# Thermodynamic exercise in GR

1. Apr 25, 2006

### Longstreet

I hope someone else finds this interesting, since it through me for a loop for a while.

Ok, so time dilation occures in gravitational fields. I used the solution for the time coordinate transformation between "free" space and the space under gravitational influence:

$$t_g = t_f\sqrt{1-\frac{2GM}{rc^2}}$$

From now on I'm just going to use beta:

$$\beta=\frac{1}{\sqrt{1-\frac{2GM}{rc^2}}}$$

Inverting it I can find a transformation for a frequency:

$$\nu_g = \beta\nu_f$$

My setup consists of two identical flat blackbodies paralell to each other (facing each other). One is in "free" space (at infinity) and the other is in the gravitiational field, and I guess there is a reflecting tube of some kind connecting them although I'm not sure that's really relevent. But, what is relevent is they are in thermal equilibrium.

First I use the above frequncy to transform the energies of the photons. Since they are linearly related to their frequency, $$E(\nu)=h\nu$$, it's fairly trivial:

$$E_g(\nu) = \beta E_f(\nu)$$

Now, I have several thermal radiation fluxes. The net power flux at both surfaces must be zero, since they're in equilibrium. $$W_f$$ is the power emmiting from the free blackbody, and $$W_f'$$ is the power being absorbed; $$W_g$$ and $$W_g'$$ are likwise for the gravitated blackbody.

I'm going to argue that $$W_g = \beta W_f$$ is the proper transformation for the power, so that $$W_g'=-\beta W_f$$ and $$W_g=-\beta W_f'$$. I'm not sure of the proper notation, but the power is the integral of the energies of the photons and the rate of photon flux. But the energy function transforms like above, which can be pulled out of the integral. The power integrals would then be exactly the same for the free and gravitated bodies except for the beta constant, which does not depend on the frequency or location on the surface of the blackbody since r doesn't really change.

edit: adding another factor of beta per pervect's observation of the rate of the photon flux.

Again, since we have equilibrium we know that:

$$W_g + W_g' = 0$$

and

$$W_f + W_f' = 0$$

I'm just going to go with the top one there pluging in the transformation:

$$W_g - \beta^2 W_f = 0$$

Since the thermal radiation power is defined by the temperature of the blackbodies:

$$W = \epsilon \sigma A T^4$$

then, assuming all the other variables are the same for the two bodies:

$$W_g = \epsilon \sigma A T_g^4$$

and

$$W_f = \epsilon \sigma A T_f^4$$

which leads to the simplified relation:

$$T_g^4 - \beta^2 T_f^4 = 0$$

which gives the relationship of the equilibrium temperatures of the blackbodies:

$$T_g = \beta^{1/2} T_f$$

So this is pretty interesting at first if you happen to be thinking about the second law of thermodynamics at the same time. Apparently we have two blackbodies in equilibrium but at different temperatures (the hot one being in the gravitation field, and the cold one in free space). Just hook up a heat engine between them and you have a prepetual motion machine.

But, after analyzing such a hypothetical cycle, I realized that since the cold source is at a higher gravitational potential, if you take some heat out of the hot side, to actually get it to the cold side would require work (since the heat is energy). That might still work if we can get enough work out of the heat engine to raise the heat back to the cold side.

I could just argue that the amount of work needed is exactly the same as is gained by photons on the way down. But, I actually first used the potential equation:

$$U_g=-(\frac{1}{\sqrt{1-\frac{2GM}{rc^2}}}-1)mc^2$$

I noticed that's just $$U_g=-(\beta-1)E$$. Going back to the energy change of the photons, it turns out to be the same. Ok, so anyway. We need that amount of work/energy to raise the heat back to the free blackbody. The efficiency of doing this action is

$$K = \frac{Q_{out}}{W_{in}}$$

Which is (substituting variables and simplifying)

$$K_u = \frac{Q_{out}}{(\beta-1)Q_{out}} = \frac{1}{(\beta-1)}$$

However, if I had used the carnot refrigerator to simply pump the heat back to the hot side, which is right next to the engine, the efficiency is (subsituting all the variables and simplifying):

$$K_{carnot} = \frac{1}{(\beta^{1/2}-1)}$$

$$K_u < K_{carnot}$$ for all beta.

It would actually be worse than a carnot refrigerator. Order is restored to the universe. I hope you had fun. If you find mistakes please let me know.

Last edited: Apr 25, 2006
2. Apr 25, 2006

### pervect

Staff Emeritus
You seem to have ignored any variation in the photon flux. Not only are the individual redshifted photons each of lower energy, but they are recived at a different rate due to time dilation.

Let us consider observer #1 in the gravity well of the sun at radius r1, and observer #2 at raidius r2 > r1.

Suppose that obsever #1 emits 1000 photons in 1-second-at-r1. Then he expects that 1000 photons will be recieved by obsever r2 per 1-second-at-r1 (because the time interval between reception and emission is constant).

All observers will agree that the same number of events occur in a certain interval on a space-time diagram, though because their clocks run at different rates they will disagree about the actual length of the space-time interval on the diagrm. Thus the observer at r2 will also agree that he recieves 1000 photons / (second-at-r1).

Thus the rate at which photons will be recieved according to obsever r2 according his local clocks is, however

(1000 photons / second at r1) * (second-at-r1)/(second-at-r2)

i.e. there is an additional factor of beta in the flux which you are not considering, power transforms as beta^2.

I haven't gone over the rest of your analysis yet.

3. Apr 25, 2006

### Longstreet

Ah. Thanks. I miss the obvious one :).

It is odd that the equilibrium temperature isn't the same as the apparent temperature. I'm just assuming there of course that the temperature transforms something like $$T_g = \beta^2 T_f$$, I think. Average velocity would be one factor of beta, and average kinetic energy would square that. But then again, if that were the case a perpetual motion machine would be possible.

Last edited: Apr 25, 2006
4. Apr 25, 2006

### pervect

Staff Emeritus
Thermodynamics and SR/GR is something that has come up on the board before, unfortunately previous discussions have been somewhat lacking.

One paper I've been reading recently talks about inverse-temperature in SR as a 4-vector.

http://arxiv.org/abs/physics/0505004

I've also recently acquired Tolman's book (Relativity, Thermodynamics, and Cosmology) - but I don't like his non-4vector approach much.

Unfortunately, I'm still looking for a more substantial discussion of the problem of relativistic temperature. The main comment I can make right at the moment is that from what I've been reading recently, entropy itself should be frame independent scalar, as one would expect from the "count the number of states" interpretation. If a system is changing with time, differing defintions of simultaneity can give you a difrerent "state count" at a particular instant, but if one has a static system with a constant number of states, that number of states should be the same regardless of the frame in which it is viewed (i.e which defintion of simultaneity is used).

Assuming that I havean't misunderstood somewhere, this observation should give one how S transforms between frames (it should stay constant).

Heat is a form of energy, so any relativistic treatment of thermodynamics has to take into account that what appears to be momentum to one observer will appear to be energy to another observer. So T = dS/dQ is going to have the issue that what appears to be energy transfer according to one observer appears to be momentum transfer to another observer.

This may not matter that much to your particular problem, though, because there isn't any relative motion. On the other hand, your problem is very similar to velocity-induced redshift, where the issue would matter.

I've got a nagging feeling that I've read something recently that's highly pertinent to this issue, but I can't recall where.

Anyway, I don't currently have a very solid answer to your question, I'm going to keep looking, and maybe someone else will have more to say.

5. Apr 25, 2006

### Longstreet

Edit (sorry for editing out the whole post. But this kind of trumps it)
edited again: I was right the first time.

Last edited: Apr 26, 2006
6. Apr 26, 2006

### pervect

Staff Emeritus
OK, it turns out there is a simple expression for the equilibrium temperature as a function of g00 for any static fluid system mentioned in Tolman's book.

"Fluids" in GR include radiation as a special case, so this would apply to the temperature of a radiation field of an equilbrium system.

The simple relation is

$$T_0 = \frac{C}{\sqrt{g_{00}}}$$

where T0 is the temperature as measured by some local observer

C is some constant

g00 is the metric coefficient for dt^2, for example 1-2M/r in the Scwarzschild metric.

Thus, at equilibrium, the lower objects (deeper in the gravity well) are hotter.

In your choice of variables, the lower object would have a temperature that's higher by a factor of beta.

There are still a few things that are puzzling me though. If we consider a case where g_00 = .25, the lower body would have twice the temperature. This should imply 16x as much radiated energy / unit time by the Steffan-Boltzman law, and I'm not sure how to account for that much more radiation.

There's no problem with the spectrum, redshfiting the spectrum of the lower body gives the correct spectral peak.

7. Apr 26, 2006

### Longstreet

I am curious where the two more factors of beta come from in terms of radiation exchange to keep equilibrium.

So, then:

$$T_g = \beta T_f$$

For carnot cycle operating between the two temps.

$$e_{carnot} = 1 - \frac{T_c}{T_h} = 1 - \frac{T_f}{T_g} = 1- \frac{T_f}{\beta T_f} = 1 - \frac{1}{\beta}$$

$$K_{carnot} = \frac{T_c}{T_h - T_c} = \frac{T_f}{T_g - T_f} = \frac{T_f}{\beta T_f - T_f} = \frac{1}{\beta - 1}$$

For mechanical transfer between the two potentials:

$$e_{u} = \frac{W_{out}}{Q_{in}}$$

now, the effective Q_in is W_out + Q_out. Technically Q_in = Q_out, but it has potential energy and that is extracted as work.

$$e_{u} = \frac{(\beta-1)Q_{out}}{(\beta-1)Q_{out} + Q_{out}} = \frac{\beta-1}{\beta} = 1 - \frac{1}{\beta}$$

Now, Q_in is the actual Q since it has no potential energy.

$$K_{u} = \frac{Q_{in}}{W_{in}} = \frac{Q_{in}}{(\beta-1)Q_{in}} = \frac{1}{\beta-1}$$

Which now exactly matches the carnot efficiencies. No increase or decrease of entropy... This is pretty interesting since one method is a ideal gas cycle, and the other is completely mechanical.

Last edited: Apr 27, 2006