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Discussion > The Moon and 255K


If anybody thinks it is worth it I would be delighted to discuss it in a pub, over several beers. Any of you lot get to London?

I would have loved to have gone to the Oxford do tonight, but could not make it. Maybe next time, as I often travel to Culham.

Feb 12, 2013 at 6:06 PM | Unregistered CommenterRoger Longstaff


I think you've answered your question when stating that the internal energy would remain constant.

Surely though this is experimentally tractable. There must be some theoretical and experimental data out there for the temperature distribution in gas centrifuges. These are routinely used for UF6 isotope enrichment. It may be difficult to find given the role of these devices in U enrichment for nuclear weapons!

Feb 12, 2013 at 6:26 PM | Unregistered CommenterPaul Dennis


I too would welcome a pub get together. I didn't realise there was one in Oxford tonight, not that I could have made it. But if given enough notice I could get either to say Cambridge, London, or Oxford.

Feb 12, 2013 at 6:28 PM | Unregistered CommenterPaul Dennis

Paul, good point about the uranium centrifuges. Does anybody know if we could get some data?

Feb 12, 2013 at 6:36 PM | Unregistered CommenterRoger Longstaff

Sorry Roger,

I don't know of any sites that deal with uranium centrifuges.

Here is a very good website that shows that the ideal gas law can be derived from the properties of an isothermal gas in a gravitational field.

Feb 12, 2013 at 7:04 PM | Unregistered CommenterPaul Dennis

Thanks for the reference Paul - very useful.

I still can not see the flaw in the "thought experiment". Also, some claim to have measured a temperature gradient experimentally:

However, for any of this to be real, the Second Law would need to state that the entropy of an isolated system in equilibrium is a maximum for an inertial frame at rest, with an accelerating inertial frame having a lower entropy and, by equivalence, a system in a gravitational field.

Apologies Rhoda - we seen to have strayed a long way from the Moon and 255K. But I found it interesting!

Feb 13, 2013 at 10:24 AM | Unregistered CommenterRoger Longstaff


I was going to suggest a look at Loschmidt's Gravito-Thermal Effect and Graeff’s experiments but I see both you and Paul have already been there.

With their perpetual motion connotations they have both been controversial.

With the density of gas, and therefore energy density capable of producing heat through work, being greatest at the base of the column and reducing exponentialy up the column, the fact we see more heat at the base of an atmosphere is not surprising. I cannot see how kinetic energy alone can cause heat although Graeff’s results seem to indicate that may be the case.

Feb 13, 2013 at 3:39 PM | Registered CommenterRKS

Thanks RKS,

Yes, Graeff's work is interesting, and seems methodical, but the "perpetual motion connotations" put me off.

I still can not see the flaw in the experiment I considered yesterday, but the more I think about it the more I think it must be wrong.

Feb 13, 2013 at 4:04 PM | Unregistered CommenterRoger Longstaff


I did some simple sums (very possibly incorrectly) and derived a formula for the temperature gradient in an isolated, acceleraing container of gas (argon in this case):

Temp gradient (deg K/m) = (2fhm)/(3k), where:

f = acceleration = 9.81 m/s^2 (same as g)
h = height of container
m = mass of argon atom
k = Boltzmann's Constant

This gives a gradient of 0.0313 K/m, which is spookily close to Graeff's experimental results.

Just for interest....

Feb 14, 2013 at 12:22 PM | Unregistered CommenterRoger Longstaff

Oops - the formula is for the temperature difference in the container, not the gradient. If you do the calc for a 1m height container you get the gradient shown above...

Feb 14, 2013 at 12:40 PM | Unregistered CommenterRoger Longstaff


I think you've made the mistake of computing the loss of kinetic energy of a single atom with height (1/2m.v^2 = m.g.h) and assume that this represents a change in temperature. The temperature of the gas is a function of the mean kinetic energy of the atoms (Ar) or molecules at any particular height. The mean kinetic energy does not change with height, or along any other field gradient that might be appropriate.

This problem is treated explicitly by Feynman in his lectures on Physics. I know that this is counter intuitive but none the less is true.

Feb 14, 2013 at 2:20 PM | Unregistered CommenterPaul Dennis

Thanks Paul,

I gave away my Feynman volumes years ago to an aspiring undergraduate - and regretted it ever since. Do you know if it is on the web?

I did not use potential energy in my formulation - there is no gravitational field. I simply used kinetic energy:

(m*V^2)/2 = (3*k*T)/2 and substituted V^2 = (3*k*T)/m into V^2 = U^2 + 2*f*h

(where U is the initial velocity at the top of the container, and V the relative velocity at the bottom, in the inertial frame of the container, with the positive thrust vector in the sense from botton to top).

This seems to show that the mean kinetic energy (or temperature) does change with height, simply due to kinematics, all in the inertial frame of the container.

Where have I gone wrong?

Feb 14, 2013 at 3:49 PM | Unregistered CommenterRoger Longstaff


I need to think about your question. We normally define temperature wrt the local reference frame and not an inertial frame. In kinetic theory we assume that all the atomic/molecular motions are random. In your thought experiment there is a drift velocity (more strictly acceleration) from the point of view of an observer rooted on the inertial reference frame. How does this change the temperature from the point of view of that observer? I'd suggest that it would be red shifted if the gas cylinder were moving away from the observer and vice versa. However, to an observer moving with the gas the temperature remains constant. This is the situation on Earth.

I'm only speculating here.

Feb 14, 2013 at 7:33 PM | Unregistered CommenterPaul Dennis

"How does this change the temperature from the point of view of that observer?"

I agree, that is the question Paul. My formulation suggests to me that there would be a temperature gradient observed in the reference frame of the container, but that would have deep consequences...

Anybody else still looking at this who could shed more light?

Feb 14, 2013 at 8:10 PM | Unregistered CommenterRoger Longstaff

I confess I haven't read a BH discussion for weeks but I've just skimmed this one and want to say thanks a bundle to Rhoda, TBYJ, Roger, Paul and all. Fascinating.

Feb 14, 2013 at 9:01 PM | Registered CommenterRichard Drake


I think you are correct that the temperatures would look different when observed from different frames of reference. However, I think that the EP says that the temperatures should be the same viewed from within an inertial frame in a 1g gravitational field and within a non-inertial frame accelerating at 1g in free space.

I thought of this originally to try to simplify the problem by eliminating potential energy and convection, but it just ended up complicating matters. I think the original question was "how can a static pressure cause a temperature gradient", so the simplest answer must be that in an isolated container (on Earth) the kinetic energy of the gas atoms (or molecules), and hence their temperature, is higher at the bottom than at the top due to the increased velocity imparted by the gravitational field (this gives the same numerical answer as I got last time, with idential mathematics).

Is this correct?

Feb 16, 2013 at 11:35 AM | Unregistered CommenterRoger Longstaff

Roger, you cannot get rid of potential energy so easily. Gravity or inertial frame will be the same excepting the effect of distance squared in the gravity case. Imagine an anvil (yes a genuine ACME anvil) on a table in your accelerated frame. Wile E. Coyote pushes the anvil off the table. It goes 'down' and when it hits the floor it makes a noise and crushes whatever was there. It had potential energy. Where did it come from? Well if the frame was accelerated by a constant force, as soon as the anvil was in free fall the rest of the frame was accelerated faster. Work was done on the frame but not the anvil. You cannot discount the work done by the accelerating force in your conservation of energy sum. And gas at the top of your cylinder has potential energy vs gas at the bottom.

Feb 16, 2013 at 11:51 AM | Registered Commenterrhoda

Thanks rhoda,

I used KE = (m*v^2)/2 = (3*k*T)/2 solely to calculate the gas temperature. Are you saying that the temperature at the top of the container (in a gravitational field) should be calculated from total enery (KE + PE), in other words adding the mgh term?

Feb 16, 2013 at 12:26 PM | Unregistered CommenterRoger Longstaff

Take a 1m container with a single atom at the top with a "thermal velocity" of, say, 500 m/s. Make its velocity vector vertically downwards. By the time it reaches the bottom its velocity will be 509.81 m/s. It will rebound endlessly with this velocity at the bottom and 500 m/s at the top. The total energy within the container (KE + PE) is conserved, and the vertical velocity profile (and hence temperature) of the atom is constantly repeating. Correct?

Feb 16, 2013 at 12:58 PM | Unregistered CommenterRoger Longstaff

No, I think when it has settled down the PE difference would be accounted for by the pressure differential. But I have no strong view except not seeing why the temperature would not tend to equalize, all parts being in contact, whereas a pressure difference seems inevitable.

Feb 16, 2013 at 12:58 PM | Registered Commenterrhoda

OK, so take my single atom (which is "hotter" at the bottom) and multiply by a few billion - and on the way down it met another atom coming up - and statistical mechanics takes over. However, with perfectly elastic collisions, I think that the velocity (and hence temperature) profile would be maintained.


Feb 16, 2013 at 1:35 PM | Unregistered CommenterRoger Longstaff

Now rambling, but.........

The Second Law would apply, and the temperature would equalise - after an almost infinite amount of time. As the momentum exchange (for the single atom) with the Earth would produce a net force (being higher at the bottom than the top) the Earth's trajectory would shift untlil the atom was at a constant velocity of 500 m/s.

With a single bound he was free! Now off to the pub...

Feb 16, 2013 at 1:59 PM | Unregistered CommenterRoger Longstaff

Thought from the pub - an identical container at the opposite side of the Earth would produce an equal and opposite momentum exchange, and hence force, therefore the Earth would remain fixed in inertial space.

Energy conserved, momentum conserved, entropy increasing and temperaures equalising - what's not to like?

Feb 16, 2013 at 3:57 PM | Unregistered CommenterRoger Longstaff