# Clarify derivation: Relating matter density parameter $\Omega_{\text{m}}$ to the scale factor $a$ at which $\rho_{\text{matter}} = \rho_{\text{rad}}$

The derivation and background here isn't really important. The reader should skip to the bottom of this post for the actual question before reading the background.

## Background

In Modern Cosmology by Dodelson and Schmidt 2nd edition, we are trying to find an equation for the Compton scattering rate $$n_{e}\sigma_{T}$$ divided by the Hubble rate $$H$$. Equation (4.43) reads $$\frac{n_{e}\sigma_{T}}{H} = 123X_{e}\left(\frac{\Omega_{\text{b}}h^{2}}{0.022}\right)\left(\frac{0.14}{\Omega_{\text{m}}h^{2}}\right)^{1/2}\left(\frac{1+z}{1000}\right)^{3/2}\left[1 + \frac{1+z}{3360}\frac{0.14}{\Omega_{\text{m}}h^{2}}\right]^{-1/2}.$$

To derive this, let us take equation (4.42) for granted as a starting point: $$\frac{n_{e}\sigma_{T}}{H} = 0.0692 \,a^{-3}X_{e}\Omega_{\text{b}}h\frac{H_{0}}{H}.$$

By the first Friedmann equation, we have $$\frac{H}{H_{0}} = \frac{\left(\frac{8\pi G}{3} \rho\right)^{1/2}}{H_{0}} = \left( \frac{8\pi G}{3H_{0}^{2}} \rho \right)^{1/2} = \left( \frac{\rho}{\rho_{\text{cr}}} \right)^{1/2}.$$

Now assuming radiation and matter only, we have

\begin{align*} \frac{\rho}{\rho_{\text{cr}}} &= \frac{\rho_{\text{r}} + \rho_{\text{m}}}{\rho_{\text{cr}}} \\ &= \Omega_{\text{r}}a^{-4} + \Omega_{\text{m}}a^{-3} \\ &= \Omega_{\text{m}} a^{-3} [1 + \frac{\Omega_{\text{r}}}{\Omega_{\text{m}}}a^{-1}] \\ &= \Omega_{\text{m}} a^{-3} [1 + a_{\text{eq}}/a] \end{align*} where $$a_{\text{eq}}$$ is the scale factor at which the density of radiation matches the density of matter. This gives $$\frac{H}{H_{0}} = \Omega_{\text{m}}^{1/2}a^{-3/2}[1 + a_{\text{eq}}/a]^{1/2}.$$

Now we put this back into (4.42) all the way above, and we have the following:

\begin{align*} \frac{n_{e} \sigma_{T}}{H} &= 0.0692 \,a^{-3} X_{e} \Omega_{\text{b}} h \frac{H_{0}}{H} \\[1.6ex] &= 0.0692 X_{e} \cdot a^{-3}\Omega_{\text{b}}h \cdot \Omega_{\text{m}}^{-1/2}a^{3/2}\left[1+a_{\text{eq}} / a\right]^{-1/2} \\[1.6ex] &= 0.0692 X_{e} \cdot \Omega_{\text{b}} h \cdot \Omega_{m}^{-1/2} \cdot a^{-3/2}\left[1+a_{\text{eq}} / a\right]^{-1/2} \\[1.6ex] &= 0.0692 X_{e} \cdot \Omega_{\text{b}} h^{2} \cdot\left(\Omega_{\text{m}} h^{2}\right)^{-1/2} a^{-3/2}\left[1+a_{\text{eq}} / a\right]^{-1/2} \\[1.6ex] &= 0.00407 X_{e} \cdot\left(\frac{\Omega_{b} h^{2}}{0.022}\right) \cdot\left(\frac{0.14}{\Omega_{m} h^{2}}\right)^{1/2} a^{-3/2}\left[1+a_{\text{eq}} / a\right]^{-1/2} \\[1.6ex] &= 0.00407 X_{e} \left(\frac{\Omega_{b} h^{2}}{0.022}\right) \cdot\left(\frac{0.14}{\Omega_{m} h^{2}}\right)^{1/2} \left(1+z\right)^{3/2}\left[1+(1+z) a_{\text{eq}}\right]^{-1/2} \\[1.6ex] &= 128.7 X_{e}\left(\frac{\Omega_{b} h^{2}}{0.022}\right) \cdot\left(\frac{0.14}{\Omega_{m} h^{2}}\right)^{1/2} \left(\frac{1+z}{1000}\right)^{3/2} \left[1+\frac{1+z}{3360} \cdot 3360 a_{\text{eq}}\right]^{-1/2}. \end{align*}

The prefactor number $$128.7$$ seems to be off from the $$123$$ prefactor in the book, but that doesn't concern me. In order to match the equation in the book, I need to take $$3360a_{\text{eq}} = \frac{0.14}{\Omega_{\text{m}}h^{2}}.$$

## Question

Is there a reason why $$3360a_{\text{eq}} = \frac{0.14}{\Omega_{\text{m}}h^{2}}?$$ How is this derived? What info do you need to obtain this? Is there some typo or is there a significance to the number $$3360$$?

The matter density scales as $$a^{-3}$$ whilst the radiation density scales as $$a^{-4}$$.
At the present epoch Dodelson uses $$\Omega_m h^2 = 0.14$$.
The radiation and matter densities are equal when $$a_{\rm eq} = \frac{\Omega_r h^2}{\Omega_m h^2}\ .$$ If we multiply both sides by 0.14, then $$\frac{0.14a_{\rm eq}}{\Omega_r h^2} = \frac{0.14}{\Omega_mh^2}\ .$$
If the coefficient of $$a_{\rm eq}$$ on the LHS is 3360, this is the equivalent of adopting $$\Omega_r h^2 = 4.166\times 10^{-5}$$. This is indeed very close to the currently estimated density parameter for photons AND neutrinos (see Where is radiation density in the Planck 2013 results? )