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### Finishing Linear Perturbation Theory

The last missing piece in the equations we’ve been discussing is the calculation of the speed of sound $v_{s}$ in these fluids. For this, we’ll focus on the baryon-photon fluid in two epochs:

- Before Decoupling ($z\sim 1200$):

The energy density of the baryon-photon fluid is given by:

$\rho =\rho _{\gamma }+\rho _{B}=bT^{4}+n_{B}m_{H}+({3 \over 2}n_{B}kT)\,\!$
where $b$ is a constant, $m_{H}$ is the mass of hydrogen, and $kT\ll m_{H}$ because $m_{H}\sim 1GeV$, which corresponds to $T\sim 10^{10}K$, which was only valid very early on. The pressure of the baryon-photon fluid is dominated by $P_{\gamma }$, so:

$P\approx P_{\gamma }={1 \over 3}\rho _{\gamma }c^{2}\,\!$
Now we can calculate the adiabatic sound speed:

$v_{s}^{2}\equiv {\partial P \over \partial \rho }{\big |}_{S}={c^{2} \over 3}{\partial \rho _{\gamma } \over \partial (\rho _{\gamma }+\rho _{B})}{\big |}_{S}={c^{2} \over 3}\left(1+{\partial \rho _{B} \over \partial \rho _{\gamma }}{\big |}_{S}\right)^{-1}\,\!$
where $S$ is entropy. Now since $\rho _{B}=n_{B}m_{H}\propto a^{-3}\propto T^{3}$, and $\rho _{\gamma }\propto T^{4}$, we can say ${\partial \rho _{B} \over \rho _{B}}=3{\partial T \over T}$, and ${\partial \rho _{\gamma } \over \rho _{\gamma }}=4{\partial T \over T}$. Therefore:

${\partial \rho _{B} \over \partial \rho _{\gamma }}{\big |}_{S}={3 \over 4}{\rho _{B} \over \rho _{\gamma }}\,\!$
Thus, our equation for the adiabatic sound speed becomes:

${v_{s}^{2}={c^{2} \over 3}\left(1+{3 \over 4}{\rho _{B} \over \rho _{\gamma }}\right)^{-1}}\,\!$
If the energy density is dominated by photons ($\rho _{B}\ll \rho _{\gamma }$), then:

${v_{s}={c \over {\sqrt {3}}}}\,\!$
Using ${\Omega _{B,0}}$, $T_{\gamma ,0}$, we can show that $\rho _{B}<\rho _{\gamma }$ for $a\leq 10^{-3}$.

After decoupling, the speed of sound in baryons are the speed of sound in neutral hydrogen (and Helium), which is an *Ideal Gas* to a good approximation. Thus, $P_{B}=n_{B}kT_{B}$. We also know $\rho _{B}=m_{H}n_{B}+({3 \over 2}n_{B}kT)$, where $kT$ is small (again). Recall that for an ideal gas ($dS=0$), that $P_{B}\propto \rho _{B}^{\gamma }$, where $\gamma ={C_{p} \over C_{V}}$. $C_{p},C_{V}$ is the specific heat for constant pressure, volume. The proof for this is as follows:

$dQ=0=dE+PdV\Rightarrow C_{V}dT+PdV=0\,\!$
Replacing $dT$, using $NkdT=PdV+VdP$, we get:

${\begin{aligned}{C_{V} \over Nk}(PdV+VdP)+PdV&=0\\{C_{V}+Nk \over Nk}PdV&={-C_{V} \over Nk}VdP\\{C_{P} \over C_{V}}{dV \over V}&=-{dP \over P}\\\end{aligned}}\,\!$
Thus, $P\propto v^{\gamma }$, where $\gamma \equiv {C_{P} \over C_{V}}$. Since we also know that $P={NkT \over V}\propto V^{-\gamma }$, so using that $\rho _{B}\propto {1 \over V}$:

${P_{B}\propto P_{B}^{\gamma }}\,\!$
We also know $T_{B}\propto V^{1-\gamma }\propto a^{3(1-\gamma )}$. We can use a previous equation to show $\gamma ={5 \over 3}$, so:

${T_{B}\propto {1 \over a^{2}}}\,\!$