Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 309 (General Relativity)

Let $q$ denote the quotient map and write $[X]$ for a point of Real projective space. For $i=1,\dots ,n+1$, take the open setIt is open in the quotient topology because its inverse image under $q$ is the open set $\{X:X_i\ne 0\}$. These sets cover, since a nonzero vector has at least one nonzero coordinate. Define the manifold chartThe hat means that the $i$th coordinate is omitted. Ratios are unchanged by multiplying $X$ by any nonzero real number, so the chart is well defined. Its inverse inserts $1$ into position $i$ and takes the resulting equivalence class. The ratios and this inverse are continuous, giving a homeomorphism $U_i\to {\mathbb{R}}^n$.

For a compact description of the smooth transition maps, write $a_i=1$ and let $a_k=X_k/X_i$ for $k\ne i$. On $U_i\cap U_j$ one has $a_j\ne 0$, andEach coordinate is a rational smooth function on the open domain $a_j\ne 0$; reversing $i,j$ gives a smooth inverse. Thus these $n+1$ charts form the standard affine atlas of real projective space.

For completeness, the underlying space is Hausdorff: normalizing $X$ identifies it with the antipodal quotient of the unit sphere, and the continuous injective map $[X]\mapsto X{X}^T/(X^{T}X)$ into the space of symmetric matrices separates any two distinct classes by disjoint open neighborhoods. It is also a second-countable space: pull back the countable bases of rational balls in these finitely many charts, and take their union. Therefore

Use the numerator in the original PDF: it contains all $n+1$ squared differentials. The local TeX ends that numerator at $n-1$, which would not give the stated Riemannian metric.

For the pullback of a Riemannian metric, substitute the embedding $\pi (y)=(y_1,\dots ,y_n,1)$. Its differential sends a tangent vector $v$ to $(v_1,\dots ,v_n,0)$, so $d{X}_i=d{y}_i$ for $i\le n$ and $d{X}_{n+1}=0$. ConsequentlyThe denominator is everywhere positive. For any nonzero tangent vector,so this is indeed a smooth positive-definite Riemannian metric on $U$.

Put $r^2=y_1^2+y_2^2$ and write the Riemannian metric as a conformal rescaling of a Riemannian metric:The base metric is flat, so its Ricci scalar is zero and its Laplacian is $\Delta ={\partial }_1^2+{\partial }_2^2$. The formula for scalar curvature under conformal rescaling simplifies in dimension two toDifferentiate explicitly:Since $e^{-2\omega }=1+r^2$, the required Ricci scalar is

Use the curvature convention $[{\nabla }_b,{\nabla }_c]V^d=R^d{}_{abc}{V}^a$, consistent with the coordinate formula later in the paper. Choose normal coordinates at an arbitrary point $p$. The Levi-Civita connection has ${\Gamma }^d{}_{ab}(p)=0$ and is torsion free, so ${\Gamma }^d{}_{ab}={\Gamma }^d{}_{ba}$. At $p$,Hence the cyclic sum isEach derivative cancels using symmetry of the two lower connection indices. Because the Riemann curvature tensor is antisymmetric in its last two indices, full antisymmetrization over $a,b,c$ equals one third of this cyclic sum. ThusThe statement is tensorial and $p$ was arbitrary, so it holds in all coordinates everywhere. This is the first Bianchi identity.

Let $D_{abc}={\nabla }_a{\nabla }_b{K}_c$. Differentiating the Killing equation gives $D_{abc}=-D_{acb}$. For a covector, the Ricci identity isCall this difference $C_{abc}$. The differentiated Killing equation then givesTaking the first minus the second plus the third yieldsThe pair symmetries of the Riemann curvature tensor and the first Bianchi identity reduce the coefficient to $2R_{cbad}$. Therefore the second covariant derivative of a Killing vector is

For an affine parameter $\lambda$ on a geodesic, its tangent obeys $V^b{\nabla }_b{V}^a=0$. ThusThe second term vanishes by the geodesic equation; the first contracts a symmetric product $V^a{V}^b$ with the antisymmetric derivative from the Killing equation. This establishes the geodesic conserved quantity from a Killing vector:

In Cartesian coordinates on Euclidean space, the Levi-Civita connection and the Riemann curvature tensor vanish. The second covariant derivative of a Killing vector therefore givesEvery first partial derivative is constant on the connected space ${\mathbb{R}}^n$. Integrating once more gives $K^{\nu }=A^{\nu }+M_{\mu }{}^{\nu }{x}^{\mu }$, with constant coefficients. The Killing equation now becomesThus the coefficient is an antisymmetric matrix. Conversely, any such constant coefficients satisfy the Killing equation. The complete Euclidean Killing vector field is thereforeThe $n$ translation parameters and $n(n-1)/2$ rotation parameters give $n(n+1)/2$ independent Killing vector fields.

The index order in $M_{\mu }{}^{\nu }{x}^{\mu }$ fixes the rotation sign. With the entries specified in the paper, the integral curves of a vector field satisfyShift to $u=x^1-A^2$, $v=x^2+A^1$. Then $\dot{u}=v$, $\dot{v}=-u$, so $u^2+v^2$ is constant. If $x(0)=x_0$, put $u_0=x_0^1-A^2$ and $v_0=x_0^2+A^1$. Solving this linear ordinary differential equation gives the full local flow:The planar motion is a circle centred at $(A^2,-A^1)$, traversed clockwise as $\lambda$ increases. A nonzero vector $(A^3,\dots ,A^n)$ produces a helical orbit of a Euclidean Killing vector; zero drift gives a circle. At zero radius the curves are straight lines in the remaining directions, or stationary points if that drift also vanishes. These formulas are defined for every real $\lambda$.

Use geometrized units $G=c=1$ and signature $(-,+,+,+)$. The original PDF starts with the d'Alembert operator $\square {\bar{h}}_{\mu \nu }=-16\pi T_{\mu \nu }$; the local TeX incorrectly transcribes its derivative indices. The harmonic condition is the Lorenz gauge in linearized gravity.

Choose the retarded solution, excluding an incoming homogeneous wave. For a spatially localized source, the Linearized Einstein equations giveLet the source size be $D$ and the characteristic angular frequency be $\omega$. Assume the weak-field approximation, nonrelativistic source velocities, and $D\ll {\omega }^{-1}\ll r$ for a radiative far-zone measurement. Then $|\mathbf{x}-\mathbf{y}|=r-\mathbf{n}\cdot \mathbf{y}+O(D^2/r)$. At leading order in $D/r$ and $\omega D$, one may replace the denominator by $r$ and the retarded time throughout the source by $t-r$, obtaining

To identify this integral, use stress-energy conservation ${\partial }_{\mu }{T}^{\mu \nu }=0$ for a symmetric source tensor. Compact support or sufficient decay permits integration by parts with no boundary flux. Because $T^{00}=T_{00}$ in this signature, the second mass moment tensor satisfiesSpatial indices are raised with ${\delta }_{ij}$, so $T^{ij}=T_{ij}$. Substitution gives the retarded quadrupole field:The assumptions are an isolated conserved source, retarded boundary conditions, weak gravity, a source small compared with the wavelength, and observation far from it. The physical radiative field follows by the transverse-traceless projector applied to this trace-reversed metric perturbation.

Write $I_a={\bar{I}}_{aa}$ and use the right-handed rotationIn the integral defining the second mass moment tensor, change variables to $\mathbf{y}=L^{-1}\mathbf{x}$. The determinant is one and $x_i=L_{ik}{y}_k$, givingSet $C=(I_1+I_2)/2$ and $\Delta =I_1-I_2$. Multiplication givesIn particular, $I_{kk}=I_1+I_2+I_3$ is constant. These expressions also show how the rotating anisotropy enters the mass quadrupole moment.

The time-dependent components of the second mass moment tensor contain $\sin (2\Omega t)$ and $\cos (2\Omega t)$. Taking two time derivatives for the retarded quadrupole field leaves that frequency unchanged. Hence the gravitational-wave frequency isHere ${\omega }_{\mathrm{GW}}$ is angular frequency and $f_{\mathrm{GW}}$ counts cycles per unit time. If $I_1=I_2$, the mass quadrupole moment is time independent, so the leading gravitational wave amplitude vanishes and there is no emitted quadrupole frequency to measure.

Remove the constant trace from the second mass moment tensor to obtain the mass quadrupole moment $Q_{ij}=I_{ij}-{\delta }_{ij}{I}_{kk}/3$. Since $I_{kk}$ is constant, ${\stackrel{...}{Q}}_{ij}={\stackrel{...}{I}}_{ij}$. In units $G=c=1$, the quadrupole formula givesIts normalization can also be seen from the gravitational-wave energy flux: $h_{ij}^{\mathrm{TT}}=2{\Lambda }_{ij,kl}{\ddot{Q}}_{kl}/r$, where $\Lambda$ is the transverse-traceless projector. With $S_{ij}={\stackrel{...}{Q}}_{ij}$ and $P_{ij}={\delta }_{ij}-n_i{n}_j$, the projection is $S_{ij}^{\mathrm{TT}}=P_{ik}{P}_{jl}{S}_{kl}-P_{ij}{P}_{kl}{S}_{kl}/2$. For a symmetric trace-free $S$,The isotropic tensor integrals $\int n_i{n}_{j}d\Omega =4\pi {\delta }_{ij}/3$ and $\int n_i{n}_j{n}_k{n}_{l}d\Omega =4\pi ({\delta }_{ij}{\delta }_{kl}+{\delta }_{ik}{\delta }_{jl}+{\delta }_{il}{\delta }_{jk})/15$ therefore give $\int |S^{\mathrm{TT}}{|}^{2}d\Omega =8\pi S_{ij}{S}_{ij}/5$. Integrating the flux yields $⟨P⟩=(8\pi {)}^{-1}\int ⟨|S^{\mathrm{TT}}{|}^2⟩d\Omega =⟨S_{ij}{S}_{ij}⟩/5$, as above.

Now differentiate the components from part (i). With $\Delta =I_1-I_2$,The off-diagonal component occurs twice in the contraction. ThusIt is already time independent, so averaging givesThis is the leading gravitational radiation from a rotating triaxial body; restoring units multiplies it by $G/c^5$. The source is treated as rotating uniformly over an averaging interval, with radiation reaction negligible at this order.

Put $h_{ab}=\delta g_{ab}$, $h=g^{ab}{h}_{ab}$, and $h^{ab}=g^{ac}{g}^{bd}{h}_{cd}$. Throughout this part use the Riemann curvature tensor convention printed in the paper. Varying $g^{ac}{g}_{cb}={\delta }^a{}_b$ gives the inverse-metric variation. The determinant identity $\delta \log |\det g|=g^{ab}{h}_{ab}$ gives the volume form variation. Together these areIn Lorentzian coordinates the latter reads $\delta \sqrt{-g}=\frac{1}{2}\sqrt{-g}h$.

For the connection, vary metric compatibility ${\nabla }_a{g}_{bc}=0$:Add the equations with derivatives $a,b$, subtract that with derivative $c$, and use symmetry of the lower indices of the Levi-Civita connection. This givesThe difference of two connections is a tensor, so this formula is covariant.

Varying the coordinate curvature formula yields the Palatini identityThe contractions of the connection variation areFinally, $\delta R=(\delta g^{ab})R_{ab}+g^{ab}\delta R_{ab}$ gives the metric variation of scalar curvature:The double divergence may equivalently be written ${\nabla }_a{\nabla }_b{h}^{ab}$: relabeling the contracted dummy indices and using symmetry of $h^{ab}$ gives the same expression. The derivative terms are a covariant divergence, useful when varying a curvature-dependent action.

Define the stress-energy tensor by metric variation with the scalar held fixed:Use compactly supported variations to remove boundary terms. The first covariant derivative of a scalar field is an ordinary derivative, so it has no connection variation. Varying the kinetic term and the volume form gives

For the curvature term set $f={\Phi }^2$. Using part (a), its variation isApplying integration by parts twice moves each pair of derivatives from the metric variation onto $f$. The result iswhere $G_{ab}=R_{ab}-R{g}_{ab}/2$ is the Einstein tensor and $\square ={\nabla }^c{\nabla }_c$ is the d'Alembert operator. Comparing with the defining variation gives the stress-energy tensor of a nonminimally coupled scalar field:The factor $2\xi$ follows from the action's coupling $-\xi R{\Phi }^2$; a convention using $-\xi R{\Phi }^2/2$ would instead give $\xi$ in that bracket. At $\xi =0$ this reduces to the massless Klein-Gordon scalar stress-energy tensor.

The Euler-Lagrange field equation obtained by varying the nonminimally coupled scalar field isOne can check stress-energy conservation directly. The divergence of the kinetic contribution is $(\square \Phi ){\nabla }_b\Phi$, because second derivatives of a scalar commute. With $f={\Phi }^2$, the contracted Bianchi identity gives ${\nabla }^a{G}_{ab}=0$, while the Ricci identity givesConsequentlySince ${\nabla }_{b}f=2\Phi {\nabla }_b\Phi$, the full stress-energy tensor satisfies

The general reason is the diffeomorphism Noether identity for a scalar field. Under the Lie derivative along a compactly supported vector field $X$, $\delta g_{ab}=2{\nabla }_{(a}{X}_{b)}$ and $\delta \Phi =X^a{\nabla }_a\Phi$. Invariance of the action under diffeomorphisms gives, after integration by parts,Arbitrariness of $X$ gives the same divergence identity. Conservation requires the scalar equation of motion, with no need to impose the Einstein field equations for the background metric.