Cosmology Lecture 17

More on Metric Perturbations

Recall our equation for metric perturbations:

$ds^{2}=c^{2}dt^{2}-a^{2}(t)({\delta _{ij}}+h_{ij})dx^{i}dx^{j}\,\!$

If we have the leisure of picking scalar modes, then we can pick the conformal Newtonian gauge:

$ds^{2}=c^{2}dt^{2}(1+2\phi )-a^{2}(t)({\delta _{ij}}+h_{ij})dx^{i}dx_{i}(1-2\Psi )\,\!$

Einstein’s equations says $ds^{2}=g_{\mu \nu }dx^{\mu }dx^{\nu }$, so to $0^{th}$ order, the Friedmann equations becomes:

${\begin{aligned}\left({{\dot {a}} \over a}\right)^{2}&={8\pi \over 3}G{\bar {\rho }}-{k \over a^{2}}\\{{\ddot {a}} \over a}&=-{4\pi \over 3}G({\bar {\rho }}+3{\bar {P}})\\\end{aligned}}\,\!$

If we want to bring this to first order, the left-hand side contains terms with $h,{\dot {h}},{\ddot {h}}$, etc. (recall that $h_{ij}={h \over 3}{\delta _{ij}}+h_{ij}^{\|}+h_{ij}^{\perp }+h_{ij}^{T}$). The right-hand side will contain terms of $\delta \equiv {\rho -{\bar {\rho }} \over {\bar {\rho }}}$, $dP$, etc. What we need is a set of equations to describe the evolution of $\delta ,dP$, etc.

The first of these is the Boltzmann (transport) Equation. If we have a function describing the phase-space distribution: $dN=f({\vec {x}},{\vec {p}},t)d^{3}xd^{3}p$ where $d^{3}p$ is the conjugate momentum to $x$. Then $f$ is described by:

${\begin{aligned}f({\vec {x}},{\vec {p}},t)&=f_{0}(p)-f_{1}({\vec {x}},{\vec {p}},t)\\f_{0}(p)&={g \over h_{p}^{3}}{1 \over e^{E \over kT}\pm 1}\\\end{aligned}}\,\!$

where $h_{p}$ is Planck’s constant (it just has a $p$ to differentiate it from our other $h$). The Boltzmann equation says:

${\partial f \over \partial t}={\dot {x}}^{i}{\partial f \over \partial x^{i}}+{\dot {p}}^{i}{\partial f \over \partial p^{i}}=\left({\partial f \over \partial t}\right)_{c}\,\!$

The subscript $c$ denotes collisional terms. If there aren’t any collisions (or they are rare), then we have the “Collision-less Boltzmann Equation” (aka. Vlasov Equation, Louisville’s Theorum), which says that:

${df \over dt}=0\,\!$

The $df \over dt$ is just what the total left-hand side of the Boltzmann equation is. In general, the Boltzmann equation is very hard to solve, so we need to exploit any symmetries we can.

If we take the velocity moments (in the non-relativistic limit):

$\int {d^{3}p\cdot f}=n\,\!$

where $n$ is the number density. The mean velocity, then, is:

$\langle {\vec {v}}\rangle ={\int {d^{3}p{\vec {v}}f} \over \int {d^{3}p\cdot f}}={\int {d^{3}p{\vec {v}}f} \over n}\,\!$

Our velocity moments are:

${\int {d^{3}pv_{i}v_{j}f} \over n}=\langle {v_{i}v_{j}}\rangle \,\!$

So we can define a velocity-moment-tensor (called a stress tensor):

${\sigma _{ij}}^{2}\equiv \langle {v_{i}v_{j}}\rangle -\langle {v_{i}}\rangle \langle {v_{j}}\rangle \,\!$

Now we will take a series of moments of our collision-less Boltzman equation:

• The zeroth moment of the collision-less Boltzmann equations is:

${\partial \over \partial t}\int {d^{3}p\,f}+\int {d^{3}p{\vec {v}}\cdot \nabla f}-\partial _{i}\phi \int {d^{3}p{\partial f \over \partial p_{i}}}=0\,\!$

Our $\nabla$ is with respect to $x$, so we can pull it out of the integral. The last integral we can do as integration by parts, but we can do it more easily if we have a well-behaved $f$, such that $f\to 0$ at large $p$. If this is the case, then the integral is 0. Using these identities, we get:

${\partial \over \partial t}n+\nabla \cdot (n\langle {\vec {v}}\rangle )=0\,\!$

This is just our continuity equation!

• The first velocity moment, $\int {d^{3}pv_{j}}$, is given by (using the Boltzmann equation):

${\partial \over \partial t}\int {d^{3}pv_{j}f}+\int {d^{3}pv_{j}v_{i}\partial _{i}f}-\partial _{i}\phi \int {d^{3}pv_{j}{\partial f \over \partial p_{i}}}=0\,\!$

We can solve the third term in this equation as:

$-\partial _{i}\phi \int {d^{3}pv_{j}{\partial f \over \partial p_{i}}}={-\partial _{i}\phi \over m}\int {d^{3}p\left({\partial \over \partial p_{i}}(p_{j}f)-f\underbrace {\partial p_{j} \over \partial p_{i}} _{\delta _{ij}}\right)}\,\!$

Again, for well-behaved $f$, ${\partial \over \partial p_{i}}(p_{j}f)$ should be 0 when integrated. Setting $m=1$, we get:

$-\partial _{i}\phi \int {d^{3}pv_{j}{\partial f \over \partial p_{i}}}=\partial _{j}\phi \int {d^{3}pf}=\partial _{j}\phi n\,\!$

Putting this into our original equation, we get:

${\begin{aligned}n{\partial \over \partial t}\langle {v_{j}}\rangle +\langle {v_{j}}\rangle [-\partial _{i}(n\langle {v_{i}}\rangle ]+\underbrace {\partial _{i}(n\langle {v_{i}}\rangle \langle {v_{j}}\rangle )} _{\langle {v_{j}}\rangle \partial _{i}(n\langle {v_{i}}\rangle )+n\langle {v_{i}}\rangle \partial _{i}(\langle {v_{j}}\rangle )}&=0\\-n\partial _{j}\phi -\partial _{i}(n{\sigma _{ij}}^{2})+n\langle {v_{i}}\rangle \partial _{i}(v_{j})&=0\\\end{aligned}}\,\!$

Where we say $\partial _{i}={\partial \over \partial t}n$ by the continuity equation.

${\partial \over \partial t}\langle {v_{j}}\rangle +\langle {v_{i}}\rangle \partial _{i}\langle {v_{j}}\rangle =-\partial _{j}\phi -\underbrace {{1 \over n}\partial _{i}(n{\sigma _{ij}}^{2})} _{new\ term}\,\!$

This equation looks a little like the equation:

${\partial {\vec {v}} \over \partial t}+({\vec {v}}\cdot \nabla ){\vec {v}}=-\nabla \phi -{1 \over n}\partial _{i}(n{\sigma _{ij}}^{2})\,\!$

we will discuss this new term next time.