这是一份ucl伦敦大学学院 PHAS0072作业代写的成功案
$$
\tilde{m}{1}{} \equiv \frac{16 \pi^{5 / 2}}{2 \sqrt{5}} \frac{v^{2}}{M_{p}}=1.1 \times 10^{-3} \mathrm{eV} $$ The ratio $\tilde{m}{1} / m{}$ enters into the efficiency factor $\kappa$, and for $\tilde{m}{1}>m{*}$,
Another interesting connection to the neutrino masses comes through the $\Delta L=$ 2 scattering diagram, whose rate does relate to the actual light masses,
$$
\Gamma_{\Delta L=2}=\frac{T^{3}}{\pi^{2} v^{4}} \sum_{i=e, \mu, \tau} m_{v i}^{2} \equiv \frac{T^{3}}{\pi^{2} v^{4}} \bar{m}^{2}
$$
PHAS0072 COURSE NOTES ：
In fact one can combine these two equations to get a single second-order differential equation for the photon energy density perturbations $\Delta_{00}^{(1)}$. Deriving Eq. (8.1) with respect to conformal time and using Eq. (8.5) to replace $\partial_{i} v_{\gamma}^{(1) i}$ yields
$$
\begin{aligned}
\left(\Delta_{00}^{(1) \prime}-4 \Psi^{(1) m}\right)+\mathcal{H} \frac{R}{1+R}\left(\Delta_{00}^{(1) r}-4 \Psi^{(1) r}\right) \
-c_{s}^{2} \nabla^{2}\left(\Delta_{00}^{(1)}-4 \Psi^{(1)}\right) &=\frac{4}{3} \nabla^{2}\left(\Phi^{(1)}+\frac{\Psi^{(1)}}{1+R}\right)
\end{aligned}
$$