the cylinders is equal.
The total kinetic energy is proportional to temperature.
KEavg = f.(1/2).(R/N).T
Rob Ellison: “The total kinetic energy is proportional to temperature.
KEavg = f.(1/2).(R/N).T
Since each molecules has three translational degrees of freedom, you can use (3/2) instead of (1/2). More importantly, if N is the number of molecules, then you can’t replace k by R/N. You replace k by R if you want the average kinetic energy per mole. If N is the Avogadro number, then you are back to energy per molecules but then R/N still is a constant.
So you have proven that there is twice as much thermal energy in the second container if there are twice as many molecules.
Rob Ellison wrote: “But k = R/N”
No. “k” is the Boltzmann constant. It is a… wait for it!… constant.
KEavg(1) = f.(1/2).R.T/n1
KEavg (2) = f.(1/2)R.T/n2
where n is the number of moles – uncompressed and compressed moles in the volume.
Rob Ellison wrote:
“KEavg(1) = f.(1/2).R.T/n1
KEavg (2) = f.(1/2)R.T/n2
where n is the number of moles – uncompressed and compressed moles in the volume.”
R/n is a variable and hence isn’t a substitute for k, which is a constant.
According to your modified law, if there is one mole in the first cylinder and two moles in the second cylinder, the average kinetic energy per molecules is twice larger in the first container. This may be the result you wanted. But since your law only depends on molar amount and temperature, it also states that if you have two cylinders with the same pressure and temperature, but with the second cylinder 10 times as big as the first one (and hence containing 10 times the molar amount of gas), then the second cylinder has ten times less kinetic energy per molecules (or equivalently, per mole) as the first one. Yet they have the same pressure and temperature. Somehow the molecules are sensitive to how big the container is. Indeed, you could insert a thin partition in the middle of a container and, *boom!*, the average energy for all the molecules would be cut in half. That’s what your law states.
P-N stated: “According to your modified law, if there is one mole in the first cylinder and two moles in the second cylinder, the average kinetic energy per molecules is twice larger in the first [denser] container.”
Actually, I think you wanted the opposite. You wanted the same total energy and hence a lower average energy in denser container. Hence your law doesn’t work for you even in that case. (And in the case that I mentioned, with two containers of merely unequal volumes, it doesn’t work for anyone).
Oops, sorry. I mixed up the two containers. You indeed get the result that you wanted in this case. But you get a bad result in the case of two containers of unequal volumes (same pressure and temperature).
Rob Ellison: “It is really the very opposite of hydrostatics – which assumes isentropy.”
But I agree that the speed distribution is everywhere isotropic. I assumed it is Boltzmann’s distribution for an ideal gas, and this is isotropic. You can’t infer from the fact that the speed distribution is isotropic that there is no vertical pressure gradient. At equilibrium there is no net molecular flux across any boundary. Molecules cross any boundary with the same average velocity in both direction. The pressure gradient, and density everywhere, are steady.
The density is uniform – it arises from ideas of random motion of particles.
http://en.wikipedia.org/wiki/Brownian_motion#mediaviewer/File:Brownian_motion_large.gif
In principle even if there were a density gradient – molecules would move from more dense to less dense regions until uniformly distributed within the box.
And you still need to account for the fact that, under your strange assumption that the hydrostatic equilibrium condition need not apply to the gas in a closed box at equilibrium, there would be no buoyancy.
As I said, put an helium filled balloon in in a thick empty (except for air at normal atmospheric pressure) hermetic safe and close the door. Let things settle down for a few hours for temperatures to equilibrate everywhere. Will the balloon fall down through the denser air? It should fall down to the bottom of the safe if there is no pressure gradient in the surrounding air. But surely that is absurd.
As I said not strange at all – but allied with the idea of random walks.
Buoyancy as always is the weight of the displaced fluid (gas or liquid) less the weight of the box and contents.
Helium? Fall? Too nonsensical for words.
Rob Ellison: “The density is uniform – it arises from ideas of random motion of particles.”
Not under gravity. A non radiating sphere of ideal gas in space would not spread out infinitely if it doesn’t have enough internal energy for all the particles to escape the gravity well. It must conserve total KE+PE. Hence the molecules tend to cluster together with the fewer more energetic ones venturing higher away from the center of gravity. (Though the occasional molecule at the far right tail of the Maxwell distribution will reach escape velocity, so we don’t reach true equilibrium until the total internal energy is so low that not one single molecules can escape with all of it).
Likewise in a box, molecules tend to pack up a little more near the bottom because many particles on the left tail of the distribution don’t have enough total energy to climb all the way to the top of the box. This is the kinetic-theoretical counterpart explanation of the barometric formula that can be derived at the macroscopic level though considering parcels of air that weigh each other down within the atmospheric column.
“In principle even if there were a density gradient – molecules would move from more dense to less dense regions until uniformly distributed within the box.”
This would violate the theorem of equipartition. It would require that the lower energy states to be under-populated so that no molecules would spend a higher amount of time in the lower part of the box. It would also mean, absurdly, that, at the macroscopic level the condition of hydrostatic equilibrium never is satisfied. It would allow a helium balloon within the box to sink into the denser air. This is a conundrum that you still haven’t faced.
I can’t be bothered reading past not under gravity. In the atmosphere we get density differences due to weight of the atmosphere decreasing as you rise. In a box random walks rule.
Rob Ellison “I can’t be bothered reading past not under gravity.”
Typical.
“In the atmosphere we get density differences due to weight of the atmosphere decreasing as you rise. In a box random walks rule.”
Molecules who walk randomly still can’t violate the law of conservation of energy. If a molecule only has total energy m*g*z then it can’t “walk” randomly above height z. The wide M-B energy distribution ensures that there are at any given time very many molecules that can’t “walk” any higher than z, whatever z, however much they would like to. When the gas is weakly interacting, then those lazy molecules change place with the zippy one (who like to bounce against the top) through collisional KE exchanges. However, at any given time, there are a number of lazy ones that don’t have enough energy to go above z. This is why there are more molecules near the bottom even tough, on the long run, every molecule gets a chance to go to the top.
This time I read the last line. The downward increase in velocity is compensated for by the exactly equal upward rebound. There are many collisions on the way up and on the way down – it all ends up as random Brownian motion and distributed molecules in the space.
Density gradients in the atmosphere do arise from gravitational acceleration of molecules but because of decreases in the mass of atmosphere above as you rise. In the box this is not a significant consideration.
Rob Ellison: “This time I read the last line. The downward increase in velocity is compensated for by the exactly equal upward rebound.”
What kind of compensation is that? KE = mv^2. The velocity is squared. The KE always is positive. It doesn’t matter if the velocity is directed downwards before the collision or upwards after the collision. When the collision is elastic, the KE is the same both before and after the collision. No compensation occurs. There is no escaping the fact that each molecule either hits the bottom harder than it hits the top, or it doesn’t have the energy to reach the top and hence it only hits the bottom. Hence the pressure is higher on the bottom.
There are many collisions on the way up and on the way down – it all ends up as random Brownian motion and distributed molecules in the space.
Density gradients in the atmosphere do arise from gravitational acceleration of molecules but because of decreases in the mass of atmosphere above as you rise. In the box this is not a significant consideration.
Sorry, the last two paragraphs above belong to Rob Ellison.
Rob Ellisson: “Buoyancy as always is the weight of the displaced fluid (gas or liquid) less the weight of the box and contents.”
Sure. This is how we calculate it. But the object that is buoyant doesn’t merely respond to an abstract calculation for the heck of it. It moves according to Newton’s second law as a response to the sum total of all the forces applied to it. One of those forces is the pull from gravity — its own weight. You also need an uplifting *force* to oppose the weight of the object. The ambient gas supplies this force. Except it can only do so if there is a pressure gradient. Else the gas would exert no net force on the balloon. The (vertical integrated) force on the top half external shell would be the same as the force on the lower external shell.
Wow. The buoyancy by definition is the difference in the force acting up (displacement) and the force acting down (weight).
It will move if the forces are unequal. You could I suppose calculate a variable density on the way up. It rises until the forces are equal.
http://upload.wikimedia.org/wikipedia/commons/7/74/Buoyancy.svg
“Wow. The buoyancy by definition is the difference in the force acting up (displacement) and the force acting down (weight).”
A force isn’t a displacement, though you can derive it from a displacement using the relation W=F*d. So, if you know the work and displacement you can calculate the force. But in the case where there is no pressure gradient in the box, you can displace the balloon freely in any direction and there is no work done (except a tiny one against viscosity and inertia).
“It will move if the forces are unequal.”
Yes, if the weight of the helium filled balloon isn’t equal to the net external force applied to it, it accelerates. But when there is no pressure gradient, there is no net external force. That’s because the net external force just is the integrated (vector) external pressure normal to the surface of the balloon.
“You could I suppose calculate a variable density on the way up. It rises until the forces are equal.”
The force that equalizes the weight must come from somewhere. If not from unequal outside pressure, from where? Merely giving it a name (buoyancy) and providing a formula doesn’t make this force materialize if the way it is normally manifested (a pressure gradient) is suppressed.
What kind of compensation is that? KE = mv^2. The velocity is squared. The KE always is positive.
But the vector is up rather than down. Motion is random – and the gas distribute uniformly. Brownian motion. That’s all I am reading. You can keep repeating yourself endlessly – but it is all talk and not in accordance with kinetic theory. So its all just nonsense repeated endlessly because P-N must be right – it’s the warmest meme after all. Superior to the science deniers. So they must defend it to the hilt – by diversions such as this is nothing else.
“But the vector is up rather than down. Motion is random – and the gas distribute uniformly.”
Yes, the vector is up rather than down. That’s precisely this reversal of the *momentum* vector that accounts for the momentum *transfer* to the bottom of the box. The vector also is reversed, though from up to down, when particles hit the top. But in that case the average magnitude of the vector is *smaller*. And some molecules never reach the top. They only reverse the vector from down to up, and hence transfer downward momentum only to the bottom of the box. They entirely give up the downward momentum gained from gravity in between the two collisions.
Earlier, I calculated exactly the amount of net momentum transfer, gained from gravity and given up to the box, that occurs for all the molecules. It is equal to the total weight of all the molecules multiplied by the time period during which this transfer occur. This means that the pressure difference between top and bottom equals the weight of the gas. And this is why the box filled with gas is heavier than the same box without any gas in it.
The net force down transferred is mg. The mass of the gas in the box times gravity. Duh. But the molecular motions are still random and the molecules tend to disperse.
http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/weighgas.html.
And the cause is the augmented downward velocity. It bounces up again with the same velocity – molecules tend to diffuse to lower density – and you haven’t disproved Brownian motion or kinetic theory. I won’t hold my breath.
http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/weighgas.html
“And the cause is the augmented downward velocity.”
You are saying that the cause of the force on the container — mg — is “augmented downward velocity”. But you also are denying that the molecules are hitting the bottom with an average velocity, or rate of collision, that are any different than the collisions with the top of the box. Hence you are denying the very cause that you are invoking.
“It bounces up again with the same velocity”
So? It still slows down on the way up and hence will not bounce as hard on the top, if at all! And this is true for all the molecules that hit the bottom.
” – molecules tend to diffuse to lower density – and you haven’t disproved Brownian motion or kinetic theory. I won’t hold my breath.”
Molecules tend to diffuse to lower density but they *also* tend to accelerate down. When a molecule moving up has converted all its kinetic energy to potential energy, it has no tendency to rise any more. It falls back down. Many molecules bounce on the bottom of the box several times in a row in a sequence of parabolic trajectories in between collisions with other molecules. Zero molecule bounces several times in a row on the top of the container without being hit by a molecule below. What you have is a *skewed* random walk that offsets the effect from diffusivity and maintains a density and pressure gradient.
Dense or light Brownian particles also exhibit skewed random walks under gravity and buoyancy.
“The net force down transferred is mg. The mass of the gas in the box times gravity. Duh. But the molecular motions are still random and the molecules tend to disperse.
http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/weighgas.html.”
It’s interesting that you would link to this hyperphysics page, quote the conclusion, and ignore the derivation. The derivation is the exact same one that I did. And you also failed to note this result that I also had derived:
“The difference between the average force on the bottom and top of the container is just the weight, mg, of the molecule.”
In case you had failed to notice when I last mentioned it, the difference in average force on the bottom and top of the container (which you denied exists) just is the pressure difference multiplied by the surface area. This is from the very definition of pressure.
It say in the above post: “◾When are journalists going to start treating climate skeptics like flat earthers? Not soon.” Is that a typo? – Substitute “stop” for “start.”
I liked reading it both ways. You ain’t seen flat, yet.
=====================
According to your modified law, if there is one mole in the first cylinder and two moles in the second cylinder, the average kinetic energy per molecules is twice larger in the first container. This may be the result you wanted. But since your law only depends on molar amount and temperature, it also states that if you have two cylinders with the same pressure and temperature, but with the second cylinder 10 times as big as the first one (and hence containing 10 times the molar amount of gas), then the second cylinder has ten times less kinetic energy per molecules (or equivalently, per mole) as the first one. Yet they have the same pressure and temperature. Somehow the molecules are sensitive to how big the container is. Indeed, you could insert a thin partition in the middle of a container and, *boom!*, the average energy for all the molecules would be cut in half. That’s what your law states.
KEavg = f.(1/2)k.T per molecule
So the gas is not compressed at all – merely a larger mass in a larger volume. In a local thermodynamic equilibrium. The temperature is uniform over a larger volume.
P-N stated: “According to your modified law, if there is one mole in the first cylinder and two moles in the second cylinder, the average kinetic energy per molecules is twice larger in the first [denser] container.”
Actually, I think you wanted the opposite. You wanted the same total energy and hence a lower average energy in denser container. Hence your law doesn’t work for you even in that case. (And in the case that I mentioned, with two containers of merely unequal volumes, it doesn’t work for anyone).
KEavg = f.(1/2).R.T per mole
These are just the molar and molecular definitions of KEavg
Total thermal energy
Uthermal = N.f.(1/2).R.T
The total thermal energy is the same for the same temperature. Which implies that the average thermal energy per mole is less in the denser material.
Kinetic theory says that energy is transferred in collisions – where there are more molecules in a volume the collisions increase in number. In thermodynamic equilibrium – the energy of these collisions is lower.
The theory of molecular kinetics hasn’t changed. .
“Uthermal = N.f.(1/2).R.T
The total thermal energy is the same for the same temperature. Which implies that the average thermal energy per mole is less in the denser material.”
I agree with the equation (where ‘f’ is the number of degrees of freedom of the molecules — kinetic, rotational, and vibrational).
But what you say contradicts the equation. ‘N’ represents the number of moles. Hence if the second container has twice the pressure, same temperature, same volume, and hence twice the molar amount, then this equation states that it has twice the total thermal energy. Thermal energy is proportional to *both* temperature and molar amount. Pressure and volume are irrelevant (only for the case of ideal or weakly interacting gases, of course).
I agree with the equation (where ‘f’ is the number of degrees of freedom of the molecules — kinetic, rotational, and vibrational).
Gee whiz – I am happy for you.
But what you say contradicts the equation. ‘N’ represents the number of moles. Hence if the second container has twice the pressure, same temperature, same volume, and hence twice the molar amount, then this equation states that it has twice the total thermal energy. Thermal energy is proportional to *both* temperature and molar amount. Pressure and volume are irrelevant (only for the case of ideal or weakly interacting gases, of course).
No – the gas cools down to a local thermodynamic equilibrium. Or if you like – we assume that compression occurs over a long time period. So it has the same total thermal energy as the less dense gas – but the average kinetic energy per molecule is lower. Necessarily.
“No – the gas cools down to a local thermodynamic equilibrium. Or if you like – we assume that compression occurs over a long time period. So it has the same total thermal energy as the less dense gas – but the average kinetic energy per molecule is lower. Necessarily.”
I know this is what you claim, but it contradicts your formula. It is contrary to physical law.
Think of it this way. Since it’s the same gas (nitrogen — N2, say), it has the same specific heat. Volumetric heat capacity is specific heat multiplied by density. If the two containers have the same volume and temperature T1, but the second one has twice the pressure (and hence twice the density), then you need twice the energy to heat up the second container as you need to heat up the first container to the same final temperature T2. (Using a coiled resistor, say.) If you thus are spending twice the amount of energy (to power the resistor), you are likewise increasing it’s internal energy by twice the amount, thus contradicting your claim that it has the same internal energy at the same temperature T2.
Ron Ellison: “Kinetic theory says that energy is transferred in collisions”
Makes sense.
” – where there are more molecules in a volume the collisions increase in number.”
True, though kinetic theory also applies to ideal gases where there are zero collisions.
“In thermodynamic equilibrium – the energy of these collisions is lower.”
You may also want to read the last two sentences of this post.
The energy isn’t “in” the collisions. It is in the degrees of freedom (translational KE, rotational E, vibrational E). The collisions merely enable molecules to transfer energy to one another and in between degrees of freedom, constrained by conservation laws (momentum, angular momentum). Those transfers don’t create or destroy energy. (They do create temporary bound states that can store some energy. But this is neglected for the case of weakly interacting gases.)
When you put two moles of a gas in a container rather than one, at the same temperature, you double the number of molecules and hence the number of degrees of freedom available to store energy. If the temperature is the same then you have the same average energy per degree of freedom, for twice as many degrees of freedom. Hence you have twice as much thermal energy. You may or may not have twice as many collisions, but that’s irrelevant. The collisions merely transfer energy in between the degrees of freedom that are the true energy “containers”. The collisions don’t contain it — except very transiently in ephemeral bound states.
And even in cases where the collisions store a relatively high proportion of the total thermal energy — as is the case in strongly interacting, or very dense, gases — then that means that the volumetric thermal energy density is even larger. That’s because bound states also have degrees of freedom and each one of those contains (while it exists), on average, E = (1/2)kT. That’s exactly contrary to your claim about densely colliding gases having low volumetric thermal energy density.
Have you ever wondered why denser gases have a higher volumetric heat capacity; or why VHC = (specific_heat * density)? That’s because they store more thermal energy in the same volume at the same temperature. Isn’t that more intuitive an logical than the reverse claim?
“Have you ever wondered why denser gases have a higher volumetric heat capacity; or why VHC = (specific_heat * density)? That’s because they store more thermal energy in the same volume at the same temperature. Isn’t that more intuitive an logical than the reverse claim?”
To be more precise, I mean “denser” in the sense of more moles per volume unit, and hence more pressurized; not because of heavier molecules.
Ron Ellison: “Kinetic theory says that energy is transferred in collisions”
Makes sense.
” – where there are more molecules in a volume the collisions increase in number.”
True, though kinetic theory also applies to ideal gases where there are zero collisions.
Ron Ellison: “Kinetic theory says that energy is transferred in collisions”
Makes sense.
” – where there are more molecules in a volume the collisions increase in number.”
True, though kinetic theory also applies to ideal gases where there are zero collisions.
I really didn’t get beyond this – elastic collisions occur with ideal gases.
You are saying that the cause of the force on the container — mg — is “augmented downward velocity”. But you also are denying that the molecules are hitting the bottom with an average velocity, or rate of collision, that are any different than the collisions with the top of the box. Hence you are denying the very cause that you are invoking.
“It bounces up again with the same velocity”
So? It still slows down on the way up and hence will not bounce as hard on the top, if at all! And this is true for all the molecules that hit the bottom.
The hyperphysics link I provided went through the weight calc.
So the molecules spend less time at the bottom – hitting faster and leaving faster – than at the top? Buoyancy of individual molecules is certainly irrelevant – gravity is minor and the mixing is dominated by a random walk. Molecules diffuse through the space – more or less evenly. It is a consequence of kinetic theory.
Merely repeating yourself again and again doesn’t change that. Any person actually trying to follow this has I assume given up long ago.
There are a coupe of germane points.
1. The distribution of molecules in a box distribute randomly across the space.
2. A compressed gas will heat up and then cool to ambient temperature.
3. An compressed gas and an uncompressed gas in thermodynamic equilibrium have the same temperature and the same total thermal energy.
4. Energy transfer occurs in collisions. In denser materials the number of collisions increase. The average kinetic energy is the total thermal energy divided by the number of molecules.
These people insist they understand science better than science deniers – and so the it all becomes endless point scoring on always quite dubious grounds. P-N is a attack gerbil with the same arrogance that anything he imagines is better that the nonsesne form science deniers. FOMBS employs far less than half understood jargon to pose as an science authority and prove that capitalism is evil. Bad faith – the essence of unethical communication is my point.
“Merely repeating yourself again and again doesn’t change that. Any person actually trying to follow this has I assume given up long ago.”
That would be my guess also.
“The hyperphysics link I provided went through the weight calc.”
Yes, they did the exact same calculation that I did earlier and that you ignored. And they explicitly concluded, just as I did, that the average force (= momentum transfered / dt) is larger on the bottom that it is on the top by exactly m*g. This precisely what I had claimed and that you denied, and it entails the same vertical pressure gradient inside or outside of a box, according to the barometric formula. And it is a result from kinetic theory under gravity.
It entails a vector addition to the -y component of velocity due to gravity. But claiming that I denied something so obvious is quite obvious misdirection and obfuscation. The quality of bad faith that is so obvious in P-N.
It is a basic physical reality that the air in cubic metre box weighs 12.7N – and exerts a pressure of 12.7Pa. The atmospheric pressure at sea level is 100,000Pa. The density difference due to hydrostatics between o and 1m is likewise negligible.
In kinetic theory – molecules disperse due to collisions randomly. The weight of the gas due to the y component of velocity and molecular dispersion are two different things.
Rob Ellison “The density difference due to hydrostatics between o and 1m is likewise negligible.”
I had claimed that the pressure *gradient* is dictated by the barometric formula in the atmosphere *and* in the box. You had denied this. You had insisted that there is no pressure gradient in the box. Now you are claiming that it is negligible. But it is still identical inside or outside the box. It is responsible for holding up an helium filled balloon both inside or outside the box.
In order to support your claim that there is no pressure gradient inside the box, you had claimed that the buoyancy force that maintains the balloon up against gravity in the box doesn’t require a pressure gradient. Now you say it’s negligible. It’s not. It’s strong enough to hold the balloon up. It’s also the basis for my claim that there is a density gradient just large enough to account for the drop-off rate that compensates the kinetic energy gain of individual molecules moving down — such as to preserve the speed distribution as a function of height.
In sum, you had denied the pressure gradient (not stated that it is negligible) when I had needed it to support my main argument and explicitly stated the magnitude of this gradient (from the barometric formula). This *tiny* gradient just is large enough to compensate for the *tiny* increase in individual molecular KE as they free-fall in the box from one level to another. It’s the same exact effect in the box or outside of it in some parcel of stable isothermal atmosphere.
Rob Ellison: “There are a coupe of germane points.
1. The distribution of molecules in a box distribute randomly across the space.
2. A compressed gas will heat up and then cool to ambient temperature.
3. An compressed gas and an uncompressed gas in thermodynamic equilibrium have the same temperature and the same total thermal energy.
4. Energy transfer occurs in collisions. In denser materials the number of collisions increase. The average kinetic energy is the total thermal energy divided by the number of molecules.”
1. False under gravity, as your hyperphysics link demonstrates clearly. If pressure is higher on the bottom under isothermal conditions, then so is the density.
2. True.
3. False.
4. True. True. True (though only for ideal or mono-atomic gases).
Again repeating things that are obvious and have no relevance and claiming that I disagree with the hyperphysics page I linked to is a lack of honesty or integrity.
1. False under gravity, as your hyperphysics link demonstrates clearly. If pressure is higher on the bottom under isothermal conditions, then so is the density.
The pressure caused on the bottom is from velocity – and not from density
2. True.
Hardly authoritative.
3. False
An compressed gas and an uncompressed gas in thermodynamic equilibrium have the same temperature and the same total thermal energy.
‘The dependence on temperature change and mass are easily understood. Owing to the fact that the (average) kinetic energy of an atom or molecule is proportional to the absolute temperature, the internal energy of a system is proportional to the absolute temperature and the number of atoms or molecules.’ http://cnx.org/contents/031da8d3-b525-429c-80cf-6c8ed997733a@8.9:102/College_Physics
As I say – authoritative – not.
.
4. True. True. True (though only for ideal or mono-atomic gases).
Energy transfer occurs in collisions. In denser materials the number of collisions increase. The average kinetic energy is the total thermal energy divided by the number of molecules.?
It is all true for regardless of degrees of freedom. Demonstrate otherwise.
We have two FOMBS thought bubbles intended to show that warmist science is superior therefore capitalism is evil. Unbelievable as that sounds – it is literally true.
The first was a box on the moon leaking molecules. Molecular projectiles on the moon were claimed to have constant kinetic energy as they rose against moon gravity. Despite the fact that kinetic energy was lost on the way up. A novel theory presented as absolute truth and defended with great energy by P-N who promises something other than verbiage some day or other.
The second was that a density gradient developed in the box – and that more molecules hit the bottom although the temperature of the surface at the top and bottom were the same. Despite kinetic theory. Again defended by P-N with great energy and with little but verbiage.
There is little but faux expertise allied to truculence in support of pretensions to moral and intellectual superiority. This is the point demonstrated again and again.
Rob Ellison: “The pressure caused on the bottom is from velocity – and not from density”
That’s true on the level of individual molecules but since the speed distribution is uniform in the box, as is average KE, the only variation must arise from a density gradient (or the equivalent drop-off effect that I mentioned). It’s also true that this hyperphysics demonstration doesn’t settle *this* issue, but it does settle the issue about the internal pressure gradient. You had insisted that there is no gradient. So we are making progress. If you now still agree that the speed distribution only is a function of temperature, then it follows that the pressure gradient can only arise from a density gradient.
“Again repeating things that are obvious and have no relevance and claiming that I disagree with the hyperphysics page I linked to is a lack of honesty or integrity.”
That’s because I had made the exact same argument (and equivalent calculation) about the rate of transfer of momentum from the molecules to the box matching the rate of momentum gained from gravity, and the resulting effect on the *pressure* differential on top and bottom. You had dismissed my argument when I made it and argued strongly against there being a pressure differential at all. You even sought after alternate explanations for buoyancy that don’t require a pressure differential!
You now purported to agree with the argument in the link and yet failed to notice that it concludes that there is a pressure differential *in the box*. And then, most lately, you suggest that you have known this since you were taught about it in engineering class. You must have forgotten then.
Rob Ellison: “3. False
An compressed gas and an uncompressed gas in thermodynamic equilibrium have the same temperature and the same total thermal energy.”
Then why is the specific heat capacity so little dependent on pressure? (For air it’s 1.014 Kj*K/kg at 1bar and 1.021 Kj*K/kg at 10bar.)
If your theory were right then we ought to expect the specific heat capacity of gases to be inversely proportional to pressure. This contradicts the well know fact that volumetric heat capacity is proportional to both specific heat capacity and molar quantity. If you have twice the pressure, then you have twice the volumetric heat capacity.
You therefore get the paradoxical result that with a compressed container (2bar), as compared with an uncompressed container (1bar), we now need twice the amount of energy to heat it up to the same higher temperature T2 (according to standard physics) and yet produce the very same change in internal energy (according to you). It looks like half the energy is destroyed.
Rob Ellison: “An compressed gas and an uncompressed gas in thermodynamic equilibrium have the same temperature and the same total thermal energy.”
Rob Ellison quotes a college physics text: “…Owing to the fact that the (average) kinetic energy of an atom or molecule is proportional to the absolute temperature, the internal energy of a system is proportional to the absolute temperature and the number of atoms or molecules.’ ”
Can’t you see that your text supports me and contradicts you?
You had been arguing that two containers with identical volume and temperature, and different pressure (and hence different molar amounts), had the same total internal energy. I had been arguing that it only is proportional to temperature and the number of molecules. The text only is consistent with your ambiguous statement if the *same* molar amount of gas is compressed to a smaller volume (and then allowed to cool at the same initial temperature). But that was not your claimed. You had claimed equality of total thermal energy for different molar amounts (and pressures) in identical volumes.
“Energy transfer occurs in collisions. In denser materials the number of collisions increase. The average kinetic energy is the total thermal energy divided by the number of molecules.?”
That was your claim.
“It is all true for regardless of degrees of freedom. Demonstrate otherwise.”
I am denying this because not all the molecular degrees of freedom count as kinetic. There are three translational degrees of freedom that store the kinetic energy m*v^2 (where v is the velocity of the molecule). The rotational degrees of freedom of diatomic molecules (two of them) and polyatomic molecules (three of them) store angular ‘kinetic’ energy but that is independent of m*v^2. Further, the other internal degrees of freedom, vibrational or potential, store energy, and hence count towards total internal (thermal) energy, but do not count towards kinetic energy.
This is why total thermal energy divided by the number of molecules doesn’t give you the average kinetic energy except in the case of weakly interacting monatomic gases where internal energy is purely kinetic (translational) and can’t be stored in any other mode (neglecting atomic excitation and radiative transfers).
This is another bit that you missed in your hyperphysics link.
“Taking an average force like this allows you to determine average forces and average pressures on the walls of a container of gas.”
Well we did cover pressure vessels in engineering school.
The pressure on walls is related to mass, velocity and the number of molecules. Energy is imparted in collisions – which depend on the energy of the particles and the number of particles. It was this quite obvious idea based on physical principles that both P-N and FOMBS insisted black and blue wasn’t true and that heat transfer depended only on average kinetic energy.
But this is all about snark and sleight of hand – it suggests that I have missed something essential which is pure diversion and dissimulation. Proving intellectual and moral superiority is the entire point and not communication in good faith.
This is the added ethical dimension of the discourse – a pretention to moral and intellectual superiority so much a part of the personality construct and so impossible to get past.
“Well we did cover pressure vessels in engineering school.
The pressure on walls is related to mass, velocity and the number of molecules. Energy is imparted in collisions – which depend on the energy of the particles and the number of particles.”
Yes, exactly. So since the speed distribution is uniform (a function of temperature only), the variable is molecular density. Molecular density decreases with height (both outside *and* inside the box — as your hypephysics link demonstrates). This is what I claimed all along, providing the vary same argument and calculation, and that you denied all along (for the case of the closed box).
“It was this quite obvious idea…”
Which you denied are have been arguing against for days. You argued that there is no pressure or density gradient in the box and that buoyant forces on a balloon inserted in the box would be the same regardless of there being a pressure gradient or not. You had forgotten about your engineering class.
“…based on physical principles that both P-N and FOMBS insisted black and blue wasn’t true and that heat transfer depended only on average kinetic energy.”
No. We said nothing at all about heat transfer. The idea of heat transfer was you own hobbyhorse that arose from the idea of collisions with wall (and your misguided conception about temperature being related to collision frequencies). I had said early on that the whole enclosure was adiabatic. You berated me for using fancy terms and later claimed that I had assumed heat sinks. But that’s irrelevant to the temperature of the gas anyway. If the (monatomic ideal) gas has temperature T, then the average kinetic energy per molecule is (3/2)kT whatever the density and regardless of there being any wall nearby, adiabatic or not.
I haven’t read any of this and don’t intend to. This all started from a statement by FOMBS that more molecules of the same average kinetic energy hitting different walls resulted in the same energy transfer to the walls.
I suggested that by the kinetic theory of gases that more molecules with the same average kinetic would impart more energy to the surface.
Cue vitriolic denunciations of science denier and the repetition some 50 times at least now that it heat transfer depends only on the average kinetic energy of the molecules and not the number of molecules. Along with a veritable storm of diversion and dissimulation.
Without any doubt P-N’s has never passed an engineering exam or indeed have a higher degree in climate relevant environmental science. Both pretend to an expertise they don’t have to advance their meme of science denier.
Rob Ellison: “I haven’t read any of this and don’t intend to.”
We have come to a point where you are force to backpedal or admit to a mistake, so I am not surprised that you would revert to your abuse mode.
“This all started from a statement by FOMBS that more molecules of the same average kinetic energy hitting different walls resulted in the same energy transfer to the walls.”
Can you quote or link to the post where FOMD argued this?
If the walls have the same temperature, then the rate of energy transfer is zero everywhere. Anyway, I don’t remember ever seeing FOMD arguing this and it seems quite unrelated to our own main disagreements, apart from your very disputable claim that the *temperature* of a gas (not the rate of heat transfer — that’s not the same thing as temperature at all) is a function of average kinetic energy *and* collision rate.
Rob Ellison: “I suggested that by the kinetic theory of gases that more molecules with the same average kinetic would impart more energy to the surface.”
Sure, and you *also* argued that this entails that the temperature of the gas is higher. It is this *second* claim that I consistently disputed. Lately you seem to have been walking back from it since you are beginning to realize that temperature only is a function of average kinetic energy, after all.
Rob Ellison: “Cue vitriolic denunciations of science denier and the repetition some 50 times at least now that it heat transfer depends only on the average kinetic energy of the molecules and not the number of molecules.”
I can’t remember arguing this ever. But if repeated it 50 times, maybe you can provide one single quote of me saying it.
P-N says everything 50 times until the horror multiplies. The premise was that there more molecules hitting the top of the box than the bottom because the box had a density gradient. But because all the molecules had the same average kinetic energy the temperature at either end was the same.
The idea is wrong on several levels – not merely a trivial thought bubble used as a rhetorical device to demonstrate the inferiority of denier science,
Rob Ellison: “P-N says everything 50 times until the horror multiplies.”
“The premise was that there more molecules hitting the top of the box than the bottom because the box had a density gradient. But because all the molecules had the same average kinetic energy the temperature at either end was the same.
The idea is wrong on several levels – not merely a trivial thought bubble used as a rhetorical device to demonstrate the inferiority of denier science.”
I found an interesting article in the American Journal of Physics that demonstrates elegantly this “wrong idea” that FOMD proposed and that I have been running with.
http://web.ist.utl.pt/ist12219/data/43.pdf
It’s in section 2b: The Kinetic Derivation of the Barometric Formula, (immediately following the Hydrostatic Derivation.)
The author *starts* from assuming the usual Boltzmann energy distribution for molecules in the lowest level of the box and lets the gas disperse freely in the whole box until steady state is achieved. The result is a density and pressure gradient according to the barometric formula *and* constant temperature across the volume. Interestingly the initial distribution has KE_avg = (3/2)kT, while the final steady-state distribution has KE_avg = (1/2)kT. The explanation is that the gas transfers 2/3 of its average kinetic energy to the gravitational field associated degree of freedom. In any case, the end result is the “wrong idea” that you can’t wrap your mind around.
I also found a more succinct derivation at hyperphysics:
http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/kintem.html#c5
(Look for the four step derivation in the Development of the Boltzmann Derivation)
They follow the treatment from Rohlf, James William, Modern Physics from a to Z0, Wiley, 1994., though they reproduce it in a rather compressed form.
In this case the demonstration proceeds from the other way around. They start by assuming the barometric formula and the equipartition of energy. This entails constraints on the levels that the particles can reach, and the time during which they sweep different volumes as a function of velocity. And the result is the Boltzman distribution with uniform vertical temperature! So, here again we have the demonstration of the “wrong idea” that kinetic steady state ensures that (1) the hydrostatic barometric formula, and (2) a uniform vertical temperature not only are consistent with one another, but can actually be derived from each other on the assumption of thermodynamic equilibrium (equipartition of energy).
Also of interest: Google “We consider a dilute gas of N atoms in a box in the classical limit.” for a ppt presentation by Christian Binek (Uni of Nebraska) that includes another derivation of the M-B distribution of ideal gases from the barometric formula — and also conclude that the energy distribution (not the density!) is constant across the volume.
The third reference that I mentioned above is from chapter 3 “Kinetic Theory” on the following course syllabus. The ppt document is linked to the title.
http://physics.unl.edu/~cbinek/Phys912.html
Here is a fourth reference:
http://assets.cambridge.org/97805218/11194/excerpt/9780521811194_excerpt.pdf
Cambridge University Press – Statistical Mechanics: A Concise Introduction for Chemists
“Another special case of the Boltzmann distribution (1.1) is the “barometric”
distribution, giving the number density ρ(h) (number of molecules per unit
volume) of an ideal gas of uniform temperature T as a function of height h
in the field of the earth’s gravity. […]”
Yet, again, vertical pressure and density gradient, free falling particles with a non-uniform vertical distribution, and uniform temperature. And this is a standard result of classical kinetic theory.
To summarize the contrarian postion on the eventual outcome of radiative forcing , if you don’t know the science, impersonate a member of the bar. If you don’t have the evidence, impersonate a statistician, and if you don’t like the verdict of the jury , try yo sound like a judge.
I found an interesting article in the American Journal of Physics that demonstrates elegantly this “wrong idea” that FOMD proposed and that I have been running with.
http://web.ist.utl.pt/ist12219/data/43.pdf
It’s in section 2b: The Kinetic Derivation of the Barometric Formula, (immediately following the Hydrostatic Derivation.)
The barometric formula is not the issue.
The question we ‘consider a dilute gas of N atoms in a box in the classical limit’ – not the atmosphere. In the box in the classical limit the density differential is vanishingly small – and overwhelmingly dominated by temperature kinetics giving random motion in 3 dimensions and a distributed gas in accordance with kinetic theory
‘For simplicity, a linear trajectory is assumed, because the velocity change originated by the
gravitational field is supposed to be small compared to the thermal velocity v.’
I also found a more succinct derivation at hyperphysics:
http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/kintem.html#c5
(Look for the four step derivation in the Development of the Boltzmann Derivation)
They follow the treatment from Rohlf, James William, Modern Physics from a to Z0, Wiley, 1994., though they reproduce it in a rather compressed form.
Uses the barometric formula to derive the Boltzmann distribution via the average kinetic energy. The Boltzman distribution – the barometric formula and the average kinetic energy are not the point at all.
He then suggests the Hyperphysics derivation ia a ‘compressed’ given in their reference. This is egregious posturing.
‘Also of interest: Google “We consider a dilute gas of N atoms in a box in the classical limit.” for a ppt presentation by Christian Binek (Uni of Nebraska) that includes another derivation of the M-B distribution of ideal gases from the barometric formula — and also conclude that the energy distribution (not the density!) is constant across the volume.’
Of course they do.
‘There are two ways to look at density: (1) the small scale action of individual air molecules or (2) the large scale action of a large number of molecules. Starting with the small scale action, from the kinetic theory of gases, a gas is composed of a large number of molecules that are very small relative to the distance between molecules. The molecules are in constant, random motion and frequently collide with each other and with the walls of a container. Because the molecules are in motion, a gas will expand to fill the container.’ http://www.grc.nasa.gov/WWW/k-12/airplane/fluden.html
‘https://www.lhup.edu/~dsimanek/ideas/kinetic.htm’
We can drop references on kinetic theory by the bucket load.
Yet, again, vertical pressure and density gradient, free falling particles with a non-uniform vertical distribution, and uniform temperature. And this is a standard result of classical kinetic theory.
The density is uniform – the molecules are not ‘free falling’ they are moving with high velocity in all directions – with a uniform temperature. Thermal energy is transferred by collisions – with other molecules and the walls.
Climate Etc readers can verify for themselves the absence of a gravito-thermal effect by modifying the (free!) Mathematica notebook “Molecular Dynamics of Lennard-Jones Particles Using the Velocity Verlet Algorithm“ to include a gravity effect. FOMBS
They really can’t.
What part of “hamiltonian flows induce symplectic isomorphisms“ troubles your dynamical intuition, Rob Ellison?
The part where Hamiltonian flows with their differentiable forms – thus symplectic manifold – have anything to do with the dynamical mechanisms at the heart of climate.
They pose as having expertise – the most inexpert of them convinced beyond doubt of the intellectual and moral superiority of the groupthink. As I say – you can’t talk to these people – you need to talk past them to the rational middle ground.
Rob Ellison: “The barometric formula is not the issue.
The question we ‘consider a dilute gas of N atoms in a box in the classical limit’ – not the atmosphere.”
Yes. That’s exactly the case that they analyze.
“In the box in the classical limit the density differential is vanishingly small”
No. In the box, the density gradient is the exact same at it is outside of the box within an atmospheric column of any height (and homogeneous temperature and molar composition). The difference between top and bottom is proportional to height, but the density gradient, pressure gradient, local buoyant forces, etc., all are the same either within the box or outside of it.
” – and overwhelmingly dominated by temperature kinetics giving random motion in 3 dimensions and a distributed gas in accordance with kinetic theory”
All my references *prove* that on the basis of a core principle of kinetic theory — the Boltzmann distribution law for energy — it can be shown that a vertical density gradient, and hence an inhomogeneous probabilistic distribution as a function of height, must be in accordance with the Boltzmann distribution for gravitational potential energy:
rho(h) = rho(h_0)*exp(-mgh(h-h_0)/kT
So, the barometric formula just happens to be the vertical density distribution that uniquely satisfies the Boltzmann distribution law for the potential energy of the molecules. Your mistake was to assume that kinetic theory dictates random distribution of particles in space regardless of their energy within an external force field. The reason why the barometric formula applies in a box of any size simply is that there is a gravitational field in the box (however small the box may be) that is the exact same as the gravitational field outside of it.
tl;dr version of my previous post:
We *start* by assuming an axiom of kinetic theory — the Boltzmann distribution law for energy — and we derive the *conclusion* that particles must be spread vertically within the gravitational field according to the vertical density distribution (probability to occupy an E = mgh potential energy state at height h) that is equal to:
rho(h) = rho(h_0)*exp(-mg(h-h_0).
So, the barometric formula for *density* just is a special case of the Boltzmann distribution for particles moving in a gravitational field. It is dictated by kinetic theory and applies to any gas filled box in a gravitational field.
Should be “rho(h) = rho(h_0)*exp(-mgh(h-h_0)/kT)” in both posts.
The velocity effects of the gravitational field is secondary by quite a lot to the temperature kinetics. It is temperature kinetics that ensures random motion and therefore random distribution of molecules in the box.
P-N’s mistake is to insist again and again on nonsense. It is the psychological dimension that is of more interest. The unswerving belief that thought bubbles by FOMBS are realistic – without the slightest rationale – or that the symplectic manifold of the Hamiltonian whatsit in support of a meme of denier science is rational discourse. And P-N leaps to the tribal defense with immense fortitude and massive verbiage.
Little else of course. The box a proxy battleground in a war of values. In itself it matters little – infinitely trivial in the scheme of climate – other than as a metaphor for the broader struggle of the Borg collective groupthink against the forces of denier science in all its manifestations with a goal of ushering in rule by enlightened planners rather than the unruly rule by consent of the governed under a fair system of law. Thus he must pontificate ponderously from on high – again and again. Hence the battle of the box must be fought to the death – by boredom – of at least one of the participants and anyone foolish enough to ponder the meaning of this thread. It is post-normal citizen science with a grab bag of extended facts to fit any contingency in a post modernist world where the collective memes – and extended memes – are defended without compromise because the myth of superior science is central to the groupthink dynamic. This is the central ethical dimension of the climate war in the 21st century – that there is no basis for communicating with them unless you join the team. But it is a team with a very inadequate appreciation of the realities of climate – hence the dilemma. .
Rob Ellison: “The velocity effects of the gravitational field is secondary by quite a lot to the temperature kinetics. It is temperature kinetics that ensures random motion and therefore random distribution of molecules in the box.”
The vertical distribution in the box isn’t random. It is *exactly* distributed according to the pressure and density gradient of the barometric formula as dictated by the Boltzmann energy distribution for potential energy at equilibrium. *This* distribution (that you now say is a negligible effect) is what I have argued for days *isn’t* inconsistent with homogeneous temperature and hence sameness of average KE at any height. You also have been disputing exactly *this* for days. Now it seems that you finally agree, but are dismissing it as “secondary” to something else that I never disputed — that the gradient is small. Sure it’s small. So? It’s the gradient from the barometric formula. I never said that it is bigger than that. It’s exactly as big as I said it is and it is consistent with homogeneous temperature. But you though that this was a “wrong idea” inconsistent with kinetic theory and deserving of scorn and ridicule. It turns out that my precisely quantified claim is a *consequence* of kinetic theory.
No it isn’t – it is determined by collisions between molecules themselves and the walls.
There is not a chance that a hydraulic engineer can avoid hydrostatics in the atmosphere or in water. I have in fact discussed this is various ways dozens of times it seems in this thread. So did he again say I denied something so basic in engineering and physics? It is another example of bad faith in discourse from the Borg collective. It is a deliberate tactic of false claims to go along with the fake expertise.
The velocity of a mass increases downward by 9.8m/s^2. The mean velocity of molecules at 0K is 300m/s.
Any accidental concentration of molecules will tend to move to less dense reasons – this is kinetic theory from classical Newtonian physics. .
As I said – to the death in the box skirmish. .
Ah… regions…
“The velocity of a mass increases downward by 9.8m/s^2. The mean velocity of molecules at 0K is 300m/s.
Any accidental concentration of molecules will tend to move to less dense reasons – this is kinetic theory from classical Newtonian physics. .”
The vertical gradient in the concentration of the molecules isn’t accidental. It occurs as mandated by the Boltzmann law of distribution of energy as applied to the gravitational potential energy. It’s not the spatial regions that the equipartition theorem says must be evenly occupied. It’s the energy states of the systems that must be occupied according to the Boltzmann distribution law. Energy is distributed among the degrees of freedom of the molecules, both kinetic *and* gravitational. Just as much as all the different possible velocities or kinetic energies aren’t realized with the same frequencies, the different gravitational potential states aren’t uniformly realized either.
You have asked several times that I prove the veracity of the claim. You argued that it must be false and that is is statistically very improbable. Now that I’ve come up with a standard proof from kinetic theory — a proof very easy to check — you dismiss the claim as “irrelevant”.
A proof that KE is constant? Don’t think so.
A proof of the hydrostatic relationship? That was probably lecture number two in engineering school. It is not proof that kinetic motion doesn’t dominate over hugely minor gravitational effects in a box. It does.
And so it goes in the annals of the Climate etc box skirmish. C’est le guerre.
Rob Ellison: “A proof that KE is constant? Don’t think so.”
The Boltzmann law of distribution only applies to systems in thermodynamic equilibrium. Constant T is assumed from the start. If it isn’t constant, then you can’t apply the law. The system is out of equilibrium and doesn’t have maximal entropy. This is why you only see one single occurrence of “T” in the statement of the Boltzmann distribution for any degree of freedom of the system. Different degrees of freedom can’t have different temperatures unless they are insulated from one another. That’s not the case in gases. The canonical expression of the Boltzmann distribution is f(E) = exp(-E/kT). There is just one temperature for the whole system.
“A proof of the hydrostatic relationship? That was probably lecture number two in engineering school. It is not proof that kinetic motion doesn’t dominate over hugely minor gravitational effects in a box. It does.”
Many laws that apply to systems in equilibrium, or to reversible isentropic processes (such as the ideal gas law applied to adiabatic compression) can be proven *either* from macroscopic thermodynamical/mechanical consideration *or* from kinetic theory. The hydrostatic relationship can be proven from both (via the density statistical distribution). My references are to the proof from pure kinetic theory as it applies to molecular energy states.
Rob Ellison wrote: “A proof of the hydrostatic relationship? That was probably lecture number two in engineering school.”
No. It’s from Statistical Mechanics: A Concise Introduction for Chemists; Chapter 1: The Boltzmann distribution law and statistical thermodynamics; Section 1.2: The Boltzmann distribution law.
They only look at molecules and energy state distributions. The vertical density follows entirely from kinetic theory and the assumption of a gravitational field. And the proof is very simple.
http://assets.cambridge.org/97805218/11194/excerpt/9780521811194_excerpt.pdf
Rob Ellison: “A proof that KE is constant? Don’t think so.”
The Boltzmann law of distribution only applies to systems…
Although not the moon molecular projectile thought bubble proof – whatever other rubbishy diversion he is on about now.
“C’est le guerre.”
C’est la guerre. (Though is isn’t really)
C’est magnifique mais – ce n’est pas la gare.
No proof of the incompetent thought bubbles? Didn’t think so.
That should be – c’est magnifique – mais ce n’est pas la gare.
Rob Ellison wrote: “No proof of the incompetent thought bubbles? Didn’t think so.”
You repeatedly asked *me* to supply a proof for *my* formulation of the claim regarding temperature, density, gravity and KE distributuons in a box. You yourself sat the goal-post thus:
“But the original and incorrect statement form FOMBS was that more molecules were hitting the bottom(sic) than the top(sic) and yet magically top and bottom were at the same temperature.”
(You reversed “top” and “bottom”). Your challenge was for me to prove this. So, I proved the correctness of this statement that you are attributing to FOMD but that is a good enough paraphrase of my own understanding of the main claim that *you* were disputing while arguing with *me*. It’s a result from kinetic theory when molecules have some gravitational potential energy EP = mgh.
Oops sorry. It was actually elsewhere that “top” and “bottom” had been swapped: “P-N says everything 50 times until the horror multiplies. The premise was that there more molecules hitting the top of the box than the bottom because the box had a density gradient. But because all the molecules had the same average kinetic energy the temperature at either end was the same.”
Wow something I say 3 times – bringing it back to the original incompetence is what it is.
The argument was that there is a gravity induced density difference between the top and bottom. 1st error.
The process of diffusion means that molecules in the box do not concentrate – molecules away from any transient areas of concentration. Despite the very minor effect of gravity on downward velocity vector in relation to the large
The 2nd error was that heat was transferred to walls the same regardless of the number of molecules hitting the wall. Heat transfer is by collision and is dependent on the number of molecules and not just the average kinetic energy.
From that we go to days of endless obfuscation with endless irrelevant argument. The reason for this is the superior science meme of the space cadets.
Rom Ellison: “Wow something I say 3 times – bringing it back to the original incompetence is what it is.”
I was merely correcting your typo for clarity.
Rob Ellison: “The argument was that there is a gravity induced density difference between the top and bottom. 1st error.
The process of diffusion means that molecules in the box do not concentrate – molecules away from any transient areas of concentration. Despite the very minor effect of gravity on downward velocity vector in relation to the large.”
The vertical density gradient in an isothermal gas (under gravity) isn’t transient. It is a steady state, as Pekka explained in his short paper. (Pekka: “Expressed in other words, the above derivation shows how it’s possible that the Maxwell-Boltzmann distribution of the same temperature can be valid at all altitudes […] In this specific case the gravitational
acceleration, the density profile and the influence of the initial vertical velocities of the molecules combine to maintain the stationary density and temperature profiles.”)
So the density gradient doesn’t change with time. It is stationary. It also is dictated by kinetic theory. The distribution of gravitational potential energy among the particles must obey the Boltzmann distribution law. In this particular case, the Bolzmann distibution for m*g*z just happens to be equal to the barometric distribution of density with height. It explains it from the standpoint of kinetic theory. The Bolzmann distibution law also applies to all thee components of the kinetic energy separately (as a function of temperature only) and thus ensures constant kinetic energy distribution with height, since the ‘z’ variable doesn’t figure in the energy term (e = m*v_z*^2).
Rob Ellison: “The 2nd error was that heat was transferred to walls the same regardless of the number of molecules hitting the wall.”
I can’t remember ever making any such claims. I don’t see what relevance it may have to the topic of the vertical density distribution at equilibrium (no heat transfer) as dictated by kinetic theory (which includes the Boltzmann distribution law).
At times you seemed to be arguing that if the average kinetic energy of the molecules of the gas were the same at the top as it is at the bottom, then, since the density is different (as I claimed), and the rate of collisions is different, then the temperature (which you alleged is dependent on both the temperature and the rate of collisions) must also be different.
Of course I would deny the validity of such an argument, and the claim that *temperature* (not heat transfer rates) depends both on average kinetic energy and collision rates per surface area. But is this really what you were arguing?
Wow.
At times you seemed to be arguing that if the average kinetic energy of the molecules of the gas were the same at the top as it is at the bottom, then, since the density is different (as I claimed), and the rate of collisions is different, then the temperature (which you alleged is dependent on both the temperature and the rate of collisions) must also be different.
The heat transfer to the surface is dependent on the number of molecules and the energy of the molecules. This would suggest that – if the density were different – which it isn’t – that the bottom is warmer than the top.
This is standard molecular dynamics. .
The heat transfer to the surface – and therefore the temperature of the surface…
Chapter 1: The Boltzmann distribution law and statistical thermodynamics…
A proof of the Boltzmann distribution? Not even that – an introduction to some ideas.
Rob Ellison: “A proof of the Boltzmann distribution? Not even that – an introduction to some ideas.”
No. It’s *not* a proof of the Boltzmann distribution law. It rather derives the “barometric distribution […] the probability of finding any specific molecule at height h” and merely notices that it is a *special case* of the Boltzmann distribution, which states that “the probability of finding a specified molecule at h varies with h as exp(−mgh/kT ). But we recognize mgh as the energy ε (potential energy in this instance) associated with the molecule’s being in that state – i.e., at the height h in the earth’s gravity. Thus, (1.6) is clearly another special case of the Boltzmann distribution law (1.1).”
Rob Ellison: “A proof of the Boltzmann distribution? Not even that – an introduction to some ideas.”
Here is another interesting reference that I just found:
http://pirila.fi/energy/kuvat/barometric_derivation.pdf
Kinetic gas theory for gas in gravitational field
Pekka Pirilä
(Sounds familiar?)
“While the above derivation doesn’t prove that other solutions do not exist, it does prove that the isothermal solution is compatible with taking gravitation into account in the *kinetic gas theory*. Expressed in other words, the above derivation shows how it’s possible that the Maxwell-Boltzmann distribution of the same temperature can be valid at all altitudes in spite of the fact that the vertical motion of the molecules is affected by the gravitation. The result is dependent on the mathematical form of the Maxwell-Boltzmann distribution through the equation (5), whose simple form is true specifically for the
Maxwell-Boltzmann velocity distribution of the vertical velocity. In this specific case the gravitational acceleration, the density profile and the influence of the initial vertical velocities of the molecules combine to maintain the stationary density and temperature profiles. The equations (1), (2) and (3) represent the stationarity requirement that particles located in certain volume with certain velocities will at a later moment be replaced by an equal number of other particles which have the same velocities when the influence of gravity on velocity is taken into account. It’s shown that the *isothermal* atmosphere with Maxwell-Boltzmann velocity distribution and barometric vertical *density profile* satisfies this requirement.” (My emphasis)
‘Expressed in other words, the above derivation shows how it’s possible that the Maxwell-Boltzmann distribution of the same temperature can be valid at all altitudes in spite of the fact that the vertical motion of the molecules is affected by the gravitation. The result is dependent on the mathematical form of the Maxwell-Boltzmann distribution through the equation (5), whose simple form is true specifically for the Maxwell-Boltzmann velocity distribution of the vertical velocity. In this specific case the gravitational acceleration, the density profile and the influence of the initial vertical velocities of the molecules combine to
maintain the stationary density and temperature profiles.’
Ah – the Boltzmann distribution is valid everywhere in the atmosphere. To which particular thought bubble does this relate? Box on the moon or density profile in a box?
Or is it just a random quote that P-N found interesting?
Rob Ellison: “Ah – the Boltzmann distribution is valid everywhere in the atmosphere. To which particular thought bubble does this relate? Box on the moon or density profile in a box?”
The claim derives from kinetic theory, and it also has the vertical density profile as part of its results. Kinetic theory also applies to a gas in an insulated box. Both the temperature profile (isothermal in the equilibrium case) and the density profile (exponential with height) are stationary and are dictated by the Boltzmann distribution law. They don’t change over time.
Rob Ellison had earlier written: “A proof of the hydrostatic relationship? That was probably lecture number two in engineering school. It is not proof that kinetic motion doesn’t dominate over hugely minor gravitational effects in a box. It does.”
http://in-the-sky.org/physics/balls.php
This is an online simulation of the Boltzmann distribution for particles in a box within a gravity field. You can see that they don’t have any tendency to disperse in the box any more than the Boltzmann distribution for the potential gravitational energy allows. The simulated molecules thus also satisfy the barometric formula *exactly* (both for pressure and density) even while confined in a very small box. Scroll down to see the plot of the (simulated) mass distribution as a function of height. Once the molecules have reached this exponential vertical density gradient, they don’t have a tendency to disperse anymore evenly than that.
The claim derives from [Newtonian molecular dynamics] and it also has the vertical density profile as part of its results. [Newtonian molecular dynamics] also applies to a gas in an insulated box. Both the temperature profile (isothermal in the equilibrium case [in a stable atmosphere]) and the density profile [linearly varying] with height) are stationary and are dictated by [average kinetic energy and the hydrostatic relationship respectively]. The Boltzmann[statistical distribution] is applicable at any height it appears – but the energies at different levels are different. Kinetic theory says that the distribution of molecules in the box is determined by collisions and the thermal energy. In the atmosphere the region of equal density – as we rise – is the region of local thermodynamic equilibrium.
“The claim derives from [Newtonian molecular dynamics] and it also has the vertical density profile as part of its results.”
It is rather difficult to derive the claim about a vertical density profile from Newtonian dynamical considerations at the molecular level alone. That’s because Newtonian laws apply just as well to situations out of equilibrium. The claim that I mentioned was the derivation of the vertical density profile *directly* from kinetic theory. It is the claim that:
rho(z) = rho(z_0)*exp(-mgh(z-z_0)/kT)
Where rho(z) is the probability density that a random particle in the box will be found at height z (above the bottom of the box, say).
Again, this is a direct result from the Boltzmann distribution *law* applied to the gravitational potential energy states of the molecules.
This is not a box – it is a section of the atmosphere which is obviously open at the top. So rather than the hydrostatic equation – m.g.h – which is a term in the equation and relates pressure and mass and gravitational acceleration at a height – it conceives of the atmosphere as a collection of particles that rise to a height depending on the initial velocity. Which varies with temperature across a statistical distribution.
The molecules diffuse as far as they can in other words. In a closed box in the classical limit they disperse evenly as a result of thermal energy. So despite the relatively small gravitational effect – a density difference doesn’t exist and has no practical significance.
The transfer of thermal energy by collision depending on the number of molecules and their kinetic has practical implications and is an obvious physical reality. It is not only dependent on the average kinetic energy as repeatedly claimed by these guys in a misunderstanding of the difference between temperature and thermal energy.
As I keep saying – the conflation of the thought bubbles – the box on the moon and the Wikipedia animation of molecular dynamics – serves only to foster the confusion they depend on.
Rob Ellison: “This is not a box – it is a section of the atmosphere which is obviously open at the top. So rather than the hydrostatic equation – m.g.h – which is a term in the equation and relates pressure and mass and gravitational acceleration at a height –”
For several days we have been discussing a box with a top. You insisted that the box was assumed to be small (which I think is irrelevant — kinetic theory applies to gases in big boxes). We discussed collisions on the top and bottom. Now you seem to be arguing that I am not entitled to make use of the energy term m*g*h because it is a term that relates to hydrostatic equilibrium and the box must be open at the top in order for it to apply.
But we are not discussing hydrostatic equilibrium (which is a marcoscopic concept). The molecules are constantly being acted upon by the force of gravity. They don’t care if the box is open at the top or not; gravity still applies. When they climb some distance dz within the box, they gain exactly m*g*dz in potential energy and lose exactly m*g*dz in kinetic energy. This is true regardless of the box being open or closed at the top. Hence this gravitational potential energy term, m*g*z, is constrained by the Boltzmann distribution law.
You are arguing that is is small. Indeed it is. It has the same value in the box as it has outside of it (since, as we’ve seen, it so happens to match the usual barometric density gradient). The vertical density gradient and the loss of kinetic energy from individual molecules as they move up both are small. But they jointly account for the density drop-off above z that compensates for the reduced KE of individual molecules that move from the bottom at least up to level z, such as to maintain the velocity distribution constant with height (as determined by the Boltzmann distribution for the kinetic energy term m*(v_z)^2.)
There are 2 thought bubbles from FOMBS – and we will discount the symplectic manifold of the differentiable Hamiltonian. The latter is a clear example of the space cadet meme of superior science in he service of the personality construct of being morally and intellectually superior to deniers. It’s not true – we can all appreciate a good Hamiltonian.
The first thought bubble was that a box of air molecules on the moon with molecules escaping in a process of effusion.
The thought bubble was that the kinetic energy for all the molecules escaping was constant at all heights – because less energetic particles reached a height and then fell back leaving only more energetic particles such that the kinetic energy was constant as the molecules rise.
There is no doubt that molecules exchange kinetic with potential energy – and the box with the balls is a reasonable simulcrum of the box on the moon scenario. What it doesn’t demonstrate is that kinetic energy is constant at all levels in the box.
Total energy = mgh + 1/2mV^2
Different balls with different initial velocities have different trajectories is basic Newtonian dynamics. What is not proved is the thought bubble that kinetic energy is constant over many balls as a result of the differing Boltzmann velocities. The rest is just frenzied obfuscation.
Dispersion in an enclosed space is very different. Molecules spontaneously move from areas of high concentration to areas of low concentration – gravity is less important by a long way than the thermal energy of the molecules in a box.
‘The collisions are, however, an essential part of kinetic gas theory as they are the mechanisms that maintains the local thermal equilibrium and the related isotropic Maxwell-Boltzmann velocity distribution. In kinetic gas theory the collisions are assumed to be momentary. Thus the particles are always in free motion, but the velocity can change discontinuously in collisions. If the velocity distribution differs from the equilibrium distribution, which is the Maxwell-Boltzmann distribution, then the collisions will bring the distribution towards the equilibrium distribution, but if the distribution is already the equilibrium distribution, then it remains unchanged.’
then it remains unchanged.’ Pekka Pirilla
It is the condition of local thermodynamic equilibrium that prevails in a stable atmosphere and certainly in the box. The velocities vary from zero to some very high velocity with a median velocity related to the temperature of the gas.
Double the mass – double the volume of the box within the classical limits – and the laws of both thermodynamics and Newtonain molecular dynamics remain the same.
Regardless – heat transfer is the result of kinetic energy of particles and the number of particles hitting the surface and not merely the average kinetic energy. A molecular dynamic as energy is transferred in collisions.
But there are 2 thought bubbles – the box on the moon intended to show bizarrely that kinetic energy is constant with height – something unproven and inconsequential – and the box on the Earth with mutually incompatible equilibrium states.
The fake claims to (symplectic and other) expertise allied with dishonest rhetorical devices in the service of the meme of science superior to deniers is obvious to everyone – eh FOMBS?
Coding one’s own thermodynamical simulations has a stupendous advantage: all the assertions of standard thermodynamics can be verified concretely.
This is a wonderful aid to integrated thermodynamical understanding!
Whereas unwillingness and/or incapacity to code simulations can lead to wasted years of futile misconception … like the too-widespread illusion of the gravity-thermal effect.
Conclusion Learn to code, test your dynamical understanding, and cure your thermodynamical illusions!
There is no royal road to thermodynamical understanding, Rob Ellison!
Bouncing balls are illustrative of Newtonian particle physics. In this case they assume a Maxwell-Boltzmann velocity distribution amongst the balls.
I think after 30 years of engineering and environmental science – complex numerical modeling of flow fields – creation and analysis – coding for high order numerical solutions for differential equation – I can cope with
Total energy = kinetic energy + potential
This is the usual rhetorical device from FOMBS which is based on empty and misleading claims whose sole aim is to show the inferiority of ‘denier’ science. It is not rational discourse as we understand it – but a core diagnostic of the psychopathology of the groupthink dynamic.
Hard-sphere simulations don’t assume a Maxwell-Boltzmann distribution: rather a Maxwell-Boltzmann distribution evolves dynamically from *ANY starting configuration.
There’s your problem right there FOMBS – almost any thermodynamic nonsense will do to confuse the ignorant punters and create the impression of inferior ‘denier’ science.
The bouncing balls I am afraid have a statistical Boltzmann distribution with a peak velocity dependent on the temperature.
Rob Ellison: “What it doesn’t demonstrate is that kinetic energy is constant at all levels in the box.”
This is a rather trivial consequence of the assumption of equilibrium. At equilibrium, there is no temperature gradient. Else, there would be a heat flux and entropy would not be maximized. In that case, the Boltzmann distribution law would not apply. In any region of the box the average kinetic energy of the molecules must be the same since it is determined by the Boltzmann distributions for the energy terms (1/2)*m*(v_x)^2, (1/2)*m*(v_y)^2, (1/2)*m*(v_z)^2. The height of the molecule, z, doesn’t figure in those energy terms. Hence kinetic energy distribution is invariant with height.
‘At times you seemed to be arguing that if the average kinetic energy of the molecules of the gas were the same at the top as it is at the bottom, then, since the density is different (as I claimed), and the rate of collisions is different, then the temperature (which you alleged is dependent on both the temperature and the rate of collisions) must also be different.’
Density difference with uniform average kinetic energy was the original premise – leading to more molecules hitting the top yet the energy transfer being equal top and bottom. Both ideas are wrong.
Rob Ellison: “Density difference with uniform average kinetic energy was the original premise – leading to more molecules hitting the top yet the energy transfer being equal top and bottom. Both ideas are wrong.”
The first idea is vindicated by kinetic theory. The second idea is vague and ambiguous. I have no idea what you mean since I’ve *never* said anything about energy transfer except for the frequent insistence that since we are assuming equilibrium, we also are assuming insulating walls. There is *no* “energy transfer”. You seem to be gesturing towards a vague argument and yet are reluctant to make it precise. Was my above paraphrase reasonable, and if it isn’t, what is it that you are saying about “energy transfer” that is of any relevance to my claims about distributions (density and KE) in the box?
Here is my paraphrase again: “At times you seemed to be arguing that if the average kinetic energy of the molecules of the gas were the same at the top as it is at the bottom, then, since the density is different (as I claimed), and the rate of collisions is different, then the temperature (which you alleged is dependent on both the temperature and the rate of collisions) must also be different.”
Do you think such an argument is valid?
The first idea is vindicated by kinetic theory. The second idea is vague and ambiguous. I have no idea what you mean since I’ve *never* said anything about energy transfer…
The first idea is based on the physical reality of decreasing pressure with height. Density in the region of local thermodynamic equilibrium is constant as it must.
Energy transfer was the original FOMBS premise – change it as you might wish. However – net energy transfer is not strictly required. The idea was that the top and the bottom were at the same temperature – despite being hit by more or less molecules with the same average kinetic energy.
Both ideas – a density difference in the box and hat transfer being solely dependent on average kinetic energy are clearly wrong. The first because dispersal is very much dominated by thermal energy rather than gravity. The second as a result of molecular dynamics. More molecules with the same kinetic energy transfers more heat.
Rob Ellison wrote: “The first idea is based on the physical reality of decreasing pressure with height. Density in the region of local thermodynamic equilibrium is constant as it must.”
No. It isn’t based on the reality of decreasing pressure. It is a direct consequence of kinetic theory applied at the level of energy distributions among molecular degrees of freedom (one among which is the freedom to move higher in the gravitational field). It’s a direct consequence of the Boltzmann distribution of the gravitational potential energy states of the molecules. The concept of pressure doesn’t figure at all in the argument. The barometric law is a macroscopic consequence of this result from kinetic theory as applied to the gas in thermodynamical equilibrium in a closed box. One can *also* derive the density gradient from the barometric formula and the law of ideal gases, but that’s simply a different derivation of the same result.
“Energy transfer was the original FOMBS premise – change it as you might wish. However – net energy transfer is not strictly required. The idea was that the top and the bottom were at the same temperature – despite being hit by more or less molecules with the same average kinetic energy.”
[…] The second as a result of molecular dynamics. More molecules with the same kinetic energy transfers more heat.
They may transfer more heat when there is a temperature difference between the wall and the gas. So what? That doesn’t mean the temperature of the gas is higher. That just means the volumetric heat capacity of the gas is higher, when it is more densely compressed. It has more heat to give up (more molecules) packed in the same volume. More heat is transferred simply because more molecules transfer the same average amount of kinetic energy. I earlier provided data for the heat capacity of air at various pressures that demonstrates this.
Simpler.
1. KE is not *proven* to be constant with height – a FOMBS corollary that is utterly ludicrous.
2. Heat transfer is the result of collisions – the number of molecules and the energy of the molecules.
3. Density is a function of pressure – which is a function of mass above a level and gravity.
A surface that receives more energy will be warmer – even at equilibrium. So the bottom will be warmer and the top cooler under this scenario – materials being the same – with a density difference and the same average kinetic energy.
This is a scenario that is utterly unrealistic. The gases in equilibrium with the different surfaces cannot have the same average kinetic energy just for a start.
Yet P-N has managed to defend it immense length. We have the example of a thought bubble that becomes an inflexible truth simply by being announced. It goes to the core of the groupthink dynamics.
Rob Ellison: Simpler.
1. KE is not *proven* to be constant with height – a FOMBS corollary that is utterly ludicrous.”
Now I understand your argument. You are saying that average KE can’t be independent of height in the case where temperature is uniform yet density varies. That’s because you believe temperature not to be a unique function of average KE, but to depend strongly on molar volumetric density.
I think average molecular KE has been fairly solidly experimentally demonstrated to be very closely equal to (3/2)kT for most gases at normal pressures, as predicted by kinetic theory. This means that average KE is dependen on T and only on T. Pressure, or volumetric molar density, is irrelevant to *average* molecular KE. It follows that, if the gas has uniform temperature, then it also has uniform average molecular KE, even if the density varies with height. Hence there is no trouble for FOMD’s corollary.
Rob Ellison: “A surface that receives more energy will be warmer – even at equilibrium.”
I can’t make any sense of this. If we are at equilibrium, then the surface can’t be “receiving more energy”. If it warms (temperature increases) as a result of receiving energy, then it isn’t at equilibrium. After it has finished warming, and equilibrium is achieved, then it doesn’t receive energy anymore.
“So the bottom will be warmer and the top cooler under this scenario – materials being the same – with a density difference and the same average kinetic energy.”
I am disputing that you can have the same average kinetic energy and not, as a consequence, have the same gas temperature. The temperature of a gas is determined entirely by the speed distribution of its molecules. Same EK_avg same T.
This is a rather trivial consequence of the assumption of equilibrium. At equilibrium, there is no temperature gradient. Else, there would be a heat flux and entropy would not be maximized. In that case, the Boltzmann distribution law would not apply.’
We talking box on the moon molecular projectiles? You have to let me know.
The claim of constant KE with height arose with moon projectiles. Thus there is nothing but –
Total energy = kinetic + gravitational potential
The moon molecules leave the box with a Maxwell-Boltzmann velocity distribution – because that’s how fast they travel in the box from thermal energy.
In any region of the box the average kinetic energy of the molecules must be the same since it is determined by the Boltzmann distributions for the energy terms (1/2)*m*(v_x)^2, (1/2)*m*(v_y)^2, (1/2)*m*(v_z)^2. The height of the molecule, z, doesn’t figure in those energy terms. Hence kinetic energy distribution is invariant with height.
Yeah right. The kinetic energy of a molecule in the box on the moon is 1.2m.V^2 – if it doesn’t leave the box the potential energy doesn’t change. If it leaves the box however then
Total energy = m.g.z + 1/2.m.V^2 = constant for any molecule
Nothing but verbiage and misdirection in the service of the groupthink meme of superior science – which must not be allowed to be compromised under any circumstance.
Rob Ellison: “We talking box on the moon molecular projectiles? You have to let me know.”
I made it clear several days ago that I am focusing on the box case, and that I would address the FOMD thought experiment in the next open thread. That’s because it confuses the issue to constantly jump back in forth between (1) the box case (at equilibrium), (2) the atmospheric column case (with extraneous out of equilibrium processes), and (3) an idealized thought experiment that aims to isolate one single intuitively puzzling feature of the z velocity distribution.
Since we are disagreeing about the result for the simplest box case, can’t we focus on that? I am assuming that the box has insulated elastic walls. It can be one millimeter or one kilometer on a side, it doesn’t matter. I also assume constant g. I am claiming that, at equilibrium, there is a vertical density gradient and both a uniform molecular speed distribution and (hence) uniform average molecular kinetic energy. Hence there also is a steady pressure gradient. You seem to think this is either incompatible with TE = PE+KE on the level of molecules, or it is unproven. I also recently linked to a 2D simulation that demonstrates the density gradient.
I don’t really think anything is clear – either deliberately or purposefully.
We consider a dilute gas in the classical limit – in which the size of the box is a function of the distance between molecules is a function of the de Broglie wavelength.
https://watertechbyrie.files.wordpress.com/2014/06/classical-limit.png
The 2-D animation is a section of atmosphere where none of the balls (molecules) achieve escape velocity.
More like the moon molecular projectiles than the box in local thermodynamic equilibrium on Earth.
In the box in local thermodynamic equilibrium at the classical limit – thermal energies dominate.
Rob Ellison “The 2-D animation is a section of atmosphere where none of the balls (molecules) achieve escape velocity.”
Not true and irrelevant. Not true since they have Maxwell speed distributions. Those are unbounded speed distributions. It’s true that they can’t escape the box because they bounce back on the top. But this hardly prevents thermodynamic equilibrium to be achieved in the box. Kinetic theory also applies to molecules in a box. And it predicts a vertical density profile in accordance with Boltzmann’s distribution over the m*g*z energy states available for them to occupy in the box. The simulation illustrates the density gradient in accordance with Boltzmann’s distribution law, and hence at equilibrium. It actually doesn’t have this density gradient initially (when you reset the simulation with the reset button) and then moves towards it quite rapidly as the energy get evenly distributed among all the available degrees of freedom, both translational and gravitational potential.
“In the box in local thermodynamic equilibrium at the classical limit – thermal energies dominate.”
You have to quantify this hand wavy “dominate” claim. It doesn’t falsify the Boltzmann distribution law. Thermal energy, in this case, is
Ui = mi*(vi_x)^2+mi*(vi_y)^2+mi*(vi_z)^2+mi*g*zi, summed over all molecules.
Since the potential energy of the molecules constitutes *part* of the thermal energy of the gas in the box, it isn’t dominated by it. It is as big as it is and not any more (to paraphrase Russell).
Not true and irrelevant. Not true since they have Maxwell speed distributions. Those are unbounded speed distributions. It’s true that they can’t escape the box because they bounce back on the top. But this hardly prevents thermodynamic equilibrium to be achieved in the box. Kinetic theory also applies to molecules in a box. And it predicts a vertical density profile in accordance with Boltzmann’s distribution over the m*g*z energy states available for them to occupy in the box. The simulation illustrates the density gradient in accordance with Boltzmann’s distribution law, and hence at equilibrium. It actually doesn’t have this density gradient initially (when you reset the simulation with the reset button) and then moves towards it quite rapidly as the energy get evenly distributed among all the available degrees of freedom, both translational and gravitational potential.
Obviously none of the bouncing balls achieve escape velocity in this animation. But at any rate – kinetic theory is quite different to Newtonian molecular dynamics.
This is not the box in the classical limit – nor does it demonstrate that kinetic energy is constant with height. The two points of error.
“In the box in local thermodynamic equilibrium at the classical limit – thermal energies dominate.”
You have to quantify this hand wavy “dominate” claim. It doesn’t falsify the Boltzmann distribution law. Thermal energy, in this case, is
Ui = mi*(vi_x)^2+mi*(vi_y)^2+mi*(vi_z)^2+mi*g*zi, summed over all molecules.
The mean velocity due to temperature at OK is 300m/s – the increase in downward velocity due to gravity is in Earth’s gravity is 9.81m/s per second. So very small. I have mentioned this several times.
But the collisions are elastic – so the velocity up is initially equal to the peak downward velocity. Without collisions it will gain exactly that amount of potential energy it lost. But this is hardly relevant as it is moving randomly at much higher velocities. Where there are accidental concentrations of molecules – the number of collisions is greater and this causes molecules to disperse.
Thermal energy is related to the average kinetic energy of the particles. In the extreme of bouncing balls – kinetic energy is converted to potential. In the real world a local thermal equilibrium is maintained with molecules moving in all directions and the parcel is stable unless heated or cooled. The parcel as a whole is acted on by buoyant forces and has a bulk kinetic energy – but is hardly relevant to the kinetic energy of the molecule.
‘Since the potential energy of the molecules constitutes *part* of the thermal energy of the gas in the box, it isn’t dominated by it. It is as big as it is and not any more (to paraphrase Russell).’
The gravitational potential energy of molecules in the box doesn’t change – KE may increase on the way down by a small delta x but increases on the way up by an equal amount. As a whole the potential energy of the mass in the box doesn’t change – obviously. Potential energy on a molecular level is intramolecular forces – assumed not to play a role by the classical assumptions – the chemical and atomic bonds and the rest energy of the mass – E = MC^2.
The thermal energy is all thermal kinetics.
In the real world there are many other processes – convection, turbulent entrainment, phase transitions, non-equilibrium dynamics, changes in optical properties, etc.
There are simple points for both the box on Earth and the box on the moon respectively.
1. That energy transfer is dependent purely on average kinetic energy is nonsense.
2.. That KE for all the bouncing balls is constant with height is based on pure waffle.
All the rest is deliberate diversion based on irrelevancies, massively uninformed thought bubbles and links to toys and elementary texts he seems to barely understand.
.
OK, one thing at a time.
Rob Ellison wrote: “The gravitational potential energy of molecules in the box doesn’t change – KE may increase on the way down by a small delta x but increases on the way up by an equal amount.”
Are you pulling my leg again? The gravitational potential energy for the sum *total* of all the molecules does not change, sure, because of the steady vertical density gradient. And of course the kinetic energy progressively gained on the way down is lost again after a molecule has bounced back up to the same initial level. It’s likewise true that the potential energy progressively lost on the way down is recovered once it has bounced back at the same level. This hardly entails that the potential energy doesn’t vary as a function of height on the way up and down, in the same proportion as KE, so as to conserve TE=PE+KE at all times for any individual molecules.
The point is, for each individual molecule, its potential energy varies as a function of height. That is why its energy must be set as m*g*z in the potential energy term of the Boltzmann distribution. This tells you how molecules *distribute* among the different available potential energy states. And the energy of those states is a function of height. This yields directly the exponential density profile (from the Boltzmann distribution law). The shape of the energy distribution within *this* (gravitational potential) degree of freedom. It is independent of the energy distributions among the three (or more) other degrees of freedom of the molecules, and hence isn’t “dominated” by them.
Another point:
Rob Ellison: “The mean velocity due to temperature at OK is 300m/s – the increase in downward velocity due to gravity is in Earth’s gravity is 9.81m/s per second. So very small. I have mentioned this several times.”
As I said, the density gradient is small, but it isn’t any smaller than it is. I claim that, according to the Boltzmann distribution law, the vertical temperature profile is:
rho(h) = rho(h_0)*exp(-mgh(h-h_0)/kT)
Notice that ‘g’ figures in the exponential. ‘g’ indeed stands for 9.81m/s.
If follows that the density gradient in the box is consistent with the pressure gradient in air from the barometric formula (also measured to apply in enclosed spaces). This is neat. It means that kinetic theory is consistent with real laboratory observations in real boxes filled with air. It’s a good theory.
I await to see experimental confirmation of your own theory that within a sealed rigid box measuring 1m*1m*1m, filled with 1kg of air at uniform temperature, the internal pressure difference between the top and bottom is anything significantly less than 9.81N/m^2. (Your theory predicts zero difference.)
I give up when the thread gets too long.
Rob Ellison wrote: “The gravitational potential energy of molecules in the box doesn’t change – KE may increase on the way down by a small delta x but increases on the way up by an equal amount.”
Are you pulling my leg again? The gravitational potential energy for the sum *total* of all the molecules does not change, sure, because of the steady vertical density gradient. And of course the kinetic energy progressively gained on the way down is lost again after a molecule has bounced back up to the same initial level. It’s likewise true that the potential energy progressively lost on the way down is recovered once it has bounced back at the same level. This hardly entails that the potential energy doesn’t vary as a function of height on the way up and down, in the same proportion as KE, so as to conserve TE=PE+KE at all times for any individual molecules.
There is no change – obviously – in bulk potential energy. The potential energy for a specific molecule has no net change and plays part in the thermal energy of the molecule. .
The point is, for each individual molecule, its potential energy varies as a function of height. That is why its energy must be set as m*g*z in the potential energy term of the Boltzmann distribution. This tells you how molecules *distribute* among the different available potential energy states. And the energy of those states is a function of height. This yields directly the exponential density profile (from the Boltzmann distribution law).
The Boltzmann has a distribution – and the peak changes with temperature. There is no gravity component in it.
The density profile obtained form speeding molecules with a Boltzmann distribution relies on bouncing molecules losing kinetic energy to gravitational potential – in which m.g.h figures.
In this model of the bouncing balls – we have decreasing total kinetic energy with height. A gravito-thermal effect?
In reality the processes are radiative, convective and turbulent and involve phase changes and mass and gravitational acceleration. So even if there were a gravito-thermal effect it is dominated by radiative and convective effects as Maxwell said long ago. P-N has problems with reality however.
The shape of the energy distribution within *this* (gravitational potential) degree of freedom. It is independent of the energy distributions among the three (or more) other degrees of freedom of the molecules, and hence isn’t “dominated” by them.
Gravitational potential energy isn’t generally discussed as a molecular degree of freedom along with translation, rotational and vibrational modes. I find the usage pretentious and impenetrable. In a box velocity changes due to gravity are a very small part of translational kinetic energies.
“In this model of the bouncing balls – we have decreasing total kinetic energy with height. A gravito-thermal effect?”
No. The distribution doesn’t change with height. Though many individual molecules, in between collisions, bounce both on the top and bottom, and always bounce faster on the bottom than on the top, you will notice that many of the slower molecules only contribute to the speed distribution of the lower levels. There is a density gradient but no KE_avg gradient.
Molecular kinetic energy decreases to zero in this model – therefore temperature decreases to absolute zero. Duh.
Rob Ellison: “Molecular kinetic energy decreases to zero in this model – therefore temperature decreases to absolute zero. Duh.
No. Average molecular kinetic energy is (3/2)kT, on this model. It is constant across the whole volume. The molecular volumetric density decreases exponentially with height, but the temperature, and molecular speed distribution, are constant.
Rob Ellison: “The mean velocity due to temperature at OK is 300m/s – the increase in downward velocity due to gravity is in Earth’s gravity is 9.81m/s per second. So very small. I have mentioned this several times.”
As I said, the density gradient is small, but it isn’t any smaller than it is. I claim that, according to the Boltzmann distribution law, the vertical temperature profile is:
rho(h) = rho(h_0)*exp(-mgh(h-h_0)/kT)
Notice that ‘g’ figures in the exponential. ‘g’ indeed stands for 9.81m/s.
You mean the density rather than temperature. Density relates to pressure.
dP = m.g.dz
If follows that the density gradient in the box is consistent with the pressure gradient in air from the barometric formula (also measured to apply in enclosed spaces). This is neat. It means that kinetic theory is consistent with real laboratory observations in real boxes filled with air. It’s a good theory.
Oh – please. By all means reference the source.
I await to see experimental confirmation of your own theory that within a sealed rigid box measuring 1m*1m*1m, filled with 1kg of air at uniform temperature, the internal pressure difference between the top and bottom is anything significantly less than 9.81N/m^2. (Your theory predicts zero difference.)
It doesn’t – the additional downward component due to gravity creates the downward pressure equal to mg.
http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/weighgas.html
I have explained this before – and linked to hyperphysics. It makes absolutely no difference to molecular dispersion in a box – which is all kinetic and determined by molecular velocities far higher than the small delta induced by gravity – and which goes equally up in the rebound.
“You mean the density rather than temperature. Density relates to pressure.”
Yes, sorry.
“It doesn’t – the additional downward component due to gravity creates the downward pressure equal to mg.”
Pressure doesn’t have the same units as a force. mg is a force.
You must mean that the air content exerts a net force on the box, from within. But you are denying that the internal pressure is any higher on the bottom than it is on the top. So how does the air content manage to exert the required force (or net pressure*area) on the box?
No – the molecules exert a force by collisions on the wall. It is a physical impact which is what distinguishes kinetic theory from the earlier theories of gases pushing the walls.
We see to have regressed 100 years.
A force on an area is a pressure of course.
“No – the molecules exert a force by collisions on the wall. It is a physical impact which is what distinguishes kinetic theory from the earlier theories of gases pushing the walls.”
You are arguing that the molecules don’t exert the same average force on the top than they do on the bottom, yet you are asserting that the gas exerts the same pressure on both the top and bottom (same area).
But the force exerted by the internal pressure on the bottom = P*A,
And the force exerted by the internal pressure on the top = -P*A.
The net force on the box is zero.
I said nothing about pressure being the result of the gas “pushing the walls” in some mysterious way. I always assumed momentum transfer. Still, it is usually understood that when you calculate the force on a flat surface that results from the presence of a pressurized fluid (gas, liquid or plasma) the net force is always exactly P*A. There isn’t an additional force on top of that which is the result of impacts from molecules. Rather, the microscopic impacts *accounts* for the macroscopic pressure.
In other words, the force exerted on a solid surface that results from molecular collisions just is the rate of momentum transfer as the molecules bounce back. This is Newton’s second law F = m*a = m*dv/dt = dp*dt. The average pressure just is this average force, or rate of momentum transfer, from the molecules, per surface area unit:
P = (dp*dt)/m^2.
So, if you have the same pressure on both the bottom and the top of the box, you don’t have any net rate of momentum transfer to the box, and the molecules therefore don’t exert any net force on the box from within.
http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/weighgas.html
http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/weighgas.html
“Taking an average force like this allows you to determine average forces and average pressures on the walls of a container of gas.”
As I said, this difference in average force delta_F on the top and bottom — mg — translates directly (in the simple case of a cubic box) as a difference of pressure delta_P = mg/A.
Rob Ellison’s design for a perpetual motion machine of the second kind
1) We build a hermetic container 10m high with a 1m*1m square base. It contains about 10kg of air at (approx) 1atm pressure.
2) We open two small holes at the top and bottom of the container and let the internal pressure equalize with the external ambient pressure. This external pressure obeys the barometric formula and hence introduce an (approx) 0.1kPa pressure differential between the top and bottom of the container.
3) We then close the two holes in order to isolate the box content from the ambient atmospheric pressure gradient and we let the pressure rapidly equalize within the container in accordance with the Ellisonian kinetic theory of confined gases.
4) Once the vertical pressure gradient within the box has been abolished by random diffusion of air molecules, it will not match the outside pressure anymore along all its length, since there still is a 0.1kPa pressure differential outside. So, we now open the two holes and let the pressure differential equalize again. We use the air flow to perform some work.
5) We go back to step #3.
You didn’t get the point of it at all? The force is the result of the increased downward velocity of a molecule in a gravity field. It results in both pressure on the surface and heat transfer.
Rob Ellison: “You didn’t get the point of it at all? The force is the result of the increased downward velocity of a molecule in a gravity field. It results in both pressure on the surface and heat transfer.”
You insisted that the pressure is the same on the top and bottom. How can the gas produce a net force on the box that is a result of “pressure” when pressure and density are equalized on all surfaces (as you allege), exactly as they would be if the gas was weightless (e.g. absent any gravity field — or in the referential of a free falling box).
The force on a square metre of a metre high column is what? 12Pa? Compared to atmospheric pressure of 100,000Pa.
The effect in a box of air is trivial and is ignored in practical calculations is probably what I said.
The theoretical box is a dilute gas in the classical limits – which I have explained elsewhere.
It is not 10m high.
Rob Ellison: “The force on a square metre of a metre high column is what? 12Pa? Compared to atmospheric pressure of 100,000Pa.”
The dynamic of the particles is the box isn’t affected by the pressure, snow flakes or white rabbits outside of the box. Every time I make a precisely quantified claim about the pressure of density gradient inside of the box, you deny that there is any such gradient, only to very later admit it reluctantly but claim that it is “negligible”. 12Pa is the pressure difference between air at the bottom of the box and the top. Over the surface of the bottom of the box, this pressure differential is equivalent to the weight of the air, and this is equivalent to the weight of a mass of 1kg. It is utterly irrelevant that the air on top of the box, all the way to the TOA, is much heavier, or that there is a molten sea of lava 100km below. The pressure gradient still accounts exactly for the density gradient in the box being *exactly* as dictated by the Boltzmann distribution over the potential gravitational energy states of the molecules.
“The effect in a box of air is trivial and is ignored in practical calculations is probably what I said.”
This is not a practical calculation. It is an exact calculation, that you are contesting the exact results of — which is the exact offset (from molecular density reduction) also discussed in Pekka’s paper that accounts for the consistency of constant average KE *distribution* in spite of the KE loss of individual rising molecules.
“The theoretical box is a dilute gas in the classical limits – which I have explained elsewhere.
It is not 10m high.”
The classical limit applies even better with a box that is 10m high, since quantum phenomena are even less relevant. And gases have no trouble expanding over a distance of 10m. If you want to make it 1mm*1mm*1mm, my demonstration works just the same.
‘Turning to the larger scale, the density is a state variable of a gas and the change in density during a process is governed by the laws of thermodynamics. Actual molecules of a gas are incredibly small. In one cubic meter the number of molecules is about ten to the 23rd power. (That’s 1 followed by 23 zero’s !!!) For a static gas, the molecules are in a completely random motion. Because there are so many molecules, and the motion of each molecule is random, the value of the density is the same throughout the container.’ http://www.grc.nasa.gov/WWW/BGH/fluden.html
Too long and too whiny – really can’t be bothered. I got as far a precisely quantified and lost the will to continue.
1. KE is not *proven* to be constant with height – a FOMBS corollary that is utterly ludicrous.”
Now I understand your argument. You are saying that average KE can’t be independent of height in the case where temperature is uniform yet density varies. That’s because you believe temperature not to be a unique function of average KE, but to depend strongly on molar volumetric density.
No – no – and no.
I am saying that the claim to KE being constant with height is hugely unlikely and supported by nothing by bafflegab.
Total thermal energy is a function of the number of molecules and the average kinetic energy. Temperature is determined by heat gained and heat lost – heat may be gained or lost by radiation, evaporation or conduction. P-N has a one dimensional understanding.
I think average molecular KE has been fairly solidly experimentally demonstrated to be very closely equal to (3/2)kT for most gases at normal pressures, as predicted by kinetic theory. This means that average KE is dependen on T and only on T. Pressure, or volumetric molar density, is irrelevant to *average* molecular KE. It follows that, if the gas has uniform temperature, then it also has uniform average molecular KE, even if the density varies with height. Hence there is no trouble for FOMD’s corollary.
It depends of course on the degrees of freedom of the gas.
KEavg = f.(3/2)kT – where f is degrees of freed
Total thermal energy = N.f(3/2)kT
So a denser gas at the same temperature has the same average kinetic temperature but a higher heat capacity. Heat transfer remains a function of the number of molecules hitting a surface and the average kinetic energy.
So two surfaces with different numbers of molecules hitting them at the same average kinetic energy will have different temperatures.
Consider 2 cylinders – one at one 1 atm and one at 2 atm at thermal equilibrium with the surroundings. If the gases had the same average kinetic energy – then the walls of the cylinder with the denser gas would be warmer. But they are in thermal equilibrium and thus the average kinetic energy of the denser gas must be lower with the same total energy in both cylinders.
But the FOMBS corollary had to do with the utterly laughable claim that the kinetic energy of molecular projectiles on the moon was constant with height.
Confusion serves their purposes – the only question is whether it is habitual lying or a disordered mind.
Let me try that again.
. KE is not *proven* to be constant with height – a FOMBS corollary that is utterly ludicrous.”
Now I understand your argument. You are saying that average KE can’t be independent of height in the case where temperature is uniform yet density varies. That’s because you believe temperature not to be a unique function of average KE, but to depend strongly on molar volumetric density.
No – no – and no.
I am saying that the claim to KE being constant with height is hugely unlikely and supported by nothing by bafflegab.
Total thermal energy is a function of the number of molecules and the average kinetic energy. Temperature is determined by heat gained and heat lost – heat may be gained or lost by radiation, evaporation or conduction. P-N has a one dimensional understanding.
I think average molecular KE has been fairly solidly experimentally demonstrated to be very closely equal to (3/2)kT for most gases at normal pressures, as predicted by kinetic theory. This means that average KE is dependen on T and only on T. Pressure, or volumetric molar density, is irrelevant to *average* molecular KE. It follows that, if the gas has uniform temperature, then it also has uniform average molecular KE, even if the density varies with height. Hence there is no trouble for FOMD’s corollary.
It depends of course on the degrees of freedom of the gas.
KEavg = f.(3/2)kT – where f is degrees of freed
Total thermal energy = N.f(3/2)kT
So a denser gas at the same temperature has the same average kinetic temperature but a higher heat capacity. Heat transfer remains a function of the number of molecules hitting a surface and the average kinetic energy.
So two surfaces with different numbers of molecules hitting them at the same average kinetic energy will have different temperatures.
Consider 2 cylinders – one at one 1 atm and one at 2 atm at thermal equilibrium with the surroundings. If the gases had the same average kinetic energy – then the walls of the cylinder with the denser gas would be warmer. But they are in thermal equilibrium and thus the average kinetic energy of the denser gas must be lower with the same total energy in both cylinders.
But the FOMBS corollary had to do with the utterly laughable claim that the kinetic energy of molecular projectiles on the moon was constant with height.
Confusion serves their purposes – the only question is whether it is habitual lying or a disordered mind.
“Consider 2 cylinders – one at one 1 atm and one at 2 atm at thermal equilibrium with the surroundings. If the gases had the same average kinetic energy – then the walls of the cylinder with the denser gas would be warmer. But they are in thermal equilibrium and thus the average kinetic energy of the denser gas must be lower with the same total energy in both cylinders.”
So, it can be true for the gas in the 1atm cylinder that KEavg = (3/2)kT. But in that case it must be false for the other cylinder. We must have KEavg < (3/2)kT in the higher pressure cyclinder.
Do you disagree with the KEavg = (3/2)kT claim from this hyperphysics page(?) :
http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/kintem.html
You must also disagree with this:
“At any given temperature, the molecules of all gases have the same average kinetic energy. In other words, if I have two gas samples, both at the same temperature, then the average kinetic energy for the collection of gas molecules in one sample is equal to the average kinetic energy for the collection of gas molecules in the other sample.”
http://www.mikeblaber.org/oldwine/chm1045/notes/Gases/Kinetic/Gases08.htm
‘For an ideal gas the heat transfer rate is proportional to the average molecular velocity, the mean free path, and the molar heat capacity of the gas. ‘
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/thercond.html
Rob Ellison: “‘For an ideal gas the heat transfer rate is proportional to the average molecular velocity, the mean free path, and the molar heat capacity of the gas. ‘”
I have no reason to dispute this abut the *speed* at which heat is exchanged. This doesn’t invalidate the relation EKavg = (3/2)kT. But your two cylinders example are inconsistent with this relation. The two cylinders at equilibrium can’t both satisfy this relation since they have, according to you, the same temperature T, but not the same average kinetic energy.
You can’t have:
EKavg(cylinder 1) != EKavg(cylinder 2) and
EKavg(cylinder 1) = (3/2)kT and
EKavg(cylinder 2) = (3/2)kT.
Rob Ellison: “KEavg = f.(3/2)kT – where f is degrees of freed”
I think this runs together the idea of thermal energy per molecules and total kinetic energy per molecules (associated only with the 3 translational degrees of freedom).
You may mean instead Uavg = f*(1/2)kT, while KEavg = (3/2)kT, where Uavg can be larger than KEavg due to internal (non-kinetic) degrees of freedom. In that case f>3. Else, in the case of monatomic gases, f=3.
“Total thermal energy = N.f(3/2)kT” (that should be 1/2)
Yes, total thermal energy U is N*Uavg. It’s the sum of the thermal energy held by all the molecules, and so it is the average per molecule * N
“So a denser gas at the same temperature has the same average kinetic temperature but a higher heat capacity.”
Yes, it has the same kinetic temperature, and hence the same average kinetic energy (3/2)kT per molecule. It has the same specific heat capacity but a higher volumetric heat capacity because its molar density is higher.
‘According to the equipartition principle the total energy of a mole of water vapor is
Utotal = Utrans + Urot + Uvib = 3/2 RT + 3/2 RT + 3RT = 6RT.
It is important to recognize that the equipartition principle is a classical idea that fails to correctly account for the true quantum energies of molecules, with particularly poor applicability to vibrations.’
http://www.fordham.edu/academics/programs_at_fordham_/chemistry/courses/physical_chemistry_i1/lectures/equipartition_6542.asp
I can’t make any sense of this. If we are at equilibrium, then the surface can’t be “receiving more energy”.
Yeah – that was my problem with FOMBS’s thought bubble.
KEavg = f.(1/2).R.T per mole
Using the molar and molecular definitions of KEavg with f as degrees of freedom.
Total thermal energy
Uthermal = N.f.(1/2).R.T
Where N is the number of molecules.
For a monotonic gas f is 3 – thus we get the well known relationship. A diatomic oxygen molecule has 6 degrees of freedom.
Molecular dynamics suggest that KEavg reduces with compression of air and I’m wondering if the partitioning of energy changes. But now I am a little tired.
The essential point remains – this is not about actual science but in defending the groupthink dynamic in which the inferiority of denier science plays such a large part.
Rob Ellison: “Molecular dynamics suggest that KEavg reduces with compression of air and I’m wondering if the partitioning of energy changes. But now I am a little tired.”
The partitioning of energy doesn’t change since it must always obey the Boltzmann law distribution for all the degrees of freedom of the system. When the gas is compressed to very high pressures, then new degrees of freedom become available when the energy involved in molecular interaction can’t be neglected anymore. But this is reflected in heat capacity, and doesn’t change the EKavg = (3/2)kT, since the new modes in interaction aren’t storing the kinetic energy of the molecule. They contribute to the distribution of total internal energy U, not KE. The magnitude of the effect can be read off the values for heat capacity for air at different pressures. It varies by about 10% at 100bar, and 20% at 1000bar, if I remember. For 2bar the change is negligible. And it has no effect on the relation EKavg = (3/2)kT, which is also valid in liquids and solids since it derives directly from the M-B distribution, and this applies as long as the particles are identical albeit distinguishable.
So much nonsense – so little credibility.
For polyatomic molecules thermal energy will also be distributed among the rotations and vibrations of the molecule. In the same way that translating molecules could move along x, y, or z, so too can each of the atoms in a molecule. Thus molecules have a total of 3N degrees of freedom, where N is the number of atoms in the molecule. Of the total 3N degrees of freedom only 3 will be translations of the whole molecule through space. The remainder are internal degrees of freedom: vibrations and rotations. Non-linear polyatomic molecules have three degrees of rotational freedom while linear polyatomic molecules have only two rotational degrees of freedom. Rotation of a linear molecule along its molecular axis does not consume thermal energy (It’s easy to roll a pencil). Each rotation is allotted 1/2 kT per rotation (or 1/2 RT per mole of rotations) according to the equipartition principle. A mole of water molecules (water is a non-linear molecule), for example, has 3 rotations and 3/2 RT of rotational energy according to the equipartition principle…
According to the equipartition principle the total energy of a mole of water vapor is
Utotal = Utrans + Urot + Uvib = 3/2 RT + 3/2 RT + 3RT = 6RT.
It is important to recognize that the equipartition principle is a classical idea that fails to correctly account for the true quantum energies of molecules, with particularly poor applicability to vibrations.’
http://www.fordham.edu/academics/programs_at_fordham_/chemistry/courses/physical_chemistry_i1/lectures/equipartition_6542.asp
“So much nonsense – so little credibility.”
So, are you disagreeing with my claim that the effect of doubling pressure from 1psi to 2psi on the specific heat capacity of air (not volumetric heat capacity) is negligible? And hence that Uavg = f(1/2)kT (let aloneEKavg = (3/2)kT) barely changes at all?
All this in defense of a couple of thought bubbles from FOMBS.
A box on the moon – and a box on the Earth with a symplectic Hamiltonian manifold.
The heat capacity is related to mass – which with gases obviously. But I would say it is trivial thought bubbles that have some other subtext – the inferiority of denier science probably. Despite rotational and vibrational degrees of freedom having a Boltzmann distribution.
Rob, I have a proposal for you. See the last paragraph of this post.
Rob Ellison wrote: “The heat capacity is related to mass ”
Indeed it is. It is related to the number of molecules and how much heat each molecules can hold on average at T. Hence the well know relations:
Uavg = f*(1/2)kT (average energy per molecule)
Uavg = f*(1/2)RT (average energy per mole)
U = N*f*(1/2)kT (number N of molecules * average energy per molecule)
U = n*f*(1/2)RT (number n of moles * average energy per mole)
f is the number of degrees of freedom of an individual molecule, and R = N_A*k, where N_A is the Avogadro constant.
Those relations should tell you at *once* that doubling the temperature of a given amount of gas doubles *both* the total *and* average energy per molecule. Or so it seems to me.
Yes, you are still not accepting this, because it contradicts your claim that gas containers at different pressure, and the same temperature, don’t have the same average kinetic energy per molecule.
More precisely, you believe that:
If we have two identical containers C1 and C2 of the same volume V and temperature T, and pressures 1atm and 2atm respectively, the container C2 has (about) twice the volumetric molar density, and therefore has (about) half the average internal energy per molecule, and also half the average kinetic energy per molecule.
And you seem to believe those claims to be consistent with the well known relations above, since you often make use of them yourself.
And this is a stumbling block for our discussion. Clearly one of us is making a very basic mistake.
Shall we ask for adjudication of this particular dispute on *any* scientific or engineering forum of your choice? Actually, if you don’t agree, I may go ahead and ask on physicsforums in the GeneralPhysics section. I will post anonymously and will not mention you, or this site. I will say that I have a dispute with a friend. But if you want to participate and “out” any one of us, that’s fine with me. If there happens to be any substantial support for your idea, I will stand corrected. I just need first to ensure that I correctly phrased your belief about the dual dependence of internal or kinetic energy on both temperature and molecular density.
Utotal = Utrans + Urot + Uvib
Temperature measures the first.
Rob Ellison wrote: “Utotal = Utrans + Urot + Uvib
Temperature measures the first.”
I agree with this. This is why if we have a diatomic molecule, and the temperature is low enough not to excite vibrational modes significantly, we have f = 5 and
Utotal = Utrans + Urot = EK + Erot = N(3/2)kT + N(2/2)kT
However this means that Utransl_avg = EKavg = (3/2)kT.
In any case, for monatomic molecules Utotal = EKtotal.
Do you agree with my paraphrase of your position for the case of monatomic molecules? Do you also believe that when C1 and C2 are filled with helium at the same temperature and 1atm vs 2atm, then EKavg is only half as large in the second container? Or are you beginning to see your error?
Boy Y’all got some stamina! Let’s say you have two identical containers. In one container you capture air at 25 C and 1 atmosphere. In the other container you capture air at 25C and 0.5 atmospheres. The two containers would have the same temperature but different energy components. So you store both samples for a while at the same location which happens to be at 25 C. Which box contains more potential energy? The one with the most mass or the one with the lower pressure?
Or how about capturing a 10 km tall column of air one horizontal and one vertical at the same average temperature and pressure under gravity. In order to compare the two you either have to stand one up or lay one down which would involve work external to the samples, but both would have the same temperature and pressure provided the containment was perfect. Then, provided they both contain the same number of molecules in the same mixture ratio they would be the same, but why would you expect your samples to be the same in a real environment?
If they were the same though, standing vertically under gravity the bottom would have a higher measured temperature than the top due to the force of gravity but the potential temperature difference isn’t useful until you invent an anti gravity device provided of course you have access to the perfect container. Since you found some means to stand up a perfect container 10 km tall, perhaps you should patent that instead :)
Molecular dynamics is right – random thought experiments not so much.
‘This is not a practical calculation. It is an exact calculation, that you are contesting the exact results of — which is the exact offset (from molecular density reduction) also discussed in Pekka’s paper that accounts for the consistency of constant average KE *distribution* in spite of the KE loss of individual rising molecules.’
So the box on the moon returns in another guise. I had to laugh. The claim was that the average KE was constant with height for moon molecular projectiles. And this constitutes the *exact* mathematical proof?
Rob Ellison: “So the box on the moon returns in another guise. I had to laugh. The claim was that the average KE was constant with height for moon molecular projectiles. And this constitutes the *exact* mathematical proof?”
Actually it is a proof of *my* claim about the distribution of molecules inside the box, which you also thought was crazy. As I said twice, I will relate my claim to FOMD’s illuminating thought experiment in another post in the next open thread. It’s tough enough to get my own points across without having to engage in exegetical intepretation of the exact meaning and assumptions of someone else.
Again, my own claim is this. We have a box 1m on a side (or 1mm, or 1km, or whatever). It has insulated walls that molecules collide elastically with. The gas is in thermodynamic equilibrium. The gravity field is g. The temperature is constant and the pressure and density vary with height in accordance with the barometric formula and the ideal gas law. All those features of the pressure and density gradient, and temperature constancy along height, are consistent with one another, and result straightforwardly from the standard kinetic theory of ideal gases alone. In particular, the constant KEavg vertical distribution is consistent with the constant gravitational acceleration of the molecules between the collisions, and individual molecules satisfy TE=PE+KE at all times.
Again you confuse molecular dynamics with kinetic theory.
It is hard enough to wade through the extreme verbiage at the best of times – this is not it. It seems not to mean anything at all..
“Again you confuse molecular dynamics with kinetic theory.”
Are you suggesting that kinetic theory doesn’t apply to real molecules, even to a good approximation? Why are the predictions of kinetic theory so very good for predicting Uavg and heat capacities of very many real gases (He, H2, O2, etc.) on the basis of the application of the Boltsmann distibution to translational and rotational energies? Why would it succeed there and fail when gravitation is involved? In any case it succeeds spectacularly, since the application of the Boltzmann distribution to m*g*z for individual molecules immediately yield the barometric pressure and density gradients that match experience for boxes of any size.
‘According to the equipartition principle the total energy of a mole of water vapor is
Utotal = Utrans + Urot + Uvib = 3/2 RT + 3/2 RT + 3RT = 6RT.
It is important to recognize that the equipartition principle is a classical idea that fails to correctly account for the true quantum energies of molecules, with particularly poor applicability to vibrations.’
http://www.fordham.edu/academics/programs_at_fordham_/chemistry/courses/physical_chemistry_i1/lectures/equipartition_6542.asp
No – I am suggesting that you make cr@p up.
‘It is important to note that the average kinetic energy used here is limited to the translational kinetic energy of the molecules. That is, they are treated as point masses and no account is made of internal degrees of freedom such as molecular rotation and vibration. This distinction becomes quite important when you deal with subjects like the specific heats of gases. When you try to assess specific heat, you must account for all the energy possessed by the molecules, and the temperature as ordinarily measured does not account for molecular rotation and vibration.’ http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/kintem.html
“When you try to assess specific heat, you must account for all the energy possessed by the molecules, and the temperature as ordinarily measured does not account for molecular rotation and vibration.”
And this is precisely why, as I had explained, in every case where f>3 (for diatomic and polyatomic gases) I insisted to write KEavg = (3/2)kT and Uavg = f*(1/2)*kT. While you used to incorrectly write KEavg = f*(1/2)*kT, which is only correct for monatomic gases. And you also used to write KEavg = Total_U/N, which also only is correct for monatomic gases where f=3, since for all other gases only part of the internal energy is kinetic, while the rest is merely internal. When f = 5, for instance, KEavg only makes up 3/5 of Uavg, since the other 2/3 is rotational.
‘KEavg only makes up 3/5 of Uavg, since the other 2/3 is rotational.’
So 60% is translational and 66% is rotational?
‘According to the equipartition principle the total energy of a mole of water vapor is
Utotal = Utrans + Urot + Uvib = 3/2 RT + 3/2 RT + 3RT = 6RT.
It is important to recognize that the equipartition principle is a classical idea that fails to correctly account for the true quantum energies of molecules, with particularly poor applicability to vibrations.’
http://www.fordham.edu/academics/programs_at_fordham_/chemistry/courses/physical_chemistry_i1/lectures/equipartition_6542.asp
“‘KEavg only makes up 3/5 of Uavg, since the other 2/3 is rotational.’
So 60% is translational and 66% is rotational?”
I meant 3/5 and 2/5, obviously. Sorry for the typo.
You earlier had written:
“Two gas cylinders at different pressures at thermal equilibrium with the surroundings will have the same thermal energy. The average will be less – you get that by dividing by N.”
So, in the case that I described, assuming a monatomic gas such as helium for simplicity, such that Uavg = EKavg, with two cylinders of equal volume C1 and C2, and with one mole of helium in C1 (N=1), and two moles of helium in C2 (N=2), You are saying that:
C1: EKavg1 per mole = (3/2)RT
C2: EKavg2 per mole = (3/2)RT/2 ?
Is that how one must interpret your claim about total and average energy? Else, what do you mean by “you get that by dividing by N”?
The energy partitioning is still wrong.
Rob Ellison: “The energy partitioning is still wrong.”
It was *your* proposal Rob, not mine. You said that when two containers have the same volume and same temperature, and the same total thermal energy (despite having different molar amounts of gas), in order to get the average energy density of “the compressed gas” one must simply divide the total internal energy by the number of molecules to get the average per molecule. You argued this in many different ways, both for KEavg and Uavg with no distinction. For instance:
(Rob Ellison) “Take these cylinders of oxygen again – they are in local thermodynamic equilibrium. This means that the total thermal energy is the same in all cases. But there are more molecules in the compressed gas – which means that the average kinetic energy per molecule is less.”
So, you’re saying that (1) the total energy — Utotal — is the *same* in both containers, despite the number of molecules being different, and that (2) one simply divide by the number of molecules N to get EKavg.
Do you still stand by those two claims?
J. H. Jeans, The Dynamical Theory of Gases, Dover Publications, 4th Edition, 1954.
S. Chapman and T. G. Cowling, The Mathematical Theory of Non-Uniform Gases: An Account of the Kinetic Theory of Viscosity, Thermal Conduction and Diffusion in Gases, Cambridge University Press, 3rd Edition, 1991.
C. Cercignani, The Mathematical theory of Dilute Gases, Springer, 1994.
L. Boltzmann, Lectures on Gas Theory, Dover Publications, Reprint Edition, 2011.
C. Truesdell and R. G. Muncaster, Fundamentals of Maxwell’s Kinetic Theory of a Simple Monatomic Gas: Treated as a Branch of Rational Mechanics, Academic Press, 1980.
Thanks Dan! I’ll try to read those tonight.
‘Do you still stand by those two claims?’
The temperature is the same yet the energy partitioning is quite different. The molecular dynamics are quite different.
But we get back to the original claim that energy transfer depends only on average kinetic energy.
There was another claim that the kinetic energy of the bouncing balls – no collisions scenario – was constant at every height.
http://in-the-sky.org/physics/balls.php
These thought bubbles are defended at great length with terrier like insistence. With nothing but verbiage. My original comment was that the thought bubble – in that case the symplectic manifold of the Hamiltonian – was symtomatic of the groupthink psychopathology. The symbolic use of less than half understood science – much less than half in FOMBS case – and the categorical inferiority of denier science.
It is so much a part of the cult of AGW groupthink construct – which is of course bound up in the personality construct of individuals. It manifests as an assumption of moral and intellectual superiority to the science deniers and an inability to reflect. Which are of course classic groupthink symptoms.
This is one of the other dimensions of climate communication as I said – the impossibility of actual communication as opposed to lobbing quibbles over the physic barricades.
‘Do you still stand by those two claims?’
Rob Ellison: “The temperature is the same yet the energy partitioning is quite different. The molecular dynamics are quite different.”
What is different from what? How does that relate to *your* own claims? Did you finally realize that they don’t make sense? Can you salvage then through reformulation?
“But we get back to the original claim that energy transfer depends only on average kinetic energy.”
I lost track of the number of times I denied saying anything like that, that I granted the opposite, or that I asked you for a direct quotation of either myself or FOMD. As much as I try to get you to clarify your own position, you strive to distort mine. Look up the “principle of charity”.
“There was another claim that the kinetic energy of the bouncing balls – no collisions scenario – was constant at every height.”
“http://in-the-sky.org/physics/balls.php”
I don’t remember the claim. A quotation for context would be useful.
There is a button for enabling collisions and the spread of energy distributions. After entropy has been maximized, you can turn off the collisions. Entropy will not fall back down.
“These thought bubbles are defended at great length with terrier like insistence. With nothing but verbiage. […]”
In the interest of saving electrons, you might consider shortening your signature.
Who is much?
Much redefining of the terms of the silly thought bubbles. .
1. That KE average is constant with height. It isn’t in the Earth’s atmosphere.
2. That energy transfer is a function only of average kinetic energy. It isn’t. It is a function of the number of molecules and kinetic energy transferred in collisions.
And we get yet more silly thought bubbles about thermometers calibrated to pressure. In the atmosphere temperature is the result mostly of radiative transfers.
But again – poor and incomplete science focusing on silly thought bubbles in the service of ideological battles is of little intrinsic interest other than as an illustration of the social dynamic.
Rob Ellison: “2. That energy transfer is a function only of average kinetic energy. It isn’t. It is a function of the number of molecules and kinetic energy transferred in collisions.”
I have consistently denied having made *this* claim (about energy transfer rates, or amounts, being only a function of temperature) every time you’ve attributed it to me. You never were able to provide a quote of me making this silly claim.
What I deny is your claim that the average kinetic energy of the molecules in a gas doesn’t determine uniquely it’s *temperature*. It’s not a claim about “heat transfer”. You are claiming that the total internal heat energy is the same in the two cylinders (same volume, same temperature, different pressures and molar amounts), and that average kinetic energy is KEavg = U_thermal/N, and hence that KEavg is only half as much in the 2atm cylinder than it is in the 1atm cylinder. It is this claim that I denied.
At a given temperature, average kinetic energy of molecules need not be any smaller to compensate for larger collision rates. The average kinetic energy of the molecules is (3/2)kT whatever the density or pressure or the gas might be (within reasonable bounds — e.g., for air, P < 1000atm, etc.)
At leas now you have an ally. Maybe captdallas will be able to explain to me where I wen't wrong without constantly attributing to me claims that I never made. Two captains, two engineers… Don't give up.
P-N, “If one considers a piece of isothermal air column enclosed in a box, then the molecules will have the same average kinetic energy at the top as they have at the bottom. Because of the barometric density gradient, the air lower in the column with have a higher volumetric heat capacity and it is only this that accounts for a higher rate of heat transfer, if there is any. There will be no heat transfer at all if the container has the same temperature as the air. In that case is irrelevant the rate of collisions is higher at the bottom than it is at the top. The air will not warm the bottom at a higher temperature just because the rate of collisions is higher, ”
First, if the container is isothermal and the gas is the same temperature as the column, you have defined the problem away. Obviously if the container is small enough you can have this ideal situation. There is no gravito-thermal effect without a height different enough to allow such a small effect to manifest itself. “How big is your box?” has been asked a few times. You are just dancing in circles until you understand that point.
Second, temperature transfer between the gas and the container is due to elastic collisional transfer. Even with the container and the gas at the same temperature there is continuous heat transfer just no net flow of heat.
Then, if the container is tall enough, there will be significantly more collisions with more energy per collision at the bottom of the container than at the top due to the downward force of gravity acting on the mass of each molecule in motion. If you force the gas to be isothermal and the container to be uniformly the same temperature in spite of the difference in collisional energy transfer, then there can be no effect, because your model doesn’t allow for an effect. However, since temperature is a result of kinetic energy transfer, logically the bottom would end up being warmer than the top. A temperature gradient would result because gravity produces a potential energy gradient. Since no energy transfer is allowed out of the system/cylinder, the bottom would warm at the expense of the top. However, also since no energy is allowed to escape the cylinder, the warmer gases at the bottom will transfer more energy per collision which will offset a portion of the effect which is why the container has to be sufficiently tall or the force of gravity has to be huge. There is nothing magical or unintuitive about it. You can define it away with assumptions, but the reality is there are more and stronger collisions at the bottom than the top. It is a completely useless thought experiment contained in an imaginary cylinder which allowed Lochsmidt to derive a hypothetical g/Cv lapse rate. In the real world the escape velocity limits the maximum velocity of the molecules and the volume is artificially “constant” meaning you should use Cp to derive lapse rate. . .
The problem is that some, like the Sky Dragons, seize on irrelevant trivia such as this and try to elevate it into something of significance when Lochsmidt was probably just having a goof.
Captdallas, I would rather not run together discussion of the hypothetical gravito-thermal effect with Rob’s cylinder/temp/heat-transfer claims. We can discuss the box thought experiment in the next open thread. I’d like to know if you still believe that the total internal energy of the air in a cylinder is independent of the pressure and molar quantity. And if you still believe that the average kinetic energy of the molecules therefore varies in inverse proportion of the molar quantity so as to preserve this independence — rather than being solely a function of temperature. I think it would be worth it resolving those issues before going into gravitational effects on air columns.
And, to be clear, the reason why I think it worthwhile to settle those basic issues first is because the alleged gravitational effect on temperature is primarily an effect on the speed, and therefore also the kinetic energy, of the molecules. If you don’t agree about the way temperature relates to molecular kinetic energy (or other forms of energy, molecular volumetric density, collision rates, etc.) then we are liable to constantly talk past one another.
And yet these were the silly thought bubbles under discussion.
Constant average kinetic energy doesn’t apply in Earth’s atmosphere. It simply does not.
*My* heat transfer mechanism involved the number of molecules with a translational kinetic energy. There is no need to go beyond that into realms where idle speculation based on verbiage is the order of the day. There is nothing
… to settle.
I replied above.
Much doesn’t care.
P-N, ” I’d like to know if you still believe that the total internal energy of the air in a cylinder is independent of the pressure and molar quantity.”
Since we are discussing kinetic energy and temperature, the KE per molecule is a function of temperature. So the total internal KE would depend on the molar quantity. More molecules at the same energy the more total energy.
“And if you still believe that the average kinetic energy of the molecules therefore varies in inverse proportion of the molar quantity so as to preserve this independence — rather than being solely a function of temperature.”
Since the “average” molecular KE is only a function of temperature it would be independent of the molar quantity. The “average” energy transferred to the container would then be directly proportional to the “average” number of collisions with the container. Doubling the number of molecules while maintaining a constant container temperature means an increase in molecular to molecule collisions due to reduced mean free path while maintaining the same molecule container collision frequency.
captdallas: “Since we are discussing kinetic energy and temperature, the KE per molecule is a function of temperature. So the total internal KE would depend on the molar quantity. More molecules at the same energy the more total energy.”
Good. So now you are agreeing with me. You are walking back from your earlier claims that:
(CD): ‘Temperature is a measure of the average energy of collisions. You can have lots of collisions with lower energy or fewer with more energy. So you are looking at the average energy per container not the average energy per molecule. The 2g container has twice the molecules at half the energy per molecule.’
You agree that both containers have the same average energy per molecule, and that they do have the same total energy.
“Since the “average” molecular KE is only a function of temperature it would be independent of the molar quantity. The “average” energy transferred to the container would then be directly proportional to the “average” number of collisions with the container. Doubling the number of molecules while maintaining a constant container temperature means an increase in molecular to molecule collisions due to reduced mean free path while maintaining the same molecule container collision frequency.”
That is a new theory. I am unsure why you want to retain the same collision frequencies at different pressures. Surely this is a gross contradiction with the very definition of pressure as average force (or equivalently, average rate of momentum transfer) per unit of surface area. If the average kinetic energy is the same (resulting from the same Maxwell distribution of speed), then so is the average amount of momentum transfer per collision. We would then have the funny result that a gas container with twice the molar amount of gas, at the same temperature, produces the same internal pressure in its walls.
Also, there is no reason that a reduction in mean free path should have any effect on the rate of collisions on the container’s walls. (Unless the pressure increase would be so large as to make van der Waals forces come into play, but that’s something else). The collisions between the molecules are elastic, so if the path from some molecule to the wall is intercepted by other molecules, then this is compensated exactly by the conservation of momentum. You can’t generally prevent a billiard ball from hitting the border through putting another ball on its path. The second ball will pick up the momentum from the first one. The general case is complicated, but the overall effect of elastic collisions is to preserve exactly the molecular flux across (or on) any surface.
P-N: “You agree that both containers have the same average energy per molecule, and that they do [not] have the same total energy.”
Sorry.
This was a correction for a post awaiting release from moderation.
P-N “We would then have the funny result that a gas container with twice the molar amount of gas, at the same temperature, produces the same internal pressure in its walls.”
Yes, it would be a funny result wouldn’t it? That is the reason for the “average” in scare quotes. The containers also have to be at the same temperature as the gas since that was a stipulation, heat was removed from the 2g container to equalize the temperatures. So the problem has to be clearly stated before you can get rid of the scare quotes. “how big is your box?”
Captdallas, there were no scare quotes in this sentence:
“Doubling the number of molecules while maintaining a constant container temperature means an increase in molecular to molecule collisions due to reduced mean free path while maintaining the same molecule container collision frequency.”
And it is troublesome. It generates a dilemma for you.
(1) In order to maintain the same “molecule container collision frequencies”, you would need to reduce average velocities (and hence average momenta p = mv) by half. Since EK = m*v^2, this means reducing average KE by about 1/4. This means that the pressure would be independent of molar amount at the same temperature (with twice the collisions, and half the momenta canceling out). The container with one mole of air would have the same pressure as the container with two moles of air. And the result about kinetic energy also contradicts the kinetic theory of gases that states that average kinetic energy is (3/2)kT, quite independently of pressure or volumetric molecular density. Putting “average” within scare quotes doesn’t fix this. It was assumed from the beginning that the temperature is the same in both containers, so that isn’t an issue.
(2) And as noted, if you do maintain EKavg constant (as you should) then you double the rate of collisions with the container walls (as should be!), which explains why pressure is doubled. You thus can’t assume that the rate of collisions is reduced at all as a result of the shorter mean free path.
You can avoid all your problems if you simply accept that EKavg = (3/2)kT, and that the rate of collisions with the walls is exactly proportionate to pressure.
http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/kintem.html
This formula has the same range of application to real gases as the ideal gas law PV = nRT (= NkT). It applies with very high accuracy to air in a very wide range of pressures and temperatures.
P-N, You are skipping past the point. T1=T2, n2=2*n1 and R is equal but you have to have V and P don’t you? 2g is not a standard value for P and the V, size of the box was never specified. You assumed one thing while Rob was specifying another. You have to be on the same page doncha?
So if the “average” energy per molecule of container 1 equals the average energy per molecule of container two, you have made another stipulation so what is P and V?
If container 1 has 2n and is at the same temperature as container two with 1n, something else has to change. Since g was included, the pressure on the bottom of the container would be larger than the top unless you make the height of the container small enough for the difference to be negligible. If you do that, there can’t be a G-T effect.
So you can assume a constant pressure and work out the influence of g or a constant volume and work out the influence of g. Loscmidt tried g/Cv while the rest of the world uses g/Cp. That is the actual problem in a nutshell.
“Doubling the number of molecules while maintaining a constant container temperature means an increase in molecular to molecule collisions due to reduced mean free path while maintaining the same molecule container collision frequency.”
That was meant to be food for thought. If the “average” energy per molecule is the same in both containers but there are twice as many molecules in one container, Magic happens if you have the same container temperatures.
So instead of just throwing out 2g, pick a volume or pick a pressure, but as it is you are posing an irrelevant problem.
“That was meant to be food for thought. If the “average” energy per molecule is the same in both containers but there are twice as many molecules in one container, Magic happens if you have the same container temperatures.”
Rob Ellison is saying the same thing almost verbatim.
But really, the “magic” is the standard physics encapsulated into EKavg = (3/2)kT, which can’t be disputed without also throwing PV = nRT out of the window. It is really a standard result of the kinetic theory of gases.
See also: http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/kintem.html#c1
I thought we were agreed about the specifications of the two containers. (1) Same volume, (2) same temperature, (3) the second one has twice the pressure and hence twice the molar amount of gas, since PV=RnT, (4) the past history is irrelevant. Maybe my later discussion of the two volumes of gas compressed together into one introduced confusions. Sorry about that.
I’ll comment further on the earlier part of your post tomorrow.
Captdallas, also, forget about g and the gravito-thermal effect. In the laboratory it is quite negligible even if it exists. It never has been detected experimentally (apart from controversial results from Graeff). I had begged you not to mix those further issues into the simple problem with the two gas containers. We can discuss gravitational effects in the next open thread. Let us first settle the basics about standard thermal physics and kinetic theory.
P-N, “Captdallas, also, forget about g and the gravito-thermal effect. In the laboratory it is quite negligible even if it exists”
I never said it was significant. btw if (PV)1 = (PV)2, then KEave=3kT/2 is the same for both containers. Now (PV)2=2(PV)1. so for the same temperature in both containers, #2 would have to lose energy to the environment.
http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/imgkin/ntcoll.gif
‘Thermal conduction in gases can be analyzed within the framework of kinetic gas theory, which you probably learned about in your freshmen or physical chemistry classes. In this theory, we assume all molecules to be identical, non-interacting, rigid spheres of given diameter and mass and moving randomly with a mean velocity v. The molecules move about in the gas and collide (like in the simulation applet). They may transfer energy between different regions of the gas if there are temperature gradients. Consider mentally dividing the gas into layers. The net flux of energy between two adjacent layers is assumed to be proportional to the energy gradient
https://watertechbyrie.files.wordpress.com/2014/06/heat-transfer.png
where ρ is the energy density, which equals nT – n is the molecular density and T is the temperature. The proportionality constant turns out to be KbvL/2, where L is the mean free path between molecular collisions and Kb is the Boltzmann’s constant. Combining these results with Fourier’s law yields the result that the thermal conductivity is KbvLn/2 for an ideal gas.’
http://zeolites.cqe.northwestern.edu/Module/heattrans.html
Temperature is the result of heat transfer to – say – a thermocouple. The transfer is dependent on energy density.
The extreme dedication to wrong ideas based on poorly understood physics is quite odd.
Yes, for some strange reason, while those simple equations ((EKavg = (3/2)NkT) and PV=nRT) agree with me and contradict you, Rob and yourself are insisting that the container with twice the molar amount, and twice the pressure, must cool down to the environment if it is initially at the same temperature as the first one. If it did, then EKavg in the second container would drop, and the temperatures would not be the same anymore in both containers — at equilibrium!
‘The Entropy change comes from the equation which incorporates the first and second laws. The energy balance is the first law, and the heat transfer is expressed as an entropy change which is a statement of the second law.’
http://lorien.ncl.ac.uk/ming/Webnotes/Therm1/revers/isothe.htm
P-N returns again and again to an equation he applies incorrectly and comes to the wrong conclusion. In fact – he starts with the conclusion based on misunderstanding something he has read – and comes to the wrong conclusion.
Rob Ellison: “P-N returns again and again to an equation he applies incorrectly and comes to the wrong conclusion. In fact – he starts with the conclusion based on misunderstanding something he has read – and comes to the wrong conclusion.”
OK. My conclusions were that EKavg1 in cylinder1 is (3/2)kT, and EKavg2 in cylinder2 also is (3/2)kT.
Also, Total_EK = N1*EKavg in cylinder1 (with N1 molecules), and twice as much in cylinder2; total_EK2 = N2*EKavg (with N2 = 2*N1).
Now show me how *you* correctly apply the equation “EKavg = (3/2)kT”, and what are the values for EKavg, and Total_EK in both cylinders that you come up with as a result of the correct application of this formula.
Cylinder1:
EKavg1 = ?
Total_EK1 = ?
Cylinder2:
EKavg2 = ?
Total_EK2 = ?
Rob Ellison: “Much redefining of the terms of the silly thought bubbles. .
1. That KE average is constant with height. It isn’t in the Earth’s atmosphere.
2. That energy transfer is a function only of average kinetic energy. It isn’t. It is a function of the number of molecules and kinetic energy transferred in collisions.”
Temperature variations with height isn’t relevant to your discussion of the two cylinders. There is no net heat transfer at thermal equilibrium, which was assumed.
I think you may have finally realized that your claims and arguments regarding the two cylinders containing oxygen at 1atm and 2atm (same volume, same temperature) were silly and invalid. This would explain why you are desperately trying to change the subject and persist in disputing imaginary claims that I never made.
The two claims were exactly that. The average KE as particles rose was constant – because – and I quote – the loss of lower energy molecules *exactly* compensated for the gain in gravitational potential – and that heat transfer to a surface was only as a result of average kinetic energy. Two separate FOMBS thought bubbles that P-N found oh so insightful.
The secret of the cylinder is that measured temperatures are a result of heat flow – and that potential energy increases with compression when the heat lost is equal to the work added.
But there is no point in a discussion with a dissimulator.
Let’s try again.
Rob Ellison: “Much redefining of the terms of the silly thought bubbles. .
1. That KE average is constant with height. It isn’t in the Earth’s atmosphere.
2. That energy transfer is a function only of average kinetic energy. It isn’t. It is a function of the number of molecules and kinetic energy transferred in collisions.”
Temperature variations with height isn’t relevant to your discussion of the two cylinders. There is no net heat transfer at thermal equilibrium, which was assumed.
I think you may have finally realized that your claims and arguments regarding the two cylinders containing oxygen at 1atm and 2atm (same volume, same temperature) were silly and invalid. This would explain why you are desperately trying to change the subject and persist in disputing imaginary claims that I never made.
The two claims were exactly that. The average KE as particles rose was constant – because – and I quote – the loss of lower energy molecules *exactly* compensated for the gain in gravitational potential – and that heat transfer to a surface was only as a result of average kinetic energy. Two separate FOMBS thought bubbles that P-N found oh so insightful.
The secret of the cylinder is that measured temperatures are a result of heat flow – and that potential energy increases with compression when the heat lost is equal to the work added.
But there is no point in a discussion with a dissimulator.
.
Rob Ellison: “The two claims were exactly that. The average KE as particles rose was constant – because – and I quote – the loss of lower energy molecules *exactly* compensated for the gain in gravitational potential – and that heat transfer to a surface was only as a result of average kinetic energy. Two separate FOMBS thought bubbles that P-N found oh so insightful.”
This is misrepresentation. What follows “and that…” is your invention. Neither FOMD (that I am aware of) neither I ever said or implied that heat *transfer* only is a result of average kinetic energy. I am saying something like this for *temperature* — e.g. that it is proportional to average the (translational) kinetic energy of the molecules. The constant of proportionality simply is (3/2)k. That you can’t seem to distinguish the concepts of heat transfer and temperature doesn’t help. Myself I am distinguishing them. So, when I am saying that *temperature* only depends on AKavg, I am not implying that heat transfer rates, or amounts of heat transferred, only depend on AKavg (or that they only depend on temperature, for that matter.) I have said this to you a dozen times already; yet you persist in the misrepresentation.
This confusion between transfer rates and temperature also is the reason why I am insisting for postponing discussion of gravitational effects on the vertical distributions of speeds, average KE, and molecular density, for the case gas columns in equilibrium. Let’s clear up misunderstandings of basic thermal physics before complicating the problem even more. We can discuss the gravitational effects on vertical distributions and profiles in the next open thread, if you wish.
Temperature is measured by heat transfer – get used to the idea. What I said was that energy transport was a function both of the number of molecules and the energy of the molecules. It was denied by both P-N and FOMBS. It doesn’t help any if the goal posts shift. .
A more dense gas at the same kinetic temperature per molecule will transfer more heat.
The other thought bubble about constant average KE is also wrong.
Rob Ellison: “What I said was that energy transport was a function both of the number of molecules and the energy of the molecules. It was denied by both P-N and FOMBS. […]”
Yet you never were able to provide a quote from either of us denying this at any time.
Rob Ellison “A more dense gas at the same kinetic temperature per molecule will transfer more heat.”
Sure. It also will have the same temperature, period. If the “kinetic temperature per molecule(sic)” is T = (2/3)EKavg/k, then the temperature of the gas is T. Hence if both gases have the same temperature T, they also have the same “kinetic temperature per molecule”.
Also, if the container has the same temperature as the gas, the rate of heat transfer is zero. If the container is colder than the gas, then the denser gas will warm it more, and faster, not because the “kinetic temperature per molecule” is larger, but rather because the collision rate is higher, and hence the transfer *rate* is higher (per unit time), and the total internal energy of the gas — N*Uavg — is larger. Uavg = (f/2)kT is the same in both gases.
Both gases have the same specific heat capacity, per kg, but the denser gas has a higher volumetric heat capacity because there are more molecules in the same volume. This is entirely consistent with both gases having the same initial temperature and the same EKavg, and same Uavg, per molecule, as dictated by the relations EKavg = (3/2)kT and Uavg = (f/2)kT.
blah blah blah blah – and yet you did say it.
And it is surely not consistent. .
Rob Ellission: “blah blah blah blah – and yet you did say it.”
Quotation?
“And it is surely not consistent.”
What *is* inconsistent is this set of three claims regarding the two cylinders:
(1) T1 = T2,
(2) EKavg = (3/2)kT,
(3) EKavg1 != EKavg2.
You haven’t show any inconsistency among my claims regarding the two cylinders.
Answer The average kinetic energy of the molecules that freely fly to any given height is independent of that height.
And I had forgotten another gem: “Temperature is the result both of the average kinetic energy of the molecules and the number of molecules hitting a surface.”
Another probable consequence of Hamiltonian-Ellisonian symplectic statistical-mechanics.
The temperature as measured is the result of molecular kinetics. And the symplectic manifold of the Hamiltonian suggests it seems that a gravito-thermal gradient does exist – http://www.mdpi.com/1099-4300/16/3/1515
This is quite different to FOMBS – and P-N’s – quite ridiculous use of science and math they don’t understand in the least for some sort of ideological point scoring. Surely that’s the point.
… molecular dynamics is what I meant… rather than kinetics
Rob, that’s exactly as I had surmised. You were badly misrepresenting FOMD too. He had said nothing about “heat transfer” being dependent only on average KE. He rather contested, just like I do, your heterodox claim that *temperature* depends on collisions rates.
For a regular thermometer to measure the temperature of a gas, it must thermally equilibrate with it (either through direct contact, or through contact with a container that has equilibrated with it). This process requires heat transfer. The result of the measurement — the temperature — only depends on the average kinetic energy of the molecules of the gas — not how fast the thermometer adjusts. You are still confusing heat with temperature.
The temperature as measured depends on heat flow – heat flow is a function of collisions of particles with the surface. It depends on both the rate of collisions and the average kinetic energy.
A surface that has more molecular collisions – in a higher density gas – with higher energy particles will be warmer. No matter how many time P-N says it isn’t – it is the case.
Note also that Rob E. defines EKavg per molecules as U_total/N even for diatomic or polyatomic molecules, though that’s not the issue here.
For an ideal gas and neglecting potential energy.
Utotal/N = Ukinetic/N = KEavg
Hard to keep up with this utter and obnoxious twit.
He first of all denies – then when I quote – moves on to claiming that I misrepresented.
Rob Ellison: “For an ideal gas and neglecting potential energy.
Utotal/N = Ukinetic/N = KEavg”
Here is a college student who made the same error as you did, and didn’t get the answer listed in his textook.
https://www.physicsforums.com/threads/determining-temperature-of-a-diatomic-gas.233933/
We were discussing oxygen, which is a diatomic molecules. At ordinary room temperature, it has effectively 5 degrees of freedom, only three of which are translational. The other two are rotational. You may wish to *call* the two rotational degrees of freedom “kinetic” if you want. That’s fine with me. It’s angular kinetic energy after all. But this energy counts towards the gas internal energy and specific heat capacity. It doesn’t count towards kinetic temperature (relevant to heat transfer through collisions) because it’s unrelated to mv^2. Hence:
KEavg = (3/2)kT —> relevant to kinetic temperature
Uavg = (5/2)kT —> relevant to total internal energy
And finally:
Uavg = Utotal/N
KEavg != Utoal/N (*unless* the gas is monatomic)
P-N might have been talking oxygen – I hardly read any of what he says. Most of it is just endless repetition of superficial nonsense.
Rob Ellison: “He first of all denies – then when I quote – moves on to claiming that I misrepresented.”
Rob Ellison: You denied that heat transfer is related to both…
P-N: No, I said *temperature* is related to EKavg only…
Rob Ellison: You denied that heat transfer is related to both…
P-N: No, I said *temperature* is related to EKavg only…
Rob Ellison: You denied that heat transfer is related to both…
P-N: No, I said *temperature* is related to EKavg only…
Rob Ellison: You denied that heat transfer is related to both…
P-N: No, I said *temperature* is related to EKavg only…
…
And finally Rob Ellison quotes me saying “temperature” and claims that he didn’t misrepresent!
“A surface that has more molecular collisions – in a higher density gas – with higher energy particles will be warmer. No matter how many time P-N says it isn’t – it is the case.”
This is your heterodox theory but it is contradicted by observation.
Observation shows that the average velocity, and average kinetic energy of a gas is independent of pressure and molecular density (withing a wide range of pressures where PV = nRT applies well enough) and that average velocity or average kinetic energy only depend on Temperature. (EKavg=(3/2)kT)
A solid will equilibrate to the same temperature as the gas with which it is in thermal contact. So, the surface will only warm to an equilibrium temperature T in proportion with the average kinetic energy of the gas molecules that impact it (at T), and regardless of the collision rate.
(In fact this is also true if the “gas” only contains photons. The amount of photons — “density,” or rate of photon absorption and emission — has not effect either on the equilibrium temperature of the solid within a radiation bath at a certain temperature. Only the energy distribution of the photons is relevant. Max Plank first explained why that is so.)
Your belief that the temperature of a gas (or the temperature of a solid surface that thermally equilibrates with it) is dependent on pressure as well as temperature also is contradicted by the *observation* that the speed of sound in a gas depends only on temperature (and the nature of the gas: its molecular weight and adiabatic constant), and not on pressure or molecular density.
If you want to calculate the speed of sound in the troposphere (or within any homogeneous gas anywhere) at any level, regardless of the local environmental lapse rate, you only need to know the temperature. Why is that? It’s because sound is a pressure wave in the gas fluid (macroscopic description) carried at the molecular level as a density wave, and the speed of propagation of this molecular density wave depends only on the average speed of the molecules, and not their volumetric molar density. Since the speed of sound depends only on temperature and not on pressure, this tells you immediately that pressure has no incidence by itself on average molecular speed at a given temperature.
In the simplest of terms, the discipline of heat transfer is concerned with only two things: temperature, and the flow of heat. Temperature represents the amount of thermal energy available, whereas heat flow represents the movement of thermal energy from place to place.
On a microscopic scale, thermal energy is related to the kinetic energy of molecules. The greater a material’s temperature, the greater the thermal agitation of its constituent molecules (manifested both in linear motion and vibrational modes). It is natural for regions containing greater molecular kinetic energy to pass this energy to regions with less kinetic energy.
Several material properties serve to modulate the heat tranfered between two regions at differing temperatures. Examples include thermal conductivities, specific heats, material densities, fluid velocities, fluid viscosities, surface emissivities, and more. Taken together, these properties serve to make the solution of many heat transfer problems an involved process. http://www.efunda.com/formulae/heat_transfer/home/overview.cfm
These things are baby physics – but yet far more nuanced than the superficial nonsense from P-N.
Rob Ellison: “A surface that has more molecular collisions – in a higher density gas – with higher energy particles will be warmer. No matter how many time P-N says it isn’t – it is the case.”
You are misrepresenting what I am denying. Yes, if the particles have more kinetic energy, then the surface will be warmer. That’s because EKavg is higher. But if they have the *same* average kinetic energy, though a lower density and lower collision rate, in that case the surface will only warm to T, whatever the gas pressure is, where T = (2/3)EKavg/k. Hence temperature depends only on average kinetic energy. As you are fond of saying, this is “baby physics”.
P-N, How could there ever be a change in temperature in your model? What I am trying to point out that energy is constantly being exchanged and that if you change the average energy or number of molecules colliding with the shell of the container that there will be a change in the net flow of energy. If you pressurize a container, there has to be a conversion of the force of the collisions to maintaining pressure and some to maintaining temperature if the energy per molecule is the same and the number of collisions double. Logically, that would be an increase in internal energy transfer, more molecule to molecule collisions required to maintain pressure which would would reduce that average momentum of the collisions with the container so that both pressure and temperature could be stable.
As it is, nothing can ever change in your model because everything is fixed, so adding gravity cannot have an impact. Is that a physical model?
captdallas: “As it is, nothing can ever change in your model because everything is fixed, so adding gravity cannot have an impact. Is that a physical model?”
The model is called the kinetic theory of gasses (as part of statistical mechanics — that also extends to solids, liquids, radiation, quantum phenomena, etc.) It is experimentally very successful, and has been so for over a century. And it is consistent with thermodynamics, the laws of which it explains as being emergent from the statistical properties of large collections of particles. You can read more about it here:
http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/kinthe.html
…or the Wikipedia page, which is also pretty good.
I’ll reply to your specific objections later on. Thanks for the civil discussion.
P-N I know what it is called and one of the first stipulations is that the distances have to be small enough to make the effect of gravity negligible.
Once the volume of the model is expanded, gravity can have an influence and you have to consider the distribution of energy in the molecules. With gravity only the more energetic can reach the top of a much larger container. That would be the next step in Kinetic Theory, determining the influence of gravity and the lapse rate. You seem to be stuck on step one.
Once you adjust your model, you should find for a ridiculously tall container there is a small G-T effect for a constant volume and a much more significant effect lapse rate effect for a constant pressure where the more energetic molecules can escape. We are dealing with an open system in reality.
“P-N I know what it is called and one of the first stipulations is that the distances have to be small enough to make the effect of gravity negligible.”
Not true. Statistical mechanics applies to objects as big as stars, white dwarfs, neutron stars, etc. It works perfectly fine under gravity. The derivation of the barometric molecular density distribution under gravity (and the resulting vertical pressure profile of the atmosphere at the macroscopic level) is a standard result of kinetic theory. Check some of the references I provided to Rob E. and that he happily ignored. The barometric density formula results *directly* from the application of the Boltzmann distribution law on the gravitational potential energy states of individual molecules.
In any case, I would prefer to postpone discussion of gravitational phenomena, and fine details of the box of gas under gravity at equilibrium, for the next open thread. My main point, which I didn’t make clear, was that the speed distribution in a gas, and the average kinetic energy (KEavg = (3/2)kT, are results that apply perfectly well at the macroscopic level, for gases in cylinders, where the vertical pressure gradient and the effect of g on vertical speeds and energy distributions aren’t relevant *to the problem under consideration* — e.g. the internal energy of the gas, the relations between temperature, pressure, density and average kinetic energy per molecule, heat capacity, etc.
P-N I have seen one paper that models the formation of galaxies on the ideal gas laws and Maxwell distribution. That model doe include gravity though. “Average” KE doesn’t mean that every molecule at the shell of the container has that “average” KE. The least energetic tend to collect at the walls of the container. Under gravity the least energetic tend to collect at the floor of the container. When the molecules have different masses the least heaviest tend to collect at the floor When the molecules have varying mass and volume, then you get a more unique effect. If you inhibit convection, you get a different effect. Modeling the real world requires increased complexity.
that should be heaviest collect at the floor.
Captdallas, before delving into huge complexities, let us try to understand (or argue) why it’s *false* to claim that in two identical cylinders filled with oxygen at two different pressures (1atm and 2atm) and the same temperature, the molecules in the denser cylinder must have a proportionally lower average KE in order to compensate for the higher collision rate with the wall and not warm it more. Let us stick with that, and closely related, issues (e.g. the speed of sound as a function of temperature only, etc.) until it is resolved. One must correctly model the very simple before one can model the complex.
P-N, I don’t believe I said that denser cylinder had to have proportionally lower energy per molecule, only that the energy transfer between the cylinder and internal molecules would have to change so that double the impact with the cylinder didn’t double the thermal energy transferred to the container. The cylinders and gases are at the same temperature with different pressures because of a shift in potential energy, pressure maintenance, and kinetic energy, temperature maintenance, between the cylinders. If you consider a perfectly spherical container, the average velocity at the walls and at the center would be closest to zero so the likelihood of the highest energy molecules is somewhere in between. There is a distribution of molecular energy, small though it may be. If you change that distribution you would have an energy gradient until the the system reaches a new equilibrium. With a large enough change it can end up in a steady state instead of an equilibrium.
You are basically feeding the Dragons by assuming “average” means fixed.
Well…
I view this a little differently. Pressurizing the cylinder causes elastic deformation of the cylinder. The gas does “work” against the cylinder walls. The cylinder metal lattice elastically deforms until the amount of force exerted by the cylinder against the gas equals the force exerted by the gas against the cylinder. This is an equilibrium – net new work done is zero (no displacement means no work).
If the gas continued to do work against the cylinder the cylinder would expand to the point of failure.
In my view, the amount of excitation of the inner surface of the metal tank can only increase a fixed amount until the energy transferred back to the gas during collision equals the energy transferred to the metal lattice..
PA I agree, but I think we are still at perfectly elastic collisions with no deformation. My main point was that there is contentious energy transfer with zero net flow, if you change the distribution there would be a net flow in some direction.
Captdallas, it’s nice to see you coming around, but coming around you certainly did ;-)
http://judithcurry.com/2014/10/21/ethics-of-communicating-scientific-uncertainty/#comment-641880
http://judithcurry.com/2014/10/21/ethics-of-communicating-scientific-uncertainty/#comment-642117
P-N, In both of those links I was pointing out that you cannot have twice the number of average KE molecule impacting the cylinder wall at the same velocity meaning twice the force without a temperature increase. Something else has to change. If you have a thermocouple being impacted with a number of molecules with the same KE and velocity, then double that, the thermocouple would measure an increase. Obliviously, to conserve energy there has to be change in the thermal energy transfer per impact. Since the cylinder temperature and size doesn’t change, the average molecular impact on the cylinder would have less momentum and there would be an increase in the molecule to molecule collisions within the cylinder. With the bouncing balls animation you can disable collision and you don’t get lower energy molecules “trapped” against the sides which creates the distribution of energy in the molecules requiring use of the Maxwell distribution.
If you consider each molecule is contained in a spherical shell, the location of maximum probable energy is at some number of radii from the center and a smaller number from the boundary of the shell. While it is not exactly analogous, with a gas flow in a pipe, the maximum velocity is in the center and the flow becomes more laminar at the sides.
So once again you are assuming that “average” means uniform or fixed and not acknowledging there is a distribution of energies.
“P-N, In both of those links I was pointing out that you cannot have twice the number of average KE molecule impacting the cylinder wall at the same velocity meaning twice the force without a temperature increase. Something else has to change.”
OK. I had misunderstood your recent comments then. You are still with Rob on this. However you aren’t agreeing with PA. He agrees with me that when you have the same average kinetic energy per molecules in both cylinders, (and different pressure) then the temperature is the same; and hence nothing “has to change”.
That’s all well. That means it’s now you and Rob against PA and I. It makes for a fairer debate. I’ll be back tonight.
P-N, “He agrees with me that when you have the same average kinetic energy per molecules in both cylinders, (and different pressure) then the temperature is the same; and hence nothing “has to change”.”
I agree that the average KE per molecule has to be the same, that is fixed by temperature but it does not mean that all molecules have the same KE. I am mystified that you cannot grasp that. A greater number of molecules with the same “average” KE doesn’t mean that “average” energy transferred per molecule at the wall of the cylinder will be the same. If you add pressure you would change the distribution of molecule to molecule collisions. Increase flow in the pipe example, and you change the distribution of molecular velocities. If you put a tiny hole in the container, only the most energetic molecules have the higher probability of escaping. So if you want to measure the average KE at the pin hole you would have to adjust your data.
In a very small container with a large density of molecules it doesn’t matter, which is why the basic Kinetic energy model can ignore gravity. Once you expand your volume, especially vertically, and decrease the number of molecules, you have to start including the small influences.
Some basic math and not just definitions helps.
https://www.youtube.com/watch?v=qSFY7GKhSRs
So what happens if you change the system by increasing N?
Not sure why my reply to this:
http://judithcurry.com/2014/10/21/ethics-of-communicating-scientific-uncertainty/#comment-642461
appeared here:
http://judithcurry.com/2014/10/21/ethics-of-communicating-scientific-uncertainty/#comment-642631
Captdallas: “P-N, In both of those links I was pointing out that you cannot have twice the number of average KE molecule impacting the cylinder wall at the same velocity meaning twice the force without a temperature increase. Something else has to change. If you have a thermocouple being impacted with a number of molecules with the same KE and velocity, then double that, the thermocouple would measure an increase.”
I think one trouble may be that you (and Rob E. as well) picture the thermocouple sensor bit, and the cylinder interior walls, as passive surfaces that can only potentially receive energy from the molecular impacts. The essential point is that when the temperature T is the same everywhere, at thermal equilibrium, the molecules at the surface of the cylinder walls, and in the thermocouple sensor, *all* have the same average kinetic energy (3/2)kT (though the molecules in the solid have special internal modes of vibrational energy storage within the lattice, while the molecules of the gas may also have rotational modes). Hence, when the individual gas molecules hit the sensor, or walls, they are exactly as likely to gain an amount of energy dE from a molecular collisions as they are to give up the same amount of energy dE in the elastic collision. This is why increasing the number of collisions has zero effect on the rate of heat transfer. It is zero on average and hence zero in total.
Agree PA, That was just an hypothetical. When the deformation is purely elastic, as you described it, in the realistic case, then the process (within the metal lattice) is reversible, the walls do not heat up while deforming and they just gain plastic energy of deformation W1 = Wp. They also push the atmosphere out and hence perform work W2 = dV*1atm. The internal energy lost by the gas is dQ = W1+W2. But it is very small since the volume change is small and dQ also equals PdV, where P is the pressure inside of the container while filling up (integrated over time). And the elastic energy, and work from the external atmospheric pressure, are likewise a minor contribution while the cylinder resettles to it’s former shape as it empties down. The big contribution to the energy flows is the heat of evaporation, the adiabatic expansion inside of the container, and the work done on the atmosphere as the gas exits the regulator.
I forgot one main energy flow: the heating of the cold cylinder by the environment. This is of course required to balance the energy of evaporation of the liquefied gas. Else the cylinder would cool down to the point where the partial pressure comes down to 1atm and no more gas would come out through the regulator.
PA: “In my view, the amount of excitation of the inner surface of the metal tank can only increase a fixed amount until the energy transferred back to the gas during collision equals the energy transferred to the metal lattice..”
Yes, your view is supported by the theorem of equipartition of energy in statistical mechanics. In a system in thermal equilibrium, energy is shared equally in between all the independent degrees of freedom of the molecules. The molecules in the metal lattice also have their own internal degrees of freedom. They also have three kinetic modes, and in addition to those, three vibrational modes associated with displacement in the lattice — potential energy. Hence they have average internal energy Uavg = (6/2)kT = 3kT. (Small deviations from this reflected in the specific heat capacities of solids are accounted for by larger scale quantum lattice modes of vibration.)
When the gas molecules also have EKavg = (3/2)kT — hence the kinetic temperature equals the temperature of the lattice — then the system is in thermal equilibrium simply because all the individual degrees of freedom of all the molecules in the system hold the same average amount of energy, in accordance with the theorem of equipartition. This is the energy distribution that maximizes entropy. There can’t be any net energy flow, regardless of the molar density of the gas or the rate of collisions. Collisions merely ensure that energy distributes randomly among the different modes and thus drive the system towards maximum entropy and thermal equilibrium. Collision rates also determine how fast this occurs, not what the equilibrium state is.
P-N: “…simply because all the individual degrees of freedom of all the molecules in the system hold the same average amount of energy [per molecule], in accordance with the theorem of equipartition.”
kT = PV/N
Where V and T are constant for a doubling of N.
P = N.m.v^2/3V
=> kT = m.v^2/3
=> kT = P1.m.v1^2/3 = P2.m.v2^2/3 = constant
Seems to imply that v1 does not equal v2.
P-N had written: “Since PV = NkT = (3/2)U, […]”
This should have been PV = NkT = (2/3)U, tough the argument remains unchanged regarding what remains the same and what must change.
Rob Ellison wrote:
**********
“kT = PV/N
Where V and T are constant for a doubling of N.
P = N.m.v^2/3V
=> kT = m.v^2/3”
**********
Since PV = NkT = (2/3)U = (2/3)N*(m*v^2)/2 (Assuming an ideal monatomic gas)
This should be P = N*(2/3)(m*v^2)/(2V), and
kT = (2/3)(m*v^2)/2
Rob Ellison: “=> kT = P1.m.v1^2/3 = P2.m.v2^2/3 = constant”
This result doesn’t make sense since kT has the dimensions of energy while your right hand side has the dimensions of [pressure]*[energy].
P-N: “This should be P = N*(2/3)(m*v^2)/(2V), and
kT = (2/3)(m*v^2)/2”
OK, That’s equivalent to what you wrote in the first step, but the next step where you only multiply the right hand side of the equation by P still doesn’t make sense.
To follow up, after the first step, you should have written:
=> kT1 = kT2 = P1V/N = P2V/N = (2/3)(EKavg1) = (2/3)(EKavg2) =
(average_in_cylinder1 m*v1^2)/3 = (average_in_cylinder2 m*v^2)/3
This is constant since when pressure doubles, so does N.
Utot = Ukinetic + U potential
Pressure increases the potential energy of the gas and so must reduce the kinetic energy considering the assumptions. Temperature lost is equal to the work done and no heat lost or gained.
Pressure is a function of the number of molecules and the velocity – so is heat transfer. At equilibrium – the heat transfer to and from the wall is equal but none the less.
“Utot = Ukinetic + U potential
Pressure increases the potential energy of the gas and so must reduce the kinetic energy considering the assumptions. Temperature lost is equal to the work done and no heat lost or gained.
Pressure is a function of the number of molecules and the velocity – so is heat transfer. At equilibrium – the heat transfer to and from the wall is equal but none the less.”
I was responding to your calculation. You used PV = nKT. This equation applies to ideal gases where the potential energy from molecule interactions and finite size is neglected. If you want to discuss van der Walls gases, then you must use the equation:
(P + a/v^2)(v – b) = RT,
where v is the molar volume and a and b are constants characteristic of the specific gas.
In any case, the relation EKavg = (3/2)kT, where T is the kinetic temperature, is valid for ideal gases, for van der Walls gases, and even for liquids and solids. This is because of the theorem of equipartition. The average energy stored in each one of the three translational degrees of freedom only is a function of temperature, and is the same as the energy stored in any other degree of freedom (e.g. kinetic rotational, lattice vibration, etc.)
“Utot = Ukinetic + U potential
Pressure increases the potential energy of the gas and so must reduce the kinetic energy considering the assumptions. Temperature lost is equal to the work done and no heat lost or gained.”
Long range van der Walls force are attractive, not repulsive. It’s only at pressures so very high that molecular distances are on the order of molecular diameters that the forces become repulsive. Hence, *less* work is required to compress a real gas (such as O2 from 1atm to 2atm) than would be required to compress an ideal gas. In any case, when the gas has been allowed to cool down to the same previous temperature T, the average kinetic energy per molecule still is going to be (3/2)kT regardless of the amount of potential energy from van der Walls forces. The relation EKavg = (3/2)kT is valid for all gases, and even liquids and solids. It’s just the average and total internal energy that depends on non-translational degrees of freedom and potentials from inter-molecular forces.
“Pressure is a function of the number of molecules and the velocity – so is heat transfer. At equilibrium – the heat transfer to and from the wall is equal but none the less.”
That doesn’t contradict the fact that EKavg = (3/2)kT, and hence only depends on temperature. That’s also true on any non-translational degree of freedom that stores some energy form (gravitational, electric potential, kinetic rotational, etc.) They all individually store an average (1/2)kT per molecules regardless of molecular volumetric density or collision rates. This is why, for instance, the specific heat of ordinary solids is very close to N*3kT/(molar mass). There are three translational degrees of freedom, and three more components of vibration within the lattice. So, in that case also, EKavg = (3/2)kT. This result is in fact guaranteed by the theorem of the equipartition of energy and applies much beyond the case of ideal gases.
P = N.m.v^2/3V
kT= PV/N = m.v^2/3
So what was the original statement. Total heat transfer is the result of molecular collisions? Do you have no clue at all?
Rob Ellison: “So what was the original statement. Total heat transfer is the result of molecular collisions? Do you have no clue at all?”
I was replying to your post: http://judithcurry.com/2014/10/21/ethics-of-communicating-scientific-uncertainty/#comment-642656
I thought your calculation purported to show that if there is a doubling of N and the average energy per molecule is constant, while pressure varies, then there ought to be a volume change as well. This would show that my assumption that EKavg is constant when T is constant is false. But your calculation is erroneous. You can’t multiply just one side of the equation by P. Whatever you wanted to demonstrate, the last equation doesn’t make sense. Maybe, fix the calculation and see what you can get.
Your first equation is exactly the same except you haven’t cancelled some of the terms – which is where you go wrong.
It takes work to compress gas. Equivalent to the temperature gained in compression.
But you are still arguing the wrong point. Heat transfer through collisions is all about energy density.
“Your first equation is exactly the same except you haven’t cancelled some of the terms – which is where you go wrong.”
You first step is correct, as I later acknowledged. It’s the second step where you go wrong. The dimensions don’t match and there is no justification for only multiplying the right hand side by P. How can one go from ‘a = b’ to ‘a = b*P’?
“But you are still arguing the wrong point. Heat transfer through collisions is all about energy density.”
How is pointing an error in your calculation — after you said “Some basic math and not just definitions helps.” — arguing the wrong point?
PV = NkT entails that if T and V are constant, and N doubles, the so does PV, and hence so does total internal energy. The first time you tried to prove the contrary, you incorrectly substituted a variable (R/N) for k, which is a constant. Now you are multiplying just one side of an equation by the pressure. You can’t do that. You have to multiply both sides by the same quantity to preserve the validity of the equation.
I am trying to bring some rationality into the discussion.
It seems a lost cause.
‘Long range van der Walls force are attractive, not repulsive. It’s only at pressures so very high that molecular distances are on the order of molecular diameters that the forces become repulsive.’
There is obviously energy in compressed gas. There are considerable latent energies that are in the separation of particles. Just how to account the energy put into compression of a real gas is a question.
Rob Ellison: “There is obviously energy in compressed gas. There are considerable latent energies that are in the separation of particles. Just how to account the energy put into compression of a real gas is a question.”
It’s stored mainly as internal energy. For monatomic gases such as helium, that’s entirely N*KEavg. For diatomic molecules, it’s N*(5/2)kT. For polyatomic molecules, it’s N*3kT (discounting vibrational modes that are accessible at high temperatures, or for big molecules).
When the gas expands and performs work on the expanding walls, the energy is taken off from the kinetic energy of the molecules that are rebounding back with a reduced average speed. It’s like throwing an elastic ball on a target that’s moving away. It comes back with reduced speed since the contact force performed work on the target owing to the displacement of the target. Inter-molecular collisions quickly redistribute the EK losses evenly among all the internal degrees of freedom of the molecules.
See figure #4 at this link:
http://www.schoolphysics.co.uk/age16-19/Thermal%20physics/Gas%20laws/text/Adiabatic_and_isothermal/index.html
Gases cool as they expand? Duh.
It seems a hopeless case. There is no energy in compressed gas – but it does work when allowed to expand? Hmmm.
Rob Ellison: “Gases cool as they expand? Duh.”
And the flipside is that the gases warm as the are compressed due to the very same kind of explanation. As the walls close in, the molecules that impact them rebound with a higher velocity and thus gain kinetic energy. This total gain (for all the molecules that impact the moving walls) is easily calculated for some bit of wall of areas A as W = P*A*dx = PdV (where dx is the normal displacement). While the proximal effect is to increase the kinetic energy of those molecules, the equilibrium result is to increase their average internal energy in the whole container by an amount N*Uavg = W, since, as I pointed out earlier, inter-molecular collisions within the gas rapidly redistribute this kinetic energy gain evenly among all the internal modes of energy storage of the molecules.
Rob Ellison: “I am trying to bring some rationality into the discussion.”
Your belief that you can *use* the relations “EKavg = (3/2)kT”, and “PV = NkT”, which are relations that *entail* that EKavg only depends on T, in order mathematically demonstrate the contrary, is irrational.
Your belief that you understand molecular dynamics is yet to be demonstrated at all conclusively. It is all a bit odd.
Rob Ellison | October 30, 2014 at 5:29 am | Reply
kT = PV/N
Where V and T are constant for a doubling of N.
P = N.m.v^2/3V
=> kT = m.v^2/3
=> kT = P1.m.v1^2/3 = P2.m.v2^2/3 = constant
Seems to imply that v1 does not equal v2.
I hate to state the obvious.
But the premise was that 1/2 the gas was removed from the cylinder.
Assuming it is done adiabatically this equation is just wrong (or being used incorrectly).
When the gas is removed the pressure goes down and the temperature goes down since the volume is fixed. The PV curves and your 2/3 rule apply to constant temperature.
PA wrote: “I hate to state the obvious.
But the premise was that 1/2 the gas was removed from the cylinder.
Assuming it is done adiabatically this equation is just wrong (or being used incorrectly).
When the gas is removed the pressure goes down and the temperature goes down since the volume is fixed. The PV curves and your 2/3 rule apply to constant temperature.”
I don’t think Rob was considering an adiabatic process, but rather a comparison between the end states of the two cylinders at equilibrium (and room temperature). “N doubles” just meant N is twice as much in the second cylinder. Also, there is no 2/3 rule. m*v^2/2 stands in for the average kinetic energy of molecules — where “v^2” is velocity squared. And the factor (3/2) comes from EKavg = (3/2)kT, while total energy is U = (3/2)PV = N*(3/2)kT. (A standard result from kinetic theory explained in the video that was linked earlier).
The trouble is the last equation that comes from nowhere and fails dimensional analysis. It equates energy with energy*pressure. I am unsure what the conclusion is supposed to be. It seems an like an argument to show that EKavg can’t be the same in the two cylinders if the temperature and volume are the same while the pressure and molar masses are different. This is what Rob has been arguing consistently, and unsuccessfully in my view, for a long time. He also argues that the total internal energy in the two cylinders is the same.
P-N, ” He also argues that the total internal energy in the two cylinders is the same.”
I believe he is arguing the total kinetic energy is the same, but total energy is double in the 2PV container due to conversion to potential energy, pressure. Since the KE per molecule is fixed by temperature, you would have to look at a section of the container wall to see how KE is converted to PE. Since you are increasing the number of molecules in this case and the average velocity of the molecules is fixed as well, the average distance traveled per molecule would decrease since there would be more interaction between molecules to offset the additional force on the on the wall. In other words, the potential energy would have to be uniformly distributed in the volume as well as the kinetic energy.
Captdallas wrote: “I believe he is arguing the total kinetic energy is the same, but total energy is double in the 2PV container due to conversion to potential energy, pressure.”
That’s not what his argument was:
Ron Ellison: “Take these cylinders of oxygen again – they are in local thermodynamic equilibrium. This means that the total thermal energy is the same in all cases. But there are more molecules in the compressed gas – which means that the average kinetic energy per molecule is less.”
Captdallas: “Since the KE per molecule is fixed by temperature, you would have to look at a section of the container wall to see how KE is converted to PE. Since you are increasing the number of molecules in this case and the average velocity of the molecules is fixed as well, the average distance traveled per molecule would decrease since there would be more interaction between molecules to offset the additional force on the on the wall. In other words, the potential energy would have to be uniformly distributed in the volume as well as the kinetic energy.”
There are two issues with this. First, this alleged effect would only occur at distances from the container walls that are in the order of a few molecular diameters. Hence it would not have any significant effect on the overall speed distribution of the molecules in the bulk of the volume of the container. The second issue, I will only broach over briefly since it is related to the effect of gravity on vertical speed distributions and we can discuss this later. It is a consequence of the theorem of the equipartition of energy. When molecules get very close to the wall, they are indeed repulsed away, and while they get close, they trade kinetic energy for potential energy. However, if you look at the speed distribution of the molecules that get to some given small distance of the wall, dz, the speed distribution, and hence the kinetic energy distribution of all the molecules that manage to get at this close distance isn’t any lower than anywhere else in the vessel, or indeed any lower than any molecules in the solid lattice of the wall. There is a scarcity of molecules very close by, but the same distribution of speed among those scarce molecules.
That’s because many of the molecules that were approaching the wall with a *lower* speed reversed their motions before getting at this short distance, dz, and this drop-off effect, from the less energetic molecules, results in a constant speed distributions of the remaining molecules that have a *higher* total energy (potential + kinetic).
This was merely a qualitative argument, but it can be shown to be quantitatively correct with the use of the Boltzmann law of distribution as applied to the energy states of molecules that get arbitrarily close to the wall. The Maxwell speed distribution (which entails KEavg = (3/2)kT) is the only possible distribution of speed that maximizes entropy at any point in the vessel, even where the potential energy is high.
He has a number of problems.
The ideal gas equations apply to the whole volume of gas, outside the cylinder as well as inside – so if you remove gas you have consider both populations.
PV = nRT applies to the tank both before and after gas removal. If T is constant and V is constant (the case of a very slow leak otherwise known as isothermal expansion)
P ∝ n
Any other result contradicts the ideal gas law and a lot of practical testing.
Since the original source of the problem is the gravito-thermal effect and the ideal gas model doesn’t allow for the influence of gravity, it is a bit difficult attempting to have someone adjust their model parameters. In the “real” world and atmosphere gravity has some weight :)
Captdallas, I’ve exchanged a hundred messages or so with Rob Ellison about the effect of gravitation on velocity and density vertical distributions in gas contained in a box. The main problem that we stumped upon constantly was his insistence that FOMD and I were committed to deny that “heat transfer” is dependent on molecular collision rates. Hence the focus of the present discussion. Since you also have an issue with this (c.f. your thermocouple example,) it is worth clearing up. For discussions of the gas in a box under gravity, wait for the next open thread. If you are impatient you can review the previous discussion in this thread and my references to the literature ;-)
Pierre-Normand | October 30, 2014 at 11:55 am | Reply
Captdallas, I’ve exchanged a hundred messages or so with Rob Ellison about the effect of gravitation on velocity and density vertical distributions in gas contained in a box. The main problem that we stumped upon constantly was his insistence that FOMD and I were committed to deny that “heat transfer” is dependent on molecular collision rates.
Say what?
Air is a insulating gas that transfers energy by convection and radiation. That is why insulation is mostly air.
The claim about collision rates seems to be based on the incorrect assumption that air transfers significant energy by conduction.
P-N: “The main problem that we stumped upon constantly was his insistence that FOMD and I were committed to deny that “heat transfer” is dependent on molecular collision rates.
PA responded: “Say what?
Air is a insulating gas that transfers energy by convection and radiation. That is why insulation is mostly air.
The claim about collision rates seems to be based on the incorrect assumption that air transfers significant energy by conduction.”
We’ve ignored radiative effects in most of the discussion and it is not very relevant for the case of oxygen in the cylinder, which is transparent to longwave radiation. But surely, conduction is relevant at the gas/solid boundary, and must occur before there is any convection within the cylinder. I don’t remember anyone arguing that convection or radiation isn’t important. Rob Ellison was insisting that a higher rate of collisions, together with the same average kinetic energy per molecule, would necessarily lead to the surface warming to a higher *temperature*. Captdallas makes the same argument. They seem to picture warming as a passive response to rates of molecular impacts and not just adjustments to energy distributions. My recent response to this specific issue appears here:
http://judithcurry.com/2014/10/21/ethics-of-communicating-scientific-uncertainty/#comment-642771
P-N, “Captdallas makes the same argument. They seem to picture warming as a passive response to rates of molecular impacts and not just adjustments to energy distributions. ”
Not completely. What effect there is is so small that it would not be apparent in a mono-atomic gas ideal model. There such a small variation in the velocity distribution in an ideal isothermal cylinder it would be negligible. Which is kinda the point of an ideal gas model.
Captdallas: “Not completely. What effect there is is so small that it would not be apparent in a mono-atomic gas ideal model. There such a small variation in the velocity distribution in an ideal isothermal cylinder it would be negligible. Which is kinda the point of an ideal gas model.”
What distinguish an ideal gas from a real gas, experimentally, principally is some properties as the heat capacity or the gas law that they obey (the ideal gas law, or van der Walls formula). The speed distribution, as a function of kinetic temperature, is, however, pretty much exactly the same. It’s the Maxwell speed distribution (it merely ignores quantum phenomena).
http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/maxspe.html
In any case, why do you think that there needs to be an adjustment (or “variation”) at all to the rate of collisions, if, at equilibrium, each molecular impact is just as likely to gain some amount of energy dE from the wall as it is to give this amount to it? The rationale for your needing an adjustment doesn’t seem to exist. You though more collisions warm the surface more. At thermal equilibrium, they don’t, whatever the rate of collisions per surface area might be. The energy already is spread around the whole system (walls included) such as to maximize entropy. The gas has no power to warm the wall, or vice versa.
P-N, “The gas has no power to warm the wall, or vice versa.”
Right, but if the wall is warmer than the gas, the wall and gas will seek a new uniform temperature and pressure. Both are just mediums, not energy sources.
Captdallas: “Right, but if the wall is warmer than the gas, the wall and gas will seek a new uniform temperature and pressure. Both are just mediums, not energy sources.”
Of course they will. But did you forget what the point of our discussion was? Rob claimed that if the temperature is the same in both cylinders, at thermal *equilibrium* (the walls and the gases have the same temperature too), then it is *not* possible that the average kinetic energy of the gases in the two cylinders be the same, since more impacts with the same average energy per impact would warm the walls of the second cylinder more. I’ve shown why this conclusion doesn’t follow from the premises.
You had agreed with him (at first) that there needed to be a compensating reduction in kinetic energy, and then, later on, you proposed that the average kinetic energy might be the same but in that case there needed to be a reduction in the rate of collision (that was your “mean free path” proposal). If you now change to topic to heat transfer with colder walls, there you don’t need any compensation anymore. You walked back entirely from Rob’s initial claim, and from your own two proposals about the need for compensation in the equilibrium case.
…Not, of course, that I would blame you at all for walking back from those unneeded proposals.
Here’s a fun way of playing with relationships. It is nowhere near as simplistic as P-N imagines.
http://intro.chem.okstate.edu/1314F00/Laboratory/GLP.htm
We get back to the original claim. That there is a density difference between the top and the bottom. There isn’t. And that heat transfer to the top and bottom surfaces relies only on average kinetic energy. Both of these ideas are incorrect – quite irrelevant to climate and obsessively pursued with the smarmy obnoxiousness so typical of the groupthink collective.
Rob, I am having problems with Java so I haven’t used your link yet. A column of air can do some wonderful things in the real world though. A lot of tall buildings end up with stack effect, hot upper and cold lower floors so the effect of gravity can’t over come buoyancy. One thing I seriously doubt, is that an isothermal situation would occur :)
You need to include the site in the site exceptions list or Java will block it.
Convection happens. Density differences in the box don’t however as motions of molecules are random.
Rob Ellison: “Convection happens. Density differences in the box don’t however as motions of molecules are random.”
Did you not agree that there was a pressure differential from top to bottom? And if there is a pressure differential, then surely the gas is more compressed where the pressure is highest, and hence denser? For a gas cylinder one meter high filled with nitrogen at 2atm, this pressure differential would be about 2Pa.
‘Now we have two cylinders of solid gas, and we expose them to a gravity field of 1 g, 10 gs, or 1000 gs.
Does the gas warm up, under the influence of gravity? Of course not. It remains at absolute zero. Misunderstanding heat, energy, temperature, pressure, and all the rest, will eventually have you believing in the gravito-thermal effect, or the ability of CO2 to warm an externally heated body.
No gravito thermal effect,..
Seems to me you need some energy for things to warm up – gravity or not. And what NASA said was that the molecules were distributed randomly throughout the box. I agree.
But – silly thought experiments notwithstanding – the symplectic manifold of the Hamiltonian for an atmosphere under gravity does suggest a gravito-thermal effect it seems. .
http://www.mdpi.com/1099-4300/16/3/1515
Who shall we assess as more credible? Flynn or Professor Christian Fronsdal at UCLA? ‘
Solid gas? I guess what he means is that it is no longer a gas.
Rob Ellison,
You are correct. I should have been clearer. Maybe I could have referred to a solidified or frozen gas, or something similar.
In relation to Professor Christian Fronsdal, are you seriously appealing to authority, and calling for consensus?
Credibility has nothing to do with fact. Apparently 97% of some odd grouping of supposed scientists believe the world is warming due to CO2 in the atmosphere. Maybe they also believe, as did the eminent physicist Lord Kelvin, that the Earth cannot possibly be older than 20,000,000 years.
So who are we to believe? Kelvin, Fronsdal, or the supremely unschooled and unqualified Mike Flynn? You choose who to believe. I choose me in relation to the existence of the gravito-thermal effect – surprise, surprise!
Live well and prosper,
Mike Flynn.
Rob, my computer doesn’t trust me enough to do that :)
“And that heat transfer to the top and bottom surfaces relies only on average kinetic energy.”
Are you back to arguing against an imaginary opponent? I asked you very many time to quote me saying anything similar about heat transfer — and that my claim is that it is the *temperature* of a gas that only is dependent on average kinetic energy (as the EKavg = (3/2)kT relation should suggest to you). When you finally offered a quote, it was about temperature, with no mention of heat transfer. It was actually me criticizing (indeed making fun of) *your* claim that “Temperature is the result both of the average kinetic energy of the molecules and the number of molecules hitting a surface.”
So, your claim that I am denying that heat transfer depends on collision rates rests entirely on my denying your claim that *temperature* depends on collision rates.
In any case, at thermal equilibrium, there is no heat transfer.
P-N, The mean free path proposal was because if you double the N then you would reduce the mean free path and change the velocity distribution. In an ideal gas container small enough to ignore gravity it wouldn’t matter, but a large enough volume would allow a larger distribution.
You however don’t wish to discuss that so the comment was poorly timed.
Then once a variation in velocities can exist, we can discuss if they would result in different measured temperatures. Off hand I say yes, since the average kinetic energy at the top of a tall enough cylinder would tend to be lower.
Captdallas wrote: “P-N, The mean free path proposal was because if you double the N then you would reduce the mean free path and change the velocity distribution. In an ideal gas container small enough to ignore gravity it wouldn’t matter, but a large enough volume would allow a larger distribution.”
What is a “larger distribution”?
If you double the density, then the mean free path is reduced, for sure. Gravity has nothing to do with it. Gravity merely creates small vertical pressure and density gradient within the cylinder. For a one meter high cylinder of gas, the density differential from top to bottom would be proportional to the pressure differential (1Pa/101kPa) = 0.001%. What significant effect would that have on the mean free path?
In any case, the mean free path doesn’t affect the speed distribution. Random elastic collisions between the molecules change their momenta individually but has zero effect on the overall distribution at any given time, in a system at equilibrium, which remains the Maxwell distribution of speed.
http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/maxspe.html
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/disfcn.html#c2
P-N, ” 0.001%” That would be a ridiculously small impact, wuddunit? I imagine a mighty tall cylinder would be needed. With that small an impact a molecule would need some room to build up a velocity, which I believe was part of Loschmidt’s original argument. Loschmidt btw, appears to have been one serious scientific PITA. I think he even had a hand in the arrow of time situation.
Captdallas wrote: “P-N, ” 0.001%” That would be a ridiculously small impact, wuddunit?”
So, do you now agree with PA and me that gases in both cylinders (oxygen at 1atm and 2atm, same temperature) have the same average kinetic energy per molecule, (3/2)kT, that the collision rate on the walls of the second cylinder is twice the rate of the first one, and that this is perfectly consistent with both cylinders, and gas contents, being at the same temperature since the collisions don’t result in any net energy transfer from the gas to the cylinder or vice versa?
How big is you cylinder and where is the pressure transducer :)
Captdallas: The result does not much depend on the size of the cylinder. Let us say, for the sake of the argument, that it is one meter high and 20cm in diameter. The pressure regulator is screwed near the top on the right side, unless you are looking at it from behind (in which case you won’t see the pressure gauge, but that’s not a problem since the main valve is closed shut). Is the average molecular kinetic energy of the gases the same? Is the collision rate on the walls the same? Is the total of internal energy the same?
Think of those questions as college physics textbook exercises that aim to test your knowledge about PV=nRT, and such. There is no gotcha.
P-N, as I have said a number of times with a small effect you need a large volume. With a small volume you are going to have extremely small variations, there will be variations, because there is constant energy transfer and motion with no change in net. If you had a thermometer with the precision to measure the small difference due to gravity, the noise would most likely be larger than the signal.
That said, average KE would be equal and the 2PV tank would have twice the internal energy. Now just for fun, how many grad students would have to measure the temperature and pressure of each tank before the average measurements equaled the ideal values?
I should add that if you tell the grad students what the ideal measurements should be before hand would that reduce the number of measurements required :)
Again – the scenario change is the problem.
There was first of all a density difference in the box – with more molecules at the bottom striking the surface with some average KE. It implies that the heat content of the bottom surface is greater than the top.
An impossibly silly scenario continued well past the point of rationality.
blah, blah, blah balh…
No problem with 2 cylinders – that was not the scenario – the problem is the box with a gravity density. A surface being struck with more molecules at the same energy is warmer.
All the rest is utter nonsense – all of which are not absolute but depend on boundary conditions.
“Again – the scenario change is the problem.
There was first of all a density difference in the box – with more molecules at the bottom striking the surface with some average KE. It implies that the heat content of the bottom surface is greater than the top.”
No. The case that I was considering was isothermal, and since I hold that KEavg = (3/2)kT, and hence only is a function of temperature, the individual collisions with both the top and bottom of the box have the same average kinetic energy, and the exact same Maxwell speed distribution for that matter. Read again all the references I provided (Pekka; Am. J. of Physics; Statistical Mechanics, CUP). They all concern a vertical density distribution (derived from the Boltzmann distribution applied to the potential energy of the molecules.) and a constant Maxwell speed distribution that only is a function of temperature. They all cover the isothermal case (though Pekka generalizes to other static temperature profiles as well, if I remember)
Molecules move randomly to fill the container. A density difference in the atmosphere is irrelevant. Energy flow as a result of molecular collisions. I have given a few dozens references – and you have misinterpreted Pekka previously I seem to remember.
“Molecules move randomly to fill the container.”
If there isn’t a pressure differential in the container, then the weight m*g of the gas — which is a force — isn’t applied to the container, which is absurd.
And if there is a pressure differential, then, since PV = nRT, and since T is constant, n/V must vary. In other words, where pressure is higher, near the bottom of the box, the gas is more densely compressed. Your online references simply are references to problems or diagrams in which this pressure gradient is neglected because it is not relevant to the problem under consideration. Were they discussing buoyancy or the condition of hydrostatic equilibrium within the vessel, they would of course mention it.
As I also mentioned many time (while providing references), kinetic theory dictates that there is a pressure gradient. This gradient derives directly from the Boltzmann distribution over the potential gravitational energies of the molecules. It is just as direct a result of kinetic theory as is the Maxwell distribution of speed.
P-N, “If there isn’t a pressure differential in the container, then the weight m*g of the gas — which is a force — isn’t applied to the container, which is absurd.”
That was kinda the point about the size of the box, the mean free path and the distribution in the box. The ideal gas equation is a great ideal model that “agrees well” with reality, but like all ideal models never agrees perfectly with reality. Once you get to very small effects like gravity and CO2 forcing, how small is small matters.
The concepts of equilibrium and iso-thermal also have real world limits. “Ideal” is a great idea that never exists.
Captdallas: “That was kinda the point about the size of the box, the mean free path and the distribution in the box. The ideal gas equation is a great ideal model that “agrees well” with reality, but like all ideal models never agrees perfectly with reality. Once you get to very small effects like gravity and CO2 forcing, how small is small matters.
The concepts of equilibrium and iso-thermal also have real world limits. “Ideal” is a great idea that never exists.”
That doesn’t make a whole lot of sense. The mass of the air in a box of dimensions 1m*1m*1m is about one kilogram and it’s a little more for pure oxygen. This means that the pressure difference between top and bottom is about 10Pa. (mg/A = 1kg*9.81m/(s^2)(m^2). The vertical pressure gradient is 10Pa/m whatever the size of the box, from simple hydrostatic considerations The vertical temperature profile is irrelevant. You can’t make the weight of the air in a box (as applied to the box) vary through changing the vertical temperature profile unless the density distributions isn’t static. Even if we postulate some gravito-thermal effect at equilibrium, however large, it makes no difference to the weight of the air in the box or to the pressure profile (which must accord with the barometric formula).
P-N: “The vertical pressure gradient is 10Pa/m whatever the size of the box, from simple hydrostatic considerations ”
I meant to say ‘whatever the size or shape of the container…’ If the shape is irregular, the local pressure integrated over the normal vector (pointing outside) to interior surface elements dA of the container still give a net force equal to the weight of the gas in the container. This is how the weight of the gas gets applied to the container and makes it more heavy as a whole.
It seems quite bizarre behaviour to insist on a trivial point with such immense persistence and repetition.
The usual idea is that molecules are randomly distributed. The weight of the gas in the box is molecular dynamics.
http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/weighgas.html
Rob Ellison: “The weight of the gas in the box is molecular dynamics.”
This claim of your’s doesn’t affirm or deny anything. It’s meaningless. The specific molecular dynamics *explains* the pressure gradient. From your own hyperphysics link:
“The *difference* between the average force on the bottom and top of the container is just the weight, mg, of the molecule.” (My emphasis)
The average force divided by the area just is the pressure of the gas. That’s from the definition of pressure (one Pascal is 1N/m^2). If there were no vertical pressure gradient, then the average force on the top would be the same as the average force on the bottom.
“Taking an average force like this allows you to determine average forces and average *pressures* on the walls of a container of gas.” (My emphasis)
You are supplying a link to an explanation of the reason why there must exist a pressure differential in support of your strange claim that there isn’t any pressure differential in the box.
P-N, “That doesn’t make a whole lot of sense.” Why? Because the Ideal gas equations aren’t 100% perfect? Once you make the container large enough small effects are more probable. The weight of the container shouldn’t change but what happens inside can get interesting because you are pushing the practical limits of the model. As I said before we are talking a very tall box. Since we are limiting all heat transfer between the box and surroundings including entropy, you would initially apply more pressure to the bottom by adding gravity that would perturb the condition . so what initial condition you select could change the results. All because the problem isn’t really physical.
For example, if the column contains various gases, ideally, the heavier gases would settle to the bottom and you would get a layer of gases by mass. But since the gases are constantly in motion you have the Brazil nut effect that can cause heavier, larger molecules to actually rise. With no energy loss, you can end up with an inversion so heat would flow with the force of gravity or you could end up with one huge lava lamp effect while the system tries to restore some stable condition.
Loschmidt was a prick
Captdallas: “Why? Because the Ideal gas equations aren’t 100% perfect? Once you make the container large enough small effects are more probable.”
Rob Ellison and I were talking about the weight of the air in a 1m*1m*1m box at equilibrium. Rob Ellison claims that the weight of the gas in the box can be applied to the box even though (according to him) there is no pressure gradient at all in the box. He is not claiming that the lack of a pressure gradient is related to the gravito-thermal effect or any other special property of real as opposed to an ideal gas. We are not talking about falling boxes or transient states. Zero pressure gradient in the box at equilibrium is not a small effect. It is a physical impossibility and an incoherent view. He supports his view with a link to an hyperphysics page that provides the explanation why there ought to be a pressure differential between top and bottom. He seems not have read the references that he provided.
“As I said before we are talking a very tall box.”
It makes no difference. You can make the box as tall as the full atmospheric column. There most definitely won’t be a uniform vertical pressure at equilibrium under gravity. The pressure gradient is given by the barometric formula whatever the size or height of the box. And the hypothetical gravito-thermal effect doesn’t change the pressure gradient either. It is merely an effect on temperature.
You are supplying a link to an explanation of the reason why there must exist a pressure differential in support of your strange claim that there isn’t any pressure differential in the box.
Odd because the hyperphysics link explains it as molecular velocities and a change in momentum. This has nothing whatever to do with density. The behavior grows ever more bizarre it seems.
Rob Elisson: “Odd because the hyperphysics link explains it as molecular velocities and a change in momentum. This has nothing whatever to do with density. The behavior grows ever more bizarre it seems.”
Yes, it indeed explains “it” — it explains the pressure differential that you had denied existed. Since we are considering an isothermal case, at equilibrium, then the gas nearer to the bottom of the box must have a higher density since it is under higher pressure. As I explained PV=nRT, entails n/V = P/RT. As pressure is higher, so is the molecular volumetric density n/V. This is also why the barometric density formula can be derived from the barometric pressure formula.
The force is a result of molecules hitting the surface. Pressure is merely force on area. This is purely the result of acceleration due to gravity – that adds a minor component to downward velocity. This has naught to do with density. Molecules disperse randomly in the box.
Rob Ellison: “The force is a result of molecules hitting the surface. Pressure is merely force on area. This is purely the result of acceleration due to gravity – that adds a minor component to downward velocity. This has naught to do with density. Molecules disperse randomly in the box.”
Well, no, not according to the known laws of physics. There ought to be a Boltzmann distribution of height for molecules that have a potential energy m*g*z in a gravitational field. Incidentally, Brownian motion of particles suspended in a liquid also satisfy the barometric formula for density, for the very same reason, and this explains rates of sedimentation. Jean Perrin observed this in 1909 with tree resin particles suspended in water seen through a microscope. See the Am. J. of Physics paper that I linked to earlier.
When there is a higher rate of momentum transfer from the gas molecules to a solid surface, at equilibrium, that can either be because there is a higher rate of collisions (higher molecular volumetric density), or because individual collisions carry more momentum on average (==> higher kinetic temperature), or some combination of both. At least you’re not denying that there is a pressure differential anymore. Now you’re only contesting the validity of the density barometric formula, and of the PV=nRT law for ideal gases.
It remains only for you to grasp the implication of the pressure difference for the molecular density, and for speed distributions at uniform temperature. Hint: since the temperature is constant, there can’t be variation in average molecular velocities as a function of height. This would contradict EKavg = (3/2)kT. The Maxwell distribution of speed only is a function of temperature and molar mass.
http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/maxspe.html
So, I guess that makes for three different law of physics for you still to be skeptical about — all of them standard results of the kinetic theory of ideal gases.
Vy = Vo + gΔt
Ring any bell at all?
http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/weighgas.html
Getting so much wrong – and behaving so bizarrely about it.
‘Einstein’s insight was that a liquid containing a large number of tiny identical particles, such as those observed in Brownian motion, was really no different from a solvent containing solute molecules. True, the Brownian particles were a lot bigger than molecules, but they were buzzing around, and would therefore bounce off the walls of a container, generating pressure. The formal analysis should be the same: the kinetic theory, with equipartition of energy, predicted they would have kinetic energy 1.5kBT. If the concentration of particles varied spatially, they would flow to even it out. ‘
So we don’t get a density difference in the box – which would set up differential temperature – with a different energy density and therefore heat transfer – with a gas at the same temperature. As I said so long ago. And as P-N has denied so many times. Very – very – odd behavior.
My reply bounced up again.
http://judithcurry.com/2014/10/21/ethics-of-communicating-scientific-uncertainty/#comment-643359
This kinematic equation applies perfectly well to individual molecules. You just seem unable to imagine its implication for distributions of speeds at a given height.
You know I have given up reading your replies anyway. I have no idea for a start what this means – which is as far as I want to get.
Gravity adds very marginally to the velocities of particles moving down and reduces velocities of particles moving up. Simples. That’s all it does. The random movement of molecules guarantees that the molecules disperse throughout the space. Just as Einstein said.
Rob Ellison: “You know I have given up reading your replies anyway.”
Why are you replying to posts that you don’t read? Why can’t you even read your own references?
This is what my reference says.
https://watertechbyrie.files.wordpress.com/2014/06/gas-pressure.png
What do you think the implications are? No don’t bother – only rhetorical.
The rhetoric is all your’s.
You reference says: “Take a *typical* gas molecule […]” (My emphasis)
An analysis relevant to our discussion would look at all the molecules, including those that don’t have enough KE to reach the top. This would make no difference for the net rate of momentum transfer to the box, but it would account for the drop-off rate, as exemplified by Perrin’s experiment about Brownian motion under gravity, or the layering of gases according to molecular mass in the mesosphere.
The source says that it can be generalized to many molecules. If we know what the molecular mass is and how many there are. And assuming random motion.
A gas in a box is very different to a liquid with a free surface – or indeed an atmosphere with a fee surface. Boundary conditions again – twee Pierre-Normand.
Rob Ellison wrote:
Your source only is interested in the amount of momentum transfer, not the density or speed distributions. It is free to abstract completely from those later considerations since the net momentum transferred from collision pairs (dp_net = dp_bottom – dp_top) just is the momentum gained by them from the force of gravity dp = m*g*dt, in between the collisions, including the duration of the two collisions. (dt is the time in between two successive collisions on the bottom.)
Since many molecules don’t impact the top at all, then the amount of momentum transfer for them becomes (dp = bottom_dp minus 0 = bottom_dp). But now, explicit consideration of those lazy molecules accounts for the density gradient, since the higher level z you consider, the fewer molecules can make it. For any value for m*g*z, some molecules will hit the bottom of the box with a KE < m*g*z and hence can't get above z following the rebound. The effect may be small, but it is just enough to satisfy the barometric density formula, as it should.
The boundary conditions imposed by a box with elastic walls don’t change the pressure profile of a parcel of gas under gravity that was already in hydrostatic equilibrium with the surrounding atmosphere before you enclosed it in the box (or closed the top of the box).
If you drop a barometer in a deep well, the pressure will increase according to the barometric formula. If you deposit an hermetic circular plug on the top of the well, that is free to move up or down like a piston but is held up against its weight by a rope, the pressure gradient in the well is going to remain exactly the same. The molecules below don’t care if they are bouncing on a solid surface rather than simply meeting an identical expanse of gas at the exact same local pressure at the top of the well — elastic collisions will do just as well as an identically distributed flow of particles in both directions. It’s basically the same boundary conditions. The free moving plug is subjected to the atmospheric pressure (101kPa, say) both from above and from below. If you then seal the plug into place, you have the equivalent of the box. The pressure gradient in the well will remain exactly what it was.
Utter nonsense.
“Utter nonsense”
Utter brain-deadness.
If that’s too complicated for you, consider a rarefied weakly interacting gas — something close to an ideal gas, at equilibrium. Do, you really believe that when a box has height H, then *all* the molecules that impact the bottom automatically have a kinetic energy equal or higher than m*g*H? How would they know how high the top is? And if they *don’t* all have such an energy, what effect might result on the vertical density distribution?
The mean molecular speed at 273K is 446m/s. Dimwit.
Wow – that came out of the Mozilla blue. I can neither confirm or deny…
Generalissimo Skippy :)
Rob Ellison wrote; “The mean molecular speed at 273K is 446m/s. Dimwit.”
This is the arithmetic mean of a Maxwell distribution of speeds that isn’t bounded below. There is no minimum speed. There also are 6*10^23 molecules per mole. So, *some* of them don’t have enough energy to reach the top. (Can you calculate how many?) Yet again you are arguing that *few* molecules that hit the bottom don’t also hit the top. Well, that’s exactly why the density gradient one obtains from the barometric density formula is small. It it the exact same gradient in the box at it is in the atmosphere outside the box (assuming about P ~= 1atm). The gradient is exactly as small as it is, and not any smaller.
What a silly argument.
http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/kintem.html#c4
Rob Ellison: “What a silly argument.”
Your link only contradicts you own claim that the speed distribution (which determines average kinetic energy per molecules) depends on anything but temperature. There is no field where to input values for pressure, volume or volumetric molar density. (Though of, course, average speeds, albeit not average kinetic energy, depends on molar mass.) Haven’t you noticed?
It also shows that the speed distribution isn’t bounded below, as I said. Gravity has a tiny effect on the variation of speed for individual molecules travelling up and down across the height of the box, and a tiny effect on the vertical density gradient. Two opposite tiny effects cancel each other out exactly and thereby *ensure* that average kinetic energy only is a function of temperature, regardless of pressure or density. Physics wins again.
It depends on the boundary conditions as I keep saying.
https://watertechbyrie.files.wordpress.com/2014/06/moles_temp.png
Molecules have a kinetic energy far in excess of the gravitational potential at 273K. A volume at higher density will have greater total kinetic energy – if temperature is constrained to a constant.
The velocities of individual molecules are not constant. They bounce off each other and impart energy. It tends to random distribution of molecules. Molecules will tend to move from regions of higher density to lower.
A tendency that *exactly* cancels out the *tiny effect* of gravity – I might say if I were inclined to rely on extreme verbosity rather than actual numbers. It echoes the silliness of the constant kinetic energy with height – as molecules that don’t have the energy to move to higher and thus the kinetic energy remains *exactly* constant. A *factoid* that the environmental lapse rate refutes.
So unless you have an actual treatment and not just your usual extreme and quite silly verbiage that *exactly* equals out this bizarre and obsessive behaviour thereby *ensuring* that the meme of superior AGW groupthink science – exemplified by the
the extreme silliness of the sympletic manifold of the Hamiltonian pose – survives another encounter with *deniers*. It is humbug of the first order.
Another reply moved up:
http://judithcurry.com/2014/10/21/ethics-of-communicating-scientific-uncertainty/#comment-643617
Nothing from Pekka’s treatment is relevant to gas in a box. We have been through that before.
We have particles moving at an average speed of 466m/s at 273K. The gravitational acceleration down – an infinitesimal increase in downward velocity has nothing to do with density.
Nothing but hopelessly inadequate verbiage do we see.
Rob Ellison: “Nothing from Pekka’s treatment is relevant to gas in a box. We have been through that before.”
Pekka’s treatment applies without modification to the gas in a box since all the same assumptions apply: (1) Maxwell distribution of speeds, (2) stationary density profile, (3) isothermal profile, (4) uniform vertical gravitational field. The finite size of the box makes no difference since collisions on the walls have no effect on the distribution of speeds.
Rob Ellison: “We have particles moving at an average speed of 466m/s at 273K. The gravitational acceleration down – an infinitesimal increase in downward velocity has nothing to do with density.”
Since my references count for nothing, what about the Fronsdal paper that you are so fond of? He also considers a *finite* volume of ideal gas (his ‘box’ is a cylinder of height h) under gravity and derives the same density gradient as everyone else. His equation 3.3 just is the barometric density formula!
Do you have anything new to say – or simply reframe all the old silly arguments.
Pekka’s considers an atmosphere in which molecules are free to move up. That is the point of it.
Similarly the case of Fronsdal – who examines the symplectic manifold of the Hamiltonian in an atmosphere under gravity. Interesting paper – one of thousands I have read.
What you fail to consider is dispersion of highly energetic molecules moving randomly in a box. They are equally as likely to move up as down – they move from areas of relative – and transient – high density to low. Extrapolating from a density difference in the atmosphere to one in a box with merely extreme and tediously obsessive and silly verbiage is laughable.
Rob Ellison: “Pekka’s considers an atmosphere in which molecules are free to move up. That is the point of it.”
Pekka considers an arbitrary level z, and the levels above z and below z that molecules travel to and from in some infinitesimal time dt. In a box, particles are free to move from any level to the next one above or below, except at the very top and bottom levels. But that doesn’t change anything to the stationarity conditions. Fronsdal obviously understands this since he “suppose[s] that the gas is confined to the section z0 < z < z0 + h of a vertical cylinder with base area A and expect the density to fall off at higher altitudes.” Did you miss that?
I’ve found another reference for you to dismiss out of hand. It’s in F. Reif, Fundamentals of Statistical and Thermal Physics; the textbook that was used in our thermodynamics class, 20 years ago. On page 210: “Molecules in an ideal gas in the presence of gravity”
They consider molecules in a container of volume V, with energy E = p^2/2m + mgz, and they derive momentum and velocity probability distributions that are “exactly the same as those obtained […] in the absence of a gravitational field.”
And then: “Finally we can find the probability P(z)dz that a molecule is located at a height z and z+dz, irrespective of its momentum or x and y position components […]. What they find is that P(z) = P(0)*exp(-mgz/kT), the “law of atmosphere”, which applies to an ideal gas in a container of constant cross section. That’s of course the barometer density formula.
And then: “Finally we can find the probability P(z)dz that a molecule is located at a height between z and z+dz, irrespective of its momentum or x and y position components […]”
I had omitted the “between”.
So right – the box is just like the atmosphere – because you can look at it as finite elements – except it has a top – at say 1m?
You are mistaking the structure of the atmosphere – which itself is vastly different to Pekka’s simplifying analysis – for the gas in the box.
Pressure exists in the atmosphere because of the weight of the overlying atmosphere and density varies with temperature and pressure. One is hydrostatics and not the result of molecular dynamics. The other is radiative and convective in a real atmosphere.
According to the kinetic gas theory every component of the velocity is distributed in accordance to the
Maxwell-Boltzmann distribution. Combining that with altitude dependent density profile ρ(z) and
temperature T(z) we can write…
Expressed in other words, the above derivation shows how it’s possible that the Maxwell-Boltzmann
distribution of the same temperature can be valid at all altitudes in spite of the fact that the vertical motion of
the molecules is affected by the gravitation. The result is dependent on the mathematical form of the
Maxwell-Boltzmann distribution through the equation (5), whose simple form is true specifically for the
Maxwell-Boltzmann velocity distribution of the vertical velocity. In this specific case the gravitational
acceleration, the density profile and the influence of the initial vertical velocities of the molecules combine to
maintain the stationary density and temperature profiles.
Now this seems to mean that assuming a stable temperature and density profiles – the stationarity conditions – the molecules have a Boltzmann distribution at every level which is the result of the density and temperature profiles and the initial velocities under gravity.
It is still not relevant to the box – where molecules bounce off the top. As much as you might imagine that the box is a model of the atmosphere – or the atmosphere the box – it is still far from demonstrated. Nor is – by the way – the constancy of average kinetic energy with height – or that the symplectic manifold of the Hamiltonian is more than a bizarre pose.
“Now this seems to mean that assuming a stable temperature and density profiles – the stationarity conditions – the molecules have a Boltzmann distribution at every level which is the result of the density and temperature profiles and the initial velocities under gravity.”
You are of course ducking as best as you can the fact that both Frønsdal (your hero) and Reif (standard textbook statistical mechanics) derive the barometric density formula for an ideal gas in thermodynamic equilibrium in a box; just as you ducked the fact that it also applies to Brownian particles suspended in a liquid, both theoretically and observationally.
You mean Professor Frønsdal at UCLA – who wrote a paper that discussed the symplectic manifold of the Hamiltonian for an atmosphere under gravity?
I think you would find if you had any clue that he doesn’t derive the barometric formula.
Other than that I don’t have a clue what you think you are saying – and nor do you it seems.
http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/barfor.html#c3
Yes – we understand the barometric formula and it’s derivation. It is not a freaking box.
Rob Ellison: “You mean Professor Frønsdal at UCLA – who wrote a paper that discussed the symplectic manifold of the Hamiltonian for an atmosphere under gravity?
I think you would find if you had any clue that he doesn’t derive the barometric formula.”
Equation 3.3 in his paper *is* the density barometric formula. And he most definitely discusses a finite cylindrical container, as my quotation show, and not just only the open terrestrial atmosphere. He also finds a density gradient in his cylindrical ‘box’.
The barometric formulas for density and pressure just are the consequences of parcels of gas having some weight both inside and outside of close containers that are placed in a gravitational field (or that are being accelerated, as in Frønsdal’ centrifuge where there is the equivalent of a radial inhomogeneous gravitational field (in the reference frame of the rotating centrifuge). This is another instance of a density and pressure gradient in an enclosed space that you ignore.
It makes no difference that a gas column be open on top as it is for an atmospheric column. The pressure gradient under conditions of hydrostatic equilibrium still is the result of the weight of the air parcels from one vertical level to the next. Kinetic theory also applies inside of boxes and yields the very same result at the molecular level (as discussed in Reif’s statistical and thermal physics textbook, and in this statistical mechanics textbook for chemists that I had linked to earlier.)
You still have no account of the buoyant force applied by air on the surface of a helium filled balloon placed inside of a sealed container. Saying that it is *calculated* as the weight of the displaced air minus the weight of the balloon+helium, as you did, doesn’t address the question *how* the force is physically applied to the balloon. The existence of a vertical pressure gradient in the container accounts for the buoyant force. If there were no pressure gradient, the balloon would fall down through the denser air. There would be no buoyant force. This is just from Newton’s second law.
And then, when a pressure gradient is granted, it trivially follows that there also will be a density gradient, from PV=nRT. This is, as you are so fond to say, “baby physics”.
Rob Ellison: “You mean Professor Frønsdal at UCLA – who wrote a paper that discussed the symplectic manifold of the Hamiltonian for an atmosphere under gravity?
I think you would find if you had any clue that he doesn’t derive the barometric formula.”
Equation 3.3 in his paper *is* the density barometric formula. And he most definitely discusses a finite cylindrical container, as my quotation show, and not just only the open terrestrial atmosphere. He also finds a density gradient in his cylindrical ‘box’.
Equation 3.3 is *not* the barometric formula – the ‘cylinder’ is open ended and is not remotely the box and he uses finite volumes to evolve the Hamiltonian for the atmosphere under gravity. No one doubts that a pressure gradient exists in the atmosphere – the idea of gravitational potential is one of the simpler building blocks.
The pseudo interpretation of Frønsdal is simplistic to the point of idiocy. This paper I introduced as an interesting but mathematically difficult paper that suggests that there is a gravito-thermal effect in an open ended cylinder. The latter is merely a simplifying assumption allowing the math to be evolved for a finite volume – isoptropic in the horizontal direction.
Frankly I stopped reading after the quote above. There is seemingly an obsession with *proving* the barometric formula. Really not necessary, relevant or all that interesting. .
http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/barfor.html#c2
Let’s fix the formatting.
Rob Ellison: “You mean Professor Frønsdal at UCLA – who wrote a paper that discussed the symplectic manifold of the Hamiltonian for an atmosphere under gravity?
I think you would find if you had any clue that he doesn’t derive the barometric formula.”
Equation 3.3 in his paper *is* the density barometric formula. And he most definitely discusses a finite cylindrical container, as my quotation show, and not just only the open terrestrial atmosphere. He also finds a density gradient in his cylindrical ‘box’.
Equation 3.3 is *not* the barometric formula – the ‘cylinder’ is open ended and is not remotely the box and he uses finite volumes to evolve the Hamiltonian for the atmosphere under gravity. No one doubts that a pressure gradient exists in the atmosphere – the idea of gravitational potential is one of the simpler building blocks.
The pseudo interpretation of Frønsdal is simplistic to the point of idiocy. This paper I introduced as an interesting but mathematically difficult paper that suggests that there is a gravito-thermal effect in an open ended cylinder. The latter is merely a simplifying assumption allowing the math to be evolved for a finite volume – isoptropic in the horizontal direction.
Frankly I stopped reading after the quote above. There is seemingly an obsession with *proving* the barometric formula. Really not necessary, relevant or all that interesting. .
http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/barfor.html#c2
Why do things float? Oh for God’s sake – lmfao.
Rob Ellison: “Equation 3.3 is *not* the barometric formula”
Equation 3.3: rho(x, y, z) = exp(-mgz/kT)
This most definitely is the barometric formula for density as a function of elevation z.
http://en.wikipedia.org/wiki/Barometric_formula#Density_equations
In Pekka’s paper, the “standard barometric formula” is:
rho(z) = exp(-mgz/kT)
“the ‘cylinder’ is open ended and is not remotely the box.”
No it is not open ended. “[…] the gas is confined to the section z0 < z < z0 + h of a vertical cylinder with base area A and expect the density to fall off at higher altitudes.” (My emphasis)
‘Confined’ doesn’t mean ‘open ended’. The height of the cylinder is ‘h’. The centrifuge also is a finite enclosure. The derivation in Reif also considers a monatomic gas in a confined container of volume V. It changes nothing essential to the derivation that the box be closed or open at the top with the weight of the atmosphere pressing in it. Elastic collisions with the top of the box preserve the speed distribution. The z component of the velocities simply are reversed rather than there being a compensating flux in the other direction. Merely closing the lid of a bottle doesn’t change the pressure, temperature or density anywhere in the bottle.
It is a confusion of energy terms for the barometric formula. It is in any inconsequential.
The finite volume is open on both ends – that is the nature of the formulation. Molecules move between levels. In general in the atmosphere and the box there is random motion. This is a difference – not one I can be bothered any longer to discuss.
The absurdity of going over this again and again with extreme verbiage only and a misapplication of papers barely understood seems pretty obvious to me.
I have long since ceased to take this with anything but disdain. What is the point? Other than someone with a less than distinguished understanding of basic physics to self importantly pontificate how physics rules.
Rob Ellison: “The finite volume is open on both ends”
OK. So when Frønsdal writes that “the gas is confined to the section z0 < z < z0 + h of a vertical cylinder" he really means to say that the cylinder is "open" on both ends. That's a strange way for him to express himself.
Yes – it takes a representative element in the column.
Rob Ellison: “The finite volume is open on both ends”
And then he immediately proceeds to say: “We may consider this an isolated system with fixed mass and fixed extension.”
Read it by all means – http://www.mdpi.com/1099-4300/16/3/1515
But this is utter madness.
Rob Ellison: “But this is utter madness.”
Agreed.
We take a finite element in a cylinder with base A – on the basis of isotropy in the horizontal direction – and density decreasing in the atmosphere? So we can assume it is a box filled with gas with a lid and a base? In a paper on the atmosphere in gravity – and the gravito-thermal effect that emerges from the Hamiltonian? Is that really the argument?
Eh…
“We take a finite element in a cylinder with base A – on the basis of isotropy in the horizontal direction – and density decreasing in the atmosphere? So we can assume it is a box filled with gas with a lid and a base? In a paper on the atmosphere in gravity – and the gravito-thermal effect that emerges from the Hamiltonian? Is that really the argument?”
No. Your idea is that enclosing a gas in a box will lead to the pressure and density gradients to disappear as the gas molecules disperse uniformly in the space of the box. This is an idea that Frønsdal clearly doesn’t endorse.
Another example is that he assumes the law ‘p/rho = RT’, and the laws of hydrodynamics, to hold for the gas in the centrifuge, and for there to be a radial pressure gradient due to the radial ‘gravitational force’ (so interpreted on the basis of Einstein’s equivalence principle). Rho just is n/V, so ‘p/rho = RT’ just is ‘PV = nRT’ as I have pointed out at least twice before (*).
If there is a pressure gradient, then there also ought to be a density gradient unless the gravito-thermal effect on the temperature profile would be so large as to offset exactly the effect from pressure on the density gradient. However, the gravito-thermal temperature gradient is supposed (according to Frønsdal) to be approximately equal to the adiabatic lapse rate, so there is no chance of that.
Of course, Frønsdal correct assumption that the laws of hydrodynamics (including hydrostatics) must hold in the confined space of the centrifuge also directly contradicts your claims about the ‘box’. Equilibrium under gravity doesn’t entail uniform pressure or density.
(*): http://judithcurry.com/2014/10/21/ethics-of-communicating-scientific-uncertainty/#comment-643335
‘No. Your idea is that enclosing a gas in a box will lead to the pressure and density gradients to disappear as the gas molecules disperse uniformly in the space of the box. This is an idea that Frønsdal clearly doesn’t endorse.’
This is something that Frønsdal clearly doesn’t address. Random motion leads to energetic molecular dispersion to fill the space uniformly. All you do is repeat that a density gradient exists in the atmosphere. So what. And such a trivial obsession at that.
Assuming a centrifuge with an outer radius of 10cm and an inner radius of 1 cm spinning at 100 rev/sec, the predicted lapse rate temperature gradient is shown in figure 2. It is assumed that the centrifuge is surrounded by a heat bath held at 15C and that the inner bore is a vacuum. Using air in the centrifuge the temperature at the inner radius is then predicted to be 2 degreec C. lower than the outer radius. Such a large effect would be easy to measure with thermocouples.
http://clivebest.com/blog/?p=4101
The gravito-thermal experiment using huge simulated gravity fields. Is this enough to make a difference to truly random motion? Frankly I am beyond caring.
Rob Ellison: “Random motion leads to energetic molecular dispersion to fill the space uniformly. All you do is repeat that a density gradient exists in the atmosphere. So what. And such a trivial obsession at that.”
You can’t keep track of your own argument. I am arguing that the density and pressure gradients exist *both* in the atmosphere and in the box for the very same kinetic-theoretical reasons (and I supplied plenty of standard textbook references in support of that claim). The core argument is that for any given vertical elevation, not all molecules have enough kinetic energy to climb above it; (and, in the isothermal case, the distribution of speeds must be the same at all levels since the Maxwell distribution of speed only is a function of temperature.)
Denying that there is a pressure gradient in the box at equilibrium also violates the condition of hydrostatic equilibrium. If would cause any air parcel in the box to free fall. But you don’t believe in air parcels because you can’t visualize them. All right then. The lack of a pressure gradient would also cause a helium filled balloon placed in the box to fall through the denser air. This is as clear a reductio ad absurdum of your heterodox theory as can be.
You are kidding right?
We just keep going back to the usual nonsense. The molecules are interchangable – they collide and impart momentum and there is enough energy a 0K to get all the molecules at an average speed of 446m/s across and back in a 1m box in millseconds. The molecules spread out to fill the box is the dominant process. That’s what gas does. Where there are no barriers it keeps spreading out until it reaches the limit of it’s kinetic energy or hits another molecule. Oddly enough – this has nothing to do with the real structure of the atmosphere.
The *core* argument is utter BS – and rather than anything substantial all we get is extreme verbiage and vague handwaving at *standard texts* and misapplied physics jargon. .
Let’s see
1. Air ‘parcels’ are normally invoked to describe convection – buoyancy. They are quite easily visualised. I wander lonely as a cloud. Gases dispersing to fill the space are not unstable.
2. This is interesting. A helium balloon in a box of gas would sink. It wouldn’t. It would float because it is lighter than the displaced air we assume is in the box.
These are utterly physically bizarre ideas but at least he can spell heterodox.
.
Help – I’m in some sort of infinite regress. It is assuming the dimensions of a cosmological argument and I don’t know how to get out.
I know – I will just ignore the twerp.
As usual…
http://judithcurry.com/2014/10/21/ethics-of-communicating-scientific-uncertainty/#comment-644163
“In science, buoyancy /ˈbɔɪ.ənsi/ is an upward force exerted by a fluid that opposes the weight of an immersed object. In a column of fluid, pressure increases with depth as a result of the weight of the overlying fluid. Thus a column of fluid, or an object submerged in the fluid, experiences greater pressure at the bottom of the column than at the top. This difference in pressure results in a net force that tends to accelerate an object upwards. The magnitude of that force is proportional to the difference in the pressure between the top and the bottom of the column, and (as explained by Archimedes’ principle) is also equivalent to the weight of the fluid that would otherwise occupy the column […]” Wikipedia — buoyancy
http://upload.wikimedia.org/wikipedia/commons/thumb/3/3c/Pressure_distribution_on_an_immersed_cube.png/1280px-Pressure_distribution_on_an_immersed_cube.png
Any object, wholy or partially immersed in a fluid, is buoyed up by a force equal to the weight of the fluid displaced by the object.
— Archimedes of Syracus
If an object weighs the same as the displaced water – it has neutral buoyancy. The downward force of the mass times acceleration is exactly equal to the upward force exerted by the fluid on the object. It the objects mass is less than the mass of displaced water – the downward force is less than the upward force and the object floats. So no the helium balloon won’t sink in a box of air.
You can’t explain buoyancy except in terms of relative density. Does Wikipedia try to do this?
I was going to ignore the twerp – but discussing buoyancy solely in terms of hydrostatics is just way too silly.
Rob Ellison: “You can’t explain buoyancy except in terms of relative density. Does Wikipedia try to do this?”
My straightforward and uncontroversial claim, which you bizarrely dispute, is that a pressure gradient is required for the buoyant force the be effective. In an ordinary fluid, when there is a pressure gradient, there is a density gradient as well. Since air compresses roughly according to PV=nRT, when P increases while T is held roughly constant, n/V, which is the molar volumetric density, must increase as well.
The density gradient accounts for the buoyant force on the immersed object at the molecular level, since while the temperature is uniform the distribution of speeds is invariant with height, and the variation in the rate of momentum transfer to the surface of the object (the force from the molecular impacts) is the result of a higher rate of collisions where the fluid is denser.
That I have to explain to a hydrological engineer that the buoyant force in a fluid depends on there being a vertical pressure gradient (the condition of hydrostatic equilibrium) boggles the mind. For you to insist that there would still be a buoyant force even in the absence of a vertical pressure gradient betrays a surprising lack of understanding of hydrostatics 101, and indeed, of high-school physics.
No – the new and quite silly *thought experiment* was that a helium balloon would fall to the bottom in container in which molecules dispersed randomly.
Rob Ellison: “No – the new and quite silly *thought experiment* was that a helium balloon would fall to the bottom in container in which molecules dispersed randomly.”
You are claiming that as a result of the random dispersion of the molecules, there can’t be any pressure or density gradient in the box. If that’s the case, then a helium balloon placed in that box would fall, since it is subjected to gravity and there would be no pressure differential in the air around the balloon to hold it up. You wave your hands vaguely towards the word ‘buoyancy’, but you are denying the very condition for there to be any buoyant force at all — i.e. the vertical pressure gradient, which is an essential part of the explanation of the buoyant force.
Depends on initial and boundary conditions – as I keep saying – and not silly verbiage based on silly little thought experiments.
http://en.wikipedia.org/wiki/Brownian_motion#mediaviewer/File:Brownian_motion_large.gif
Rob Ellison wrote: “Depends on initial and boundary conditions – as I keep saying – and not silly verbiage based on silly little thought experiments.”
What kind of strange boundary conditions would cause a solid object immersed in a fluid to be subject to a positive buoyant force when there is no pressure gradient at all within the fluid. You are just waving your hands furiously… again.
What sort of strange boundary conditions result in Brownian motion in a box? It is all utterly irrelevant and pointless – at any rate – and intended only to mislead and obfuscate.
The original comment in this incredibly long string of trivial nonsense was a molecules in the box animation from Wikipedia and a pose about symplectic manifolds of the Hamiltonian. It is all a pose in support of meaningless drivel – obviously almost endlessly it seems.
Rob Ellison: “What sort of strange boundary conditions result in Brownian motion in a box?”
Of course, as I pointed out before. (1) Jean Perrin observed through a microscope in 1909 that Brownian particles in a bottle also distribute vertically according to the barometric density formula — “law of the vertical distribution of an emulsion.” He discusses it in his 1926 Nobel Lecture. (2) George Stokes demonstrated why it has to be so:
“Consider, for instance, particles suspended in a viscous fluid in a gravitational field. Gravity tends to make the particles settle, whereas diffusion acts to homogenize them, driving them into regions of smaller concentration. Under the action of gravity, a particle acquires a downward speed of v = μmg, where m is the mass of the particle, g is the acceleration due to gravity, and μ is the particle’s mobility in the fluid. George Stokes had shown that the mobility for a spherical particle with radius r is \mu=\tfrac{1}{6\pi\eta r}, where η is the dynamic viscosity of the fluid. In a state of dynamic equilibrium, the particles are distributed according to the barometric distribution.” (My Emphasis) Wikipedia — Brownian motion.
(3) Reif, Fundamentals of Thermal and Statistical Physics, p.210, demonstrates it for particles of an ideal monatomic gas in a container.
And (4) the online simulation on this page illustrates this neatly, as well as providing the kinetic-theoretical explanation:
http://in-the-sky.org/physics/balls.php
Finally, however much smoke you may be generating in order to obscure and dismiss all evidence, and ignore the logical consequences of your own claims, you haven’t begun to explain how a helium filled balloon can be held up against gravity if the pressure of the fluid it is immersed into is uniform. You haven’t come to grip with Newton’s second law yet.
Brownian motion is illustrative of the diffusion process. That particles tend to settle and don’t because of random molecular motion.
As far as the balloon is concerned – perhaps he should move onto Newton’s third law. Although the use of a further *thought experiment* to demonstrate the validity of a former *thought experiment* is an example of extreme verbiage rather than a rigourous treatment.
Much like the average kinetic energy being *exactly* equal at every level or the Wikipedia gas in a box animation being illustrative of the symplectic manifold of the Hamiltonian. Just verbiage with no significance.
Oh – and bouncing balls. A *simulation* involving balls and kinetic and potential energy in a gravity well. Repeated several times and gloated over by FOMBS. So phucking what is the relevant question.
Rob Ellison, this latest post of yours is especially incoherent, even by your own standards. I take it you simply can’t provide any justification for your astounding suggestion that a buoyant force is applied to an immersed solid even when the pressure in the fluid doesn’t vary with height. I guess this conversation is over until you have something intelligible to add.
I may post some calculations later about the magnitude of the two gravitational effects on falling molecules (i.e. (1) the slow-down of rising molecules, and (2) the ‘drop-off effect’ on the speed distribution) and how they exactly cancel each other. This may be more intuitive than the direct derivation of the barometric formula, and of isothermality, from the Boltzmann law of distribution of energy at equilibrium, or Pekka’s derivation of the barometric formula on the assumptions that the vertical density profile is stationary, and the temperature doesn’t vary with height, although those two equivalent derivations ought to be proof enough of my initial claim.
“Frankly I stopped reading after the quote above. There is seemingly an obsession with *proving* the barometric formula. Really not necessary, relevant or all that interesting.”
It most definitely is relevant to your claim that the density in the box is uniform. This is what the whole discussion has been about. You wish to deny that there is a density drop-off with height that can compensate for the effect of gravitational acceleration on the speed distributions. This is the essence of your second main objection. The only other relevant claim of yours is that a uniform temperature is inconsistent with there being a density gradient when the average kinetic energy doesn’t vary with height. This is exemplified by your discussion of the two cylinders; which you recently dismissed as being irrelevant too. You now have come to say of both of your own main arguments that they are irrelevant. They aren’t irrelevant. They are hopelessly flawed.
Let me add that the fundamental mechanical-statistical reason why the velocity and spatial probability distributions for ideal gas molecules are what they are (the M-B velocity distribution, and the barometric density formula) is because the probabilities to find the molecules to have some particular velocity or spatial location is proportional to the number of cells in phase space that a molecule with some energy between E and E+dE can occupy wile having those velocities or locations. It is this and only this that governs the distributions at thermodynamical equilibrium.
Since the cells in phase space that correspond to higher levels in the gravitational field (higher values for z) distribute over a smaller range of velocities along the three dimensions x, y, and z, for a given range of total energy PE+KE between E and E+dE, the probability to find molecules at the higher levels is correspondingly smaller. It’s only in the absence of external force fields that molecules distribute randomly in space. You are misconstruing this special case as a general principle of the kinetic theory of ideal gases.
http://judithcurry.com/2014/10/21/ethics-of-communicating-scientific-uncertainty/#comment-643368
“So we don’t get a density difference in the box – which would set up differential temperature – with a different energy density and therefore heat transfer – with a gas at the same temperature. As I said so long ago. And as P-N has denied so many times. Very – very – odd behavior.”
We can easily get a density difference without a temperature difference — just as two cylinders can contain gases at two different pressure and yet have the same temperature.
Einstein makes only one single mention of the force of gravity in his Brownian Motion paper:
“…at least when the force of gravity (which does not interest us here) is ignored.”
And the effect Perrin experimentally verified four years later isn’t rendered any less real by your desire to ignore it.
“George Stokes had shown that the mobility for a spherical particle with radius r is \mu […], where η is the dynamic viscosity of the fluid. In a state of dynamic equilibrium, the particles are distributed according to the barometric distribution.” — Wikipedia
Rob Ellison: “[…] – molecules tend to diffuse to lower density – and you haven’t disproved Brownian motion or kinetic theory. I won’t hold my breath.”
Brownian motion makes my point.
Rob Ellison: “…with some average KE.”
I may have misread you.
In any case, your claim about the need for a reduced average KE at higher pressure in order to maintain the same temperature at the gas/solid boundary despite the higher collision rate is independent of any claim about a gravito-thermal effect or the lack thereof. Many physicists would either deny the gravito-thermal effect, or be agnostic about it, but I think few would agree that EKavg can be any different from (3/2)kT just because the pressure is higher, of agree that the the second cylinder could have twice the molar amount and nevertheless the same U, (same V and same T).
Again the cylinder is irrelevant – the impossible scenario was the density in a box and the temperature at the surfaces.
Gravito-thermal is a result based on the symplectic manifold of the Hamiltonian – FOMBS other throwaway posturing – in a recent peer reviewed paper. .
… It is precisely this latter claim that your two cylinder case dramatizes. It was also your own thought experiment (though you adapted it from Flynn)…
I discussed changing densities – energy density – and energy flows – mostly in relation to how temperatures are measured.
https://watertechbyrie.files.wordpress.com/2014/06/heat-transfer.png
For instance. Repeating things endlessly and misrepresenting me constantly is not all that endearing.
“Repeating things endlessly and misrepresenting me constantly is not all that endearing.”
What did I misrepresent? You didn’t issue any correction. You only said, concerning your very own claims (about the two cylinders), that they weren’t *relevant*. And then you immediately reassert one of them in another setting (the gas filled box with a density gradient). You claim is that temperature depends on density, and collisions rate on the solid surface, and not just average kinetic energy. I didn’t misrepresent this, did I?
Rob Ellison: “Repeating things endlessly…”
Maybe you should stop repeating endlessly vague generalities that nobody takes issue with and you should instead start responding to specific objections to your claims, or retracting those that are false.
Huh? I assert that there are no density differences and have endlessly.
And for the rest it depends on the boundary conditions – as I keep saying.
https://watertechbyrie.files.wordpress.com/2014/06/moles_temp.png
My response jumped here, for some reason:
http://judithcurry.com/2014/10/21/ethics-of-communicating-scientific-uncertainty/#comment-643045
captdallas: “P-N, as I have said a number of times with a small effect you need a large volume.”
What small effect are you talking about? Gravity is acting on the two cylinders equally. Rob’s problem with the two cylinders has nothing to do with gravity. He is claiming that the internal energy is the same in both cylinders. I am claiming that it is twice as large in the second (2atm). He is claiming that the average kinetic energy is twice as much in the first cylinder (1atm). I am arguing that it is exactly the same. He is arguing that twice the rate of collisions with the same average kinetic energy per molecules would lead to the cylinder walls warming up to a higher temperature. I am arguing that only the speed distribution (and KEavg) determines the temperature.
“That said, average KE would be equal and the 2PV tank would have twice the internal energy.”
Yes, I am glad to see you agreeing with this. This will simplify the discussion about the vertical speed distribution in a gravity field in the next open thread, if you decide to get involved.
In relation to Professor Christian Fronsdal, are you seriously appealing to authority, and calling for consensus?
Credibility has nothing to do with fact. Apparently 97% of some odd grouping of supposed scientists believe the world is warming due to CO2 in the atmosphere. Maybe they also believe, as did the eminent physicist Lord Kelvin, that the Earth cannot possibly be older than 20,000,000 years.
So who are we to believe? Kelvin, Fronsdal, or the supremely unschooled and unqualified Mike Flynn? You choose who to believe. I choose me in relation to the existence of the gravito-thermal effect – surprise, surprise!
A reference to someone who actually has credibility is not an appeal to authority. An appeal to Flynn is pissing in the wind.
“Sigh”.
I was trying to point out that no one cares how fast the air molecules move – that isn’t a significant contributor to gaseous air heat transfer.
I think what is happening is someone is trying to redefine thermodynamic equilibrium.:
http://en.wikipedia.org/wiki/Thermodynamic_equilibrium
“Two systems are in thermal equilibrium when their temperatures are the same”.
If the air in the tank is the same temperature as the tank – it isn’t transferring energy to the tank. Period, end of story.
PA: “He has a number of problems.
The ideal gas equations apply to the whole volume of gas, outside the cylinder as well as inside – so if you remove gas you have consider both populations.”
Yes, if both populations are at all time at (very nearly) the same pressure and temperature.
In the original setup we were merely considering the rather simpler case of two different cylinders that had had time to get in thermal equilibrium with the surrounding. Rob E. claimed that if they had the same temperature and different pressures then the second cylinder could not have the same average kinetic energy per molecules (as seems to agree captdallas) and, it had to have the same total internal energy. The two ideas are “problems” that he has, as you say.
Rob E. also has an argument that appeals to the idea of compressing (“stuffing”) a second mole of gas (or whatever amount is double the content of the first cylinder) adiabatically into the second cylinder (that also has one mole in it, initially) and then letting the excess heat that has resulted from the compression work dissipate away. His view is that the second cylinder ends up with the same total internal energy as the first cylinder with twice the number of molecules, and hence half the average KE energy per molecule.
“PV = nRT applies to the tank both before and after gas removal. If T is constant and V is constant (the case of a very slow leak otherwise known as isothermal expansion)
P ∝ n”
Agreed.
Also mutually inconsistent are those three claims, where n1 and n2 are the molar quantities of oxygen in the two cylinders, and U1 and U2 are the total internal energies.
(1) Both cylinders have the same total internal energy U1 = U2
(2) The average thermal energy per mole in the two cylinders are U1/n1 and U2/n2 respectively.
(3) n2 = 2*n1
Yet, you believe that the three of them can be true at the same time. As you had written:
“Take these cylinders of oxygen again [see (*) below] – they are in local thermodynamic equilibrium. This means that the total thermal energy is the same in all cases. But there are more molecules in the compressed gas – which means that the average kinetic energy per molecule is less.”
(*) Reference to this post by Mike Flynn:
http://judithcurry.com/2014/10/21/ethics-of-communicating-scientific-uncertainty/#comment-640729
Note also that Rob E. defines EKavg per molecules as U_total/N even for diatomic or polyatomic molecules, though that’s not the issue here.
http://judithcurry.com/2014/10/21/ethics-of-communicating-scientific-uncertainty/#comment-640851
Captdallas: “Now (PV)2=2(PV)1. so for the same temperature in both containers, #2 would have to lose energy to the environment.”
Not true. The second container has twice the molecules with the same average kinetic energy and hence twice the total kinetic energy. Hance it is to be expected that, at the same temperature, PV for the second container (P = 2atm) is twice as much as PV for the first container.
Rob Ellison: “Much redefining of the terms of the silly thought bubbles. .
1. That KE average is constant with height. It isn’t in the Earth’s atmosphere.
2. That energy transfer is a function only of average kinetic energy. It isn’t. It is a function of the number of molecules and kinetic energy transferred in collisions.”
Temperature variations with height isn’t relevant to your discussion of the two cylinders. There is no net heat transfer at thermal equilibrium, which was assumed.
I think you may have finally realized that your claims and arguments regarding the two cylinders containing oxygen at 1atm and 2atm (same volume, same temperature) were silly and invalid. This would explain why you are desperately trying to change the subject and persist in disputing imaginary claims that I never made.
‘According to the equipartition principle the total energy of a mole of water vapor is
Utotal = Utrans + Urot + Uvib = 3/2 RT + 3/2 RT + 3RT = 6RT.’
This is the total kinetic energy – and temperature is involved in one. But there is also intermolecualar potential energy that increases with compression.
Nor do the statistical energies explain the physical mechanisms of the molecular dynamics in play.
Simple declarations repeated endlessly notwithstanding.
Rob Ellison: “But there is also intermolecualar potential energy that increases with compression.”
You need a whole lot of compression for that. The specific heat capacity for air (isobar) at 300°K, for different pressures is:
c_p kJ/(kg*K)
1atm: 1,007,
10atm: 1,021,
100atm: 1,158
1000atm: 1,303
So for your compressed 2atm cylinder, it makes a less than 2% difference.
Also, the effect is actually the *opposite* from what you need. You had insisted that the compressed cylinder has the *same* total internal energy at twice the pressure, and hence twice the molar amount. This was your ground for inferring that it has half the KEavg (which you calculated as KEavg = U_thermal/N).
But since the heat capacity is for kilograms of air, the compressed cylinder must have twice this capacity, and this suggests (approx) twice the thermal energy U. This is not what you wanted. The fact that it increases *even more* with pressure isn’t very consequential for your claim, but it hardly helps it.
Ref: http://www.thermopedia.com/content/553/
There is obviously potential energy in compressed gas – although the physical dynamics are a bit obscure.
Bizarre thought bubbles and extreme verbiage, misrepresentation and obfuscation notwithstanding.
Rob Ellison wrote: “There is obviously potential energy in compressed gas – although the physical dynamics are a bit obscure.”
Yes, obviously. That’s what the specific heat capacities at high pressures show. But you had insisted that a compressed gas has the *same* total thermal energy despite the larger molar amount, remember?
“Take these cylinders of oxygen again – they are in local thermodynamic equilibrium. This means that the total thermal energy is the same in all cases. But there are more molecules in the compressed gas […]”
“A compressed gas and an uncompressed gas in thermodynamic equilibrium have the same temperature and the same total thermal energy.”
“P-N confuses the mean energy of the particle – which is related to temperature of course to the total energy of the system – with the total energy of the system. Two gas cylinders at different pressures at thermal equilibrium with the surroundings will have the same thermal energy. The average will be less – you get that by dividing by N.”
However, if the uncompressed 1atm cylinder has 1kg of air and the compressed 2atm cylinder has 2kg of air, then you are saying that they have the same total thermal energy. You explained that this was the reason why the 2atm cylinder doesn’t warm up more through molecular collisions despite the higher pressure: it has the same temperature because the higher pressure is compensated by a lower average EK. Twice the molar amount, half the average EK, hence the same total thermal energy.
This seems to conflict with the fact that the energy required to warm both cylinders by 10°C, say, would be twice as much for the second container, since the specific heat capacity (at constant volume) is
c_v = 0.72kJ/(kg*K).
The first would gain Q*c_v*1kg = 72kJ and the second one Q*c_v*2kg = 144kJ. But since they still have the same volume and same temperature, you seem committed to say that they have the same internal energy. Something has to give.
‘Macroscopically, the thermal energy of a system at a given temperature is proportional to its heat capacity. .’
Wikipedia
But there are obvious differences in molecular dynamics. Despite progressive science denialism of the worst kind.
Rob Ellison wrote: “‘Macroscopically, the thermal energy of a system at a given temperature is proportional to its heat capacity. .’
Wikipedia
But there are obvious differences in molecular dynamics. Despite progressive science denialism of the worst kind.”
But they are the *same* gas in both cylinders. And I quoted figures about the heat capacity of air that show that you can’t have more than a 2% difference from 1atm to 2atm for air. Yet if you add nearly twice the amount of heat to the compressed cylinder, both cylinder will make the same *gain* in total thermal energy (according to you), since they will end up at the same temperature.
So you have that 72kJ added to the first cylinder, and ~144kJ (+-2%) added to the second cylinder, produce the exact same thermal energy gain dU in both cylinders. That’s a huge violation of the law conservation of energy.
P-N’s accounting is deficient – work it out for yourself using the correct assumptions and not made up cr@p.
But it is not even a complete list of energies.
Robert Ellison: “But it is not even a complete list of energies.”
I’m using c_v, hence heating the cylinders at constant volume. There is no work done. I am assuming that the heat dQ being added up to the gas going into the thermal energy of the gas. Where else could heat expended to warm a gas go if not into its thermal energy? Has the gas any other way to store energy that doesn’t count as thermal energy? Will air at 2atm being warmed up by a mere 10°C turn into a plasma?
To the external environment – and yes – duh. All that motivated reasoning and he still gets it wrong.
“To the external environment – and yes – duh. All that motivated reasoning and he still gets it wrong.”
There is a cylinder filled with 2kg or air pressurized at 2atm. We warm it at constant pressure. The heat capacity is c_v = 0.72kJ/(kg*K), which means we that 0.72kJ is required to warm 1kg of air 1°K. So, to raise the temperature of 2kg of air ten degrees higher, one must expend 14.4kJ. This is twice as much as required to raise the temperature of the gas in the first cylinder. But in the second case half the energy is lost to the external environment. How?
‘Kyoto has permitted different groups to tell different stories about themselves to themselves and to others, often in superficially scientific language. But, as we are increasingly coming to understand, it is often not questions about science that are at stake in these discussions. The culturally potent idiom of the dispassionate scientific narrative is being employed to fight culture wars over competing social and ethical values. Nor is that to be seen as a defect. Of course choices between competing values are not made by relying upon scientific knowledge alone. What is wrong is to
pretend that they are.’ http://www.lse.ac.uk/researchandexpertise/units/mackinder/pdf/mackinder_wrong%20trousers.pdf
There were 2 thought bubbles the moon molecular projectiles and the box animation form Wikipedia demonstrating the symplectic manifold of the Hamiltonian.
It is all such nonsense – as a justification for prattling about the inferiority of denier science. In this pseudo scientific meme there is not room for even the slightest doubt or the smallest concession. It is a rhetorical game in which no points are ever to be allowed to the science denier.
Ethics of communication don’t enter the equation – it is all Allinsky with the stories told each other superficially in the objective idiom of science. The groupthink dynamic poisons all communications – scientific, legal and journalistic.
In the FOMBS construct it is the symplectic manifold of the Hamiltonian – therefore capitalism is evil and democracy has failed – all moral STEM professionals and the Pope agrees and it is a pleasure to educate CE denizens in the absence of a gravito-thermal effect, the weak statistics of cycles and the precision of Navier-Stokes solutions for turbulent flow over a wing. There is not a chance in hell that any real discourse is possible with these people. .
Well…
I apply engineering analysis to global warming theory and there is no way to make the numbers add up.
1. Can’t reach doubled CO2. 550-577 PPM appears to be a practical limit.
2. Most of the resisting forces increase by the difference from the equilibrium level, which means the higher the CO2 forcing the less effect it has.
3. The feedback to CO2 is at best neutral – that makes CO2 a passenger on the bus instead of the driver.
About the only thing they have partially right is CO2 increases should cause slight to moderate warming on paper.
The CAGW crowd displays a non-engineer surrealist viewpoint that is not conductive to coherent rational processing of information.
But back to your problem…
CAGW theory is a tool for driving anti-fossil fuel/anti-affluence policy. The policy came first. The CAGW theory was a convenient tool which served its purpose but is getting dull enough that it will be replaced by another tool.
The CAGW crowd is policy driven and filter out/distort data that doesn’t support the policy. They are adverse to rational discussion (the debate is over), repeatedly use the logical fallacy of “consensus”, have been rather ruthless in eradicating objective/skeptical views from climate science field, filter out any data they don’t want to hear, and adjust the reminder…
They have their fingers in their ears. Discourse isn’t possible unless they can lip read and their eyes are open.
PA wrote: “I apply engineering analysis to global warming theory and there is no way to make the numbers add up. … They have their fingers in their ears. Discourse isn’t possible unless they can lip read and their eyes are open.”
Yes, this is Rob’s argument also. He can’t get his very simple technical points across because I am an alarmist and I can’t think like an engineer. He also claims that my thinking is highly motivated since his argument destroys FOMD’s assumptions about heat transfer. But maybe you can sway my opinion.
Rob is arguing that if two cylinders of identical volume and temperature contain air at different pressure, 1atm and 2atm, say, then the air molecules in the second cylinder will tend to collide more often with the wall of the cylinder. This would tend to transfer more energy to the cylinder walls and warm them at a higher temperature than the first one if the average kinetic energy of the molecules were the same. So, it must be the case that the average kinetic energy of the molecules in the second cylinder is lower, in order to compensate for the higher rate of collision due to density. He also argues that the total internal heat energy is the same, though shared among more molecules in the second container, and this explains how the average kinetic energy per molecule is lower in the second cylinder. In fact, it is about half as much. EKavg = (Total_thermal_U)/N.
I am arguing that this argument is logically inconsistent. If the cylinders are at the same temperature, then they have the same average kinetic energy per molecule. This is given by the relation EKavg = (3/2)kT, and it is (very nearly exactly) the same in both cylinders. And this is logically consistent.
What is our own opinion as an engineer? Would the walls of the second cylinder really warm up more if I were right about the average kinetic energy being the same in spite of the much higher rate of molecular collisions?
P-N, “I am arguing that this argument is logically inconsistent. If the cylinders are at the same temperature, then they have the same average kinetic energy per molecule,”
Which is wrong. If you take a single volume of air and compress it, it will warm due to heat of compression. You cool that volume to the same initial temperature and the energy per molecule has to decrease. Rob pretty clearly said that the more compressed cylinder had to cool to its external environment to be at the same temperature as the first. That is how you can get two cylinders at different pressures and the same temperature to begin with.
If the temperature of the gas is the same in both containers then the kinetic energy per molecule is the same. If it is the container we are interested in and the container is cooler than the gas then the container with more gas will warm more because it contains more total energy. If they warm the container the same then the gases are not at the same temperature and don’t have the same kinetic energy per molecule. PV=nRT (pressure volume moles constant temperature) You can increase the pressure by adding molecules without increasing the temperature but this requires more total energy. I’m not sure you aren’t all right but looking at this from different perspectives.
captdallas wrote: “Which is wrong.”
What is wrong? That the average kinetic energy of the molecules in the two cylinders is (3/2)kT in both of them when they are at the same temperature? Rob is saying that KEavg is not the same in both cylinders, though he is defining KEavg (per molecule) as U_thermal/N, where N is the
number of molecules and U the total thermal energy of the gas.
“If you take a single volume of air and compress it, it will warm due to heat of compression. You cool that volume to the same initial temperature and the energy per molecule has to decrease. Rob pretty clearly said that the more compressed cylinder had to cool to its external environment to be at the same temperature as the first. That is how you can get two cylinders at different pressures and the same temperature to begin with.”
This explains how the molecules in the compressed cylinder lose kinetic energy after it has cooled back from a higher temperature (whatever the cause of the higher temperature). It gives no support to the idea that the two different cylinders with two different pressures, at the same temperature, have different average kinetic energy per molecule; let alone that they have the same amount of total internal energy irrespective of molar quantity.
steven wrote: “If the temperature of the gas is the same in both containers then the kinetic energy per molecule is the same. […]”
Thanks steven. Agreed with everything you wrote. That is also my understanding.
P-N, “What is wrong? That the average kinetic energy of the molecules in the two cylinders is (3/2)kT in both of them when they are at the same temperature.”
It is wrong because you are not solving the same problem Rob posed. Temperature is a measure of the average energy of collisions. You can have lots of collisions with lower energy or fewer with more energy. So you are looking at the average energy per container not the average energy per molecule. The 2g container has twice the molecules at half the energy per molecule. Sometimes it is better to start simple, restate the actual problem and use the simplest approach before jumping into quantum physics.
captdallas: “It is wrong because you are not solving the same problem Rob posed.”
We are looking at the same problem, and evaluating the end state after the “compressed gas” has cooled at the same temperature. I don’t think the gas remembers that it was warmer in the past, though I understand it’s part of the argument for the conclusion. Still, I am evaluating the end state: two containers of identical volume, same temperature, and pressures 1atm (“uncompressed”) and 2atm (“compressed”). I am saying that they have the same average kinetic energy per molecule but I now see that you really are agreeing with Rob.
“Temperature is a measure of the average energy of collisions. You can have lots of collisions with lower energy or fewer with more energy.”
Yes, this seems to agree with Rob. I am denying this. I am rather going with standard physics textbooks and saying that (kinetic) temperature is a measure of average kinetic energy per molecules. This is standard classical statistical and thermal physics, not QM.
“So you are looking at the average energy per container not the average energy per molecule.”
Do you mean the volumetric density of kinetic energy? Or the gross molecular flux of KE per surface area units? The latter, not the former, would relate rather more to the collisions with the surface, though it seems irrelevant to temperature, in my opinion. (It is relevant to rates of heat transfer).
“The 2g container has twice the molecules at half the energy per molecule.”
So you also are agreeing with Rob that both containers have the same total thermal energy content.
This would mean that the container with twice the molar amount of air (hence twice the mass), at the same temperature, though twice the pressure, has the same total internal energy content.
Are you really believing this to be true?
“Sometimes it is better to start simple, restate the actual problem and use the simplest approach before jumping into quantum physics.”
Agreed.
P-N, if you don’t like the basic, boring explanation of temperature then use entropy, which container has less entropy? The one with more molecules per unit area or the one with less? If you don’t like collisions how about molecular vibration in the temperature sensor. No matter how you slice it once you contain a gas you are looking at the container temperature.If you strap a thermocouple to the outside of the container you will measure the temperature of the container allowing for insulation of course. Not one gas molecule has to impact that thermocouple for you to “measure” the temperature of the gas inside the container. The thermocouple really doesn’t care how many degrees of freedom are available.
captdallas: “P-N, if you don’t like the basic, boring explanation of temperature then use entropy, which container has less entropy?”
It’s not a matter of my not liking your definition of temperature. I am noting that it conflicts with the standard definition that one can find in statistical mechanics textbooks. For the case of ideal gases, or most simple monatomic or diatomic molecules (He, Ar, N2, O2), and normal ranges of temperature and pressure (100K-400K, 0.01atm-100atm), then the relation EKavg = (3/2)kT for kinetic temperature is verified experimentally and kinetic temperature thus defined is a very close match to the thermodynamical definiton as 1/T = dS/dE.
My main point is that your definition of temperature and your claims about total energy content don’t mesh very well. Suppose we start with two identical volumes of gas V at room temperature, one mole each, with total internal energy U for each. We place them both into a big container with a partition between the two volumes. The total energy for the combined system now is 2U, and the volume 2V. We then compress the big container adiabatically back to the original volume V (while the partition stays at the midpoint). This requires some external work dW that is entirely transformed into internal energy dQ. So the internal energy of the gas now is 2U+dQ. We let the gas warm back to the original room temperature. The gain in internal energy thus is lost back to the environment. The internal energy is back to 2U.
However, both Rob and yourself are committed to say that the internal energy is back to U, as it was for each initial volume V before the compression. There is an amount of energy U that has vanished from the universe.
“We let the gas [cool] back to the original room temperature.”
Sorry.
Pierre-Normand | October 27, 2014 at 6:28 am |
Yes, this is Rob’s argument also. He can’t get his very simple technical points across because I am an alarmist and I can’t think like an engineer. He also claims that my thinking is highly motivated since his argument destroys FOMD’s assumptions about heat transfer. But maybe you can sway my opinion.
Rob is arguing that if two cylinders of identical volume and temperature contain air at different pressure, 1atm and 2atm, say, then the air molecules in the second cylinder will tend to collide more often with the wall of the cylinder. This would tend to transfer more energy to the cylinder walls and warm them at a higher temperature than the first one if the average kinetic energy of the molecules were the same.
Well… I’m pretty good with an acetylene cutting torch and from experience the acetylene and oxygen cylinders are both pretty cold.
Molecular collisions are in most cases elastic. There is some cylinder deformation (elastic) when the cylinder is filled. So the act of filling does transfer some energy to metal lattice that is released when the cylinder is emptied.
We are going to ignore corrosion and other effects that don’t appear related to the discussion.
After the heat from filling diffuses out of the cylinder… there are only elastic collisions. Elastic collisions against the cylinder wall do not do work since the wall doesn’t move. If the collisions did do work the air inside the cylinder would cool to absolute zero.
On an atomic level… elastic collision with atoms of a metal lattice would transfer some energy…
I’m going with: the filling of a cylinder transfers a fixed amount of energy due to average elastic deformation of the cylinder and the average doesn’t change.
PA: “Well… I’m pretty good with an acetylene cutting torch and from experience the acetylene and oxygen cylinders are both pretty cold.”
I don’t suppose they are at anything else but room temperature after they’ve been stored for awhile. Acetylene is stored as a liquid, and I assume that’s the case for your oxygen cylinders too?
When the gas is released, the partial pressure over the liquid is reduced, and some liquid evaporates to restore the partial pressure on top of the liquid surface at that temperature. The latent heat of evaporation then cools the surface of the liquid, and the whole cylinder progressively cools down as a result of conduction in the liquid and cylinder walls, convection, etc.
If the cylinder only contains highly pressurized gas above the boiling point, then, as the gas is released, it can’t be replenished by evaporation, and the remaining gas in the cylinder expands adiabatically (work is primarily done against the air at atmospheric pressure at the exit of the regulator, I would assume?) and it therefore cools down for that reason.
“[…]Elastic collisions against the cylinder wall do not do work since the wall doesn’t move. If the collisions did do work the air inside the cylinder would cool to absolute zero.”
Yes, if the walls would move out (e.g. if the cylinder would just go on deforming plastically,) then the gas would expand adiabatically and would cool down while performing work W = PdA through pushing against the receding walls (while the walls would tend to warm up due to irreversible plastic deformation and thus give back some heat to the gas!)
“On an atomic level… elastic collision with atoms of a metal lattice would transfer some energy…”
My contentions (against Rob) is that at thermal equilibrium (when the temperature of the wall is T = EKavg(2/3)/k, (where EKavg is the average kinetic energy of the molecules of the gas, and k is the Boltzmann constant) then there is no *net* energy transfer, whatever the gas pressure, and molecular collision rate might be. No net energy transfer means no heat transfer.
“I’m going with: the filling of a cylinder transfers a fixed amount of energy due to average elastic deformation of the cylinder and the average doesn’t change.”
Average kinetic energy? Agreed. Also agreed with the skipped parts.
Correct understanding by Pierre-Normand, minor extension by FOMD!
It’s interesting to contemplate a bizarre world in which this principle is *NOT* true. For example, if temperature of a gas depended (somehow!) on the gravitational potential and/or the air density, then thermometers would come marked with calibration constants relating to altitude.
That’s because the calibration of an ordinary lab-thermometer would change as we transported it from, for example, New York City (high atmospheric pressure, low in Earth’s gravity well) to Denver (low atmospheric pressure, high in Earth’s gravity well).
http://www.sciencelabsupplies.com/images/magictoolbox_cache_from_database/061e573f61ad4f788267ad21acb7d776.jpg
Lot’s of things change when we move from New York City to Denver (eggs take longer to hard-boil, for example), but thermometer calibrations are *NOT* among them.
Concretely, if I breath on a laboratory thermometer, it registers my breath temperature as 98.6F, both in New York City and in Denver even though the atmospheric density is considerably less in Denver, and the gravitational potential considerably higher. And I *DON’T* have to correct these measurements for daily fluctuations in barometric pressure.
Conclusion The independence of lab-thermometer calibration on atmospheric pressure and gravimetric altitude is one more piece of evidence that the Loschmidt gravito-thermal effect is a cognitive illusion, commonly founded upon an imbalanced appreciation of abstract thermodynamic arguments relative to concrete statistical mechanical computations.
It is a pleasure to help dispel your thermometric illusions, by reminding you of principles that you (and everyone) *ALREADY* know, Rob Ellison!
Hi Fan,
Yes, that’s a very good point about the lack of thermometer calibration as a function of ambient pressure. Rob’s two cylinders with gases at different pressures would also register different temperatures on ordinary thermometers put in them if they truly measured kinetic temperature (average KE of molecules). Though RE would himself argue that the gases “transfer heat” at the same rate to the thermometers in both cylinders since he would insist that, in that case too, the temperatures registered by the thermometers depend on the rate of collisions and not just average KE. So that will not persuade him.
The defender of the gravito-thermal effect would not have much of a problem with this since it is argued that the effect establishes a *real* vertical temperature gradient at equilibrium that closely matches the dry adiabatic lapse rate. So the thermometers would not need any special calibration, as they indeed don’t need any with the normal environmental lapse rates established through normal atmospheric processes rather than ‘thermo-gravitationally’. They measure what they ought to measure — local temperature.
FOMD, it takes slightly longer to boil an egg or make rice at a mile higher altitude because the boiling point of water changes slightly due to the lower pressure. Thermometers were calibrated to the freezing and boiling points of water – at sea level. Fahrenheit used the freezing point of heavy brine for ~0 F, 32F for pure water and 212F for boiling, at sea level. So if you re-calibrate the thermometers for altitude you would be “adjusting” for part of the minor effect being discussed.
http://docs.engineeringtoolbox.com/documents/1344/boiling_point_water_elevation_feet.png
If the thermometer were invented during the last glacial maximum, the scale would be different but the zero would be about the same. That won’t change the physics, just the reference.
P-N, “Yes, that’s a very good point about the lack of thermometer calibration as a function of ambient pressure. ”
Then why not call it the Gravito-Energy effect if y’all have problems understanding temperature? A molecule going up against the force of gravity will transfer less energy to a upper fixed surface than a molecule going down with the force of gravity will to a lower fixed surface. Since gravity is an acceleration, the greater the distance between the fixed surfaces the greater the differences in the energy transfer. More molecules decreases the average distance, the mean free path, reducing the gravity related energy transfer per collision. Since molecules have small mass, they really have to be hauling a$$ to transfer significant energy per molecule. In the real atmosphere, if they haul too much a$$, they don’t fall back to the same location if they fall at all.
So the Gravito-Energy Effect would be greatest where the mean free path is the longest which would be the higher altitude, which contains the fewest molecules.
From an engineering perspective, few molecules, assuming all molecules have the same specific heat capacity, means less total energy transfer. So to use the Gravito-Energy Effect, you would need a much larger surface at the higher point to conductively transfer the same amount of energy at the lower point. That would be a waste of time and money so you don’t see gravito-energy machines that use gases, you see hydro electric plants.
With the longer mean free path at higher altitude you also would have increased radiant heat loss. You also don’t see a lot of high altitude radiant heat exchangers used to remove sea level heat.
What you do see at higher altitude is increased use of evaporative coolers and larger fans.
Now it is interesting that that with increased altitude it is easier for water to evaporate but there is very little change in the energy required to freeze water. That would be a “non-linear” response. Since you can only decrease pressure so far, that non-linear response has a closer limit of zero with decreasing pressure than it does with increasing pressure which is limited by infinity. So Earth is operating at a small end of a large range which often gets to be “ASSUMED” linear, pretty much like you and Fan assume small changes are meaningless unless of course they are changes you embrace. That is a common trait y’all share with Dougy the Sky Dragon.
So instead of chasing the dragon, why don’t y’all consider the validity of C-C assumption in the models that aren’t working all that well?
captdallas wrote: “Then why not call it the Gravito-Energy effect if y’all have problems understanding temperature? A molecule going up against the force of gravity will transfer less energy to a upper fixed surface than a molecule going down with the force of gravity will to a lower fixed surface.”
If one considers a piece of isothermal air column enclosed in a box, then the molecules will have the same average kinetic energy at the top as they have at the bottom. Because of the barometric density gradient, the air lower in the column with have a higher volumetric heat capacity and it is only this that accounts for a higher rate of heat transfer, if there is any. There will be no heat transfer at all if the container has the same temperature as the air. In that case is irrelevant the rate of collisions is higher at the bottom than it is at the top. The air will not warm the bottom at a higher temperature just because the rate of collisions is higher, though the average kinetic energy of the molecules is the same. But Rob would dispute this, hence his insistence that the average kinetic energy of the denser air ought to be lower. I don’t think he has realized that this commits him to say that the molecules must therefore be slowing down as they fall within the gravity field of the Earth.
P-N, “If one considers a piece of isothermal air column enclosed in a box, then the molecules will have the same average kinetic energy at the top as they have at the bottom.”
Consider a tall container with just one molecule bouncing from bottom to top under gravity. Because of the acceleration of gravity, the molecule will impact the bottom harder than the top. Once that example is “isothermal”, the molecule would be at rest on the bottom. As long as “isothermal” doesn’t mean absolute zero, there will be motion and gravity will have some impact on that motion until all the molecules are closely packed on the bottom.
As I said earlier, the garvito-thermal effect is very small since molecules have very small mass and limited mean free path if there is any significant concentration. That doesn’t mean that gravity isn’t still a part of nature.
captdallas: “Consider a tall container with just one molecule bouncing from bottom to top under gravity. Because of the acceleration of gravity, the molecule will impact the bottom harder than the top. Once that example is “isothermal”, the molecule would be at rest on the bottom. As long as “isothermal” doesn’t mean absolute zero, there will be motion and gravity will have some impact on that motion until all the molecules are closely packed on the bottom.”
Temperature is an ill defined concept for one single molecules. Maybe discussion of this case (gas under gravity in a box) should be postponed to the next open thread. Let me just note that when there is a velocity distribution among many particles, some of them fall down before reaching the top.
A fan of *MORE* discourse | October 27, 2014 at 7:51 am | Reply
Pierre-Normand asserts [correctly] “If [and only if] the cylinders are at the same temperature, then they have the same average kinetic energy per molecule.”
Correct understanding by Pierre-Normand, minor extension by FOMD!
Well… a simpler argument is the definition of the ideal gas law (which also applies to “pretty good” gases.
PV = nRT
P – absolute pressure
V – absolute volume
n – number of moles of gas
R – ideal gas constant
T – absolute temperature (K)
A cylinder at 1/2 of the pressure, with 1/2 the moles of gas, by definition has 1/2 of the energy… so two cylinders at the same temperature have the same energy per molecule.
Further, Joule’s 2nd Law states that the energy of an ideal gas is independent of pressure and volume and only depends on temperature.
PA wrote: “A cylinder at 1/2 of the pressure, with 1/2 the moles of gas, by definition has 1/2 of the energy… so two cylinders at the same temperature have the same energy per molecule.
Further, Joule’s 2nd Law states that the energy of an ideal gas is independent of pressure and volume and only depends on temperature.”
Yes, we are on the same page. I didn’t know this was called Joule’s second law. Thank you. Now, how can one convince Rob of those facts? Let me preemptively clarify that “the energy of an ideal gas” refers to energy-per-mole, or average energy-per-molecule, for Rob may be tempted to interpret this law as supporting his contention that the two cylinders (‘the two gases’) have the same energy.
And yet these were the silly thought bubbles under discussion.
Constant average kinetic energy doesn’t apply in Earth’s atmosphere. It simply does not.
*My* heat transfer mechanism involved the number of molecules with a translational kinetic energy. There is no need to go beyond that into realms where idle speculation based on verbiage is the order of the day. There is nothing to settle.
Rob Ellison: “*My* heat transfer mechanism involved the number of molecules with a translational kinetic energy. There is no need to go beyond that into realms where idle speculation based on verbiage is the order of the day. There is nothing to settle.”
Your mechanism isn’t at issue. It’s your claim that EKavg = U_thermal/N that’s at issue. Also you claim that if EKavg is the same in two cylinders, while the pressure is different, the more pressurized cylinder would warm to a higher temperature due to the higher collision rate.
Those two claims of yours support one another. If one is true, then the other one (or something very much like it) must be true as well. It’s with them that I take issue.
Rob, we have two cylinders with the same volume, same *temperature*, and hence the *same* average kinetic energy per molecules, and twice the pressure (P=2atm) and molar quantity (n=2) of gas in the second container. I am saying this is possible, and it is a stable situation. Those two cylinders, and the gases in them, will all remain in thermal equilibrium in a room at the same temperature T. The reason the average kinetic energy is the same in both cylinders is because T is the same in both and EKavg = (3/2)Kt. This relation is independent of pressure.
You are saying this is a “wrong idea” because, since there is twice the rate of collisions from the molecules on the internal solid surface of the second cylinder, and there is the same average internal energy per molecules, then the wall of the second cylinder will have to warm up more than the first owing to the higher rate of collisions.
You had produced you argument here, among many other places:
http://judithcurry.com/2014/10/21/ethics-of-communicating-scientific-uncertainty/#comment-640769
(The last two paragraphs “Take these cylinders of oxygen again […]”)
I am saying that it won’t warm up more. There is no net energy transfer. The gases and the cylinders already are at equilibrium at temperature T. At equilibrium, when everything is at the same temperature, there is no heat transfer or spontaneous entropy change. The total internal energy of the gas in the second cylinder will remain twice as large as the total internal energy in the first simply because there are twice as many molecules holding this energy, with the same average energy per molecules. Nothing will warm anything else up. Everything is going to stay at the same temperatures and there will be no net heat transfer anywhere.
My conclusions:
(1) Gases in both cylinders have the same T and hence the same EKavg. How is that an incorrect application of EKavg = (3/2)kT?
(2) The cylinder with twice as many molecules has twice as much total kinetic energy. Total_KE = N*KEavg. Why is that incorrect?
Rob Ellison: “Much redefining of the terms of the silly thought bubbles. .
1. That KE average is constant with height. It isn’t in the Earth’s atmosphere.
2. That energy transfer is a function only of average kinetic energy. It isn’t. It is a function of the number of molecules and kinetic energy transferred in collisions.”
Temperature variations with height isn’t relevant to your discussion of the two cylinders. There is no net heat transfer at thermal equilibrium, which was assumed.
I think you may have finally realized that your claims and arguments regarding the two cylinders containing oxygen at 1atm and 2atm (same volume, same temperature) were silly and invalid. This would explain why you are desperately trying to change the subject and persist in disputing imaginary claims that I never made.
The two claims were exactly that. The average KE as particles rose was constant – because – and I quote – the loss of lower energy molecules *exactly* compensated for the gain in gravitational potential – and that heat transfer to a surface was only as a result of average kinetic energy. Two separate FOMBS thought bubbles that P-N found og so insightful.
The secret of the cylinder is that measured temperatures are a result of heat flow – and that potential increases with compression with no added or lost heat.
But there is no point in a discussion with a liar.
Rob Ellison: “The secret of the cylinder is that measured temperatures are a result of heat flow – and that potential energy increases with compression when the heat lost is equal to the work added.”
This is too hand wavy. When you compress the gas in the second cylinder, the temperature increase, and internal energy increase, that result from adiabatic compression (dW = dQ) isn’t relevant to the end state after you allowed the gas to cool back down to room temperature, the same temperature as the first 1atm cylinder. It’s the final values for T, P, V and n only that are relevant to calculating EKavg and Total_EK at the terminal equilibrium state. The cylinders don’t have secrets. The molecules don’t care about the past history about how they came to end up in the cylinders or how warm the gas might have been before cooling down. You still haven’t explained how one can have at the same time (at thermal equilibrium):
(1) T1 = T2,
(2) EKavg = (3/2)kT,
(3) EKavg1 != EKavg2.
This is what is called an inconsistent triad.
P-N: “(dW = dQ)”
I meant dW = dU. However this will translate to dU = dQ, the heat lost when the gas cools back down to room temperature.
Rob Ellison: “P-N might have been talking oxygen – I hardly read any of what he says. Most of it is just endless repetition of superficial nonsense.”
You replied to this bit of text that you had quoted:
“Note also that Rob E. defines EKavg per molecules as U_total/N even for diatomic or polyatomic molecules, though that’s not the issue here.”
Notice the mention of diatomic molecules? That’s what you used to incorrectly do with molecules with f > 3 (want me to quote you?), but glad that I was able to help and you and that you now *finally* realize that it’s only true that EKavg = Utotal/N for the case of monatomic molecules.
Very nice and instructive video.
“So what happens if you change the system by increasing N?”
Since PV = NkT = (3/2)U, then, if you increase N, something else has to change. So it depends what else you choose to keep constant. In the cylinder case, it is stipulated that T and V are the same, while P is twice as much in the second cylinder. So, N has to be twice as much, and, also, U has to be twice as much. This ought to be obvious since you now have twice as many molecules with the same average energy each.
As an aside, there is an interesting discussion of the environmental lapse rate at Nick Stokes’s place:
http://www.moyhu.blogspot.ca/2014/10/calculating-environmental-lapse-rate.html#more
Your first equation makes no sense – which is where you go wrong.
It takes work to compress gas. Equivalent to the temperature.
But you are still arguing the wrong point. Heat transfer through collisions is all about energy density.
Rob Ellison: “Your first equation makes no sense – which is where you go wrong.”
My first equation is your first equation, with terms that cancel out, as you yourself pointed out.
“It takes work to compress gas. Equivalent to the temperature.”
Work is energy. Energy isn’t equivalent to temperature. That doesn’t make sense. The work is explained in figure #4 in my previous link.
It seems a hopeless case. There is no energy in compressed gas – but it does work when allowed to expand? Hmmm.
Rob Ellison: “It seems a hopeless case. There is no energy in compressed gas – but it does work when allowed to expand? Hmmm.”
Who said there is no energy in compressed gases? You thought it was mainly from potential energy associated with van der Walls forces. But that would entail gross violations of PV = nRT (see the formula for van der Walls gases) and gross deviations from known specific heat capacities for ordinary gases (e.g. He, H2, O2, N2, CO2, NH4, etc). There is an effect for van der Walls forces but it’s small. And it accounts *negatively* for energy storage. It reduces heat capacities. Real gases are akin to outstretched rather than compressed springs (regarding the van der Walls contribution to stored energy and expansion-compression works). I said the energy is stored as internal energy — and that’s almost entirely kinetic and rotational energy for those gases at low pressures (e.g. < 100atm) and ordinary temperatures.
Ron Ellison: “Take these cylinders of oxygen again – they are in local thermodynamic equilibrium. This means that the total thermal energy is the same in all cases. But there are more molecules in the compressed gas – which means that the average kinetic energy per molecule is less.”
No. Just no. This statement violates the ideal gas law.
Units of energy are joules (kg*m2/s2).
nRT = mol * 8.3144622 J K-1 mol -1 * K = joules
By definition the energy of a gas is the product of the # of moles times the ideal gas constant times the absolute temperature.
New definitions of the energy of a gas such as suggested by RE have not been accepted by the physics/engineering community.
PA: “By definition the energy of a gas is the product of the # of moles times the ideal gas constant times the absolute temperature.
New definitions of the energy of a gas such as suggested by RE have not been accepted by the physics/engineering community.”
I am glad to hear that there hasn’t occurred a schism between those two communities recently as I was beginning to doubt my sanity. And I also am grateful that you extended you stay into this discussion a little bit beyond your initial drive-by shooting!
PA wrote: “By definition the energy of a gas is the product of the # of moles times the ideal gas constant times the absolute temperature.”
Though, to nitpick, that should be (3/2)nRT for monatomic gases, and it gets more complicated for other gases, especially when vibrational modes get involved. But Rob’s claim was about total kinetic energy anyway. (And he didn’t always distinguish KEavg from Uavg in the case of diatomic and polyatomic gases.)
http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/eqpar.html#c2
“If you now change to topic to heat transfer with [warmer] walls,…”
Take two cylinders of gas – each containing 1 kg.
One is at 150 bar, one at 1 bar.
Both are at the same temperature.
If you lay both on the ground, and knock the valves off, you will discover that the highly compressed gas is capable of doing far more work than the less compressed gas.
Now allow both cylinders (restored to original conditions, of course), to cool to absolute zero.
Both now possess exactly the same internal work capacity due to pressure and temperature – ie nothing at all. And all by simply allowing the gas to rid itself of excess EMR and achieve its natural condition.
What happened to all the energy in the 150 bar cylinder? It certainly just didn’t vanish, and will return if you allow the cylinder to heat to its original temperature. We leave them at absolute zero for the moment.
Now we have two cylinders of solid gas, and we expose them to a gravity field of 1 g, 10 gs, or 1000 gs.
Does the gas warm up, under the influence of gravity? Of course not. It remains at absolute zero. Misunderstanding heat, energy, temperature, pressure, and all the rest, will eventually have you believing in the gravito-thermal effect, or the ability of CO2 to warm an externally heated body.
No gravito thermal effect, no CO2 warming. No phlogiston, luminiferous aether, N Rays, or flat Earth either.
Live well and prosper,
Mike Flynn.
Mike, That was pretty good until the time you mentioned CO2.
Pierre-Normand,
Thanks. We can agree to disagree about CO2. I’m easily convinced – just point me to an experiment showing the CO2 warming effect. All the University bench experiments show a diminution – an impeding of energy flow between objects separated by CO2, with concommitant losses.
So far so good. Nature seems to be on my side. I believe physics supports me, which is why I have no trouble in denigrating climatology in general, and self proclaimed climate scientists in particular.
I find it amusing that people choose to take offense, when they could just as easily not be offended. Humans are an interesting lot.
Live well and prosper,
Mike Flynn.
Mike Flynn,
“University bench experiments show a diminution – an impeding of energy flow between objects separated by CO2, with concommitant losses.”
Indeed that’s what happens with the climate system. An impediment of energy flow between the Earth surface and space. This is why the IR spectrum measured by satellites shows big chunks taken out in the CO2 spectrum, and the emission temperature being from the troposphere.
Pierre-Normand,
Why discriminate against the other frequencies? Energy comes in infinite wavelengths. Sum them all – in and out. You will discover that the Earth cools.
All this talk about particular wavelengths is obfuscation by people who should know better. Fact trumps supposedly logical argument. Look at Sir Isaac Newton, Lord Kelvin, Einstein, and any number of other first rate minds, who led themselves, unknowingly, down a variety of garden paths.
I don’t see too many first rate minds calling themselves climatologists, so I have to forgive them even more for being singularly dense, and refusing to accept fact. In any case, who really cares? None of the billions, the research, the interminable propaganda, has resulted in anything of value to man or beast so far. Maybe tomorrow or next year, which can Sayang?
Sorry. No global warming. I’m right – you’re wrong. Have fun with Rob Ellison. I’m off!
Live well and prosper,
Mike Flynn.
Pierre-Normand,
That should be “Who can say?” Sorry. Rushing.
Live well and prosper,
Mike Flynn.
Mike Flynn wrote: “Why discriminate against the other frequencies?”
We are discriminating among the frequencies because nature Herself is discriminating among them. The radiation from the surface depends on the surface emissivity and the Planck distribution associated with the surface temperature. Unless the surface would warm to nearly 2000 Kelvin degrees, no significant amount of energy can be radiated away in the visible spectrum or any other higher energy radiations. So, greenhouse gases, including water vapor, are bound to absorb big chunks of it and re-emit at a lower temperature (due to the lapse rate — it’s colder at the altitude where the absorbed radiation finally escapes to space)
What small effect are you talking about? Gravity is acting on the two cylinders equally. Rob’s problem with the two cylinders has nothing to do with gravity. He is claiming that the internal energy is the same in both cylinders. I am claiming that it is twice as large in the second (2atm). He is claiming that the average kinetic energy is twice as much in the first cylinder (1atm). I am arguing that it is exactly the same. He is arguing that twice the rate of collisions with the same average kinetic energy per molecules would lead to the cylinder walls warming up to a higher temperature. I am arguing that only the speed distribution (and KEavg) determines the temperature.
No problem with 2 cylinders – that was not the scenario – the problem is the box with a gravity density. A surface being struck with more molecules at the same energy is warmer.
All the rest is utter nonsense – all of which are not absolute but depend on boundary conditions.
“No problem with 2 cylinders – that was not the scenario – the problem is the box with a gravity density. A surface being struck with more molecules at the same energy is warmer.”
It is precisely this latter claim that your two cylinder case dramatizes. It was also your own thought experiment (though you adapted it from Flynn). It is precisely this claim (and the equally strange claim about sameness of U) that PA, Captdallas and I, and probably most physicists and engineers, would dispute. It is just plain incompatible with the well known relation KEavg = (3/2)kT. If the gas is at T, then so is the solid surface that it is in thermal equilibrium with. Simple.
Rob Ellison: “Huh? I assert that there are no density differences and have endlessly.
And for the rest it depends on the boundary conditions – as I keep saying.”
Yes, and you also repeatedly insisted, and argued, that there is no pressure difference (must I quote you again?). Indeed, as I’ve just shown, arguing for no density difference ought to commit you to argue for no pressure difference as well. Though you didn’t make the connection through PV = nRT, as I just did, you argued that it is momentum transfer that accounts for the gas to weight on the box *rather* than a pressure gradient. But the claim makes no sense. It is *because* of differential rates of momentum transfer that there is a pressure difference, as I’ve shown mathematically, and your own hyperphysics link later confirmed.
As for the boundary condidtions, they are simple and agreed on: Elastic collisions both on top and bottom with exact reversal of particle momenta along the vertical axis, and a uniform vertical gravitational field g.
“Vy = Vo + gΔt
Ring any bell at all?
http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/weighgas.html
Getting so much wrong – and behaving so bizarrely about it.”
This kinematic equation applies perfectly well to individual molecules. You just seem unable to imagine its implication for distributions of speeds at a given height.
The hyperphysics explanation of the weight of gases (that I also had provided to you before, without the simplification that misleads you) simplifies the situation by only considering molecules that have enough kinetic energy to reach all the way up and bounce on the top of the box with some finite amount of kinetic energy before coming back down with the same finite amount. (That’s fine for the purpose of the demonstration that they are making.) Maybe that’s what threw you off. If all the molecules would do that — bouncing on both the bottom and top — it would indeed entail that the molecules have a smaller average speed the higher up you go. They would not have a Maxwell distribution of speed, though. Maximum entropy would not be achieved.
You forgot that for any given height some molecules fall back down before reaching it, and this skews the speed distribution *up*, such as to cancel the effect from the slowing down of individual molecules that are rising against gravity. Don’t you remember the “drop-off” effect that I had explained to you from the very beginning of this discussion (and indeed was the whole point of FOMD thought experiment?) It’s also explained very well in the Am. J. of Physics paper, and in Pekka’s paper. We had discussed this for a very long time before you finally acknowledged having grasped the point (*). And now you’ve completely forgotten the whole point of the discussion… again.
(*) http://judithcurry.com/2014/10/21/ethics-of-communicating-scientific-uncertainty/#comment-640786
“The random movement of molecules guarantees that the molecules disperse throughout the space. Just as Einstein said.”
Einstein didn’t say this. You can’t even read your own quotations. He suggested that Brownian particles and molecules behave similarly. He also indicated that he would discount the effects of gravity in his paper. Brownian particles have been shown to distribute according to the barometric formula — experimentally by Perrin, and theoretically by Stokes. Einstein says that the behavior of molecules is no different. What can you conclude?
Rob, Pierre,
Do either of you remember what the original argument was about, because nobody else here does.
phatboy: Do either of you remember what the original argument was about, because nobody else here does.
The original argument was that if a gas is in thermodynamic equilibrium in a box, then there ought to be a progressive reduction of the average molecular speed (and average kinetic energy) of the molecules as one go up, since the molecules slow down as they move up against gravity. This is an argument supported by Doug C. and Rob E. I am arguing that it is flawed because it fails to account for the effect of the vertical density gradient on molecular speed distributions.
Rob E. also is arguing that if the gas has a lower density (e.g. at the top of the box), then it can’t have the same temperature and the same average kinetic energy per molecule (or else, he argues, the gas would warm the bottom of the of the box at a higher temperature). This claim doesn’t have any basis in physics. Temperature only is determined by the average kinetic energy of the molecules.
Rob E. also is arguing that there are no vertical pressure or density gradients in a gas within an enclosed space. This claim is refuted by basic hydrostatic considerations.
Rob E.’s argument also is refuted there:
http://pirila.fi/energy/kuvat/barometric_derivation.pdf
http://web.ist.utl.pt/ist12219/data/43.pdf
Rob Ellison: “Molecules have a kinetic energy far in excess of the gravitational potential at 273K.”
When the molecule travels a vertical distance dz, the variation in kinetic energy just is the variation in potential energy. They’re identical. You aren’t entitled to consider only the effect from one and neglect the effect from the other on the velocity distribution at given heights.
“A *factoid* that the environmental lapse rate refutes.”
The environmental lapse rate is affected by radiation, convection, horizontal winds, and latent heat releases. The tropospheric columns are very much out of equilibrium. So, the ELR doesn’t have any bearing on the present problem.
“So unless you have an actual treatment […]”
Pekka provides an “actual treatment.” He demonstrates that the stationary density gradient of the barometric formula *derives* from — and hence is consistent with — the conditions of local thermodynamic equilibrium (Maxwell speed distributions depending only on temperature) for the isothermal case. The mathematics is very simple and only involves partial derivatives of simple products and exponential functions.
The conclusion is: “The equations (1), (2) and (3) represent the stationarity requirement that particles located in certain volume with certain velocities will at a later moment be replaced by an equal number of other particles which have the same velocities when the influence of gravity on velocity is taken into account. It’s shown that the isothermal atmosphere with Maxwell-Boltzmann velocity distribution and barometric vertical density profile
satisfies this requirement.”
The American Journal of Physics article provides three more derivations of the same result. You can’t just hand-wave away mathematical results that you don’t like.
P-N: “Pekka provides an “actual treatment.” He demonstrates that the stationary density gradient of the barometric formula *derives* from — and hence is consistent with — the conditions of local thermodynamic equilibrium (Maxwell speed distributions depending only on temperature) for the isothermal case.”
This sentence should be taken out and shot. I meant to say that the barometric density formula derives from (1) Maxwell distribution of speeds (local thermodynamic equilibrium), (2) the stationarity condition on the vertical density profile and (3) the assumption that the column is isothermal.
If follows trivially that those four conditions are consistent.
Rob Ellison,
Possibly the TOA – which you steadfastly insist exists in a real sense – is the top of the box.
If I propose theory, you claim the reality defense.
If I propose reality, you take refuge in theory.
There is no gravito-thermal effect in reality, in or theory based on physics within the bounds of present knowledge.
If you wish the approbation of your peers, either propose something novel, or come up with a different interpretation of facts, fitting better with observation than present explanation.
No global warming, no gravito-thermal effect, no luminiferous aether.
So sad, too bad, better luck next time.
Like well and prosper,
Mike Flynn.
Rob Ellison: “The finite volume is open on both ends”
And then he immediately proceeds to say: “We may consider this an isolated system with fixed mass and fixed extension.”
Read it by all means – http://www.mdpi.com/1099-4300/16/3/1515
But this is utter madness.
Rob Ellison wrote: “…there is enough energy a 0K to get all the molecules at an average speed of 446m/s across and back in a 1m box in millseconds.”
Not even in the right ballpark. This must be a typo? At zero Kelvin, the standard deviation of the velocity along the vertical axis = sqrt(kT/m) = zero. At 300°K, for nitrogen molecules, it’s around 300m/s (Note that the one-dimensional velocity distribution is smaller than the distribution for the modulus of the speed vector, since it’s a projection along an axis). Since the M-B velocity distribution is Gaussian, the cumulative probability distribution for the molecules travelling at less than 4.5m/s (and hence that can’t make it to the top from the level one meter below) is P(Z <= 0.009) = 0.050359. That's the case regardless of their initial velocity being up or down. So this makes for 0.0718% of the molecules from the lowest level that can't make it to the top. This is much more than needed to account for the barometric density variation (~0.012% for a height of one meter). You can now integrate for the intermediate level contributions to get the correct number.
http://stattrek.com/online-calculator/normal.aspx
http://www.mide.com/products/slamstick/air-pressure-altitude-calculator.php (That's for air rather than nitrogen, but close enough)
Rob Ellison: "The molecules spread out to fill the box is the dominant process."
'He who refuses to do arithmetic is doomed to talk nonsense.‘ — John McCarthy
“2. This is interesting. A helium balloon in a box of gas would sink. It wouldn’t. It would float because it is lighter than the displaced air we assume is in the box.”
The balloon doesn’t know that. Buoyancy only works because there is a pressure gradient in the fluid being displaced. This is how there can be a reaction force (Newton’s third law) to the force required to hold or move the displaced liquid. Abolish the pressure gradient and you likewise abolish the reaction force (and violate the law of conservation of energy if you move the balloon vertically). Buoyancy isn’t a magical effect nor a force at a distance like gravity. It is a force of contact. It requires molecular impacts. No pressure means no net momentum transfer since there wouldn’t be an asymmetrical distribution of molecular impacts around the balloon. It is strange that a hydrological engineer wouldn’t be able to conceive that.
What sort of strange boundary conditions result in Brownian motion in a box? It is all utterly irrelevant and pointless – at any rate – and intended only to mislead and obfuscate.
The original comment in this incredibly long string of trivial nonsense was a molecules in the box animation from Wikipedia and a pose about symplectic manifolds of the Hamiltonian. It is all a pose in support of meaningless drivel – obviously almost endlessly it seems.
Brownian motion is illustrative of the diffusion process. That particles tend to settle and don’t because of random molecular motion.
As far as the balloon is concerned – perhaps he should move onto Newton’s third law. Although the use of a further *thought experiment* to demonstrate the validity of a former *thought experiment* is an example of extreme verbiage rather than a rigourous treatment.
Much like the average kinetic energy being *exactly* equal at every level or the Wikipedia gas in a box animation being illustrative of the symplectic manifold of the Hamiltonian. Just verbiage with no significance.
Can’t help myself.
Both a helium and a lead balloon exhibit the same amount of buoyancy in zero gravity – for example in the ISS. None.
Gravity causes a pressure gradient. Without it, objects have mass, but no weight. It can be observed that in zero gravity, flames burn in a rather peculiar fashion, as the hot combustion products are surrounded by denser air, but have no directional buoyancy without gravity.
So gravity results in helium balloons rising, candle flames assuming their characteristic shape, and so on.
Still no gravito-thermal effect, and unless such can be demonstrated experimentally, still about as factual as phlogiston or caloric.
Warmists have the ability to believe anything that they think is really, really, true. CO2 creating warming, gravito-thermal effects, heat creep, treemometers, the Earth created cold, non-moving continents . . .
Some parasitic organisms perform useful functions, albeit in a roundabout way at times. Warmists at least provide an example of the ease with which people of all types can be lead to accept nonsense as fact, so I suppose they are useful in that regard. Nothing else, as far as I know. If I am wrong, I am sure a Warmist will correct me factually.
Live well and prosper,
Mike Flynn.