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## Homework Statement

The result

[tex]n_{0 \gamma} = \left( \frac{k_BT_{0r}}{hc} \right)^3 \int_0^{\infty} \frac{8 \pi x^2 dx}{e^x-1} = 2 \frac{\zeta(3)}{\pi^2} \left( \frac{k_BT_{0r}}{hc} \right)^3[/tex]

is obtained for photons by integrating over the Planck distribution appropriate for bosons. In the case of neutrinos (or other fermions), show that the number-density in thermal equilibrium at a temperature [itex]T_{0 \nu}[/itex] is

[tex]n_{0 \nu} = 3 \frac{\zeta(3)}{2 \pi^2} \left( \frac{k_BT_{0r}}{hc} \right)^3[/tex]

## Homework Equations

Bose-Einstein distribution:

[tex]f_{BE}(E) = \frac{1}{e^{E/kT}-1}[/tex]

Fermi-Dirac distribution:

[tex]f_{FD}(E) = \frac{1}{e^{(E-E_F)/kT}+1}[/tex]

Density of states for bosons:

[tex]g_{BE} = \frac{8 \pi VE^2}{c^3h^3}[/tex]

Density of states for fermions:

[tex]g_{FD}(E) = \frac{4 \pi (2m)^{3/2}}{h^3} VE^{1/2}[/tex]

## The Attempt at a Solution

First of all, I guess that [itex]T_{0r}[/itex] in the second formula should be [itex]T_{0 \nu}[/itex].

I guess that in the first case, for bosons, the Bose-Einstein distribution is used together with the density of states for bosons to give the correct formula by also using [itex]n_{BE}(E) = g_{BE}(E)f_{BE}(E)[/itex].

But now, in the case of fermions, I have tried to use the Fermi-Dirac distribution together with the density of states for fermions, exactly the same procedure as I used for bosons, but this gives me

[tex]n_{0 \nu} = \frac{4 \pi (2m)^{3/2}}{h^3} \int_0^{\infty} \frac{(xkT+E_F)^{1/2} kT dx}{e^x + 1}[/tex]

so this cannot be the correct procedure?

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