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

Spatially project <stress-energy conservation>,
$$
\perp^\nu{}_\alpha\nabla_\mu T^\mu{}_\nu=0,
$$
and substitute the decomposition from part a. The $\rho n^\mu n_\nu$ term contributes $\rho a_\alpha$. The two momentum terms combine into
$$
\mathcal L_nj_\alpha-Kj_\alpha,
$$
because the two contractions with the full extrinsic curvature cancel and
$$
\nabla_\mu n^\mu=-K.
$$
Finally, part b gives
$$
\perp^\nu{}_\alpha\nabla_\mu S^\mu{}_\nu
=D_\mu S^\mu{}_\alpha+S^\mu{}_\alpha a_\mu.
$$
Thus
$$
\rho a_\alpha+\mathcal L_nj_\alpha-Kj_\alpha
+D_\mu S^\mu{}_\alpha+S^\mu{}_\alpha a_\mu=0.
$$
The <momentum equation in a 3+1 decomposition> is
$$
\boxed{
\mathcal L_nj_\alpha
=-D_\mu S^\mu{}_\alpha
-S^\mu{}_\alpha a_\mu
+Kj_\alpha-\rho a_\alpha}.
$$
Therefore
$$
\boxed{c_1=-1,\qquad c_2=-1,\qquad c_3=1,\qquad c_4=-1}.
$$

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 357 / 2 / d / Solution

Local energy-momentum conservation is the covariant equation $\nabla_\mu T^{\mu\nu}=0$. Its normal and spatial projections give energy and momentum evolution equations in a spacetime foliation.


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