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I'm interested in the following question: What could we say about the thermodynamical properties of the Universe, using thermodynamics alone (without using general relativity), assuming that the Universe can be described by an homogeneous perfect fluid? I'm wondering about the validity of the following thermodynamical argument.

Firstly, I'm considering an homogeneous perfect fluid of energy density $\rho$, pressure $p$ and temperature $T$, inside a volume $V$. The energy of the fluid element is $E = \rho \, V$. I'm considering the fluid's entropy as a function of the independent variables $T$ and $V$: $S(T, V)$ (instead of the usual $S(E, V)$). Since $dE = \rho \, dV + V \, d\rho$, the thermodynamics first law $dE = T \, dS - p \, dV$ give the following two equations (this part is easy to verify): \begin{align} \frac{\partial S}{\partial T} &= \frac{V}{T} \, \frac{\partial \rho}{\partial T}, \tag{1} \\[2ex] \frac{\partial S}{\partial V} &= \frac{V}{T} \, \frac{\partial \rho}{\partial V} + \frac{\rho + p}{T}. \tag{2} \end{align} From the constraint \begin{equation*} \frac{\partial^2 S}{\partial V \, \partial T} = \frac{\partial^2 S}{\partial T \, \partial V}, \end{equation*} we easily find a relation for the three functions $\rho(T, V)$, $p(T, V)$ and $S(T, V)$: \begin{align} \frac{\partial p}{\partial T} = \frac{V}{T} \, \frac{\partial \rho}{\partial V} + \frac{\rho + p}{T} \equiv \frac{\partial S}{\partial V}. \tag{3} \end{align} Presently, the two parameters $T$ and $V$ are independent variables, while $\rho$, $p$, $S$ are unknown functions of these variables. An equation of state could be added as a relation between $p$ and $\rho$. In cosmology, the linear relation $p = w \rho$ is usually assumed.

Next, I want to apply the general equations (1), (2) and (3) to cosmology. Since all thermodynamical quantities are actually evolving in time, we have a single independent variable: cosmological time $t$, or equivalently the cosmological scale factor $a$. An element of cosmological fluid has its volume varying as $V \propto a^3$. I consider the following two cases, and I'm not sure they are valid reasoning. This is why I'm asking a question here.

Case 1: The state doesn't have an explicit dependance on the temperature. I suppose that the energy density $\rho$ and pressure $p$ doesn't depend on the $T$ variable in (1), (2) and (3). Their time evolution is parameterized with the volume $V$ only. In this special case, we get the following three equations from (1)-(3): \begin{align} \frac{\partial S}{\partial T} &= 0, \tag{4} \\[2ex] \frac{\partial S}{\partial V} &= \frac{V}{T} \, \frac{\partial \rho}{\partial V} + \frac{\rho + p}{T} \equiv \frac{\partial p}{\partial T} = 0, \tag{5} \\[2ex] \frac{\partial \rho}{\partial V} &= -\, \frac{\rho + p}{V}. \tag{6} \end{align} Equations (4) and (5) imply that the entropy of the fluid element doesn't vary: $$\tag{7} S(T, V) = \text{cst}. $$ Equation (6) can be solved for the linear relation $p = w \rho$: $$\tag{8} \rho \propto V^{-1-w} \propto a^{- d}, $$ where $d = 3 + 3 w$, which is a known result in standard cosmology, using general relativity and the local conservation of the fluid's energy-momentum tensor. Notice that this method doesn't allow to find the temperature as a function of the volume (or the scale factor, since $V \propto a^3$). The function $T(V)$ cannot be found in this case, without adding some constraint to the system of equations.

Case 2: The state doesn't have an explicit dependance on the volume. I now suppose that the energy density $\rho$ and pressure $p$ doesn't depend on the $V$ variable in (1), (2) and (3). The cosmic evolution is now parameterized with the temperature $T$ only. In this special case, we get the following equations: \begin{align} \frac{\partial S}{\partial T} &= \frac{V}{T} \, \frac{\partial \rho}{\partial T}, \tag{9} \\[2ex] \frac{\partial S}{\partial V} &= \frac{\rho + p}{T}, \tag{10} \\[2ex] \frac{\partial p}{\partial T} &= \frac{\rho + p}{T}. \tag{11} \end{align} Equations (9) and (10) can be integrated: $$\tag{12} S(T, V) = \frac{(\rho + p) V}{T} + \text{cst}, $$ where $\rho \equiv \rho(T)$ and $p \equiv p(T)$. Using the linear relation $p = w \rho$, we can integrate (11) for $w \ne -1$ and $w \ne 0$ (notice that (11) was found by assuming $\partial_V p = 0$ and $\partial_T p \ne 0$): $$\tag{13} \rho \propto T^{\frac{1 + w}{w}}. $$ This case alone doesn't allow me to find the volume $V$ as a function of temperature, so we can't find back the relation (8).

Still, if I do use (8) from case 1: $\rho \propto a^{- d}$, then (13) is equivalent to $$\tag{14} T \propto V^{- w} \propto a^{- 3 w}. $$ This result is consistent with standard cosmology in the case of the radiation fluid ($w = \frac{1}{3}$, so $T \propto a^{-1}$). Taken together, the results from both cases 1 and 2 are all consistent with the standard cosmology using general relativity.

I'm wondering about the validity of all the previous thermodynamical argument. In cosmology, there's only one variable left in the system (unless I'm missing something): cosmological time $t$, or equivalently the scale factor $a$, or volume $V$, or temperature $T$. Then how could we specify the other variables ($\rho$, $p$, $S$, ...), as functions of the unique independent variable chosen (say $V$ or $T$), from thermodynamics alone? Am I doing a mistake in the thermodynamical reasoning shown above? (the calculations themselves are all easy and straightforward).

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  • $\begingroup$ Too many unjustified assumptions. The most important is that equilibrium Thermodynamics can be applied. Is the Universe at equilibrium? $\endgroup$ Jan 5, 2023 at 22:09
  • $\begingroup$ @GiorgioP, please, could you just make a list of all the assumptions, as an answer? Yes, equilibrium thermodynamics can be assumed for an universe that expands not too fast. This is already in the full general relativistic treatment (from the local conservation of energy-momentum). And what other assumptions are there? $\endgroup$
    – Cham
    Jan 5, 2023 at 22:34

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