Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 56 (General Relativity)

A covariant derivative on the tangent bundle is an affine connection: it assigns a vector field ${\nabla }_{X}Y$ to vector fields $X,Y$, is linear over smooth functions in $X$, is real-linear in $Y$, and obeys the Leibniz rule ${\nabla }_X(aY)=X(a)Y+a{\nabla }_{X}Y$. Thus it differentiates a vector field while taking account of how its local basis changes. It extends to tensor fields by the Leibniz rule, compatibility with tensor contraction, and ${\nabla }_{X}a=X(a)$ for a scalar.

In a local basis $e_{\mu }$, the connection components are defined by ${\nabla }_{e_{\rho }}{e}_{\nu }={\Gamma }^{\mu }{}_{\nu \rho }{e}_{\mu }$. In a coordinate basis, $e_{\rho }={\partial }_{\rho }$, soA covariant derivative of a tensor field has tensorial transformation properties, but the connection components themselves do not: changing the coordinate basis introduces second derivatives of the coordinate change. They describe the connection in a chosen basis, rather than components of a tensor.

The torsion tensor of an affine connection is $\mathcal{T}(X,Y)={\nabla }_{X}Y-{\nabla }_{Y}X-[X,Y]$, where the last term is the Lie bracket of vector fields. A torsion-free connection has $\mathcal{T}=0$. Since a coordinate basis commutes, this is equivalent there to ${\Gamma }^{\mu }{}_{\nu \rho }={\Gamma }^{\mu }{}_{\rho \nu }$; in a general basis its commutator must also be included.

The Levi-Civita connection is the unique torsion-free connection with metric compatibility, $\nabla g=0$. To derive its Christoffel symbols, lower the first index: ${\Gamma }_{\sigma \nu \rho }=g_{\sigma \mu }{\Gamma }^{\mu }{}_{\nu \rho }$. Metric compatibility givesWrite down this identity with the derivative indices $\rho ,\nu ,\sigma$, add the first two identities and subtract the third. Symmetry in the last two indices, supplied by the torsion-free connection, cancels all terms except twice ${\Gamma }_{\sigma \nu \rho }$. Raising the first index yieldsThis calculation proves uniqueness. Conversely, these Christoffel symbols are symmetric in $\nu ,\rho$ and satisfy $\nabla g=0$ by substitution, proving existence as well.

Work on a connected regular radial interval, where $f$ is smooth and nonzero; replacing $f$ by $|f|$ allows us to take $f>0$. The only nonzero Christoffel symbols areAlong the circle, write a tangent vector in the single-valued orthonormal basis $E_r={\partial }_r$, $E_{\varphi }=f^{-1}{\partial }_{\varphi }$ as $V=u{E}_r+v{E}_{\varphi }$. The equations for parallel transport reduce toConsequently $u+iv$ acquires the factor $e^{-2\pi i{f}^{\prime }(r_0)}$ after one circuit. The basis itself agrees at $\varphi =0$ and $2\pi$, so no extra basis change is missing. The holonomy is the identity on every tangent vector precisely whenThe first condition makes this a regular circle of the Riemannian metric; the second is the actual transport condition. In particular, zero slope is sufficient but not necessary. This is holonomy around circular fibres.

If every regular circle has trivial holonomy, the preceding calculation says that the continuous function $f^{\prime }$ takes values in $\mathbb{Z}$. On a connected radial interval it must therefore be constant. Integrating gives the complete familyConversely, every such function has identity holonomy around every circle. With the opposite sign convention for $f$, replace the positivity condition by nonvanishing; the Riemannian metric is unchanged by changing its sign.

The slope parameter is discrete. Thus the printed parameter count can mean one integer and one real parameter, but it cannot mean two freely varying real parameters. Allowing $f=ar+b$ with arbitrary real $a$ gives a locally flat family, not trivial circular holonomy: for example $a=1/2$ gives a rotation by $\pi$. This distinction is captured by flat circular metrics with trivial holonomy.

For $n\ne 0$, put $\rho =f(r)/|n|$ and $\theta =|n|\varphi$. Then $d{\rho }^2=d{r}^2$ andThese give a local isometry wherever $\rho >0$. For $n=0$, $f=b>0$, and the local coordinates $X=r$, $Y=b\varphi$ similarly give the Euclidean plane metric tensor. Equivalently, the Gaussian curvature of the original Riemannian metric is $-f^{\prime \prime }/f=0$ throughout this family.

A local isometry need not be a global isometry. If $|n|=1$, the displayed polar map is injective modulo $\varphi \sim \varphi +2\pi$ and identifies the regular domain with a Euclidean annulus, possibly a punctured disc or the punctured Euclidean plane, according to the radial interval. If $|n|>1$, it is a $|n|$-fold covering of that annulus: distinct angles differing by $2\pi /|n|$ have the same image. This developing map cannot be made injective by a different choice of Euclidean coordinates, since two local Euclidean isometries on a connected flat domain differ by a rigid motion once their value and differential are fixed at one point. For $n=0$, the domain is a flat cylinder; its circular curves are closed geodesics, whereas the Euclidean plane has no nonconstant closed geodesic.

All these metrics are locally Euclidean; they are not all globally Euclidean. Even the regular $|n|=1$ domain is an annular region rather than the entire Euclidean plane. A regular axis can be added without a conical singularity only when $|n|=1$; with the full radial range this completion becomes the Euclidean plane. Trivial holonomy alone therefore does not supply a global injective Euclidean chart.

Describe the geodesic variation by a smooth map $F(s,\tau )$, with each curve $\tau \mapsto F(s,\tau )$ a timelike geodesic parametrized by proper time. Its tangent vector is $T=F_{\ast }{\partial }_{\tau }$ and its deviation vector is $S=F_{\ast }{\partial }_s$. Thus $S$ measures the first-order separation of neighboring members at the same value of proper time.

Mixed coordinate derivatives on the parameter surface commute. Consequently, along the variation,For an embedded two-dimensional variation this is the usual Lie bracket of vector fields on its image; more generally the same identity is read along the map $F$ using its commuting parameter derivatives. Since the Levi-Civita connection is torsion-free, it follows that ${\nabla }_{T}S={\nabla }_{S}T$.

Use precisely the stated convention for the Riemann curvature tensor, and let $D={\nabla }_T$. The geodesic equation says ${\nabla }_{T}T=0$ because proper time is an affine parameter. The commuting variation and the torsion-free connection giveThus the geodesic deviation equation isHere $(R(X,Y)Z{)}^a=R^a{}_{bcd}{Z}^b{X}^c{Y}^d$. Declaring this index convention fixes the sign: writing $R(S,T)T$ instead would require a minus sign. A deviation vector satisfying this equation is a Jacobi field.

For every member of the geodesic variation, proper time normalization gives $g(T,T)=-1$. By metric compatibility, the geodesic equation, and ${\nabla }_{T}S={\nabla }_{S}T$,The last equality uses normalization across the whole variation, not merely constancy along one curve. Hence $g(S,T)$ is constant along each timelike geodesic, provingThis is orthogonality of proper-time deviation vectors.

Contracting the assumed Riemann curvature tensor gives the Ricci tensorThe contracted Bianchi identity and metric compatibility now implyThus ${\nabla }_{b}R=0$, so the scalar curvature is constant on each connected component of spacetime. The sectional curvature is consequently the constant $\kappa =R/12$. This is the four-dimensional case of the Schur theorem in pseudo-Riemannian geometry.

Let $\kappa =R/12$, which is constant by the preceding argument. Substitution into geodesic deviation givesusing $g(T,T)=-1$ and the established orthogonality. With $V=DS$, metric compatibility yields $\dot{f}=2g(S,V)$ andThereforeBoth $S$ and $DS$ are orthogonal to $T$, so their squared norms are nonnegative. The minus sign in this conserved deviation energy in constant curvature reflects $D^{2}S=+\kappa S$, rather than an ordinary oscillator when $R>0$.

Differentiate $f=g(S,S)$ twice, use geodesic deviation, and eliminate $g(DS,DS)$ with the conserved $K$:The most informative way to interpret this linear differential equation is to express $S$ in a parallel-propagated orthonormal frame along the timelike geodesic. Its three spatial components satisfy ${\ddot{S}}^i=\kappa S^i$.

If $R>0$, put $H=\sqrt{R/12}$. With constant spatial vectors $A,B$ in that frame,Thus generic deviations grow in length like $e^{H\tau }$, and their squared separation grows like $e^{2H\tau }$. Initially comoving neighbors, $DS(0)=0$, have $S(\tau )=S(0)\cosh (H\tau )$ and exhibit this growth. The unqualified printed assertion is too strong: the nonzero Jacobi field $S=B{e}^{-H\tau }$ has $K=0$ and shrinks exponentially. Such a Jacobi field is realizable by varying initial position and initial unit timelike velocity, with initial relative velocity $DS(0)=-HB$.

If $R<0$, put $\omega =\sqrt{-R/12}$. ThenThe deviations are bounded and oscillatory, rather than exponentially growing. The squared separation is periodic with period $\pi /\omega$ or is constant. Initially comoving neighbors have $S(\tau )=S(0)\cos (\omega \tau )$ and focus at $\tau =\pi /(2\omega )$; arbitrary vector initial data need not focus together. For completeness, $R=0$ gives $S=A+B\tau$ and $f=|A+B\tau {|}^2$. These conclusions concern infinitesimal geodesic deviation on the domain of the variation, not arbitrarily large finite separations.

Keep only first order in the metric perturbation and use the Minkowski metric to raise and lower indices; set $G=c=1$. The Levi-Civita connection hasThe quadratic products of Christoffel symbols in the Riemann curvature tensor can be dropped at this order. Contracting the remaining derivatives giveswhere $\square ={\eta }^{\rho \sigma }{\partial }_{\rho }{\partial }_{\sigma }=-{\partial }_t^2+{\nabla }^2$ is the d'Alembert operator. In four dimensions the trace-reversed metric perturbation satisfies $\bar{h}=-h$ and $h_{\mu \nu }={\bar{h}}_{\mu \nu }-\frac{1}{2}{\eta }_{\mu \nu }\bar{h}$. Writing the Linearized Einstein equations in terms of it givesIn Lorenz gauge in linearized gravity, the first, second and fourth terms vanish. Therefore $G_{\mu \nu }^{(1)}=8\pi T_{\mu \nu }$ becomesThe divergence of this equation also gives ${\partial }^{\mu }{T}_{\mu \nu }=0$, the leading-order stress-energy conservation required by the contracted Bianchi identity.

For a nonzero, nonconstant vacuum plane gravitational wave in linearized gravity with real wave covector $k$, the Linearized Einstein equations and Lorenz gauge in linearized gravity requireThe zero-amplitude solution places no restriction on $k$; a constant mode $k=0$ is not a propagating wave. Under the sign convention for the infinitesimal change specified here, the trace-reversed metric perturbation changes byThus the residual gauge symmetry of linearized gravity is characterized by $\square {\xi }_{\nu }=0$. The proposed plane-wave form obeys this for every constant complex $X_{\nu }$, because $k$ is null. Changing the sign used to name $\xi$ would reverse both gauge formulas, with no physical consequence.

Rotate the spatial axes so the wave propagates along $+z$, choosing $k_{\mu }=(-\omega ,0,0,\omega )$ with $\omega >0$. The Lorenz gauge in linearized gravity says $H_{3\nu }=-H_{0\nu }$; in particular $H_{03}=-H_{00}$, $H_{33}=H_{00}$ and $H_{3i}=-H_{0i}$ for $i=1,2$. The amplitude change isChooseIndeed, $\delta H_{00}=i\omega (X_3-X_0)$ and $\delta H_{0i}=-i\omega X_i$ cancel the time components, while $\delta H_{11}=\delta H_{22}=-i\omega (X_0+X_3)$ cancels the transverse trace. The preserved Lorenz gauge in linearized gravity then cancels all longitudinal components. This is an explicit plane-wave reduction to transverse-traceless gauge, yieldingThe trace-reversed metric perturbation has zero trace in this transverse-traceless gauge, so $h^{\prime }={\bar{h}}^{\prime }$. The null dispersion relation is ${\omega }^2=|\mathbf{k}{|}^2$, giving the speed of light. Only components perpendicular to the direction of propagation remain, and the two independent amplitudes give the plus and cross gravitational wave polarizations.

These are physical transverse tidal distortions, rather than just a convenient display of the metric perturbation: in transverse-traceless gauge, $R_{0i0j}^{(1)}=-\frac{1}{2}{\partial }_t^2{h}_{ij}^{\prime }$, so a freely falling detector has ${\ddot{S}}^i=\frac{1}{2}{\ddot{h}}_{ij}^{\prime }{S}^j$ at first order. There is no longitudinal tidal acceleration. The plus polarization stretches one transverse axis while compressing the other; the cross polarization does the same along axes rotated by $\pi /4$.

The equivalence principle permits Riemann normal coordinates at any event, in which the first derivatives of the metric tensor, and hence the Christoffel symbols, vanish. A putative universal local gravitational stress-energy tensor constructed from the metric and its first derivatives would vanish in that freely falling description. If it were a genuine tensor, it would then vanish in every description, even where other coordinates assign nonzero gravitational energy density. The gravitational field therefore does not have a local stress-energy tensor analogous to that of matter, with coordinate-independent energy density and the usual energy-balance interpretation.

Gravitational energy is not represented by a universal local covariant stress-energy tensor. This does not assert that all local curvature tensors vanish or are forbidden: the Riemann curvature tensor measures tidal gravity, and higher-curvature tensors can be constructed. They do not provide the sought ordinary local gravitational energy density. Boundary charges, quantities associated with a background, and suitable averaged wave energies have different definitions and evade that local requirement. This is the absence of a local gravitational stress-energy tensor.

Expand the Einstein tensor about the Minkowski metric, keeping a second-order metric perturbation as well:Here $G^{(2)}[h,h]$ includes all quadratic terms from the Einstein tensor, including the inverse-metric corrections in its contractions. In vacuum, the first-order Linearized Einstein equations give $G^{(1)}[h]=0$, while the next order gives $G^{(1)}[j]=-G^{(2)}[h,h]$. Define the quadratic gravitational effective stress-energy tensor byIt is symmetric because the Einstein tensor is symmetric, and is quadratic in $h$ and its derivatives. The linearized contracted Bianchi identity is the off-shell identity ${\partial }^{\mu }{G}_{\mu \nu }^{(1)}[j]=0$. The second-order vacuum equation therefore impliesOne can obtain the same conclusion directly by expanding the full contracted Bianchi identity to second order: every correction to the flat divergence multiplies $G^{(1)}[h]$ and vanishes on the first-order vacuum solution. Thus conservation holds for vacuum solutions of linearized gravity, rather than for an arbitrary off-shell $h$. Relative to the fixed Minkowski background, $t$ is a tensor under background changes of coordinates, but this does not make it invariant under the perturbative gauge freedom.

The quadratic gravitational effective stress-energy tensor depends on the split between the fixed background and the metric perturbation. A residual gauge symmetry of linearized gravity changes $h$ without changing the first-order tidal field, but generally changes $G^{(2)}[h,h]$ and hence the local value of $t$. At second order the compensating change of $j$ restores the same physical geometry; assigning only the quadratic term to an energy density loses this compensation.

Moreover, a stress-energy superpotential can change the local energy expression while preserving its divergence and, with suitable boundary behavior, its integrated charges. Symmetry and flat conservation do not make $t$ a unique, gauge-independent local gravitational energy density. Averaging in an appropriate short-wavelength regime can lead to a useful physical gravitational-wave energy flux, but that is a qualified approximation, not a cure for the local definition requested here.

A Killing vector field $\xi$ generates a local one-parameter family of isometries of the metric tensor. Equivalently, its tensor Lie derivative of the metric vanishes:The latter is the Killing equation; the equivalence uses metric compatibility and the torsion-free connection.

Let $u$ be the tangent vector to a geodesic with affine parameter $\lambda$. Its geodesic equation is ${\nabla }_{u}u=0$. For a Killing vector field $\xi$, metric compatibility givesThusThis geodesic conserved quantity from a Killing vector is independent of whether the geodesic is timelike, spacelike or null. For a timelike geodesic one can use proper time; for a null geodesic choose an affine parameter. An arbitrary nonaffine tangent rescales the displayed quantity and need not make it constant.

The metric tensor of the planar warped spacetime depends only on $z$. The three translations therefore preserve it, givingThey are linearly independent Killing vector fields everywhere. The associated geodesic conserved quantities from Killing vectors are $-A^2\dot{t}$, $A^2\dot{x}$ and $A^2\dot{y}$, for an affine parameter; the conventional positive energy is $E=A^2\dot{t}$. No restriction on the positive function $A$ is needed for these three symmetries.

The displayed $e^a$ are one-forms, so they form an orthonormal coframe, dual to the orthonormal frame $E_0=A^{-1}{\partial }_t$, $E_1=A^{-1}{\partial }_x$, $E_2=A^{-1}{\partial }_y$, $E_3={\partial }_z$. Use ${\eta }_{ab}=\mathrm{diag}(-1,1,1,1)$ and abbreviateFor $i=0,1,2$, $d{e}^i=a{e}^3\wedge e^i$, while $d{e}^3=0$. The Levi-Civita connection is fixed by Cartan's first structure equation together with Lorentzian connection-form antisymmetry, ${\omega }_{ab}=-{\omega }_{ba}$, where ${\omega }_{ab}={\eta }_{ac}{\omega }^c{}_b$. Its nonzero connection 1-forms areIn particular, the time-index pair has the same sign with one index raised; incorrectly imposing ${\omega }^a{}_b=-{\omega }^b{}_a$ would lose the Lorentzian sign. All diagonal and all other entries vanish.

Apply Cartan's second structure equation. For $i=0,1,2$,For distinct $i,j$ in $\{0,1,2\}$, putting $s_0=-1$ and $s_1=s_2=1$ gives ${\Theta }^i{}_j={\omega }^i{}_3\wedge {\omega }^3{}_j=-s_j{a}^2{e}^i\wedge e^j$. Thus a complete independent set of curvature 2-forms isAll diagonal entries vanish. The remaining entries follow from ${\Theta }_{ab}=-{\Theta }_{ba}$: ${\Theta }^j{}_i=-s_i{s}_j{\Theta }^i{}_j$ for $i,j\le 2$, and ${\Theta }^3{}_i=-s_i{\Theta }^i{}_3$. These relations and the displayed six entries specify all sixteen curvature 2-forms. This computation is Cartan curvature of a planar warped spacetime.

Extract the Riemann curvature tensor from ${\Theta }^a{}_b=\frac{1}{2}{R}^a{}_{bcd}{e}^c\wedge e^d$ using the curvature 2-forms just computed. Contraction $R_{bd}=R^a{}_{bad}$ gives the diagonal Ricci tensor in the orthonormal coframe:For example, the $00$ component receives $a^2$ from each of the $1$ and $2$ directions and $q$ from direction $3$. The signs in the spatial components reflect the same Lorentzian index lowering used for the connection 1-forms.

For a null vector $k$ in this frame, $(k^0{)}^2=(k^1{)}^2+(k^2{)}^2+(k^3{)}^2$. ThereforeThe Einstein field equations imply $8\pi T_{ab}{k}^a{k}^b=R_{ab}{k}^a{k}^b$, since the metric term vanishes for a null vector; an included cosmological-constant term would also vanish. Choose $k^a=(1,0,0,1)$ and apply the null energy condition. It follows thatConversely, this inequality makes the same contraction nonnegative for every null vector, so it is exactly the null energy condition for a planar warped spacetime when its matter stress-energy tensor is defined by the Einstein field equations. Null vectors tangent to the planar slices saturate the condition. As checks, constant $A$ gives flat spacetime and $A=e^{bz}$ gives constant negative sectional curvature with $a^{\prime }=0$, also saturating it.