Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 309 (General Relativity)

Use signature $(-,+,+,+)$, so the future-directed unit normal obeys $n^{\mu }{n}_{\mu }=-1$. The spatial projection tensor and the induced metric areIndeed $P^{\mu }{}_{\alpha }{n}^{\alpha }=0$ and $P^{\mu }{}_{\rho }{P}^{\rho }{}_{\alpha }=P^{\mu }{}_{\alpha }$: this is an idempotent linear map onto the tangent space of the spatial hypersurface. For a covariant tensor both slots must be projected:Contracting either slot with $n$ supplies the vanishing factor $P^{\mu }{}_{\alpha }{n}^{\alpha }$. Thus both normal contractions vanish, without requiring symmetry of $T$.

Take the orientation induced by the future-directed unit normal and positive lapse function $\alpha$. In coordinates adapted to the 3+1 decomposition of spacetime, any component ${ϵ}_{\ell ijk}$ with all four indices spatial vanishes: only three distinct spatial indices are available. The shift vector therefore contributes nothing, andHere $[ijk]$ is the Levi-Civita symbol, and the last equality uses the metric determinant in a 3+1 decomposition, $|g|={\alpha }^2\gamma$. The right side is precisely the spatial metric volume tensor, with component $\sqrt{\gamma }$ at $ijk=123$, the appropriate permutation sign, and zero for repeated indices. HenceMore invariantly, the spatial volume form is the restriction to the hypersurface of the contraction of the spacetime volume form with $n$.

For the Vacuum Einstein equations with zero cosmological constant, $R_{\alpha \beta }=0$, and its tensor contraction gives $R=0$. Every Ricci tensor and scalar curvature term in the Weyl tensor consequently disappears:Thus the Weyl tensor contains all vacuum curvature. Here vacuum means Ricci-flat; a nonzero cosmological constant would retain a constant-curvature contribution to the Riemann curvature tensor.

The electric part of the Weyl tensor has one normal in each antisymmetric pair of the Weyl tensor. ThereforeFor the magnetic part of the Weyl tensor, contracting the first slot gives $n^{\alpha }{n}^{\lambda }{ϵ}_{\alpha \lambda \mu \nu }=0$ by antisymmetry of the metric volume tensor; contracting the second gives $n^{\beta }{n}^{\rho }{C}^{\mu \nu }{}_{\beta \rho }=0$ by antisymmetry of the last curvature pair. Consequently both tensors are entirely spatial:These spatial projection tensor identities actually hold for any Weyl tensor; the vacuum assumption is needed subsequently to replace it by the Riemann curvature tensor.

All free indices below are spatial, and $K={\gamma }^{\mu \nu }{K}_{\mu \nu }$. Contract the vacuum Ricci tensor using $g^{\mu \nu }={\gamma }^{\mu \nu }-n^{\mu }{n}^{\nu }$:Reversing both antisymmetric curvature pairs identifies the last term as $-E_{\alpha \beta }$. The spatial Gauss–Codazzi equations for a spatial hypersurface then giveFor the magnetic part of the Weyl tensor, moving the normal to the first slot of the metric volume tensor introduces a minus sign:The remaining curvature indices are spatial, so the normal projection in the Gauss–Codazzi equations for a spatial hypersurface givesThe interchange $\mu \leftrightarrow \nu$ makes the two spatial covariant derivative terms equal. Thus the constants, with the paper's orientation and extrinsic-curvature convention, areIn particular, the sign of the magnetic expression must include the minus from moving $n$ past the first volume-tensor index. These formulas reconstruct the spatial electric part of the Weyl tensor and magnetic part of the Weyl tensor from the induced metric and the extrinsic curvature of a spatial hypersurface.

Let $\lambda$ be an affine parameter and $X=d{x}^{\mu }/d\lambda {\partial }_{\mu }$ the tangent to the geodesic. Since ${\nabla }_{{\partial }_{\mu }}{\partial }_{\nu }={\Gamma }^{\rho }{}_{\mu \nu }{\partial }_{\rho }$, the product rule for the covariant derivative givesIndependence of the coordinate basis vectors yields the coordinate geodesic equationThe affine parameter condition matters: an arbitrary reparametrization instead allows a multiple of the tangent on the right, as in the unparametrized geodesic equation.

Set $f=1-2M(v)/r$, so $f_r=2M/r^2$ and $f_v=-2M^{\prime }/r$, where $M^{\prime }=dM/dv$. Dots in this solution mean derivatives with respect to proper time $\tau$. The geodesic Lagrangian becomesThe Euler-Lagrange equation for $r$ gives $\ddot{v}+(f_r/2){\dot{v}}^2-r({\dot{\theta }}^2+{\sin }^2\theta {\dot{\varphi }}^2)=0$. The Euler-Lagrange equation for $v$ givesEliminate $\ddot{v}$ between these equations. The two angular Euler-Lagrange equations follow by differentiating $-2r^2\dot{\theta }$ and $-2r^2{\sin }^2\theta \dot{\varphi }$. Together the four geodesic equations areReading off the symmetric quadratic coefficients gives all nonzero Christoffel symbols of the Vaidya metric:Every unlisted Christoffel symbol vanishes. The lower-index symmetry is that of the Levi-Civita connection. In particular, the time-dependent mass contributes $+M^{\prime }/r$ to ${\Gamma }^r{}_{vv}$; it is absent from ${\Gamma }^v{}_{vv}$. There are eleven independent nonzero entries, or fifteen when both lower-index orders are counted.

For radial motion, take an affine parameter $\lambda$ and use dots for $d/d\lambda$. The radial null geodesics of the Vaidya metric satisfy the two radial geodesic equationsThe radial squared norm is $-f{\dot{v}}^2+2\dot{v}\dot{r}$. If $v$ is constant it vanishes, and the geodesic equation reduces to $\ddot{r}=0$. Thus constant-$v$ radial curves are null geodesics, with $r=a\lambda +b$ an affine parametrization. Future ingoing motion has $a<0$.

For the other family let $dr/dv=f/2$. Its tangent $k={\partial }_v+(f/2){\partial }_r$ is a null vector. Using the Christoffel symbols of the Vaidya metric, direct differentiation givesHence these are also null geodesics, but $v$ is generally not affine. For completeness, put $p=dv/d\lambda$ and chooseThis makes $\ddot{v}=-(M/r^2){\dot{v}}^2$. Differentiating $\dot{r}=(f/2)\dot{v}$ then gives $\ddot{r}=(f_v/2){\dot{v}}^2=-M^{\prime }{\dot{v}}^2/r$; substitution verifies the radial $r$ equation as well. Thus the null condition and both radial geodesic equations hold, not just the null condition alone.

Rearranging the outgoing radial null geodesics of the Vaidya metric equation along $r=r_0(v)$ givesFor the prescribed parametrization by $w$, the chain rule yieldsSince $n>1$, $v/w=1+2\beta w^{n-1}\to 1$ as $w\to 0^{+}$. The asymptotic equivalence $w\sim v$ therefore impliesMore precisely $w=v+O(v^n)$ and $M(v)=(\beta /2)v^n[1+O(v^{n-1})]$, using Big O notation. The positivity of $dv/dw$ justifies treating $w$ as a single-valued function of $v$ locally.

Write $V=T_p\mathcal{M}$ and $V^{\ast }$ for its dual space. A type-$(1,1)$ tensor is equivalently a linear map $A:V\to V$: the corresponding scalar-valued multilinear map is $(X,\eta )\mapsto \eta (A(X))$. Thus the composition definesIt is linear in $X$ because $A$ and $B$ are linear maps, and linear in the covector $\eta$ by evaluation. This proves directly that the composition is a tensor of type $(1,1)$.

Choose a basis $e_{\alpha }$ of the tangent space and its dual basis $e^{\beta }$. With the paper's component order $A(e_{\alpha })=A_{\alpha }{}^{\gamma }{e}_{\gamma }$,Equivalently, in the usual output-first matrix notation, $C^{\beta }{}_{\alpha }=B^{\beta }{}_{\gamma }{A}^{\gamma }{}_{\alpha }$: the matrix product is $BA$, not $AB$. This composition of mixed tensors contracts one output index with one input index. Under the tensor component transformation law, the two Jacobian factors at that contracted index cancel, leaving exactly one factor for each free index.

The Lie derivative of a function is differentiation along the flow, while the Lie derivative of a vector field compares a vector with its pullback by that flow. Their index-free expressions areHere $X(f)$ is the directional derivative of the smooth function, and $[X,Y]$ is the Lie bracket of vector fields, characterized by $[X,Y](f)=X(Y(f))-Y(X(f))$. To see the second expression directly, if ${\varphi }_t$ is the local flow of $X$, differentiating $(d{\varphi }_t{)}^{-1}Y\circ {\varphi }_t$ at $t=0$ gives $dY(X)-dX(Y)=[X,Y]$. This uses the flow definition of the Lie derivative of a tensor field.

On smooth functions, the Lie derivative of a function satisfiesOn an arbitrary vector field $W$, the Lie derivative of a vector field givesThe middle equality follows from the Jacobi identity for the Lie bracket of vector fields. One can verify that identity without assuming the result being proved: view vector fields as derivations acting on smooth functions, expand their commutators, and cancel the six compositions. Equality as derivations implies equality as vector fields.

For the second identity, let $A,B,C$ be any operators on either smooth functions or vector fields. Associativity of composition givesEach term cancels in the sum. Applying this to $A={\mathcal{L}}_X$, $B={\mathcal{L}}_Y$, $C={\mathcal{L}}_Z$ proves both requested identities on both kinds of arguments:The commutator identity for Lie derivatives expresses that the Lie derivative of a tensor field represents the Lie bracket of vector fields.

Use $t$ as an affine parameter on each member of the geodesic variation, so ${\nabla }_{T}T=0$. The parameter directions commute, giving $[T,S]=0$ along the variation. The torsion-free connection then givesConsequently the covariant derivative and curvature definition giveThis is the geodesic-deviation equation with the stated curvature sign:The component expression defines the slot order: $\beta$ is the vector acted on, and $\gamma ,\delta$ are the two derivative directions of the Riemann curvature tensor. Both the commuting parameter directions and affine parametrization are essential; a general connecting vector field or a nonaffine parameter would add terms.

Let $\delta h_{\mu \nu }={\partial }_{\mu }{V}_{\nu }+{\partial }_{\nu }{V}_{\mu }$. Substitution into the linearized Levi-Civita connection cancels four terms and leavesTo first order in the metric perturbation, the products of Christoffel symbols are quadratic and may be discarded. Therefore the change in the Riemann curvature tensor isThis is the gauge invariance of the linearized Riemann tensor. Lowering the first index with the Minkowski metric does not change the conclusion; the change from using the full metric tensor instead is of second order because the background curvature is zero. HenceThe result is invariance at linear order about flat spacetime, not componentwise invariance of arbitrary nonlinear curvature under finite coordinate changes. Directly in the four-term formula, the eight third derivatives of $V$ cancel in pairs by commuting partial derivatives.

Put $A,B\in \{1,2\}$ and write $h_{AB}=H_{AB}{e}^{ik\cdot x}$, where $H_{11}=H_{+}$, $H_{22}=-H_{+}$ and $H_{12}=H_{21}=H_{\times }$. Physical fields are the real parts. Use $c=1$ here, and take the initial separation at equal time, so $L^0=0$. For propagation in the positive $x^3$ direction one may set $k_{\mu }=(-\omega ,0,0,\omega )$; only $k_0^2={\omega }^2$ enters the answer.

The linearized Levi-Civita connection in this transverse-traceless gauge has the potentially nonzero componentswith all others zero to this order. In particular ${\Gamma }^{\mu }{}_{00}=0$, so a particle initially at rest has fixed spatial coordinates in transverse-traceless gauge, and $t$ equals its proper time. Expanding the Riemann curvature tensor givesThe plus sign in the last expression agrees with $R(T,S)T$ in the geodesic deviation convention above.

For the separation measured in a parallel-propagated orthonormal frame, write $S^{\hat{\mu }}=L^{\hat{\mu }}+{\xi }^{\hat{\mu }}$. The frame removes the connection terms from the time derivative, so the geodesic deviation equation reduces to ${\ddot{\xi }}^{\hat{A}}=R^{\hat{A}}{}_{\hat{0}\hat{0}\hat{B}}{L}^{\hat{B}}={\ddot{h}}_{AB}{L}^{\hat{B}}/2$ at first order. The observable transverse relative acceleration isThese are the plus polarization and cross polarization tidal patterns; there is no longitudinal acceleration. The curvature is evaluated along the reference worldline, and replacing $S$ by $L$ on the right changes only second-order terms. This is the infinitesimal-separation approximation intrinsic to geodesic deviation.

There is an important component distinction if the symbols $L^{\mu }+{\xi }^{\mu }$ are instead interpreted literally in the transverse-traceless gauge coordinate basis. Expanding the left side, rather than replacing a covariant derivative by an ordinary derivative, givesThus in TT coordinatesWith zero initial coordinate displacement and velocity, ${\xi }_{\mathrm{coord}}=0$ throughout. This does not remove the physical gravitational wave strain. To first order an orthonormal coframe in spacetime along the reference worldline is ${\vartheta }^{\hat{A}}=({\delta }^A{}_B+h_{AB}/2)d{x}^B$; its dual frame is parallel propagated because ${\partial }_0({\delta }^A{}_B-h_{AB}/2)+{\Gamma }^A{}_{0B}=0$. Henceafter choosing the initial flat-frame reference. The TT coordinate separation and measured separation distinction accounts exactly for the apparently different accelerations. In a region initially free of the wave, the measured displacement is ${\xi }^{\hat{A}}=h_{AB}{L}^{\hat{B}}/2$ with the corresponding initial conditions.

Take the oscillation to be $z(t)=L\cos \omega t$; a different initial phase does not affect a period average. In the mass-density normalization of the quadrupole formula, the second mass moment tensor of the point mass has only $I_{33}=m{z}^2$ nonzero. Its trace-free mass quadrupole moment is thereforeThe constant part does not radiate. Since $d^3({\cos }^2\omega t)/d{t}^3=4{\omega }^3\sin 2\omega t$, the tensor contraction isUsing $⟨{\sin }^2(2\omega t)⟩=1/2$ in the quadrupole formula gives the averaged positive radiated powerThe mass quadrupole moment oscillates at twice the source's angular frequency, and the source energy loss has the opposite sign. The expression uses the leading slow-motion weak-field approximation, $\omega L\ll c$. With SI stress-energy tensor components, the mass-density integrand is $T_{00}/c^2$; the displayed $c^{-5}$ quadrupole formula uses that mass normalization. This is the point-mass contribution requested by the model: an apparatus maintaining an accelerating mass would also contribute its own mass quadrupole moment.