# Cosmology Lecture 15

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## Linear Terms in the Expanding Universe

Recall that we are solving for the (linear) perturbing parameters for an expanding universe using 3 equations:

${\displaystyle {\dot {\delta }}+\nabla \cdot {\vec {v}}_{1}+{{\dot {a}} \over a}{\vec {r}}\cdot \nabla \delta =0\,\!}$
${\displaystyle {\partial {\vec {v}}_{1} \over \partial t}+{{\dot {a}} \over a}({\vec {r}}\cdot \nabla ){\vec {v}}_{1}+{{\dot {a}} \over a}{\vec {v}}_{1}=-v_{s}^{2}\nabla \delta -\nabla \phi _{1}\,\!}$
${\displaystyle \nabla ^{2}\phi _{1}=4\pi G\rho _{0}\delta \,\!}$

(recall that ${\displaystyle \rho _{0}\delta =\rho _{i}}$ by definition of ${\displaystyle \delta }$). The trick to solving these three equations is to go into Fourier space and change from physical coordinates to comoving coordinates (to get rid of the Hubble expansion component). We have 3 fields (${\displaystyle \delta ,{\vec {v}}_{1},\phi _{1}}$) that we need to move into Fourier space:

${\displaystyle {\begin{pmatrix}\delta ({\vec {r}},t)\\{\vec {v}}_{1}({\vec {r}},t)\\\phi _{1}({\vec {r}},t)\\\end{pmatrix}}=\int {d^{3}re^{i{\vec {k}}\cdot {\vec {r}} \over a(t)}{\begin{pmatrix}\delta _{k}(t)\\{\vec {v}}_{k}(t)\\\phi _{k}(t)\end{pmatrix}}}\,\!}$

where ${\displaystyle {\vec {k}}}$ is the comoving wavenumber, and ${\displaystyle {\vec {r}}=a{\vec {x}}}$ are the physical coordinates. Thus:

{\displaystyle {\begin{aligned}{\dot {\delta }}({\vec {r}},t)&=\int {d^{3}re^{i{\vec {k}}\cdot {\vec {r}} \over a(t)}({\dot {\delta }}_{k}-i{{\dot {a}} \over a}{{\vec {k}}\cdot {\vec {r}} \over a}\delta _{k}})\\\nabla \delta ({\vec {r}},t)&=\int {d^{3}re^{i{\vec {k}}\cdot {\vec {r}} \over a(t)}{i{\vec {k}} \over a}\delta _{k}}\\\nabla {\vec {v}}_{1}({\vec {r}},t)&=\int {d^{3}re^{i{\vec {k}}\cdot {\vec {r}} \over a(t)}{i{\vec {k}}\cdot {\vec {v}}_{k} \over a}}\\\end{aligned}}\,\!}

and so on. Using the properties listed above, we get for the first equation:

{\displaystyle {\begin{aligned}{\dot {\delta }}_{k}-i{{\dot {a}} \over a}{{\vec {k}}\cdot {\vec {r}} \over a}\delta _{k}+{i{\vec {k}}\cdot {\vec {v}}_{k} \over a}+{{\dot {a}} \over a}{\vec {r}}{i{\vec {k}} \over a}\delta _{k}&=a\\\end{aligned}}\,\!}
${\displaystyle {{\dot {\delta }}_{k}+{i{\vec {k}}{\vec {v}}_{k} \over a}=0}\,\!}$

The second equation becomes:

{\displaystyle {\begin{aligned}{\dot {\vec {v}}}_{k}=i{{\dot {a}} \over a}{{\vec {k}}\cdot {\vec {r}} \over a}{\vec {v}}_{k}+{{\dot {a}} \over a}({\vec {r}}\cdot {i{\vec {k}} \over a}){\vec {v}}_{k}+{{\dot {a}} \over a}{\vec {v}}_{k}&=-v_{s}^{2}{i{\vec {k}} \over a}\delta _{k}-{i{\vec {k}} \over a}\phi _{1k}\\\end{aligned}}\,\!}
${\displaystyle {{\frac {d}{dt}}(a{\vec {v}}_{k})+iv_{s}^{2}{\vec {k}}\delta _{k}+i{\vec {k}}\phi _{k}=0}\,\!}$

And finally, Poisson’s equation is:

${\displaystyle {{k^{2} \over a^{2}}\phi _{k}=-4\pi G\rho _{0}\delta _{k}}\,\!}$

To solve these equations, we first combine (1) and (2). Taking ${\displaystyle a^{2}\times }$(1), we get:

${\displaystyle a^{2}{\dot {\delta }}_{k}+i{\vec {k}}(a{\vec {v}}_{k})=0\,\!}$

then taking the derivative with respect to time:

{\displaystyle {\begin{aligned}a^{2}{\ddot {\delta }}_{k}+2a{\dot {a}}{\dot {\delta }}_{k}+i{\vec {k}}(-iv_{s}^{2}{\vec {k}}\delta _{k}-\overbrace {i{\vec {k}}\phi _{k}} ^{=0})&\\\end{aligned}}\,\!}
${\displaystyle {{\ddot {\delta }}_{k}+2{{\dot {a}} \over a}{\dot {\delta }}_{k}+\left({v_{s}^{2}k^{2} \over a^{2}}-4\pi G\rho _{0}\right)\delta _{k}=0}\,\!}$

We can rewrite this in terms of the comoving Jeans wavenumber ${\displaystyle k_{J}\equiv {\sqrt {4\pi G\rho _{0}a^{2} \over v_{s}^{2}}}}$:

${\displaystyle {{\ddot {\delta }}_{k}+2{{\dot {a}} \over a}{\dot {\delta }}_{k}+{v_{s}^{2} \over a^{2}}(k^{2}-k_{J}^{2})\delta _{k}=0}\,\!}$

The unstable regime for this equation is where ${\displaystyle k\ll k_{J}}$:

${\displaystyle {{\ddot {\delta }}_{k}+2{{\dot {a}} \over a}{\dot {\delta }}_{k}-4\pi G\rho _{0}\delta _{k}=0}\,\!}$

Sometimes you’ll find this equation written with fewer dots (time derivatives). This is because people will occasionally redefine their time to be “conformal time”, which is normal time divided by the expansion parameter.

• Suppose we have a flat, matter dominated universe where ${\displaystyle \Omega _{m}=1}$. Then rewriting ${\displaystyle \rho _{0}\to \rho }$ so we don’t get confused about perturbation parameters vs. present-day values, we get:
${\displaystyle 4\pi G\rho ={3 \over 2}H^{2}={3 \over 2}\left({{\dot {a}} \over a}\right)^{2}={2 \over 3t^{2}}\,\!}$
${\displaystyle {{\ddot {\delta }}_{k}+{4 \over 3t}{\dot {\delta }}_{k}-{2 \over 3t^{2}}\delta _{k}=0}\,\!}$

This equation has a polynomial solution. If ${\displaystyle \delta _{k}(t)\propto t^{\alpha }}$, then:

{\displaystyle {\begin{aligned}\alpha (\alpha -1)+{4 \over 3}\alpha -{2 \over 3}&=0\\(\alpha +1)(\alpha -{2 \over 3})&=0\\\end{aligned}}\,\!}

Thus, there are 2 solutions–a growing mode:

${\displaystyle {\delta _{k}^{+}\propto t^{2 \over 3}\propto a}\,\!}$

and a decaying mode:

${\displaystyle {\delta _{k}^{-}\propto t^{-1}}\,\!}$

These are very important results for understanding how structure forms in an expanding universe.

Now we can solve for ${\displaystyle \phi _{k}(t)}$: ${\displaystyle \phi _{k}\propto a^{2}\rho _{0}\delta _{k}\propto a^{2}a^{-3}a=constant}$. The perturbed gravitational potential is constant.

We also want to solve for ${\displaystyle {\vec {v}}_{1}({\vec {r}},t)\to {\vec {v}}_{k}(t)}$. To do this, we’ll decompose ${\displaystyle {\vec {v}}={\vec {v}}_{\perp }+{\vec {v}}_{\|}}$, where ${\displaystyle \nabla \cdot {\vec {v}}_{\perp }=0}$ is called the divergence free “rotational” or “transverse” mode. Then using:

{\displaystyle {\begin{aligned}&{\dot {\delta }}_{k}+i{{\vec {k}} \over a}\cdot {\vec {v}}_{k}=0\\&{\partial \over \partial t}(a{\vec {v}}_{k})+iv_{s}^{2}{\vec {k}}\delta _{k}+i{\vec {k}}\phi _{k}=0\\\end{aligned}}\,\!}

we find that ${\displaystyle i{\vec {k}}{\vec {v}}_{\perp ,k}=0\,\Rightarrow \,{\vec {v}}_{\perp ,k}\perp {\vec {k}}}$. For ${\displaystyle {\vec {v}}_{\|}}$ (the curl-free “longitudinal” mode), ${\displaystyle \nabla \times {\vec {v}}_{\|}=0}$, so ${\displaystyle {\vec {k}}\times {\vec {v}}_{\|,k}=0\,\Rightarrow \,{\vec {v}}_{\|,k}\propto {\vec {k}}}$. Thus in k-space:

${\displaystyle \nabla \cdot {\vec {v}}=\nabla \cdot {\vec {v}}_{\|}=i|{\vec {k}}|\cdot |{\vec {v}}_{\|,k}|\,\!}$

For the ${\displaystyle \perp {\vec {k}}}$ component, equation 2 gives us ${\displaystyle {\partial \over \partial t}(a{\vec {v}}_{\perp ,k})=0}$, so ${\displaystyle |{\vec {v}}_{\perp ,k}|\propto {1 \over a}}$. Likewise, for the ${\displaystyle \|\,{\vec {k}}}$ component, we get:

${\displaystyle {\partial \over \partial t}(a{\vec {v}}_{\|})+iv_{s}^{2}{\vec {k}}\delta _{k}+i{\vec {k}}\phi _{k}=0\,\!}$

Using (1), this becomes:

${\displaystyle {\dot {\delta }}_{k}+{i{\vec {k}}{\vec {v}}_{\|,k} \over a}=0\,\!}$
${\displaystyle {{\vec {v}}_{\|,k}={-a \over ik}{\dot {\delta }}_{k}\propto t^{2 \over 3}t^{-1 \over 3}\propto t^{1 \over 3}\propto {\sqrt {a}}}\,\!}$

So we worked out linear perturbations for a matter dominated universe.