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In Frank van den Bosch's Theory of Galaxy Formation Lecture 6 Page 41 he says that

The stagnation of growth in pressureless matter perturbations during radiation dominated era is known as the Meszaros effect

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The Meszaros effect is simply a manifestation of the fact that the Hubble drag term during the radiation dominated era is larger than during the matter dominated era.

Where the Hubble drag term is the red term in the linearized fluid equations $$\frac{\partial^2\delta}{\partial t^2} +{\color{red}{ 2 H \frac{\partial \delta}{\partial t}}}=\left(4\pi G\bar\rho + \frac{c_s^2}{a^2}\nabla^2\right)\delta$$ I can see how the perturbations stagnate during the radiation dominated era from the following qualitative argument in the slides and the calculation in the previous pages. However, I am confused by the second statement in the quote relating the Meszaros effect with the Hubble drag. For a single component universe flat universe we can solve the Friedmann equation to get the time dependence of the scale factor as (see Eq. (1.3.123) of Daniel Baumann's notes) $$a(t) \propto \begin{cases}t^{2/3}&\textrm{Matter Dominated}\\ t^{1/2} &\textrm{Radiation Dominated}\end{cases}$$ This then tells us $$H_{\rm RD}\propto H_{\rm MD} \propto t^{-1}$$

So shouldn't the Hubble drag terms in the radiation dominated era and matter dominated era be comparable?

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The Meszaros effect is simply a manifestation of the fact that the Hubble drag term during the radiation dominated era is larger than during the matter dominated era.

I think that way of putting things is misleading at best. For clarity, here's the growth equation again, now with more colors. I'm also assuming $c_s=0$, e.g. we are thinking about either baryons far above the Jeans scale or dark matter. $${\color{green}{ \frac{\partial^2\delta}{\partial t^2}}} +{\color{red}{ 2 H \frac{\partial \delta}{\partial t}}}=\color{blue}{4\pi G\bar\rho\delta}$$

The reason the growth of perturbations $\delta$ stagnates during radiation domination is that the green and red terms both dominate over the blue term. The green and red terms are of order $H^2\sim G\rho_r$, where $\rho_r$ is the radiation density, while the blue term is of order $G\bar\rho$, where $\bar\rho$ is the subdominant matter density.

As you note, in comparison to the green term, the red "Hubble drag" term does not become larger during radiation domination. In fact it becomes smaller, because $H=2/(3t)$ during matter domination and $H=1/(2t)$ during radiation domination. That's a noteworthy observation in itself because it is the reason perturbations can grow to any significant degree during radiation domination. For example, we can rewrite the equation above as $${\color{green}{ \frac{\partial^2\delta}{\partial t^2}}} +{\color{red}{ \frac{\alpha}{t} \frac{\partial \delta}{\partial t}}}=\frac{1}{t^\alpha}\frac{\partial}{\partial t}\left(t^\alpha\frac{\partial \delta}{\partial t}\right)=\color{blue}{4\pi G\bar\rho\delta},$$ where $\alpha=1$ for radiation domination and $\alpha=4/3$ for matter domination. If the blue term is again negligible, then $$\frac{\partial \delta}{\partial t}\propto t^{-\alpha}.$$ Radiation domination implies $\delta\propto\log t+\text{const}$, while matter domination implies $\delta\propto t^{-1/3}+\text{const}$. So if peculiar gravitational forces (blue term) are absent in both cases,$^*$ perturbations actually grow faster during radiation domination than during matter domination, due to the different Hubble drag terms!

That's why I find it misleading to attribute the stagnation of perturbation growth to differences in Hubble drag. Growth just stagnates because the peculiar gravity term is negligible. Reduced Hubble drag during the radiation epoch is the reason perturbation growth continues at all.


$^*$ Note that absence of peculiar forces during matter domination isn't completely unphysical -- it's just not realized in our universe as far as we know. For example, baryons don't cluster below the Jeans scale. In a hypothetical universe where dark matter is subdominant to baryons, growth of dark matter perturbations at small scales would be described in this way.

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  • $\begingroup$ Thanks for the answer! Quick clarifying question: why can we say that the red and green term are of order $H^2$? $\endgroup$
    – delon
    Commented Mar 31, 2023 at 1:48
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    $\begingroup$ @delon $\mathrm{d}\delta/\mathrm{d}t\sim \delta/t\sim H\delta$ (you can contrive functions for which the first '$\sim$' doesn't hold, but for power laws or logarithms it does). $\endgroup$
    – Sten
    Commented Mar 31, 2023 at 6:48

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