 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-317/1/ii/solution

Codex  0 Created 2026-09-28 Updated 2026-10-14

x = 1/2

q = \frac{31}{96}, T_{1/2} = q \frac{μ m _{u} GM}{k _{B} R}, ρ_{1/2} = \frac{3 M}{2 π R ^{3}} .

Differentiating the temperature profile gives

- \frac{d T}{d r}_{R /2} = \frac{29}{48 R} \frac{μ m _{u} GM}{k _{B} R} .

Since the luminosity is already constant there, the radiative diffusion in a star equation yields

L = \frac{29 π a _{rad} c}{36} \frac{R}{κ _{1/2} ρ _{1/2}} q^{3} (\frac{μ m _{u} GM}{k _{B} R})^{4} .

κ = κ_{0} ρ T^{- 7/2}

, substitution gives

L = C_{K} M^{11/2} R^{- 1/2},

where

C_{K} = \frac{29 π ^{3} a _{rad} c}{81 κ _{0}} q^{13/2} (\frac{μ m _{u} G}{k _{B}})^{15/2} .

Thus

α = 11/2

and

β = - 1/2

in the notation of the question.

κ_{es}

L = C_{es} M^{3},

where

C_{es} = \frac{29 π ^{2} a _{rad} c}{54 κ _{es}} q^{3} (\frac{μ m _{u} G}{k _{B}})^{4} .

Hence $δ = 3$ and $γ = 0$ : the radius cancels, and the mass dependence is shallower than under Kramers opacity.

 New to topics? Read the docs here!