# Lorentz Invariance of Entropy

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?

• Entropy is a scalar, right? Jul 3 '16 at 0:51
• Yeah, but temperature as well and people argue that it does transform in a certain way. Jul 3 '16 at 0:52
• 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? Jul 3 '16 at 0:59
• 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 physics.stackexchange.com/questions/142413/…. You need a different way to count states, specifically one that involves integrating over a Lorentz invariant measure. Jul 3 '16 at 1:00
• Thank you CuriousOne and Luke Pritchett. Those comments really shine some light on my doubt. Jul 3 '16 at 1:10