Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 309 (General Relativity)

The proper time elapsed along the smooth timelike curve isThe expression is invariant under every orientation-preserving reparametrization of the curve.

Write ${\dot{x}}^{\mu }=d{x}^{\mu }/ds$ and $L=\sqrt{-g_{\mu \nu }{\dot{x}}^{\mu }{\dot{x}}^{\nu }}$. Varying $T=\int Lds$ with fixed endpoints and then choosing proper time as parameter, for which $L=1$, gives the Euler-Lagrange equationExpanding the derivative, raising the free index with the inverse metric, and using the symmetry of ${\dot{x}}^{\alpha }{\dot{x}}^{\beta }$ gives the geodesic equationThese are the Christoffel symbols of the Levi-Civita connection.

For the quadratic geodesic Lagrangianthe Euler-Lagrange equation isRaising $\mu$ and symmetrizing the coefficient of the two velocities produces exactlyThe quadratic action fixes an affine parameter; proper time is an affine parameter for a timelike geodesic.

Put $u=t-z$. Wherever $|u|>U$, the profile $A(u)$ vanishes and the displayed metric is exactly the Minkowski metric. Thus the two flat regions in the $(t,z)$-plane are the half-planes $t-z>U$ and $t-z<-U$, separated by the stripAt fixed $t$ this strip is $t-U\le z\le t+U$, so it has longitudinal width $2U$. A surface $u=\text{constant}$ obeys $z=t-u$ and travels in the positive $z$ direction at the speed of light. The curvature is confined to that moving strip, identifying the solution as a gravitational-wave pulse represented by a plane-fronted gravitational wave.

With $u=t-z$, the quadratic geodesic Lagrangian isIts transverse Euler-Lagrange equations areThe longitudinal equations areafter using the difference of those same equations. In particular,This conserved quantity also follows from the Killing vector field ${\partial }_t+{\partial }_z$ and the geodesic conserved quantity from a Killing vector.

The initial data and $\ddot{u}=0$ give $u=\tau$. To first order in $ϵ$, replace $x$ and $y$ on the right-hand sides of the transverse equations by $x_0$ and $y_0$. Twice integrating with the initial rest conditions givesFor $\tau >U$, define $A_0={\int }_{-\infty }^{\infty }a(s)ds$ and $A_1={\int }_{-\infty }^{\infty }s,a(s)ds$. ThenHenceThese permanent changes are forms of displacement memory and velocity memory.

There is a discrepancy in the question's final instruction. The longitudinal equation actually givesand thereforeThus $z$ does not vanish to first order for a general allowed profile. It vanishes after the pulse under the additional hypothesis $A_0=0$ used in part (iv), but it need not vanish while that pulse is passing. The proper-time normalization independently gives the same relation $\dot{z}=xyA+O({ϵ}^2)$.

Now $A_0=0$, so every mass is again at rest after the pulse and $z=0+O({ϵ}^2)$. Put $k=ϵA_1>0$. The final transverse coordinates areThe unit vectors along $x=y$ and $x=-y$ are eigenvectors with eigenvalues $1-k$ and $1+k$, respectively. Consequently an initial circle becomes, to first order, an ellipse compressed along the line $x=y$ and stretched along $x=-y$. This persistent deformation is displacement memory.

For any vector field $Y$, apply the Leibniz rule to the scalar $\omega (Y)$:Using ${\mathcal{L}}_{V}Y=[V,Y]$ and expanding the Lie bracket in a coordinate chart leavesThe connection terms cancel because the Levi-Civita connection is torsion-free. Applying this formula to each slot of the metric tensor and using metric compatibility givesEquivalently, one may prove both identities at a point in normal coordinates; since both sides are tensors, the result then holds in every coordinate system.

The trace of the electromagnetic stress-energy tensor in four spacetime dimensions isFor its covariant divergence, the source-free Maxwell equations eliminate the derivative of the first factor. Contracting the Bianchi identity ${\nabla }_{[a}{F}_{bc]}=0$ with $F^{ac}$ giveswhich cancels the derivative of the trace term. HenceFor $J^a=T^{ab}{V}_b$, stress-energy conservation and symmetry of $T^{ab}$ now imply

At the chosen event use the orthonormal frame in spacetime from the hint. The electromagnetic energy density isand the energy flux is the Poynting vector $\mathbf{S}=\mathbf{E}\times \mathbf{B}$. Depending on the index convention, the required contraction is $\rho -w{S}_1$ or $\rho +w{S}_1$. In either case,Rescaling $V$ and rotating the spatial frame covers arbitrary future timelike $V$ and future causal $W$. Thus the Maxwell field satisfies the dominant energy condition.

Since $t=u+r$, the hypersurface ${\Sigma }_t$ is an ordinary constant-Minkowski-time slice. Its future unit normal isin the $(u,r,\theta ,\varphi )$ basis. Lowering the index gives $n_a=(-1,-1,0,0)=-{\nabla }_{a}t$. Restricting the metric to $dt=du+dr=0$ gives the Euclidean spherical metricso its induced Riemannian volume form is

For $V=r{\partial }_r$, Cartan's magic formula giveswhereas contraction and exterior differentiation both vanish for $du,d\theta ,d\varphi$. ThusApplying the Lie derivative to
$g=-d{u}^2-2,du,dr+r^{2}d{\Omega }^2$ yieldsHere $n_a=-(du+dr{)}_a$ and $V_a=-r(du{)}_a$, so comparison with $2g$ givesUsing tracelessness from part (b)(i),by the dominant energy condition, because $n$ is future timelike and $V$ is future null.

The integrand defining $\epsilon (t)$ is nonnegative by the dominant energy condition, henceApply the spacetime divergence theorem to $J^a=T^{ab}{V}_b$ on the slab $S$. The flux through the cylinder at $r\to \infty$ vanishes by the stated decay. With the Lorentzian boundary orientation, the two spacelike boundary contributions givePart (ii) makes the right-hand side nonnegative, so for $t_1\ge t_0$,Therefore $\epsilon$ is a nonnegative monotone decreasing energy functional.

Put $h_{ab}=\delta g_{ab}$. Varying $g^{ac}{g}_{cb}={\delta }_b^a$ givesThe determinant identity $\delta \log |\det g|=g^{ab}{h}_{ab}$ givesFinally, varying the formula for the Christoffel symbols and rewriting partial derivatives covariantly gives the tensorThese are the basic metric variation identities.

Varying the Ricci tensor and contracting givesSubstitution of part (i), followed by metric compatibility, reduces the divergence toThus the metric variation of scalar curvature isand consequently

Vary the gravitational volume factor and use part (a)(ii):Twice applying integration by parts moves both derivatives in $\delta R$ from the compactly supported $h_{ab}$ onto $f^{\prime }(R)$. Combining the result with the stated matter variation and requiring every coefficient of $h_{ab}$ to vanish gives the f(R) gravity equationTherefore

For Minkowski spacetime, $R_{ab}=0$ and $R=0$, while $f^{\prime }(R)$ is constant. In vacuum $T_{ab}=0$, so every derivative term vanishes and the remaining term is $-\frac{1}{2}{\eta }_{ab}f(0)=0$. Hence Minkowski spacetime is a vacuum solution of this modified gravity theory.

Because $R=O(ϵ)$ andthe Taylor series gives $f(R)=R+O({ϵ}^3)$ and $f^{\prime }(R)=1+O({ϵ}^2)$. The extra derivative terms in the f(R) equation are therefore at least second order. At first order the theory reduces to the Linearized Einstein equationsIn harmonic coordinates, equivalently the Lorenz gauge in linearized gravity ${\partial }_{\mu }{\bar{h}}^{\mu }{}_{\nu }=0$, the trace-reversed metric perturbationsatisfiesThus Linearized f(R) gravity about Minkowski spacetime has the same weak-field predictions as general relativity under these assumptions. Differences can first appear through nonlinear curvature terms or in backgrounds about which $f^{\prime \prime }(R)$ is nonzero.

Let $v=\sqrt{2m/r}$. The proposed coframe giveswhich expands to the stated metric. Hence it is the orthonormal coframe in Painlevé–Gullstrand coordinates with signature $(-,+,+,+)$.

SetDirect exterior differentiation givesSolving Cartan's first structure equation and imposing ${\omega }_{\mu \nu }=-{\omega }_{\nu \mu }$ gives the independent mixed-index connection 1-formsThe remaining forms follow from Lorentz-signature antisymmetry: ${\omega }^1{}_0={\omega }^0{}_1$, ${\omega }^2{}_0={\omega }^0{}_2$, ${\omega }^3{}_0={\omega }^0{}_3$, and ${\omega }^j{}_i=-{\omega }^i{}_j$ for spatial indices.

WriteFor a diagonal curvature operator, ${\Theta }^{\mu \nu }$ contains only $f^{\mu }\wedge f^{\nu }$. Consequently $R^{\mu }{}_{\nu \rho \sigma }$ can be nonzero only when the plane indexed by $\rho ,\sigma$ is the same as the plane indexed by $\mu ,\nu$. In the contractionthis condition cannot hold for $\nu \ne \sigma$. Thereforewhich is the statement that a diagonal curvature operator implies diagonal Ricci tensor.

Substituting the forms from part (b) into Cartan's second structure equation gives the independent mixed-index curvature 2-formsEvery form is proportional to its corresponding basis 2-form, so the curvature operator is diagonal. The diagonal Ricci tensor components are contractions of these sectional curvature coefficients. In each case the coefficients cancel in the pattern $2-1-1=0$, with the Lorentzian sign included when the time direction is contracted. HenceThis is consistent with Schwarzschild spacetime being a vacuum solution away from $r=0$.

The curvature 2-forms show that orthonormal curvature components grow as $m/r^3$. The geodesic deviation equation converts them into the relative acceleration of nearby parts of an observer. As $r\to 0$, the Schwarzschild tidal force stretches radial separations and compresses transverse separations with unbounded magnitude. An extended body therefore undergoes destructive tidal deformation before reaching the Schwarzschild singularity.