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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/iii/paper-53/1/a/solution

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Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 53 / 1 / a / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-07

Assume

K > 0

and a positive central mass density. For this polytrope of index one, the spherical hydrostatic pressure support equation becomes

2 K ρ \frac{d ρ}{d r} = - \frac{G m ρ}{r ^{2}}, r^{2} \frac{d ρ}{d r} = - \frac{G m}{2 K} .

Differentiate the second equation and use mass conservation. With

k^{2} = 2 π G / K

, this gives

\frac{1}{r ^{2}} \frac{d}{d r} (r^{2} \frac{d ρ}{d r}) + k^{2} ρ = 0.

The regular central solution, with

ρ (0) = ρ_{c}

and

ρ^{'} (0) = 0

, is

ρ (r) = ρ_{c} \frac{s i n k r}{k r} .

This is also the index-one solution of the Lane-Emden equation. The mass density remains positive up to its first zero, so the free surface is at

k R = π

R = \frac{π}{k} = (\frac{π K}{2 G})^{1/2} .

Taking a later zero would include a region of negative mass density and would not describe a physical star.

Integrating the mass gives

M = \frac{4 π ρ _{c}}{k ^{3}} \int_{0}^{π} ξ sin ξ d ξ = \frac{4 π ^{2} ρ _{c}}{k ^{3}} .

Consequently

\frac{ρ}{ρ _{c}} = \frac{3 M}{4 π R ^{3} ρ _{c}} = \frac{3}{π ^{2}} .

The radius is independent of the central mass density, whereas the mass is proportional to it; this is the special polytropic mass-radius relation at index one.

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