 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-2/6/b/iii/solution

\partial_{t} u^{B} = \frac{1}{2} \nabla \times \partial_{t} B = - \frac{1}{2} \nabla \times (\nabla \times E) = \frac{1}{2} \nabla^{2} E,

where the last equality uses

\nabla \cdot E = 0

. Combining with part (ii),

\partial_{t} (u_{i}^{E} + u_{i}^{B}) = \frac{1}{2} \nabla^{2} E_{i} - \frac{c ^{2}}{2} \nabla^{2} B_{i} .

The divergences of the symmetric gradients satisfy

\partial_{j} S_{ij}^{E} = \frac{1}{2} (\partial_{i} \partial_{j} E_{j} + \partial_{j} \partial_{j} E_{i}) = \frac{1}{2} \nabla^{2} E_{i}, \partial_{j} S_{ij}^{B} = \frac{1}{2} \nabla^{2} B_{i} .

Thus the local identity is

\partial_{t} (u_{i}^{E} + u_{i}^{B}) = \partial_{j} (S_{ij}^{E} - c^{2} S_{ij}^{B})

. Integrate over the fixed volume and apply the divergence theorem componentwise:

\frac{\partial}{\partial t} \int_{V} (u_{i}^{E} + u_{i}^{B}) d V = \oint_{A} (S_{ij}^{E} - c^{2} S_{ij}^{B}) d A_{j} .

Here

d A_{j} = n_{j} d A

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