Let
$$
\perp^\mu{}_\alpha=\delta^\mu{}_\alpha+n^\mu n_\alpha,
\qquad n^\mu n_\mu=-1.
$$
The identity decomposes as
$$
\delta^\mu{}_\alpha=\perp^\mu{}_\alpha-n^\mu n_\alpha.
$$
Applying this decomposition to both indices of $T_{\mu\nu}$ gives
$$
T_{\alpha\beta}
=(\perp^\mu{}_\alpha-n^\mu n_\alpha)
(\perp^\nu{}_\beta-n^\nu n_\beta)T_{\mu\nu}.
$$
The four terms are, from the definitions, $S_{\alpha\beta}$, $j_\alpha n_\beta$, $n_\alpha j_\beta$, and $\rho n_\alpha n_\beta$. Hence the <3+1 decomposition of the stress-energy tensor> is
$$
\boxed{
T_{\alpha\beta}
=\rho n_\alpha n_\beta+j_\alpha n_\beta
+n_\alpha j_\beta+S_{\alpha\beta}}.
$$
It also makes explicit that
$$
n^\alpha j_\alpha=0,\qquad
n^\alpha S_{\alpha\beta}=n^\beta S_{\alpha\beta}=0.
$$

Solution

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

Relative to a future unit normal $n^\mu$, the stress-energy tensor decomposes as
$$
T_{\mu\nu}=\rho n_\mu n_\nu+j_\mu n_\nu+n_\mu j_\nu+S_{\mu\nu},
$$
where $j_\mu$ and both indices of $S_{\mu\nu}$ are spatial.


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