# Equilibrium temperature distribution in gravitational field

1. Feb 9, 2013

### Jano L.

I came across Tolman's paper on the equilibrium distribution of temperature in gravitational field. He argues that in contrast to classical thermodynamics situations, in gravitational field, a fluid in thermodynamic equilibrium will not have the same temperature everywhere (as measured by thermometer in various heights). He comes to conclusion that the temperature decreases as height increases.

What is the basic reason behind that? Do you know some simple explanation/argument for this conclusion, or perhaps a counter-argument?

The paper:

http://prola.aps.org/abstract/PR/v35/i8/p904_1

2. Feb 9, 2013

### Staff: Mentor

I haven't read the paper, but I would assume that it is due to gravitational redshift. Consider two concentric spherical blackbodies separated by vacuum and in thermal contact with only each other. The lower body will emit a high-temperature black body spectrum, but the upper body will receive a low-temperature black body spectrum due to gravitational redshift of the radiation. And conversely for the spectrum emitted by the upper body.

3. Feb 9, 2013

### Naty1

sounds logical: My guess.....at increased pressure you'd expect a smidgen higher temperature....that is, lower down in the gravitational potential well where gravity compresses a fluid more....so temp, say as a result of Heisenberg uncertainty, is a bit higher as a result of additional confinement/oscillation movement??

4. Feb 9, 2013

### Staff: Mentor

This happens, but it's not just a result of gravity; it's a result of the equation of state of the fluid. Also it has nothing to do with relativity; it happens in a non-relativistic fluid, such as the Earth's atmosphere. Tolman's paper is talking about relativistic effects.

The pressure-temperature relationship from the equation of state is purely classical; there are no quantum effects involved. (More precisely, there's no need to invoke the uncertainty principle, and no need to use quantum statistics.)

5. Feb 9, 2013

### Naty1

unfortunately, I thought of that after I posted....

Is there an intuitive explanation from the equation of state results..??

6. Feb 9, 2013

### Staff: Mentor

I don't think so; I think the relativistic effect Tolman is talking about is in addition to the non-relativistic equation of state effect. The paper appears to be behind a paywall, so I can only read the abstract; but he appears to be saying that, for example, even if we assume an (unrealistic) equation of state where temperature is independent of density (so that non-relativistically, temperature would be constant with height even though density increases as you go deeper into the gravity well), the temperature would *still* have to vary with height because of gravitational redshift/blueshift.

7. Feb 10, 2013

### Jano L.

Yes, Oppenheimer told Tolman the same thing (see the footnote in the paper). I am not entirely satisfied with that argument yet, since it is not clear to me what happens with the spectrum as the radiation rises. Can you show or post a reference showing that the spectral curve in greater heights is given by Planck formula, only with different temperature? I've got trouble deriving that, as I do not know how to calculate how EM energy changes when it rises in gravitational field...

8. Feb 10, 2013

### Bill_K

Redshifted blackbody radiation is still blackbody, thus the CMB is still blackbody after 13+ billion years.

This follows from photon conservation - the number of photons 1/(exp(hv/kT) -1) in each normal mode is conserved, only the frequency changes. So the distribution remains the same but with a different T.

9. Feb 10, 2013

### Jano L.

I think for Doppler effect, this may be so. If the photon conservation is assumed, then temperature has to change in the same way the frequency does.

But is this so even if the redshift is due to gravitation? In order to perform the calculation with normal modes, these have to be enclosed in a box in an inertial system.

But the question is what happens to radiation when it passes different levels of gravitational potential in a non-inertial system (the star, or planet). I do not know how to calculate with normal modes in such situation. I was looking rather for a non-quantum argument, based on relativity/electrodynamics.

It may seem that it suffices to calculate the Doppler redshift and extend the result to gravity, but I do not think this is so, because there is quite a difference in the physical processes.

When considering the radiation received by body A from a source B that is running away, there can be no thermodynamic equilibrium between the bodies A and B, because they both see red-shifted radiation and the mutual ping-pong of the scattered waves will lead to steady decrease of frequency of radiation.

Whereas in the case of gravity, the upper levels of the atmosphere receive red-shifted light from levels below, and the levels down below receive blue-shifted light from the levels above, so there can be heat equilibrium between the two in principle.

I think my question can be restated in this way: is the spectrum of radiation the same function of $\nu/T$ when the radiation rises in gravitational field and why?

10. Feb 10, 2013

### Staff: Mentor

Redshift is redshift. The partitioning of it into Doppler or gravitational redshift is rather arbitrary in general.

If a Doppler-redshifted blackbody spectrum is still a blackbody spectrum then a gravitationally-redshifted blackbody spectrum is also still a blackbody spectrum. (However, I have not personally worked through the proof of either claim).

11. Feb 11, 2013

### Jano L.

I think I have to go deeper into relativity to understand why this happens. Thanks to everyone.