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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/iii/paper-54/2/a/solution

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Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 54 / 2 / a / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-06

The magnetic field satisfies

\nabla \cdot B = 0

. Expanding the divergence of its Maxwell stress tensor gives

\partial_{j} M_{ij} = \frac{1}{μ _{0}} (B_{j} \partial_{j} B_{i} - \frac{1}{2} \partial_{i} B^{2}) = \frac{[( \nabla \times B ) \times B ] _{i}}{μ _{0}} .

For the Newtonian gravitational field

g = - \nablaΦ

, Poisson equation for Newtonian gravity gives

\partial_{j} g_{j} = - 4 π Gρ

, and the gradient representation gives

\partial_{j} g_{i} = \partial_{i} g_{j}

. Consequently

\partial_{j} (g_{i} g_{j} - \frac{1}{2} g^{2} δ_{ij}) = g_{i} \partial_{j} g_{j} + g_{j} (\partial_{j} g_{i} - \partial_{i} g_{j}) = - 4 π Gρ g_{i} .

The negative of the Newtonian gravitational stress tensor, in this force-stress convention, therefore supplies

ρ g_{i}

. Adding the pressure stress supplies

- \partial_{i} p

, so

ρ D_{t} u_{i} = \partial_{j} T_{ij}, T_{ij} = - p δ_{ij} - \frac{g _{i} g _{j} - g ^{2} δ _{ij} /2}{4 π G} + \frac{B _{i} B _{j} - B ^{2} δ _{ij} /2}{μ _{0}} .

Each summand is symmetric. Absence of external gravitational sources is needed to represent the full gravitational force by this self-gravitating stress.

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