At $T_\nu/a\ll m_\nu$, the <Taylor expansion> of the relativistic energy is $\sqrt{p^2+m_\nu^2}=m_\nu+p^2/(2m_\nu)+O(p^4/m_\nu^3)$. The number density is
$$
n_\nu=\frac1{\pi^2}\left(\frac{T_\nu}{a}\right)^3
\int_0^\infty\frac{x^2}{e^x+1}\,dx
=\frac{3\zeta(3)}{2\pi^2}\left(\frac{T_\nu}{a}\right)^3.
$$
The ratio of the next momentum moment to this one is
$$
\frac{\int_0^\infty x^4/(e^x+1)\,dx}
{\int_0^\infty x^2/(e^x+1)\,dx}
=\frac{15\zeta(5)}{\zeta(3)}.
$$
Therefore
$$
\boxed{\rho_\nu\simeq n_\nu m_\nu
\left[1+\frac{15\zeta(5)}{2\zeta(3)}
\left(\frac{T_\nu}{am_\nu}\right)^2\right]},
$$
so $\boxed{F=15\zeta(5)/(2\zeta(3))\simeq6.47}$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 310 / 3 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-310/3/c/solution