I hope someone else finds this interesting, since it through me for a loop for a while.(adsbygoogle = window.adsbygoogle || []).push({});

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:

[tex]

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

[/tex]

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

[tex]\beta=\frac{1}{\sqrt{1-\frac{2GM}{rc^2}}}[/tex]

Inverting it I can find a transformation for a frequency:

[tex]\nu_g = \beta\nu_f[/tex]

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, [tex]E(\nu)=h\nu[/tex], it's fairly trivial:

[tex]E_g(\nu) = \beta E_f(\nu)[/tex]

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

I'm going to argue that [tex]W_g = \beta W_f[/tex] is the proper transformation for the power, so that [tex]W_g'=-\beta W_f[/tex] and [tex]W_g=-\beta W_f'[/tex]. 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:

[tex]W_g + W_g' = 0[/tex]

and

[tex]W_f + W_f' = 0[/tex]

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

[tex]W_g - \beta^2 W_f = 0[/tex]

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

[tex]W = \epsilon \sigma A T^4[/tex]

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

[tex]W_g = \epsilon \sigma A T_g^4[/tex]

and

[tex]W_f = \epsilon \sigma A T_f^4[/tex]

which leads to the simplified relation:

[tex]T_g^4 - \beta^2 T_f^4 = 0[/tex]

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

[tex]T_g = \beta^{1/2} T_f[/tex]

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:

[tex]U_g=-(\frac{1}{\sqrt{1-\frac{2GM}{rc^2}}}-1)mc^2[/tex]

I noticed that's just [tex]U_g=-(\beta-1)E[/tex]. 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

[tex]K = \frac{Q_{out}}{W_{in}}[/tex]

Which is (substituting variables and simplifying)

[tex]K_u = \frac{Q_{out}}{(\beta-1)Q_{out}} = \frac{1}{(\beta-1)}[/tex]

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):

[tex]K_{carnot} = \frac{1}{(\beta^{1/2}-1)}[/tex]

[tex]K_u < K_{carnot}[/tex] 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.

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# Thermodynamic exercise in GR

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