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

LetThe identity decomposes asApplying this decomposition to both indices of $T_{\mu \nu }$ givesThe 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 isIt also makes explicit that

Define the acceleration of the normal congruence byBecause $S^{\sigma \gamma }$ is spatial, differentiating
$n_{\sigma }{S}^{\sigma \gamma }=0$ givesInsertinto the contracted derivative and project the free index:Rearranging proves

The Lie derivative of the spatial covector $j_{\alpha }$ along $n^{\mu }$ isTo show that it is spatial, contract with $n^{\alpha }$. Differentiating $n^{\alpha }{j}_{\alpha }=0$ along $n^{\mu }$ giveswhereasThe terms cancel, soA spatial projector therefore acts trivially:

Spatially project stress-energy conservation,and substitute the decomposition from part a. The $\rho n^{\mu }{n}_{\nu }$ term contributes $\rho a_{\alpha }$. The two momentum terms combine intobecause the two contractions with the full extrinsic curvature cancel andFinally, part b givesThusThe momentum equation in a 3+1 decomposition isTherefore

Sinceand $j_{\mu }{n}^{\mu }=0$, the scaling rule for a covector Lie derivative has no extra term:The acceleration is the spatial lapse gradient,Multiplying the momentum equation by $\alpha$ therefore givesFor spatial components, the shift term can be writtenso explicitly