Speed of sound in a gas (for adiabatic perturbations in cosmology)

Question

In the book "Introduction to Cosmology" by Barbara Ryden, equation 4.57 gives the sound speed for adiabatic perturbations in a gas with pressure P and energy density $\epsilon$. The equation is as follows: $c_{s}^2= c^2 \frac{\ dP}{d\epsilon}$, where $c$ is the speed of light $c_{s}$ is the speed of sound. The general equation for speed of sound in a medium is given by: $c_{s}^2= \frac{\ dP}{d\rho}$ ,where $\rho$ is the mass density of the medium. For interstellar medium with non-relativistic gas with mass the energy density is approximately given by: $\epsilon \approx \rho c^2$. So for such gases the first equation can be obtained by substituting for $\epsilon$ from 3rd equation. But how one derive the first equation for a relativistic gas of photons? It seems that the author of the book treats the equation valid for all cases, both relativistic and non-relativistic. Is there any other general way to derive the first equation without invoking relativistic or non-relativistic differentiation ?

Answer 1

The general expression for the speed of sound is $$ c_s^2 = \frac{\partial P}{\partial \epsilon} $$ (in units where $c=1$). If there is a conserved particle number, then the derivative has to be taken at constant entropy per particle $s/n$. This relation can be derived from the equations of relativistic fluid dynamics, combined with thermodynamic identities, and the derivation can be found in text books that cover relativistic fluids, for example Weinberg's book on general relativity. (Note that weinberg, as well as some other authors, use the symbol $\rho$ to denote the energy density, not the mass density.)

The non-relativistic expression is $$ c_s^2 = \left.\frac{\partial P}{\partial\rho}\right|_{s/n} $$ where $\rho$ is the mass density of the fluid. This follows from the non-relativistic limit of the first formula, or directly from the usual (non-relativistic) Euler equation.

For a gas of photons both $P$ and $\epsilon$ are proportional to $T^4$, but with a different coefficient of proportionality, $P=\epsilon/3$. As a result $c_s^2=1/3$ (in units of $c$). This result applies to any gas of weakly interacting massless particles.

Note that sound is a collisional mode. This means that a cold gas of photons cannot support sound, but a hot gas of $\gamma,e^\pm$ has a sound mode.

Answer 2

I feel like this should be a comment, as I don't have a definite answer (more like musings), but the comments are size limited.

For an isolated adiabatic system we know that $$dE=-PdV,$$ where $E$ is the total internal energy, $P$ is the pressure (by definition), and $V$ is the volume. The energy density $\epsilon=E/V$, and so we may write $$\epsilon dV + Vd\epsilon = -PdV.\tag{1}$$ Dividing by the total mass of the system $M$ and noting that the density $\rho=M/V$ we may the write $$\rho d\epsilon = (P+\epsilon)d\rho, \tag{2}$$ or, rearranging, $$\frac{d\epsilon}{d\rho} = \frac{P+\epsilon}{\rho}.$$

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Added Detail:

Since $\rho=M/V$, we may also write $V=M/\rho$. Note that $M$ is a constant because the system is isolated. Then we may write $$dV=d\left(\frac{M}{\rho}\right) = Md(\rho^{-1}) = -\frac{M}{\rho^2}d\rho.$$ Substituting this form into Eq. (1) then yields $$-\epsilon \frac{M}{\rho^2}d\rho + \frac{M}{\rho}d\epsilon = P\frac{M}{\rho^2}d\rho.$$ Multiplying the equation by $\rho^2/M$ then gives $$-\epsilon d\rho + \rho d\epsilon = P d\rho,$$ which implies $$\rho d\epsilon = P d\rho + \epsilon d\rho.$$ Factoring $d\rho$ on the right-hand side yields Eq. (2).

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Now consider the definition of the speed of sound using the same isolated adiabatic system: $$c_s^2 \equiv \frac{dP}{d\rho} = \frac{d\epsilon}{d\rho}\frac{dP}{d\epsilon}=\frac{P+\epsilon}{\rho}\frac{dP}{d\epsilon}. $$ Thus, the definition given in the textbook should hold as long as $$\frac{P+\epsilon}{\rho}=c^2.$$

Since we are dealing with cosmology, I would assume that $P$ is usually considered small, and if $\epsilon=\rho c^2$ then we have about the right answer. However, I don't really know cosmology very well, so I can't comment on the accuracy of these two assumptions.