# Cosmology Lecture 18

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## Still More on Metric Perturbations

Recall that we have our Boltzmann equation, which describes metric perturbations. To gain physical intuition into this equation, started taking velocity moments of this equation (in the non-relativistic limit).

$0^{th}$ moment:

$\int {d^{3}pf}=n={p \over m}\,\!$ $1^{st}$ moment:

${\int {d^{3}p{\vec {v}}f} \over \int {d^{3}pf}}=\langle {\vec {v}}\rangle \,\!$ $2^{nd}$ moment:

$\langle v_{i}v_{j}\rangle ={1 \over n}\int {d^{3}pv_{i}v_{j}f}\,\!$ and we defined ${\sigma _{ij}}^{2}=\langle v_{i}v_{j}\rangle -\langle v_{i}\rangle \langle v_{j}\rangle$ . We can iterate to calculate any $n^{th}$ moment. We then used these moments inside the Boltzmann equation.

$0^{th}$ moment of Boltzmann (the continuity equation):

${\partial \over \partial t}n+\nabla \cdot (n\langle {\vec {v}}\rangle )=0\,\!$ $1^{st}$ moment of Boltzmann:

${\partial \over \partial t}\langle v_{j}\rangle +\langle v_{i}\rangle \partial _{i}\langle v_{j}\rangle =-\partial _{j}\phi -{1 \over n}\partial _{i}(\overbrace {n{\sigma _{ij}}^{2}} ^{anisotropic \atop pressure})\,\!$ In the Newtonian limit (${\dot {p}}_{i}=-\partial _{i}\phi \cdot m$ ), dropping the $\langle \rangle$ ’s, and using $\rho {\sigma _{ij}}^{2}\sim P{\delta _{ij}}+\tau _{ij}$ (where $\tau _{ij}$ is the non-diagonal piece), we have:

${\partial {\vec {v}} \over \partial t}+({\vec {v}}\cdot \nabla ){\vec {v}}=-\nabla \phi -{1 \over \rho }\nabla P\,\!$ Which is just Euler’s Equation for a perfect fluid.

$2^{nd}$ moment of Boltzmann. T his is messy, so we’ll use that for a perfect gas, ${\sigma _{ij}}^{2}\Leftrightarrow {P \over \rho }\propto T$ , which gives us:

$diffusion\ term={3 \over 2}{dT \over dk}+(\nabla \cdot {\vec {v}})T+{1 \over \rho }\nabla \cdot {\vec {q}}\,\!$ where $q$ is the heat flux.

$n^{th}$ moment: Recall the linearized continuity and Euler’s equations (in comoving coordinates):

${\begin{cases}{\dot {\delta }}_{k}+{{\vec {k}}+{\vec {v}}_{k} \over a}=0\\{\partial \over \partial t}(a{\vec {v}}_{k})+iv_{s}^{2}{\vec {k}}\delta _{k}+i{\vec {k}}\phi _{k}=0\end{cases}}\,\!$ Defining $\theta =\nabla \cdot {\vec {v}}$ (and so $\theta _{k}=i{\vec {k}}{\vec {v}}_{k}$ ):

${\begin{cases}{\dot {\delta }}_{k}+{1 \over a}\theta _{k}=0\\{\dot {\theta }}_{k}+{{\dot {a}} \over a}\theta _{k}-{v_{s}^{2}k^{2} \over a}\delta _{k}-{k^{2} \over a}\phi _{k}=0\end{cases}}\,\!$ So compare the Boltzmann equation:

${\partial f \over \partial t}+{\dot {x}}_{i}{\partial f \over \partial x_{i}}+{\dot {p}}_{i}{\partial f \over \partial p_{i}}=\left({\partial f \over \partial t}\right)_{c}\,\!$ with the full theory (the coupled Boltzmann and Einstein equations in a conformal Newtonian gauge):

$ds^{2}=c^{2}dt^{2}(1+2\phi )-a^{2}(t)(1-2\Psi )dx^{i}dx_{i}\,\!$ This gives us the Geodesic equation (which is the GR equation of motion):

${{\dot {p}}=p{\dot {\Psi }}-\epsilon n_{i}\partial _{i}\phi }\,\!$ where the ${\dot {\Psi }}$ term is specific to GR, $\epsilon ={\sqrt {p^{2}+m^{2}}}$ , and $p_{i}=pn_{i}$ . In GR, the Boltzmann equation is written using the energy-momentum tensor:

$T_{\mu \nu }=\int {d^{3}p(-g)^{-{\frac {1}{2}}}{P_{\mu }P_{\nu } \over p^{0}}f}\,\!$ where $f$ is the phase space distribution. The first-order part of the energy-momentum conservation equation ($T_{i\mu }^{\mu \nu }=0$ ) gives us the following moments:

{\begin{aligned}&{\dot {\delta }}_{k}+{(1+w) \over a}\theta _{k}-3(1+w){\dot {\Psi }}_{k}=-3{{\dot {a}} \over a}(v_{s}^{2}-w)\delta _{k}\\&{\dot {\theta }}_{k}+(1-3w){{\dot {a}} \over a}\theta _{k}+{{\dot {w}} \over 1+w}\theta _{k}-{v_{s}^{2} \over 1+w}{k^{2} \over a}\delta _{k}+{k^{2} \over a}\sigma _{k}-{k^{2} \over a}\phi _{k}=0\\\end{aligned}}\,\! where $\sigma _{k}\equiv (k_{i}k_{j}-{1 \over 3}{\delta _{ij}}){\sigma _{ij}}_{k}$ . There are, of course, higher moments as well. Compare the above result to the result we got using just the linearized fluid equations. That simplistic result was almost right, except for the factor of ${1 \over 3}$ . We can look at different components of this energy-momentum conservation equation:

• CDM: here we are pressure-less, $v_{s}=0$ , $w=0$ , so we have:
{\begin{aligned}{\dot {\delta }}_{c}+{1 \over a}\theta _{c}-3{\dot {\Psi }}&=0\\{\dot {\theta }}_{c}+{{\dot {a}} \over a}\theta _{c}-{k^{2} \over a}\phi &=0\\\end{aligned}}\,\! • Baryons: we have a non-relativistic fluid, $v_{s}^{2}\ll c^{2}$ , and ${\sigma _{ij}}\approx 0$ :
{\begin{aligned}{\dot {\delta }}_{b}+{1 \over a}\theta _{b}-3{\dot {\Psi }}&=0\\{\dot {\theta }}_{b}+{{\dot {a}} \over a}\theta _{b}-{v_{s}^{2}k^{2} \over a}\delta _{b}-{k^{2} \over a}\phi -\underbrace {{4 \over 3}{{\bar {\rho }}_{\gamma } \over {\bar {\rho }}_{b}}n_{e}\sigma _{T}(\theta _{\gamma }-\theta _{b})} _{{momentum\ transfer \atop between} \atop photons\ and\ baryons}&=0\\\end{aligned}}\,\! • Photons: $w={1 \over 3}$ , $v_{s}^{2}={c^{2} \over 3}$ :
{\begin{aligned}{\dot {\delta }}_{\gamma }+{4 \over 3a}\theta _{\gamma }-4{\dot {\Psi }}&=0\\{\dot {\theta }}_{\gamma }\dots \sigma _{\gamma }\dots -n_{e}\sigma _{T}(\theta _{b}-\theta _{\gamma })&=0\\\end{aligned}}\,\! • Neutrinos: no Thomson scattering, collision-less, etc.

### Coupled Baryon-Photon Fluid

Before photon decoupling, baryons and photons are tightly coupled via Thomson scattering (it behaves like a single fluid). Recall that Thomson scattering is the low-energy limit of Compton scattering, which is the general mechanism for photon-electron scattering. We are in this low-energy regime because $kT_{\gamma }\ll m_{e}c^{2}\ll m_{p}c^{2}$ after about 1 second. The cross-section for Thomson scattering is $\sigma _{T}=.665\cdot 10^{-24}cm^{2}$ , and

${d\sigma \over d\Omega }={3 \over 15\pi }\sigma _{T}(1+\cos ^{2}\theta )\,\!$ The optical depth to Thomson scattering is given by:

$\tau =\int {n_{e}\sigma _{T}ds}\,\!$ where $n_{e}$ is the electron density, and $ds$ is distance along the line of sight. $\tau \gg 1$ is optically thick (opaque), and $\tau \ll 1$ is optically thin (transparent).

We can calculate when photons decouple (when $\tau$ from Thomson scattering $\sim 1$ ):

$\tau =\int {n_{e}\sigma _{T}c{dt \over dz}dz}\,\!$ where integral over distance is written in terms of redshift. We can use the Friedmann equation to replace $dt \over dz$ , and we’ll use that the free electron density is given by:

$n_{e}(z)=x_{e}(z)n_{H}(z)=x_{e}(z)(1+z)^{3}n_{H,0}\,\!$ where $n_{H}$ is the # density of hydrogen and $x_{e}$ is the ionization fraction. We can use the Y parameter to calculate $n_{H},0$ :

$Y={4n_{He} \over n_{H}+4n_{He}}={2n_{neutron} \over n_{B}}={n_{B}-n_{H} \over n_{B}}\,\!$ which tells us that $n_{H}=(1-Y)n_{B}$ . Thus:

$n_{e}(z)=x_{e}(z)(1+z)^{3}(1-Y)\underbrace {\Omega _{B,0} \over m_{\rho }} _{n_{B,0} \over \rho _{crit,0}}\underbrace {3H_{0}^{2} \over 8\pi G} _{\rho _{crit,0}}\,\!$ We’ll actually do this out to get $\tau$ as a function of $z$ and cosmological parameters on PS#8. Next time we’ll see if we can calculate the sound speed $v_{s}$ .