Wednesday, April 29, 2015

neutrinos everywhere!

please estimate how many cosmic neutrinos are in the observable Universe. are there more or less photons than neutrinos?

3 comments:

  1. both photons and neutrinos are ultrarelativistic particles. integrating over the respective distributions gives for the density of photons $n_\gamma$
    \begin{equation}
    n_\gamma = g_\gamma\frac{\zeta(3)}{\pi^2}\left(\frac{kT}{\hbar c}\right)^3
    \end{equation}
    and for the density of neutrinos $n_\nu$
    \begin{equation}
    n_\mu = \frac{3}{4}\frac{g_\nu}{g_\gamma} n_\gamma
    \end{equation}
    then, the ratio between the densities is given by
    \begin{equation}
    \frac{n_\nu}{n_\gamma} = \frac{3}{4}\frac{g_\nu}{g_\gamma}\left(\frac{T_\nu}{T_\gamma}\right)^3
    \end{equation}
    which can be evaluated using the relation of the two background temperatures
    \begin{equation}
    T_\nu = \left(\frac{4}{11}\right)^{1/3} T_\gamma
    \end{equation}
    due to the fact that the photon background has a higher temperature because of $e^+e^-$-annihilation, and the ratio of the statistical weights $g_\nu/g_\gamma=2\times 3 / 2$ for two spin states of both the photon and the neutrino, as 3 neutrino families. putting everything together yields
    \begin{equation}
    \frac{n_\nu}{n_\gamma} = \frac{3}{4}\frac{g_\nu}{g_\gamma}\frac{4}{11} = \frac{9}{11}<1
    \end{equation}
    i.e. about 20% fewer neutrinos than photons: the smaller temperature overcompensates the larger number of neutrino families.

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  2. the total number of neutrinos could be estimated like this: the number density $n_\nu$ is roughly $10^8$ per cubic metre, and the volume $4\pi/3\chi_H^3$ contains then $10^{83}$ neutrinos in total.

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  3. and $10^{83}$ is very close again to Eddington's number (the number of protons in the Universe), which is $10^{80}$!

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