Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 357 (Gravitational Waves and Numerical Relativity)

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

Set $A=1-2M/R=(R-2M)/R$ andSubstitution into the Schwarzschild metric givesThe cross coefficient and radial coefficient simplify toTherefore

With $R=r+M$ and $dR=dr$,Defining the conformal factor byalso gives $R^2={\psi }^4{r}^2$. HenceThe constant-time spatial metric is therefore conformally flat.

Comparison with the 3+1 decomposition of spacetime gives the diagonal spatial metricThe mixed metric coefficient is ${\beta }_r=g_{tr}=M/r$. Raising its index with ${\gamma }^{rr}=r^2/(r+M{)}^2$ yields the shift vectorNowand $g_{tt}=-{\alpha }^2+{\beta }_i{\beta }^i=-(r-M)/(r+M)$. The positive lapse function is consequently

The spatial metric is stationary and only ${\beta }^r=b(r)=Mr/(r+M{)}^2$ is nonzero. The evolution equation therefore saysthe Lie derivative of the spatial metric along the shift vector. For the radial component,soFor the angular components,and spherical symmetry suppliesAll off-diagonal components vanish.

Contracting with the inverse spatial metric gives the mean curvatureThus

Stationarity makes ${\partial }_t\alpha =0$, whileThe left side of the Bona--Masso slicing condition is thereforeIts right side isEquality requiresSince $\alpha =r/(r+M)$, this is the Stationary Schwarzschild Bona--Masso slicing functionConsequently ${\lim }_{\alpha \to 1}f(\alpha )=0$, as expected in the asymptotically flat region $r\to \infty$.

The retarded and advanced null coordinates invert toThusand the Minkowski metric becomes

The coordinate basis transforms by the chain rule:or equivalentlyIn the $(z,t)$ diagram, ${\partial }_t$ points vertically upward and ${\partial }_z$ horizontally right. The vector ${\partial }_u$ points along the future-left null ray and ${\partial }_v$ along the future-right null ray; their factors of one half affect length in the coordinate drawing but not direction.

Using these derivative relations, the Minkowski wave operator in null coordinates is

The phase isBecause the transverse-traceless gauge perturbation has no component alongonly the third term of the linearized Riemann curvature operator contributes to the required contraction:The derivatives and transverse polarization contraction areTherefore the Newman--Penrose scalar Psi4 isIts real and imaginary parts encode the plus and cross gravitational wave polarizations, up to the stated Riemann curvature tensor and complex null tetrad conventions.

Only the transverse coordinates occur in the perturbation, so its tensor components are unchanged by replacing $(t,z)$ with $(u,v)$. The plane gravitational wave in linearized gravity isEvery component depends on $u$ alone. The Minkowski wave operator in null coordinates therefore givesThus the field satisfies the vacuum Linearized Einstein equations.

Using the stated Christoffel symbols in the definition of the Riemann curvature tensor, the two nontrivial contractions areHenceand the vacuum Einstein field equations reduce toThis is the diagonal Rosen coordinates for a plane gravitational wave equation.

For the plus-polarized wave, setComparison of the transverse metric components gives, to linear order,orThereforeThe Vacuum Einstein equations are consequently satisfied at linear order. A real gravitational wave is obtained by taking the real part of the complex plane-wave notation.