Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 309 (General Relativity)

Under parallel transport, the gyroscope spin obeys $u^b{\nabla }_b{s}^a=0$, and its freely falling circular orbit is a geodesic, $u^b{\nabla }_b{u}^a=0$. Metric compatibility therefore givesThus $s^a{u}_a$ is constant along the orbit.

Initially $s^a$ is radial, whereas the circular-orbit four-velocity has only $t$ and $\varphi$ components. The diagonal Schwarzschild metric therefore gives $s^a{u}_a=0$ at $t=0$. By part i,throughout the orbit.

In a coordinate basis, parallel transport isAlong the circular orbit, $dr=d\theta =0$ and $d\varphi /dt=\Omega$. Dividing by $dt/d\tau$ gives

On the equatorial plane, the only potentially relevant angular connection coefficient is ${\Gamma }_{\varphi \varphi }^{\theta }=-\sin \theta \cos \theta$, which vanishes at $\theta =\pi /2$. The $\theta$ transport equation is consequently $d{s}^{\theta }/dt=0$. Since the initially radial spin has $s^{\theta }(0)=0$,

Writing $f=1-2M/R$, orthogonality from part a givesThe needed Christoffel symbols areThe radial and azimuthal transport equations reduce toHenceWith the stated initial direction,

Metric compatibility and parallel transport implyUsing $s^t=R^2\Omega s^{\varphi }/f$ givesSubstitution of part d makes this independent of $t$ precisely whenthe relativistic circular-orbit form of Kepler third law. One orbit takes $T=2\pi /\Omega$, during which the spin phase advances by ${\Omega }^{\prime }T=2\pi \sqrt{1-3M/R}$. Relative to the radial direction, the spin therefore lags by the geodetic precession angle

Varying the inverse metric in $H_{abc}{H}^{abc}$ produces three identical terms. With $T_{ab}=-(2/\sqrt{-g})\delta S_H/\delta g^{ab}$,In four dimensions use the Hodge star operator to define the dual covector $v^a=(1/3!){ϵ}^{abcd}{H}_{bcd}$. Then $H^2=-6v^2$ and $H_{acd}{H}_b{}^{cd}=2(v_a{v}_b-g_{ab}{v}^2)$, so equivalently

Under an infinitesimal diffeomorphism generated by ${\xi }^a$,Insert these variations intoand integrate derivatives of ${\xi }^a$ by parts. Diffeomorphism invariance and arbitrariness of ${\xi }^b$ give the Noether identityOn the covector equation of motion $E^a=0$, this reduces to stress-energy conservation, ${\nabla }_a{T}^a{}_b=0$.

The projection $X\mapsto X_{\parallel }$ is tensorial. Although a covariant derivative is not $C^{\infty }$-linear in its second argument, the extra term in ${\nabla }_{X_{\parallel }}(f{Y}_{\parallel })$ is proportional to $Y_{\parallel }$, whose contraction with the normal vanishes. Thus $K(X,Y)$ is $C^{\infty }$-linear in both arguments and defines a $(0,2)$ tensor on the hypersurface.

Since $n_a{Y}_{\parallel }^a=0$,Therefore the extrinsic curvature is

A hypersurface normal is locally proportional to the gradient of a defining function. The Frobenius theorem therefore impliesTaking the antisymmetric part of the formula in part i gives $K_{[ab]}=0$, henceEquivalently, torsion freedom gives $K(X,Y)-K(Y,X)=-n\cdot [X,Y]=0$ because the bracket of tangent vector fields is tangent.

The affine geodesic equation givesUnit normalization implies $n^b{\nabla }_a{n}_b=0$, and decomposition of the first index givesConsequently, at the intersection point,

Choose any tangent vector $V^a\in T_p\Sigma$ and let the affinely parametrized geodesic with initial tangent $V^a$ start at $p$. If $\Sigma$ is totally geodesic, then $U^a{n}_a$ remains zero. Its initial derivative is therefore zero. Part iii, with $U^a{n}_a=0$ at $p$, givesThis holds for every tangent $V$. Since $K$ is symmetric, the polarization identity implies

Assume a spatially compact source of size $d\ll r=|\mathbf{x}|$, internal speeds much smaller than one, and wavelength much larger than $d$. In the radiation zone,soDefine the mass quadrupole momentTwice using ${\partial }_{\mu }{T}^{\mu \nu }=0$, discarding boundary terms, givesHence

In Lorenz gauge, ${\partial }_{\mu }{\bar{h}}^{\mu \nu }=0$. A leading outgoing far-zone field depends on retarded time $u=t-r$, so ${\partial }_j=-{\hat{x}}_j{\partial }_u+O(r^{-1})$. Taking $\nu =i$ and using ${\bar{h}}^{0i}=-{\bar{h}}_{0i}$ givesDropping a nonradiative integration constant and using part i,

With $u=t-r$, direct integration givesTherefore the time-dependent spatial field isThe gravitational-wave frequency is $2\omega$. On the positive $z$-axis this matrix is transverse and traceless. It is a rotating combination of the plus polarization and cross polarization, with amplitude proportional to $1/z$, exactly as for a plane wave propagating in the $z$ direction.

The quadrupole formula in units $G=c=1$ isHere $I_{kk}=m{a}^2/3$ is constant, so its trace subtraction has no third derivative. If $B=4m{a}^2{\omega }^3/3$, thenThus ${\stackrel{...}{I}}_{ij}{\stackrel{...}{I}}_{ij}=2B^2$ andRestoring units multiplies this by $G/c^5$.

For the displayed frame,and all other pairings vanish. Hence the proposed one-forms are the dual coframe. Substitution intoreproduces the metric, proving that the frame is an orthonormal frame with signature $(-,+,+)$.

The exterior derivatives of the coframe areInsert the proposed connection into Cartan's first structure equation and use ${\omega }_{\mu \nu }=-{\omega }_{\nu \mu }$. The three equations giveThus

Using Cartan's second structure equation givesSince ${\Theta }_{\mu \nu }=R_{\mu \nu \rho \sigma }{e}^{\rho }\wedge e^{\sigma }/2$, the independent nonzero lowered Riemann curvature tensor components arewith all others determined by the symmetries of the Riemann curvature tensor.

Contraction in the orthonormal frame yieldsHenceFor $u^{\mu }=(1,0,0)$, pressureless matter has $T_{00}=\rho$ and vanishing spatial components. The spatial Einstein equations $G_{ii}+\Lambda {\eta }_{ii}=0$ give $\Lambda =-1/2$, while the time component gives $1=8\pi \rho$. Thus, for $G_N=1$,If the convention is $G_{ab}+\Lambda g_{ab}=T_{ab}$, the corresponding density is simply $\rho =1$.

Because the metric coefficients are independent of $t$ and $y$,are Killing vector fields. The mapssatisfy ${\varphi }_s\circ {\varphi }_r={\varphi }_{s+r}$. Moreover, $dx$ is unchanged and both $e^{x}dy$ and $dt+\sqrt{2}{e}^{x}dy$ are invariant, so ${\varphi }_s^{\ast }g=g$. Differentiating at $s=0$ produces the third Killing fieldGiven two points, first use ${\varphi }_s$ to match their $x$ coordinates, then translations generated by ${\partial }_y$ and ${\partial }_t$ to match $y$ and $t$. The isometry group therefore acts transitively, so the spacetime is a homogeneous space.