Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 50 (General Relativity)

For a smooth map between manifolds $\varphi :\mathcal{M}\to \mathcal{N}$, the four constructions go in the directions dictated by composition and the differential of a smooth map.

The pullback of a smooth function $f\in C^{\infty }(\mathcal{N})$ is ${\varphi }^{\ast }f=f\circ \varphi$, a function on $\mathcal{M}$. The pushforward of a curve $\lambda :I\to \mathcal{M}$ is ${\varphi }_{\ast }\lambda =\varphi \circ \lambda$, a curve in $\mathcal{N}$ with the same parameter interval.

For a tangent vector $V\in T_p\mathcal{M}$, its pushforward is ${\varphi }_{\ast }V=d{\varphi }_p(V)\in T_{\varphi (p)}\mathcal{N}$. Intrinsically, treating a tangent vector as a derivation on functions,For a covector $\eta \in T_{\varphi (p)}^{\ast }\mathcal{N}$, the pullback of a covector is the dual linear map:No inverse map is required. A pushed-forward field for a general map is a field along that map; it need not assign a unique vector to an image point with several preimages.

Write $\epsilon =n_a{n}^a\in \{-1,+1\}$. The non-null hypersurface projection is $P^a{}_b={\delta }^a{}_b-\epsilon n^a{n}_b$, reproducing both signs in the question. It annihilates the normal and is the identity on tangent vectors to the hypersurface.

Represent $V$ by the tangent to a curve $\lambda$ through $p$. The curve $\varphi \circ \lambda$ lies in $\Sigma$, so its tangent $d{\varphi }_{p}V$ is tangent to $\Sigma$. Orthogonality therefore gives $n_a({\varphi }_{\ast }V{)}^a=0$, andThis uses the stated smooth hypersurface assumption; the dimension relation alone would not make the image of an arbitrary smooth map a regular hypersurface.

A type $(0,s)$ covariant tensor is a multilinear form on $s$ vectors. For arbitrary $V_1,\dots ,V_s\in T_p\mathcal{M}$, the pullback of a covariant tensor isProjection of all covariant slots means $(P\omega )(X_1,\dots ,X_s)=\omega (P{X}_1,\dots ,P{X}_s)$. Since every pushed-forward vector is already tangential,Equality on all arguments proves ${\varphi }^{\ast }\omega ={\varphi }^{\ast }(P\omega )$. No antisymmetry is needed: this holds for every covariant tensor, not only for differential forms.

The pushforward of a contravariant tensor applies $d{\varphi }_p$ to each of its $r$ vector slots. Choose a basis $e_i$ of $T_p\mathcal{M}$ and expandIts pushforward is $T^{i_1\cdots i_r}(d\varphi e_{i_1})\otimes \cdots \otimes (d\varphi e_{i_r})$. Each factor is tangent to the hypersurface and fixed by $P$. ConsequentlyThe basis expansion proves the result for arbitrary tensors, rather than only a single decomposable tensor product.

The two coordinate tangent vectors of the embedding areTheir Euclidean inner products give the pullback of a Riemannian metric, equivalently the induced metric:Locally set $u=\rho \phi$. The line element becomes $d{u}^2+d{\hat{z}}^2$, so the induced Riemann curvature tensor vanishes identically. This is the intrinsic flatness of a circular cylinder. Its bending in the ambient space is extrinsic curvature, not intrinsic curvature; periodicity of the angular coordinate does not change the local flatness.

Choose the outward normal vector $n=(\cos \phi ,\sin \phi ,0)$ and the positive conventionThe ambient connection vanishes in Cartesian coordinates. Since ${\partial }_{\phi }n=e_{\phi }/\rho$ and ${\partial }_{\hat{z}}n=0$, the extrinsic curvature components are $K_{\phi \phi }=\rho$, $K_{\phi \hat{z}}=K_{\hat{z}\hat{z}}=0$. Taking the trace using the induced metric,Reversing the normal or using the negative extrinsic-curvature convention gives $-1/\rho$. The two principal curvatures are $1/\rho$ and zero in the chosen convention; the averaged mean curvature would be $K/2$, so it must not be confused with the requested trace.

In geometrized units, a mass is converted to a length by multiplying its SI value by $G/c^2$, or to a time by multiplying by $G/c^3$. Using the supplied constants, the solar mass becomesThus the characteristic scales are kilometres and a few microseconds. With potential zero at infinity, the Newtonian gravitational potential at the surface, in units of $c^2$, isIts very small magnitude is the relevant weak-field measure. In SI potential units the same number corresponds to about $-1.91\times 10^{11}{\mathrm{m}}^2{\mathrm{s}}^{-2}$.

Power is energy per time. In geometrized units, energy has the dimension of length and time is converted to length using $c$, so the dimensionless geometrized luminosity isEquivalently, $c^5/G$ is the corresponding SI power unit. The lifetime for emitting the entire rest energy at the given constant luminosity isThe same result follows from $M_{\odot }^{(\mathrm{time})}/L_{\odot ,\mathrm{geom}}$. The stipulated full-mass radiation time is of order $10^{13}$ years; this is the constant-luminosity energy budget asked for, not a stellar-evolution calculation.

Here $I_{ij}$ denotes the unreduced second mass moment tensor, distinct from its trace-free mass quadrupole moment. In coordinates referred to the chosen origin,This expression uses $c=1$. In SI units the leading nonrelativistic mass density is $T^{00}/c^2$. For slowly moving point masses, $T^{00}\simeq {\sum }_A{m}_{(A)}{\delta }^{(3)}(\mathbf{y}-{\mathbf{y}}_{(A)}(t))$, soKinetic corrections to the energy density are higher order in the velocity. The second mass moment tensor is not the mechanical moment of inertia tensor, which instead has components ${\delta }_{ij}{I}_{kk}-I_{ij}$.

In the center of mass frame, put the two positions at $(0,0,z(t))$ and $(0,0,-z(t))$, with $z>0$ during infall. Their separation is $2z$, so Newtonian gravity givesMultiplying by $\dot{z}$ and integrating givesThe negative square root is essential for the falling branch. All second mass moment tensor components except $I_{zz}=2m{z}^2$ vanish. Differentiating three times,Substitution yields ${\stackrel{...}{I}}_{zz}=-m^2\dot{z}/z^2$, hencewith every other component of ${\stackrel{...}{I}}_{ij}$ zero. The distance entering each acceleration is $2z$, which accounts for the factor one quarter in the equation of motion.

The trace-free mass quadrupole moment is diagonal:The same constant factors multiply its third derivatives, giving ${\stackrel{...}{Q}}_{ij}{\stackrel{...}{Q}}_{ij}=(2/3)({\stackrel{...}{I}}_{zz}{)}^2$. The quadrupole formula and the preceding infall result therefore giveFor infall from infinity, $P=m^5/(15z^5)$ and $|\dot{z}|=\sqrt{m/(2z)}$. To find the emitted energy, integrate power over time, using $dt=|dz|/|\dot{z}|$:The total initial mass is $2m$, so the radiated fraction is $1/840\simeq 1.19\times 10^{-3}$. The Sun would emit that fraction of its rest energy inunder the stipulated constant luminosity. This is the head-on quadrupole radiation from equal masses prediction with the prescribed stopping rule. At the endpoint $|\dot{z}|=1/2$ and $m/z=1/2$, so extending the slow-motion weak-field formula that far is an extrapolation, not a controlled strong-field prediction.

Treat $V$ as a derivation. For an arbitrary smooth function $f$ on $\mathcal{N}$, the definition of the differential of a smooth map and the chain rule giveComparison with $({\varphi }_{\ast }V{)}^{\alpha }{\partial }_{\alpha }f$ for all $f$ provesThe Jacobian is an $n\times m$ matrix; it need not be square or invertible.

Evaluate the pullback of a covector on the coordinate basis vector ${\partial }_i$. Its pushforward is $({\partial }_i{y}^{\alpha }){\partial }_{\alpha }$, soEquivalently, pulling back the coordinate one-forms gives ${\varphi }^{\ast }(d{y}^{\alpha })=d(y^{\alpha }\circ \varphi )=({\partial }_i{y}^{\alpha })d{x}^i$. The same Jacobian enters the vector pushforward and the covector pullback, with the different index contractions reflecting their duality.

With vanishing extrinsic curvature, the momentum constraint is identically satisfied and the vacuum Hamiltonian constraint reduces to $\mathcal{R}=0$. To evaluate that Ricci scalar, write ${\psi }_i={\partial }_i\psi$, raise these intermediate indices with ${\delta }^{ij}$, and use ${\gamma }^{ij}={\psi }^{-4}{\delta }^{ij}$. The Levi-Civita connection isSubstitution into the stated curvature convention givesTracing with ${\gamma }^{ij}$ cancels the gradient-square terms:Here the Laplacian and norm on the right are those of the flat Euclidean metric. Thus the time-symmetric conformally flat vacuum initial data constraints becomeThe equivalence uses $\psi \ne 0$. This identity is also the three-dimensional specialization of scalar curvature under conformal rescaling; locally the conformal exponent is $2\log |\psi |$, so either fixed nonzero sign of $\psi$ gives the same metric.

On the punctured region $r>0$, choose a positive constant $a$ and setFor a radial function in three dimensions, $\Delta \psi =r^{-2}{\partial }_r(r^2{\partial }_r\psi )$. Since $r^2{\partial }_r\psi =-a$, this harmonic function has $\Delta \psi =0$ at every $r>0$, is nonconstant and positive, and approaches one at infinity. It defines the requested time-symmetric conformally flat vacuum initial data on that punctured or exterior region. One often writes $a=M/2$, giving the time-symmetric spatial metric of the Schwarzschild metric in isotropic coordinates.

A puncture or inner boundary is necessary if global regularity was intended. A smooth harmonic function on all of ${\mathbb{R}}^3$ that tends to one at infinity must equal one everywhere: applying the maximum principle for harmonic functions to larger and larger balls bounds $|\psi -1|$ by its arbitrarily small boundary values. The nonconstant example is therefore not a smooth solution through $r=0$, and no such globally regular example exists under those stronger assumptions.

Vary the geodesic Lagrangian before imposing the normalization of proper time. With $u^{\mu }=d{x}^{\mu }/d\tau$,The Euler-Lagrange equation isequivalently the affinely parametrized geodesic equation. For the static weak metric, its leading spatial connection coefficient isAll terms with two spatial velocities are suppressed by $v^2$. Therefore $d^2{x}^i/d{\tau }^2\simeq -{\partial }_i\Phi (dt/d\tau {)}^2$.

Because $t$ is cyclic, the same Euler-Lagrange equation gives $(1+2\Phi )dt/d\tau =\text{constant}$. On converting to coordinate time, the additional term $(d^{2}t/d{\tau }^2)(d{x}^i/dt)$ is of order $v^2\nabla \Phi$ and can be dropped. Henceto leading order, the Newtonian limit of general relativity. Restoring SI units gives $\Phi ={\Phi }_{\mathrm{SI}}/c^2$, so the acceleration is $-\nabla (c^2\Phi )$. Substituting $L=1$ from proper-time normalization before varying would incorrectly erase the dynamics.

Use $A_{\beta }={\partial }_{\mu }{h}^{\mu }{}_{\beta }$ and $\square ={\eta }^{\mu \nu }{\partial }_{\mu }{\partial }_{\nu }$. The linearized Ricci tensor and scalar obeyThe background Ricci tensor and Ricci scalar vanish, so varying the metric in the scalar-curvature term contributes nothing at first order beyond this trace. The Einstein tensor perturbation is consequentlyAll contractions use the Minkowski metric. Direct differentiation gives ${\partial }^{\alpha }\delta G_{\alpha \beta }=0$, the linearized contracted Bianchi identity, which checks the relative signs. No time derivatives or gauge-dependent terms have been discarded.

Let the symmetric metric perturbation $h^{\alpha \beta }$ be varied with compact support. Integrating the mixed derivative product by parts makes its integral equal to that of $A_{\sigma }{A}^{\sigma }$. Thus the supplied massless Fierz-Pauli action has, up to a boundary term, densityFor example, the equivalence of the mixed term follows from commuting flat-space partial derivatives after moving one derivative off $h^{\rho \sigma }$. Here $\delta h={\eta }_{\alpha \beta }\delta h^{\alpha \beta }$ and $\delta A^{\nu }={\partial }_{\mu }\delta h^{\mu \nu }$.

Integrating each variation by parts, the four terms contribute respectivelyThe symmetrization in the second term is required because the varied field is symmetric. Comparing this coefficient with the Einstein tensor perturbation above givesThus arbitrary compactly supported variations give $\delta G_{\alpha \beta }=0$, exactly the vacuum Linearized Einstein equations. The minus sign and overall normalization of the action do not change those equations; boundary conditions justify the discarded total derivatives.