Define the acceleration of the normal congruence by
$$
\boxed{a_\sigma=n^\rho\nabla_\rho n_\sigma}.
$$
Because $S^{\sigma\gamma}$ is spatial, differentiating
$n_\sigma S^{\sigma\gamma}=0$ gives
$$
n_\sigma\nabla_\rho S^{\sigma\gamma}
=-(\nabla_\rho n_\sigma)S^{\sigma\gamma}.
$$
Insert
$$
\delta^\rho{}_\sigma
=\perp^\rho{}_\sigma-n^\rho n_\sigma
$$
into the contracted derivative and project the free index:
$$
\begin{aligned}
\perp^\gamma{}_\alpha\nabla_\mu S^\mu{}_\gamma
&=\perp^\gamma{}_\alpha
\perp^\rho{}_\sigma\nabla_\rho S^\sigma{}_\gamma
-\perp^\gamma{}_\alpha
n^\rho n_\sigma\nabla_\rho S^\sigma{}_\gamma\\
&=D_\mu S^\mu{}_\alpha+a_\sigma S^\sigma{}_\alpha.
\end{aligned}
$$
Rearranging proves
$$
\boxed{
D_\mu S^\mu{}_\alpha
=\perp^\gamma{}_\alpha\nabla_\mu S^\mu{}_\gamma
-a_\sigma S^\sigma{}_\alpha}.
$$


Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-357/2/b/solution