Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 357 / 1 / a

Expanding the spatial projection tensor givesRearrangement yields the normal-plus-spatial decompositionIt also makes $n^{\mu }{\Theta }_{\mu }=0$ immediate.

The definition of the extrinsic curvature of a spatial hypersurface and ${\perp }^{\rho }{}_{\mu }={\delta }^{\rho }{}_{\mu }+n^{\rho }{n}_{\mu }$ givewhere $a_{\nu }=n^{\rho }{\nabla }_{\rho }{n}_{\nu }$ is the normal acceleration. HenceDifferentiating $n^{\nu }{n}_{\nu }=-1$ shows $n^{\nu }{a}_{\nu }=0$, so the acceleration is spatial as required.

The inverse spacetime metric decomposes asContracting the covariant derivative of $Z_{\nu }$ and recognizing the fully projected contraction as its spatial covariant derivative gives

Insert $Z_{\mu }={\Theta }_{\mu }+n_{\mu }\Theta$ from part (i) into the projected divergence:The term containing ${\nabla }_{\mu }\Theta$ vanishes because ${\perp }^{\mu \nu }{n}_{\nu }=0$. The remaining contraction isby the definition of the extrinsic curvature of a spatial hypersurface. Therefore