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.


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 $$ 3360a_{\text{eq}} = \frac{0.14}{\Omega_{\text{m}}h^{2}}. $$


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 $3360$?


1 Answer 1


The number 3360 is the redshift at which the densities of matter and radiation (all ultrarelativistic particles) are equal.

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? )


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