Past exam of the general relativity course of the University of Oxford / 2022 (A16729W1)

Source: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/10/Physics%20Papers%202021.pdf?cookie_passthrough=1

Source bundle pages 139–144. The paper cover establishes 2022; the attachment filename uses 2021.

For the Levi-Civita connection, the all-lowered Riemann curvature tensor has the algebraic symmetriesThe first Bianchi identity is the cyclic identityThese statements concern the Riemann curvature tensor at each point and are unchanged by reversing its overall curvature sign convention.

A semicolon denotes the covariant derivative for the Levi-Civita connection. For example, the four lower indices of the Riemann curvature tensor contribute four connection terms:Here ${\partial }_{\eta }=\partial /\partial x^{\eta }$ is a partial derivative, ${\Gamma }_{b\eta }^a$ are Christoffel symbols, and repeated indices are summed from $0$ to $3$. Write $g_{\mu \nu }$ for the metric tensor, $g^{\mu \nu }$ for its inverse metric, $R_{\mu \nu }=g^{\lambda {\nu }^{\prime }}{R}_{\lambda \mu {\nu }^{\prime }\nu }$ for the Ricci tensor, and $R=g^{\mu \nu }{R}_{\mu \nu }$ for the Ricci scalar. This tensor contraction agrees with the paper's $R_{\mu \nu }=-R^a{}_{\mu \nu a}$.

By metric compatibility, the inverse metric may pass through each covariant derivative. Contracting the second Bianchi identity with $g^{\lambda \nu }$ giveswhere ${\nabla }_{\eta }$ is the same covariant derivative as the semicolon. Next contract with $g^{\mu \kappa }$. Antisymmetry of the first pair gives $g^{\mu \kappa }{R}_{\lambda \mu \kappa \eta }=-R_{\lambda \eta }$, soThus the contracted Bianchi identity is ${\nabla }^{\mu }{R}_{\mu \eta }=\frac{1}{2}{\partial }_{\eta }R$. Raising the remaining index and using metric compatibility provesThe Einstein tensor $G^{\mu \nu }$ therefore has identically zero covariant divergence, without imposing any field equation. This makes the Einstein field equations compatible with stress-energy conservation, ${\nabla }_{\mu }{T}^{\mu \nu }=0$, where $T^{\mu \nu }$ is the stress-energy tensor. The connection terms describe how energy and momentum are compared at neighboring points; ordinary componentwise partial derivatives would not give a coordinate-independent conservation law in curved spacetime.

Take the tensor contraction of the Einstein field equations with the inverse metric $g^{\mu \nu }$. In four-dimensional spacetime, $g^{\mu \nu }{g}_{\mu \nu }=4$, henceHere $T$ is the tensor contraction of the stress-energy tensor. Substituting this result back into the Einstein field equations, while keeping both free indices lowered, gives the trace-reversed stress-energy tensor:The factor $1/2$ uses the dimension four: in dimension $n>2$, the Trace-reversed Einstein field equations in D dimensions have $1/(n-2)$ in place of $1/2$.

Use $x^0=ct$ and metric signature $(-,+,+,+)$. With the source's curvature sign convention, the Ricci tensor isIts time-time component is thereforeFor a slowly moving, pressure-negligible source, the stress-energy tensor is approximated by relativistic dust: $T^{\mu \nu }=\rho U^{\mu }{U}^{\nu }$, where $\rho$ is mass density and $U^{\mu }$ is the four-velocity. In its approximate rest local inertial frame, $U^{\mu }\simeq (c,0,0,0)$ and $U_0\simeq -c$. Thusand the trace-reversed stress-energy tensor satisfies

Independently, introduce the metric perturbation $g_{\mu \nu }={\eta }_{\mu \nu }+h_{\mu \nu }$, with $|h_{\mu \nu }|\ll 1$ and constant Minkowski metric ${\eta }_{\mu \nu }$. The Christoffel symbols are first order in $h$, so their products are discarded in the weak-field approximation. With a comma denoting a partial derivative, the linearized Riemann curvature tensor isTaking the tensor contraction gives, to this order,In the static Newtonian limit, all time partial derivatives vanish. The constant ${\eta }_{00}$ also has zero partial derivatives, leavingHere ${\nabla }^2={\sum }_i{\partial }_i^2$ is the spatial Laplace operator in the approximately Cartesian background coordinates. This sign follows the paper's curvature sign convention.

For the Newtonian gravitational potential $\Phi$, the weak static metric tensor has $g_{00}\simeq -1-2\Phi /c^2$. The Newtonian limit derived above consequently givesMatching the Poisson equation for Newtonian gravity, ${\nabla }^2\Phi =4\pi G\rho$, fixesHere $G$ is the gravitational constant and $c$ is the speed of light. This negative coupling agrees with the source formula sheet. The familiar positive coupling uses the opposite curvature sign convention for the Ricci tensor; the physical Poisson equation for Newtonian gravity is the same.

Choose normal coordinates ${\xi }^{\alpha }$ centered at an event $p$. At $p$, the metric tensor in these coordinates is the constant Minkowski metric ${\eta }_{\alpha \beta }=\mathrm{diag}(-1,1,1,1)$, with ${\xi }^0$ measured in length units, and its first partial derivatives vanish. In arbitrary coordinates $x^{\mu }$, the chain rule for the line element givessoThis construction is pointwise. At nearby events, the metric tensor in the chosen normal coordinates is $g_{\alpha \beta }^{(\xi )}(\xi )={\eta }_{\alpha \beta }+O(|\xi {|}^2)$, with the quadratic corrections determined by the Riemann curvature tensor. Replacing $g^{(\xi )}$ by one constant Minkowski metric throughout an open region would make that region flat, so it cannot describe general curved spacetime exactly.

For a contravariant vector field $A^{\lambda }$, the covariant derivative isThe Levi-Civita connection is the unique torsion-free connection satisfying metric compatibility. Its Christoffel symbols in a coordinate basis areThere is generally no symmetry interchanging the upper index with a lower one. The Christoffel symbols themselves are not the components of a tensor.

An ordinary partial derivative of the components of a vector field includes the apparent change caused by the varying coordinate basis. The connection term corrects for that change, so ${\nabla }_{\nu }{A}^{\lambda }$ obeys the tensor component transformation law. It gives coordinate-independent ways to express parallel transport, geodesics and stress-energy conservation in curved spacetime.

Each lower index contributes a minus connection term to the covariant derivative of the metric tensor:Substitute the Christoffel symbols of the Levi-Civita connection and use the inverse metric identity $g_{\rho \nu }{g}^{\rho \sigma }={\delta }_{\nu }^{\sigma }$:Adding cancels the first and third terms and leaves exactly ${\partial }_{\lambda }{g}_{\mu \nu }$. HenceThis is metric compatibility: the Levi-Civita connection preserves the inner product under parallel transport.

The metric tensor in a basis ${\mathbf{e}}_a$ is $g_{ab}={\mathbf{e}}_a\cdot {\mathbf{e}}_b$. Differentiating this inner product with the product rule and using the assumed basis-derivative relation yieldsThus the coefficients satisfy metric compatibility. For a coordinate basis with zero torsion tensor, symmetrizing this relation in $a,c$, adding the corresponding two equations and subtracting the third givesThey are therefore the Christoffel symbols of the Levi-Civita connection. Metric compatibility by itself does not establish zero torsion tensor; that also uses the coordinate basis and its commuting coordinate derivatives.

For an embedded curved surface, the stated ambient basis-derivative relation needs a qualification: the Gauss formula is ${\partial }_c{\mathbf{e}}_a={\Gamma }_{ac}^b{\mathbf{e}}_b+h_{ac}\mathbf{n}$, where $\mathbf{n}$ is a unit normal and $h_{ac}$ is the second fundamental form. The normal term has zero inner product with each tangent basis vector, so the displayed identity for ${\partial }_c{g}_{ab}$ still holds. For a concrete counterexample to dropping that term, on a two-sphere one has ${\partial }_{\theta }{\mathbf{e}}_{\theta }=-R\mathbf{n}$, whereas every ${\Gamma }_{\theta \theta }^b$ is zero. Thus the intrinsic connection describes the tangential projection of the ambient derivative.

In standard spherical coordinates, parametrize the two-sphere byThe coordinate basis transforms from the fixed Cartesian basis according to ${\mathbf{e}}_a=(\partial x^i/\partial q^a){\mathbf{e}}_i$. ThereforeTheir inner products are ${\mathbf{e}}_{\theta }^2=R^2$, ${\mathbf{e}}_{\theta }\cdot {\mathbf{e}}_{\varphi }=0$ and ${\mathbf{e}}_{\varphi }^2=R^2{\sin }^2\theta$. Thus the induced metric isThe source's intermediate display places an extra $R^2$ outside entries already labelled $g_{\theta \theta },\dots$. Interpreting those entries as the actual metric tensor components would double-count the factor: for example, $R=2$ gives $g_{\theta \theta }=4$ directly from ${\mathbf{e}}_{\theta }^2$, not $16$. Its final matrix is the correct induced metric; the entries of its intermediate matrix must instead denote dimensionless coefficients.

The only nonconstant entry of the metric tensor is $g_{\varphi \varphi }=R^2{\sin }^2\theta$, whose partial derivative is ${\partial }_{\theta }{g}_{\varphi \varphi }=2R^2\sin \theta \cos \theta$. The nonzero Christoffel symbols areAlong a fixed latitude with $0<\theta <\pi$, parallel transport givesSet $X=V^{\theta }$ and $Y=\sin \theta V^{\varphi }$. Since $\theta$ is fixed, this linear ordinary differential equation becomesIt is a rotation at constant rate $-\cos \theta$. Its general solution, also valid at the equator where the rate is zero, isThe initial data give $A=1$ and $B=0$. At $\theta =\pi /3$, a complete circuit has $\varphi \cos \theta =\pi$, soThe physical squared length uses the metric tensor, rather than the Euclidean sum of coordinate components:It is constant and equals $R^2$ for these data. More generally, metric compatibility gives $dg(V,V)/d\varphi =({\nabla }_{{\partial }_{\varphi }}g)(V,V)+2g({\nabla }_{{\partial }_{\varphi }}V,V)=0$. The final change of direction is holonomy, exemplified by parallel transport around a latitude of the unit sphere; parallel transport preserves length without requiring the direction to return unchanged.

A photon follows a null geodesic, whose spatial path approaches straight lines far from the deflecting mass. Extend the incoming asymptote forward: this is the reference path the photon would follow if the mass were removed while its incoming direction and transverse offset were held fixed. The impact parameter $b$ is the perpendicular distance from the mass to that incoming asymptote. It is not, in general, the distance of closest approach.

The positive deflection angle $\delta$ is the angle between the outgoing propagation direction and the continued incoming direction. The photon bends toward the mass, so a ray passing above it has a downward outgoing slope in the figure. In the weak-field approximation for an isolated spherical mass, Schwarzschild light deflection gives $\delta \simeq 4GM/(b{c}^2)$ for distant endpoints. This angle differs from the apparent image angle measured from the observer's line of sight to the mass.

By spherical symmetry, choose the orbital plane to be $\theta =\pi /2$. The geodesic Lagrangian of the Schwarzschild metric isFor a massive test particle, dots denote proper time derivatives and $2\mathcal{L}=-c^2$. The cyclic coordinates $t$ and $\varphi$ give, through the Euler-Lagrange equations, two constants:At infinity, $\mathcal{E}={\gamma }_{\infty }=1/\sqrt{1-v_{\infty }^2/c^2}$ is the Lorentz factor. The constant $\ell$ is the specific angular momentum, and the energy per unit rest mass is $E_{\infty }/m=c^2{\gamma }_{\infty }$. The normalization of the four-velocity givesUse $dr/d\varphi =\dot{r}{r}^2/\ell$ and define the energy-normalized angular momentum for a Schwarzschild geodesicThenFor an incoming massive test particle with impact parameter $b$, $L_{\infty }=m{\gamma }_{\infty }{v}_{\infty }b$, so $J=b{v}_{\infty }$. In particular, $J$ is not the same as the specific angular momentum $\ell$ unless ${\gamma }_{\infty }=1$.

A photon has no proper time parametrization and no finite Lorentz factor. For it, use an affine parameter in the same geodesic Lagrangian and impose $2\mathcal{L}=0$. The ratio $J=c^2{L}_{\infty }/E_{\infty }=bc$ remains well defined, and the null result isEquivalently this is the ultrarelativistic limit of the massive formula at fixed $J$, with $1/{\gamma }_{\infty }^2\to 0$. The printed massive formula needs this interpretation before it is applied to a photon.

Let $m_g=GM/c^2$, $u=1/r$, and use a prime for $d/d\varphi$. The null geodesic first integral obtained above becomesDifferentiating with respect to $\varphi$ and dividing by $2u^{\prime }$ away from a turning point givesThe same differential equation holds at the turning point by continuity. When $m_g=0$, the general solution is $u_0=C_1\cos \varphi +C_2\sin \varphi$. Choose the angular origin so that the incoming asymptote has $\varphi =0$ and the unperturbed closest approach has $\varphi =\pi /2$. The impact parameter then fixesThis is the straight reference path, not a curved orbit.

For perturbation theory with $m_g/b\ll 1$, write $u=u_0+u_1+O(m_g^2/b^3)$, where $u_1=O(m_g/b^2)$. Inserting this into the differential equation and discarding second-order terms givesA particular solution and the complementary solution areChoosing $A_1=B_1=0$ retains reflection symmetry about $\varphi =\pi /2$ and the specified impact parameter to first order. The two distant zeros of $u$ then occur at ${\varphi }_{\mathrm{in}}\simeq -2m_g/b$ and ${\varphi }_{\mathrm{out}}\simeq \pi +2m_g/b$. Their separation exceeds $\pi$ by $4m_g/b$, consistently giving the Schwarzschild light deflection $\delta \simeq 4GM/(b{c}^2)$.

In the thin-lens approximation for gravitational lensing, consider a ray crossing the lens plane at transverse distance $b$ above the aligned source-observer axis. With the observer at axial position $0$, the black hole at $L$ and the source at $D$, the segment joining the lens plane to the source has slope $-b/(D-L)$, while the segment joining the observer to the lens plane has slope $b/L$. The observer sees the ray at apparent angle $\theta \simeq b/L$ to the axis. When following the photon toward the observer, the two small angles to the axis have magnitudes $b/(D-L)$ and $b/L$. The physical bend is their sum:Substitute the Schwarzschild light deflection, $\delta \simeq 4GM/(b{c}^2)$, and use $b\simeq L\theta$:Every azimuth around the axis is equivalent by rotational symmetry. Isotropic emission therefore supplies rays at all these azimuths, and the ideal image is an Einstein ring centered on the black hole, with Einstein angular radius $\theta$ and angular diameter $2\theta$. A finite source size or imperfect alignment broadens or breaks the ideal ring.

The stated expression uses small angles, distant endpoints and weak-field approximation $b\gg 2GM/c^2$, with ordinary Euclidean distances. For cosmological gravitational lensing, use angular diameter distances $D_l,D_s,D_{ls}$ in ${\theta }^2=4GM{D}_{ls}/(c^2{D}_l{D}_s)$; generally $D_{ls}\ne D_s-D_l$.

For the hypothetical Hubble flow, Hubble's law gives $v=H_{0}d$, where $d=4\times 10^5\mathrm{km}$ and $H_0$ is the Hubble constant. The accompanying constants sheet is not present in the supplied PDF, so retain its dependence explicitly. Taking $H_0=70\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$ as an illustrative input, and using one million parsecs equal to $3.08568\times 10^{19}\mathrm{km}$,A year of $365.25$ days contains $3.15576\times 10^7\mathrm{s}$. ThereforeEquivalently, $H_0=100h\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$ gives $v=4.09h\mathrm{cm}\mathrm{y}{\mathrm{r}}^{-1}$. This is a counterfactual recession rate for independently comoving bodies, rather than a prediction for the actual Earth-Moon orbit.

The Earth-Moon pair is a gravitationally bound system, so its separation does not obey $\dot{r}=H_{0}r$. The Hubble flow describes the motion of the approximately homogeneous cosmological matter distribution; it is not a compulsory extra velocity assigned to the members of every local orbit. Thus the numerical proximity of the counterfactual rate to $3.8\mathrm{cm}\mathrm{y}{\mathrm{r}}^{-1}$ is not evidence for that explanation.

The observed recession arises principally from tidal dissipation. The Earth rotates faster than the Moon orbits, so the displaced tidal response transfers angular momentum from the Earth's spin into the lunar orbit. For a nearly circular orbit, increasing orbital angular momentum increases its size. A model of this effect must include the appropriate tidal response, rather than add a universal Hubble flow recession.

A residual local cosmological tidal acceleration can exist, but its background scale is $|(\ddot{a}/a)r|\sim H_0^{2}r$, not an imposed first-order secular drift $H_{0}r$. At $r=4\times 10^8\mathrm{m}$,using $G(M_E+M_M)\simeq 4\times 10^{14}{\mathrm{m}}^3{\mathrm{s}}^{-2}$. The cosmological scale is about $10^{-24}$ of the mutual gravitational acceleration. Its detailed coefficient depends on how the local system is embedded in the cosmological background. Cosmological expansion can be neglected for the lunar recession calculation at the stated precision; this conclusion follows from binding and the tiny residual local cosmological tidal acceleration, rather than from comparing two recession speeds.

Let $t_0$ be the initial epoch, let $r(t)$ be the physical radius of the outer matter shell, and put $r(t_0)=r_0$. The background Einstein-de Sitter universe has critical density ${\rho }_b=3H_i^2/(8\pi G)$ at that epoch. Its mass inside $r_0$ is $M_b=M-\delta M$, where $M$ is the actual conserved mass. Hence the background Friedmann equation impliesFor spherical collapse of pressureless matter, with no shell crossing, the shell theorem gives $\ddot{r}=-GM/r^2$. Its conservation of energy first integral isConsequentlyFor $\delta M>0$, the initially expanding branch reaches cosmological turnaround at $r_{\mathrm{ta}}=A/B=M{r}_0/\delta M$. Since $dt=dr/\dot{r}$ on that branch, the elapsed cosmic time isThis is spherical turnaround from an unperturbed Hubble velocity. The integral measures the comoving matter observers' time interval in the pressureless model. The integration variable is the physical radius, not the radius divided by the background scale factor. If an absolute cosmic time is wanted, add $t_0$ to the displayed interval.

Set $u=Br/A$, retaining $r_0$ in $B$. Replacing only the lower integration limit by zero givesSubstitution of the constants yieldsTo quantify the approximation, write $ϵ=\delta M/M\ll 1$. The true lower limit is $u_0=B{r}_0/A=ϵ$, and an elementary integration by substitution, $u={\sin }^2\chi$, givesSince $\arcsin \sqrt{ϵ}-\sqrt{ϵ(1-ϵ)}=\frac{2}{3}{ϵ}^{3/2}+O({ϵ}^{5/2})$, the approximate time is too large by a relative fraction $\frac{4}{3\pi }{ϵ}^{3/2}+O({ϵ}^{5/2})$. This explains why changing the lower limit introduces little error for a small excess mass. The cosmological turnaround time diverges as $\delta M\to 0^{+}$, consistently with the absence of finite cosmological turnaround in an exactly critical Einstein-de Sitter universe.

There are two first-order effects: an intrinsic temperature change from adiabatic compression, and the gravitational redshift of the emerging photons. Evaluate the excess Newtonian gravitational potential at the boundary of the spherical fluctuation, with the exterior reference potential subtracted:In the weak-field approximation, gravitational time dilation gives $d{\tau }_{\mathrm{local}}/dt\simeq 1+\delta \Phi /c^2$. A local emitter's successive wave crests therefore arrive farther apart at a reference observer outside the potential well. Their frequency changes by $\delta \nu /\nu =\delta \Phi /c^2<0$. Since a redshifted black-body radiation spectrum has its temperature scaled by the same factor as its frequency, the gravitational contribution is

For the intrinsic effect, compare the local scale factor $R+\Delta R$ with the unperturbed value $R$ at the reference epoch. An equilibrium photon gas has energy density $u=a_{\mathrm{rad}}{T}^4$ and pressure $p=u/3$. For adiabatic compression, the first law of thermodynamics applied to a comoving volume $V\propto R^3$ givesThus $TR$ is constant along this adiabatic process, and the initial intrinsic contribution is $\Delta T/T=-\Delta R/R$. The smaller local expansion associated with the fluctuation leaves its radiation hotter; this is the compression or local expansion-history contribution, distinct from the clock-rate effect on the escaping radiation.

Multiplying the two first-order factors, and cancelling the common subsequent background redshift, givesFor the overdense region produced by adiabatic compression, $\Delta R<0$, so the final term is positive. It competes with the negative gravitational redshift term. Their sum is the intrinsic-plus-endpoint part of the Sachs-Wolfe combination; computing a full observed Cosmic microwave background anisotropy can also require Doppler CMB anisotropy and the Integrated Sachs-Wolfe effect.^[ref]