Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 65 / 2 / ii / Solution

Let $m_{\chi }$ be the dark matter particle mass, so its number density is $n_{\chi }=\rho /m_{\chi }$. Interpret $ϵ$ as total emitted power per particle, and assume isotropic optically thin dark-matter decay radiation; volume emissivity is $j_d=ϵ\rho /m_{\chi }$. At projected radius $R$, a line-of-sight coordinate $z$ has $r^2=R^2+z^2$. Writing $A=V^2/(2\pi G)$, the surface brightness isHere $z=R\tan \theta$ turns the integral into $\pi /R$. For the isothermal outer halo,If emissivity was defined per unit solid angle, the factor $1/(4\pi )$ is already included in $ϵ$; this only changes the normalization. The radial law holds between any core and outer truncation. For example, a phenomenological cored profile $\rho =A/(r^2+r_c^2)$ gives $I_d=ϵA/[4m_{\chi }\sqrt{R^2+r_c^2}]$: finite central surface brightness, but the same $R^{-1}$ outer law. This cored interpolation is not being asserted to be the exact isotropic constant-dispersion equilibrium.