Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 52 (General Relativity)

Choose local coordinates $y^1,\dots ,y^{D-1}$ on the hypersurface $\Sigma$, and let ${\Phi }_s$ be the local flow of the vector field $X$. The map $F(s,y)={\Phi }_s(p(y))$ follows the integral curves of a vector field starting at $p(y)\in \Sigma$. At $s=0$, its differential maps the $s$ direction to $X$ and the remaining directions to a basis of $T_p\Sigma$. Transversality makes these vectors linearly independent, so the inverse function theorem makes $F$ a local manifold chart.

Set $x^0=s$ and transport the $y^i$ unchanged along each integral curve of a vector field. Then the adapted coordinates satisfyThe flow-box theorem gives this construction locally. The parameter is the flow parameter, and need not be arc length; a global manifold chart would require additional hypotheses.

In a stationary spacetime chart, take $V={\partial }_0$. Its squared length is $g_{00}<0$, so $V$ is a timelike vector. Since its components are constant, the Lie derivative of a tensor field obeys $({\mathcal{L}}_{V}g{)}_{\mu \nu }={\partial }_0{g}_{\mu \nu }=0$. For the Levi-Civita connection,so $V$ satisfies the Killing equation.

Conversely, a timelike Killing vector field is nowhere zero. Apply the flow-box theorem to choose coordinates with $V={\partial }_0$. The Killing equation then gives ${\partial }_0{g}_{\mu \nu }=0$, and $g_{00}=g(V,V)<0$. Thus local stationarity is equivalent to the existence of a timelike Killing vector field.

Use the standard meaning of static spacetime: a stationary spacetime chart whose time-space components also vanish. The printed static definition needs to inherit the timelike requirement $g_{00}<0$; without it, the requested timelike conclusion does not follow.

The preceding argument shows that $V={\partial }_0$ is a timelike Killing vector field. Its metric-dual one-form is $v=V_{\mu }d{x}^{\mu }=g_{00}d{x}^0$. Its exterior derivative satisfiesFor a torsion-free connection, antisymmetrizing the covariant derivative of a one-form gives its exterior derivative. Consequently the Killing vector has hypersurface orthogonality:Geometrically, the hypersurfaces $x^0=\mathrm{constant}$ are orthogonal to $V$.

Write $F=V_{\rho }{V}^{\rho }<0$ and $A_{\mu \nu }={\nabla }_{\mu }{V}_{\nu }$. The Killing equation makes $A_{\mu \nu }$ antisymmetric. Expanding hypersurface orthogonality and contracting with $V^{\alpha }$ givesHere metric compatibility gives ${\partial }_{\mu }F=2V^{\rho }{A}_{\mu \rho }$. Thereforeand the normalized Killing one-form satisfiesThus the required exponent is $\boxed{n=-2}$. The notation $|V{|}^2$ in this paper denotes the negative scalar $F$, rather than a positive squared norm; the answer uses only its real reciprocal. More generally, for a multiplier $h(F)$ the antisymmetric derivative is $(h/F+h^{\prime })(V_{\nu }{\partial }_{\mu }F-V_{\mu }{\partial }_{\nu }F)$, so $h=C/F$ is the universally valid choice. If $F$ is constant, other exponents can work in that particular metric.

The normalized Killing one-form $\eta =F^{-1}{V}_{\mu }d{x}^{\mu }$ is a closed differential form. The Poincaré lemma gives a local scalar potential $\varphi$ with $\eta =d\varphi$, hence $V_{\mu }=F{\partial }_{\mu }\varphi$. Choose the adapted chart of part (i), so $V={\partial }_0$. ThenThe closed differential form condition supplies the integrability of these spatial derivatives. For a fixed $g$ and $V$, $f$ is determined locally up to an additive constant; its derivatives cannot be chosen freely despite the wording in the paper.

Use ${\bar{x}}^0=\varphi$, ${\bar{x}}^i=x^i$. The Jacobian determinant is one, so this is a local coordinate change. Holding ${\bar{x}}^i$ fixed gives $V={\partial }_{\bar{0}}$, while its metric-dual one-form is $v=Fd{\bar{x}}^0$. Thus ${\bar{g}}_{00}=F<0$ and ${\bar{g}}_{0i}=0$. The Killing equation still holds, so ${\partial }_{\bar{0}}{\bar{g}}_{\mu \nu }=0$. The metric is locally static:A closed differential form need not be globally exact; neither a global potential nor a global static chart follows without additional topological assumptions.

The contracted Bianchi identity gives ${\nabla }^a{R}_{ab}=\frac{1}{2}{\nabla }_{b}R$. Combining the proposed equation with stress-energy conservation therefore imposes ${\nabla }_{b}R=0$. Taking its trace gives $R=\kappa T$, with $T=g^{ab}{T}_{ab}$, so for nonzero $\kappa$ the matter trace must also be constant. Ordinary matter need not satisfy this extra condition. For example, if asymptotic flatness forces the constant scalar curvature to vanish, the proposal would require $T=0$ everywhere.

The problem is an unjustified restriction on the matter trace, rather than the absence of every possible solution. The divergence-free Einstein tensor avoids it:The Einstein field equations $G_{ab}+\Lambda g_{ab}=\kappa T_{ab}$, with constant cosmological constant $\Lambda$, are consequently compatible with stress-energy conservation.

Contract the Christoffel symbol formula for the Levi-Civita connection. The first and third derivative terms cancel after relabelling their summed indices:The Jacobi determinant derivative formula gives ${\partial }_{\mu }g=g{g}^{\rho \sigma }{\partial }_{\mu }{g}_{\rho \sigma }$. In a Lorentzian chart $g<0$, and hence ${\partial }_{\mu }\sqrt{-g}=\frac{1}{2}\sqrt{-g}{g}^{\rho \sigma }{\partial }_{\mu }{g}_{\rho \sigma }$. Thus the trace is the logarithmic derivative of the volume density:In particular, the covariant divergence formula for a vector field is ${\nabla }_{\mu }{A}^{\mu }=(-g{)}^{-1/2}{\partial }_{\mu }(\sqrt{-g}{A}^{\mu })$.

Vary the inverse metric tensor while holding the matter fields fixed. Differentiating $g_{a\mu }{g}^{\mu b}={\delta }_a{}^b$ gives $\delta g_{ab}=-g_{a\mu }{g}_{b\nu }\delta g^{\mu \nu }$. The Jacobi determinant derivative formula consequently implies the inverse metric volume variationSince the matter Lagrangian in this question contains no derivatives of the metric tensor, its metric variation is algebraic. The product rule givesMultiplying by $-2/\sqrt{-g}$ proves the equivalence of the stress-energy definitions:The variations of the symmetric metric tensor are understood symmetrically. If the matter action contained metric derivatives, the definition would instead require a functional derivative and the corresponding integration by parts.

For the minimally coupled scalar field, keep the metric tensor fixed and vary $\varphi$. A scalar's first covariant derivative equals its ordinary derivative, soApply integration by parts for tensor fields. The boundary contribution is proportional to $\delta \varphi$ and vanishes under the given boundary condition, leavingThe interior variation is arbitrary. Thus the scalar equation of motion isHere the d'Alembert operator contains the Levi-Civita connection in the second covariant derivative, even though the first derivative of the scalar field is ordinary.

The metric dependence of the minimally coupled scalar field Lagrangian is $\partial L/\partial g^{\alpha \beta }=-\frac{1}{2}{\nabla }_{\alpha }\varphi {\nabla }_{\beta }\varphi$. Part (ii) therefore gives the stress-energy tensorUse metric compatibility and the symmetric Hessian of a scalar field to computeThis is the scalar stress-energy divergence identity, an instance of the diffeomorphism Noether identity for a scalar field. On the scalar equation of motion it gives stress-energy conservation, $\boxed{{\nabla }_{\mu }{T}^{\mu }{}_{\alpha }=0}$.

For each fixed coordinate function $x^{\alpha }$, the expression $\square x^{\alpha }$ is a scalar: $\alpha$ labels four chosen functions and is not a free tensor index. The equation is therefore coordinate invariant when those functions are held fixed. However, the demand that the currently chosen coordinates themselves be harmonic coordinates is a chart condition, not a four-vector equation invariant under arbitrary changes of coordinate functions.

Indeed, for new functions ${\bar{x}}^{\alpha }=f^{\alpha }(x)$ the d'Alembert operator givesThe second term prevents a vector transformation law. An affine redefinition preserves harmonic coordinates; an arbitrary nonlinear one does not. For example, a Cartesian spatial coordinate $x$ in Minkowski spacetime has $\square x=0$, whereas $\square (x^2)=2$, and $x^2$ is a valid new coordinate on $x>0$. The equations are scalar equations for specified functions, while harmonic gauge is chart dependent.

Treat each $x^{\alpha }$ as a scalar field. Its second covariant derivative is ${\nabla }_{\beta }{\nabla }_{\gamma }{x}^{\alpha }=-{\Gamma }^{\alpha }{}_{\gamma \beta }$, so $\square x^{\alpha }=-H^{\alpha }$, where $H^{\alpha }=g^{\beta \gamma }{\Gamma }^{\alpha }{}_{\beta \gamma }$.

On the other hand, metric compatibility givesUsing the Christoffel trace identity in the last term yieldsSince the volume density never vanishes in a valid chart, the harmonic coordinate conditions are equivalent:

Contract the printed Riemann curvature tensor using $R_{ab}=R^{\mu }{}_{a\mu b}$. Only the derivative terms can contain second derivatives of the metric tensor:Substituting the Christoffel symbol formula, its second derivative part isSet $H_b=g_{b\lambda }{H}^{\lambda }=g^{\mu \nu }{\partial }_{\mu }{g}_{\nu b}-\frac{1}{2}{g}^{\mu \nu }{\partial }_b{g}_{\mu \nu }$. The first, second and fourth terms are the second derivative terms of $\frac{1}{2}({\partial }_a{H}_b+{\partial }_b{H}_a)$. Derivatives of the inverse metric tensor produce only products of first derivatives. Thuswhere $Q$ contains at most first metric derivatives. In harmonic coordinates, $H^{\lambda }$ vanishes throughout the chart, so $H_b$ and its derivatives vanish. The vacuum equations have the wave principal partThese are quasilinear partial differential equations with the principal part of a wave equation, because $g^{\mu \nu }$ has Lorentzian signature. This is the harmonic reduction of the vacuum Einstein equations.

First, the inverse metric tensor printed in the PDF has a missing factor: its lower-left entry must be $g^{i0}={\alpha }^{-2}{\beta }^i$, symmetric with $g^{0i}$. This correction follows either from symmetry or direct block multiplication and does not change the evolution equation used here.

Let $D_t={\partial }_t-{\beta }^k{\partial }_k$. Contract the spatial metric tensor evolution equation with ${\gamma }^{ij}$. Both shift-gradient terms contribute ${\partial }_k{\beta }^k$, givingThe Jacobi determinant derivative formula applies to every derivative of $\gamma =\det ({\gamma }_{ij})$ and gives $D_t\gamma =\gamma {\gamma }^{ij}{D}_t{\gamma }_{ij}$. Therefore the spatial determinant evolves asEquivalently, the spatial volume evolution identity is $D_t\sqrt{\gamma }=\sqrt{\gamma }({\partial }_k{\beta }^k-\alpha K)$, with $K$ the trace of the extrinsic curvature of a spatial hypersurface.

Use the equivalent density form of harmonic slicing from part (ii). With $\sqrt{-g}=\alpha \sqrt{\gamma }$, positive lapse function $\alpha$, and the corrected inverse metric tensor, its time component becomesThus $D_t(\sqrt{\gamma }/\alpha )=(\sqrt{\gamma }/\alpha ){\partial }_i{\beta }^i$, orThe spatial volume evolution identity gives $D_t\log \sqrt{\gamma }={\partial }_i{\beta }^i-\alpha K$. Subtraction yields $D_t\log \alpha =-\alpha K$, hence the harmonic lapse evolution isThe shift vector need not vanish; its transport term is essential in a general 3+1 decomposition of spacetime.

A spacetime is a smooth four-dimensional manifold equipped with a smooth Lorentzian metric, conventionally of metric signature $(-,+,+,+)$, and a choice of time orientation. The usual manifold assumptions include the Hausdorff space and second-countable space conditions. A physical model also specifies matter fields and requires the Einstein field equations and matter equations.

Diffeomorphism invariance of general relativity means that relabelling events by a smooth invertible map, while transforming the metric tensor and every matter field together, preserves the form of the equations. Passively, a coordinate change gives new components for the same geometric fields. Actively, pulling all fields back by a diffeomorphism gives another representative of the same physical geometry, subject to any prescribed boundary conditions or boundary symmetries.

The geometry and matter together carry the physical content; coordinate labels do not. A diffeomorphism need not be an isometry of a fixed metric, and unrelated metrics are not automatically physically equivalent.

Use the connection convention ${\nabla }_{\alpha }{Y}^{\rho }={\partial }_{\alpha }{Y}^{\rho }+{\Gamma }^{\rho }{}_{\mu \alpha }{Y}^{\mu }$, with the derivative index last. This is consistent with the printed curvature formula and the final formula in part (iv). The paper's $T_{\mu \nu }{}^{\rho }={\Gamma }^{\rho }{}_{\mu \nu }-{\Gamma }^{\rho }{}_{\nu \mu }$ is then the negative of the geometric torsion tensor defined by $\mathcal{T}(X,Y)={\nabla }_{X}Y-{\nabla }_{Y}X-[X,Y]$. Keep the paper's component convention throughout this question.

Set $A_{\mu \nu \rho }=g_{\rho \lambda }({\Gamma }^{\lambda }{}_{\mu \nu }-S^{\lambda }{}_{\mu \nu })$. The difference of affine connections is a tensor. Since the Levi-Civita connection $S$ is symmetric and metric compatible, expanding the nonmetricity tensor givesHere $T_{\mu \nu \rho }=g_{\rho \lambda }{T}_{\mu \nu }{}^{\lambda }$ is antisymmetric in its first two slots, while $N_{\mu \nu \rho }$ is symmetric in its first two slots. Cyclically permuting these identities and eliminating the other components givesRaise the last slot to obtain the connection decompositionwhereThe PDF is missing the factor $1/2$ in its definition of $K$. With the printed $K$, the decomposition would instead contain $-K/2$. In these formulas $T^{\rho }{}_{\mu \nu }$ raises the first slot of the fully lowered tensor; it must not be confused with $T_{\mu \nu }{}^{\rho }$, which raises the last slot.

The Levi-Civita connection is the unique affine connection that has both vanishing torsion tensor and metric compatibility. The affine connection decomposition shows exactly how an arbitrary affine connection departs from it: the contorsion tensor contributes the metric-compatible torsion correction, and the disformation tensor contributes the nonmetricity correction. The torsion and nonmetricity tensor, together with the metric tensor, determine the difference uniquely.

In the paper's convention the actual contorsion correction is $-K$, with the corrected half factor from part (i), and the disformation correction is $W$. Both corrections are tensors, although the separate connection coefficients $\Gamma$ and $S$ are not tensors. If both torsion and nonmetricity vanish, then $\boxed{\Gamma =S}$. Vanishing nonmetricity alone permits torsion, and vanishing torsion alone permits nonmetricity.

Let ${\gamma }_X(s)$ be the affinely parametrized geodesic of the affine connection with ${\gamma }_X(0)=p$ and ${\dot{\gamma }}_X(0)=X$. For fixed $t$, the curve $\eta (u)={\gamma }_X(tu)$ has initial velocity $tX$. The geodesic equation is homogeneous of degree two in the velocity, so its left-hand side for $\eta$ is $t^2$ times the left-hand side for ${\gamma }_X$, and is zero. Uniqueness of the geodesic initial value problem gives $\eta ={\gamma }_{tX}$.

By the definition of the affine exponential map, rescaling the initial velocity rescales affine parameter:At $t=0$ both sides equal $p$. The statement is made where the geodesics exist; without geodesic completeness, the affine exponential map is defined only on a suitable neighbourhood of zero in $T_{p}M$, not necessarily all of $T_{p}M$. An affine parameter distance need not be a metric length.

Choose a basis of $T_{p}M$ and use the affine exponential map to define affine normal coordinates. Its differential at zero is the identity, so the inverse function theorem gives a chart near $p$. By part (iii), radial geodesics have coordinates $x^{\mu }(s)=s{X}^{\mu }$. Their geodesic equations at $p$ imply ${\Gamma }^{\lambda }{}_{\mu \nu }(p)X^{\mu }{X}^{\nu }=0$ for every $X$. The polarization identity therefore gives ${\Gamma }^{\lambda }{}_{(\mu \nu )}(p)=0$.

The disformation tensor $W_{\mu \nu }{}^{\lambda }$ is symmetric in $\mu ,\nu$, whereas $T_{\mu \nu }{}^{\lambda }$ is antisymmetric. Also $T_{\nu }{}^{\lambda }{}_{\mu }=-T^{\lambda }{}_{\nu \mu }$. Taking the symmetric part of the corrected affine connection decomposition consequently gives the normal-coordinate identityThus the final identity in the PDF is correct after the half-factor repair in part (i). In general ${\Gamma }^{\lambda }{}_{\mu \nu }(p)=\frac{1}{2}{T}_{\mu \nu }{}^{\lambda }(p)$ in the paper's convention, rather than zero. Neither $S$ nor the first derivatives of the metric tensor need vanish in these affine normal coordinates. With the Levi-Civita connection, both torsion and nonmetricity vanish and the usual vanishing of the Christoffel symbols is recovered.