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Discussion > The Moon and 255K

You are right - if modellers are so fundamentally dishonest that they hide their codes and methodology, with models that can neither be verified nor validated, then there is nothing that can be done. However, I tend to think that most GCM modellers are misguided, rather than fundamentally dishonest (I have witnessed "target fixation" several times in the past). Perhaps this is naive...? However, the point still remains - the mathematics is either correct, or it is not - and I would have thought that this point could easily be settled.

Feb 6, 2013 at 9:50 AM | Unregistered CommenterRoger Longstaff

It could, if that was the purpose of the modelling. But the purpose is, and always has been, political. Models are used to demonstrate data that they can't get out of real world observations. SO we agree that GCMs are pretty much worthless except for looking at scenarios. For choosing which scenario is more likely, they are pointless.

But that's not the reason they exist, they exist to prop up a political position. Remove that prop by showing them to be false... you think the edifice will tumble, I say they will just get another prop.

Feb 6, 2013 at 9:57 AM | Unregistered CommenterTheBigYinJames

TBY, I have been thinking about your question: "I still can't see how a static pressure can cause a raised temperature" and devised a thought experiment.

Take a thermally isolated cylinder (say, 10m long, 1m diameter) containing a monatomic gas (say, Argon at STP) and put it in a spacecraft in a weightless environment (either Earth orbit, or deep space). The Second Law of Thermodynamics tells us that the gas will reach thermal equilibrium, or maximum entropy, where there is no temperature gradient. Correct?

Now, fire the spacecraft's rocket engine giving a constant acceleration of 1g. What will happen to the gas? It seems to me that a pressure gradient would be established in the direction of relative motion, with a higher pressure at the base of the cylinder (located toward the base of the spacecraft). This would give the argon molecules a higher kinetic energy at the base, and hence a higher temperature, thereby establishing a temperature gradient along the axis of the accelerating cylinder when equilibrium had been achieved. Is this correct so far?

If it is correct then we can apply the equivalence principle of General Relativity in order to state than an identical effect will be observed in a gravitational field, effectively explaining "how a static pressure can cause a raised temperature".

The problem is, of course, that this appears to defy the Second Law, as the entropy of the gas has been reduced in the accelerating frame of reference and/or the gravitational field. However, some have speculated that gravity has an "entropic value" such that the total entropy of the system (including either the gravitational field or the accelerating frame of reference) is a maximum at equilibrium, and equivalent to the gas at rest in field free space.

Does this make sense? Any comments or observations?

Feb 12, 2013 at 10:49 AM | Unregistered CommenterRoger Longstaff

Correction: "Argon molecules" should be "Argon atoms" in the above. (Argon has been used to avoid any complications with vibrational states of diatomic or triatomic molecules).

Feb 12, 2013 at 12:41 PM | Unregistered CommenterRoger Longstaff

Roger (and anyone else who is interested),

I suggest you take out your copy of Feynmann's lectures on Physics volume 1 and read (from memory) chapters 39-41. Feynmann deals explicitly with a gas in a gravitational field (or any other field). He demonstrates that (i) the pressure varies exponentially with height (ii) the temperature is constant throughout the column (i.e. the gas is isothermal).

Feb 12, 2013 at 1:16 PM | Unregistered CommenterPaul Dennis

Yep, gas heats when it is compressed, but it does not possess heat because of its pressure. However, what we have on a planet is an atmosphere in touch with a surface of variable temperature. We are going to get convection and therefore wind regardless of radiative components of the atmosphere. We are going to have heat moving to colder areas be they polar or at higher altitude and therefore a reduction in radiation and an equalisation of surface temps. We are going to have weather. Without GHE. How much weather and of what nature I could not begin to guess. And I wonder whether N&Z could guess it either from the limited number of planetary parameters they use.

Feb 12, 2013 at 1:57 PM | Registered Commenterrhoda

Paul, thanks for this, but I can not understand where the flaw in my argument about the accelerating spacecraft lies (and, if this were to be correct, why the EP does not show that the gas can not be isothermal).

This is a genuine question - please help.

Feb 12, 2013 at 2:37 PM | Unregistered CommenterRoger Longstaff


I wonder if the confusion arises from the concept of temperature. The temperature of a gas is given by the mean kinetic energy of the ensemble of molecules in the specified volume. This mean kinetic energy does not change with height in a gravitational field. As one goes up in height there are fewer molecules because one loses those that do not have enough kinetic energy to get across the potential energy barrier that is m.g.h . However, the mean energy of those that do remains constant.

I know this is counter intuitive but is true and very elegantly demonstrated by Feynman.

Feb 12, 2013 at 2:55 PM | Unregistered CommenterPaul Dennis

Sorry Paul - I have mental block on this (and I stupidly gave my Feynmann volumes years ago).

You say "The temperature of a gas is given by the mean kinetic energy of the ensemble of molecules in the specified volume. This mean kinetic energy does not change with height in a gravitational field", but the temperature of the atmosphere DOES change with height - the lapse rate.

I still would like to understand the flaw in my spacecraft argument. It should be easily sorted out with am analytical approach using the kinetic theory of gasses (or a numerical model, or even an experiment with a ground based centrifuge). If nobody can explain it I will try an analytical approach, when I get time, but I am sure there must be a simple explanation.

Feb 12, 2013 at 3:10 PM | Unregistered CommenterRoger Longstaff

Roger, in your experiment you have suddenly applied a 1g acceleration to the gas - ie. you added energy to it. Are you surprised that its temperature increases?

Feb 12, 2013 at 3:38 PM | Unregistered CommenterBitBucket

BB, the EP says that the acceleration is identical to instantaneously placing it into the Eath's gravitation field (which is conservative).

(Like you, I come here for a good argument).

Feb 12, 2013 at 3:45 PM | Unregistered CommenterRoger Longstaff

You are comparing a gas that has just been placed in the earth's field (ie. accelerated; ie. heated) with one that was already in that field.

Feb 12, 2013 at 3:57 PM | Unregistered CommenterBitBucket

Placing it in gravity field can't be done instantaneously, you would be moving it in the field and thus changing its potential energy, ie doing work.

Feb 12, 2013 at 4:02 PM | Registered Commenterrhoda

No, I am taking a gas, at equilibrium in a thermally isolated cylinder in free space, and instantaneously either:

1) accellerating it at 1g, or,

2) placing it in a 1g gravitational field.

What's the difference?

Feb 12, 2013 at 4:04 PM | Unregistered CommenterRoger Longstaff

Good point rohda. So simply flood the gravitationally bound cylider with Argon with the same energy as it would have had in space, and see what happens at equilibrium.

Feb 12, 2013 at 4:08 PM | Unregistered CommenterRoger Longstaff


the lapse rate (more specifically the dry adiabatic lapse rate) results from movement of air. If the column is static, and there are no radiatively active gas components the column is isothermal (as described by Feynman and others). If parcels of air ascend or descend then under adiabatic conditions (i.e. no energy entering or leaving the system) one has to compensate the work done by, or on the gas (the p x delta V term) by a change in the internal energy of the gas (a reduction in temperature of the gas does work and expands, or increase in temperature if work is done on the gas and it contracts). The key eqation is:

dQ = dU + dW = 0 (for an adiabatic process)

where U is the internal energy, W is work and Q is the energy entering or leaving the parcel of gas. We can write this equation in terms of the heat capacity of the gas Cv (at constant volume), number of mols, and temperature and volume changes as:

dQ = n.Cv.dT + P.dV

Noting that (by taking the derivative of the ideal gas law (PV = nRT)):

PdV + V.dP = n.R.dT

and that:

R = Cp - Cv (the heat capacity at constant pressure minus the heat capacity at constant volume)

one finds (after re-arranging):

dT/dP (lapse rate) = V/(n.Cp)

To reiterate, the lapse rate only comes about by the fact that work is done on the gas as it either expands or contracts. In the absence of movement, as in a static gas column, the gas is isothermal.

Feb 12, 2013 at 4:13 PM | Unregistered CommenterPaul Dennis

Paul, so where is the flaw in my argument?

Feb 12, 2013 at 4:42 PM | Unregistered CommenterRoger Longstaff


As I understand it the N&Z argument in NOT that pressure alone causes atmospheric temperature, rather that excitation by solar radiation of energised molecules in the lower atmosphere causes the rise in temperature. The kinetic energy being a product of pressure and volume as per the gas laws. The insolation for star formation being supplied by the 2.7K cosmic background radiation.

Feb 12, 2013 at 4:55 PM | Registered CommenterRKS

RKS, my argument (or more exactly question) has nothing to do with N&Z. I find their work very interesting, but I do not agree with all of it (for example, the question of planetary rotation). But they were certainly right about Holder's Inequality, and their empirical work with the planets is very impressive.

Feb 12, 2013 at 5:09 PM | Unregistered CommenterRoger Longstaff


the flaw in your argument is to think that pressure has anything to do with kinetic energy of the gas molecules (or atoms in your example). The kinetic energy is a function of temperature alone.

Feb 12, 2013 at 5:13 PM | Unregistered CommenterPaul Dennis

"RKS, my argument (or more exactly question) has nothing to do with N&Z. I find their work very interesting, but I do not agree with all of it (for example, the question of planetary rotation). But they were certainly right about Holder's Inequality, and their empirical work with the planets is very impressive.

Feb 12, 2013 at 5:09 PM | Roger Longstaff"


I mentioned it as one possible explanation for temperature rise in a pressurised gas although your question is about a non radiated gas.

Feb 12, 2013 at 5:17 PM | Registered CommenterRKS

"the flaw in your argument is to think that pressure has anything to do with kinetic energy..."

"Pressure and kinetic energy - Pressure is explained by kinetic theory as arising from the force exerted by molecules or atoms impacting on the walls of a container" From wiki:

Feb 12, 2013 at 5:25 PM | Unregistered CommenterRoger Longstaff

I nearly cross posted with Roger on this one but to put it in more detail - From the kinetic theory of gases-

PV = 2/3K

Thus, the product of pressure and volume per mole is proportional to the average (translational) molecular kinetic energy.

Feb 12, 2013 at 5:33 PM | Registered CommenterRKS


that's right. The temperature of a gas rises as one does work on a gas to increase it's pressure, P.dV. For example when one is pumping up a bicycle tyre the gas in the pump is compressed and warms. However, if the gas is static as in a gas column in a gravitational field then there is no work term and no change in temperature. Thus pressure in the gas column changes with height but the temperature is constant.

As I see it your experiment in the rocket is exactly the same as the column in the gravitational field and so the gas would be isothermal. In fact it wouldn't matter what the nature of the field was. For example if the atoms were all ionised and in an electric field we would end up with the same result.

I think, perhaps, we all need to sit down in a pub with a thick notepad and have the discussion over some beers.

Feb 12, 2013 at 5:45 PM | Unregistered CommenterPaul Dennis

Thanks RKS,

I would value your opinion on the space experiment - do you think it would result in both a pressure and temperature gradient at equilibrium? It is an important question, because if this did happen, General Relativity would take care of all the rest (the Equivalence Principle has been proven to umpteen decimal places....).

Paul, just read your reply. I would modify it to say that in a thermally isolated gas there can be no change in energy (not temperature, in the case of an accelerated frame of reference or a gravitational field). The internal energy must remain constant.

At least we are now in the lager zone....

Feb 12, 2013 at 5:55 PM | Unregistered CommenterRoger Longstaff