 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ia/paper-3/12b/b/solution

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

The measure, Gaussian factor,

r^{- 1}

, coordinates, and derivatives all transform covariantly under rotations, so

T_{ijk l}

β = γ

Contracting

i = j

and using

\nabla^{2} (1/ r) = 0

away from the origin gives

3 α + 2 β = 0.

Next contract

i = k

and

j = l

. Since

\sum_{i} x_{i} \partial_{i} \partial_{j} (r^{- 1}) = \frac{2 x _{j}}{r ^{3}},

one obtains

3 α + 12 β = 2 \int_{R^{3}} \frac{e ^{- r^{2}}}{r}, d V = 4 π .

Solving yields

α = - \frac{4 π}{15}, β = γ = \frac{2 π}{5} .

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