Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 309 (General Relativity)

Put $D=\mathrm{diag}(a_1,\dots ,a_{n+1})$. The invertible linear map $x\mapsto u=Dx$ takes the ellipsoid to the unit sphere $S^n$. Write $u=(v,s)$ with $v\in {\mathbb{R}}^n$, and let $N=(0,1)$ and $S=(0,-1)$ be its two poles. The following two stereographic projections define manifold charts on the ellipsoid:Their ranges are ${\mathbb{R}}^n$. With $r^2=|w{|}^2$, their inverses, followed by $D^{-1}$, areThese are continuous and smooth functions, and directly satisfy the defining constraint. On the overlap, the smooth transition map isIt is its own inverse and is smooth away from zero. The two domains are open in the subspace topology and cover the ellipsoid. Its subspace topology is inherited from Euclidean space, so the ellipsoid is Hausdorff and has a second-countable space topology. Thus these charts form a smooth atlas and make it an $n$-dimensional smooth manifold, with an explicit diffeomorphism to $S^n$. For the degenerate dimension $n=0$, each chart consists of a single point with range ${\mathbb{R}}^0$, and their overlap is empty; the same conclusion holds.

At $p\in \mathcal{Q}$, the differential of a smooth map defines the pushforward of a vector field byThis is a tangent vector at $\psi (p)$, and varies smoothly as a section of the pullback tangent bundle. In local coordinates its components are $X^i(p){\partial }_i{\psi }^a(p)$.

For a general smooth map between manifolds, this is a vector field along a map, rather than an intrinsic vector field on all of $N$. A vector field $Y$ on the image can be defined only if these vectors agree whenever two points have the same image, and the resulting field must also be smooth; a projectable vector field must meet that requirement. For a diffeomorphism, there is no ambiguity and $Y_q=d{\psi }_{{\psi }^{-1}(q)}{X}_{{\psi }^{-1}(q)}$. For example, $\psi (x)=x^2$ on $\mathbb{R}$ and $X={\partial }_x$ give $2x{\partial }_y$, with opposite nonzero values at the two preimages of $y>0$. Thus the source's notation must be read with this qualification.

On a branch where $(\rho ,\varphi )$ are valid local coordinates, the inclusion map is $j(\rho ,\varphi )=(x(\rho ),\rho \cos \varphi ,\rho \sin \varphi )$, with $x^2=1-a^2{\rho }^2$. Its differential givesStrictly this is first a vector field along a map on the surface. The displayed formula extends it canonically to a smooth vector field on all of ${\mathbb{R}}^3$. Its integral curves of a vector field through $(x_0,y_0,z_0)$ obey $\dot{x}=0$, $\dot{y}=-z$, $\dot{z}=y$, henceThese are circles around the $x$-axis, traversed with angular speed one; points on the axis are stationary. Since $K(x^2+a^2(y^2+z^2))=0$, curves starting on the ellipsoid stay there. The local angular chart can expire while the ambient flow map and the geometric curve remain well defined.

The induced metric is obtained from the ambient Euclidean metric by the pullback of a Riemannian metric operation. Differentiating $x^2+a^2{\rho }^2=1$ on a branch with $x\ne 0$ gives $dx=-a^2\rho d\rho /x$. Also $d{y}^2+d{z}^2=d{\rho }^2+{\rho }^{2}d{\varphi }^2$. ConsequentlyThe assumption that $(\rho ,\varphi )$ are coordinates already excludes the equator $x=0$ as well as the poles $\rho =0$. On such a chart, $0<\rho <1/|a|$ and $E>0$.

The conformal flattening of a surface of revolution supplies the particularly simple conformal factorSince ${\sigma }^{\prime }>0$, these are isothermal coordinates with a flat rescaled metric. An angular coordinate is understood on a local branch; the flat metric may equally be viewed locally as a cylinder metric. The apparent divergence of $E$ at the equator is a coordinate failure, not a singularity of the induced metric. For example, $(x,\varphi )$ give $g=[1+x^2/(a^2(1-x^2))]d{x}^2+(1-x^2)d{\varphi }^2/a^2$, regular at $x=0$.

For a timelike tangent $v=dx/du$, put $L=\sqrt{-g(v,v)}>0$ and $\mathcal{L}=L^2/2=-g(v,v)/2$, and use the Euler-Lagrange equations. For the quadratic geodesic Lagrangian, they reduce toHere the first equation has been multiplied by an irrelevant minus sign. For the length Lagrangian, its derivatives are $L^{-1}$ times those of $\mathcal{L}$, and differentiating that factor yields insteadThis is the unparametrized geodesic equation, the key to why timelike length and energy have the same geodesic images. Define proper time locally by $d\tau /du=L>0$ and put $w=dx/d\tau =v/L$. The product rule then gives ${\nabla }_{w}w=L^{-2}[{\nabla }_{v}v-(\dot{L}/L)v]=0$ and $g(w,w)=-1$. Conversely an affinely parametrized timelike geodesic has constant $L$, by metric compatibility, and therefore satisfies both equations.

Thus the two variational problems have the same oriented timelike geodesic images, after reparametrization. They do not have exactly the same parametrized solutions with an arbitrary fixed $u$: in Minkowski spacetime, $t(u)=u^2$, $\mathbf{x}=0$, $u>0$, satisfies the length equation but not the quadratic equation. The timelike assumption excludes $L=0$, where this argument and the length derivative would fail.

Use spatial indices $i,j=1,2,3$, set ${\varphi }_i={\partial }_i\varphi$, and assume $\varphi$ is smooth. The inverse metric has $g^{tt}=-c^{-2}{e}^{-2\varphi /c^2}$ and $g^{ij}={\delta }^{ij}$. The Christoffel symbols of the Levi-Civita connection follow from ${\Gamma }^{\alpha }{}_{\beta \gamma }=g^{\alpha \delta }({\partial }_{\beta }{g}_{\gamma \delta }+{\partial }_{\gamma }{g}_{\beta \delta }-{\partial }_{\delta }{g}_{\beta \gamma })/2$:All other components vanish. Although $g_{tt}=-c^2-2\varphi +O(c^{-2})$ diverges, its affine connection has a smooth limit on compact subsets:This is the Newtonian connection from an exponential lapse. It is torsion-free, and its geodesic equation, using $t$ as an affine parameter when $\dot{t}\ne 0$, is $d^2{x}^i/d{t}^2=-{\partial }_i\varphi$. Thus $\varphi$ has the role of a Newtonian gravitational potential. There is no finite limiting nondegenerate Lorentzian metric in these fixed coordinates; $-g(c)/c^2$ tends to $d{t}^2$, a rank-one tensor. The limiting connection preserves $dt$ and the contravariant spatial tensor with components $h^{ij}={\delta }^{ij}$, $h^{t\mu }=0$, which describe the degenerate temporal/spatial structures of this limit.

Fix the Riemann curvature tensor conventionFor the limiting affine connection, every product of two nonzero connection coefficients is zero, because its only nonzero coefficients have spatial upper index and both lower indices $t$. Its only possibly nonzero curvature components, apart from antisymmetry in the last two indices, areThe Ricci tensor is thereforeAll other components vanish, so the limiting connection is a connection with vanishing Ricci tensor exactly when $\varphi$ satisfies the Laplace equation on spatial Euclidean space. This is Ricci-flatness, not necessarily vanishing curvature: $\varphi =x^2-y^2$ is a harmonic function but has nonzero Hessian matrix and hence a nonflat limiting connection. Reversing the curvature sign convention reverses the displayed Ricci sign but leaves the equivalence unchanged.

Use $x^0=t$ and the Levi-Civita symbol ${ϵ}^1{}_{23}=1$. Define a smooth affine connection byThe lower indices are symmetric, so it is a torsion-free connection. With $t$ itself as parameter, the time component of the geodesic equation is identically satisfied, while its spatial components becomeThe last term is $2(\mathbf{B}\times \dot{\mathbf{x}}{)}^i$, with the wedge in the source interpreted as the three-dimensional cross product. Thus the spacetime lifts $(t,\mathbf{x}(t))$ are actually affinely parametrized geodesics, which proves the requested unparametrized claim. Conversely a geodesic with nonzero constant $dt/ds$ can be parametrized by $t$ and obeys this particle equation; purely spatial geodesics with $dt/ds=0$ are not such trajectories. No field equations or metric compatibility are needed. This construction is the affine connection for a velocity-linear force.

For a smooth vector field $X$, solve the initial-value problem $d{\varphi }_t(p)/dt=X({\varphi }_t(p))$, ${\varphi }_0(p)=p$. Local existence, uniqueness and smooth dependence for ordinary differential equations give a local flow. Uniqueness yields ${\varphi }_{t+s}(p)={\varphi }_t({\varphi }_s(p))$ whenever both sides are defined, and ${\varphi }_{-t}$ is the inverse of ${\varphi }_t$ on the corresponding domains. Thus each local flow map is a diffeomorphism between open subsets.

A group of diffeomorphisms of the entire manifold for every real $t$ requires a complete vector field. This follows, for example, on a compact manifold without boundary: local existence times can be made uniform on a finite cover, and repeated continuation prevents finite-time escape. Without completeness, the source's group is only local. The vector field $x^2{\partial }_x$ on $\mathbb{R}$ has flow $x\mapsto x/(1-tx)$ and blows up at $t=1/x$ for $x>0$, so smoothness alone cannot imply a global one-parameter group.

Define the Lie derivative of a differential form by ${\mathcal{L}}_X\alpha =(d/dt{)}_{t=0}{\varphi }_t^{\ast }\alpha$, using the local flow. The pullback of a differential form respects the wedge product of differential forms and commutes with the exterior derivative. Therefore ${\mathcal{L}}_X$ is a derivation of an algebra of degree zero and commutes with $d$.

Let $D_X={\iota }_{X}d+d{\iota }_X$, where ${\iota }_X$ is the interior product of a differential form. The graded product rules for $d$ and ${\iota }_X$ show that their anticommutator $D_X$ is also a degree-zero derivation of an algebra: the two mixed terms cancel. Since $d^2=0$, it commutes with $d$. On a function $f$, ${\iota }_{X}f=0$ and $D_{X}f={\iota }_{X}df=Xf={\mathcal{L}}_{X}f$. Hence $D_X(df)=d(Xf)={\mathcal{L}}_X(df)$.

In a manifold chart, every differential form is a sum of functions times products $d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_p}$. The two derivations agree on its generators $f$ and $d{x}^i$, and therefore agree on every such form. This proves Cartan's magic formula, including the case $p=0$:

Use Cartan's magic formula. The exterior derivative and interior product of a differential form giveThus $d({\iota }_X\alpha )=-2xdx-2ydy+2zdz$, and the first two components cancel:The rotation in the $xy$-plane leaves $ydx-xdy$ invariant, while the dilation in $z$ doubles the infinitesimal weight of $zdz$, providing a geometric check of this Lie derivative.

Integrating the two vector fields gives their complete flow maps:The first is a simultaneous rotation about the z-axis through angle $t$ and a dilation by $e^t$ in the $z$-direction. The second is an isotropic dilation about the origin. Both satisfy the group law for all real $t$ and have inverse given by time $-t$. In particular ${\varphi }_t$ is not a pure rotation unless restricted to the plane $z=0$.

The Jacobian determinant of ${\varphi }_t$ is $e^t$, while every coordinate and every coordinate differential is multiplied by $e^t$ under ${\psi }_t$. Hence the pullback of a differential form operation givesTo compare directly with Cartan's magic formula, note $d\mu =0$ and computeThe first two differentiated terms vanish because their coordinate differentials repeat; the last contributes $dz\wedge dx\wedge dy=\mu$. AlsoTheir sum is $2\alpha$, as required. Equivalently the divergence of $X$ relative to the standard volume form is one.

An electromagnetic field tensor obeys the homogeneous Maxwell equations $dF=0$. Together with its invariance under the Killing vector field $V$, Cartan's magic formula gives $d({\iota }_{V}F)={\mathcal{L}}_{V}F-{\iota }_{V}dF=0$. The local potential of a closed differential one-form construction, the one-form Poincaré lemma, therefore supplies a scalar field $\Phi$ on a sufficiently small contractible neighborhood withFor an explicit local proof, write ${\iota }_{V}F={\alpha }_i(x)d{x}^i$ on a coordinate ball centered at zero and take $\Phi (x)={\int }_0^1{x}^i{\alpha }_i(sx)ds$. Since $d\alpha =0$, differentiating under the integral gives ${\partial }_j\Phi ={\int }_0^1[{\alpha }_j(sx)+s{x}^i{\partial }_i{\alpha }_j(sx)]ds={\alpha }_j(x)$. There need not be a global potential when this closed one-form has nonzero periods. The homogeneous Maxwell equation is essential; an arbitrary invariant two-form would not suffice.

Use proper time for the massive particle, let $u$ be its tangent, and assume $m\ne 0$. The Killing equation makes $u^a{u}^b{\nabla }_a{V}_b=0$. The Lorentz force equation then impliesConsequently the local conserved quantity isThe sign follows from contracting $F$ in its first argument, exactly as in the source. Adding a constant to $\Phi$ merely shifts the conserved quantity. This is a charged-particle conserved quantity from a Killing symmetry.

Choose signature $(-,+,+,+)$, with $e^0$ timelike. The orthonormal coframe in spacetime reconstructs the Lorentzian metric asBecause $A$ and $B$ are nowhere zero on the domain, the coframe is invertible and the metric nondegenerate; either sign of each function is allowed. In retarded and advanced null coordinates $u=t-z$, $v=t+z$ the metric is $-dudv+A(u{)}^{2}d{x}^2+B(u{)}^{2}d{y}^2$, a diagonal example of Rosen coordinates for a plane gravitational wave. Multiplication by $-1$ would give the opposite overall signature convention; all subsequent signs here use the displayed one.

Put $u=t-z$, $\ell =du=e^0-e^3$, $p=A^{\prime }/A$, $q=B^{\prime }/B$, with primes denoting $d/du$. The exterior derivatives are $d{e}^0=d{e}^3=0$, $d{e}^1=p\ell \wedge e^1$, $d{e}^2=q\ell \wedge e^2$. For a Levi-Civita connection, the connection 1-forms obey both Cartan's first structure equation and metric compatibility. With lower indices ${\omega }_{ab}={\eta }_{ac}{\omega }^c{}_b$, the result isAll other lower-index forms vanish or follow from ${\omega }_{ab}=-{\omega }_{ba}$. Raising the first time index changes its sign: for example ${\omega }^0{}_1={\omega }^1{}_0=p{e}^1$, whereas ${\omega }^1{}_3=-p{e}^1$ and ${\omega }^3{}_1=p{e}^1$.

For direct verification, ${\omega }^1{}_0\wedge e^0+{\omega }^1{}_3\wedge e^3=-p\ell \wedge e^1=-d{e}^1$, and the same holds for index two. For indices zero and three each term wedges $e^1$ or $e^2$ with itself, so it is zero. These forms are metric-compatible, and uniqueness follows from the existence and uniqueness of the Levi-Civita connection. The printed uniqueness statement implicitly includes metric compatibility: the torsion equation alone permits adding nonzero forms $C^a{}_{bc}{e}^c$ with $C^a{}_{bc}=C^a{}_{cb}$, since their wedge with $e^b$ vanishes.

Keep $\ell =e^0-e^3$ and set $P=p^{\prime }+p^2=A^{\prime \prime }/A$, $Q=q^{\prime }+q^2=B^{\prime \prime }/B$. Differentiation gives $d(p{e}^1)=P\ell \wedge e^1$ and $d(q{e}^2)=Q\ell \wedge e^2$. In Cartan's second structure equation, the quadratic terms for these components vanish. The apparently possible transverse component cancels:Thus the potentially nonzero mixed-index curvature 2-forms areAll other components vanish. Equivalently, the independent lower-index forms are ${\Theta }_{01}={\Theta }_{13}=-P\ell \wedge e^1$ and ${\Theta }_{02}={\Theta }_{23}=-Q\ell \wedge e^2$, with the rest supplied by lower-index antisymmetry. These signs use ${\Theta }^a{}_b=\frac{1}{2}{R}^a{}_{bcd}{e}^c\wedge e^d$ and the curvature convention in question 2(c). This is the curvature of a diagonal plane wave in Rosen coordinates.

Let $E_a$ denote the frame dual to the coframe. For the convention just stated, the Ricci tensor is obtained from the curvature 2-forms through the Ricci one-forms ${\mathcal{R}}_b={\sum }_a{\iota }_{E_a}{\Theta }^a{}_b=R_{bd}{e}^d$. In particular ${\iota }_{E_1}(\ell \wedge e^1)=-\ell$ and likewise for index two. It follows thatThusThe scalar curvature vanishes even before imposing vacuum because $du$ is null, but the full tensor vanishes exactly whenThis is the Vacuum Einstein equations in this family. The nonzero-coframe hypothesis makes division by $A$ and $B$ legitimate. A nonzero $P=-Q$ can leave the Riemann curvature tensor nonzero even in vacuum.

Use linearization around Minkowski spacetime after constant rescaling of $x,y$: set $A=1+\epsilon a(u)$, $B=1+\epsilon b(u)$, $u=t-z$. To first order,Hence $s=a+b$ is affine in $u$, while $h=a-b$ is the freely propagating profile. The affine transverse trace in a diagonal linearized plane wave is pure gauge: it has zero linearized curvature and is locally a coordinate artifact. To see this without imposing extra boundary conditions, use a first-order linearized coordinate gauge transformation $h_{\mu \nu }\mapsto h_{\mu \nu }-{\partial }_{\mu }{\xi }_{\nu }-{\partial }_{\nu }{\xi }_{\mu }$, withThe $tx,ty,zx,zy$ changes cancel and the remaining time/longitudinal changes are proportional to $s^{\prime \prime }=0$. The transverse perturbation becomeswith all time and longitudinal components zero. It is a transverse-traceless gauge plane gravitational wave travelling in the positive $z$-direction, with plus polarization and no cross polarization in these transverse axes.

For freely falling nearby particles, geodesic deviation gives opposite transverse tidal accelerations, since $R_{txtx}=-\epsilon a^{\prime \prime }$ and $R_{tyty}=-\epsilon b^{\prime \prime }=+\epsilon a^{\prime \prime }$ in vacuum. Thus the transverse geodesic deviation eigenvalues have opposite signs; the stretching and compressing roles interchange when the curvature profile changes sign. If $h^{\prime \prime }=0$ as well, there is no linearized tidal wave: that special profile is also locally flat to this order. The exact diagonal metric describes a fixed transverse polarization through its opposite vacuum curvature eigenvalues; the linearized calculation identifies it as plus relative to the chosen axes.