Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 49 (General Relativity)

Let ${\varphi }_s$ be the local flow of the smooth vector field $X$, with ${\varphi }_0=\mathrm{id}$ and $\frac{d}{ds}{\varphi }_s(p)=X_{{\varphi }_s(p)}$. The flow definition of the Lie derivative of a tensor field isFor a tensor field of type $(r,s)$, the mixed tensor pullback transports the tensor at ${\varphi }_s(p)$ back to $p$: it applies $(d{\varphi }_s{)}_p^{-1}$ to every contravariant factor and $(d{\varphi }_s{)}_p^{\ast }$ to every covariant factor. Thus every difference quotient belongs to the same tensor space at $p$. This definition gives another tensor field of the same type and uses no choice of affine connection. The tensor Lie derivative is defined wherever the local flow exists, including points where $X$ vanishes.

For a diffeomorphism, the differential and its inverse preserve the natural pairing of a vector with a covector. Consequently the mixed tensor pullback commutes with every tensor contraction $C$:Differentiating at zero gives ${\mathcal{L}}_X(CT)=C({\mathcal{L}}_{X}T)$. The mixed tensor pullback also preserves tensor products, soThe ordinary product rule for differentiation therefore yieldsThis proves the contraction and Leibniz rule properties for all smooth generators, without needing straightening coordinates.

The suggested flow-box theorem applies locally only where $X\ne 0$. For example $X=x{\partial }_x$ vanishes at $x=0$, so it cannot equal a coordinate basis vector there. Its local flow is nevertheless ${\varphi }_s(x)=e^{s}x$ and ${\varphi }_s^{\ast }dx=e^{s}dx$, giving ${\mathcal{L}}_{X}dx=dx$ even at that zero. This Lie derivative at a zero of its generator illustrates why the general flow proof is needed to cover every point.

For a smooth function, pullback of a smooth function is composition. The chain rule givesFor a vector field $Y$, use any local coordinates. To first order,Multiplying this inverse differential by $Y({\varphi }_s(x))$ givesThusThis is the Lie bracket of vector fields, whose action on a function is $X(Yf)-Y(Xf)$.

Write the covector field as $\omega ={\omega }_{\mu }d{x}^{\mu }$. From the contraction and product rule properties,Take $Y={\partial }_{\mu }$. The Lie bracket of vector fields is $[X,{\partial }_{\mu }]=-({\partial }_{\mu }{X}^{\nu }){\partial }_{\nu }$, soEquivalently, the covariant-factor contribution comes from ${\mathcal{L}}_{X}d{x}^{\mu }=d{X}^{\mu }=({\partial }_{\nu }{X}^{\mu })d{x}^{\nu }$. The formula holds in any coordinate basis and is the one-form case of the coordinate tensor Lie derivative.

Expand $T=T^{\mu }{}_{\nu }{\partial }_{\mu }\otimes d{x}^{\nu }$. Using the tensor Lie derivative of functions, vectors and covectors and its Leibniz rule givesThe contravariant slot has a minus sign and the covariant slot a plus sign.

For the commutator identity for Lie derivatives, put $D=[{\mathcal{L}}_X,{\mathcal{L}}_Y]-{\mathcal{L}}_{[X,Y]}$. The commutator of two tensor derivations is itself a tensor derivation, and $D$ commutes with tensor contractions. On a function, $Df=XYf-YXf-[X,Y]f=0$. On a vector field $Z$,by the Jacobi identity, which follows here by expanding the commutators of the operators $X,Y,Z$ acting on functions. For a type $(1,1)$ tensor, $T(Z)$ is a vector field, and contraction compatibility givesAs this holds for every $Z$, $DT=0$. ThereforeThe derivation argument also establishes the identity for arbitrary tensor types by applying it to covector–vector pairings and then to tensor products.

Use $G=c=1$, select the retarded source field with no added homogeneous radiation, and work to leading order in the weak-field approximation and long-wavelength source approximation. The retarded fundamental solution of the Linearized Einstein equations in Lorenz gauge in linearized gravity isFor $r\gg d$, the denominator is $r$ to leading order. The delay is $t-r+\mathbf{n}\cdot {\mathbf{x}}^{\prime }+O(d^2/r)$, where $\mathbf{n}=\mathbf{x}/r$. In the usual slowly evolving source regime, its characteristic time $\tau$ obeys $d/\tau \ll 1$, so the source-size part of the delay is negligible:The stress-energy conservation equation ${\partial }_{\mu }{T}^{\mu \nu }=0$ is required at the approximation order used; it follows from the divergence of the gauge-fixed field equation. Compact support removes the integration by parts surface terms. Since $T^{00}=T_{00}$ and $T^{ij}=T_{ij}$ in this signature,Combining these identities gives the retarded quadrupole fieldFor an ordinary nonrelativistic bound source, $\tau \sim d/v$ makes $d/\tau \sim v\ll 1$. More generally the size relative to the variation timescale must also be small; speed alone does not exclude a rapidly varying small-amplitude motion.

The radiation boundary condition for linearized gravity is essential for a statement about the full field. Without it, the displayed implication is false: take $T_{\mu \nu }=0$ and add ${\bar{h}}_{11}=A\cos (k(t-z))$, ${\bar{h}}_{22}=-{\bar{h}}_{11}$, all other components zero. This weak plane gravitational wave satisfies both the homogeneous wave equation and Lorenz gauge in linearized gravity, while every $I_{ij}$ is zero. The proved formula is for the retarded source contribution, with the standard slow-source qualification. Its transverse–traceless projection gives the physical radiative strain.

The separation of the stars is $2R$. Newton's law of universal gravitation and circular acceleration giveTo leading nonrelativistic order, the mass density is the sum of the two translated point-mass Dirac delta distributions. Thuswith all components involving $z$ zero. The trace is $2M{R}^2$, a constant. Define the trace-free mass quadrupole moment $Q_{ij}=I_{ij}-{\delta }_{ij}{I}_{kk}/3$; its third derivatives equal those of $I_{ij}$. If $K=8M{R}^2{\Omega }^3$, thenBoth off-diagonal entries count in the contraction. Therefore ${\stackrel{...}{Q}}_{ij}{\stackrel{...}{Q}}_{ij}=2K^2=128M^2{R}^4{\Omega }^6$, independent of phase. The quadrupole formula gives the equal-mass circular-binary quadrupole luminosityRestoring units, ${\Omega }^2=GM/(4R^3)$ andHere $R$ is each star's radius about the centre of mass, not the separation. The rest-frame prescription supplies the leading mass density; a moving star does not still have zero momentum density and spatial stress. The calculation uses the assumed quadrupole formula and the Newtonian orbit rather than imposing those rest-frame zeros on the moving binary.

At fixed masses, the equal-mass circular-binary quadrupole luminosity grows as $R^{-5}$. The binary's Newtonian binding energy is $-G{M}^2/(4R)$, so reducing the separation increases both the binding and the radiated power. Equivalently, its luminosity scaling iswhich makes the importance of orbital compactness explicit.

Ordinary extended stars cannot remain separate at very small orbital radii: contact, mass transfer and tidal disruption intervene. A neutron star or black hole can remain a compact orbiting object down to separations of order a few gravitational radii, allowing high orbital speeds, rapidly changing mass quadrupole moments and strong gravitational waves. Thus compact, tightly bound binaries are especially efficient emitters. The Newtonian quadrupole formula explains the scaling; precision predictions near merger require relativistic dynamics, where that approximation itself ceases to be reliable.

Write $h_{ab}=\delta g_{ab}$ and $C^a{}_{bc}=\delta {\Gamma }^a{}_{bc}$. The difference between two affine connections is a tensor, so its infinitesimal change $C$ is a type $(1,2)$ tensor field. Both affine connections are torsion-free, giving $C^a{}_{bc}=C^a{}_{cb}$. Varying metric compatibility givesLower the first index of $C$. Add the versions with derivatives $b,c$ and subtract the one with derivative $d$; the lower-slot symmetry cancels the unwanted terms, leavingThereforeAll covariant derivatives use the original Levi-Civita connection. The PDF has ${\nabla }_b{h}_{dc}$ as its second term; the converted TeX's ${\nabla }_d{h}_{dc}$ is a transcription error.

Adopt the printed Riemann curvature tensor convention and contract $R_{ab}=R^c{}_{acb}$. At an arbitrary point choose normal coordinates for the original metric tensor. There the original connection coefficients vanish. Varying the coordinate curvature formula leavesAt that point these partial derivatives equal the covariant derivatives of the tensor $C$. Both sides are tensors, so the result is valid in every coordinate system:Contracting its first and third indices proves the Palatini identityThe normal-coordinate argument is applied independently at every point; it does not assume a flat background or set derivatives of the original connection to zero.

Variation of the inverse metric relation $g^{ac}{g}_{cb}={\delta }^a{}_b$ gives $\delta g^{ab}=-h^{ab}$. Hence the metric variation of scalar curvature isPut $h=g^{ab}{h}_{ab}$. The connection variation gives the contractionsUsing metric compatibility and the Palatini identity,Renaming dummy indices givesNo interchange of covariant derivatives on a tensor is needed in this derivation, so no hidden curvature-commutator term is discarded.

Take compactly supported metric variations, or impose boundary conditions that remove the integration by parts terms. Let $F=f^{\prime }(R)$ and $\square ={\nabla }^a{\nabla }_a$. Since $\delta \sqrt{-g}=\frac{1}{2}\sqrt{-g}{g}^{ab}{h}_{ab}$, the gravitational action varies asApplying integration by parts twice to the derivative terms givesFor the usual stress-energy tensor definition $T_{ab}=-2(-g{)}^{-1/2}\delta S_m/\delta g^{ab}$, varying the covariant metric tensor gives $\delta S_m=\frac{1}{2}\int \sqrt{-g}{T}^{ab}{h}_{ab}{d}^{4}x$. Thus the action's stated normalization impliesThere is no implicit $1/(16\pi )$ in this gravitational action. This is the normalization of the metric f(R) field equation, rather than the commonly normalized version with $8\pi T_{ab}$ on the right.

Finally, the chain rule gives ${\nabla }_a{\nabla }_{b}F=f^{\prime \prime }{\nabla }_a{\nabla }_{b}R+f^{(3)}{\nabla }_{a}R{\nabla }_{b}R$ and $\square F=f^{\prime \prime }\square R+f^{(3)}(\nabla R{)}^2$. Substitution yieldsThis is a metric f(R) gravity variation: the connection is always the Levi-Civita connection of the varied metric tensor, not an independent variable.

The pure gravitational action is invariant under diffeomorphisms. An infinitesimal diffeomorphism generated by a compactly supported vector field $\xi$ changes the metric by its tensor Lie derivative,Using symmetry of $E^{ab}$ and the variational expression already derived,As ${\xi }_b$ is arbitrary, the off-shell metric divergence identity isThis Noether identity follows for every metric tensor without imposing either the gravitational field equation or the matter equations. It is a consequence of the metric action's diffeomorphism invariance, not a conclusion requiring a long component calculation.

In four dimensions, Lovelock's theorem restricts a local natural symmetric divergence-free metric tensor with at most second derivatives of the metric to a constant linear combination of the Einstein tensor and the metric. A generic nonlinear f(R) gravity equation contains ${\nabla }_a{\nabla }_b{f}^{\prime }(R)$: the Ricci scalar already has second metric derivatives, so this term generally introduces fourth metric derivatives. It therefore violates the second-order hypothesis, although it remains covariant, symmetric and divergence-free.

The exception must be stated. For $f(R)=aR+b$, the same tensor reduces towhich is precisely of Lovelock form. In particular, $f(R)=R$ is a counterexample to a blanket claim that the theorem never applies. The intended exclusion concerns generic nonlinear $f$, with $f^{\prime \prime }$ not identically zero. Restricting attention to a special constant-curvature solution of a nonlinear theory does not turn its off-shell equations into a universally second-order metric tensor.

A four-dimensional vacuum Einstein solution with the fixed cosmological constant $\Lambda$ has $R_{ab}=\Lambda g_{ab}$ and $R=4\Lambda$, constant. Thus every derivative of $f^{\prime }(R)$ vanishes andThe metric tensor is nondegenerate, so this is zero precisely whenThis is the necessary and sufficient Einstein metric condition in f(R) gravity for the specified $\Lambda$, and works for every such Einstein metric, even when its Weyl tensor is nonzero. There is no need to divide by $f^{\prime }(4\Lambda )$; the degenerate case where both $f$ and $f^{\prime }$ vanish at that curvature is included. At $\Lambda =0$, the condition is simply $f(0)=0$.

If the intention is to demand the inclusion for every real $\Lambda$ simultaneously, the stronger functional condition is $R{f}^{\prime }(R)=2f(R)$ for all $R$. On each nonzero half-line it integrates to $f(R)=C{R}^2$; smoothness across zero makes the constants equal. Thus the all-$\Lambda$ version gives $f(R)=C{R}^2$, including $C=0$. This stronger reading is distinct from fixing one cosmological constant.

Treat the printed $e^{\mu }$ as an orthonormal coframe, with frame metric $\eta =\mathrm{diag}(-1,1,1,1)$. Work on a patch $z\ne 0$ and $F=1-\alpha z^3>0$, so the given real coframe exists. Define $f=\sqrt{F}$ and $A=z{f}^{\prime }-f$. The exterior derivatives areThe connection 1-forms satisfying Cartan's first structure equation and metric compatibility areAll other forms vanish. With a Lorentzian frame, it is the lowered forms ${\omega }_{ab}={\eta }_{ac}{\omega }^c{}_b$ that are antisymmetric; the two mixed time–space forms are equal. These displayed forms solve the torsion-free structure equation, and uniqueness of the Levi-Civita connection identifies them as the required connection.

Apply Cartan's second structure equation. For example,The remaining derivatives give ${\Theta }^1{}_3=Af{e}^1\wedge e^3$ and ${\Theta }^1{}_2=-f^2{e}^1\wedge e^2$, with analogous forms for index 2. Put $q=\alpha z^3$ and $B=1+q/2$. From $f^2=1-q$,Therefore all six independent curvature 2-forms areThe other six are fixed by ${\Theta }^b{}_a=-{\eta }_{aa}{\eta }_{bb}{\Theta }^a{}_b$ for $a\ne b$, with no sum: equal for time–space pairs and opposite for spatial pairs. All diagonal forms are zero. The Lorentzian connection-form antisymmetry is essential to these signs.

Use the convention ${\Theta }^a{}_b=\frac{1}{2}{R}^a{}_{bcd}{e}^c\wedge e^d$ and $R_{ab}=R^c{}_{acb}$. Each independent two-form has only one coordinate-plane wedge, so the Ricci tensor is diagonal. Contracting the six coefficients givesBut $2B+F=2(1+q/2)+(1-q)=3$, soIn coordinate-independent form, $R_{\mu \nu }=-3g_{\mu \nu }$. Thus the vacuum Einstein field equations $G_{\mu \nu }+\Lambda g_{\mu \nu }=0$ hold withThe curvature radius is one in the metric's normalization. The parameter $\alpha$ changes the individual curvature 2-forms but cancels from their Ricci contraction. For $\alpha =0$ this is Anti-de Sitter spacetime; for nonzero $\alpha$ it is the planar Einstein metric with cubic radial function, with a nontrivial Weyl tensor rather than universally constant sectional curvature. The result is local on regular coordinate patches; a zero of $F$ invalidates this particular static coframe, not the tensor equation in a suitable regular extension.