I'm trying to obtain $(1)$ from $(2)$.
$$\frac{dr}{d \ln a} = \frac{\gamma}{H} + r \frac{d \ln g_*}{d \ln a} \tag{1}$$
$$ -HT \left( \frac{\partial f_s}{\partial T} \right)_{E/T} = \frac{\sin^2 2\theta \hspace{1mm} \Gamma f_a}{2}\tag{2}$$
where:
- $f_a= \frac{1}{e^{E/T}+1}$
- $n_i= 2\int f_i \hspace{1mm} d^3p /(2\pi)^3$ and $r=n_s/n_a$
- $\gamma \hspace{1mm} n_a= \int \frac{d^3p}{(2\pi)^3} \frac{\sin^2 2\theta}{2}f_a$
- $H= \frac{d \ln a}{dt}$ is the Hubble parameter.
- $T=\frac{t^{-1/2}1.15}{k_B}$ , better time-temperature relations might exist that I am unaware of.
- $g_*a^3t^3= \text{constant}$
I'm told in the article that I must do this by "Changing the time variable from t to $a$,(...) and integrating Eq. (2) over momenta, one finds $(1)$. "
My attempt:
Substitute $T$ in terms of $t$ and integrate both sides of the eq in terms of momentum, $d^3 p$:
$$\frac{t^{-1/2} 1.15}{k_B} \frac{\partial f_s}{\partial T} = - \frac{1}{H} \frac{\sin^2 2\theta}{2} \gamma f_a \tag{3}$$
$$2 \int \frac{d^3 p}{(2\pi)^3} \frac{\partial f_s}{\partial T} \frac{t^{-1/2} 1.15}{k_B} = -\frac{1}{H} \int \frac{d^3 p}{(2\pi)^3} \sin^2 2\theta \Gamma f_a \tag{4}$$
Apply the definition of $n_i$:
$$\frac{\partial n_s}{\partial T} \frac{t^{-1/2} 1.15}{k_B} = -\frac{\gamma n_a}{H} \tag{5}$$
and move the $n_a$ to the left-side and use the definition of $r$:
$$ \frac{t^{-1/2} 1.15}{k_B} \frac{dr}{dT} = -\frac{\gamma}{H}\tag{6}$$
Rewrite $dT$ in terms of $dt$ which we use to get to $d \ln a$
$$dT= -\frac{1}{2} \frac{t^{-3/2}1.15}{k_B} dt \tag{7} $$
To have the denominator in terms of $\ln a$ I start by
$$dT = -\frac{1}{2} \frac{t^{-3/2}1.15}{k_B} \frac{d \ln a}{H} \tag{8} $$
Further simplify this to:
$$2tH \frac{dr}{d \ln a}= \frac{\gamma}{H}$$
This is where I get stuck, I believe I must use the equation above, which is derived from $g_* a^3 T^3 = \text{constant}$, and written in a form similar to the one below:
$$d \ln g_* + 3 d \ln a + 3 d \ln T = 0$$
I'm not entirely sure if this is the path I should be taking.
Could someone hint how to proceed from here, I've been trying to solve this for a long time.