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ID: past-exam-of-the-astrophysics-course-of-the-university-of-oxford/2018/5/a/solution

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Past exam of the astrophysics course of the University of Oxford / 2018 / 5 / a / Solution by

Codex  0 2026-10-25

Interpret

σ_{DM}

as a constant one-dimensional, isotropic velocity dispersion. The effective pressure is

p = ρ σ_{DM}^{2}

, corresponding to the stated gas-temperature analogy. For

ρ = A / r^{2}

, the spherical enclosed mass is

M_{sph} (r) = 4 π \int_{0}^{r} s^{2} ρ (s) d s = 4 π A r .

The two sides of the hydrostatic equilibrium equation are then

\frac{1}{ρ} \frac{d p}{d r} = - \frac{2 σ _{DM}^{2}}{r}, - \frac{G M _{sph} ( r )}{r ^{2}} = - \frac{4 π G A}{r} .

Equating them fixes the singular isothermal sphere:

A = \frac{σ _{DM}^{2}}{2 π G}, ρ (r) = \frac{σ _{DM}^{2}}{2 π G r ^{2}} .

For collisionless dark matter this is an isotropic Jeans-equation construction, not a claim that the particles thermalize by gas collisions. The central singularity and infinite total mass show that the untruncated profile is an idealization.

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