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In a context of cosmology, I need help about a differential equation that I can't get to demonstrate:

The growth of density fluctuations obeys a second order differential equation. At early enough times, when those fluctuations are still small, the fluid equations can be linearised. During matter domination, considering matter as a pressureless ideal fluid, the equation for the evolution of the density contrast becomes:

$$\ddot{\delta}_{\mathrm{m}}(\boldsymbol{k}, z)+2 H \dot{\delta}_{\mathrm{m}}(\boldsymbol{k}, z)-\frac{3 H_{0}^{2} \Omega_{\mathrm{m}, 0}}{2 a^{3}} \delta_{\mathrm{m}}(\boldsymbol{k}, z)=0\quad\quad(1)$$

In the LCDM scenario with no massive neutrinos, this equation can be written in terms of the redshift as:

$$\delta_{\mathrm{m}}^{\prime \prime}(\boldsymbol{k}, z)+\left[\frac{H^{\prime}(z)}{H(z)}-\frac{1}{1+z}\right] \delta_{\mathrm{m}}^{\prime}(\boldsymbol{k}, z)-\frac{3}{2} \frac{\Omega_{\mathrm{m}}(z)}{(1+z)^{2}} \delta_{\mathrm{m}}(\boldsymbol{k}, z)=0\quad\quad(2)$$

It is this second equation that I can't to find, especially the second term:

$$\left[\frac{H^{\prime}(z)}{H(z)}-\frac{1}{1+z}\right] \delta_{\mathrm{m}}^{\prime}(\boldsymbol{k}, z)$$

where prime refers to the derivative with respect to $z$:

For this instant, If I express the first term as a function of redshift $z$ and not time, I find:

$$\ddot{\delta}_{\mathrm{m}}(\boldsymbol{k}, z) = \delta_{\mathrm{m}}^{\prime \prime}(\boldsymbol{k}, z)\,\,H(z)^2\,(1+z)^2$$

So to get the first term of equation $(2)$, i.e $$\delta_{\mathrm{m}}^{\prime \prime}(\boldsymbol{k}, z)$$, I have to divide $\ddot{\delta}_{\mathrm{m}}(\boldsymbol{k}, z)$ by $H(z)^2\,(1+z)^2$

This way, I find easily the third term of (eq 2), i.e:

$$-\frac{3}{2} \frac{\Omega_{\mathrm{m}}(z)}{(1+z)^{2}} \delta_{\mathrm{m}}(\boldsymbol{k}, z)$$

with $$\Omega_{\mathrm{m}}(z)=\frac{\Omega_{\mathrm{m}, 0}(1+z)^{3}}{E^{2}(z)}$$

and $H(z) = H_{O}\,E(z)$

My main issue is that second term of (eq 2) makes appear second derivatives on scale factors $a(t)$ and I don't where it could come from.

Any help is welcome, I am frustrated since I am not far from the demonstration.

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    $\begingroup$ In my previous comment I pointed out a mistake in the way you write the 2nd time derivative making a mistake myself. Here's the 1st time derivative: $\frac{d}{dt}=(1+z)H(z)\frac{d}{dz}$, and here the 2nd time derivative:$\frac{d^2}{dt^2}=(1+z)^3H^2(z)\left[\left(\frac{1}{(1+z)^2}+\frac{H'(z)}{H(z)}\right)\frac{d}{dz}+\frac{d^2}{dz^2}\right]$. $\endgroup$ – anonymous Oct 18 at 0:50
  • $\begingroup$ @anonymous thanks a lot ! $\endgroup$ – youpilat13 Oct 18 at 8:51

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