# Cosmology Lecture 19

### Finishing Linear Perturbation Theory

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

• Before Decoupling (${\displaystyle z\sim 1200}$):

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

${\displaystyle \rho =\rho _{\gamma }+\rho _{B}=bT^{4}+n_{B}m_{H}+({3 \over 2}n_{B}kT)\,\!}$

where ${\displaystyle b}$ is a constant, ${\displaystyle m_{H}}$ is the mass of hydrogen, and ${\displaystyle kT\ll m_{H}}$ because ${\displaystyle m_{H}\sim 1GeV}$, which corresponds to ${\displaystyle T\sim 10^{10}K}$, which was only valid very early on. The pressure of the baryon-photon fluid is dominated by ${\displaystyle P_{\gamma }}$, so:

${\displaystyle P\approx P_{\gamma }={1 \over 3}\rho _{\gamma }c^{2}\,\!}$

Now we can calculate the adiabatic sound speed:

${\displaystyle v_{s}^{2}\equiv {\partial P \over \partial \rho }{\big |}_{S}={c^{2} \over 3}{\partial \rho _{\gamma } \over \partial (\rho _{\gamma }+\rho _{B})}{\big |}_{S}={c^{2} \over 3}\left(1+{\partial \rho _{B} \over \partial \rho _{\gamma }}{\big |}_{S}\right)^{-1}\,\!}$

where ${\displaystyle S}$ is entropy. Now since ${\displaystyle \rho _{B}=n_{B}m_{H}\propto a^{-3}\propto T^{3}}$, and ${\displaystyle \rho _{\gamma }\propto T^{4}}$, we can say ${\displaystyle {\partial \rho _{B} \over \rho _{B}}=3{\partial T \over T}}$, and ${\displaystyle {\partial \rho _{\gamma } \over \rho _{\gamma }}=4{\partial T \over T}}$. Therefore:

${\displaystyle {\partial \rho _{B} \over \partial \rho _{\gamma }}{\big |}_{S}={3 \over 4}{\rho _{B} \over \rho _{\gamma }}\,\!}$

Thus, our equation for the adiabatic sound speed becomes:

${\displaystyle {v_{s}^{2}={c^{2} \over 3}\left(1+{3 \over 4}{\rho _{B} \over \rho _{\gamma }}\right)^{-1}}\,\!}$

If the energy density is dominated by photons (${\displaystyle \rho _{B}\ll \rho _{\gamma }}$), then:

${\displaystyle {v_{s}={c \over {\sqrt {3}}}}\,\!}$

Using ${\displaystyle {\Omega _{B,0}}}$, ${\displaystyle T_{\gamma ,0}}$, we can show that ${\displaystyle \rho _{B}<\rho _{\gamma }}$ for ${\displaystyle a\leq 10^{-3}}$.

• 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, ${\displaystyle P_{B}=n_{B}kT_{B}}$. We also know ${\displaystyle \rho _{B}=m_{H}n_{B}+({3 \over 2}n_{B}kT)}$, where ${\displaystyle kT}$ is small (again). Recall that for an ideal gas (${\displaystyle dS=0}$), that ${\displaystyle P_{B}\propto \rho _{B}^{\gamma }}$, where ${\displaystyle \gamma ={C_{p} \over C_{V}}}$. ${\displaystyle C_{p},C_{V}}$ is the specific heat for constant pressure, volume. The proof for this is as follows:

${\displaystyle dQ=0=dE+PdV\Rightarrow C_{V}dT+PdV=0\,\!}$

Replacing ${\displaystyle dT}$, using ${\displaystyle NkdT=PdV+VdP}$, we get:

{\displaystyle {\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, ${\displaystyle P\propto v^{\gamma }}$, where ${\displaystyle \gamma \equiv {C_{P} \over C_{V}}}$. Since we also know that ${\displaystyle P={NkT \over V}\propto V^{-\gamma }}$, so using that ${\displaystyle \rho _{B}\propto {1 \over V}}$:

${\displaystyle {P_{B}\propto P_{B}^{\gamma }}\,\!}$

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

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