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

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

First take the spacetime trace of the generalized field equation. Since $n_{\mu }{Z}^{\mu }=-\Theta$ and spacetime has dimension four, this givesNext contract twice with the unit normal. The result isAdding the trace equation to twice this normal projection cancels $T$. The Scalar Gauss equation then converts the curvature terms to

For the derivative terms, part (iii) givesDifferentiating $n^{\nu }{Z}_{\nu }=-\Theta$ along $n^{\mu }$ yieldsCombining these identities produces

Because $\Theta$ is a scalar field and $n^{\mu }={\alpha }^{-1}(1,-{\beta }^i)$,The normal acceleration is spatial, so $Z_{\mu }{a}^{\mu }={\Theta }_{\mu }{a}^{\mu }$. Part (iv) also gives $D^{\mu }{Z}_{\mu }=D^{\mu }{\Theta }_{\mu }-K\Theta$. Solving the preceding constraint for ${\partial }_t\Theta$ givesThus the constants in this Z4 formulation evolution equation are