Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 309 / 2 / b / i

Solution

T^{a}_{a} = F^{a c} F_{a c} - \frac{1}{4} δ_{a}^{a} F_{c d} F^{c d} = 0.

\nabla_{[a} F_{b c]} = 0

with

F^{a c}

gives

F^{a c} \nabla_{a} F_{b c} = \frac{1}{4} \nabla_{b} (F_{c d} F^{c d}),

which cancels the derivative of the trace term. Hence

\nabla_{a} T^{a}_{b} = 0, T^{a}_{a} = 0 .

For

J^{a} = T^{ab} V_{b}

T^{ab}

now imply

\nabla_{a} J^{a} = T^{ab} \nabla_{a} V_{b} = \frac{1}{2} T^{ab} (\nabla_{a} V_{b} + \nabla_{b} V_{a}) = \frac{1}{2} T^{ab} (L_{V} g)_{ab} .

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