Guest Post by Willis Eschenbach
Over at the Notrickszone, there’s much buzz over a new paper entitled Molar Mass Version of the Ideal Gas Law Points to a Very Low Climate Sensitivity, by Robert Holmes. The Notrickszone article is headlined with the following quotation from the paper:
“In particular, formula 5 (and 6) as presented here, totally rules out any possibility that a 33°C greenhouse effect of the type proposed by the IPCC in their reports can exist in the real atmosphere.”
– Holmes, 2017
And here’s the abstract:
Abstract: It has always been complicated mathematically, to calculate the average near surface atmospheric temperature on planetary bodies with a thick atmosphere. Usually, the Stefan Boltzmann (S-B) black body law is used to provide the effective temperature, then debate arises about the size or relevance of additional factors, including the ‘greenhouse effect’. Presented here is a simple and reliable method of accurately calculating the average near surface atmospheric temperature on planetary bodies which possess a surface atmospheric pressure of over 10kPa.
This method requires a gas constant and the knowledge of only three gas parameters; the average near-surface atmospheric pressure, the average near surface atmospheric density and the average mean molar mass of the near-surface atmosphere. The formula used is the molar version of the ideal gas law. It is here demonstrated that the information contained in just these three gas parameters alone is an extremely accurate predictor of atmospheric temperatures on planets with atmospheres >10kPa. This indicates that all information on the effective plus the residual near-surface atmospheric temperature on planetary bodies with thick atmospheres, is automatically ‘baked-in’ to the three mentioned gas parameters.
Given this, it is shown that no one gas has an anomalous effect on atmospheric temperatures that is significantly more than any other gas. In short; there can be no 33°C ‘greenhouse effect’ on Earth, or any significant ‘greenhouse effect’ on any other planetary body with an atmosphere of >10kPa.
Instead, it is a postulate of this hypothesis that the residual temperature difference of 33°C between the S-B effective temperature and the measured near-surface temperature is actually caused by adiabatic auto-compression.
Dang … “adiabatic auto-compression” as a permanent energy source. Is it patented yet?
Please forgive my sarcasm, I just get tired of endless claims of endless energy … onwards. Here is a look at the various planetary atmospheres:
And finally, here is his math that leads to his mystery formula. From the paper:
Molar Mass Version of Ideal Gas Law Calculates
Planetary Surface Temperatures
The ideal gas law may be used to more accurately determine surface temperatures of planets with thick atmospheres than the S-B black body law [4], if a density term is added; and if kg/m³ is used for density instead of gms/m³, the volume term V may be dropped. This formula then may be known as the molar mass version of the ideal gas law. The ideal gas law is;
P V = n R T (1)
Convert to molar mass;
P V = m/M R T (2)
Convert to density;
PM / RT = m / V = ρ (3)
Drop the volume, find for density;
ρ = P / (R T / M) (4)
Find for temperature;
T = P / (R ρ/M) (5)
[VARIABLES]
V = volume
m = mass
n = number of moles
T = near-surface atmospheric temperature in Kelvin
P = near-surface atmospheric pressure in kPa
R = gas constant (m³, kPa, kelvin⁻¹, mol⁻¹) = 8.314
ρ = near-surface atmospheric density in kg/m³
M = near-surface atmospheric mean molar mass gm/mol⁻¹
Now, I agree with all of that. Well, other than the strange form of the last equation, Equation 5. I’d simplify it to
T =P M / (ρ R) (5)
But that’s just mathematical nitpicking. The underlying math is correct. That’s not the problem. The problem is where it goes from there. The author makes the following claim:
In short, the hypothesis being put forward here, is that in the case of Earth, solar insolation provides the ‘first’ 255 Kelvin – in accordance with the black body law [11]. Then adiabatic auto-compression provides the ‘other’ 33 Kelvin, to arrive at the known and measured average global temperature of 288 Kelvin. The ‘other’ 33 Kelvin cannot be provided by the greenhouse effect, because if it was, the molar mass version of the ideal gas law could not then work to accurately calculate planetary temperatures, as it clearly does here.
I’m sorry, but the author has not demonstrated what he claims.
All that Robert Holmes has shown is that the atmospheres of various planets obey, to a good approximation, the Ideal Gas Law.
… So what?
I mean that quite seriously. So what? In fact, it would be a huge shock if planetary atmospheres did NOT generally obey the Ideal Gas Law. After all, they’re gases, and it’s not just a good idea. It’s a Law …
But that says exactly NOTHING about the trajectory or the inputs that got those planetary atmospheres to their final condition. Whether the planet is warmed by the sun or by internal radioactivity or whether the warming is increased by GHGs is NOT determinable from the fact that the atmospheres obey the Ideal Gas Law. They will ALWAYS generally obey the Ideal Gas Law, no matter how they are heated.
And more to the point, this does NOT show that greenhouse gases don’t do anything, as he incorrectly claims in the above quote.
Look, we could start up ten million nuclear reactors and vent all their heat to the atmosphere. The planet would assuredly get warmer … but the atmosphere wouldn’t stop obeying the Ideal Gas Law. The variables of density and temperature and mean near-surface atmospheric molar mass would simply readjust to the new reality and the Ideal Gas Law would still be satisfied. You could still use his Equation 5 version of the Ideal Gas Law to calculate the temperature from the other variables, regardless of whether or not the atmosphere is heated by nuclear reactors.
So I’m sorry, but the underlying premise of this paper is wrong. Yes, planetary atmospheres generally obey the Ideal Gas Law, duh, why wouldn’t they … and no, that doesn’t mean that you can diagnose or rule out heating processes simply because the atmosphere obeys the Ideal Gas Law. They will always obey the law regardless of how they are heated, so you can’t rule out anything.
Best of another sunny day to everyone,
w.
MY USUAL POLITE REQUEST: When you comment, please QUOTE THE EXACT WORDS YOU ARE TALKING ABOUT so we can all understand what you have an issue with.
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“In short; there can be no 33°C ‘greenhouse effect’ on Earth, or any significant ‘greenhouse effect’ on any other planetary body with an atmosphere of >10kPa.”
The with atmosphere temperature of 288 K is a number pulled out of IPCC’s/WMO’s consensual collective [pruned].
The without atmosphere temperature of 255 K is actually a theoretical S-B BB calculation for the OLR of 240 W/m^2 at ToA. Uhh, that’s THEORETICAL & WITH an atmosphere.
The 33 C difference between these two fabricated, estimated, calculated, hypothetical, made up numbers is useless, meaningless crap. S-B & PV=nRT have zip to say about it.
The earth without an atmosphere would be like the moon, barren, no ice snow or water, dusty pock marked, blazing hot on the lit side, wicked cold on the dark. Earth and every other planet is colder with their albedos/atmospheres than without.
The earth’s surface is warmer than ToA because of Q=UAdT, the universal thermal blanket insulation equation.
— nickreality65
February 13, 2018 at 7:21 am
“In short; there can be no 33°C ‘greenhouse effect’ on Earth, or any significant ‘greenhouse effect’ on any other planetary body with an atmosphere of >10kPa.”
The with atmosphere temperature of 288 K is a number pulled out of IPCC’s/WMO’s consensual collective [pruned].
The without atmosphere temperature of 255 K is actually a theoretical S-B BB calculation for the OLR of 240 W/m^2 at ToA. Uhh, that’s THEORETICAL & WITH an atmosphere.–
Yes with atmosphere one can’t, simply take amount emitted and pretend you are dealing with something like a ideal thermal conductive blackbody, as in Earth emits on average 240 watts per square meter and if ideal thermal conductive blackbody emitted 240 watts it would have uniform temperature of 255 K.
I think one can use an ideal thermally conductive blackbody at 1 AU from sun to get a rough idea of how warm a planet should be at 1 AU from the sun. And so Earth should be around 5 C.
Or ideal thermally conductive blackbody would emit uniformly, about 340 watts. And 340 watts
is about 278 K about 5 C.
And one can do same thing with Venus- ideal thermally conductive blackbody at Venus distance should emit uniformly about 675 watts and have uniform temperature about 330 K [57 C].
And allowing for fact that Earth and Venus have an atmosphere. Earth is about 5 C and Venus is about 57 C.
That earth surface can become as warm as around 70 C, is not surprising as blackbody surface can become about 120 C- or the lunar surface does get this hot.
That Venus rocky surface can become around 737 K (464 C) is surprising or can’t be explained
by highest blackbody surface at Venus distance from the sun- which should limited to around
467 K [194 C].
What this should tell you about venus is that the sunlight is not directly warming the rocky surface-
but idiots think that it does.
So question to begin with, is what surface on Venus is being warmed by the sunlight?
And one can assume that this surface can not be warmed higher than about 194 C.
The alternative is that Venus temperature of about 737 K is not being warmed by the sunlight.
Earth
1,368 discular ISR, .30 albedo means 957.6 absorbed ASR discular, emits 239.4 (957.6/4) over entire spherical ToA. This does NOT represent physical reality.
Yes, 342 W/m^2 emits at about 278 K, but 342 W/m^2 ISR as averaged over spherical ToA does not exist.
1,368 W/m^2 emits at 394 K. Lit side of moon approaches that temp as would earth.
Venus with 2,614 W/m^2 ISR and .77 albedo absorbs 601.2 W//m^2 discular ASR & emits over entire ToA (601.2 / 4) OLR 150.3 W/m^2 and 226.9 K. Venus ToA emits at a lower temp than Earth.
This K-T type heat balance does not work!
http://writerbeat.com/articles/15855-Venus-amp-RGHE-amp-UA-Delta-T
https://www.linkedin.com/feed/update/urn:li:activity:6324679342349705216
” nickreality65
February 13, 2018 at 3:53 pm
Earth
1,368 discular ISR, .30 albedo means 957.6 absorbed ASR discular, emits 239.4 (957.6/4) over entire spherical ToA. This does NOT represent physical reality.”
Something can be hot and not absorb any energy. The lunar surface at 120 C isn’t absorbing much energy. Or a blackbody surface which is insulated can reach an equilibrium temperature [not be absorbing energy] of 120 C at 1 AU. Or one can have block of material, have all sides insulated except side facing the sun which is blackbody surface- and chunk material can reach 120 C.
So start with block at say 10 C, and when it’s warming it’s absorbing energy, and increasing in temperature and it have some heat gradient. and once surface reaches 120, it will not have heat gradient and block will not absorb any more energy- it will not get warmer- it’s at equilibrium temperature at 1 AU from the sun- in terms of blackbody surface [btw, if it’s a non blackbody, it could get hotter than 120 C]. And at Venus distance from sun it’s .about 467 K [194 C]
On earth one does not normally get a equilibrium temperature, desert sand at 70 C is fairly close.. An greenhouse or parked car can also get pretty close. As can a solar pond reach near equilibrium temperature,
So with solar energy the distance from the sun determines equilibrium temperature of blackbody surface.
Now with ideal thermally conductive blackbody sphere, the idea is it’s absorbing all the sunlight
and equilibrium temperature is related to the area which radiate- which 4 times diameter of sphere. So you will have surface of this sphere in sunlight with sun at zenith forever, and that surface is about 5 C- and on night of planet it’s also 5 C. So you have all energy absorbed and equal amount of IR radiant energy is emitted.
Now it put a piece what paper above that surface, the paper would warm to higher temperature than 5 C, blacken piece of paper can warm to about 120 C. And say use some aluminum foil, And it could be about, say 200 C. Or insulated aluminum 6061-T6 has equilibrium temperature of 450 C according to Space mission Analysis and design, third edition.
And OSR has -51 C
OSR is quartz over silver. The reflective silver has quartz glass over it, The glass has high emissivity and silver is reflective and conducts heat well to the quartz glass which radiates the heat- so can quite cold in 1367 watts per square of sunlight.
Anyhow, aluminum 6061-T6 does not absorb much energy from sunlight [0.379 ] but has a low emissivity of IR:.[0.0346]. And if painted with black paint, it’s 0.975 and emissivity of IR of 0.874
and cool to 136 C as it’s equilibrium temperature
Anyhow with atmosphere almost the opposite of OSR, the atmosphere covering earth doesn’t radiate much and silver is replace something which absorbs endless amounts of sunlight energy- ie, the ocean.And ocean evaporate most of it’s energy absorbed rather than radiating it.
Trick February 13, 2018 at 5:57 am
Our measurements are nowhere near accurate enough to determine a global imbalance of 0.6 W/m2. The CERES data have been ARTIFICIALLY ADJUSTED to give a global imbalance of 0.6 W/m2, for the dumbest reason in the world … care to know why?
Because James Hansen’s “peer-reviewed” paper said the earth was in an imbalanced state, based on his climate models and such … I discussed this over a decade ago here.
w.
“I discussed this over a decade ago..”
Willis has some catch up reading to do, CERES data has moved past being ARTIFICALLY ADJUSTED as in that linked piece.
CERES calibration science efforts have progressed a lot since that piece based on models in Hansen 2004. The EBAF data calibration through Ed2.5 was indeed based on later publication of that work Hanson 2005 in Science. Starting with Ed2.6 CERES Team started using ARGO data 2006-2010 for CERES data calibration and the latest Ed4.0 uses ARGO 2006-6/2015.
With the latest calibration efforts detailed in Loeb 2016 updated from Loeb 2012 and others, CERES Team shows daytime LW flux 1/2003 to 12/2014 now calibrated accurate to C.I. +/-0.4 W/m^2 per decade in the data Willis downloads now.
P.I. Dr. Norman Loeb used to have his personal website with all the relevant papers freely available & a couple years ago I took the time to read thru them. Very impressive work. I can’t find his website anymore or would link it. To confirm all this to top of your pyramid triangle level, the CERES site can show you to Table 5-1 in this link:
https://ceres.larc.nasa.gov/documents/DQ_summaries/CERES_EBAF_Ed4.0_DQS.pdf
Free access N.G. Loeb (NASA Langley Research Center, Va.) 2016 CERES Instrument Calibration:
http://www.mdpi.com/2072-4292/8/3/182
Thanks for that link, Trick. Looking there, I find:
Not exactly as you described it, but that’s no surprise. And for reasons that I’ve posted before, I don’t trust the OHC error estimates …
I also don’t trust their “ice warming and melt, and atmosphere and lithosphere warming”. They claim that we can measure these to both a precision and an accuracy of a few hundredths of a degree per decade … not buying that.
As to your ugly assertion that I have some “catch up reading to do”, I could have been doing nothing but reading my whole life and I’d still have “catch up reading” to do. There are on the order of 2.5 million new scientific papers published every year. No one can possibly keep up. There are papers I haven’t read, just as there are papers you haven’t read. So you can gently place your charming comments where the solar constant equals zero …
w.
“As to your ugly assertion..”
Willis’ response in part reaches only step 1 in his pyramid. Of course, lotsa’ reading but on this topic it is limited to the specialist papers since 2005. You know, a handful, 2 will do Loeb 2012, 2016. See Table 4 in the 2016 Loeb paper linked for the improved C.I. past your 2015 clip & the supporting reasons. Appears though you aren’t interested in climbing further up your own pyramid of response types.
Too many rabbit holes for me to counter here.
I see it be very simple and have described it simply enough for open minds.
Objectors are desperately introducing irrelevant side issues to avoid the inevitable conclusion that convective changes neutralise radiative imbalances.
That is not my invention.
See here:
http://www.public.asu.edu/~hhuang38/mae578_lecture_06.pdf
I have simply described the step by step mechanism by which it works.
From Stephen’s link: ”The atmosphere is very close to hydrostatic balance most of the time…For many applications, it is enough to replace the vertical momentum equation by the hydrostatic equation.”
So even that link disagrees with Stephen’s imaginary descending columns over half the globe. The link does agree with fundamental meteorology that there is little to no PE available for conversion to KE most of the time in the general atm. for Stephen’s unobserved in nature constantly descending (or spiraling down) columns. Since: “The atmosphere is very close to hydrostatic balance most of the time…”
No desperation Stephen, only a reading of fundamental meteorology papers with an open mind for learning from them. Try it!
The link is referring to averages. The atmosphere is divided into multiple rising and falling columns.
Note the diagram which shows the circular up and down convective motion between surface and tropopause.
I see that Don132 has declared absolute victory on behalf of the “gravitationists” (February 12, 2018 at 5:38 am) and proposed that Willis and Anthony should admit their total defeat at the hands of Frolly (a.k.a. Robert Holmes, the paper’s author) and remove his paper from the “Bad Science” category. Frolly/Holmes has responded to this (February 12, 2018 at 2:12 pm) in condescending manner by crowing smugly:
“I left you guys to sort this out and started writing another paper, but then saw you all lose your way going down Willis’s rabbit-holes.
Glad to help out!”
What vain self-conceit and self-delusion! Don132 proclaims Frolly/Holmes’s argument to be “irrefutable”. But meanwhile, back in reality, Willis had already refuted it in his lead article at the top of the page by pointing out, quite simply and clearly, that Frolly/Holmes had merely demonstrated that greenhouse atmospheres obey the ideal gas law just as non-greenhouse atmospheres do, which we should expect them to do anyway since it is a universal law that applies automatically to all kinds of gases irrespective of their radiative and chemical natures. So how does Frolly/Holmes’s “irrefutable argument” prove that greenhouse gases in a planet’s atmosphere have no effect on its surface temperature? It doesn’t, simply because it doesn’t test for that condition. Consequently, all that Frolly/Holmes’s argument has demonstrated is that the ideal gas law is irrelevant to the question!
Only, Frolly/Holmes and his fellow “gravitationists” don’t see it. And it’s not just that they have somehow misunderstood Willis’s simple argument, or that they have understood it but are holding serious objections to it that they think are valid and reasonable. If their complete non-addressing of it in their comments here is anything to tell by, they are not even conscious of its existence! How can that be, when Willis’s argument has been there at the top of the page for them to read all the time and when he has bent over backwards to explain it further and to relate to their objections in the comments below? Why is it that after so much concentrated effort has been expended on communicating his argument to them, they still haven’t even grasped what it is? Are they simply dull-minded? Are they mentally blocked, perhaps? Or have they just not been paying it any attention?
Anyway, whatever the answer to this question might be, I’m inclined to agree with Don132 that Frolly/Holmes’s paper does not belong in the “Bad Science” category. I think it belongs in the “No Science At All” category, because that’s essentially what it is.
Cassio, what you say is, to put it bluntly, absurd. What you want to do is say that your paradigm of radiative heating of the atmosphere is true, regardless of whether or not it fails a test of consistency with known and certain facts! It’s true because it’s internally consistent– surprise!– and because you say so. I don’t think so!
I think Willis is a great guy and I have a great deal of admiration and respect for him, but it’s crystal clear that he’s continually made arguments by assuming the conclusion and using that to prove his conclusion: classic, blatant, circular reasoning, and it’s painful to watch that continue, just as it’s painful to hear someone play the violin and continually miss hitting the notes just right. Apparently they can’t hear it; either they’re tone-deaf or they’re not paying attention.
My advice: pay attention!
I find it incredible that you feel you can post such thoughtless nonsense.
The gravitationists HAVE won. There is no way out.
“The gravitationists HAVE won. There is no way out.”
Not sure what you mean by gavitationist. However, if you are convinced the IGL wins out, then pick an object for which NASA has not measured the atm. density with a probe. Compute the global avg. surface temperature. Try an exoplanet for example. Use only IGL compute, global avg. surface T.
Radiation balance estimates are what they use today; none of that is allowed as input. You can’t even use the object’s orbit or sunload as that is not input to IGL in any form. All you have is P, R, density to get T. Remember R is gas specific.
Cassio
Thanks for your input, which was liiuminating.
I actually thought that WUWT was a skeptical site which pursued advances in climate science through logical debate.
From your post here, it is now difficult to distinguish between this site and many others who have abandoned scientific inquiry in favor of ad-hominems and pseudoscience.
Your inability to understand something, does not mean that the line of argument is wrong, or that the people advancing it are deluded.
The only saving grace I can see is that there are folk here who are open-minded and ready to debate science in order to advance knowledge.
” A realistic example would be two identical non GHG atmospheres to one of which a potent GHG would be added, P and M remain the same but T increases and therefore density and volume change to follow the IGL.”
What if space aliens came and took all our saltwater?
Or can’t add water to a planet like Earth or it would planet with average depth of water about 3000 meter deep- or you need deep ocean basins to provide the planet with that much water a land surface. But one could compare planet completely covered with ocean and planet without any oceans.
Since our planet, has average ocean temperature of 17 C, I would say planet covered with ocean would average temperature of 20 C or warmer, and of course there could not be permanent ice caps at the poles- or need a land surface to make a polar ice cap. And any polar sea ice formed in winter would float around.
And earth without ocean and ocean basin, would be less than 5 C. With hot daytime tropics and much colder and one goes towards the poles.
Mars poles in winter it can get to about -120 C, and poles on earth without water and with no sunlight for 6 month should also get as cold. And regions without the sun getting more than 20 degree above horizon, would have sun going a lot atmosphere and hitting the surface at low angle
and be unable to warm surface during the day.
Or with clear weather and pointing a solar panel at the sun [rather lying level to surface] you getting about 200 watts, rather than 1000 watts with sun near zenith. So quite cold where sun isn’t getting high over horizon.
And the hot tropical day, is going to cool to lower temperatures than hot deserts on earth cool during the night.
.
Cassio February 13, 2018 at 2:55 pm
Is this true? Is “Frolly” merely a sock-puppet for Robert Holmes?
Tell me it’s not so, someone. That would be … well … wildly against site policy …
w.
Willis
Is science also against site policy?
frolly February 13, 2018 at 3:38 pm
Don’t be daft. WUWT was voted three times the best science blog on the web.
More to the point, are you a sock-puppet for Robert Holmes? Because if so, that is definitely against site policy, viz (emphasis mine):
So … are you or aren’t you Robert Holmes pretending to be someone else?
w.
I fear that it is true, Willis, if Frolly himself is to be believed. See his self-outing at February 12, 2018 at 2:12 pm.
Cassio February 13, 2018 at 3:53 pm
Man, that is the lowest of the low! Pretending to be someone else, someone who supports and agrees with your ideas, is the action of a deceptive charlatan.
w.
Stop with the cheap tricks, please. Frolly said who he was. I suspect you’re just looking for an excuse to dismiss an argument you cannot win.
Play the ball, not the man!!
Willis,
No.
I am not Robert Holmes ‘pretending to be someone else’!
I am Robert Holmes who also runs the YouTube site 1000Frolly.
Sometimes my computer signs me in as Frolly, as it did here; there is no ‘pretense’.
If I was ‘pretending to be someone else’ I would hardly admit it myself would I?
Would not be surprised if you carried out your threat to delete all my posts – since you have clearly lost the argument here.
If Willis and Anthony delete your comments and this argument, that would truly be the lowest of the low. I frankly cannot believe that Willis would stoop so low.
I know from experience that sometimes I’m signed in as another alias; I take care to remain anonymous, and consistently Don132, on this site. But I understand how it can happen.
132? Beethoven’s string quartet Opus 132.
Play the ball, not the man!!
Don132
Very many people posing here are anonymous, and I can understand it in this field.
I had to be anonymous on YouTube because of all the real threats of violence I was getting from climate activist/fanatic’s against my family.
Now the kids are older and leaving home I am not so worried and don’t care so much if my real name gets out.
Frolly, that is a perfectly reasonable explanation that Willis and Anthony should accept. I am fairly well-known locally– as my real self– for my anti-consensus views, and although there are many good people here I have two young children and find myself holding back out of concern for the consequences they might face if I speak up too forcefully or too loudly.
You have not attempted to deceive anyone! You already said plainly who you are, well before you were supposedly outed by “Cassio.”
This is getting ugly. I’m counting on Willis and Anthony to take the high road.
frolly February 13, 2018 at 5:24 pm
Don132 February 13, 2018 at 5:36 pm
Neither Anthony nor I say that nobody should be anonymous. Some people have valid reasons. And some don’t. And neither of us said that Frolly shouldn’t be anonymous.
What we say is that nobody should run a sockpuppet as frolly did. That’s active deception. In that regard, I note that Frolly is once again fooling people who joined the thread late and missed his microscopic confession upthread … but nooo, he won’t give it up.
No big surprise there.
w.
frolly February 13, 2018 at 4:22 pm
So this is the “my computer made me do it” defense? Right, no dog, blame it on your computer … it signed you on as “frolly” and you never noticed us all responding to “frolly”, you just overlooked that detail?
That’s your excuse?
Really?
Frankly, Frolly, I have no clue what you would do and would not do. You are certainly not the first sockpuppet in my own experience to out themself, so I know someone can certainly be a sockpuppet and then reveal it for any of a dozen reasons … why not you?
I didn’t threaten to “delete all your posts”, that’s a damn lie. Plus I don’t have the authority to do it—I’m a guest AUTHOR, not a guest MODERATOR. Finally, you’re lost in your paranoia, I don’t WANT to delete your posts.
All I did was to quote the site policy verbatim, in case you were unaware of it. I had no intention, and I have no desire, to erase the evidence of your duplicity and of your incredible beliefs. Why would I do that? They’re my evidence that you don’t have a clue and you have no problem being a sockpuppet … I don’t want that to disappear in any sense.
w.
Willis, this is not the 1st time false outrage has been used here to attack a winning argument. Extra low base lines may be needed for the triangle….Grin. Brett
I cannot tell you how disappointed I’d be if Willis used Frolly’s supposed “deception”– when he outed himself!– to dismiss the arguments here. Rest assured that what transpires here is seen and cannot be swept away. I encourage everyone to save the complete webpage as a record, in case we do find the arguments deleted.
Don132 February 13, 2018 at 5:23 pm
Paranoid much? I have no desire to see him banned, and WUWT doesn’t erase stuff in general, that’s all you. But heck, fill your whole hard drive up uselessly, fine by me …
As to what you try to dismiss as his “supposed ‘deception'”, “frolly” made a number of comments clandestinely supporting himself. I note that you, I and others answered him and there was much discussion BEFORE he outed himself.
That’s scummy sock-puppetry that is specifically forbidden in the site policy.
Finally, you seem to think that the fact that he confessed to his deception somehow exonerates him … funny how the justice system doesn’t work that way. All confession does is maybe get you a bit of consideration from the judge, IF the person confessing is repentant.
Which Robert is not, and which you support.
So no, Don, or Jane, or whatever your real name is … I’m not trying to get Robert banned. I’m just pointing out that he’s engaging in very deceptive practices.
And finally, obviously when a man is desperate enough to invent a fake person to pretend to support him, he has very little faith in his scientific arguments …
w.
Willis, Frolly did not ” confess” after Cassio outed him; he stated who he was well before then: https://wattsupwiththat.com/2018/02/06/ideal-gases/comment-page-1/#comment-2742226
I’m glad you’re not resorting to cheap tricks. And no, it doesn’t take much memory to copy a webpage. I don’t know exactly how you operate here at WUWT but I know from experience that sometimes if they don’t like what you say, the comments get deleted. So yes, maybe a bit paranoid. This is an important argument and I do not want to see it disappear.
You are making far, far too much of Holme’s use of a pseudonym; he has explained himself.
And yes, my name is Don. I don’t say who I am to protect my young children.
Willis
“Paranoid much?”
.
You are the paranoid one, or delusional anyway.
Making up hateful, fake stories about people in an attempt to discredit their science only because its demonstrably far better than yours.
.
“frolly” made a number of comments clandestinely supporting himself..”
.
There was only two posts (on the 11th Feb) before I ‘outed myself’ and they was not specifically supporting myself either – and occurred only because my computer had switched itself over to Frolly, which I did not notice then.
As soon as I noticed it – the next day – I outed myself, so that everyone here would know that Frolly was me, Robert Holmes.
.
“I’m just pointing out that he’s engaging in very deceptive practices…when a man is desperate enough to invent a fake person to pretend to support him, he has very little faith in his scientific arguments …”
.
This is so delusional its really very funny.
All you are pointing out to everyone here is that you are not only deluded and paranoid, but that you fear the line of argument being advanced here so much that your only line of defense is to attack the man instead of his science.
Somehow I have seen this a million times before on my YouTube account, used by other fanatical AGW believers, so I can recognize this fraudulent way of ‘debating’ easily.
To be frank (Oops, I don’t mean that I am Frank as well) I don’t give a rats arse about WUWT’s very odd intrigues, plainly run by very small unimaginative men.
Ok enough of this, you’ve had you say (or rant in this case) time to choose an identity and stick with it, and then get back to the real issue of the article, not this distraction and name-calling. Further posts like this go to the bit-bucket.
The fact that you have “frolly” at all suggests that you have a need to hide on occasion. Otherwise it wouldn’t exist. So be one or the other.
Willis,
The internet is a truly brutal place and folk’s desire for anonymity here is well founded.
Posting using WordPress is challenging if you want to preserve your anonymity elsewhere because, although WordPress allows you to change your handle, this cannot always be checked prior to posting.
I find it best to only use one persona per browser, for example Firefox, Chrome, IE and Brave will all act independently and so this forces WordPress to request separate logins for each browser, this way you can keep your anonymity (and keep your family safe) if you need to.
Philip Mulholland
Philip Mulholland February 13, 2018 at 7:47 pm Edit
Thanks, Philip, but that’s not quite true. The internet is a truly brutal place and some folk’s desire for anonymity here is well founded. However, there are undoubtedly others for whom it’s just a way to avoid taking responsibility for their words.
What the anonymous often fail to understand is that it costs them credibility. They can say anything at all, walk away, change their screen name, and bear absolutely no responsibility for what they have said.
You and I, on the other hand, must choose our words much more carefully, because we need to stand behind them even years later. And since we have to back our words, that gives them more credibility than the same words from some random anonymous internet popup.
w.
Willis,
You are the most disingenuous and ignorant person!
As anyone can see, I used my OWN name right from the start here!
When my computer accidentally used my U-tube screen name Frolly, I immediately corrected this the next day!
You are simply using this accident as a diversionary tactic, just like the planet with its ‘Argon atmosphere’ and its 1,000 suns. When are you going to admit that you actually have no answers to Don132, Stephen Wilde, the Reverend Badger, Roger Clague or me?
My reply to Willis, yesterday, seems to have been lost.
As I pointed out then, Holmes stated who he was long before Cassio supposedly outed him, as Cassio himself acknowledged. So Holmes “confessed” to nothing.
Also, my name really is Don. I use an alias to protect my young children. Whether my fears are unfounded or real, these are my children and I’m not taking any chances.
Robert Holmes February 13, 2018 at 11:58 pm
You used your own name AT the start, not FROM the start.
Yes … after posting three more comments before correcting it. Four comments with a number of people replying to you as “Frolly”, and you want me to believe you corrected it “immediately”??? Pull the other one!
I’m using a proof that your scientific claims are wrong as a “diversionary tactic”? Dude, what are you smoking?
I’ve answered your questions quite patiently. I’ve made many good-faith efforts to show you exactly where you are wrong, including in the head post. I have offered explanations. I’ve given you two complete proofs that you are wrong. What more do you want??
The problem is not that I “have no answers”. It is that you are unwilling to engage with the answers that I have given.
But hey, rock on. You’re obviously beyond my poor ability to add or detract … none so blind as those who will not see.
Regards and regrets,
w.
Willis
I want to expose your lies and nonsense so much I examined the entire train, took me two hours.
You said; “You used your own name AT the start, not FROM the start.”
.
I used my name FROM the start, and here are my post to prove it;
Robert Holmes February 6, 2018 at 6:21 pm
Robert Holmes February 6, 2018 at 6:46 pm
Robert Holmes February 6, 2018 at 9:35 pm
Robert Holmes February 6, 2018 at 10:21 pm
Robert Holmes February 7, 2018 at 4:23 am
Robert Holmes February 7, 2018 at 3:59 pm
Robert Holmes February 7, 2018 at 4:23 pm
Robert Holmes February 7, 2018 at 4:41 pm
Robert Holmes February 7, 2018 at 5:08 pm
Robert Holmes February 7, 2018 at 7:00 pm
Robert Holmes February 7, 2018 at 7:15 pm
Robert Holmes February 7, 2018 at 9:09 pm
Robert Holmes February 7, 2018 at 9:27 pm
Robert Holmes February 7, 2018 at 9:36 pm
Robert Holmes February 7, 2018 at 9:40 pm
Robert Holmes February 7, 2018 at 10:25 pm
Robert Holmes February 7, 2018 at 11:14 pm
Robert Holmes February 7, 2018 at 11:27 pm
Robert Holmes February 8, 2018 at 2:53 pm
Robert Holmes February 8, 2018 at 3:08 pm
Robert Holmes February 8, 2018 at 4:29 pm
Robert Holmes February 8, 2018 at 4:59 pm
Twenty-two posts to be exact, and no ”sock-puppet” to be seen. Then I gave you guys a break while I had a computer update, which the computer obviously did not like much – because the next post I made was accidentally under “Frolly” here;
frolly February 11, 2018 at 2:07 pm
Which I corrected as soon as I noticed, this was 24hrs and 5 minutes later here;
frolly February 12, 2018 at 2:12 pm
“I left you guys to sort this out and started writing another paper, but then saw you all lose your way going down Willis’s rabbit-holes.
Glad to help out!
1000Frolly aka Robert Holmes”
And I see you have posted yet another diversionary rabbit-hole for us to run down.
Yes this was really an unnecessary diversion and looks like another attempt to evade Badger’s [Holmes’] irrefutable argument.
>>
Yes this was really an unnecessary diversion and looks like another attempt to evade Badger’s [Holmes’] irrefutable argument.
<<
The Badger/Holmes argument is not irrefutable–it’s a magic trick. If planets E1 and E2 have the same surface temperature and E2 has no GHGs, then the atmosphere of E2 is undefined by those facts. The real test would be to make the atmosphere of E2 have the same total mass as E1. Because the energy flows will be different, then the surface temperature of E2 can’t be the same.
For starters, E1 (Earth) has three windows (due mostly to water vapor), one between 5 centimeters and 11 meters (used by radio astronomers), one between 300 nanometers and 1100 nanometers (that our eyes take advantage of), and one between 8 microns and 12 microns (that the planet uses to cool itself–the so-called IR window). Although some like to extend the IR window to 14 microns so it looks like CO2 has more of an effect (14.99 microns).
E2 would have no windows and would be free to radiate at all wavelengths. That alone would change the surface temperature.
Jim
Jim,
Its no trick, its a real scenario which could theoretically be carried out in reality.
Certainly GHG have an effect, no-one here is denying that. What we are saying is that negative feedbacks in the climate system reduce the warming effect to zero.
Try this short version of the thought experiment for simplicity;
What I am doing here, is pointing out that if the GHG in E1 do cause anomalous warming, then they can ONLY do this by altering the three gas parameters in a way in which non-GHG can not..
BUT since a NON-GHG mix (E2) can also be created which matches these exact three E1 parameters, (and IF the MM version of the IGL is correct – and so E2 must form the same temperature as E1) then the GHG’s cannot be creating any net anomalous warming by altering the three gas parameters in an unusual way.
Looks like you missed some other instances. In any event, pick one persona and stick with it, as our site policy on multiple identities is very clear. Also, tone down your angry rhetoric.
“frolly” now goes to the bit bucket.
Feel free to be as upset as you wish.
Willis is correct. It seems Mr. Holmes has had at least 4 separate identities here in addition to “frolly” on further investigation, I find we have these variances:
Our site policy on multiple identities is very clear. From now on, “frolly” goes direct to the bit bucket.
>>
What I am doing here, is pointing out that if the GHG in E1 do cause anomalous warming, then they can ONLY do this by altering the three gas parameters in a way in which non-GHG can not..
<<
True, I agree with you here.
>>
BUT since a NON-GHG mix (E2) can also be created which matches these exact three E1 parameters, (and IF the MM version of the IGL is correct – and so E2 must form the same temperature as E1) then the GHG’s cannot be creating any net anomalous warming by altering the three gas parameters in an unusual way.
<<
The problem here is that to get to E1’s same parameters, you must alter E2’s atmosphere to get those parameters (it’s what you and Badger claimed). That would make E2’s atmospheric mass different from E1’s atmospheric mass. I doubt that with different energy flows you can make E1 = E2. And since E1’s atmosphere is different from E2’s atmosphere, those GHGs must be having an effect.
Jim
Jim Masterson February 14, 2018 at 2:57 pm:
“The problem here is that to get to E1’s same parameters, you must alter E2’s atmosphere to get those parameters (it’s what you and Badger claimed). That would make E2’s atmospheric mass different from E1’s atmospheric mass. I doubt that with different energy flows you can make E1 = E2. And since E1’s atmosphere is different from E2’s atmosphere, those GHGs must be having an effect.”
Jim, that cannot be.
Assume that E1’s atmosphere is warmed by GHG. How warm will it be? Let’s use Frolly’s formula to find out. This formula will take the pressure, density, and molar mass and find the temperature, and do this pretty accurately. Then the only thing we’re doing is taking the pressure, density, and molar mass of E1 and transferring them to E2– EXACTLY, except without GHGs. We can do this, so we are in essence mimicking E1’s atmosphere.So then what will E2’s temperature be? According to Frolly’s formula, it will be exactly what E1’s temp is.
I don’t understand how we’d be altering the mass of E2. Wouldn’t it be EXACTLY the same as E1’s?
Once again, I believe you are probably assuming that GHGs must have an effect, and that effect cannot be accounted for in E2 except by altering the mass of E2, since it has no GHG. Do you see that you are assuming the conclusion in order to prove the conclusion?
>>
I don’t understand how we’d be altering the mass of E2. Wouldn’t it be EXACTLY the same as E1’s?
<<
What do you mean by “EXACTLY the same?” The atmosphere of E1 doesn’t equal the atmosphere of E2 because one contains GHGs and the other doesn’t. So you need to pick another parameter to compare them. I choose total atmospheric mass. Given that, the surface of E2 is free to radiate at all wavelengths–that alone should make E2’s surface temperature different.
Yes, I think GHGs have an effect. My experience with deserts and tropical jungles tells me that these two regions have different temperature profiles–especially at night. The only main difference is the presence of moisture in the air–a GHG.
With equal total atmospheric mass and T1 not equal to T2, then to make the Holmes version of the IGL work out the density and molar mass of E2 must be different. Altering the density and molar mass of E2 so it matches the density and molar mass of E1 says absolutely nothing about the total atmospheric mass of E2. Neither you, Holmes, or Badger can demonstrate that the total atmospheric mass of E1 equals E2 in such a case. If you can, show me.
>>
Once again, I believe you are probably assuming that GHGs must have an effect, and that effect cannot be accounted for in E2 except by altering the mass of E2, since it has no GHG. Do you see that you are assuming the conclusion in order to prove the conclusion?
<<
Assuming that GHGs don’t have an effect is also assuming the conclusion to prove a conclusion. Except, I have a reason to make my assumption–GHGs actually do interfere with the flow of energy.
I also expect E2 to have lots of dust devils–like Mars and dry regions here on Earth.
Jim
Well my desktop PC logs me in as Stephen Wilde and my ipad as wildeco2014. I don’t regard that as a deception.
Anyway, to summarise the science so far:
i) The ideal gas law gives a useful prediction of the surface temperature of a planet with an atmosphere regardless of whether GHGs are present or not and regardless of their quantity.
ii) That prediction invariably gives a higher temperature than that predicted by the radiation only S-B equation
iii) The only ways that energy in an atmosphere can get back to the surface in order to raise it above S-B are by conduction and/or radiation.
iv) If an atmosphere is non radiative then the only way is by conduction so the fact of back conduction heating a surface cannot be denied.
v) If one then adds a radiative gas then the DWIR energy reaching the surface below alters the density patterns at the surface so as to compromise the efficiency of conduction between atmosphere and surface. Conduction is more effective when the gases are denser or where the density variability in the horizontal plane is greater.
vi) It follows that more DWIR means less conduction and the surface temperature remains set at that predicted by the ideal gas laws.
vii) The ideal gas laws predict a maximum surface temperature that cannot be further enhanced by radiative gases in the atmosphere.
The above is a duplicated post that I thought had not gone through.
Well my desktop PC logs me in as Stephen Wilde and my ipad as wildeco2014. I don’t regard that as a deception.
Anyway, to summarise the science so far:
i) The ideal gas law gives a useful prediction of the surface temperature of a planet with an atmosphere regardless of whether GHGs are present or not and regardless of their quantity.
ii) That prediction invariably gives a higher temperature than that predicted by the radiation only S-B equation
iii) The only ways that energy in an atmosphere can get back to the surface in order to raise it above S-B are by conduction and/or radiation.
iv) If an atmosphere is non radiative then the only way is by conduction so the fact of back conduction heating a surface cannot be denied.
v) If one then adds a radiative gas then the DWIR energy reaching the surface below alters the density patterns at the surface so as to compromise the efficiency of conduction between atmosphere and surface. Conduction is more effective when the gases are denser or where the density variability in the horizontal plane is greater.
vi) It follows that more DWIR means less conduction and the surface temperature remains set at that predicted by the ideal gas laws.
vii) The ideal gas laws predict a maximum surface temperature that cannot be further enhanced by radiative gases in the atmosphere.
wildeco2014 February 14, 2018 at 1:10 am
Well my desktop PC logs me in as Stephen Wilde and my ipad as wildeco2014. I don’t regard that as a deception.
Anyway, to summarise the science so far:
i) The ideal gas law gives a useful prediction of the surface temperature of a planet with an atmosphere regardless of whether GHGs are present or not and regardless of their quantity.
The IGL is incapable of predicting the surface temperature, all it says is if you measure three of the parameters you know what the fourth value is (that’s not a prediction). Measure only two of them and you have no idea what the other two are, just what their relationship is.
Typically you’d need to solve the heat balance equation to determine the other two.
For example Titan has a surface pressure of 1.45 bar and a Molar mass of about 28, well in IGL range what can you say about its temperature?
Robert (Holmes), I’m unwilling to leave it there, so lets set all of that aside and go back to the science.
Assume we have two identical planets with identical atmospheres containing GHGs. They are both warmed to 255K by the sun. On planet E1, the greenhouse effect works, and this boosts the temperature by 33K to a final temperature of 288K.
On planet E2, on the other hand, the greenhouse effect doesn’t work for unknown reasons. Fortunately, on this planet adiabatic auto-compression kicks in as Robert Holmes postulates. And purely by chance this also boosts the temperature by the same 33K to a final temperature of 288K.
Now, both planets end up at 288K, with identical molar mass, density, and pressure. As a result, the temperature as calculated by the IDL is right around 288K for both planets.
So here is the important question:
Using just the Ideal Gas Law, can you tell which planet gains additional warmth from GHGs and which planet gains additional warmth from adiabatic auto-compression?
I say no. The IGL cannot tell us anything about the CAUSES of the temperature. It can only allow us to calculate the temperature, and even then only when we know the three other variables. But it gives us no information about CO2, adiabatic auto-compression, or any other cause for the temperature. It can only give us the temperature.
So what say you to that question? Can the IDL used BY ITSELF tell us which planet, E1 or E2, is warmer because of GHGs and which is warmer because of autocompression?
Thanks in advance for your answer,
w.
Both planets are capable of gaining additional; warmth from both processes but the more DWIR there is the less adiabatic auto-compression there is so that the IGL is complied with.
You see, in the process of warming the surface DWIR makes the surface gases less dense so that proportionately less conduction can occur which feeds through the convective overturning cycle to produce less auto-compression.
The important fact for the AGW theory is that one cannot exceed the surface temperature predicted by the gas laws by increasing GHGs.
Stephen, that’s interesting but I fear you haven’t come close to answering my question. It was:
Using just the Ideal Gas Law, can you tell which planet (E1 or E2) gains additional warmth from GHGs and which planet gains additional warmth from adiabatic auto-compression?
If you say “yes”, please explain how.
w.
Stephen, I’m sure you’re right; I just can’t understand what you’re saying! Pretend we’re bright ninth graders and slow down a bit! What are the basic concepts?
Yes, because the gas laws will operate with auto-compression if GHGs are not available.
Thus the one with the GHGs switched off gets all additional warmth from auto-compression.
The one with GHGs still working gets some of the additional warmth from DWIR and the rest from a correspondingly reduced amount of auto- compression.
The proportions cannot be calculated from the IGL but the maximum attainable temperature from the two processes combined is calculated from the IGL.
The figure of 33K is not attained by chance, it is a consequence of the gas laws fixing the amount of additional surface energy required to hold the atmosphere off the ground in hydrostatic equilibrium against gravity. The IGL does not care whether the necessary energy comes from DWIR or auto-compression but will adjust to the individual situation by varying surface densities as necessary to switch between the two.
Willis Eschenbach February 14, 2018 at 1:15 am:
“So here is the important question:
Using just the Ideal Gas Law, can you tell which planet gains additional warmth from GHGs and which planet gains additional warmth from adiabatic auto-compression?”
Willis, it doesn’t matter. Badger’s argument (OK, the one Holmes/Frolly used, but which he attributes to Reverend Badger; I’ll consistently call it “Badger’s argument” from now on unless someone objects) is not concerned with WHY something is heating; you’re reading too much into it. It’s only concerns proving why something CANNOT be heating from a particular source, in this case, GHGs. Please read the argument carefully. Let’s not get confused about what the argument is actually saying.
Willis,
I agree with you, there is a problem of too many variables. So I want to go back to the basics, the real basics of planetary atmospheres. How many variables can we identify that are completely independent?
Well #1 on my list is the size of the planet. Planetary mass determines the strength of the gravity at the solid surface upon which the atmosphere rests (the datum level for the potential energy equation). The planet’s gravity at the surface determines the escape velocity; it is the escape velocity that determines the minimum molecular weight of a species of gas that can be retained in the atmosphere under a given loading of incoming solar radiation.
So that means that #2 on my list is the average orbital distance of the planet from its parent star. This distance (and orbital trajectory) determines the maximum surface temperature that a solid planet without an atmosphere can achieve. Mercury, with an escape velocity of 4.3 km/s, does not retain an atmosphere. The moon Titan, orbiting the Sun at the distance of Saturn, has a lower escape velocity of 2.64 km/s and so retains a cold nitrogen rich atmosphere that is denser than the Earth’s.
The next independent variable on my list #3 (and possibly the most contentious one) is the speed of the planet’s rotation. The speed of a planet’s daily rotation determines the latitudinal reach of the primary atmospheric cell, the Hadley Cell. On Earth, a fast rotating planet, the Hadley Cell is constrained by the “Coriolis force” and does not extend beyond the tropics due to the conservation of planetary angular (rotational) momentum. Poleward moving air at the top of the Hadley Cell is eventually forced down and returns to the surface. This forced descent allows the development in the Earth’s atmosphere of two more high latitude cells, the Ferrel Cell and the Polar Cell. On Venus however, a slowly rotating planet that does not have an equatorial bulge, the Hadley Cell extends across each hemisphere from the equator to its poles. It is the latitudinal reach of a planet’s Hadley Cell that determines the amount of solid surface area that can achieve direct ground to space radiative cooling. So in essence the faster a terrestrial planet spins the more it is able to cool.
So now let’s look at the gaseous constituents of planetary atmospheres and how these are affected by variables #1 & #2. The closer a planet is to a star, and the smaller its mass and escape velocity, the higher will be its atmospheric mean molecular weight. This is because the solar radiation loading will effectively “boil off” the lower molecular weight gases with elapsed time. Therefore Venus, with a lower escape velocity of 10.36 km/s than the Earth’s 11.18 km/s, and its closer orbit to the sun, can only retain the high molecular weight gas carbon dioxide rather than the low molecular weight gases such as nitrogen and critically water vapour. Add to this the slow rotational speed of Venus and its consequent poor ability to loose atmospheric heat by ground thermal radiation to space and we have a credible explanation for the dense carbon dioxide nature of the Venusian atmosphere.
So back let us go back to the use of the Ideal Gas Law to predict the average surface temperature of a terrestrial planet. Gravity is an independent variable that determines the maximum particle velocity of a species of gas that can be retained under a given thermal loading. This then means that planetary atmospheres evolve with time to have a molecular weight that is in balance with these forces.
The higher a planet’s mass, the greater its escape velocity and so the greater can be the mass of its retained atmosphere and consequently the higher its surface pressure can be. This is the potential energy part of the hydrostatic balance equation. The greater the radiation loading of a planet the higher will be its atmospheric mean molecular weight. This is the kinetic energy part of the hydrostatic balance equation. Therefore we can use two variables of surface atmospheric pressure (potential energy) and surface mean molecular weight (kinetic energy) to predict the third unknown, the average surface temperature of a terrestrial planet.
h/t Stephen Wilde
Thank you Stephen. This is the explanation I remember from my meteorology course in Environmental Science too many years ago.
Vertical structure of the atmosphere – Hydrostatic balance
Philip,
Everything in a good logical order; I might lift this whole and put in my next paper!
Next your explanation, we have the IPCC/Wilis saying “No; Man-Made CO2 Controls 98% of Everything!
Robert,
No worries. See if you can include the term “Comparative Planetology” in the title of your paper.
Philip
Willis,
This is just another obvious straw man you are throwing up – and is yet another diversionary tactic you are using. No-one has even claimed that the IGL itself can tell us the causes of a temperature change.
However, we can learn something about what caused a change in temperature from the MM version of the IGL, by the changes which must occur in the three gas parameters – under a known set of inputs.
For example, let’s say we instantly add to the E1 (Earth’s current) atmosphere, 0.03% of a NON-GHG mixture with an average molar mass of 44, and a density ~60% higher than air.
This quantity is 100% known; we know it’s molar mass, we can estimate pretty well how this addition will change the averages of the three gas parameters, as I did in the paper.
And the net change in atmospheric temperature comes directly out of this in formula 5.
There is NO reason to suppose that the addition of 0.03% of another gas with very similar properties (CO2) instead, could have an ‘anomalous’ effect, and multiply the above changes caused by the non-GHG up to 100-fold, as the IPCC certainly claims it would. This is because the MM version of the IGL makes NO DISTINCTION between gases, and cares only about how their addition changes the averages of pressure, density and molar mass.
BUT; as Don 132 said, the version of the IGL can tell us where the heat could NOT have come from. This is where the afore-mentioned E1 and E2 thought experiment comes in.
For your pleasure, I include Badger’s thought experiment here (just tidied up a little);
Consider; two very Earth-like rocky planets with Earth-like atmospheres orbiting at the same distance (1AU) from the Sun. We provide one with an atmosphere identical in every way to the present Earth’s; let this planet be E1. Now the other planet E2, is going to be identical in every way to E1 except for the composition of the atmosphere. The atmosphere of E2 will be very similar to E1’s atmosphere but will contain no greenhouse gases. It will be an almost identical mixture of the same gases as E1, but these non-greenhouse gases will be designed and mixed in such a way, that E2’s atmosphere possesses exactly the same measured atmospheric pressure, density and molar mass on the near-surface as E1 does.
Clearly the existing greenhouse gas hypothesis for Earth predicts that E1 should have a much higher (33K?) surface temperature than E2 because of its greenhouse gases. The hypothesis presented here, using formula 5, predicts that both planets will have identical temperatures. Notably, the predicted temperature figure for both planets, calculated from formula 5, is the same temperature as that predicted by the greenhouse gas hypothesis for the planet with the greenhouse gases, E1.
How could the possibility be eliminated, that a simple formula such as formula 5, (which contains no reference to the percentage of greenhouse gases in an atmosphere) accurately predicts the temperature of a planet with a very specific percentage of greenhouse gases, such as planet E1? Perhaps it would be informative to have a look at the atmospheres of other planetary bodies, some with up to 96% greenhouse gases in their atmospheres (Venus), and some others with none (Jupiter, Saturn). A simple formula with no reference to greenhouse gases could not be expected to predict the atmospheric temperature of eight such widely differing planetary atmospheres, by the measurement of just three gas parameters. And yet it does.
The only way that is possible, if the greenhouse gas hypothesis is correct, is that changes in the greenhouse gases’ percentage in an atmosphere must alter the pressure/density/molar mass in such a way as to make formulae 5 fit.
Yet, it would be theoretically possible to change the pressure/density/molar mass in exactly the same way numerically – by using non-greenhouse gases to reach the same parameter result – and the same predicted temperature.
Therefore, the greenhouse gas hypothesis must be incorrect.
The molar mass version of the ideal gas law is clear in that since these two planets have the same density, pressure and molar mass, they must also have the same temperature. Yet one of them contains greenhouse gases and the other does not. To conclude, either the molar mass version of the ideal gas law is correct (and both planets are the same temperature), or significant net warming from greenhouse gases is correct (and the planet with GHG is warmer) – both cannot be correct.
Thank you, Holmes, for continually beating them over the head with logic, which they’ve apparently spent a long time ignoring.
Robert,
An important phrase is ‘significant net warming from greenhouse gases’.
My description shows how a net warming from GHGs is avoided because auto-compression reduces to neutralise such warming when GHGs are present.
If one had sufficiently intense DWIR to account for ALL warming above S-B then radiation from above and conduction from below would merge to create an isothermal atmosphere with no auto-compression at all.
We can see an example of such a radiatively induced isothermal scenario in the section of our atmosphere just above the tropopause.
For a few kilometers vertically our atmosphere is isothermal as a result of ozone’s radiative ability cancelling out the lapse rate slope and preventing convection. A similar section exists between the mesosphere and the thermosphere because the radiative molecules in the thermosphere also heat up directly from receipt of incoming solar radiation.
In contrast, the non GHG atmosphere has the steepest lapse rate slope, the most vigorous convection and most auto-compression.
That is the means by which atmospheres with highly variable radiative characteristics all manage to observe the IGL.
Are E1 and E2 the same in terms of moderating the swings in temperature? The “greenhouse” effect could be the charge/discharge effect of atmosphere, and oceans. Less OLWR at T^4 means warming the average temperature.
Let T0 be the avg temp of an no atm earth.
Let T1 be the avg temp of an argon atm earth
Let T2 be the avg temp of the current earth
Let T3 be the avg temp of current earth with double the CO2.
At the current time, we cannot assert that we know T1 or T3. MM equations alone cannot answer the question: Will the earths avg temp be higher/lower with no GHGs atm or more GHGs atm? We cannot know because we cannot compute the shifts in wind, convection, conduction, dust, avg insolation, etc.
We can say that the likely change of T2 to T3 is very small, as low as 0.03 C, and unlikely to be as high as 1.2 C, the theoretical amount.
There is also the possibility of an average cooler earth.
None of the above violates the IGL.
Robert Holmes February 14, 2018 at 4:11 am
For your pleasure, I include Badger’s thought experiment here (just tidied up a little);
Consider; two very Earth-like rocky planets with Earth-like atmospheres orbiting at the same distance (1AU) from the Sun. We provide one with an atmosphere identical in every way to the present Earth’s; let this planet be E1. Now the other planet E2, is going to be identical in every way to E1 except for the composition of the atmosphere. The atmosphere of E2 will be very similar to E1’s atmosphere but will contain no greenhouse gases. It will be an almost identical mixture of the same gases as E1, but these non-greenhouse gases will be designed and mixed in such a way, that E2’s atmosphere possesses exactly the same measured atmospheric pressure, density and molar mass on the near-surface as E1 does.
Clearly the existing greenhouse gas hypothesis for Earth predicts that E1 should have a much higher (33K?) surface temperature than E2 because of its greenhouse gases. The hypothesis presented here, using formula 5, predicts that both planets will have identical temperatures. Notably, the predicted temperature figure for both planets, calculated from formula 5, is the same temperature as that predicted by the greenhouse gas hypothesis for the planet with the greenhouse gases, E1.
It does no such thing, you have assumed that they have the same temperature by setting three parameters the same. The correct way to do it is to set the atmospheres to have the same pressure and molar mass then predict the temperature and density.
By assuming that three variables are the same if you correctly ‘predict’ the measured temperature all you’ve done is show that it’s an ideal gas.
The only way that is possible, if the greenhouse gas hypothesis is correct, is that changes in the greenhouse gases’ percentage in an atmosphere must alter the pressure/density/molar mass in such a way as to make formulae 5 fit.
Which is exactly how the IGL works, in order to obey thermodynamic laws the temperature goes up and therefore to follow IGL the volume increases (and hence the density decreases).
The molar mass version of the ideal gas law is clear in that since these two planets have the same density, pressure and molar mass, they must also have the same temperature. Yet one of them contains greenhouse gases and the other does not. To conclude, either the molar mass version of the ideal gas law is correct (and both planets are the same temperature), or significant net warming from greenhouse gases is correct (and the planet with GHG is warmer) – both cannot be correct.
But it’s an assumption which you have made, not an observation!
“Which is exactly how the IGL works, in order to obey thermodynamic laws the temperature goes up and therefore to follow IGL the volume increases (and hence the density decreases).”
If density decreases then conduction becomes less efficient and less auto-compression occurs whereupon the surface temperature drops back again.
Net effect of GHGs = zero.
Phil. February 14, 2018 at 7:12 am:
“you have assumed that they [E1,E2] have the same temperature by setting three parameters the same. The correct way to do it is to set the atmospheres to have the same pressure and molar mass then predict the temperature and density.”
OK, what about Holme’s other argument
Robert Holmes February 14, 2018 at 4:11 am:
“However, we can learn something about what caused a change in temperature from the MM version of the IGL, by the changes which must occur in the three gas parameters – under a known set of inputs.
For example, let’s say we instantly add to the E1 (Earth’s current) atmosphere, 0.03% of a NON-GHG mixture with an average molar mass of 44, and a density ~60% higher than air.
This quantity is 100% known; we know it’s molar mass, we can estimate pretty well how this addition will change the averages of the three gas parameters, as I did in the paper.
And the net change in atmospheric temperature comes directly out of this in formula 5.
“There is NO reason to suppose that the addition of 0.03% of another gas with very similar properties (CO2) instead, could have an ‘anomalous’ effect, and multiply the above changes caused by the non-GHG up to 100-fold, as the IPCC certainly claims it would. This is because the MM version of the IGL makes NO DISTINCTION between gases, and cares only about how their addition changes the averages of pressure, density and molar mass.”
Stephen Wilde February 14, 2018 at 7:25 am
“Which is exactly how the IGL works, in order to obey thermodynamic laws the temperature goes up and therefore to follow IGL the volume increases (and hence the density decreases).”
If density decreases then conduction becomes less efficient and less auto-compression occurs whereupon the surface temperature drops back again.
Net effect of GHGs = zero.
Again that’s entirely your assumption, and the heat transfer isn’t only due to GHGs, it could be changed by albedo (e.g. clouds) or solar.
Stephen,
“My description shows how a net warming from GHGs is avoided because auto-compression reduces to neutralise such warming when GHGs are present.”
.
You raise very good points Stephen, well you are the meteorologist here!
I knew there must be strong negative feedbacks which effectively eliminated any ‘extra’ warming from GHG – but were not sure of the details. And you have detailed these very well!
Phil,
“It does no such thing, you have assumed that they have the same temperature by setting three parameters the same. The correct way to do it is to set the atmospheres to have the same pressure and molar mass then predict the temperature and density.
By assuming that three variables are the same if you correctly ‘predict’ the measured temperature all you’ve done is show that it’s an ideal gas.”
.
No.
I have done more than ‘show its an ideal gas’.
What I am doing here, is pointing out that if the GHG in E1 do cause anomalous warming, then they can ONLY do this by altering the three gas parameters in a way in which non-GHG can not..
BUT since a NON-GHG mix (E2) can also be created which matches these exact three E1 parameters, (and IF the MM version of the IGL is correct – and so must form the same temperature as E1) then the GHG’s cannot be creating any anomalous warming by altering the three gas parameters in an unusual way.
Stephen 7:25am: ”Net effect of GHGs = zero.”
Not observed. When GHGS in the form of clouds arrive on clear, calm nights, the temperature warms – there is no conduction becoming less efficient and less auto-compression occurring whereupon the surface temperature drops back again. T remains warmer until the clouds depart.
When the clouds clear to a starry night, the temperature invariably does drop again. This is observed by being outdoors and in every weather station data, radiometer data at which Stephen and Robert care to look. The night clouds cause the surface density to decrease then when they depart density increasse. Pressure may vary randomly during these events.
IGL and GHGs working at the same time in observations. No problem. IGL and GHGs are not mutually exclusive.
Trick,
Sure when its cloudy, it stays warmer longer – it also stays cooler longer when the sun is blocked!
This is a local effect which probably balances out globally.
The net effect of GHG can still be net zero planet-wide.
You should read this excellent paper, which outlines just how powerful the negative feedback from clouds is;
Cederlöf, M. (2014). Using seasonal variations to estimate earth’s response to radiative forcing.
”Sure when its cloudy, it stays warmer longer – it also stays cooler longer when the sun is blocked!”
Thus Robert now admits there is a greenhouse effect locally on Earth since calm, cloudy nights with added GHGs are also in his experience warmer than clear sky nights. The IGL and GHG effect are then clearly not mutually exclusive given this evidence and instrumental observations.
The Cederlöf pdf also discusses this GHG radiative effect globally.
Trick,
“Thus Robert now admits there is a greenhouse effect locally on Earth since calm, cloudy nights with added GHGs are also in his experience warmer than clear sky nights. The IGL and GHG effect are then clearly not mutually exclusive given this evidence and instrumental observations.”
.
I don’t know why some of you seem to have got the impression that I do not think that the greenhouse effect exists, or that greenhouse gases do not have different properties to non-GHG gases! No-one here is saying that.
What we are saying (quite clearly I thought) is that any warming from those greenhouse gases is subjected to a 100% negative feedback in the climate system.
What makes us think that? Many reasons, already detailed here and in my paper. The upshot of all this is that the null hypothesis for climate is wrong, and has been for decades, since it was first assumed that there was a net warming from GHG.
What we need to do is to re-define the null hypothesis of climate to reflect empirical science.
This will be addressed in my next paper, which I have already submitted and it is due out soon.
Willis Eschenbach
February 14, 2018 at 1:15 am
Assume we have two identical planets with identical atmospheres containing GHGs. They are both warmed to 255K by the sun. On planet E1, the greenhouse effect works, and this boosts the temperature by 33K to a final temperature of 288K.
Eschenbach starts by assuming his radiative theory is correct . The radiative theory has 2 wrong assumptions
1. T caused by energy flux ( flow) J/s/m^2
2. This energy is partly from the sun and partly from CO2 in the atmosphere
The question asked by Holmes is what causes the surface temperature.
It not energy flow, J/s/m^2. It is energy density J/m^3 and mass density, d.
Gravity cause energy density , pressure, p and also
Gravity cause mass density, d.
There is evidence for the gravity theory
The gravity theory is simpler .
Radiative warming has 2 very different sources energy
To summarise the science so far:
i) The ideal gas law gives a useful prediction of the surface temperature of a planet with an atmosphere regardless of whether GHGs are present or not and regardless of their quantity.
ii) That prediction invariably gives a higher temperature than that predicted by the radiation only S-B equation
iii) The only ways that energy in an atmosphere can get back to the surface in order to raise it above S-B are by conduction and/or radiation.
iv) If an atmosphere is non radiative then the only way is by conduction so the fact of back conduction heating a surface cannot be denied.
v) If one then adds a radiative gas then the DWIR energy reaching the surface below alters the density patterns at the surface so as to compromise the efficiency of conduction between atmosphere and surface. Conduction is more effective when the gases are denser or where the density variability in the horizontal plane is greater.
vi) It follows that more DWIR means less conduction and the surface temperature remains set at that predicted by the ideal gas laws.
vii) The ideal gas laws predict a maximum surface temperature that cannot be further enhanced by radiative gases in the atmosphere.
To avoid criticism please be aware that my desktop PC logs in with my real name but my ipad uses wildeco2014
I have mentioned that previously.
Stephen Wilde February 14, 2018 at 2:06 am
I’m sure you did, but I hadn’t come across it. How about you fix the ipad so it uses your real name? Because from my perspective, and I’m sure for many other people, it’s been a sockpuppet. I seriously thought that someone had entered the conversation and was agreeing with you … and that’s not on.
w.
Willis Eschenbach February 14, 2018 at 2:24 am
Willis, I figure out many months ago that Stephen was wildeco2014. It wasn’t that hard. It was pretty obvious.
Not convenient to change my ipad for various reasons so you will have to live with it.
Don132 February 14, 2018 at 2:52 am
Well, aren’t you special! Do you want a participation trophy?
For us ordinary mortals, and particularly for people who just drop in to read occasionally, it is not “obvious” in the slightest.
w.
Stephen Wilde February 14, 2018 at 2:54 am
I love this planet! It’s not “convenient” for you to stop being a sockpuppet so we all have to put up with you breaking site policy? You’re as bad as Robert Holmes with his “my computer made me do it”.
Dude, you are VERY lucky I don’t run this website. I’d bounce you permanently for continued sockpuppetry in a millisecond.
w.
(Stephen Wilde February 14, 2018 at 2:06 am Edit
To avoid criticism please be aware that my desktop PC logs in with my real name but my ipad uses wildeco2014
I have mentioned that previously.) MOD
Our site policy on multiple identities is very clear. Pick one persona and stick with it, whether it’s “convenient” or not. That goes for Robert Holmes too.
Otherwise, these alternate identities go straight to the bit bucket.
Sockpuppetry involves an intention to deceive. Such intention is not apparent where users simply have different devices with different log in names.
That said, if the Mods do object then let them say so and I’ll just use my desktop.
Here is a helpful tool presented by Frank: http://physics.weber.edu/schroeder/md/InteractiveMD.html
In this tool, you can see what happens when you introduce a gravity field to an atmosphere.
You can turn gravity on or off and play with it. Here are the settings I’m using:
Number of atoms: 1000
Box size: 100 (volume=10000)
Gravity = 0.02
Time step = 0.016
Steps per frame = 25
Assume that the height of the box represents the height of the troposphere.
Assume that the gas laws are in force, and do not confuse temperature with heat (although in the model it should be clear even if you do confuse them!)
What do you see? Are you not witnessing a gravity-induced temperature gradient? Or, to put it another way, aren’t you witnessing a pressure-induced temperature gradient?
Check the x10 button near the gravity setting if necessary.
That website says that the gravitational effect is ‘utterly negligible’.
So let me ask you this, Phil. Regardless of how accurate the gravity slider is for the scale used, imagine that the box is the height of the troposphere and we don’t care if the gravity slider matches earth or not, we only want to know the effect of any gravitational force on the atoms in the atmosphere. Whether we apply lots of gravity or a little gravity, do you not see a gravity-induced temperature gradient?
Anyone else? It’s a cool tool to try out and I think it’ll be helpful in sorting things out.
Don132 February 14, 2018 at 8:38 am
Nope. I see a gravity-induced PRESSURE gradient, which is a very different thing.
w.
Willis Eschenbach February 14, 2018 at 9:17 am
“Nope. I see a gravity-induced PRESSURE gradient, which is a very different thing.”
You’re right, Willis.
Now, would someone please refresh my memory regarding what happens to a gas as the pressure decreases? I believe it has something to do with the ideal gas laws.
Willis Eschenbach February 14, 2018 at 9:17 am
“Nope. I see a gravity-induced PRESSURE gradient, which is a very different thing.”
We’ve been through this before and I don’t believe anyone has refuted the conclusion I stated: as a gas parcel expands it necessarily loses pressure, and according to the gas laws it necessarily losses heat (as distinguished from the temperature, or velocity, of the atoms.) Even in an “isothermic” atmosphere, a temperature gradient must exist. Why? Because although the temperature (velocity) of the atoms stays the same, there are fewer collision between those atoms: the heat content of any parcel of air must decrease as the pressure of that parcel goes down, even if the temperature of the atoms remains the same.
That is what our molecular model shows.
Don132 February 14, 2018 at 9:53 am
Willis Eschenbach February 14, 2018 at 9:17 am
“Nope. I see a gravity-induced PRESSURE gradient, which is a very different thing.”
You’re right, Willis.
Now, would someone please refresh my memory regarding what happens to a gas as the pressure decreases? I believe it has something to do with the ideal gas laws.
It expands!
>>
Don132
February 14, 2018 at 10:26 am
. . . and according to the gas laws it necessarily losses heat . . . .
<<
There’s no heat (or mass) loss for an adiabatic process–look up the definition.
Jim
Jim Masterson February 14, 2018 at 10:34 am:
“There’s no heat (or mass) loss for an adiabatic process–look up the definition.”
Wikipedia: In thermodynamics, an adiabatic process is one that occurs without transfer of heat or matter between a thermodynamic system and its surroundings. In an adiabatic process, energy is transferred to its surroundings only as work.
I believe Stephen has already answered this objection.
So you’re contending that as a parcel of air loses pressure, its heat remains the same? What exactly are you claiming?
>>
So you’re contending that as a parcel of air loses pressure, its heat remains the same? What exactly are you claiming?
<<
I’m simply using the thermodynamic definitions of terms. The thermodynamic definition of heat is the transfer of energy across a system boundary due to a temperature difference. So a bucket of very hot water contains no heat. However, if the surrounding environment is cooler than the bucket of hot water (the bucket and water is our system), then there should be an energy loss across the system boundary. That energy loss can be identified as heat. Heat and work are boundary phenomenon.
The term that people here seem to be groping for is “internal energy.” When your packet of air rises, it is neither losing or gaining heat. However, its temperature will most likely fall because of pressure-volume changes. The product of pressure and volume is work and has the units of energy.
Jim
Jim Masterson February 14, 2018 at 11:33 am:
Thank you. It seems, then, that even an isothermal gas must cool as pressure decreases. Is anyone debating this, or are we agreed?
Does the “molecular action model” (linked to earlier) demonstrate the establishment of a temperature gradient through gravity? Does it demonstrate a pressure gradient? If a gas cools with decreasing pressure, then why would lower pressure not lead to lower temperature of a parcel of gas, even if this gas maintains internal energy (ie, is isothermal)?
I will brush up on my understanding of “temperature.”
Would Willis be so kind as to explain why a pressure gradient in a gas does not imply a temperature gradient?
Don132 February 14, 2018 at 8:38 am
“So let me ask you this, Phil. Regardless of how accurate the gravity slider is for the scale used, imagine that the box is the height of the troposphere and we don’t care if the gravity slider matches earth or not, we only want to know the effect of any gravitational force on the atoms in the atmosphere. Whether we apply lots of gravity or a little gravity, do you not see a gravity-induced temperature gradient?”
And here is Willis’ reply, which seems to deny a temperature gradient: Willis Eschenbach February 14, 2018 at 9:17 am:
“Nope. I see a gravity-induced PRESSURE gradient, which is a very different thing.”
Don132 February 14, 2018 at 2:06 pm
Sure. Because it would violate the Second Law. If it did, all we’d have to do is build tall insulated cylinders filled with air, and run a heat engine off the temperature difference between the top air and the bottom air … which is a perpetual motion machine forbidden by the laws of thermodynamics.
Your question was:
Don132 February 14, 2018 at 9:53 am Edit
Temperature drops, as you point out implicitly, as the pressure decreases. Now, imagine a tall insulated cylinder filled with air in an existing gravity field.
Where in it is the pressure decreasing?
The answer is “Nowhere”. The pressure at any given point inside is unchanging, whatever it might be. So your argument is inoperative.
Now, “heat” is defined as the spontaneous flow of energy from warm to cold. And thermodynamic laws apply inside the cylinder as well as outside. As a result, IF there is a temperature imbalance anywhere in the air in the cylinder, heat will flow spontaneously to that parcel of air. This will continue until the air inside the cylinder is perfectly isothermal.
Regards,
w.
”If it did, all we’d have to do is build tall insulated cylinders filled with air..”
Good idea. That has been done for real after being suggested in the 1880s. The various Ocean Thermal Energy Conversion (OTEC) installations that produce electricity fill them with a better fluid and place them in the ocean. They are demonstrated to work, no violation of 2LOT.
“…heat will flow spontaneously to that parcel of air. This will continue until the air inside the cylinder is perfectly isothermal.”
This was Maxwell’s suggestion in the late 1800s, however Bohren 1998 Chapter 4.4 proves isothermal T(z) is a slightly lower entropy solution than the non-isothermal solution which is obtained by maximizing entropy for the ideally insulated cylinder.
Willis Eschenbach February 14, 2018 at 5:25 pm:
Don: Would Willis be so kind as to explain why a pressure gradient in a gas does not imply a temperature gradient?
Willis: “Sure. Because it would violate the Second Law. ”
But would it, or are you simply making an assertion? For now I’m just going to make that observation. I believe that Stephen has already answered that argument, although I do not yet fully understand it (I’m working on it!) so I’m not qualified to repeat it here. I suspect you’re throwing up a smokescreen, because later on you say: “Temperature drops, as you point out implicitly, as the pressure decreases.” No Willis, I pointed that out explicitly.
Willis continues: “Now, imagine a tall insulated cylinder filled with air in an existing gravity field.
“Where in it is the pressure decreasing?
“The answer is “Nowhere”. The pressure at any given point inside is unchanging, whatever it might be. So your argument is inoperative.
“Now, “heat” is defined as the spontaneous flow of energy from warm to cold. And thermodynamic laws apply inside the cylinder as well as outside. As a result, IF there is a temperature imbalance anywhere in the air in the cylinder, heat will flow spontaneously to that parcel of air. This will continue until the air inside the cylinder is perfectly isothermal.”
That’s a wonderful thought-experiment but I’m afraid it proves nothing. It’s does not clarify anything, either; in fact, it confuses things.
The pressure inside the cylinder at each point remains the same: true enough (maybe!) But thanks to our wonderful molecular motion model (which Frank pointed us too; I trust we all understand what this is) we can see that even as the gravity at any point remains constant, the number of atoms at each point does not. We can visualize our molecular motion model as your cylinder, only fatter. It is the height of the troposphere. We can claim that when we apply gravity, at each point the pressure remains the same! And, as you say, we can also acknowledge that heat is transferred to each atom/molecule in turn until the cylinder is isothermal– but what does “isothermal” mean? It means that the internal energy of all the atoms and molecules that are the same mass, etc, is the same. Does that mean that the temperature of the atmosphere is the same at each point? No. Why not? Because there are more atoms on the bottom of the cylinder than at the top, and we can measure the temperature of the cylinder at the top and then at the bottom to confirm this.
So to answer your question regarding where the pressure is decreasing in your imaginary cylinder, the answer is, paradoxically, nowhere and everywhere. For the cylinder as a whole, the pressure remains the same, and so we can make the dubious claim that at each point the pressure remains the same. But since gravity is at play, the pressure at the bottom must necessarily be greater than at the top: the amount of pressure from point to point is constantly changing, even if we make the assertion (which seems to me not quite accurate) that the pressure at each point is unchanging.
Willis, who is hand-waving here?
Don132 February 15, 2018 at 1:16 am
Let’s build a cylinder (internal diameter 1 m^2) all the way into space, top can be open so radiation can still leave to space. At the bottom a heating element mimicking solar energy warming Earths surface.
First at 1m we have a plate perfectly containing the air we let in at the bottom.
Letting in air at surface pressure, the plate has to be weighted with ~10.000 kgs to contain the atmospheric pressure at the surface.
Now we move the plate to ~5500 m. again letting in air at the bottom. The plate now has to weighted with ~5.000 kgs to contain the pressure of the atmosphere.
Remove the plate completely and the cylinder will fill up.
Obviously we have a pressure gradient, contained by the weight of the air above the height we measure the pressure. The air is in Hydrostatic Equilibrium against gravity at every level.
This means pressure is lower at each higher altitude, but remains constant OVER TIME at each altitude,
NO expansion.
The temperature gradient depends on the amount of energy the heater at the bottom supplies, and how fast this energy is transported to space.
Introduce a second heater inside the cylinder around 20 km up, and the temperature increases around this heater, mimicking solar UV interacting with oxygen/ozone in the stratosphere.
>>
Willis Eschenbach
February 14, 2018 at 5:25 pm
Because it would violate the Second Law.
<<
Except that atmospheres, which are usually modeled as “closed systems,” need not obey the second law. The second law only applies to “isolated systems” or systems where isolation can be assumed. The Universe obeys the second law, but the Universe is assumed to be an isolated system.
Jim
Phil. February 14, 2018 at 7:27 am:
“There’s an optional uniform downward force, controlled by the Gravity slider. The magnitude of this force, however, is not meant to be realistic. Earth’s gravitational constant is utterly negligible in the units used here (a little over 10−13 for argon).”
Do we care if the magnitude is realistic, so long as we can observe the effects of any magnitude of gravity?
Don: FWIW, you can use the molecular dynamics simulator to see how the volume of the atmosphere increases. With a strong gravitational field and a low temperature, essentially all of the molecules will stay in the bottom half of the box. As you increase the temperature, the height of the average molecule will rise, increasing the volume (area in this 2D demo) and the density at any height will go done. (Once a significant number of molecules to bouncing off of the top of the box, you are simulating a closed container, not an atmosphere held in place by gravity.)
Perhaps this will allow you to see that those who create problems with a defined a surface pressure and a defined surface density simultaneously defined a surface temperature. The surface temperature doesn’t arise alone from the surface pressure, it arises from a combination of surface pressure and surface density. If more GHGs warm the surface, then the surface density will fall, the top of the atmosphere will rise and the height of the average molecule will increase.
Upthread I’d asked Robert the following question, which to date he has not answered:
It is an important question because Robert has claimed that he can use the Ideal Gas Law by itself to establish that the warming of the earth is NOT from the poorly-named “greenhouse effect”, viz:
My question is designed to find out just how that is supposed to work.
Onwards,
w.
PS—Distractions are useless, I’m a patient man. Just answer the question. It’s a simple “yes/no”. Can the IGL by itself distinguish between planets E1 and E2?
Willis,
“My question is designed to find out just how that is supposed to work.”
.
I wish it were.
I made no answer because your question is yet another attempt to muddy the waters by creating another thought experiment, even using the same E1 and E2 (but meaning different things) that ‘our side’ used before.
You are doing this because you have no answers to our thought experiment.
On top of that, your thought experiment is disingenuous because it is un-physical – our is not.
Robert Holmes February 14, 2018 at 2:08 pm
To be clear, Robert, I didn’t understand your thought experiment so I let it be.
Whoa, whoa. E1 is the earth as I think it exists, with the greenhouse effect working and the adiabatic autocompression not working.
And E2 is the earth as you think it exists, with the greenhouse effect not working and the adiabatic autocompression working.
Which of those are you claiming is “unphysical”?
Look, Robert, I don’t care if you don’t want to answer a very simple question. Here it is. Two earths. One the way you say it works, one the way I say it works.
Can the Ideal Gas Law by itself distinguish between the two earths?
Answer or not, your choice, but don’t bother me with excuses if you are unwilling to answer.
Now, if you’d like to state your thought experiment equally clearly, I’ll give a shot at answering it … but please recall, I am not the one defending a novel and untested theory here. That would be you.
It’s not my theory, it’s yours, so whether or not I answer questions about your theory is meaningless. But when you refuse to answer questions about your own theory, I fear that’s a very bad sign.
As such, it would behoove you to answer simple questions about your theory, and stop with your excuses for you not answering.
w.
I will answer your hypothetical.
Your hypothetical is un-physical and has no sensible answer, because on one planet the GHE works, and on the other it doesn’t.
This invalidates the exercise – making it meaningless and self-contradictory.
Why?
Because we know from the accuracy of the MM version of the IGL across 8 disparate planetary atmospheres in the calculation of temperatures, (from 0% GHG up to 96% GHG) that if GHE gases create anomalous temperatures, then this effect MUST affect one, two or all three of the gas parameters in anomalous ways on all planetary bodies which do have some GHG. i.e.; GHG must be ‘special’ and different to other gases in relation to their effects on the parameters in the MM version of the IGL.
Yet it is clear that a non-GHG mix could theoretically be chosen in such a way as to mimic all three parameters on all 8 of the bodies – and so therefore their resultant average temperatures.
In this way, either the MM version of the IGL is correct or anomalous warming from the GHG is correct; both cannot possibly be correct.
Also remember; to my knowledge no-one here has claimed that the GHE does not ‘work’ and that GHG are ‘not’ GHG. What we are saying is that there is no anomalous net warming from GHG such as CO2 in the troposphere. The proof is above.
Robert Holmes February 14, 2018 at 8:05 pm
Also remember; to my knowledge no-one here has claimed that the GHE does not ‘work’ and that GHG are ‘not’ GHG. What we are saying is that there is no anomalous net warming from GHG such as CO2 in the troposphere. The proof is above.
This is not a proof, it’s an assertion!
Robert Holmes February 14, 2018 at 8:05 pm
Well, you may indeed answer it, but so far you haven’t. I know, because it’s a yes-or-no question, and you haven’t taken either position.
Here’s my “hypothetical” for those just joining the thread. Two planets, identical atmospheres, same gases, same density, same pressure, same molar mass, same temperature.
One planet works as I claim, where the greenhouse effect is real and the “adiabatic auto-compression isn’t real and has no effect.
The other works as you claim, where the greenhouse effect isn’t real and has no effect, and the “adiabatic auto-compression” is real. So the “hypothetical” has been and still is:
Can we tell the difference between the two planets USING NOTHING BUT THE IGL, as you claim above, viz:
And if you can distinguish between the planets based solely on the IGL as you claim above, please let us in on the secret.
w.
It is a process of deduction.
The IGL tells you approximately what the surface temperature enhancement should be for a given set of parameters.
For your two planets that is assumed to be identical.
The only two ways that the atmosphere could heat the surface are by radiation or by auto compression and you can apply logic from there.
Starting from that information you then need to know whether DWIR is present in the two atmospheres.
For a planet with no DWIR the gas laws tell you that it must all be auto compression because there is no other option.
If DWIR is present then the gas laws tell you that there needs to be an apportionment between the surface heating effects of DWIR and auto compression because there are two active processes ongoing.
If DWIR provides the full amount of the surface temperature enhancement then the gas laws tell you that auto compression must drop to zero because there is no other option.
Logically there are no other possible scenarios otherwise you end up with a surface temperature enhancement different to that indicated by the application of the IGL.
Whether that satisfies your definition of the IGL doing the telling is neither here nor there.
Anthony,
I understand the rule on sockpuppetry, and agree with it.
However there was no intention to deceive on my part (or I believe on Stephen’s part).
Anyway, henceforth I will ensure that any posts I am silly enough to put on Willis’s WUWT trains, will only be in my name.
Robert, I would hope that you’d continue with this argument, as you’re the only one who can answer for yourself and you are the primary target.
And, I need help.
I was about to give up on this thread because the points I have made several times are just not sinking in. I’ll give it a bit longer.
I replied ‘yes’ to Willis’s question as to whether one can tell from the IGL what is causing the warming.
The reason I gave was that if there are no GHGs then ALL surface warming must be down to auto compression otherwise the IGL will not work.
If there are GHGs then the surface warming is a result partly of DWIR from those GHGs and the rest is down to auto compression otherwise the IGL will not work.
Also, if the GHGs could ever account for ALL the warming then auto compression would cease because convection would cease.
I would like to hear Willis’s response.
The thing is that if the IGL is to work in the whole variety of situations then the only way it can happen is if auto compression declines when DWIR increases.
I have given a plausible mechanism for that namely that heating from DWIR would reduce surface air density which would compromise conduction and therefore auto compression.
The opponents to the gravito-thermal hypothesis have yet to respond to that suggestion.
It seems to me that such a proposition satisfactorily accommodates both the radiative AND gravito-thermal hypotheses into a single scenario that accords with observations and various well known physical laws.
If someone has an objection free of bias then let’s hear it.
Stephen, it’s my belief that you are so familiar with the material and so quick-minded that I simply can’t follow you, for the most part. This is my fault.
But for now, our collective task is to demonstrate the flaws in Willis’ argument– or, alternatively, to concede defeat if necessary. So far it seems to me that the most reasonable and logical arguments are given by the gravitationists.
I’ve noticed that a common problem is that people gloss over/dismiss arguments without giving them any real consideration, and I’m sure that’s happening with your comments.
So I hope you’ll stick with it. For my part, I’ll try to break down the arguments and the flaws in logic. At present it seems to me that at least we’re getting them to concede that an atmosphere under gravitational influence forms a pressure gradient that leads to a temperature gradient– which is some progress. I believe that your explanations are the key to resolving this particular dispute, although your arguments are glossed over and dismissed by those who assume you must be wrong.
I wish I knew more, but I’m just a dumb philosopher (but learning every day.) On the other hand, it seems that we are in desperate need of a philosopher.
Don
“I’ve noticed that a common problem is that people gloss over/dismiss arguments without giving them any real consideration, and I’m sure that’s happening with your comments.”
.
The fact that Anthony/Willis listed my paper (and even Prof. Svensmark’s!) under ‘Bad Science’ with the ‘shrinking beetles’ and the ‘THE CONCEPTUAL PENIS’ tells you everything you need to know about their bias.
Really wasting our time here.
The battle against AGW and the CO2 fraud will be fought in the literature and then in the media, not here where people have made up their minds already, and have chosen the wrong side.
Yes, that has been happening all along and dealing with multiple objectors in parallel is very difficult.
When they make points I sit back and think ‘what if they are right, what follows on from that’ and usually I find a flaw or a different way of looking at it that provides a different conclusion.
Unfortunately they tend not to give my points similarly thoughtful consideration because they are stuck in a paradigm and cannot get out.
The reason that I continue is that the objections have steadily crystallised my own ideas such that they are now becoming simpler and more accessible but they still require level of broad background knowledge that few contributors have so they find it difficult to believe what I tell them , especially about meteorology which is a highly specialised and rarely studied discipline.
Robert:
“The battle against AGW and the CO2 fraud will be fought in the literature and then in the media, not here where people have made up their minds already, and have chosen the wrong side.”
I believe we are close to either forcing a concession, or else making it crystal clear that WUWT is simply playing games with logic: avoiding checkmate by moving the king off the board and to the other side, and pretending nothing happened. I still have faith that most people will realize what’s going on.
Stephen Wilde February 14, 2018 at 2:32 pm
Thanks for the reply, Steph. I understand that, but I still don’t see how you are saying that the Ideal Gas Law can distinguish between the two.
Here’s the problem. All of the variables in the IGL are the same in both earths—T1, the temperature in one Earth is equal to T2, the temperature of the other Earth. And the same is true for pressures P1 and P2, and so on for all variables.
Since all variables in the IGL are IDENTICAL for the surface of the two earths … how can the IGL tell what is causing the warming?
Regards,
w.
It is a process of deduction.
The IGL tells you approximately what the surface temperature enhancement should be for a given set of parameters.
For your two planets that is assumed to be identical.
The only two ways that the atmosphere could heat the surface are by radiation or by auto compression and you can apply logic from there.
Starting from that information you then need to know whether DWIR is present in the two atmospheres.
For a planet with no DWIR the gas laws tell you that it must all be auto compression because there is no other option.
If DWIR is present then the gas laws tell you that there needs to be an apportionment between the surface heating effects of DWIR and auto compression because there are two active processes ongoing.
If DWIR provides the full amount of the surface temperature enhancement then the gas laws tell you that auto compression must drop to zero because there is no other option.
Logically there are no other possible scenarios otherwise you end up with a surface temperature enhancement different to that indicated by the application of the IGL.
Whether that satisfies your definition of the IGL doing the telling is neither here nor there.
–Stephen Wilde
February 14, 2018 at 2:32 pm
I was about to give up on this thread because the points I have made several times are just not sinking in. I’ll give it a bit longer.
I replied ‘yes’ to Willis’s question as to whether one can tell from the IGL what is causing the warming.
The reason I gave was that if there are no GHGs then ALL surface warming must be down to auto compression otherwise the IGL will not work.
If there are GHGs then the surface warming is a result partly of DWIR from those GHGs and the rest is down to auto compression otherwise the IGL will not work.
Also, if the GHGs could ever account for ALL the warming then auto compression would cease because convection would cease.
I would like to hear Willis’s response.
The thing is that if the IGL is to work in the whole variety of situations then the only way it can happen is if auto compression declines when DWIR increases.
I have given a plausible mechanism for that namely that heating from DWIR would reduce surface air density which would compromise conduction and therefore auto compression.–
Well water vapor does reduce density.
The obvious is that water vapor is greatest GHGs, and obviously it affects lapse rate, but I would say it’s due to water vapor not being an ideal gas- it condenses at earth pressure and temperature- or not to do with the radiant properties of water vapor.
The there is idea that GHGs can actually increase the kinetic energy gases- or actually warm atmospheric gas.
And this would occur a lower elevation and it seems to me to be minor effect, though again water vapor would have the largest effect.
–The opponents to the gravito-thermal hypothesis have yet to respond to that suggestion.–
What seems to missing from GHE theory and gravito-thermal hypothesis is warming effect of the water of the ocean. Or earth oceans have average temperature of 17 C and land surface average is 10 C- that doesn’t seem to me to be explained by either.
I would say if you add ocean of water to Mars, Mars average temperature would increase, obviously Mars water vapor of about 210 ppm would increase, but I would say it’s the ocean which is doing most of the warming. Or if seal the oceans so they don’t evaporate and don’t increase Mars global level of water vapor, it would increase the average temperature of the surface of Mars- particularly in tropical zone of Mars- or would less effective say, about 45 degree latitude or higher. Though similar to solar pond which works at 45 degree or higher latitude on Earth, they would likewise work on Mars..
Well since I wrote about the ‘Hot Water Bottle Effect’ some years ago I agree with your concern about accounting for the oceans but that doesn’t affect the background baseline for a gaseous atmosphere because the weight of the atmosphere itself controls the amount of solar energy that the oceans can store.
Best keep off that for present purposes because things would get side tracked otherwise.
I was trying to find the difference of solar pond at higher latitude. And sort found it- but in brief search I found something on “natural solar ponds” rather than human built ones, which I though was interesting:
https://books.google.com/books?id=nbH1QDIv11gC&pg=PA101&lpg=PA101&dq=Comparison+of+Caribbean+Solar+Ponds+with+Inland+Solar+Lakes+of+British+Columbia&source=bl&ots=fF_bCN5owC&sig=2ZclRQ5t_9eU1hD-XXxw1GNJbH0&hl=en&sa=X&ved=0ahUKEwiOxrSg6KbZAhVP32MKHbSKB8YQ6AEILjAC#v=onepage&q=Comparison%20of%20Caribbean%20Solar%20Ponds%20with%20Inland%20Solar%20Lakes%20of%20British%20Columbia&f=false
If that doesn’t, tiny:
https://tinyurl.com/y7zly6zg
“Well since I wrote about the ‘Hot Water Bottle Effect’ some years ago I agree with your concern about accounting for the oceans but that doesn’t affect the background baseline for a gaseous atmosphere because the weight of the atmosphere itself controls the amount of solar energy that the oceans can store.
Best keep off that for present purposes because things would get side tracked otherwise.”
But I live to be side tracked.
So can you tell me the temperature of solar pond on Mars?
Put greenhouse with .1 psi, air pressure over a solar pond, at say the equator of Mars.
My problem with the greenhouse effect theory is it doesn’t predict, jack.
So call it pseudoscience, because science predicts.
So with your on track idea, predict,
https://www.nasa.gov/audience/forstudents/5-8/features/nasa-knows/what-is-jupiter-58.html
http://hockeyschtick.blogspot.co.nz/2014/12/how-gravity-continuously-does-work-on.html
A tunnel might be drilled right through Earth, and insulated perfectly against geothermality. The atmosphere would follow the lapse rate to the centre from both ends. For instance, 7000km times 7K is 49000K, and there happens to be no net gravity either at the true centre. That just leaves insolation, pressure and its mass backing. Game over. Of course, the tunnel has an opposing force from the other end acting for gravity’s pseudoforce.
GHG effects are not obvious, which is why we can discount them, but any deviation from IGL would have an explanation. Geothermality would show up as what you must subtract to get that 49000K. Jupiter above would be a case in point, though IIRC measurements taken show a normal IGL lapse rate as far as they go. However, a big enough gas giant is a sun once it ignites atomically. To posit a GHE inside IGL is not rational. It is what it is, the IGL, and it stands alone. Any deviations can be accounted for by variations from the assumed nil effect of molecular volume in the pretty empty spaces of gases.
“A tunnel might be drilled right through Earth, and insulated perfectly against geothermality. The atmosphere would follow the lapse rate to the centre from both ends. For instance, 7000km times 7K is 49000K, and there happens to be no net gravity either at the true centre. That just leaves insolation, pressure and its mass backing. Game over. Of course, the tunnel has an opposing force from the other end acting for gravity’s pseudoforce.”
A 7000 km radius sphere is 1.44×10^12 cubic km
A 3000 km radius sphere as 1.13×10^11 cubic km
1.44×10^12 cubic km – 1.13×10^11 cubic km = 1.327 ×10^12 cubic km cubic km.
1.327 ×10^12 = 13,27 ×10^11
13,27 ×10^11 / 1.13×10^11 is 11.74
If sphere had uniform density, a 7000 km radius sphere with 3000 km radius hole
in middle of it is 11.74 times more massive
than same density sphere of 3000 radius.
Gravity is related to mass.
Or when goes down 4000 km of 7000 sphere, one will have more mass below you
but you will also have considerable amount of mass above you, which will lessen
the acceleration of gravity.
A 1000 km radius sphere is 4.19×10^9 cubic km.
So at 6000 depth, one still has more mass below you, but difference is less with mass above
you. And at 7000 depth it’s balanced with zero gravity.
Now how big is hole in terms of diameter. Let’s make it small: 10 meter diameter.
so area of 10 meter is 78.53975. So 1000 meter depth is 78539.75 cubic meters
and 1000 km is 78,539,750 cubic meters. Hmm
let’s go with 100 km. And make simply lets say its close to Venus: 737 K and density of 65. kg/m3
And keep it at 65 kg per cubic meter below 100 km.
So 1000 km at 65. kg/m3 is 78,539,750 times 65:
5105 million tonnes of air. So unless density increase a lot [and it would, probably not by more than say by factor of 100] not enough mass to worry about. But much bigger diameter hole could be a different matter. Or wondering if enough in comparison to Earth entire atmosphere- but anyhow. Hmm if hole was mad 1000 or 2000 km, would it “self-drill”?
Now I have a basic problem with idea in general- I don’t think you increasing the velocity of the gas molecules, rather one increasing the density of a gas. And Co2 is become a liquid like thing, under enough pressure:
“More specifically, it behaves as a supercritical fluid above its critical temperature (304.25 K, 31.10 °C, 87.98 °F) and critical pressure (72.9 atm, 7.39 MPa, 1,071 psi), expanding to fill its container like a gas but with a density like that of a liquid.” wiki
Or doesn’t work with liquid but is property or nature of gases.
But a problem with that is you can’t make oxygen or Nitrogen a liquid with pressure.
But whole idea of increase density is that molecules of gas will increase their amount collision with themselves [if going at same average velocity] and things which aren’t gases like liquids and solids.
So, I could guess that at some increase in density, that a gas doesn’t get in increase in the amount of collision with itself and other things.
But it seems the nature of O2 and N2 of it having weird aspect of not becoming a liquid with just pressure [you can easily make liquid oxygen and nitrogen a liquid by cooling them], tend to indicate they might need to be quite dense before reaching this- imagined point.
Oh well it’s late, but it’s been fun..
— You’re right, Willis.
Now, would someone please refresh my memory regarding what happens to a gas as the pressure decreases?
Temperature drops, as you point out implicitly, as the pressure decreases. Now, imagine a tall insulated cylinder filled with air in an existing gravity field.
Where in it is the pressure decreasing?–
The pressure decreases as you go up the tall cylinder- both density and pressure.
An column of air decreases in pressure and in density as you go up- and increase pressure
and increases density as go down [obviously].
And how slowly or quickly depend upon amount of gravity- no gravity is none, 2 gees is more than 1 gee Earth. And don’t need “natural gravity” it works with artificial gravity- with centrifuges one making high levels of artificial gravity. Also things like a spin cycle of a washing machine.
Generally gravity is regarded as weak force though works over large distance. So any dramatic effect will be with very tall cylinder- km vs meters. But more manageable distance are possible with the higher gee artificial gravity machines.
gbaikie February 14, 2018 at 6:36 pm Edit
That’s the pressure on me as I go up or down inside the cylinder, not the air pressure. The air doesn’t change temperature because the pressure on me is changing.
The issue is that there is no location inside the cylinder where the air pressure itself is changing, so there is no ongoing pressure, density, or temperature change.
Thanks,
w.
-That’s the pressure on me as I go up or down inside the cylinder, not the air pressure.–
The pressure on you is same as the air pressure
“The air doesn’t change temperature because the pressure on me is changing.”
Right, it’s not pressure affecting temperature, it’s air density which affects air temperature;
Or air temperature is Kinetic energy. And Kinetic energy – mass time 1/2 times velocity squared.
So density is mass of air molecules and velocity = average velocity of molecules. of a volume of air.
And pressure does not have factor in: KE = mass 1/2 times velocity squared,
But there is relationship of pressure to air density- pressure is mass of air molecules above you-. it’s weight- and weigh depends upon gravity.
So kinetic energy of air is average velocity of the air molecules and mass of volume of air- say cubic meter of air which at sea level about 1.2 kg and the pressure is the 10,000 kg of weight weight which is above this cubic meter of air.
“The issue is that there is no location inside the cylinder where the air pressure itself is changing, so there is no ongoing pressure, density, or temperature change.”
Air pressure is lower the higher elevation of the air in a very tall cylinder.
Or 50 km tall cylinder has 14.7 psi at bottom and close to 0 psi at the top
If it’s 5,5 km tall, it’s about 14.7 / 2 = 7.35 psi
” For example, in Earth’s atmosphere the pressure at a height of 5.5 kilometers is only 50% of the surface pressure.”
And:
“About 99% of the total atmospheric mass is concentrated in the first 20 miles (32 km) above Earth’s surface.”
Or pressure at 32 km is low. Or about 14.7 times .01 = .147 psi
So pressure is related to density of air.
There two ways to increase air temperature- increase mass or increase average velocity
of gas molecules.
If push gas molecules into smaller volume, you increase mass per volume.
That increase temperature. If have volume of air and increase the volume- less molecules per a volume, you decrease air temperature.
Or you keep the volume the same and heat the air- increase the average velocity of air molecules.
Or to cool the air- lower the average velocity of molecules, which can slowed by air heating the cooler container walls.
With a fixed volume controlled by pressure vessel, if heat gas, it increased pressure, and when you cool air, it decreases pressure.
Earth’s atmosphere is not a pressure vessel, though if you wish one call it a one sided “pressure vessel” with gravity forcing gas to the earth surface [which is the one side]
A pressure vessel with have uniform pressure- Earth’s “one side vessel” doesn’t.
But if you have a very tall pressure vessel, Earth gravity will cause a pressure vessel to have uneven pressure- it will have more pressure at sea level than at 5.5 km height- don’t think it will have twice the pressure- but it will be more pressure.Though if pressure is 14.7 psi at sea level and same temperature as earth’s air temperature, it will be twice the pressure as compared 5.5 km.[or one can assume a pressure vessel has pressurized gas in it- which could be 50 or 100 or whatever psi, Also one could different gases in vessel which alters it- but whatever gas and whatever pressure it will have higher pressure at sea level as compared to at 5.5 km higher in the pressure vessel ],
Willis Eschenbach February 14, 2018 at 8:12 pm:
“That’s the pressure on me as I go up or down inside the cylinder, not the air pressure. The air doesn’t change temperature because the pressure on me is changing.
The issue is that there is no location inside the cylinder where the air pressure itself is changing, so there is no ongoing pressure, density, or temperature change.”
That’s word-play that gets us absolutely nowhere, except to confuse the issue. Do you really think anyone is so unperceptive as to buy that?
I think people are perceptive enough to notice what is actually going on. So let me try to break through to you another way, by noting that gases cool AS THEY EXPAND. They don’t take up different temperatures because they are at different pressures. We can have three tanks of gas, each at a different temperature, but all at the same pressure. The gas only changes temperature WHEN IT EXPANDS.
So please point out to us all, just where in the tall insulated cylinder the gas is expandING … that’s right, nowhere.
I suspect that you are confusing what happens when you turn on gravity (the upper part expands and cools, the lower part compresses and warms) and what happens in the long run (neither the upper or lower part are expanding or compressing, so no warming or cooling).
In any case, if you think some part of the air in the cylinder is expanding, please point it out to me so we can discuss it. Otherwise, I continue to hold that THERE IS NO EXPANSIVE OR COMPRESSIVE COOLING OR HEATING INSIDE THE CYLINDER.
w.
Willis Eschenbach February 15, 2018 at 10:08 am:
“So please point out to us all, just where in the tall insulated cylinder the gas is expandING … that’s right, nowhere.”
Don’t gases cool as the pressure decreases? How could they not? At the very bottom of the cylinder the atoms (for example) are all bunched together so there’s a great deal of internal energy there and hence the collective energy (temperature) is relatively high; at the top of the cylinder there are almost no atoms, so the collective internal energy there (which must equal the temperature) is much lower.
Put your hand in the thermosphere, in which the molecules are moving very fast (high internal energy) yet the pressure is very low. What do you feel?
When a gas expands, it lowers the pressure. How else does that expanding mechanism work? Please explain.
The three tanks of gas you use for illustration aren’t the same as an atmosphere. We’ve already examined how the heat will conduct out in the case of a gas compressed in, say, a bicycle tire. In your insulated cylinder, is that possible? With the three tanks of gas, are they all insulated?
Did no-one read my paper?
This has been solved experimentally;
Graeff, R. W. (2007). Viewing The Controversy Loschmidt–Boltzmann/Maxwell Through Macroscopic Measurements Of The Temperature Gradients In Vertical Columns Of Water. Preprint. Additional Results Are on the Web Page.
A temperature gradient was found to form in vertical sealed tubes of both water and air.
Don132 February 15, 2018 at 10:36 am
Absolutely … but there is no location inside the cylinder where the pressure is either increasing or decreasing. It is in steady state, with no parcel of gas anywhere which is increasing or decreasing in pressure.
All you’re doing is measuring the pressure at different points within the cylinder … but that does NOT mean that a gas parcel in the cylinder is changing pressure.
You also say:
You seem to think that it is impossible for the gas to be isothermal, that it violates the Ideal Gas Law or something. It is not impossible. All that is required is that the Ideal Gas Law be satisfied everywhere. The IDL says that PV = n RT, where P is pressure, V is volume, n is number of atoms, R is a constant, and T is temperature.
So let us compare a cubic centimeter of air at the top and bottom of the cylinder. We get two calculations:
P(b) V(b) = n(b) R T(b)
and
P(t) V(t) = n(t) R T(t)
where (b) means the bottom and (t) means the top.
Now, let’s assume for the discussion that the column is isothermal, same temperature top and bottom. In that case T(b) = T(t), which we’ll call just “T”.
Solving both equations for T and setting them equal we get
P(b) V(b) / n(b) R = T = P(t) V(t) / n(t) R
Since the volumes are equal and the constant “R” is the same on both sides, this simplifies to
P(b) / n(b) = P(t) / n(t)
Now, what has to happen to make that condition true? All it needs is for n to vary proportionally to P, so that if P doubles, n doubles, and if P goes to half, n goes to half.
But this is exactly what happens in the real world. We know that from the ideal gas law itself—if all other variables are held constant, n varies as P. Other things being equal, twice the pressure means twice the number of atoms in one cc of the gas.
So an isothermal insulated vertical column of gas in a gravitation field does NOT violate the IGL. Pressure varies from top to bottom, as does “n”, the number of atoms, but the temperature is the same top to bottom.
w.
Trouble is that using a cylinder is not a good shape to use to model the vertical profile for the atmosphere of a planet.
It appears that you might need to use something else, more likely an inverted cone.
General relationships between pressure, weight and mass of a hydrostatic fluid
Use what you like, cone or cylinder, it will still end up isothermal.
w.
It won’t for a sphere with convective overturning.
https://www.nature.com/articles/ngeo2020
“A minimum atmospheric temperature, or tropopause, occurs at a pressure of around 0.1 bar in the atmospheres of Earth1, Titan2, Jupiter3, Saturn4, Uranus and Neptune4, despite great differences in atmospheric composition, gravity, internal heat and sunlight. ”
This might be about right. But I would say it has to do with air density rather than pressure,
but around 1/10 of 1 atm should have enough air density.
Or at higher elevation with less air density, one will have faster molecules, but there not enough molecule colliding with themselves or anything else to have measurable temperature. Or space has very fast velocity molecules but space has no temperature. Or the thermosphere of Earth is said to be very hot- but all it means is the gas molecules are going very fast [have high average velocity] but they do not warm or cool things- they have no temperature.
Or the kinetic energy of gases of atmosphere above 0.1 bars can be ignored, in terms warming the lower atmosphere. Or a warmer lower atmosphere doesn’t warm the atmosphere above 0.1 bars. I would tend to also ignore the radiant effect of gases higher than 0.1 bars upon lower atmosphere or surface. Though would not rule out all possible effect.
For example it’s thought that if you “blow up” impactor before hits earth, so that instead of rock hitting, the impactor’s “dust cloud ” hits earth, that could heat up earth a lot- and basically the dust cloud of what was once a rock, would be heating atmosphere above 0.1 bar. Not sure it would heat the Earth as much as it’s imagined it would, but I think would heat earth by more than a noticeable amount.
Or say something like supernova too close to earth would also heat the upper atmosphere. So some kind of extreme effect would have an effect.
“In all of these bodies, the tropopause separates a stratosphere with a temperature profile that is controlled by the absorption of short-wave solar radiation, from a region below characterized by convection, weather and clouds”
I don’t think so, Jupiter, Uranus, Neptune, and Saturn are heated by their internal heat.
I’ve listened to the arguments of Willis and others regarding that the IGL does not tell us how a planet is heated. I even put that into a logical argument, and for a moment I thought I’d have to concede Willis’ point.
So let’s review Badger’s argument:
Major premise: GHGs raise the temperature of an atmosphere significantly.
_______________________________
Minor premise: We have a formula derived from the IGLs that can predict the temperatures of all atmospheres reasonably accurately.
Minor premise: We have two identical planets except that one has GHGs and the other does not.
Minor premise: The formula derived from the IGLs gives the same temperature for both planets.
_______________________________
Conclusion: Either the IGL are false, or else GHGs do not raise the temperature of an atmosphere significantly.
Let’s take this argument and assume that volcanoes, and not GHGs, are heating the planet up– we assume, say, they’re making it 10C warmer:
Major premise: Volcanoes raise the temperature of an atmosphere significantly.
_______________________________
Minor premise: We have a formula derived from the IGLs that can predict the temperatures of all atmospheres reasonably accurately.
Minor premise: We have two identical planets except that one has volcanoes and the other does not.
Minor premise: The formula derived from the IGLs gives the same temperature for both planets.
_______________________________
Conclusion: Either the IGL are false, or else volcanoes do not raise the temperature of an atmosphere significantly.
So is Willis right? Because if volcanoes really are raising the temperature of the planet, our argument says they cannot be! Badger’s argument is wrong!
And then I realized the answer: If we say that the pressure, density, and molar mass of the volcanoe-less planet (E2) is due to other causes– it’s heated more by solar insolation, for example and that’s why the pressure, density, and molar mass are the same as E1– then in that case we don’t have identical planets but for the fact that one has volcanoes, do we? And in fact if it really is the case that if we have IDENTICAL planets except for the fact that we SUPPOSE one is heated by volcanoes, and the IGL gives us the same temperature, then it MUST be false that volcanoes are heating the planet!
Yes, that may sound absurd, but that is the only logical conclusion, and it proves, I think, that Badger’s argument is sound. Our assumption that volcanoes were heating the planet was wrong; there is no escape from that conclusion.
Don
This may seem like a curved ball but the molar mass equation that Robert Holmes uses to calculate the Mean Surface Temperature of the Earth can also be used to determine how the Snowball Earth ended.
Using Robert’s equation and values and assuming a rise in carbon dioxide in the Snowball Earth atmosphere to ~0.12 bar of atmospheric CO2 (Hoffman et al. 1998) and a consequent mean surface molecular weight of 32.44 g/mol, then the average surface temperature would rise to:-
T= 113.3*32.44/8.314*1.37 = 322 Kelvin.
The result is that the Earth’s atmosphere would rise in temperature to a surface average of 49C. This pressure induced average annual surface temperature would release the Earth from the Snowball without any need to invoke the radiative greenhouse mechanism at all.
OK, now my head is spinning.
I will lay this out for comments. Thanks to Phil and Frank for pointing this out to me in a way that I could understand.
Suppose that GHGs really do warm the atmosphere. With Badger’s argument, that means that E1 has warmed significantly, and consequently it’s pressure, density, and MM have been altered from a non-ghg state. Now consider E2, without GHGs. If it’s true that GHGs have warmed E1, then E2’s pressure, density, and MM will not match E1’s because it’s cooler. HOWEVER, in Badger’s argument we are asserting FROM THE START that the two planets have the same pressure, density, and MM; in other words, we are forbidding E2’s temperature to drop from the very onset!
That seems to me to be a fatal flaw in Badger’s argument: it is assuming the conclusion. That does not mean that Holmes, etc., are wrong. It only means, for now, that Badger’s argument seems to be flawed.
And no, I’m not trying to get both sides pissed at me. It is what it is.
Well that’s why you need auto compression to decline when DWIR increases.
If that happens then the situation is stable at the IGL temperature and all the parameters remain identical for the two planets.
Stephen Wilde February 15, 2018 at 6:47 am:
“Well that’s why you need auto compression to decline when DWIR increases.
If that happens then the situation is stable at the IGL temperature and all the parameters remain identical for the two planets.”
Stephen, please point me to where you explain this so I can read it again.
Are you agreeing that Badger’s argument fails, even if the general ideas you’re presenting are true? Or are you still saying that Badger’s argument is correct?
Don
“To summarise the science so far:
i) The ideal gas law gives a useful prediction of the surface temperature of a planet with an atmosphere regardless of whether GHGs are present or not and regardless of their quantity.
ii) That prediction invariably gives a higher temperature than that predicted by the radiation only S-B equation
iii) The only ways that energy in an atmosphere can get back to the surface in order to raise it above S-B are by conduction and/or radiation.
iv) If an atmosphere is non radiative then the only way is by conduction so the fact of back conduction heating a surface cannot be denied.
v) If one then adds a radiative gas then the DWIR energy reaching the surface below alters the density patterns at the surface so as to compromise the efficiency of conduction between atmosphere and surface. Conduction is more effective when the gases are denser or where the density variability in the horizontal plane is greater.
vi) It follows that more DWIR means less conduction and the surface temperature remains set at that predicted by the ideal gas laws.
vii) The ideal gas laws predict a maximum surface temperature that cannot be further enhanced by radiative gases in the atmosphere”
Badger just needs to alter his submission to the effect that introducing GHGs does not cause any net warming because convection changes to neutralise their effect.
In that situation both planets retain identical IGL parameters but will show differences in lapse slopes and the vigour of convection.
Stephen Wilde February 15, 2018 at 7:53 am:
“Badger just needs to alter his submission to the effect that introducing GHGs does not cause any net warming because convection changes to neutralise their effect.”
I’m not sure those who hold the GHG position will allow that! I think the whole argument needs to be recast, or else another argument needs to be constructed.
Some of your points below are obvious but some need more elaboration. You’ve done this earlier but right now I don’t have time to look through comments.
Then they have to show that when DWIR warms a surface it does not reduce density such as to slow down the rate of conduction.
They also have to show that auto-compression PLUS DWIR are together able to create a surface temperature higher than that predicted from the IGL.
They currently deny auto compression altogether yet a GHG free atmosphere still fits the IGL so how do they suggest that is achieved without DWIR
Not easy, and until they do they should not be indulged by any serious researcher.
Stephen Wilde February 15, 2018 at 9:14 am:
“They currently deny auto compression altogether yet a GHG free atmosphere still fits the IGL so how do they suggest that is achieved without DWIR”
Yes, I think this is the next area of attack because it seems to me that those who say that gravity by itself can’t induce a temperature gradient are wrong. So: either I don’t understand, or they’re getting concepts jumbled up, or I’m getting concepts jumbled up, or a little of everything.
DWIR=downwelling infrared?
The goal isn’t to “win”– at least, not for me. The goal is to sort it out. Yes, I do have a prejudice, but I think I’ve demonstrated that I can set that aside and simply follow the logic.
It may be, as suggested earlier, that everyone is “right” up to a point. Wouldn’t that be nice?
Well the concept of DWIR (Down Welling Infrared Radiation from GHGs) AND the concept of auto-compression can both be accommodated if they are mutually exclusive and they must be otherwise the IGL could not hold in all the varied scenarios.
However, it will still meet opposition because it drives a stake through the idea that there must be net warming from GHGs.
I agree, if they are saying that gravity by itself can’t induce a temperature gradient, they are clearly wrong. That this is the case is obvious from the thermal gradient observed on all planetary bodies from a pressure of 10kPa upwards;
Robinson, T. D., & Catling, D. C. (2014). Common 0.1 [thinsp] bar tropopause in thick
atmospheres set by pressure-dependent infrared transparency. Nature Geoscience, 7(1), 12-15.
Robert Holmes February 14, 2018 at 1:37 pm
Phil,
“It does no such thing, you have assumed that they have the same temperature by setting three parameters the same. The correct way to do it is to set the atmospheres to have the same pressure and molar mass then predict the temperature and density.
By assuming that three variables are the same if you correctly ‘predict’ the measured temperature all you’ve done is show that it’s an ideal gas.”
.
No.
I have done more than ‘show its an ideal gas’.
What I am doing here, is pointing out that if the GHG in E1 do cause anomalous warming, then they can ONLY do this by altering the three gas parameters in a way in which non-GHG can not..
BUT since a NON-GHG mix (E2) can also be created which matches these exact three E1 parameters, (and IF the MM version of the IGL is correct – and so must form the same temperature as E1) then the GHG’s cannot be creating any anomalous warming by altering the three gas parameters in an unusual way.
So you have a fixed pressure and Molar mass please explain how you can change the density without changing either of those parameters?
Everyone genuinely interested in getting to the truth of this matter should read Harry Huffman’s article: http://theendofthemystery.blogspot.co.uk/2010/11/venus-no-greenhouse-effect.html
But some words of instruction and caution, before you start reading it;
1. Approach it with an enquiring mind, fully open and receptive to new arguments and reasoning. For some, the words may cause some mental anguish, but do not stop until you reach the end of the article: I assure you that the discomfort will be worth it. Do not duck your responsibilities as a free, independent thinker – do not attempt to read the article’s discussion thread to seek self-assurance.
2. Once you’ve read the article do nothing else (but have a cup of tea) and contemplate what you have read. If necessary, re-read the article again in order to allow you to complete your contemplation of what you have read.
3. Judge the article as it is presented to you – no previous concepts or beliefs should be allowed to obstruct your thinking. Be assured, the author has been entirely truthful i.e. he is not attempting to deceive you or allow you to hold a position that differs from his own. All the figures are from actual empirical data. Mankind has successfully landed craft on the surface of Venus, six times. Many more Venus missions have successfully returned atmospheric data. The figures used in the article are NOT theoretical. They are 162 million miles away from a “Thought Experiment”. All you have to do, on your own, with your own mind only, is decide whether what he has said is the TRUTH (remember there’s a difference between ‘truthful’ and ‘truth’).
4. Until you have decided whether what you have read (in just that article) is the TRUTH or is a FALSEHOOD, do not seek to read any other opinions or commentaries on the article or of the author. Remember YOU are an independent thinker. YOU are NOT a lemming.
5. Once you have decided (using only your own rational thinking mind) tell someone of your decision. Be proud to own your decision – if you find you are not proud enough of your decision, perhaps you have no capacity for independent thought / decision making (but beware that this unfulfilment will inevitably leave you with some cognitive dissonance, so you MUST decide – and you are urged do so with your own independent mind, unassisted by anyone else’s thoughts or opinions: just you and the article: TRUTH or FALSEHOOD?
6. Either way that you decide, the cliff-edge is fast approaching when Nature itself will reveal the truth or falsehood of the IPCC’s Greenhouse theory and their computer model projections. Don’t you think you owe it to yourself to be on the correct side of the argument before that time comes? That’s a rhetorical question by the way – your opinions and commentaries are not needed by free-thinking individuals.
Oh dear.
The only thing that Harry said that is relevant here is to point out that within the atmosphere of Venus the temperature is the same as the Earth’s at the same atmospheric pressure after accounting for distance from the sun.
That was not unique to him.
Until now nobody ever explained in a step by step manner how that outcome was achieved.
So at 55km on Venus the pressure is ~0.5 atm and a temperature of 27ºC, at 50 km it’s ~1atm and a temperature of 75ºC
If you are saying that the proposition about Venus’s temperature and pressure relationship is wrong please supply a detailed link.
For example:
Blumenthal, Kay, Palen, Smith (2012). Understanding Our Universe. New York: W.W. Norton & Company. p. 167. ISBN 9780393912104.
Is there a problem with this?
“The Magellan spacecraft recorded temperature and pressure data for Venus’ atmosphere.
At ~1000 mb it recorded ~66°C – ~339 K.
At ~1000 mb Earth’s surface temperature is ~ 15°C – ~288K.
Venus’ average orbital radius is ~108.21 million km.
Earth’s average orbital radius is ~149.6 million km.
The Inverse Square ratio for temperatures using the Stefan-Boltzmann law is the square root of Venus’ orbital radius divided by the square root of Earth’s orbital radius multiplied by Venus’ temperature gives Earth’s temperature.
Hence the square root of 108.21/149.595 = ~0.8505 times 339 K = ~288.3 K for similar pressure of ~1000 mb.”