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ID: past-exam-of-the-formation-of-structure-in-the-universe-course-of-the-university-of-cambridge/2016/paper-1/2/a/solution

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Past exam of the formation of Structure in the Universe course of the University of Cambridge / 2016 / Paper 1 / 2 / a / Solution by

Codex  0 2026-10-20

For the binding energy of a truncated singular isothermal sphere, build the mass distribution from concentric shells. Its enclosed mass and mass element are

M (r) = \int_{0}^{r} 4 π s^{2} ρ (s) d s = \frac{2 k}{G} r, d M = \frac{2 k}{G} d r, M_{0} = \frac{2 k r _{0}}{G} .

(1)

The interaction of a newly added shell with the interior contributes

- GM (r) d M / r

. Counting each pair once,

W = - \int_{0}^{r_{0}} \frac{GM ( r )}{r} d M (r) = - \frac{4 k ^{2}}{G} \int_{0}^{r_{0}} d r = - \frac{4 k ^{2} r _{0}}{G} .

(2)

Eliminating

k

gives

W = - \frac{G M _{0}^{2}}{r _{0}} .

(3)

The central density is singular, but the mass and this energy integral are finite. The result assumes the stated truncation and no additional gravitating component.

The gravitational potential of a truncated singular isothermal sphere is a separate quantity from this shell-assembly energy allocation. With zero potential at infinity, both interior and exterior shells contribute:

Φ_{grav} (r) = - G [\frac{M ( r )}{r} + \int_{r}^{r_{0}} \frac{d M ( s )}{s}] = - 2 k [1 + ln \frac{r _{0}}{r}], 0 < r \leq r_{0} .

(4)

It matches

- G M_{0} / r_{0} = - 2 k

at the edge. The equivalent total energy is

W = \frac{1}{2} \int ρ Φ_{grav} d V

; the half prevents double counting. By contrast,

\int ρ [- Φ_{grav}] d V

counts the escape work of test parcels in a fixed potential and equals

- 2 W

. Confusing these quantities would change a local feedback threshold.

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