I am trying to understand the following claim from a professor, in the context of studying the evolution of the fluid that fills the universe according to Cosmology:
If we take the linear growth equation:
$$\partial^2_\tau\delta(\vec{k},\tau)+\mathcal{H}(\tau)\partial_\tau\delta(\vec{k},\tau)-\dfrac{3}{2}\Omega_m(\tau)\mathcal{H}^2(\tau)\delta(\vec{k},\tau)=0\ \ \ \ \ \ \ \ \ \ (1)$$
and rewrite it in terms of derivatives with respect to the scale factor $a$, we get:
$$-a^2\mathcal{H}^2\partial^2_a\delta+\dfrac{3}{2}\mathcal{H}^2[\Omega_m(a)-2]a\partial_a\delta+\dfrac{3}{2}\Omega_m\mathcal{H}^2\delta=0\ \ \ \ \ \ \ \ \ \ \ \ \ (2)$$
where the notation used is the following:
- $\tau$ denotes conformal time, where $d\tau=dt/a$ with $t$ being coordinate time and $a$ the scale factor that quantifies the expansion of the universe.
- $\rho=\bar{\rho}(1+\delta)$ is the total density of the cosmological fluid, where $\bar{\rho}$ is the background density and $\delta$ the density contrast. We are considering $\rho\simeq\rho_m$, that is, a matter-dominated universe.
- $\mathcal{H}=\dfrac{\partial_\tau a}{a}$ is the conformal Hubble parameter.
I understand very well where equation (1) comes from, but I am having trouble getting to (2). In order to prove (2) from (1), the first thing I have done is to write the first and second order partial derivatives with respect to conformal time $\tau$ in terms of the partial derivatives with respect to the scale factor $a$, obtaining (I omit here the boring calculations):
$$\begin{cases}\partial_\tau=a\mathcal{H}\partial_a \\ \partial^2_\tau=(\partial^2_\tau a)\partial_a+a^2\mathcal{H}^2\partial^2_a\end{cases}$$
If I use the second expression in order to rewrite (1), what I get is:
$$\partial^2_\tau\delta+\mathcal{H}\partial_\tau\delta-\dfrac{3}{2}\Omega_m\mathcal{H}^2\delta=0\ \ \Rightarrow\ \ [(\partial^2_\tau a)\partial_a+a^2\mathcal{H}^2\partial^2_a]\delta+a\mathcal{H}^2\partial_a\delta-\dfrac{3}{2}\Omega_m\mathcal{H}^2\delta=0\ \ \Rightarrow$$
$$\Rightarrow\ \ -a^2\mathcal{H}^2\partial^2_a\delta-[(\partial^2_\tau a)+a\mathcal{H}^2]\partial_a\delta+\dfrac{3}{2}\Omega_m\mathcal{H}^2\delta=0$$
The middle term seems to be the problematic one. In order to recover (2), I would need to prove that:
$$-\bigg[\dfrac{1}{a}(\partial^2_\tau a+\mathcal{H})\bigg]=\dfrac{3}{2}\mathcal{H}^2(\Omega_m-2)$$
But… how? I suspect I might need to use the second Friedmann equation, since its left hand side looks suspiciously similar to $(1/a)(\partial^2_\tau a)$, but with the derivative with respect to coordinate time instead of conformal time:
$$\dfrac{1}{a}\partial^2_t a=\dfrac{4\pi G}{3}\bigg(\rho+\dfrac{3P}{c^2}\bigg)$$
but I don't know how to proceed to get the correct result. Help, please?
Edit: I am now quite sure that equation (2) does in fact not contain any typos, since its Fourier space version is used several times later on in the notes.
Best Answer
We have the linear growth equation $$\partial_\tau^2 \delta+ \mathcal H \partial_\tau \delta - \frac{3}{2}\Omega_m(\tau) \mathcal H^2 \delta = 0\tag{A}$$ And we want to rewrite this in terms of derivatives with respect to scale factor $a$ $$-a^2 \mathcal H^2 \partial_a^2 \delta + \frac{3}{2} \mathcal H^2 \left[ \Omega_m(a) -2\right] a \partial_a \delta + \frac{3}{2}\Omega_m \mathcal H^2 \delta = 0 $$ Where we have that $$d\tau = dt / a\tag{B}$$ $$\mathcal H = \frac{1}{a} \frac{ \partial a} { \partial \tau } = \frac{ \partial a} { \partial t } =a H \tag{C}$$
Assume we are in a universe with only matter and cosmological constant. We can start with the usual Friedmann equation $$\frac{H^2} {H_0^2} = \Omega_{m,0} a^{-3} + \Omega_{\Lambda,0}\tag{1}$$ Rearrange so that $$\begin{align} 1 &= \Omega_{m,0} a^{-3} \frac{H_0^2} {H^2} + \Omega_{\Lambda,0} \frac{H_0^2} {H^2} \\ &\equiv \Omega_m(a)+\Omega_\Lambda(a)\tag{2} \end{align} $$ Where in the second line I have introduced the definition of what people usually mean when they write $\Omega_m(a)$ and $\Omega_\Lambda(a)$ $$\Omega_m(a) = \frac{\Omega_{m,0}} {a^3} \frac{H_0^2} {H^2}\tag{3a}$$
$$\Omega_\Lambda(a) = \Omega_{\Lambda,0} \frac{H_0^2} {H^2} \tag{3b}$$
Something we will need is $\partial_\tau \mathcal H$ so lets compute that first $$\begin{align} \partial_\tau \mathcal H &= \frac{ \partial } { \partial \tau } (a H)\\ &= H \frac{ \partial a} { \partial \tau } + a \frac{ \partial H} { \partial \tau } \\ \textrm{(Use Eq. (B))} &= H a \frac{ \partial a} { \partial t } + a^2 \frac{ \partial H} { \partial t } \\ &= H^2 a^2 \left(1 + \frac{1}{H^2} \frac{ \partial H} { \partial t } \right)\\ \textrm{(Use Eq. (C))} &= \mathcal H^2\left( 1 + \frac{1}{H^2}\frac{ \partial H} { \partial t } \right) \tag{4} \end{align} $$ We can do some manipulations to get $\partial_t H$. Start by taking the time derivative of the Friedmann equation (Eq. (1)) and let $\dot A = \partial A / \partial t$ $$\begin{align} \frac{2 H \dot H} {H_0^2} &= -3a^{-4}\Omega_{m,0} \frac{ \partial a} { \partial t } \\ \Rightarrow \dot H &= H_0^2 \Omega_{m,0}\times \left(-\frac{3 a^{-4}}{2 H} \frac{ \partial a} { \partial t } \right)\\ \textrm{(Use Eq. (3a))} &= H^2 a^3 \Omega_m(a) \times \left( - \frac{3a^{-4}} {2H} \frac{ \partial a} { \partial t } \right)\\ \dot H&= -\frac{3}{2}\Omega_m(a) H^2 \tag{5} \end{align} $$ Plugging Eq.(5) into Eq.(4) gives us $$\partial_\tau \mathcal H = \mathcal H^2 \left( 1 - \frac{3}{2}\Omega_m(a) \right)\tag{6}$$
Now we can rewrite $\partial_\tau$ in terms of $\partial_a$ $$\begin{align} \partial_\tau &= a \mathcal H\partial_a\tag{7a}\\ \partial_\tau^2 &= a \mathcal H \partial_a \left( a \mathcal H \partial_a \right)\\ &= (a \mathcal H)^2 \partial_a^2 + a\mathcal H^2 \partial_a + a^2\mathcal H(\partial_a \mathcal H) (\partial_a)\\ \textrm{(Use Eq.(7a))} &= (a \mathcal H)^2\partial_a^2 +a \mathcal H^2 \partial_a + a^2 \mathcal H \left( \frac{1}{a \mathcal H} \partial_\tau \mathcal H \right) \partial_a\\ \textrm{(Use Eq.(6))} &=(a \mathcal H)^2\partial_a^2 +a \mathcal H^2 \partial_a + a^2 \mathcal H \left( \frac{1}{a \mathcal H} \times \mathcal H^2\left[ 1 - \frac{3}{2}\Omega_m(a) \right] \right) \partial_a\\ \partial_\tau^2&= (a \mathcal H)^2 \partial_a^2 + a \mathcal H^2 \partial_a + a \mathcal{H} ^2 \left[ 1 - \frac{3}{2}\Omega_m(a) \right]\partial_a\tag{7b} \end{align} $$
Plugging in Eq.(7) into the linear growth equation (Eq.(A)) then gives us what we want $$\begin{align} 0&=\color{red}{ \partial_\tau^2 \delta}+ \color{blue}{ \mathcal H\partial_\tau \delta} - \frac{3}{2}\Omega_m \mathcal H^2 \delta\\ &=\color{red}{ (a \mathcal H)^2 \partial_a^2 \delta + a \mathcal H^2\partial_a \delta + a \mathcal H^2 \left[1 - \frac{3}{2}\Omega_m(a) \right]\partial_a\delta} + \color{blue} { a\mathcal H^2\partial_a \delta} - \frac{3}{2}\Omega_m \mathcal H^2 \delta\\ &= (a \mathcal H)^2 \partial_a^2 \delta -\frac{3}{2} a\mathcal H^2 \left[ \Omega_m(a) -2\right]\partial_a\delta -\frac{3}{2}\Omega_m \mathcal H^2\delta \end{align} $$
$$\boxed{- a^2 \mathcal H^2 \partial_a^2 \delta + \frac{3}{2} a\mathcal H^2 \left[ \Omega_m(a) -2\right]\partial_a\delta + \frac{3}{2}\Omega_m \mathcal H^2\delta = 0}$$