Past exam of the general relativity II course of the University of Oxford / 2015 (A12212W1)

Source: released examination PDF, bundle pages 92–95.

We use metric signature $(-,+,+,+)$ and set the speed of light to one. In question 2, $M$ and $Q$ have the geometric normalization of the Reissner-Nordstrom metric, with $M>0$ and $r>0$. Every numbered part is treated, including all three cases in question 2(b).

A Killing vector field $K$ generates local isometries: its flow of a vector field ${\Phi }_s$ satisfies ${\Phi }_s^{\ast }g=g$. Equivalently, the Lie derivative of the metric tensor vanishes. For the Levi-Civita connection, metric compatibility converts the coordinate expression into the Killing equation:The Levi-Civita connection is an implicit extra assumption in this question. The printed assumption of vanishing torsion tensor alone does not imply metric compatibility. For example, on Minkowski spacetime, take the affine connection whose only nonzero Christoffel symbol is ${\Gamma }^0{}_{00}=1$. The vector field $K={\partial }_0$ preserves the Minkowski metric, but $K_0=-1$ gives ${\nabla }_0{K}_0=1$, contradicting the Killing equation. The following calculation, and the covariant formulas in part (b), use the Levi-Civita connection; this is the role of metric compatibility in Killing identities.

For the stated curvature sign convention, the Ricci identity on a covector has a minus sign:Write $A_{\mu \nu \rho }={\nabla }_{\mu }{\nabla }_{\nu }{K}_{\rho }$. Differentiating the Killing equation gives $A_{\mu \nu \rho }=-A_{\mu \rho \nu }$. Alternately using this relation and the Ricci identity givesThe first Bianchi identity and antisymmetry of the last two indices of the Riemann curvature tensor reduce the three curvature terms to $2R^{\sigma }{}_{\mu \nu \rho }{K}_{\sigma }$. Consequently the second covariant derivative of a Killing vector is

For a Conformal Killing vector field, the same metric compatibility calculation givesThus its Conformal Killing equation is ${\nabla }_{(\mu }{X}_{\nu )}=\lambda g_{\mu \nu }$. Taking the tensor contraction with the inverse metric, and using $g^{\mu \nu }{g}_{\mu \nu }=4$, determines the conformal Killing factor:Parentheses include the factor $1/2$ in tensor index symmetrization.

Use an affine parameter $s$ on the geodesic, so its tangent vector $U$ obeys ${\nabla }_{U}U=0$. The product rule, metric compatibility, and the Conformal Killing equation giveHenceIt is conserved on every null geodesic with an affine parameter, but generally not on other geodesics. On a timelike geodesic parametrized by proper time, $g(U,U)=-1$, so the derivative is $-\lambda$. A Killing vector field, or a geodesic on which $\lambda$ vanishes, supplies the exceptional conservation. This is the conformal Killing geodesic identity.

For an arbitrary parameter with ${\nabla }_{U}U=fU$, the right-hand side acquires $f(X\cdot U)$. Thus even for a null geodesic, the simple conserved expression presupposes an affine parameter.

In Cartesian coordinates on Minkowski spacetime, the Conformal Killing equation becomesDifferentiate the equations with index pairs $(\nu ,\rho )$, $(\mu ,\rho )$ and $(\mu ,\nu )$ by ${\partial }_{\mu }$, ${\partial }_{\nu }$ and ${\partial }_{\rho }$, respectively. Adding the first two and subtracting the third, Clairaut's theorem yields the flat conformal Killing integrability formulaApply ${\partial }^{\rho }$ and use ${\partial }^{\rho }{X}_{\rho }=4\lambda$. With the d'Alembert operator $\square ={\eta }^{\alpha \beta }{\partial }_{\alpha }{\partial }_{\beta }$, this givesA tensor contraction with ${\eta }^{\mu \nu }$ gives $2\square \lambda =-4\square \lambda$, so $\square \lambda =0$. Returning to the preceding equation proves the stronger statementThus the conformal Killing factor is an affine function, not an arbitrary solution of the wave equation.

Write the affine function obtained in part (iii) as $\lambda =c+2b_{\mu }{x}^{\mu }$, with constant $c$ and $b_{\mu }$. A Conformal Killing vector field with precisely this conformal Killing factor isDirect partial differentiation verifies its Conformal Killing equation. Subtracting it from any other solution leaves a Killing vector field $Y$. The formula in part (iii), now with $\lambda =0$, says ${\partial }_{\mu }{\partial }_{\nu }{Y}_{\rho }=0$. Therefore $Y$ is affine, and the Killing equation makes its linear coefficient antisymmetric after lowering an index. Every solution is consequentlyThe four constant $a^{\mu }$ generate translations, the six independent ${\omega }_{\mu \nu }$ generate Lorentz transformations, $c$ generates a dilation, and the four $b^{\mu }$ generate special conformal transformations. Their constant, linear and quadratic terms are independent; within the linear terms, the antisymmetric coefficient cannot equal a nonzero multiple of the identity. Hence the conformal Killing algebra of Minkowski spacetime hasFor $\lambda =1$, the canonical choice is the dilation generator $D=x^{\mu }{\partial }_{\mu }$. Its flow of a vector field solves $d{x}^{\mu }/ds=x^{\mu }$, givingOne may add any Killing vector field without changing $\lambda =1$, so this choice is not unique. The displayed dilation gives a global one-parameter group of diffeomorphisms. The count of fifteen concerns independent Conformal Killing vector fields; some special conformal transformations have only local flows on all of affine Minkowski spacetime.

A nonzero vector field has hypersurface orthogonality when it is locally normal to a family of hypersurfaces. In terms of its associated differential one-form $K^{♭}=g_{\mu \nu }{K}^{\nu }d{x}^{\mu }$, the Frobenius theorem gives the equivalent conditionslocally with nonzero $f$ and $dh$; square brackets denote tensor index antisymmetrization.

All coefficients of the Reissner-Nordstrom metric are independent of $t$, and $K={\partial }_t$ has constant coordinate components. Thus the Lie derivative obeys ${\mathcal{L}}_{K}g={\partial }_{t}g=0$: $K$ is a Killing vector field. Moreover,Thus $K$ has hypersurface orthogonality, with the $t=\text{constant}$ hypersurfaces providing the orthogonal family wherever this chart is regular. It is timelike where $F>0$, since $g(K,K)=-F$. Its hypersurface orthogonality also holds in a regular extension across a horizon; singular static coordinates do not invalidate the Frobenius theorem test.

For the three cases, distinguish a curvature singularity from a coordinate singularity. The Reissner-Nordstrom Kretschmann scalar isIt diverges at $r=0$ when either $M$ or $Q$ is nonzero. Any positive root of $F$ instead admits Eddington-Finkelstein coordinates: set $v=t+\int dr/F$ to obtainThe determinant of its $(v,r)$ block is $-1$, including at $F=0$. Hence positive roots of $F$ are removable coordinate singularities of the static chart. The apparent degeneracy at the poles of the angular chart is also a coordinate singularity.

If $M^2<Q^2$, thenThere is no positive root and no horizon hiding $r=0$ in the extended Reissner-Nordstrom spacetime. Its Kretschmann scalar diverges there, so $r=0$ is a naked singularity. It is a timelike curvature singularity, as the normal to nearby $r=\text{constant}$ hypersurfaces has positive squared norm $g^{rr}=F$.

If $M^2=Q^2$ with $M>0$, thenThe curvature singularity remains at $r=0$. At $r=M$ the static chart has a coordinate singularity, removed by the Eddington-Finkelstein coordinates above. The Kretschmann scalar there equals $8/M^4$, which is finite. The coincident horizons form a degenerate Killing horizon, and its future outer branch is the event horizon of the extremal black hole. A double zero of $F$ does not remove the horizon.

For $0<Q^2<M^2$ there are two distinct positive roots:Both roots are coordinate singularities of the static chart. In the standard extension, $r_{+}$ is the outer event horizon and $r_{-}$ is an inner Cauchy horizon. The curvature singularity is $r=0$. Between the horizons $F<0$, so the radial direction is timelike; for $0<r<r_{-}$, $F>0$ again and the central curvature singularity is timelike. These are the three radial regions of the global structure of a charged spherical black hole.^[ref]

For completeness, if $Q=0$ and $M>0$, the Schwarzschild spacetime limit has just the positive horizon $r=2M$ and a spacelike curvature singularity at $r=0$; the formal root $r_{-}=0$ of $r^{2}F$ is not a regular inner horizon. If $M=Q=0$, the metric tensor is flat and the radial origin is regular in Cartesian coordinates.

A regular hypersurface $h=0$, with $dh\ne 0$, is a null hypersurface if its normal is null:The raised normal is then also tangent to the hypersurface, and its induced metric is degenerate. This is a local geometric property.

In an asymptotically flat spacetime, choose its future null infinity ${\mathcal{I}}^{+}$. The black hole region and future event horizon areHere $J^{-}({\mathcal{I}}^{+})$ is the causal past of future null infinity: the region from which a future-directed causal curve can escape to arbitrarily distant observers. An event horizon therefore depends on the global future causal structure, and an arbitrary null hypersurface need not be an event horizon.

In the extremal case, the squared norm of the Killing vector field isFor $M=|Q|>0$, it is negative for every regular $r\ne M$ and vanishes precisely at $r=M$ in the extended spacetime. ThusThe static chart cannot itself test the geometry at its horizon. In the regular Eddington-Finkelstein coordinates, the normal to $h=r-M=0$ is $n=dr$, andAt $r=M$ this gives $n^{♯}={\partial }_v=K$, a nonzero null vector tangent to the surface. Therefore $r=M$ is a null hypersurface, even though $F$ has the same sign on its two sides.

On the equatorial path, the Lagrangian per unit mass isBoth $t$ and $\varphi$ are cyclic coordinates. The Euler-Lagrange equations therefore conserve their canonical momenta:ConsequentlyHere $E$ is the total conserved energy in the specified gauge of the electromagnetic four-potential. The paper's $l$ is the angular momentum per unit mass; the physical angular momentum is $ml$. The equatorial choice is consistent with the $\theta$ Euler-Lagrange equation, since its angular force term contains $\sin \theta \cos \theta =0$ at $\theta =\pi /2$ and the electromagnetic four-potential has no angular component.

With $\tau$ the proper time of the massive test particle, the four-velocity has squared norm $-1$:The conserved quantities in part (i) giveMultiplying the normalization by $F$ and substituting gives the charged-particle radial equation in Reissner-Nordstrom spacetime:At a radial turning point $r_{\ast }>0$, the left side is zero, soThe parenthesis on the left is strictly positive. Hence $F(r_{\ast })\ge 0$. The factorization in part (b) now yieldsThere is no radial turning point between the horizons. Indeed, for $r_{-}<r<r_{+}$, the right-hand side of the radial equation is strictly positive, independently of the electric charge $e$ of the test particle. This is the charged-particle turning-point exclusion in Reissner-Nordstrom spacetime.

The inequalities are necessary, not a claim that a radial turning point exists for every $E,l,e$. At a regular horizon crossing a timelike vector cannot be tangent to a null hypersurface, so a massive test particle actually has $\dot{r}\ne 0$ there. Algebraic equality at $r_{\pm }$ would require $E=eQ/r_{\pm }$ and does not describe a regular finite-proper time radial turning point on the horizon. In the Schwarzschild spacetime limit $Q=0$, the positive-radius inner alternative disappears.

The scale factor $a(t)>0$ specifies the common time dependence of proper distance in cosmology between comoving observers. Their comoving coordinates remain fixed, while their separation is multiplied by $a(t)$. Thus $\dot{a}>0$ means expansion and $\dot{a}<0$ means contraction. The coordinate $t$ is cosmic time, measured by the proper time of a comoving observer.

The parameter $k$ fixes the sign of the spatial curvature of an FLRW universe. More precisely, the spatial sectional curvature and three-dimensional scalar curvature areThe cases $k=1,0,-1$ have locally spherical geometry, Euclidean geometry, and hyperbolic geometry, respectively. For $k=\pm 1$, $a$ is the curvature radius of a constant-curvature space in the displayed normalization. Local curvature alone does not fix the global topology.

The Friedmann-Lemaître-Robertson-Walker metric implements the cosmological principle: on the scales being modelled, the spatial slices have spatial homogeneity and cosmological isotropy. There is a family of comoving observers whose orthogonal slices share one cosmic time and one scale factor. Each spatial slice is a locally maximally symmetric space, with time-independent normalized $k$.

The matter is represented by a perfect fluid in general relativity, at rest relative to the comoving observers, with energy density $\varrho (t)$ and pressure $p(t)$ depending only on cosmic time. Its stress-energy tensor isThere is no preferred spatial direction, so there is no anisotropic stress or spatial heat flux in that rest frame. Gravity obeys the Einstein field equations; the equations in this paper contain no separate cosmological constant, although a component with $p=-\varrho$ can represent one. An equation of state relating $p$ and $\varrho$ closes the evolution equations. Spatial homogeneity and cosmological isotropy are idealizations of the cosmological background; they do not assert that individual local structures are uniform.

Set $H=\dot{a}/a$, the Hubble parameter. The first supplied component of the Einstein field equations gives the Friedmann acceleration equationSubstitution into the second supplied equation givesTherefore the first Friedmann equation isDifferentiate this equation with respect to cosmic time:Insert the Friedmann acceleration equation on the left and cancel $8\pi Ga/3$, which is nonzero for $a>0$. This givesThis is the cosmological continuity equation, expressing stress-energy conservation for the perfect fluid in general relativity. No division by $\dot{a}$ was made, so the derivation remains valid at a cosmological bounce or an instant of maximum expansion.

For $a>0$, the assumed inequality and the Friedmann acceleration equation imply $\ddot{a}\le 0$. Thus $a$ is a concave function. Let $a_0=a(t_0)>0$ and $v_0=\dot{a}(t_0)>0$. At every earlier time in the same regular expanding interval, $\dot{a}(t)\ge v_0$. Integration gives the concavity bound on past FLRW evolution:The right-hand side reaches zero at $t=t_0-a_0/v_0$. Hence an everywhere regular positive scale factor cannot extend arbitrarily far into the past. With the usual implicit continuation assumption that the cosmological solution remains regular while $a>0$, it reaches $a(t_B)=0$ at a finite past boundary satisfyingPast expansion lasts at most the present inverse Hubble parameter. The inequality is the essential argument, and equality is attained by a linearly growing scale factor.

There is a literal qualification to the claim that $a$ must reach zero: the printed hypothesis alone does not exclude an earlier curvature singularity with positive limiting $a$. In arbitrary fixed units, take $k=0$ and $a(t)=1+\sqrt{t}$ for $t>0$, and define $\varrho ,p$ using the two Friedmann equations and the Friedmann acceleration equation. Then $\dot{a}>0$, $\ddot{a}<0$, and $\varrho +3p=-3\ddot{a}/(4\pi Ga)>0$, but $a\to 1$ while the energy density and pressure diverge as $t\downarrow 0$. Thus the rigorous unconditional conclusion is the finite bound on a regular past interval; reaching zero additionally requires the stated continuation assumption. Nor does a zero of $a$ alone prove a curvature singularity, as the Milne model illustrates.

For the equation of state $p=-\varrho$, the cosmological continuity equation gives $a\dot{\varrho }=0$. Hence $\varrho ={\varrho }_0$ is constant. First take the physically intended positive energy density and defineThe Friedmann equation and the original Friedmann acceleration equation becomeSolve the second equation as $a=A{e}^{H_{\Lambda }t}+B{e}^{-H_{\Lambda }t}$. The first equation then requires $4H_{\Lambda }^{2}AB=k$. Choosing the time origin separately in each solution gives all positive branches:For $k=-1$, there is also the time reverse $a=H_{\Lambda }^{-1}\sinh [H_{\Lambda }(t_B-t)]$ on $t<t_B$, which contracts to a future zero. For $k=+1$, the cosmological bounce joins contraction to expansion at the positive minimum $a(t_b)=H_{\Lambda }^{-1}$. Retaining $\ddot{a}=H_{\Lambda }^{2}a$ at the minimum excludes a spurious permanently static solution that would satisfy the first-order constraint alone.

These are vacuum-energy FLRW solutions: the expanding flat branch, closed slicing of de Sitter spacetime, and open slicing of de Sitter spacetime. To compare them at the same energy density, the figure sets $T=H_{\Lambda }(t-t_c)$, takes $A_0=H_{\Lambda }^{-1}$, and aligns the open boundary and closed minimum at $T=0$. Thus the three plotted functions are $H_{\Lambda }a=e^T$, $\cosh T$, and $\sinh T$, the last only for $T>0$.

Only the expanding $k=-1$ branch has a finite past zero. The expanding $k=0$ branch tends to zero only as $t\to -\infty$, and the $k=+1$ branch is bounded below. All three describe portions of de Sitter spacetime, with Ricci scalar $R=12H_{\Lambda }^2$ and Kretschmann scalar $24H_{\Lambda }^4$. In particular the $k=-1$ zero is a boundary of that slicing, not a curvature singularity. The inequality in part (d) does not apply, because $\varrho +3p=-2{\varrho }_0<0$.

For zero energy density, the remaining solutions are $a=A_0>0$ for $k=0$, and the expanding or contracting Milne model $a=\pm (t-t_B)>0$ for $k=-1$; no positive real solution exists for $k=+1$. If negative energy density is admitted mathematically, write ${\alpha }^2=-8\pi G{\varrho }_0/3>0$. Only $k=-1$ is possible, with $a={\alpha }^{-1}\sin [\alpha (t-t_B)]$ for $0<\alpha (t-t_B)<\pi$, an open slicing of anti-de Sitter spacetime. These limiting and negative-density cases do not alter the positive-density comparison above.

Take the expanding flat branch $a(t)=A{e}^{H_{\Lambda }t}$, with $A>0$. A radial null geodesic in FLRW spacetime satisfies $d\rho /dt=\pm 1/a(t)$. For the incoming past light cone of the spacetime event $(t_0,\rho =0)$, integration giveswith arbitrary angular direction. Equivalently, conformal time $\eta =-1/(H_{\Lambda }a)$ gives $\rho ={\eta }_0-\eta$.

Let the reception time tend to infinity. The largest comoving radial distance from which a signal emitted at time $t$ can reach the central comoving observer isThus the cosmological event horizon has constant proper distance in cosmology:An emitted spacetime event is seen at a finite time if and only if $a(t)\rho <H_{\Lambda }^{-1}$. Equality would require $t_0=\infty$, while an event with $a(t)\rho >H_{\Lambda }^{-1}$ never reaches the observer. This is a bound on events at their emission time; a fixed comoving observer can have earlier visible events and later invisible ones.

The boundary really is a null hypersurface. For $f(t,\rho )=\rho -[H_{\Lambda }a(t){]}^{-1}$, its normal has components ${\partial }_{t}f=1/a$ and ${\partial }_{\rho }f=1$, soThe finite future integral, and hence this observer event horizon, uses the expanding branch. For the contracting flat solution $a=A{e}^{-H_{\Lambda }t}$, the same integral diverges and there is no future observer event horizon of this kind; the static zero-density solution likewise has none.