I'm considering standard FLRW cosmology for a simple perfect fluid of pressure $p = w \, \rho$, with $-1 < w \le 1$. In the special case of $w =\frac{1}{3}$ (radiation), the temperature is $T \propto a^{-1}$, where $a$ is the cosmological scale factor. This relation comes from solving the nul-geodesic equation (for photons), or from quantum mechanics which gives $E \propto \lambda^{-1}$ for a photon, and $\rho \propto a^{-4}$ combined with Stefan's law $\rho \propto T^4$.

But then, what is the similar relation for fluids of parameter $w \ne \frac{1}{3}$?

For dust (cold by definition), we have $T \approx 0$. But what is $T$ for a fluid of any $w$?


1 Answer 1


The distribution function of particles in phase space for a homogeneous, isotropic universe is denoted as $f(q)$, where $q$ is the magnitude of the momentum. It can be taken to be the Fermi-Dirac or Bose-Einstein distribution, which in natural units read $f(q) = (\exp ( (E - \mu)/T) \pm 1)^{-1}$ respectively, and which depend on the temperature $T$, on the chemical potential $\mu$ and on the energy $E = E(q) = \sqrt{q^2 + m^2}$. The ultra-relativistic case, $w=1/3$, corresponds to $q \gg m$, so $E \approx q$, while the non-relativistic case $w=0$ corresponds to $q \ll m$, so $E \approx m + q^2 /2m \approx m$.

From this distribution function, we can recover the number density, energy density, entropy density and pressure for that particle species with some integrals - see equations 5 through 8 of Husdal et al (2016).

We can then ask about intermediate values, $0 < w < 1/3$ - I don't believe talking about "temperature" makes a lot of sense for $w<0$ or $w>1/3$, since you couldn't get those EOS from the regular FD/BE distribution formula; at the very least you'd have to determine how the distribution function for particles with those EOS depends on temperature.

Now, a restatement of the fact that the entropy in a comoving volume is conserved is that the quantity $a^3 ( \rho + P) / T$ is a constant - see Kolb and Turner, 1994, equation 3.70. We can then say that $ T \propto a^{3} (1+w) \rho$; this is the usual $T \propto 1/a$ in the ultra-relativistic case.

The advantage of "getting here" this way is that we can generalize: we can express $\rho = \rho (T)$ and $P = P(T)$ by solving the relevant integrals, and we are left with an equation which relates $T$ and $a$.

I must admit I have not done this computation, so I cannot give a precise result; I believe it will depend on the specifics of the particle species considered because we lose the "scale invariance" inherent to relativistic particles.

Edit: a proof of the relation $T \propto a^3 (1+w) \rho$, based on section 3.4 in Kolb and Turner.

The entropy differential reads $$ \text{d}{S} = \frac{1}{T} (\underbrace{\text{d}{(\rho(T) V)}}_{ \text{d}{E}} + P(T) \text{d}{V}) = \frac{1}{T} (V \text{d}{\rho (T)} + (P(T) + \rho(T) ) \text{d}{V}) $$ so its partial derivatives are $$ \frac{\partial S}{\partial V} = \frac{1}{T} (\rho (T) + P(T)) \qquad \text{and} \qquad \frac{\partial S}{\partial T} = \frac{V}{T} \frac{\text{d}\rho (T)}{\text{d}T} $$ and therefore the second partials being equal can be expressed as $$ \begin{align} \frac{\partial ^2 S}{\partial T \partial V} &= \frac{\partial ^2 S}{\partial V \partial T} \\ \frac{\partial}{\partial T} \left(\frac{1}{T}(\rho (T) + P(T))\right) &= \frac{\partial}{\partial V} \left(\frac{V}{T} \frac{\partial\rho (T)}{\partial T} \right) \\ - \frac{1}{T^2} (\rho + P) + \frac{1}{T} \left(\frac{\partial \rho }{\partial T} + \frac{\partial P}{\partial T}\right) &= \frac{1}{T} \frac{\partial \rho }{\partial T} \\ \frac{\partial P}{\partial T} &= \frac{1}{T} (\rho +P) \end{align} $$ which allows us to verify that $$ \begin{align} \frac{\text{d}}{\text{d}t} \left(\frac{a^3}{T} (\rho + P )\right) &= \frac{1}{T} \frac{\text{d}}{\text{d}t} (a^3 (\rho + P)) + a^3 (\rho + P) \frac{\text{d}}{\text{d}t} \left( \frac{1}{T}\right) \\ &= \frac{1}{T} a^3 \dot{P} - a^3 (\rho + P) \frac{\dot{T}}{T^2} \\ &= \frac{a^3}{T} \frac{\text{d}P}{\text{d}T} \dot{T} - a^3 (\rho + P) \frac{\dot{T}}{T^2} \\ &= \frac{a^3}{T} \frac{(\rho + P)}{T} \dot{T} - a^3 (\rho + P) \frac{\dot{T}}{T^2} = 0 \,. \end{align} $$

  • $\begingroup$ This is interesting. The relation $T \propto a^3 (1 + w) \rho$ may be what I’m looking for. But how do you justify that $(\rho + p) a^3 /T$ is a constant if entropy stays conserved? Can you add a proof of this in your answer? $\endgroup$
    – Cham
    Jul 9, 2021 at 18:26
  • 1
    $\begingroup$ That calculation is beautifull! In the last steps, we have to use the local conservation of energy: $$\dot{\rho} + 3H(\rho + p) = 0.$$ I don’t know, but I find this proof surprising. And where is the entropy supposed to be conserved? Nowhere I see the assumption that $S = cste$. $\endgroup$
    – Cham
    Jul 9, 2021 at 20:56
  • $\begingroup$ Yes, right, that is a further step: you can show that the differential of the conserved quantity is equal to the differential of S, therefore up to an additive constant they are the same. $\endgroup$ Jul 9, 2021 at 21:08
  • $\begingroup$ I’m wondering about the temperature found for all fluids with $w < 0$ and $w> \frac{1}{3}$. I don’t clearly see any contradiction in the theory. The formula $T \propto a^{-3w}$ is really intruiging and I don’t know what to think, especially for $\frac{1}{3} < w< 1$. Stiff fluid ($w = 1$) can’t have a temperature? What about its entropy then? $\endgroup$
    – Cham
    Jul 10, 2021 at 12:11
  • $\begingroup$ To be honest, I don't know what temperature would mean in those cases - there may very well be a way to discuss it, I just am not familiar with the definition you would use. That was a part of your question, but I don't want to try and guess about something I'm not familiar with, so unless someone else can provide an answer here you might be better off asking that as a separate question. $\endgroup$ Jul 10, 2021 at 14:32

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