The trace of the <electromagnetic stress-energy tensor> in four <spacetime dimensions> is
$$
T^a{}_a=F^{ac}F_{ac}-\frac14\delta^a_aF_{cd}F^{cd}=0.
$$
For its <covariant divergence>, the source-free <Maxwell equations> eliminate the derivative of the first factor. Contracting the <Bianchi identity> $\nabla_{[a}F_{bc]}=0$ with $F^{ac}$ gives
$$
F^{ac}\nabla_aF_{bc}=\frac14\nabla_b(F_{cd}F^{cd}),
$$
which cancels the derivative of the trace term. Hence
$$
\boxed{\nabla_aT^a{}_b=0,
\qquad T^a{}_a=0}.
$$
For $J^a=T^{ab}V_b$, <stress-energy conservation> and symmetry of $T^{ab}$ now imply
$$
\nabla_aJ^a=T^{ab}\nabla_aV_b
=\frac12T^{ab}(\nabla_aV_b+\nabla_bV_a)
=\boxed{\frac12T^{ab}(\mathcal L_Vg)_{ab}}.
$$

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-309/2/b/i/solution