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So I am working on Relativistic Thermodynamics and I have skimmed through a few papers by Hamity, Callen, etc. People are not in agreement as far as I know, but I don't understand why people don't derive the transformation laws from first principles. So for example for the entropy I would go about it this way (correct me if I am wrong): The entropy is defined as the logarithm of number of states which have a given Energy E for a particular Hamiltonian (omitting the logarithm): $$\int\,dp\,dx\,\,\delta(H(x,p)-E)$$ Isn't it obvious from this that the Entropy defined in this way cannot be Lorentz Invariant? Because $$dp\rightarrow \gamma dp$$ $$dx\rightarrow \gamma dx$$ and $$H(x,p)-E\rightarrow \gamma(H(x,p)-E)$$ So the Dirac Delta is not affected. Am I thinking wrong here?

onephys
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  • Entropy is a scalar, right? – CuriousOne Jul 03 '16 at 00:51
  • Yeah, but temperature as well and people argue that it does transform in a certain way. – onephys Jul 03 '16 at 00:52
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    Temperature only transforms properly for massless fields, one can't define it for relatively moving massive ones. I would say the same is true for entropy, unless you agree that for massive systems it's always the scalar measured in the center of mass system? – CuriousOne Jul 03 '16 at 00:59
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    Might it have something do with the fact that relativistic Hamiltonians aren't described by conjugate x and p coordinates? There isn't a useful position operator in relativistic QM. See, eg http://physics.stackexchange.com/questions/142413/position-operator-in-qft. You need a different way to count states, specifically one that involves integrating over a Lorentz invariant measure. – Luke Pritchett Jul 03 '16 at 01:00
  • Thank you CuriousOne and Luke Pritchett. Those comments really shine some light on my doubt. – onephys Jul 03 '16 at 01:10
  • @LukePritchett: I thought of doing something like this: $\int \frac{dp}{p^0} \delta(H-E)$ – onephys Jul 04 '16 at 01:35

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