Time since hot big bang - relativistic species

[BOAS]

Homework Statement

According to the standard assumptions, there are three species of (massless) neutrinos. In the temperature range of 1MeV < T < 100MeV, the density of the universe is believed to have been dominated by the black-body radiation of photons, electron-positron pairs, and three neutrinos all of which were in thermal equilibrium.

1. Neglecting any change in the degrees of freedom at T > 100MeV, show using the Friedmann equation for a flat radiation-dominated universe $H^2 = \frac{8\pi G}{3} \rho_R$ that for temperatures T > 1MeV the time since the start of the hot big bang is given by $t(T) = (\frac{A}{g_∗})^{1/2} \frac{M_P}{T^2}$ where $M_P$ is the reduced Planck mass, and A is a constant that you should give explicitly.

Homework Equations

The Attempt at a Solution

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I have found an expression for $t(T)$ but have a factor of $ln(a)$ spoiling my constant $A$ and I don't know how I can get rid of it.

Starting with the Friedmann equation:

$H^2 = \frac{8\pi G}{3} \rho_R$

I know that the energy density for several relativistic species in thermal equilibrium is given by $\rho_R = \sum_i \rho_i = \frac{\pi^2}{30} g_* T^4_{\gamma}$, where $g_*$ is the effective degrees of freedom.

Substituting this into the Friedmann equation yields

$H^2 = \frac{8\pi G}{3} \frac{\pi^2}{30} g_* T^4_{\gamma}$

I take the square root and use the definition of the Hubble parameter

$\frac{da}{dt} \frac{1}{a} = \sqrt{\frac{8\pi G}{3} \frac{g_*}{30}} \pi T_{\gamma}^2$

Simply integrating this causes the following problem

$t = \frac{ln(a)}{\pi T^2_{\gamma}} \sqrt{\frac{3}{8 \pi G} \frac{30}{g_*}}$

which we can write as

$t = ln(a) \sqrt{\frac{90}{\hbar c g_* \pi^2}} \frac{M_P}{T^2_{\gamma}}$

However, I can't write that in the form I'm asked to, since ln(a) is not a constant.

I think I have made an error by integrating the equation in the way that I did, but I don't know how to go about this.

Thank you for any help you can give!

 

[BOAS]

I think I have a solution, but it depends on the photons obeying $T = T_0(1 + z)$

I say that $da = d(1+z)^{-1} = d(T_0 / T) = - T_0 T^{-2} dT$

then $da/a = -T^{-1} dT$

substituting this into my expression $\frac{da}{dt} \frac{1}{a} = \sqrt{\frac{8\pi G}{3} \frac{g_*}{30}} \pi T_{\gamma}^2$,

I find that $t(T) = \frac{1}{2} \sqrt{\frac{90}{\hbar c \pi^2 g_*}} \frac{M_P}{T^2}$