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.
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.
