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.

  • $\begingroup$ Are you sure that Equation 2 is correct ? See arXiv:astro-ph/0305286v2 and arXiv:1005.0233. You can also try to simplify Eqn.2 (such as cancel $H^2$ and divide by $a^2$ etc.) $\endgroup$
    – seVenVo1d
    Commented Oct 5, 2023 at 17:28
  • $\begingroup$ Also remember that $\Omega_k(a) \neq \Omega_{k0}{\rm exp}(3\int_a^1[1+w(a')] d{\rm ln}a')$. But: $$ \Omega_k(a) \equiv \frac{\Omega_{k0}{\rm exp}(3\int_a^1[1+w(a')] d{\rm ln}a')}{E^2(a)}$$ where $E(a) = H(a)/H_0$ $\endgroup$
    – seVenVo1d
    Commented Oct 5, 2023 at 17:29
  • $\begingroup$ @seVenVo1d There might be some kind of errata, but this all came from a professor's notes for a summer school on particle physics and cosmology, so... I don't know. Since I couldn't figure it out, I thought it was me who was wrong, not him. In equation (6) of the first paper you mention, I suppose the derivatives are with respect to coordinate time, not conformal time, but still, there is a factor of 2 in the middle term that doesn't appear in equation (1) of my question, and I don't understand why. I'm so lost. And I don't understand where the equations in your second comment come from... $\endgroup$ Commented Oct 6, 2023 at 17:55
  • $\begingroup$ It just a general equation...umm I am not sure how can I help you more.. $\endgroup$
    – seVenVo1d
    Commented Oct 9, 2023 at 13:37
  • 1
    $\begingroup$ @Sten Thank you, you are right. According to the answer below by delon, the equation is also true in a universe with dark energy, not only in a matter-dominated one. It's strange because many times this professor considers $\rho\simeq\rho_m$ (so, a matter-dominated universe), but then includes $\Omega_m$ in the equations. It seems a bit inconsistent now that I think about it... $\endgroup$ Commented Oct 12, 2023 at 16:45

1 Answer 1


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 $$\begin{align} 0&=\color{red}{ \partial_\tau^2 \delta}+ \color{blue}{ \mathcal H\partial_\tau \delta} - \frac{3}{2}\Omega_m \mathcal H^2 \delta\\ &=\color{red}{ (a \mathcal H)^2 \partial_a^2 \delta + a \mathcal H^2\partial_a \delta + a \mathcal H^2 \left[1 - \frac{3}{2}\Omega_m(a) \right]\partial_a\delta} + \color{blue} { a\mathcal H^2\partial_a \delta} - \frac{3}{2}\Omega_m \mathcal H^2 \delta\\ &= (a \mathcal H)^2 \partial_a^2 \delta -\frac{3}{2} a\mathcal H^2 \left[ \Omega_m(a) -2\right]\partial_a\delta -\frac{3}{2}\Omega_m \mathcal H^2\delta \end{align} $$

$$\boxed{- a^2 \mathcal H^2 \partial_a^2 \delta + \frac{3}{2} a\mathcal H^2 \left[ \Omega_m(a) -2\right]\partial_a\delta + \frac{3}{2}\Omega_m \mathcal H^2\delta = 0}$$

  • $\begingroup$ Thank you very much! I'm trying to slowly digest this. Just a quick question at first glance, can't we write $dH/d\tau$ instead of $\partial H/\partial\tau$, since $H=H(\tau)$ depends only on time? $\endgroup$ Commented Oct 11, 2023 at 12:58
  • $\begingroup$ Also, when you derive the Friedmann equation (1) with respect to time, you seem to be assuming that $\Omega_\Lambda$ is a constant and so its time derivative is zero, but I believe that $\Omega_\Lambda$ depends on $a$ and therefore is time-dependent. I think that in order to get what we need we have to assume a matter-dominated universe, so that $\Omega_\Lambda\simeq 0$ in equation (1). This seems to be an implicit hypothesis in the notes I'm reading. Am I right here? $\endgroup$ Commented Oct 11, 2023 at 16:40
  • $\begingroup$ No, $\Omega_{\Lambda,0}$ in Eq. (1) is truly a constant by definition (made a small edit to make this more clear). See the discussion around Eq. (1.3.120) in these lectures notes. I'm not sure what notes you're reading so I don't know if the author is considering a matter-dominated universe but this result is true in a universe where $\Omega_{\Lambda,0} \ne 0$ $\endgroup$
    – delon
    Commented Oct 11, 2023 at 17:18
  • $\begingroup$ Thank you again for this wonderful answer, it has helped me so much. Your knowledge on this is impressive. Enjoy the bounty, you truly deserve it! :) $\endgroup$ Commented Oct 12, 2023 at 13:15
  • 1
    $\begingroup$ And the Cosmology notes you linked there are great, thank you for that too! I didn't know they existed, but I have Baumann's book on Cosmology. It's one of the best books I've ever read and where I learned all I know about Cosmology. Have a great day and thanks again! $\endgroup$ Commented Oct 12, 2023 at 17:02

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