 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-3/14b/d/solution

 Top articles  Latest articles New article in topic

Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 3 / 14B / d / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-03

Since

R = a ξ_{1}

, the scaling in part (b) gives

ρ_{c}^{(1 - n) / n} = \frac{4 π G}{( n + 1 ) K} \frac{R ^{2}}{ξ _{1}^{2}} .

Substitute this into

M = 4 π (R / ξ_{1})^{3} ρ_{c} Y_{n}

. After collecting powers,

M = A_{n} K^{n / (n - 1)} R^{(3 - n) / (1 - n)},

where

A_{n} = 4 π Y_{n} (\frac{n + 1}{4 π G})^{n / (n - 1)} ξ_{1}^{(3 - n) / (n - 1)} .

This is the polytropic mass-radius relation.

An

n = 3

polytrope models a white dwarf whose electrons are an ultrarelativistic degenerate gas. For

n = 3

, the radius exponent vanishes and the mass is independent of both

R

and

ρ_{c}

. This fixed limiting value is the Chandrasekhar mass: increasing the central density can no longer support a larger mass, so a cold nonrotating white dwarf above the limit cannot remain in hydrostatic equilibrium.

 Read the full article

 New to topics? Read the docs here!