 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-3/36d/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Let

σ

be the stated Maxwell stress tensor. Taking its divergence and using

(A \cdot \nabla) A - \frac{1}{2} \nabla∣ A ∣^{2} = - A \times (\nabla \times A)

gives

\partial_{j} σ_{ij} = ϵ_{0} [E \times (\nabla \times E)]_{i} - ϵ_{0} E_{i} \nabla \cdot E + \frac{1}{μ _{0}} [B \times (\nabla \times B)]_{i} - \frac{1}{μ _{0}} B_{i} \nabla \cdot B .

The Maxwell equations in vacuum with sources are

\nabla \cdot E = \frac{ρ}{ϵ _{0}}, \nabla \cdot B = 0, \nabla \times E = - \partial_{t} B, \nabla \times B = μ_{0} J + μ_{0} ϵ_{0} \partial_{t} E .

Substitution, together with antisymmetry of the cross product, yields

\partial_{j} σ_{ij} = - ρ E_{i} - (J \times B)_{i} - ϵ_{0} [E \times \partial_{t} B + \partial_{t} E \times B]_{i} = - [ρ E + J \times B]_{i} - \partial_{t} [ϵ_{0} E \times B]_{i} .

g = ϵ_{0} E \times B = \frac{S}{c ^{2}},

and

j = 1 \sum 3 \frac{\partial σ _{ij}}{\partial x _{j}} + \frac{\partial g _{i}}{\partial t} = - (ρ E + J \times B)_{i} .

σ_{ij}

is the flux of

i

-momentum in the

j

direction in the convention of the question,

g

is field momentum per unit volume, and

ρ E + J \times B

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