# Cosmology Lecture 16

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Recall that for $k\ll k_{J}$ :

${\ddot {\delta }}_{k}+2{{\dot {a}} \over a}{\dot {\delta }}_{k}-4\pi G\rho \delta _{k}=0\,\!$ We did different cases for this equation:

• For flat $\Omega _{m}=1$ , we found two solutions:
{\begin{aligned}\delta _{+}&\propto a\propto t^{2 \over 3}\\\delta _{-}&\propto t^{-1}\\\end{aligned}}\,\! We also found that $\phi _{2}$ was independent of $a$ , and that ${\vec {v}}_{\perp }\propto {1 \over a}$ , and ${\vec {v}}_{\|}\propto t^{1 \over 3}\propto {1 \over a^{\frac {1}{2}}}$ .

• Suppose we have an open universe:
{\begin{aligned}a(\theta )&={\Omega _{0} \over 2(1-\Omega _{0})}(\cosh \theta -1)\\t(\theta )&={\Omega _{0} \over 2H_{0}(1-\Omega _{0})^{3 \over 2}}(\sinh \theta -\theta )\\\end{aligned}}\,\! Using the equation we derived previously:

${\ddot {\delta }}_{k}+2{{\dot {a}} \over a}{\dot {\delta }}_{k}-4\pi G\rho \delta _{k}=0\,\!$ and we can do some algebra to come up with the equation:

$(\cosh \theta -1){d^{2}\delta \over d\theta ^{2}}+\sinh \theta -{d\delta \over d\theta }-3\delta =0\,\!$ This equation again has two solutions:

{\begin{aligned}\delta _{+}&\propto {\frac {3\sinh \theta (\sinh \theta -\theta )}{(\cosh \theta -1)^{2}}}-2\\\delta _{-}&\propto {\sinh \theta \over (\cosh \theta -1)^{2}}\\\end{aligned}}\,\! • For a closed universe (we will play with this on the homework), we get the following equation:
$(1-\cosh \theta ){d^{2}\delta \over d\theta ^{2}}+\sin \theta {d\delta \over d\theta }-3\delta =0\,\!$ This has solutions:

{\begin{aligned}\delta _{+}&\propto 2-{\frac {3\sin \theta (\theta -\sin \theta )}{(1-\cos \theta )^{2}}}\\\delta _{-}&\propto {\frac {\sin \theta }{(1-\cos \theta )^{2}}}\\\end{aligned}}\,\! ### Linear Perturbation Theory of Gravitational Instabilities

We’re just going to do a sketch of the full story, which grew out of Lifshitz (1946). Additionally, look at Weinberg section $15.10$ , Dodelson Ch. 4, and Peebles (1980). The full theory is described by linearized, coupled Einstein equations (for metric perturbations), and the Boltzmann equations (for density perturbations).

### Metric Perturbations (in “Synchronous” Gauge)

Recall our space-time interval in expanding space:

$ds^{2}=c^{2}dt^{2}-a^{2}(t)(\delta _{ij}))dx^{i}\cdot dx^{j}\,\!$ where $i,j=1,2,3$ . We are going to make perturbations to this equation. To do this, we are going to define a “gauge”. This is analogous to the vector ${\vec {A}}$ in E&M (called the Coulomb gauge), where:

${\begin{matrix}\nabla \cdot {\vec {B}}=0&\Rightarrow &{\vec {B}}=\nabla \cdot {\vec {A}}\end{matrix}}\,\!$ and $\nabla \cdot {\vec {A}}=0$ . For metric perturbations, we will create a metric perturbation tensor $h_{ij}$ , which is a $3\times 3$ symmetric tensor with 6 degrees of freedom. This is added to our equation for the space-time interval:

$ds^{2}=c^{2}dt^{2}-a^{2}(t)(\delta _{ij}+h_{ij}))dx^{i}\cdot dx^{j}\,\!$ We then decompose this perturbation tensor into its trace and traceless components:

$h_{ij}=\underbrace {{h \over 3}{\delta _{ij}}} _{trace}+\underbrace {h_{ij}^{\|}+h_{ij}^{\perp }+h_{ij}^{T}} _{traceless}\,\!$ We will make the following definitions:

• $trace(h_{ij})=h$ (which has 1 degree of freedom)
• $h_{ij}^{T}$ is transverse: $\partial _{i}h_{ij}=0$ (2 degrees of freedom)
• The divergences of the vectors $h_{ij}^{\|}$ (with 1 degree of freedom) and $h_{ij}^{\perp }$ (with 2 degrees of freedom) are longitudinal (curl-free) and transverse (div free) respectively. We will write the longitudinal component as (recall $\nabla \times {\vec {W}}=0\Leftrightarrow \epsilon _{ijk}\partial _{j}W_{k}=0$ ):
$\epsilon _{ijk}\partial _{j}\partial _{\ell }h_{\ell k}^{\|}=0\,\!$ and the transverse component:

$\partial _{j}\partial _{i}h_{ij}^{\perp }=0\,\!$ These definitions together give us:

{\begin{aligned}h_{ij}^{\|}&=(\partial _{i}\partial _{j}-{1 \over 3}\delta _{ij}\nabla ^{2})\mu \\h_{ij}^{\perp }&=\partial _{i}\partial _{j}+\partial _{j}A_{i}\\\end{aligned}}\,\! where $\mu$ is some scalar field and $A_{i}$ is some vector field where $\partial _{i}A_{i}=0$ .

Now we will discuss three different modes of perturbations:

• Scalar Mode, described by $(h,\mu )$ (recall $tr(h_{ij})=h$ , $\mu =h_{ij}^{\|}$ ), which accounts for galaxy formation.
• Vector Mode, described by $A_{i}$ (which is $h_{ij}^{\perp }$ )
• Tensor Mode, described by $h_{ij}^{T}$ , which accounts for gravitational waves/radiation.

How did we get 6 degrees of freedom? Einsteins equations tell us that $G_{\mu \nu }\propto T_{\mu \nu }$ , where $G_{\mu \nu }$ is the metric ($\mu ,\nu =0,1,2,3$ ), and $T_{\mu \nu }$ is the energy-momentum tensor. Since these tensors are symmetric, we have 10 degrees of freedom here. However, energy conservation (the Bianchi identity) states that $\partial _{\mu }T^{\mu \nu }=0$ , which imposes 4 more constraints, leaving us with 6 degrees of freedom.

It should also be said that we have other gauge choices besides the one we defined. For example, the Conformal Newtonian gauge (which is restricted for scalar modes) is:

$ds^{2}=c^{2}dt^{2}(1+2\phi )-a^{2}(t)(1-\Psi )dx^{i}dx_{i}\,\!$ and has 2 scalar fields ($\phi ,\Psi$ ) which are related by gauge transformation to ($h,\mu$ ).