Solutions for nonrelativistic-matter perturbations

[happyparticle]

Homework Statement

Solutions for nonrelativistic-matter perturbations

Relevant Equations

##\delta'' + \frac{2+3y}{2y(1+y)}\delta' - \frac{3}{2y(1+y)}\delta = 0##

##y=\frac{a}{a_{eq}}##

I'm studying the nonrelativistic-matter perturbations if the expansion of the Universe is driven by a combination of components.

I'm currently Following this document (The growth of density perturbations) from Caltech. However, the author doesn't explain how he has found the solutions for the following expression that can be found in page 5.

##\delta'' + \frac{2+3y}{2y(1+y)}\delta' - \frac{3}{2y(1+y)}\delta = 0##

##y=\frac{a}{a_{eq}}##

Moreover, I found those notes (University of Cambridge Part II Mathematical Tripos) from David Tong at Cambridge explaining the transfer function (p.146).

Thus, I'm wondering if there is a relationship between the initial conditions and the transfer function. However, the first document doesn't mention the initial conditions for the solutions.

My thoughts are that we can find 2 independents solutions, one for small wavelengths and one for long wavelength as explained by Tong. Thus, #T(k) \approx C# for long wavelengths and #T(K) \approx C ln(a?)# for short for small wavelengths.

Furthermore, if the transfer function is defined as follow.

##\delta(\vec{k},t_0) = T(k)\delta(\vec{k},t_i)##

Does it means that #T(k)# acts as the constant in the particular solution?

I might be totally wrong. There is a lot of guesses here, since I'm unsure to understand.

 

[ergospherical]

It's the Meszaros equation for the growth of matter perturbations. When you combine the continuity and Euler equations for pressureless perturbations you get$$\delta_{m}'' + \mathcal{H} \delta_m' - 4\pi G a^2 \bar{\rho}_m \delta_m = 0$$where ##\mathcal{H} \equiv aH## (in cosmological perturbation theory we generally work with this so-called conformal Hubble parameter). The Friedmann equation is$$\mathcal{H}^2 = \frac{8\pi G}{3}a^2(\rho_m + \rho_r)$$You have already come across the parameter ##y = a/a_{eq}##, where ##a_{eq}## is the scale-factor at matter-radiation equality. Because ##\rho_m = \rho_{eq} y^{-3}## and ##\rho_r = \rho_{eq} y^{-4}## (remembering the scalings), check that you can in fact write the Friedmann equation in the form$$\mathcal{H}^2 = \mathcal{H}_0^2 \frac{\Omega_{m0}^2}{\Omega_{r0}}(y^{-1} + y^{-2})$$where

##\Omega_{m0} = \frac{8\pi G}{3\mathcal{H}_0^2} \rho_{m0} = \frac{8\pi G}{3\mathcal{H}_0^2}\rho_{eq} a_{eq}^3##
and
##\Omega_{r0} = \frac{8\pi G}{3\mathcal{H}_0^2} \rho_{r0} = \frac{8\pi G}{3\mathcal{H}_0^2} \rho_{eq} a_{eq}^4##

as per usual. Now you are almost ready to calculate ##\delta_m'## and ##\delta_m''##. Remember that the prime indicates derivation with respect to conformal time ##d\eta = dt/a##, so you can check that$$\frac{d}{d\eta} = y \mathcal{H} \frac{d}{dy}$$OK, so now you can use this to figure out ##\delta_m'## and ##\delta_m''##, by making use of the expression above for ##\mathcal{H}## in terms of ##y##. Be warned that you will need to do some slightly tedious simplification.

 

[PeterDonis]

@happyparticle is this a homework problem?

 

[happyparticle]

@happyparticle is this a homework problem?

Not exactly. However, it's a question about a specific example from the document.
I guess it is like a homework question.

@ergospherical Thank you again.
Your explanation helped me.