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When internal energy $E$ is expressed as $E(S, V, p)$, $p$ is momentum, $S$ entropy, $V$ volume. We write its differential form as:

$dE = T\text{d}S - P\text{d}V + v\text{d}p$ $= \left( \dfrac{\partial E}{\partial S}\right)_{V, \, p} \text{d}S + \left( \dfrac{\partial E}{\partial V}\right)_{S, \, p} \text{d}V + \left( \dfrac{\partial E}{\partial p}\right)_{V, \, S}\text{d}p$

where, the thermdynamic system under consideration is moving with velocity $v$ and has momentum $p$.

For an observer attached to local frame, the energy is E', and is given by

$dE' = T'dS' - P'dV'$ $= \left( \dfrac{\partial E'}{\partial S'}\right)_{V'} \text{d}S' + \left( \dfrac{\partial E'}{\partial V'}\right)_{S'} \text{d}V'$

In the formulation of relativistic thermodynamics, entropy is considered as invariant, that is $S = S'$, and volume is assumed to be varied as $V = V'/\gamma$. The last parameter is $\gamma = \frac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}$.

From this, we can derive relativistic expression from temperature, as was done in this paper.

https://www.researchgate.net/publication/330164616_Lorentz_transformations_of_the_thermodynamic_quantities

I am not able to understand its derivation in this paper, particularly, equations (24) to (30). How did they arrive at $T = T'/\gamma$?

Thank you.

Parag
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  • Not a duplicate, but related: https://physics.stackexchange.com/q/502789 – Chiral Anomaly Apr 17 '21 at 18:41
  • Do you have a reference for your initial equality? – DanielC Apr 17 '21 at 19:29
  • Hi Daniel, I am trying to understand the derivation given in the research paper of the link. The initial equality $V = V'/\gamma$ comes from the fundamentals of special relativity, independent of thermodynamics. – Parag Apr 17 '21 at 19:33
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    @Chiral, thank you, the question you mention has another referenced question. https://physics.stackexchange.com/questions/83488/is-temperature-a-lorentz-invariant-in-relativity – Parag Apr 17 '21 at 19:42

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