 Solution

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

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

Put

Y = \nabla_{ρ} ϕ \nabla^{ρ} ϕ .

2 \nabla^{μ} T_{μν} = (\nabla^{μ} \nabla_{μ} ϕ) \nabla_{ν} ϕ + \nabla^{μ} ϕ \nabla_{μ} \nabla_{ν} ϕ + A \nabla_{ν} Y + 2 Bϕ \nabla_{ν} ϕ = m^{2} ϕ \nabla_{ν} ϕ + \frac{1}{2} \nabla_{ν} Y + A \nabla_{ν} Y + 2 Bϕ \nabla_{ν} ϕ = (A + \frac{1}{2}) \nabla_{ν} Y + (m^{2} + 2 B) ϕ \nabla_{ν} ϕ .

For this to vanish for arbitrary solutions and local field data, the two coefficients must vanish. Hence

A = - \frac{1}{2}, B = - \frac{m ^{2}}{2} .

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