2
$\begingroup$

I'm an undergraduate student currently following a particle astrophysics lecture. I was reading some stuff about the neutrinos decoupling before the $e^+ e^-$ annihilations. It's possible to quickly get an estimation of the decoupling temperature of the neutrinos by using the relation between the interaction rate and the Hubble parameter :

\begin{align} & \Gamma_{\nu} \sim G_{F}^2 T^5 \\ & H(T) = \sqrt{\frac{8 \pi G}{3} g_* \frac{\pi^2}{30} T^4} \sim \sqrt{g_*} \frac{T^2}{m_{pl}} \end{align} where $g_*$ is the number of relavistic degrees of freedom.

Using these 2 equations, the decoupling temperature $T_{\nu D}$ is given by

\begin{align} T_{\nu D} = \left(\frac{\sqrt{g_*}}{G_F^2 m_{pl}}\right)^{1/3} \sim g_*^{1/6} MeV \end{align}

Since we are at the MeV range $g_* = 2 + \frac{7}{8} (4+6) =10.75$ and we see that $T_{\nu D}$ is of order MeV.

Now this is the tricky part for me. The temperature obtained here in the instantaneous decoupling approximation is unique. I read that in order to get the decoupling temperatures for $\nu_e$ and $\nu_{\tau,\mu}$, you need to solve the Boltzmann equations for neutrino spectra. Dealing with the collisions term is not easy but after different approximations (...) you can derive something like this (Dolgov 2002) :

\begin{align} H x \frac{\partial{f_{\nu_\alpha}} }{\partial{x}} = - \frac{80 G_F^2(g_L^2 + g_R^2) m^9}{3 \pi ^3 x^5} y f_{\nu_\alpha} \end{align} with the neutrino distribution : \begin{align} f_{\nu_{\alpha}} = \frac{1}{e^{y} +1} \end{align}

where $y=p a$, $x=m a$ and $m$ some mass scale (here taken 1 MeV).

$a$ is the scale factor of the Universe and it can be taken as $a \sim 1/T$.

$g_R = \sin ^2 \theta_w = 0.23$ and $g_L = \pm 1/2 + \sin ^2 \theta_w$ (+ for $\nu_e$)

Taking the average momentum value $y=3$ , one finds $T_{\nu_e D} = 1.87$ MeV and $T_{\nu_{\nu , \tau} D} = 3.12$ MeV.

My issue is that I can't manage to get these values. I have just derived the distribution and used the values of the Fermi constant and Planck mass in MeV. I think there's something missing and which is dealing with natural units...

PS : Sorry for my bad English

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.