# What's the relation between temperature and scale factor for arbitrary EoS $-1 < w \le 1$?

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

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}

• 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?
– Cham
Jul 9, 2021 at 18:26
• 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 $S = cste$.
– Cham
Jul 9, 2021 at 20:56
• 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. Jul 9, 2021 at 21:08
• 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?
– Cham
Jul 10, 2021 at 12:11
• 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. Jul 10, 2021 at 14:32