# Cosmology Lecture 19

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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:

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:

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