# Compute the fermion number density

Consider fermion DM with g internal degrees of freedom and the statistical distribution results in:

$$f(E) = \frac{g}{ \exp[(E − µ)/T] + 1}$$

with $$g = 1$$ when $$E < µ$$ and $$g = 0$$ when $$E > µ$$

When I integrate $$f(E)$$ the phase-space, in the relativistic limit, distribution I get this:

$$n = \frac{gT^3}{2\pi^2} \int_0^{\infty} \frac{x^2}{(e^x +1)} dx =\frac{3 \zeta(3)gT^3}{4\pi^2}$$ (1)

But the answer I'm looking for is

$$n = \frac{g m^3 v^3}{6\pi^2}$$ (2)

where $$v$$ is the fermi velocity $$v = \sqrt{2E_f/m}$$

I don't see how to get from (1) to (2), what am I missing? I'm sure it must be something dumb..

• Yes, I mean zeta, thanks – Silvia_Arer Jun 30 at 5:37
• Have you computed $E_f$ and thus $v$ as a function of $T$? – G. Smith Jun 30 at 5:58
• I get $n = \frac{m^3 v^3}{3 \pi^2}$ when computing $\E_f$ and then computing $\v^3$ and then clear n. – Silvia_Arer Jun 30 at 6:16
• I think your condition for $g$ is only true at $T=0$. – ProfM Jun 30 at 7:00
• You seem to have forgotton $\mu$ in your integral. It will shift the limits of integration Also why is $g$ determined by $\mu$? This is not the usual case. – mike stone Jun 30 at 12:12