 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-1/38b/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

D

R_{μν} - \frac{1}{2} g_{μν} R = κ T_{μν} .

Taking the trace gives

(1 - \frac{D}{2}) R = κ T, R = - \frac{2 κ}{D - 2} T .

R_{μν} = κ (T_{μν} - \frac{1}{D - 2} g_{μν} T) .

For the values from part (a), let

Y = \nabla_{ρ} ϕ \nabla^{ρ} ϕ

. Then

T_{μν} = \frac{1}{2} \nabla_{μ} ϕ \nabla_{ν} ϕ - \frac{1}{4} g_{μν} (Y + m^{2} ϕ^{2}),

and

T = \frac{2 - D}{4} Y - \frac{D}{4} m^{2} ϕ^{2} .

Substitution cancels the

Y g_{μν}

terms and yields

R_{μν} = \frac{κ}{2} [\nabla_{μ} ϕ \nabla_{ν} ϕ + \frac{m ^{2}}{D - 2} g_{μν} ϕ^{2}] .

For the convention

κ = 8 π G

, the prefactor is

4 π G

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