Cambridge Part III MathsLent 2016

Black Holes

based on a course given by written up byHarvey Reall Josh Kirklin

Please send errors and suggestions to jjvk2@cam.ac.uk.

In this course we take G = c = 1 and Λ = 0.

Contents1 Spherical stars 3

1.1 Spherical symmetry and time independence . . . . . . . . . . . . . . . . . . . . . . 31.2 Static, spherically symmetric spacetimes . . . . . . . . . . . . . . . . . . . . . . . . 41.3 The TOV equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.4 Maximum mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2 The Schwarzschild solution 82.1 Gravitational redshift . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.2 Geodesics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.3 Eddington-Finkelstein coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.4 Finklestein diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.5 Gravitational collapse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.6 Black hole region . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.7 Detecting black holes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.8 Orbits around black holes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.9 White holes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.10 Kruskal extension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.11 Einstein-Rosen bridge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.12 Extendability and singularities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

3 The Initial Value Problem 173.1 Predictability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173.2 Initial value problem in GR . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183.3 Strong cosmic censorship . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

4 The Singularity Theorem 214.1 Null hypersurfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214.2 Geodesic deviation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214.3 Geodesic congruences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

– 1 –

Contents

4.4 Null geodesic congruences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224.5 Expansion, rotation and shear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234.6 Gaussian null coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244.7 Trapped surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254.8 Raychaudhuri’s equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264.9 Conditions on energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264.10 Conjugate points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274.11 Causal structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

5 Asymptotic Flatness 305.1 Conformal compactification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305.2 Asymptotic flatness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335.3 Definition of a black hole . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355.4 Weak cosmic censorship conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . . 375.5 Apparent horizon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

6 Charged Black Holes 376.1 The Reissner-Nordstrom solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . 376.2 Eddington-Finklestein coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . 386.3 Kruskal-like coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 386.4 Extreme Reissner-Nordstrom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416.5 Majumdar-Papapetrou solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

7 Rotating Black Holes 427.1 The Kerr-Newman solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 427.2 The Kerr solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 437.3 Maximal analytic extension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 437.4 Ergosphere/Penrose process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

8 Mass, Charge, and Angular Momentum 468.1 Charges in curved spacetime . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 468.2 Komar integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 478.3 Hamiltonian formulation of GR . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 488.4 ADM energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

9 Black Hole Mechanics 509.1 Killing horizons and surface gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . 509.2 Interpretation of the surface gravity . . . . . . . . . . . . . . . . . . . . . . . . . . 519.3 Zeroth law of BH mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 519.4 First law of BH mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 529.5 Second law of BH mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

10 QFT in Curved Spacetime 5410.1 Free scalar field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5510.2 Bogoliubov transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5610.3 Particle production in a non-stationary spacetime . . . . . . . . . . . . . . . . . . . 5710.4 Rindler spacetime . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5810.5 Wave equation in Schwarzschild solution . . . . . . . . . . . . . . . . . . . . . . . . 5910.6 Hawking radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6010.7 Black hole thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

– 2 –

1 Spherical stars 1 Spherical stars

10.8 Black hole evaporation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

1 Spherical starsLecture 115/01/16 Consider a gas of cold fermions. This gas will resist compression due to degeneracy pressure

resulting from the Pauli principle. For example, in a white dwarf star, gravity is balanced byelectron degeneracy pressure. Using Newtonian gravity, we can find an upper limit on the mass ofa stable white dwarf, known as the Chandrasekhar limit:

MWD . 1.4M

In a neutron star, gravity is balanced by neutron degeneracy pressure. Neutron stars are tiny; aneutron star with the mass of the sun has a radius of approximately 10 km (for comparison, thesun has a radius of R ≈ 7× 105 km. At the surface of a neutron star, the gravitational potential|φ| ≈ 0.1. Recall that in order to be able to apply Newtonian gravity, we must have |φ| 1.0.1 6 1, so it is important to consider general relativity when reasoning about neutron stars.In this section we will establish that M . 3M for any cold star.

1.1 Spherical symmetry and time independenceConsider the round metric on S2:

dΩ2 = dθ2 + sin2 θ dφ2

Equipped with this metric, if we exclude reflections, S2 has SO(3) as its isometry group. Thismotivates the following:

Definition. A spacetime is spherically symmetric if its isometry group has an SO(3) subgroupwhose orbits are 2-spheres.

Definition. The area-radius function r :M→ R is defined by:

A(p) = 4πr(p)2, r(p) ≥ 0

where A(p) is the area of the SO(3) orbit through p.

A consequence of this is that the induced metric on the SO(3) orbit through p is r(p)2 dΩ2.

Definition. (M, g) is stationary if it permits a timelike Killing vector field (KVF).

Suppose we have a stationary spacetime with timelike Killing vector k. Let Σ be a spacelike 3dimensional hypersurface, and let xi, i = 1, 2, 3 be coordinates on Σ.We define coordinates for the manifold in the following way: from each point (x1, x2, x3) extend anintegral curve of k; the point (t, xi) is a parameter distance t along this curve.

– 3 –

1 Spherical stars 1.2 Static, spherically symmetric spacetimes

(0, xi)k

(t, xi)

Σ

In the chart (t, xi), we can write:

k = ∂

∂t

Then, using the defining property of Killing vectors, we have thatthe metric is independent of t. Thus, we can write:

ds2 = g00(xk) dt2 + 2g0i(xk) dt dxi + gij(xj) dxi dxj

and we have g00 < 0 since k is timelike.Suppose we have a surface Σ given by f(x) = 0, where f :M→ R, df |Σ 6= 0. Then df is normalto Σ. Suppose n is another 1-form that is normal to Σ. Then we can write n = g df + fn′, where gis a function and n′ is some 1-form. We have:

dn = dg ∧ df + g d2f︸︷︷︸=0

+ df ∧ n′ + f dn′

=⇒ dn |Σ = (dg − n′) ∧ df =⇒ n ∧ dn |Σ = 0In fact, the converse is true:Theorem 1 (Frobenius). If n is a 1-form such that n∧ dn = 0, then there exist functions f, g suchthat n = g df , so that n is normal to surfaces of constant f .If n is a 1-form of this type, we say it is hypersurface-orthogonal.

Definition. (M, g) is static if it contains a hypersurface-orthogonal timelike KVF.

Suppose we are in a static spacetime, and define coordinates t, xi as before. Σ is a surface of constantt, so we have k ∝ dt, kµ ∝ (1, 0, 0, 0). Also note that kµ = gµνk

ν = gµν( ∂∂t)ν = (g00, g10, g20, g30).

Hence we can deduce that gi0 = 0, and can write the metric as:ds2 = g00(xk) dt2 + gij(xk) dxi dxj

where as before g00 < 0. In this metric we have a discrete isometry (t, xi) → (−t, xi). A staticmetric must be time-independent and invariant under time reversal. A simple case of a stationarybut not static metric is that associated with a rotating star. If we reverse time the star spins in theother direction.

1.2 Static, spherically symmetric spacetimesIf we have a spacetime that is both stationary and spherically symmetric, then the isometry groupmust contain:

R︸︷︷︸time

translation

× SO(3)︸ ︷︷ ︸S2 orbits

It can be shown that with this condition the spacetime must also be static.Let Σt ⊥ ka be a foliation of the spacetime, and use coordinates (r, θ, φ) on each surface, whereθ, φ are the usual spherical coordinates and r is the area-radius function as defined earlier. Thenwe must have:

ds2 |Σt = e2Ψ(r) dr2 + r2 dΩfor some function Ψ(r). Note that we have no dr dθ or dr dφ terms because they would violatespherical symmetry. If we define t as above we can then write the entire metric as:

ds2 = −e2Φ(r) dt2 + e2Ψ(r) dr2 + r2 dΩfor some other function Φ(r).

– 4 –

1 Spherical stars 1.3 The TOV equations

1.3 The TOV equationsConsider now the matter inside a stationary and spherically symmetric star. We will model thestar as a perfect fluid, which means we have the following energy-momentum tensor:

Tab = (ρ+ P )uaub + ρgab

where ρ is the energy density, P is the pressure, and ua is the 4-velocity of the fluid. Since thestar is stationary, we can assume the fluid is at rest, so ua = e−Φ

(∂∂t

)a(since u is a unit vector

pointing in the t direction). Also, since we have spherical symmetry we can assume that ρ and Pare functions of r only.Lecture 2

18/01/16 Make the following definition:

e2Ψ(r) =(

1− 2m(r)r

)−1

Note that since e2Ψ(r) > 0, we have m(r) < r2 . Using the Einstein field equations G = 8πT it is

now possible to derive the Tolman-Oppenheimer-Volkoff equations:

dmdr = 4πr2ρ (TOV1)

dΦdr = m+ 4πr3P

r(r − 2m) (TOV2)

dPdr = −(P + ρ)m+ 4πr3P

r(r − 2m) (TOV3)

We now have three equations, but four unknowns (m,Φ,ρ and P ). In order to solve this system, wewill need a fourth equation, and the one most commonly chosen is an equation of state relating Pand ρ. In a cold star, we can assume that the temperature T (ρ, P ) = 0 and we can solve this to getP explicitly in terms of ρ:

P = P (ρ)

This is called a barotropic equation of state.We will assume that ρ, P > 0. We will also assume that dP

dρ > 0; this is a stability condition1.Let the radius of the star be R.

Outside the star (r > R) we can assume ρ = P = 0. (TOV1) then gives that m(r) = M a constant.(TOV2) further provides that Φ = 1

2 log(1− 2M

r

)+ Φ0, where Φ0 is another constant. Note

that since gtt = −e2Φ → e−2Φ0 as r → ∞, we can eliminate Φ0 by making a change ofcoordinates t→ eΦ0t, so w.l.o.g. we assume that Φ0 = 0. Hence we have the Schwarzschildmetric:

ds2 = −(

1− 2Mr

)dt2 +

(1− 2M

r

)−1dr2 + r2 dΩ2

By taking r to be large and comparing with Newtonian gravity, we can deduce that M is infact the mass of the star. Note that this metric has a problem. It is singular at the so-calledSchwarzschild radius r = 2M . Thus a static, spherically symmetric star must have R > 2M(in a normal star, R 2M).

1Consider dPdρ < 0. Then if ρ increases by a small amount in a region R, P decreases in R, but then this causes

more fluid to flow into R, increasing ρ further.

– 5 –

1 Spherical stars 1.4 Maximum mass

Inside the star (r < R), we now have matter to deal with. Integrating (TOV1), we have:

m(r) = 4π∫ r

0ρ(r′)r′2 dr′ +m∗

where m∗ is a constant. Consider a constant t hypersurface. The induced line element onthis hypersurface is ds2 = e2Ψ dr2 + r2 dΩ2. The proper radius (i.e. distance to r = 0) of apoint is given by

∫ r0 e

Ψ(r′) dr′. In order for our spacetime to be a manifold, we require that itis locally flat at r = 0, and this requires that the proper radius tends to the area-radius asr → 0. Note that

∫ r0 e

Ψ(r′) dr′ ∼ eΨ(0)r as r → 0, so we require that eΨ(0) = 1, or equivalentlym(0) = 0. From this we deduce that m∗ = 0.If we match this expression on the boundary of the star to the Schwarzschild solution for theexterior, we see that m(R) = M , or:

M = 4π∫ R

0ρ(r)r2 dr (∗)

The volume form on a constant t hypersurface is eΨr2 sin θ dr ∧ dθ ∧ dφ, and so the energy ofthe matter in the star for t constant is:

E = 4π∫ R

0ρeΨr2 dr

Note that since m is increasing, so is eΨ and hence eΨ ≥ 1 for all 0 ≤ r ≤ R. Thus we haveE > M . The reason for this is that we have a gravitational binding energy E −M .If we evaluate m(r)

r < 12 at r = R we see that M

R < 12 . In fact it is possible to improve this:

(TOV3) =⇒ dPdr ≤ 0 =⇒ dρ

dr ≤ 0, and from this we can deduce:

m(r)r

<29

(1− 6πr2P (r) +

[1 + 6πr2P (r)

] 12)

(†)

Setting r = R and noting P (R) = 0, we obtain the so-called Buchdahl inequality: MR < 4

9 .In general, we must solve this system of equations numerically. (TOV1) and (TOV3) are apair of coupled first order ODEs for m(r) and ρ(r), from which we can obtain a unique solutiongiven m(0) = 0 and specifying ρ(0) = ρc, the central density. From (TOV3) we have that Pis decreasing in r, so R(ρc) is determined by fixing P (R) = 0. Then, using (∗) we can obtainM(ρc). Finally, using (TOV2) and the boundary condition that Φ(R) = 1

2 log(1− 2M

R

)we

can deduce Φ(r).To summarise, given an equation of state, static, spherically symmetric, cold stars are a1-parameter family labelled by ρc.

1.4 Maximum massWe wish to find a limit on the maximum mass of a star.

ρc

M

Mmax

– 6 –

2 The Schwarzschild solution

In general, Mmax depends on the equation of state, but here we run into a problem: we do notknow the equation of state in certain conditions, namely ρ > ρ0, where ρ0 is typically on the orderof the density of an atomic nucleus.

coreρ > ρ0

envelopeρ < ρ0

r = r0

r = R

Remarkarbly, it is still possible to find an upperbound on the mass of a star. We do this bysplitting the star into two regions: an envelope,in which we know the equation of state (soρ < ρ0), and a core, in which we do not (ρ > ρ0).Since dρ

dr < 0, the envelope does in fact envelopethe core.Let m0 = m(r0); we call this the core mass.Since the minimum density in the core is ρ0, wehave m0 ≥ 4

3πr30ρ0. Additionally, we can apply

(†) at r = r0 to obtain:

m0r0

<29

(1− 6πr2

0P0 +[1 + 6πr2

0P0] 1

2)

where P0 = P (ρ0). This is a decreasing functionof P0, so m0

r0< 4

9 .Lets plot these two constraints:

43πr

30ρ0

49r0

r0

m0

m0max

core mass must be in this region

We see that we have an upper bound on the core mass. Solving for this upper bound, we find:

m0 <

√16

23πρ0

If ρ0 ≈ nuclear density, then we have m0 . 5M.Now we can extend our solution to the envelope. m0 and r0 together uniquely determine theenvelope, as we can solve (TOV1) and (TOV3) starting at r = r0 and using the known equation ofstate for ρ < ρ0. From this we obtain M as a function of m0 and r0, and so can find the maximalvalue of M when m0, r0 take values in the region in the graph above.Numerically, we can find that M is maximised when m0 is maximised, and that the maximummass is M ≈ m0 ≈ 5M.In fact, it is possible to improve this limit by imposing that the speed of sound is physical, i.e. lessthan the speed of light:

√dPdρ ≤ 1. Using this gives M . 3M.

– 7 –

2 The Schwarzschild solution 2 The Schwarzschild solution

2 The Schwarzschild solutionLecture 320/01/16 We showed earlier that the only static, spherically symmetric solution of the vacuum EFEs is the

Schwarzschild solution:

ds2 = −(

1− 2Mr

)dt2 +

(1− 2M

r

)−1dr2 + r2 dΩ2

t, r, θ, φ are known as Schwarzschild coordinates. We will assume M > 0.In fact:

Theorem 2 (Birkhoff). Any spherically symmetric solution of the vacuum Einstein equations isisometric to the Schwarzschild solution.

So in particular, spherical symmetry and a vacuum implies a static spacetime (for r > 2M).

2.1 Gravitational redshiftConsider two fixed observers A,B in a Schwarzschild spacetime. A sends two photons to B,seperated by a time ∆t.

r

t

A B

∆t

∆t

Because ∂∂t is an isometry of the spacetime, the second

photon’s path is the same as that of the first, but trans-lated by ∆t. Consider the 4-velocity of a fixed observer.We have:

−1 = uµuµ = gtt

( dtdτ

)2= −

(1− 2M

r

)( dtdτ

)2

Hence we have dτ =√

1− 2Mr dt. Therefore the proper

time intervals between the photons at A and B are:

∆τA =√

1− 2MrA

∆t, ∆τB =√

1− 2MrB

∆t

So we have:∆τB∆τA

=

√1− 2M

rB√1− 2M

rA

If we suppose that the photons were sent at two subsequent wavecrests, then ∆τ is the period ofthe waves, equal to λ, the wavelength (since c = 1). We define the redshift z by:

1 + z = λBλA

=

√1− 2M

rB√1− 2M

rA

For rB > rA, we have z > 0, so light is redshifted as it climbs out of the gravitational field. ForrB 2M :

1 + z =√

11− 2M

rA

Note that this →∞ as rA → 2M .For a star, we have the Buchdahl inequality, R > 9

4M , so plugging this into the above, we find thatthe maximum redshift from the surface of a spherical star is z = 2.

– 8 –

2 The Schwarzschild solution 2.2 Geodesics

2.2 GeodesicsSuppose xµ(τ) is an affinely parametrised geodesic, and let its 4-velocity be uµ = dxµ

dτ . We haveKilling fields k = ∂

∂t and m = ∂∂φ , so along geodesics we have two conserved quantities:

E = −k · u =(

1− 2Mr

) dtdτ and h = m · u = r2 sin2 θ

dφdτ

If our geodesic is timelike and we choose τ to be proper time, we can identify E as the energy perunit mass and h as the angular momentum per unit mass associated with the geodesic.In the null case, we can define the impact parameter b =

∣∣∣ hE ∣∣∣, and identify this as the limit of thedistance between the geodesic and the star perpendicular to the geodesic as r → 0.

φ = constant

r = 0

null geodesic

bb

equationÃľeieieieieieeExercises:

1. Derive the Euler-Lagrange equation for θ(τ). Show that one can choose coordinates such thatθ(τ) = π

2 , so that motion is contained in the equatorial plane.

2. Rearrange the definition of proper time:

gµνuµuν = σ =

1 timelike0 null−1 spacelike

to obtain 12

(drdτ

)2+ V (r) = 1

2E2, where V (r) = 1

2

(1− 2M

r

) (σ + h2

r2

).

2.3 Eddington-Finkelstein coordinatesConsider radial null geodesics (σ = 0) in r > 2M . Since φ is constant, we have h = 0 and so V = 0.Since we are dealing with a null geodesic, we are free to scale τ such that E = 1. Hence we have:

dtdτ =

(1− 2M

r

)−1,

drdτ = ±1

where the sign in the second equation depends on whether the geodesic is outgoing or ingoing. Onething of note is that an ingoing geodesic reaches r = 2M in finite τ . The same is not true of t:

dtdr = ±

(1− 2M

r

)−1

so t→ ∓∞ as r → 2M .Define r∗ = r + 2M log

∣∣ r2M − 1

∣∣, dr∗ = dr1− 2M

r

(∗).

– 9 –

2 The Schwarzschild solution 2.4 Finklestein diagram

r∗

r2M

We have dtdr∗ = ±1, so t ∓ r∗ is a constant. Define v = t+ r∗ (†),

a constant along ingoing radial geodesics. The ingoing Eddington-Finklestein coordinates are v, r, θ, φ. In these coordinates, the lineelement is given by:

ds2 = −(

1− 2Mr

)dv2 + 2 dv dr + r2 dΩ2

This is smooth for all r > 0. In matrix form, the metric is:

gµν =

−(1− 2M

r

)1 0 0

1 0 0 00 0 r2 00 0 0 r2 sin2 θ

We have g = det gµν = −r4 sin2 θ, so gµν is non-degenerate for all r > 0 and furthermore it isLorentzian for all r > 0.In summary, spacetime can be extended through r = 2M to a new region r < 2M .Exercise: for 0 < r < 2M , define r∗ by (∗) and t by (†). Show that the metrix in coordinatest, r, θ, φ is the Schwarzschild metric with 0 < r < 2M .So for a ingoing radial null geodesic inside r = 2M we have dr

dτ = −1, so it reaches r = 0 in finite τ .Consider RabcdRabcd. Some work will lead to:

RabcdRabcd ∝ M2

r6 →∞ as r → 0

This quantity is a scalar, so it diverges in any coordinate system. We call r = 0 a curvaturesingularity. There are infinite tidal forces at r = 0. Note that r = 0 is not a part of the ourspacetime, because gab is not defined there.For r > 2M we have the “static” KVF ∂

∂t . In Eddington-Finklestein coordinates xµ, we have:

k = ∂xµ

∂t

∂

∂xµ= ∂

∂v

Also, k2 = gvv = −(1− 2M

r

), so k is null at r = 2M , and spacelike at r < 2M . Only r > 2M is

static.

2.4 Finklestein diagramLecture 422/01/16 Consider outgoing radial null geodesics in r > 2M . We have t− r∗ = constant, so:

v = 2r∗ + constant = 2r + 4M log∣∣∣∣ r2M − 1

∣∣∣∣+ constant (∗)

Exercise: Consider null geodesics in ingoing Eddington-Finklestein coordinates and show that thesefall into 2 families: ingoing with v = constant, and outgoing either of the form (∗) or r = 2M .Let t∗ = v − r. We can draw the radial null geodesics in a Finklestein diagram:

– 10 –

2 The Schwarzschild solution 2.5 Gravitational collapse

r

curv

atur

esin

gula

rity

t∗

2M

ingoing

outgo

ing

In r < 2M , r decreases along both families, and we reach r = 0 in finite τ . In fact we will showlater that r decreases along any timelike or null curve in r < 2M . Equipped with this knowledgewe can make a rough definition of a black hole as a region of space from which no signal can reach“infinity”.

2.5 Gravitational collapseThe surface of a collapsing star follows a timelike geodesic, and we can plot this on a Finklesteindiagram:

r

curv

atur

esin

gula

rity

t∗

2M

stellar interior (not Schwarzschild)

It will be shown in the first example sheet that the total proper time along a timelike curve withr ≤ 2M can’t exceed πM , so a star collapses from r = 2M to r = 0 within a proper time πM (thisis about 10−5 s for M = M. When this happens, a distant observer never sees the star crossr = 2M – it just redshifts away.

– 11 –

2 The Schwarzschild solution 2.6 Black hole region

2.6 Black hole region

Definition. A non-zero vector is causal if it is timelike or null. A curve is causal if its tangentvector is everywhere causal.

Definition. A spacetime is time-orientable if is admits a time orientation, i.e. a causal vectorfield T a.

Definition. A future-directed causal vector is one that lies in the same lightcone as the timeorientation T a; a past-directed causal vector is one that does not.

Given a time orientation T a, we always have a second inequivalent time orientation −T a.For r > 2M Schwarzschild, the obvious choice of time orientation is k = ∂

∂t . However, k = ∂∂v is

not causal for r < 2M . In ingoing Eddington-Finklestein coordinates, ± ∂∂r is null, since grr = 0.

We can choose either of these as a time orientation, but we want to pick a sign that agrees with kfor r > 2M . We have:

k ·(± ∂

∂r

)= ±gvr = ±1

Thus we use − ∂∂r as the time orientation.

Lemma 1. Let xµ(λ) be a future-directed causal curve. If r(λ0) ≤ 2M , then r(λ) ≤ 2M for allλ ≥ λ0.Proof. Let V µ = dxµ

dλ . V µ is future-directed. Thus we have:

0 ≤(− ∂

∂r

)· V = −grµV µ = −V v = −dv

dλand so:

V 2 = −(

1− 2Mr

)(dvdλ

)2+ 2 dv

dλdrdλ + r2

(dΩdλ

)2

=⇒ − 2 dvdλ

drdλ = −V 2︸ ︷︷ ︸

≥0

−(

1− 2Mr

)(dvdλ

)2+ r2

(dΩdλ

)2

︸ ︷︷ ︸≥0

Thus if r ≤ 2M , we have dvdλ

drdλ ≤ 0.

Suppose r ≤ 2M and drdλ > 0. Then since dv

dλ ≤ 0 we must have dvdλ = 0, and hence V 2 = 0 = dΩ

dλ .The only non-zero component of V is V r = dr

dλ > 0, so V is a positive multiple of ∂∂r , but this

implies that V is past-directed, which is a contradiction.Hence we have dr

dλ ≤ 0 if r ≤ 2M , and we can show similarly drdλ < 0 if r < 2M . Hence if r(λ0) < 2M ,

then r(λ) is monotonically decreasing for λ ≥ λ0.

2.7 Detecting black holesBlack holes have two recognizable qualities:• Unlike in the case of cold stars, there is no upper bound on the mass of a black hole.

• Black holes are very small for a given mass.One interesting case is that of the supermassive black holes. No one knows how they form...

– 12 –

2 The Schwarzschild solution 2.8 Orbits around black holes

2.8 Orbits around black holesLecture 525/01/16 Comsider timelike geodesics, and recall the orbital equation of a Schwarzschild black hole:

12

(drdτ

)2+ V (r) = 1

2E2 where V (r) = 1

2

(1− 2M

r

)(1 + h2

r2

)

It can easily be shown that V ′(r) = 0 if r = r± = h2±√h4−12h2M2

2M . Lets plot V :

r

V

12

r− r+

r = r+ is a stable circular orbitr = r− is an unstable circular orbit

It is a simple exercise to show that 3M < r− < 6M < r+. We call r = 6M the innermost stablecircular orbit (ISCO).Suppose r = r±, then we have a circular orbit dr

dτ = 0, and we can show:

E2

2 = V (r) =⇒ E = r − 2Mr

12 (r − 3M)

12≈ 1− M

2r for r 2M

Hence we have that the energy of a distant orbit is approximately m− Mm2r . m is the rest mass

energy of the orbiting particle, and Mm2r is the gravitational binding energy of its orbit.

r=

6Mr = 2M

When a star orbits around a black hole, the black hole robsthe star of matter, forming an accretion disc around theblack hole. As a first approximation we will assume thatparticles in this accretion disc follow stable circular orbitsof the form above. Friction between the particles causestheir E to decrease, and hence their r to also decrease,and so these particles will fall towards the ISCO, wherethey will then fall into the black hole.At r → ∞ we have E = 1, and at the ISCO we haveE =

√89 . Thus the proportion of lost to friction (and

then radiated away as x-rays) is 1−√

89 ≈ 6%.

2.9 White holesConsider again the region r > 2M , and let u = t− r∗. Wedefine the outgoing Eddington-Finklestein coordinates asu, r, θ, φ. In these coordinates, the line element is given by:

ds2 = −(

1− 2Mr

)du2 − 2 dudr + r2 dΩ2

– 13 –

2 The Schwarzschild solution 2.10 Kruskal extension

Since gµν is smooth and det g 6= 0 for all r > 0, we can extend this to 0 < r ≤ 2M , and again wehave a curvature singularity at r = 0. However it is important to note that this is not the samer < 2M region as before! To see this, consider outgoing radial null geodesics which have u constantand dr

dτ = +1. These have r increasing through r = 2M , in direct contradiction with the previouscase. r < 2M is not a black hole.Exercise: show (i) k = ∂

∂u , (ii) the time orientation equivalent to k for r 2M is + ∂∂r .

r < 2M is known as a white hole region; it is a region into which no signal from infinity can enter.In a certain sense a white hole is the time reversal of a black hole: u 7→ −v is an isometry mappingoutgoing Eddington-Finklestein coordinates to ingoing Eddington-Finklestein coordinates, but itdoes not preserve the time-orientation.

2.10 Kruskal extensionConsider r > 2M Schwarzschild spacetime. We define Kruskal-Szekeres coordinates (U, V, θ, φ) by

U = −e−u/4M < 0 and V = ev/4M > 0.

We haveUV = −er∗/2M = −er/2M

(r

2M − 1), (∗∗)

which is monotonic and thus determines r = r(U, V ). Similarly,

V

U= −et/2M

determines t = t(U, V ).Exercise: show that the metric in Kruska-Szekeres coordinates is given by

ds2 = −32M3e−r(U,V )/2M

r(U, V ) dU dV + r(U, V )2 dΩ2 .

We use (∗∗) to define r(U, V ) for U ≥ 0 or V ≤ 0. We can analytically extend the spacetime withdet g 6= 0 through U = 0 or V = 0 to new regions with U ≥ 0 or V ≤ 0. r = 2M corresponds totwo surfaces U = 0 and V = 0, intersecting at U = V = 0. r = 0 corresponds to a hyperbola with 2branches. Ingoing and outgoing geodesics are described by V , U constant respectively. We can plotthse features on a Kruskal diagram:

r constant →

t constant →

ingoing

outgo

ing

radial null geodesicsIII

IIIIV

r = 0

r = 0

VU

– 14 –

2 The Schwarzschild solution 2.11 Einstein-Rosen bridge

Note that points in this diagram correspond to U = V = constant two dimensional surfaces inspace.There are four regions on the Kruskal diagram:

I: This is just the r > 2M Schwarzschild spacetime that we are used to.

II: This is the black hole region of ingoing Eddington-Finklestein coordinates.

III: This is the white hole region of outgoing Eddington-Finklestein coordinates.

IV: This is new; it is an asymptotically flat region isometric to region I.

Exercise: showk = 1

4M

(V

∂

∂V− U ∂

∂U

)and k2 = −

(1− 2M

r

).

Thus k is timelike in regions I and IV, spacelike in II and III, and null at V = 0 or U = 0. We canplot its integral curves:

III

IIIIV

VU

U = 0 and V = 0 are both independently fixed by k. k = 0 on U = V = 0, a region known as thebifurcation 2-sphere.Lecture 6

27/01/16

2.11 Einstein-Rosen bridgeConsider a constant t slice of Kruskal spacetime. We define the coordinate ρ on this slice by

r = ρ+M + M2

4ρ such that ρ >M

2 in I, ρ < M

2 in IV.

In isotropic coordinates (t, ρ, θ, φ), the line element is

ds2 = −

(1− M

2ρ

)2

(1 + M

2ρ

)2 dt2 +(

1 + M

2ρ

)4(dρ2 + ρ2 dΩ2).

Note that ρ→ M2

4ρ interchanges I and IV.

– 15 –

2 The Schwarzschild solution 2.12 Extendability and singularities

I

IV

S2

ρ→∞

ρ→ 0

On a constant t surface, the induced metric is

ds2 =(

1 + M

2ρ

)4(dρ2 + ρ2 dΩ2).

If we take ρ > 0, we see that the surface is a Riemannianmanifold with topology R × S2. We can visualise thissurface by embedding into four dimensional Euclideanspace. We have two asymptotically flat regions at ρ→∞,ρ → 0 connected by a “throat” with minimum radiusr = 2M at ρ = M

2 .

2.12 Extendability and singularities

Definition. A spacetime (M, g) is said to be extendable if it is isometric to a proper subset ofanother spacetime (M′, g), called an extension of (M, g).

Example. r > 2M Schwarzschild spacetime is extendable, with for example Kruskal spacetimeas an extension. The isometry in question is the identity map. Kruskal spacetime on the otherhand is inextendible, and is in fact a maximal analytic extension of (M, g).

There are many types of singularities.

• A scalar curvature singularity is a region in which a scalar constructed from the Riemanntensor Rabcd blows up.

• More generally, a curvature singularity is a region in which there does not exist a chart suchthat the components of the Riemann tensor Rµνρσ are finite.

• It is possible to have singularities which have nothing to do with the curvature tensor. Forexample consider M = R2 with polar coordinates (r, φ), where we identify φ ∼ φ+ 2π, withline element given by

ds2 = dr2 + λ2r2 dφ2

for some constant λ > 0. There are two cases:– In the case λ = 1, this is just Euclidean space, and r = 0 is just a coordinate singularity.– If λ 6= 1, then set φ′ = λφ to obtain

ds2 = dr2 + r2 dφ′2 .

Locally, this is isometric to Euclidean space. It is flat, so Rabcd = 0 and there is nocurvature singularity. However the range of the angular coordinate has changed; ouridentification has changed to φ′ ∼ φ+ 2πλ. Consider a circle with r = ε. The ratio ofthe circumference to the radius of this circle is 2πλε/ε = 2πλ. As we take ε, we wouldexpect this ratio to approach 2π if the geometry were locally flat at r = 0, but this is notthe case. g is not smooth at r = 0. This type of singularity is called a conical singularity.

– 16 –

3 The Initial Value Problem 3 The Initial Value Problem

Definition. p ∈ M is a future endpoint of a future-directed causal curve γ : (a, b) → M iffor any neighbourhood O of p there exists a t0 such that γ(t) ∈ O for all t > t0. We say γ isfuture-inextendable if it has no future endpoint.

Example. Let (M, g) be Minkowski space, and γ : (−∞, 0) → M, γ(t) = (t, 0, 0, 0). Then(0, 0, 0, 0) is a future end point of γ. If however (M, g) is Minkowski\(0, 0, 0, 0), then γ isfuture-inextendable.

Definition. A geodesic is complete if an affine parameter extends to ±∞. A spacetimeis geodesically complete if all inextendable geodesics are complete. If a spacetime is bothinextendable and geodesically incomplete, then we say it is singular.

3 The Initial Value Problem3.1 Predictability

Definition. A partial Cauchy surface Σ is a hypersurface for which no two points are connectedby a causal curve in M.

Definition. The future domain of dependence of Σ is the set

D+(Σ) = p ∈M s. t. every past-inextendible causal curve through p intersects Σ.

We define the past domain of dependence D−(Σ) in a similar way. The domain of dependenceis D(Σ) = D(Σ+) ∪D(Σ−).

A causal geodesic in D(Σ) is uniquely determined by its velocity at q ∈ Σ (this is because thegeodesic equation is a hyperbolic PDE).

Example. Consider 2d Minkowski space, and the surface Σ = (0, x) s. t. x > 0. ThenD+(Σ) = (t, x) s. t. 0 ≤ t < x, D−(Σ) = (t, x) s. t.−x < t ≤ 0.

Definition. (M, g) is called globally hyperbolic if it contains a Cauchy surface i.e. a partialCauchy surface whose domain of dependence is all of M.

Example. Minkowsi spacetime is globally hyperbolic (for instance t constant is a Cauchysurface). Kruskal spacetime is globally hyperbolic (for instance U + V constant is a Cauchysurface).2d Minkowski spacetime with (0, 0) removed is not globally hyperbolic.

– 17 –

3 The Initial Value Problem 3.2 Initial value problem in GR

Theorem 3. Suppose (M, g) is a globally hyperbolic spacetime. Then:

1. There exists a global time function i.e. a map t :M→ R such that (dt)a is future-directedand timelike.

2. t constant surfaces are Cauchy and all have the same topology Σ.

3. M = R× Σ.

Exercise: show U+V is a global time function for Kruskal spacetime. U+V = 0 is an Einstein-Rosenbridge, so Σ = R× S2 and M = R2 × S2.

xi(p)

p

t = 0

If we have a time function t, we will often carry out a3+1 split to obtain a set of local coordinates. Suppose wehave coordinates xi on the surface t = 0, and a timelikevector field T a. For a point p near t = 0, we let xi(p) bethe coordinate of the point where the integral curve ofT a through p intersects t = 0. Thus we have a coordinatechart t, xi. In these coordinates, the metric is given by

ds2 = −N(t, x)2 dt2 + hij(t, x)(dxi +N i(t, x) dt)(dxj +N j(t, x) dt),

where N and N i are known as the lapse function and shift vector respectively, and hij is the metricon a constant t surface.

3.2 Initial value problem in GRLecture 729/01/16 We can view the Einstein equations as an initial value problem. We are given Σ, a 3d Riemannian

manifold with metric hab and extrinsic curvature Kab. In addition we have the Hamiltonianconstraint

R′ −KabKab +K2 = 16πρ,

where R′ is the Ricci scalar of h, K = K aa and ρ = Tabn

anb (with n a unit normal to Σ), and themomentum constraint

DbKba −DaK = 8πh b

a Tbcnc,

where Db is the Levi-Civita connection with regard to h.

Theorem 4 (Choquet-Bruhat and Geroch 1969). Given initial data satisfying vacuum constraints(i.e. the right hand sides of the above equal to 0), there exists a unique (up to diffeomorphism)spacetime M, g, known as the maximal Cauchy development of Σ, hab,Kab, such that the followingare satisfied:

1. (M, g) obeys the vacuum Einstein equations.

2. (M, g) is globally hyperbolic with Cauchy surface Σ.

3. The induced metric and extrinsic curvature of Σ are hab and Kab respectively.

4. Any other spacetime obeying 1-3 is isometric to a subset of (M, g).

Note that (M, g) may be extendable, but the solution will be non-unique outside of D(Σ).

– 18 –

3 The Initial Value Problem 3.2 Initial value problem in GR

Example. Consider Σ = (x, y, z) s. t. x > 0, hµν = δµν , Kµν = 0. Then (M, g) is the regionof Minkowski spacetime with |t| < x, and this is clearly extendable.

In the preceding example, (M, g) was extendable because (Σ, hab) was extendable, but this neednot be the case.

Example. Consider M < 0 Schwarzschild. The metric is

ds2 = −(

1 + 2|M |r

)dt2 +

(1 + 2|M |

r

)−1dr2 + r2 dΩ2 .

There is a curvature singularity at r = 0 but, unlike the positive mass case, there is no eventhorizon. We choose the initial data (Σ, hab,Kab) to be that given by the surface t = 0; this isinextendable (but not geodesically complete since it is singular at r = 0). For outgoing radialnull geodesics at small r, we have

∂t

∂r=(

1 + 2|M |r

)−1≈ r

2|M | ,

so we can write t ≈ t0 + r2

4|M | . If t0 > 0 then the geodesic never intersects Σ, so Σ is not aCauchy surface for M. The boundary of D(Σ) is given by those geodesics with t0 = 0.

r

ΣD(Σ)

The solution outside of D(Σ) is not determined by the data on Σ; in particular, it need not bethe negative mass Schwarzschild that we started with.

This time, the maximal development was extendable because the initial data was singular, butagain this is not necessary.

Example. Consider 4d Minkowski spacetime, and start with data on the surface Σ = −t2 +x2 + y2 + z2 = −1, t < 0. This is one sheet of a hyperboloid. The maximal development of Σis the interior of the past lightcone of the origin, and this is extendable.

– 19 –

3 The Initial Value Problem 3.3 Strong cosmic censorship

t

y

xΣ

D(Σ)

Now, the maximal development is extendible because the initial data is “asymptotically null”. Toavoid this problem, we will use asymptotically flat initial data.

Definition. (Σ, hab,Kab) is an asymptotically flat end if:

• Σ is diffeomorphic to R3 \B where B is a closed ball centred on the origin in R3.

• If we pull back R3 coordinates to give coordinates xi on Σ, then the metric is hij =δij +O(1/r) and the extrinsic curvature is Kij = O(1/r2).

• hij,k = O(1/r2) etc.

Definition. A set of initial data is said to be asymptotically flat with N ends if it is a unionof a compact set with N asymptotically flat ends.

Example. Consider M > 0 Schwarzschild spacetime. Σ = t = constant, r > 2M is anasymptotically flat end. Σ is part of an Einstein-Rosen bridge, which is asymptotically flat with2 ends (in fact it is the union of two copies of Σ and the bifurcation two-sphere at r = 2M).

3.3 Strong cosmic censorshipThe strong cosmic censorship conjecture (by Penrose) is the following:

Given vacuum initial data (Σ, hab,Kab) that is geodesically complete and asymptoticallyflat, then generically the maximal Cauchy development is inextendable.

The conjecture has been shown to be true for nearly flat data, but there are non-generic counterex-amples.

– 20 –

4 The Singularity Theorem 4 The Singularity Theorem

4 The Singularity Theorem4.1 Null hypersurfaces

Lecture 801/02/16

Definition. A null hypersurface is a hupersurface N whose normal n is everywhere null.

Example. Consider a constant r surface in Schwarzschild spacetime in ingoing Eddington-Finklestein coordinates. The metric is

gµν =

0 1 0 01 1− 2M

r 0 00 0 1

r2 00 0 0 1

r2 sin2 θ

.The normal 1-form is n = dr, and we have n2 = gµνnµnν = grr = 1− 2M

r , so r = 2M is a nullhypersurface.

Note that iff Xa is tangent to N , then either Xa is spacelike or it is parallel to na. Thus na istangent to N and in particular the integral curves of na lie within N .

Lemma 2. The integral curves of na are null geodesics (they are referred to as the generators ofN ).

Proof. Define N by f = constant for some function f . We have df 6= 0 on N and n = hdf forsome h. Let N = df . The integral curves of n and N are the same up to reparametrisation, sowe focus on N . Since N is null, we have NaN

a = 0 on N , and so d(NaNa) is normal to N . Thus,

∇a(N bNa) = 2αNa for some α on N . The left hand side is

2N b∇aNb = 2N b∇a∇bf = 2N b∇b∇af = 2N b∇bNa.

Hence on N , Na satisfies the geodesic equation N b∇bNa = αNa.

Example. Consider Kruskal spacetime with coordinates (U, V, θ, φ). N = dU is null everywhere,so we have a family of null hypersurfaces U = constant. In this case, we have N b∇bNa =12∇a(N

2) = 0, so Na is tangent to affinely parametrised geodesics. It is easy to show thatNa = − r

16M3 er/2M

(∂∂V

)a, so if we let N = U = 0 then V is an affine parameter for generators

of N , and similarly U is an affine parameter for generators of V = 0.

4.2 Geodesic deviation

Definition. A 1-parameter family of geodesics is a function γ : I×I ′ →M where I, I ′ are openintervals in R, such that λ 7→ γ(s, λ) is a geodesic with affine parameter λ and (s, λ) 7→ γ(s, λ)is smooth and 1-to-1 with a smooth inverse.

– 21 –

4 The Singularity Theorem 4.3 Geodesic congruences

U U U

S

S

Let γ(s, λ) take coordinates xµ(s, λ); we define the vector fields

Sµ = ∂xµ

∂sand Uµ = ∂xµ

∂λ.

We also define the deviation vectors (?) from the coordinates s, λon the image of γ,

S = ∂

∂sand U = ∂

∂λ.

We have [S,U ] = 0, so we can write U b∇bSa = Sb∇bUa, and thus obtain the geodesic deviationequation

U c∇c(U b∇bSa) = RabcdUbU cSd.

A solution Sa to this equation along γ is called a Jacobi field.

4.3 Geodesic congruences

Definition. Let U ∈ M be open. A geodesic congruence is a family of geodesics such thatexactly one geodesic passes through each p ∈ U .

Let Ua be the tangent vector of a geodesic congruence, normalised such that U2 = −1 if timelike,0 if null, and 1 if spacelike. We define the velocity gradient Ba

b = ∇bUa. By the above we haveU b∇bSa = Ba

bSb. It is easy to see that UaBa

b = 0 = BabU

b. We also have

U ·∇(U · S) = (U ·∇Ua)︸ ︷︷ ︸=0

Sa + UaU ·∇Sa = UaBab︸ ︷︷ ︸=0

Sb = 0,

so U · S is constant along any geodesic in the congruence.Suppose we redefine our affine parameter λ→ λ′ = λ−a(s). We have S′a = Sa + da

dsUa. Sa and S′a

point to the same geodesics, so we have a kind of gauge freedom here. Note that U ·S′ = U ·S+ dadsU

2,so in the spacelike and timelike cases, we can choose a(s) such that U · S = 0 at λ = 0, and henceeverywhere.

4.4 Null geodesic congruencesThings are less easy if U is null. Pick some spacelike hypersurface Σ that is tranverse to U (i.e.not tangent). Pick a vector field Na such that N2 = 0 and N · U = −1 on Σ, and additionallyU ·∇Na = 0. It can be shown that this implies that N2 = 0 and N · U = −1 everywhere. Nowwrite

Sa = αUa + βNa + Sa

where U · S = N · S = 0 (this means that Sa is either spacelike or zero). We have U · S = −β, soβ is constant along each geodesic. Note that αUa + Sa is orthogonal to Ua and βNa is parallelytransported along Ua.

Example. Consider a null hypersurface N and pick a congruence containing the generators ofN . For a 1-parameter family of generators, we have Sa tangent to N , and so β = −U · S = 0.

– 22 –

4 The Singularity Theorem 4.5 Expansion, rotation and shear

We can write Sa = P ab Sb where P ab = δab = NaUb + UaNb. P ab is a projection (P abP bc = P ac )

onto T⊥ = vectors ⊥ to Ua, Na ⊂ Tp(M), a 2d space. We have U ·∇P ab = 0.

Lemma 3. If U · S = 0, then U ·∇Sa = Bab S

b, where Bab = P acB

cdP

db .

Proof. We have:

U ·∇Sa = U ·∇(P ac Sc)= P acU ·∇Sc

= P acBcdS

d

= P acBcdP

de S

e (using B · U = U · S = 0)= P acB

cdP

db︸ ︷︷ ︸

=Bab

P be Se︸ ︷︷ ︸

=Sb

(since P = P 2)

4.5 Expansion, rotation and shearLecture 903/02/16

Definition. Expansion θ, rotation ωab and shear σab are defined in the following way:

θ = Baa ω = B[ab] σab = B(ab) −

12Pabθ

We have Bab = 1

2θPab + σab + ωab, and it can be shown that θ = gabBab = ∇aUa.

Lemma 4. If a congruence contains the generators of a null hypersurface N , then ω = 0 on N .Conversely, if ω = 0 then Ua is everywhere hypersurface orthogonal.

Proof. Since B · U = U ·B = 0, we can write

Bbc = Bb

c + U bNdBdc + UcB

bdN

d + U bUcNdBdeN

e.

We haveU[aωbc] = U[aBbc] = U[aBbc] = U[a∇cUb] = −1

6(U ∧ dU)abc,

so since Ua is orthogonal to N and hence U ∧ dU = 0 on M, we can conclude

0 = U[aωbc]

∣∣∣N

= 13(Uaωbc + Ubωca + Ucωab)

∣∣∣∣N.

Contracting with Na, we see that ωbc = 0 on N as required.For the converse, we have ω = 0 =⇒ U ∧ dU = 0, which by Frobenius’ theorem implies U ishypersurface orthogonal.

Consider geodesics in N with tangent vector Sa. By foliating N into a family of constant λ surfaces,we can visualise how expansion and shear act on these geodesics.

– 23 –

4 The Singularity Theorem 4.6 Gaussian null coordinates

expansion shear

λ = λ0

λ = λ1

With expansion, geodesics move apart from one another for positive θ, and closer together fornegative θ. If there is a shear, then geodesics move closer together in one direction, but furtherapart in the other.

4.6 Gaussian null coordinatesWe can define a vector field V on N by

V 2 = 0, V · U = 1 and V · ∂∂yi

= 0.

N

(λ, yi)

S

yi

V(r, λ, yi)

To construct Gaussian null coordinates near N ,we assign the coordinates (r, λ, yi) to a pointaffine parameter distance r along a null geodesicstarting at (λ, yi) ∈ N with tangent V a there.Recall that U = ∂

∂λ , so V = ∂∂r is tangent to

affinely parametrised null geodesics, so grr =0. Exercise: the geodesic equation reduces togrµ,r = 0. Therefore we have grλ = grλ|r=0 =U · V |N = 1 and gri = gri|r=0 = V · ∂

∂yi

∣∣∣N

= 0.Also, since gλλ|r=0 = U2|N = 0 we have gλλ =rF , and similarly since gλi|r=0 = U · ∂

∂yi

∣∣∣N

= 0we have gλi = rhi, where F and hi are somesmooth functions. Hence we have the followingline element:

ds2 = 2 dr dλ+rF dλ2+2rhi dλ dyi+hij dyi dyj

The induced line element on N is therefore

ds2 |N = 2 dr dλ+ hij dyi dyj .

Using this metric, we can move the index downstairs on Uµ|N = (0, 1, 0, 0) to obtain Uµ|N =

– 24 –

4 The Singularity Theorem 4.7 Trapped surfaces

(1, 0, 0, 0), and hence from U ·B = B · U = 0, we have Brµ = Bµ

λ = 0. Also

θ = Bµµ = Bi

i = ∇iU i = ∂iUi + ΓiiµUµ

= Γiiµ = 12(gµi,λ + gµλ,i − giλ,µ)

= 12h

ij(gij,λ + gjλ,i︸︷︷︸=0

− giλ,j︸︷︷︸=0

)

= 12h

ij∂λhij = ∂λ√h√h,

where h = dethij . Thus we have∂

∂λ

√h = θ

√h.

√h is the area element on a surface of constant λ in N , so we can see why θ is called the “expansion”.

4.7 Trapped surfacesLet S be a 2d spacelike (orientable) surface. Given a point p ∈ S, there exist two independentfuture-directed null vectors orthogonal to S at p, say U2, U1 (up to scaling). Thus there are twofamilies of null geodesics starting on S and orthogonal to S, and two null hypersurfaces N1 and N2generated by these families. The two families are outgoing and ingoing light rays from S. Let theexpansion on these two surfaces be θ1 and θ2 respectively.

Definition. A compact orientable spacelike 2-surface S is trapped if θ1, θ2 < 0 everywhere onS. It is marginally trapped if θ1, θ2 ≤ 0 everywhere on S.

Example. Consider S = U = U0, V = V0 ∼ S2 in Kruskal spacetime. The generators of Niare the radial null geodesics with either U = constant or V = constant.

VU

III

IIIIV

S(U0,V0)

N1N2

Each null surface contains geodesic tangent vectors of the form (dU)a ∝ rer

2M(∂∂V

)a= Ua1 and

– 25 –

4 The Singularity Theorem 4.8 Raychaudhuri’s equation

(dV )a ∝ rer

2M(∂∂U

)a= Ua2 respectively. We have

θ1 = ∇aUa1 = 1√−g

∂µ(√−gUµ1 )

= r−1er

2M ∂V (re−r

2M rer

2M )= 2e

r2M ∂V r.

Using UV = −er

2M(r

2M − 1)

we can find ∂V r, and so can obtain θ1 = −8M2

r U . Similarly wehave θ2 = −8M2

r V . We set U = U0 and V = V0 to find the expansion of the two surfaces.If S is in region I, we have θ1 > 0 and θ2 < 0, so this is not a trapped surface. However, if S isin the black hole region II, then θ1, θ2 < 0, and so S is trapped.

4.8 Raychaudhuri’s equationLecture 10

05/02/16 Lemma 5. Raychaudhuri’s equation holds:

dθdλ = −1

2θ2 − σabσab + ωabωab −RabUaU b

Proof. We have

dθdλ = U ·∇(Ba

bPba ) = P baU ·∇Ba

b

= P baUc∇c∇bUa

= P baUc(∇b∇cUa +RadcbU

d)= P ba (∇b(U c∇cUa︸ ︷︷ ︸

=0

)− (∇bU c)∇cUa) + P baRadcbU

cUd

= −BcbP

baB

ac −RcdU cUd

= −BcaB

ac −RabUaU b

= −12θ

2 − σabσab + ωabωab −RabUaU b.

4.9 Conditions on energyOften we will impose energy conditions on the energy momentum tensor.

Dominant energy condition

This states that −T ab V b is a future-directed causal vector (or zero) for all future-directed timelikevectors V . The DEC implies that if Tab = 0 in a closed S ∈ Σ, then Tab = 0 in D+(S). Saidanother way: nothing can travel faster than the speed of light.

– 26 –

4 The Singularity Theorem 4.10 Conjugate points

Example. Consider the energy-momentum tensor of a scalar field

Tab = ∂aφ∂bφ−12gab(∂φ)2.

We define ja = −T ab V b. We have

ja = −(V · ∂φ)∂aφ+ 12V

a(∂φ)2 =⇒ j2 = 14 V 2︸︷︷︸

<0

((∂φ)2

)2

︸ ︷︷ ︸≥0

≤ 0,

so j is causal or zero. Also

V · j = −(V · ∂φ)2 + 12V

2(∂φ)2 = −12(V · ∂φ)2︸ ︷︷ ︸≤0

+ 12V

2︸ ︷︷ ︸<0

[∂φ− V · ∂φ

V 2 V

]2

︸ ︷︷ ︸⊥V a =⇒ ≥0

≤ 0,

so j is future-directed. Therefore scalar fields obey the DEC.

Weak energy condition

This states that TabV aV b ≥ 0 for all causal vectors V a. Note that DEC =⇒ WEC.

Null energy condition

This states that TabV aV b ≥ 0 for all null vectors V a. Note that WEC =⇒ NEC.

Strong energy condition

This states that (Tab − 12gabT

cc )V aV b ≥ 0 for all causal vectors V a. Alternatively, using Einstein’s

equations, RabV aV b ≥ 0, i.e. “gravity is attractive”.The SEC is independent to the other three; it is neither implied by, nor does it imply, any of theDEC, WEC or NEC.

4.10 Conjugate points

Lemma 6. If both the Einstein equations and the NEC are obeyed, then the generators of a nullhupersurface N obey dθ

dλ ≤ −12θ

2.

Proof. We have ω = 0 and σabσab ≥ 0, since vectors in T⊥ are spacelike. We also have by the NECthat RabUaU b = 8πTabUaU b ≥ 0 (using U2 = 0). Hence by Raychaudhuri’s equation, we havedθdλ ≤

12θ

2 as required.

This has a simple corollary: if θ = θ0 at p ∈ γ, where γ is a generator of N , then θ → −∞ withinaffine parameter 2

|θ0| , provided γ extends that far.

Proof. Without loss of generality, set λ = 0 at p. We have ddλθ−1 ≥ 1

2 , so θ−1 − θ−10 ≥ 1

2λ. Thuswe have θ ≤ θ0

1+λθ0/2 , which approaches −∞ for some λ ≤ 2|θ0| .

– 27 –

4 The Singularity Theorem 4.11 Causal structure

Definition. Two points p and q along a geodesic γ are said to be conjugate if there exists aJacobi field Sa along gamma such that Sa = 0 at p, q but Sa 6≡ 0.

Note that two points are conjugate iff the number of geodesics connecting them is more than one.

Theorem 5. Consider a null conguence containing all null geodesics through p. If θ → −∞ at qon a null geodesic γ through p, then q is conjugate to p along γ.

Theorem 6. Let γ be a causal curve containing points p, q. Iff γ is a null geodesic with no pointconjugate to p along γ between p and q, then there does not exist a smooth 1-parameter family ofcausal curves γs connecting p, q such that γ0 = γ and γs is timelike for s > 0.

Definition. Suppose we have a 2d spacelike surface S, with N normal to S and γ a generatorof N . We say that p is conjugate to S along γ if there exists a Jacobi field Sa such that Sa = 0at p and Sa|S is tangent to S.

Note that p is conjugate to S along γ iff θ → −∞ along γ in a congruence containing generators ofN .

4.11 Causal structure

Definition. Let U ⊂ M, the chronological future I+(U) of U is the set of all points q suchthat there exists a future-directed timelike curve from U to q. The causal future J+(U) is theset of all points q such that there exists a future-directed causal curve from U to q. We definethe chronological and causal pasts I−(U) and J−(U) similarly.

Example. In Minkowski space, I+(p) is the interior of the future lightcone of p, and J+(p) isthe interior and surface of the future lightcone, including p.

Lecture 1108/02/16 We have that I±(U) are open (since small deformations of timelike curves are also timelike). Recall

that the closure S of a set S is the union of S with its limit points. In Minkowski space, we haveI±(p) = J±(p), but this is not true in general; for example, consider 2d Minkowski space with apoint x deleted. If x is contained in the boundary of the future lightcone of p, then I±(p) 6= J±(p).Recall that the interior int(S) of S is the set of all interior points of S, i.e. those points q such thatq ∈ V ⊂ S for some neighbourhood V . The boundary S of S is defined as S \ int(S).

Theorem 7. Let p ∈M. There exists a convex normal neighbourhood of p (i.e. an open U ∈ Mcontaining p such that for all q, r ∈ U there is a unique geodesic from q to r that staus in U).Furthermore, the chronological future of p in U is the set of all points along future directed timelikegeodesics in U from p, with boundary equal to the set of all points along future directed null geodesicsin U from p.

A corollary of this is: if q ∈ J+(p) \ I+(p), then there exists a null geodesic from p to q.

Lemma 7. If S ⊂M, then (i) J+(S) ⊂ I+(S), and (ii) I+(S) = int(J+(S)).

Note that I+(S) ⊂ J+(S), so (i) implies that I+(S) = J+(S), and (ii) implies that J+(S) = I+(S).

– 28 –

4 The Singularity Theorem 4.11 Causal structure

Definition. S ⊂M is achronal if no two points in S are connected by a timelike curve in M.

Theorem 8. Suppose U ∈ M, then J+(U) is an achronal 3d submanifold of M.

Proof of achronality. Assume p, q ∈ J+(U), with q ∈ I+(p).

J+(U)

p

q

s

r

U

Since I+(p) is open, there exists a point r near q such that r ∈ I+(p) and r ∈ J+(U). Similarly,since I−(r) is open, there exists a point s near p such that s ∈ I−(r) and s ∈ J+(U). But then wehave a causal curve from U to s to r that leaves J+(U), which is a contradiction.

Theorem 9. Suppose U ∈ M is closed and suppose r ∈ J+(U) and r 6∈ U . Then r lies on a nullgeodesic λ, where λ lies entirely in J+(U) and is either past-inextendable or has a past endpoint onU .

Theorem 10. Let S be a 2d spacelike orientable compact submanifold of a globally hyperbolicspacetime. Then every p ∈ J+(S) lies on a future-directed null geodesic starting on and orthogonalto S, with no point conjugate to S between S and p.

Definition. The future Cauchy horizon of a partial caucy surface H+(Σ) = D+(Σ)\I−(D+(Σ)).H−(Σ) is defined similarly.

Note that H±(Σ) are null hypersurfaces.

Theorem 11 (Penrose singularity theorem, 1965). We assume the following:

• (M, g) is a globally hyperbolic spacetime with non-compact Cauchy surface Σ.

• (M, g) satisfies the Einstein equations and the NEC.

• M contains a trapped surface T .

Let θ0 = maxT θ where we take the maximum over both sets of null geodesics orthogonal to T (noteθ0 < 0).Then we have: at least one null geodesic orthogonal to T is future inextendable and has affine lengthless than or equal to 2

|θ0| .

Proof. Suppose all future-inextendable null geodesics orthogonal to T have affine length greaterthan 2

|θ0| . By the corollary to Lemma 6, θ → −∞ along all such geodesics. By Theorem 5, on eachgeodesic there is a point q that is conjugate to T that is affine parameter distance λ ≤ 2

θ0from T .

Now let p ∈ J+(T ), p 6∈ T . Since trapped surfaces are by definition 2d, spacelike, orientable andcompact, we can apply Theorem 10 to deduce that p lies on a future-directed null geodesic starting

– 29 –

5 Asymptotic Flatness 5 Asymptotic Flatness

on and orthogonal to T with not point conjugate to T between T and p. Hence, using the above, pmust be affine parameter less than or equal to 2

|θ0| from T . So we have

J+(T ) ⊂

points along geodesics ⊥ to T with affine parameter distance ≤ 2θ0

.

Note that the former set is closed while the latter is compact, so we can deduce that J+(T ) iscompact. Let α : J+(T )→ Σ be a function that carries back a point p ∈ J+(T ) along the orbits ofsome timelike vector field onto Σ (since Σ is a Cauchy surface, α is well-defined). Since J+(T ) isachronal (Theorem 8), α is a one to one map. Also, α is continuous and is thus a homeomorphism.Since J+(T ) is closed, so then is α(J+(T )). Also, since J+(T ) is a manifold, α(J+(T )) must alsobe open. Since M is connected, so too is Σ, and hence we must have α(J+(T )) = Σ. But Σ isnon-compact, while J+(T ) is compact, so this is a contradiction.

5 Asymptotic FlatnessLecture 12

10/02/16 5.1 Conformal compactification

Definition. Given a spacetime (M, g), a conformal transformation modifies the metric g →g = Ω2g, where Ω :M→ R is greater than 0.

To conformally compactify a spacetime, we will want Ω → 0 “at infinity”. More precisely, wechoose Ω such that (M, g) is extendable onto (M, g), a so-called unphysical spacetime, and on theboundary of M in M (“infinity” in M), we require Ω = 0.

M

M

Ω = 0

For example, consider 3+1 Minkowski space, with the metric ds2 = −dt2 + dr2 + r2 dω2, wheredω2 is the round metric on the unit 2-sphere. Reparametrise into retarded time u = t − r andadvanced time v = t+ r, to obtain ds2 = − dudv + 1

4(u− v)2 dω2. Note that since r ≥ 0, we have−∞ < u ≤ v <∞. If we define new coordinates p and q by u = tan p and v = tan q, we see that weobtain a finite range −π

2 < p ≤ q < π2 and ds2 = (2 cos p cos q)−2

[−4 dp dq + sin2(q − p) dω2

]. We

see that we can carry out a conformal transformation with Ω = 2 cos p cos q to obtain an unphysicalmetric ds2 = −4 dpdq + sin2(q − p) dω2. Finally, we reparametrise with T = q + p and χ = q − pand get

ds2 = −dT 2 + dχ2 + sin2 χdω2 .

dχ2 + sin2 χdω2 is the round metric of the unit 3-sphere, and we see that we obtain just theEinstein static universe, except for one crucial difference. In the ESU, T ∈ (−∞,∞) and χ ∈ [0, π),but here we have T ∈ (−π, π) and χ ∈ [0, π). Thus we choose (M, g) = ESU as an extension of

– 30 –

5 Asymptotic Flatness 5.1 Conformal compactification

(M, g), and identify “infinity” of (M, g) as the points i±, i0 and the null hypersurfaces I± given byT = ±(π − χ):

χ = 0

T

χ = π

T = −π: i−

T = π: i+

i0

I+

I−

ESU

M

If we denote each S2 at a given T and χ by a single point, we obtain the so-called Penrose diagramfor Minkowski space:

T

χ

i−

i+

i0

I−

I+

r = constant

t = constant

radial null geodesicr = 0

This is a bounded subset of R2 with the flat metric. The boundary is the union of infinity with theaxis (r = 0). The different components of infinity have different names: I− and I+ are past andfuture null infinity respectively, i− and i+ are past and future timelike infinity respectively, and i0

is spatial infinity.Suppose we have a massless scalar field obeying ∇a∇aψ = 0. Exercise: show that the general

– 31 –

5 Asymptotic Flatness 5.1 Conformal compactification

spherically symmetric solution is

ψ(t, r) = 1r

(f(u) + g(v)) = 1r

(f(t− r) + g(t+ r)) .

We need this to be smooth at r = 0 so we set g(x) = −f(x). Thus ψ(t, r) = 1r (f(u)− f(v)) =

1r (F (p)− F (q)), where F (x) = f(tan x). Now let F0(q) = (rψ)I− . Since p = −π

2 on I−, we haveF0(q) = F (−π/2)− F (q). Therefore,

ψ(t, r) = 1r

(F0(q)− F0(p)) ,

so the solution is uniquely determined everywhere by its behaviour on I−.Suppose we had instead considered 2D Minkowski space, ds2 = −dt2 + dr2. Now the range ofr is (−∞,∞). Proceeding as before, we have −∞ < u, v < ∞ and so −π

2 < p, q < π2 . Thus

T, χ ∈ (−π, π), and the Penrose diagram is now a square.

i0R

i+

i0L

i−

I+RI+

L

I−L I−R

We can carry out a similar procedure in the Kruskal spacetime. We define P = P (U) and Q = Q(V )such that P,Q ∈

(−π

2 ,π2), and we find a conformal factor Ω such that g extends onto M, and

M∈M has boundary at P = ±π2 or Q = ±π

2 . We find that null infinity has four components: I±

in region I, and I±′ in region IV.

– 32 –

5 Asymptotic Flatness 5.2 Asymptotic flatness

III

IIIIV i0

i+i+′

i0′

i−i−′

I+I+′

I−′ I−

t = constantr = constant

r = constant

r = 0

r = 0

We can also draw a Penrose diagram for gravitational collapse:

r = 0

r=

0

i0

i+

I+

I−

Lecture 1312/02/16

5.2 Asymptotic flatness

Definition. A time orientable spacetime (M, g) is asymptotically flat at null infinity if thereexists a (M, g) such that:

1. There exists a positive Ω :M→ R such that (M, g) is an extension of (M,Ω2g) (wherewe regard M as a subset of M on which g = Ω2g).

2. We can extend M within M to obtain a manifold with boundary M∪ ∂M (a manifoldwith boundary is just like a manifold, but charts areM→ R/2 = (x1, . . . , xn) : xn ≤ 0).

– 33 –

5 Asymptotic Flatness 5.2 Asymptotic flatness

3. Ω smoothly extends to M such that Ω|∂M = 0 and dΩ|∂M 6= 0.

4. We can choose I± such that ∂M = I+ ∪ I−, I+ ∩ I− = ∅, and I± ' R× S2.

5. No future or past directed causal curves in M intersect I− or I+ respectively.

6. I± are “complete” (to be defined shortly).

Example. Consider the Schwarzschild solution in outgoing Eddington-Finklestein coordinates:

g = −(

1− 2Mr

)du2 − 2 du dr + r2 dω2

Setting r = 1x and carrying out a conformal transformation with Ω = x, we obtain

g = −x2(1− 2Mx) du2 + 2 dudx+ dω2 .

We can extend g smoothly across I+ = x = 0. I+ is parametrised by (u, θ, φ), so I+ ' R×S2.

It can be shown that under a conformal transformation, the Ricci tensor transforms as

Rab = Rab + 2Ω−1∇a∇bΩ + gabgcd(Ω−1∇c∇dΩ− 3Ω−2∂cΩ∂dΩ

).

Multiplying by Ω gives

ΩRab = ΩRab + 2∇a∇bΩ + gabgcd∇c∇dΩ︸ ︷︷ ︸

A

− 3Ω−1gabgcd∂cΩ∂dΩ︸ ︷︷ ︸

B

.

We assume that ΩRab is 0 at I+ (as would be the case for a sufficiently quickly reached vacuum atinfinity) and observe that since A is smooth at I+, then so must B be. Since Ω−1 diverges at I+,we must thus have gcd∂cΩ∂dΩ

∣∣∣I+

= 0. Hence dΩ|I+ is null. Since I+ is a surface of constant Ω, wethen have that I+ is a null hypersurface in (M, g).Note that we have some gauge freedom in our choice of conformal factor. We could just as well havechosen Ω′ = ωΩ, where ω :M→ R is some positive function on M∪ ∂M. It is possible to choosea gauge such that ∇a∇bΩ

∣∣∣I+

= 0 (†). Assume that we take this gauge and define na = gab(dΩ)b.

We have ∇anb∣∣∣I+

= 0, so Bba – the expansion and shear of the generators of I+ are zero.

We can define a natural set of coordinates in the neighbourhood of I+ in the following way.Firstly we have spherical coordinates (θ, φ) on the spheres at each point of I+, with metricgab|S2 = dθ2 + sin2 θ dφ2. na is tangent to I+, so we can use the distance u along its integral curvesas a third coordinate on I+. Now there is a unique null direction perpendicular to ∂

∂θ and ∂∂φ and

not tangent to I+, and we can use Ω as a coordinate for this direction (since dΩ 6= 0 at I+). Thuswe have coordinates (Ω, u, θ, φ).

– 34 –

5 Asymptotic Flatness 5.3 Definition of a black hole

(Ω, u, θ, φ)

(u, θ, φ)

na

S2 I+

Note that since dΩ and na are null, we have guu = gΩΩ = 0. Also, δµu = nµ = gµν(dΩ)ν = gµΩ.Examining the gauge condition (†) we can deduce that the spherical part of the metric does notdepend on u. Thus we have

g|Ω=0 = 2 du dΩ + dθ2 + sin2 θ dφ2 .

If we define r = 1Ω , then we obtain

g = Ω−2g = −2 dudr + r2(dθ2 + sin2 θ dφ2) + . . .︸︷︷︸subleading as r →∞

= −dt2 + dx2 + dy2 + dz2 .

Definition. I+ is complete if in the gauge given by (†), the generators of I+ are complete.

5.3 Definition of a black holeWe have I+ ⊂ M. Consider the causal past of I+: J−(I+) ⊂ M. M∩ J−(I+) is the set of allpoints that can send a signal to I+.

Definition. Let (M, g) be a spacetime that is asymptotically flat at null infinity. We definethe following:

• The black hole region is B =M\ [M∩ J−(I+)].

• The future event horizon is H+ = B =M∩ J−(I+).

• The white hole region is W =M\ [M∩ J+(I−)].

• The past event horizon is H− = W =M∩ J+(I−).

Example. In the Kruskal spacetime, we have:

B = u ≥ 0 W = v ≤ 0H+ = u = 0 H− = v = 0

– 35 –

5 Asymptotic Flatness 5.3 Definition of a black hole

IIIIII

IV

I+

I−

black hole

white hole

Note that H± are null hypersurfaces and their generators have no future endpoints.

Definition. An asymptotically flat spacetime (M, g) is strongly asymptotically predictableif there exists some open set V ∈ M such that M∩ J−(I+) ⊂ V , and (V , g) is globallyhyperbolic.

Theorem 12. Suppose we have a strongly asymptotically predictable spacetime (M, g) with Cauchysurfaces Σ1, Σ2 for V such that Σ2 ⊂ I+(Σ1), and B a conncected component of B ∩ Σ1. ThenJ+(B) ∩ Σ2 is contained within a connected component of B ∩ Σ2.

Proof. Suppose the black hole does split (bifurcate). Then there exist open sets O,O′ ⊂ Σ2 suchthat O ∩O′ = ∅, J+(B) ∩ Σ2 ∈ O ∪O′, J+(B) ∩O 6= ∅ 6= J+(B) ∩O′.

Σ1

Σ2

B

O O′

Note that B ∩ I−(O) 6= ∅ 6= B ∩ I−(O′), and B ⊂ I−(O) ∪ I−(O). If p is a point in B such thatp ∈ I−(O) and p ∈ I−(O′), then we can divide the future-directed timelike geodesics from p into 2sets depending on which of O, O′ they reach. Thus we can divide the timelike vectors at p into 2disjoint sets, but this is a contradiction since the future light cone is connected. So there does notexist such a p, so B ∩ I−(O)∩ I−(O′) = ∅, and B = [B ∩ I−(O)]∪ [B ∩ I−(O′)] is a disjoint union,contradicting the connectedness of B.

– 36 –

6 Charged Black Holes 5.4 Weak cosmic censorship conjecture

5.4 Weak cosmic censorship conjectureIn the Kruskal spacetime, the white hole singularity is “naked”, i.e. it is visbile to I+. For M < 0Schwarzschild we have the Penrose diagram

and we see that the r = 0 singularity is also naked here.It is reasonable to ask whether gravitational collapse could lead to a naked singularity. In fact wehave the following conjecture (weak cosmic censorship):

Suppose (Σ, hab,Kab) is a geodesically complete, asymptotically flat initial data set,and that matter fields obey hyperbolic equations and the dominant energy condition.Then, generically, the maximal development is asymptotically flat (so I+ is complete)and strongly asymptotically predictable.

Note that the strong and weak censorship conjectures are logically distinct.

5.5 Apparent horizon

Theorem 13. If (M, g) is a strongly asymptotically predictable spacetime satisfying the null energycondition and Einstein’s equations that contains a trapped surface T , then T ⊂ B.

Consider a 1-parameter family of Cauchy surfaces Σt. When we say the black hole region at time tor the event horizon at time t, we are referring to the sets Bt = B∩Σt and Ht = H∩Σt respectively.

Definition. The trapped region at time t is τt = p ∈ Σt s. t.∃ trapped S with p ∈ S ⊂ Σt.The apparent horizon is At = τt.

Note that we have τt ∈ Bt, and At ∈ Bt. Also At is on or inside Ht, and At is marginally trapped.

6 Charged Black HolesLecture 14

15/02/16 6.1 The Reissner-Nordstrom solutionRecall the Einstein-Maxwell action

S = 116π

∫d4x√−g(R− F abFab) where F = dA .

This action gives the following equations of motion:

Rab −12Rgab = 2

(FacF

cb −

14F

cdFcdgab

), ∇bFab = 0

– 37 –

6 Charged Black Holes 6.2 Eddington-Finklestein coordinates

Theorem 14. The unique spherically symmetric solution with non-constant area-radius function ris the Reissner-Nordstrom solution, given by

ds2 = −(

1− 2Mr

+ e2

r2

)dt2 +

(1− 2M

r+ e2

r2

)−1

dr2 + r2 dΩ2

andA = −Q

rdt− P cos θ dφ

where e =√Q2 + P 2.

The Reissner-Nordstrom solution has three parameters. We identify them as the mass M , electriccharge Q, and magnetic charge P of the black hole. It is static with a timelike KVF k = ∂

∂t , and itis asymptotically flat at null infinity.To make the metric easier to write down, we define ∆ = r2 − 2Mr + e2 = (r − r+)(r − r−) wherer± = M ±

√M2 − e2. Then

ds2 = −∆r2 dt2 + r2

∆ dr2 + r2 dΩ2 .

Consider M2 < e2. Then ∆ > 0 for all r > 0, and we have a naked curvature singularity at r = 0.Thus we will ignore this case and assume from now on that M > e (confer M < 0 Schwarzschild).

6.2 Eddington-Finklestein coordinatesWe define r∗ by dr∗ = r2

∆ dr. If we want we can solve this to obtain

r∗ = r + 12κ+

log∣∣∣∣r − r+r+

∣∣∣∣+ 12κ−

log∣∣∣∣r − r−r−

∣∣∣∣+ constant

where κ± = r±−r∓2r2±

. Let u = t− r∗ and v = t+ r∗. Then we can write the metric in ingoing E-Fcoordinates (v, r, θ, φ) as

ds2 = −∆r2 dv2 + 2 dv dr + r2 dΩ2 .

This is smooth for all r > 0, and det g 6= 0, so we can extend spacetime to the region 0 < r < r+.There is a curvature singularity at r = 0. Since dr is null when grr = ∆

r2 = 0, we have that r = r±are null hypersurfaces. It can be shown that r decreases along any future-directed causal curve inr− < r < r+. ?Hence we have a black hole region r ≤ r+ with event horizon H+ given by r = r+.We can carry out a similar argument in (u, r, θ, φ) to show that there is a white hole region withr < r+.

6.3 Kruskal-like coordinatesNow let U± = −e−κ±u and V ± = ±eκ±v. In r < r+ we have coordinates (U+, V +, θ, φ) and themetric takes the form

ds2 = −r+r−κ2

+r2 e−2κ+r

(r − r−r−

)1+ κ+|κ−| dU+ dV + + r2 dΩ2

where r(U+, V +) is defined by

−U+V + = e2κ+r(r − r+r+

)(r−

r − r−

) κ+|κ−|

.

– 38 –

6 Charged Black Holes 6.3 Kruskal-like coordinates

Initially U+ < 0 and V + > 0, but we can analytically continue to U+ ≥ 0 or V + ≤ 0.

r constant →

t constant →

III

IIIIV

V +

r=r+

U+

r =r+

Note ka = 0 at U+ = V + = 0. Ingoing radial null geodesics have r → r− (U+V + → −∞) in finiteaffine parameter.Unlike the Schwarzschild solution, we can continue to extend the spacetime. In region II, lett = v − r∗ and u = t− r∗ = v − 2r∗. We use coordinates (U−, V −, θ, φ) where U−, V − < 0. Themetric is

ds2 = −r+r−κ2−r

2 e−2|κ−|r

(r+ − rr+

)1+ |κ−|κ+ dU− dV − + r2 dΩ2

where r(U−, V −) is defined by

−U+V + = e2|κ−|r(r − r−r−

)(r+

r+ − r

) |κ−|κ+

.

We analytically continue to U−, V − ≥ 0 to find new regions V, VI and III′. Lecture 1517/02/16

VIIII′

IIV

V −

r=r−

U−

r =r−

r = 0 r = 0

Note that III′ is isometric to III, so we can use coordinates (U+′, V +′) an extend out to new regionsI′, II′ and IV′.

– 39 –

6 Charged Black Holes 6.3 Kruskal-like coordinates

I′II′

III′IV′

V +

r=r+

U+

r =r+

We can repeat this process indefinitely. We can also draw a Penrose diagram of the extendedspacetime:

VI′V′

I′II′

IV′

VIIII′

V

III

IIIIV r =

r+

r =r−

r=

0

I+

I−

i+

i−

– 40 –

6 Charged Black Holes 6.4 Extreme Reissner-Nordstrom

6.4 Extreme Reissner-Nordstrom

0 < r < M

0 < r < M

0 < r < M

i0

i0

I+

I+

I−

I−

H −, r =

M

H+ ,r=M

Σ (t constant)

r > M

r > M

The Reissner-Nordstrom solution with M = eis known as extreme RN. In this case r+ = r− =M and the metric takes the simple form

ds2 = −(

1− M

r

)2dt2+

(1− M

r

)−2dr2+r2 dΩ2 .

We can introduce Eddington-Finklestein coor-dinates in r > M with dr∗ = dr

(1−M/r)2 andv = t+ r∗ to obtain

ds2 = −(

1− M

r

)2dv2 + 2 dv dr + r2 dΩ2 .

Extreme RN has the following Penrose diagramas shown to the left.We have H± = H±(Σ). The proper length ofa line from r = r0 > M to r = M with t, θ, φconstant diverges:∫ r0

M

dr1−M/r

If we set r = M(1 + λ) where λ is small, weobtain

ds2 ≈ −λ2 dt2 +M2 dλ2

λ2︸ ︷︷ ︸AdS2

+M2 dΩ2︸ ︷︷ ︸S2

.

The spacetime looks like an infinite throat aswe approach r = M :

S2

6.5 Majumdar-Papapetrou solutionsWe introduce a new radial coordinate ρ = r −M , and assume that the magnetic charge vanishes,P = 0. Then the metric takes the form

ds2 = −H−2 dt2 +H2(dρ2 + ρ2 dΩ2)

where H = 1 + Mρ . This is a special case of the Majumdar-Papapetrou solution

ds2 = −H(x)−2 dt2 +H(x)2(dx2 + dy2 + dz2), A = H−1 dt ,

– 41 –

7 Rotating Black Holes 7 Rotating Black Holes

where x = (x, y, z) and H is harmonic, ∇2H = 0. The fact that the equation for H is linear allowsa large family of exact solutions to exist.An interesting case arises when we set H = 1 +

∑Ni=1

Mix−xi

. This leads to a static solution with Nextreme RN black holes. The physical interpretation of this is that since Mi = Qi there is an exactbalance of electrostatic repulsion and gravitational attraction.

7 Rotating Black HolesLecture 16

19/02/16 We need to slightly modify our definitions of stationary and static.

Definition. An asymptotically flat spacetime is stationary if there exists a KVF ka that istimelike in a neighbourhood of I±, and static if ka is also hypersurface orthogonal.

It is conventional to normalise such that k2 → −1 at I±.

Definition. An asymptotically flat spacetime is stationary and axisymmetric if:

• It is stationary.

• There exists a KVF ma that is spacelike near I±.

• ma generates a 1-parameter group of isometries isomorphic to U(1).

• [k,m] = 0.

We can choose coordinates such that k = ∂∂t , m = ∂

∂φ , φ ∼ φ+ 2π.

Theorem 15. If (M, g) is a static, asymptotically flat, vacuum spacetime that contains a blackhole and is regular on and outside H+, then (M, g) is isometric to Schwarzschild spacetime.

Theorem 16. If (M, g) is a stationary, non-static, asymptotically flat spacetime that is an analyticsolution of the Einstein-Maxwell equations and is regular on and outside H+, then (M, g) isstationary and axisymmetric.

Theorem 17. If (M, g) is a stationary, axisymmetric, asymptotically flat, vacuum spacetime thatis regular on and outside connected H+, then (M, g) belongs to the Kerr family of solutions.

7.1 The Kerr-Newman solutionIn Boyer-Lindquist coordinates, the Kerr-Newman solution is as follows:

ds2 = −∆− a2 sin2 θ

Σ dt2 − 2a sin2 θr2 + a2 −∆

Σ dt dφ

+ (r2 + a2)2 −∆a2 sin2 θ

Σ sin2 θ dφ2 + Σ∆ dr2 + Σ dθ2

A = − 1Σ[Qr(dt− a sin2 θ dφ) + P cos θ(a dt− (r2 + a2) dφ)

]where Σ = r2 + a2 cos2 θ, ∆ = r2 − 2Mr + a2 + e2, e =

√Q2 + P 2

– 42 –

7 Rotating Black Holes 7.2 The Kerr solution

This solution is stationary and axisymmetric with KVFs k = ∂∂t and m = ∂

∂φ . It also permits adiscrete isometry t→ −t, φ→ −φ. It has four parameters: the mass M , electric charge Q, magneticcharge P , and angular momentum J = aM (these names will be justified in the next section).Under φ→ −φ, we have a→ −a, so w.l.o.g. we can assume that a ≥ 0.

7.2 The Kerr solutionThe Kerr solution is the Kerr-Newman solution with Q = P = e = 0. In this case we have∆ = (r − r+)(r − r−), where r± = M ±

√M2 − a2. If M2 < a2 we get a naked singularity, so we

will assume M > |a|. The metric is singular at θ = 0, π in the ordinary spherical way. We also havesingularities at ∆ = 0 and at Σ = 0. These are just coordinate singularities. To show this, definenew coordinates v and χ by

dv = dt+ r2 + a2

∆ dr and dχ = dφ+ a

∆ dr (χ ∼ χ+ 2π).

In these coordinates we have k = ∂∂v and m = ∂

∂χ . Kerr coordinates are (v, r, θ, χ), and the metricbecomes

ds2 = −∆− a2 sin2 θ

Σ dv2 + 2 dv dr − 2a2 sin2 θr2 + a2 −∆

Σ dv dχ

− 2a sin2 θ dχ dr + (r2 + a2)2 −∆a2 sin2 θ

Σ sin2 θ dχ2 + Σ dθ2 .

This metric can be analytically extended into 0 < r < r+.

Lemma 8. r = r+ is a null hypersurface with normal xia = ka + ΩHma where ΩH = a

r2+a2 .

(This can be proven by showing that ξµ dxµ|r=r+∝ dr, so ξa ⊥ r = r+ and ξµξ

µ|r=r+= 0.)

r ≤ r+ is the spacetime’s black hole region B, and r = r+ is its event horizon H+.Lecture 1722/02/16 In Boyer-Lindquist coordinates, ξ = ∂

∂t + ΩH∂∂φ , and we have ξµ∂µ(φ − ΩHt) = 0, so φ =

ΩHt+ constant on orbits of ξa. Thus the generators of H+ rotate with angular velocity ΩH withrespect to a stationary observer at ∞. We say that ΩH is the angular velocity of the black hole.

7.3 Maximal analytic extensionKerr coordinates are analagous to the ingoing E-F coordinates used for the RN black hole. We cansimilarly find outgoing coordinates to find a white hole region, and then construct a Kruskal likeextension containg the regions I, II, III, IV. Similar to RN, we can extend II to new regions V, VI,III′, etc. One key difference is that r = 0 becomes a “ring” singularity, leading to a new flat regionof space.The Kerr solution has no spherical symmetry, so we cannot draw a Penrose diagram. Instead wewill just refer to submanifolds. One submanifold is the “axis” given by θ = 0 or π. This is twodimensional, since we fix θ, and φ is meaningless on the axis. We can construct a Penrose-likediagram for the axis:

– 43 –

7 Rotating Black Holes 7.4 Ergosphere/Penrose process

VI′V′

I′II′

IV′

VIIII′

V

III

IIIIV r =

r+

r =r−

r=

0

I+

I−

i+

i−

r < 0

Note that the Kerr solution just describes the final state of a rotating black hole; it has nothing todo with collapse.If M = a, we obtain extreme Kerr, similar to extreme RN.

7.4 Ergosphere/Penrose processWe have

k2 = gtt = −(

1− 2Mr

r2 + a2 cos2 θ

).

Therefore k is timelike in region I when r2−2Mr+a2 cos2 θ > 0, i.e. when r > M+√M2 − a2 cos2 θ

(this is the root in region I).There is a region outside the event horizon in which k is spacelike, and this is known as theergosphere:

M +√M2 − a2 = r+ < r < M +

√M2 − a2 cos2 θ

The boundary of the ergosphere is called the ergosurface.

– 44 –

7 Rotating Black Holes 7.4 Ergosphere/Penrose process

θ = 0

θ = π

ergosurfaceergosphere

Note that stationary observers follow timelike geodesics parallel to k, and these can’t exist in theergosphere.Consider now a particle with 4-momentum P a = µua, where µ is the particle’s rest mass. Then theenergy of the particle with respect to an observer at infinity is given by E = −k · P . Suppose theparticle decays inside the ergoregion into two particles with momenta P1, P2 respectively.

P

P1

P2

We have E = E1 + E2 where Ei = −k · Pi. Since k is spacelike, we can have E1 < 0. ThenE2 = E + |E1| > E. Particle 1 must fall into the black hole, but particle 2 can escape to infinity.Hence a particle leaves the ergoregion with more energy than when it entered – this is the Penroseprocess. Since E1 < 0, the mass of the black hole decreases.There is a bound on the amount of energy that can be extracted. Since P and ξ are both causal,we have −P · ξ ≥ 0, so −P · (k+ ΩHm) ≥ 0. Thus E−ΩHL ≥ 0, where L = m ·P is the conservedangular momentum, so L ≤ E/ΩH . After the black hole settles, we have δM = E and δJ = L, so

δJ ≤ δM

ΩH= 2M(M2 +

√M4 − J2)

JδM.

It can be shown that this is equivalent to δMirr = 0, where Mirr =[

12

(M2 +

√M4 + J2

)] 12 is the

“irreducible mass”. We have

M2 = M2irr + J2

4M2irr≥M2

irr ≥ initial M2irr,

so the mass can not decrease below the RHS.It can be shown that A = 16πM2

irr is the area of the event horizon (i.e. the area of the intersectionof H+ with a partial Cauchy surface), so δA ≥ 0 (this is a special case of the second law of blackhole mechanics).

– 45 –

8 Mass, Charge, and Angular Momentum 8 Mass, Charge, and Angular Momentum

8 Mass, Charge, and Angular Momentum8.1 Charges in curved spacetimeConsdier a general orientable manifold (M, g) of n dimensions. We have the volume form ε,satisfying

εa1...apcp+1...cnεb1...bpcp+1...cn = ±p!(n− p)!δa1[b1. . . δ

apbp],

where + corresponds to a Riemannian signature, and − corresponds to a Lorentzian signature.

Definition. The Hodge dual of a p-form X is an (n− p)-form ?X defined by

(?X)a1...an−p = 1p!εa1...an−pb1...bpX

b1...bp .

It is simple to see that ?(?X) = ±(−)p(n−p)X. We also have

(? d ? X)a1...ap−1 = ±(−)p(n−p)∇bXa1...ap−1b.

In Euclidean 3D space, we have:

∆f = df∇ ·X = ?d ? X∇×X = ?dX

Lecture 1824/02/16 Recall the Maxwell equations

∇aFab = −4πjb and ∇[aFbc] = 0,

where j is the current density 4-vector. In terms of forms we have

d ? F = −4π ? j and dF = 0.

Definition. Let Σ be a spacelike hypersurface. The electric charge Q on Σ is given by

Q = −∫

Σ?j = 1

4π

∫σ

d ? F = 14π

∫∂Σ?F.

Example. Consider Minkowski space with spherical coordinates (t, r, θ, φ), with orientationgiven by the volume form ε = r2 sin θ dt ∧ dr ∧ dθ ∧ dφ. Let Σ = t = 0, with orientationdr∧dθ∧dφ, and ΣR = t = 0, r ≤ R with the same orientation. We have ∂σR = t = 0, r = Rwith orientation dθ ∧ dφ. Consider A = − q

r dt. We have F = dA = − qr2 dt ∧ dr. Thus

(?F )θφ = r2 sin2 θF tr = q sin θ, so

Q(ΣR) = 14π

∫S2R

?F = 14π dθ dφ q sin θ = q.

– 46 –

8 Mass, Charge, and Angular Momentum 8.2 Komar integrals

Definition. Let (Σ, hab,Kab) be an asymptotically flat end, S2r = xixi = r2. Then the

electric charge Q and magnetic charge P of the spacetime are given by

Q = 14π lim

r→∞

∫S2r

?F

P = 14π lim

r→∞

∫S2r

F

Exercise: show this agrees with Q and P in the Kerr-Newman solution.

8.2 Komar integralsLet (M, g) be a stationary spacetime. There exists a conserved energy-momentum current Ja =−Tabkb, d ? j = 0. Hence we can define the total energy of matter on a spacelike hypersurface Σ as

E[Σ] = −∫

Σ?J.

This quantity is conserved. Suppose Σ and Σ′ bound a region R of spacetime.

Σ′

Σ

R

Then we haveE[Σ′]− E[Σ] = −

∫∂R?J = −

∫R

d ? J = 0.

Lemma 9. A KVF ka obeys ∇a∇bkc = Rcbadkd.

Hence we have

(? d ? dk)c = −∇b(dk)cb= −∇b(∇ckb −∇bkc)= 2∇b∇bkc= −2Rcdkd

= 8πJ ′c,

where J ′a = −2(Tab − 1

2Tgab)kb. Thus d ? dk = 8π ? J , and

−∫

Σ?J ′ = − 1

8π

∫Σ

d ? dk = − 18π

∫∂Σ? dk . (∗)

Exercise: consider a static, spherically symmetric perfect fluid star, and let Σ = t = constant, r ≤r0 where r0 > R = star radius. Show that the RHS of (∗) is the parameter M in the Schwarzschildsolution. Show that the LHS of (∗) is the total energy of matter in the star in the Newtonian limit(p ρ, |Φ| 1, |Ψ| 1).

– 47 –

8 Mass, Charge, and Angular Momentum 8.3 Hamiltonian formulation of GR

Definition. Let (Σ, hab,Kab) be an asymptotically flat end in a stationary spacetime. TheKomar mass is

MKomar = − 18π lim

r→∞

∫S2r

? dk .

If we have axisymmetry, then the Komar angular momentum is

JKomar = 116π lim

r→∞

∫S2r

? dm.

It can be shown that for the Kerr-Newman solution, we have MKomar = M and JKomar = J .

8.3 Hamiltonian formulation of GRSet 16πG = 1, and assume global hyperbolicity, so that we can write

ds2 = −N2 dt+ hij(dxi +N i dt)(dxj +N j dt).

We have the following action in the 3+1 formulation:

S =∫

dt d3xL =∫

dt d3xN√H(

(3)R−KijKij +K2

)(3)R is the Ricci scalar of hij , and K is the trace of Kij the extrinsic curvature associated witht = 0, given by

Kij = 12N (hij −DiNj −DjNi),

where D is the covariant derivative with respect to hij . ijk indices are raised and lowered with h.Note that S is a function of N , N i and hij only – it is independent of ∂tN and ∂tN i. If we vary Nwe obtain the Hamiltonian constraint, and if we vary N i we obtain the momentum constraint.The momentum conjugate to h is given by

πij = δS

δhij=√h(Kij −Khij).

The Hamiltonian is thus

H =∫

d3x (πij hij − L) =∫

d3x√h[NH+N iHi],

whereH = −(3)R+ h−1πijπij −

12h−1π2, π = hijπij

andHi = −2hikDj

(h−

12πjk

).

N and Ni play the role of Lagrange multipliers, so on a solution H = Hi = 0. The equations ofmotion are

hij = δH

δπij, πij = − δH

δhij.

Using this Hamiltonian we’d like to be able to define an energy, but there is a problem: on anysolution, H = 0.Lecture 19

26/02/16

– 48 –

8 Mass, Charge, and Angular Momentum 8.4 ADM energy

To fix this, we will need to introduce a boundary term into the Hamiltonian. We assume that thet = constant is an asymptotically flat end, so we can write:

hij = δij +O

(1r

)πij = O

( 1r2

)δhij = O

(1r

)δπij = O

( 1r2

)Comparing with the expression for the metric in terms of N and N i, we see that we must haveN = 1 +O

(1r

)and N i → 0 as r →∞. Under these assumptions, we find that the variation of the

Hamiltonian takes the formδH =

∫d3x (. . . )δhij − δEADM︸ ︷︷ ︸

surface term

whereEADM = lim

r→∞

∫S2r

dAni(∂jhij − ∂ihjj).

dA is the area element on S2r and ni is the outward unit normal to S2

r . We can thus replace theHamiltonian H by H ′ = H + EADM. Since they differ by a surface term, the equations of motiongive the same result, but now we have no runaway boundary terms when varying hij . Thus H ′ isthe true Hamiltonian of GR.

8.4 ADM energySince H = 0 on a solution, the energy of the spacetime must be given by EADM. Return to G = 1units and we have:

Definition. The Arnowitt-Deser-Misner energy of an asymptoticall flat end is

EADM = 116π lim

r→∞

∫S2r

dAni(∂jhij − ∂ihjj).

In a stationary spacetime, if the constant t surface is orthogonal to ka as r →∞, then it can beshown that EADM = MKomar.Similarly, we have:

Definition. The ADM 3-momentum of an asymptotically flat end is

Pi = 18π lim

r→∞

∫S2r

dA (Kijnj −Kni).

Theorem 18. Let (Σ, hab,Kab) be a geodesically complete asymptotically flat end obeying the DEC.Then EADM ≥

√PiPi, with equality iff (Σ, hab,Kab) is a surface in Minkowski spacetime.

Definition. The ADM mass of an asymptotically flat end is

MADM =√EADM − PiPi.

– 49 –

9 Black Hole Mechanics 9 Black Hole Mechanics

9 Black Hole Mechanics9.1 Killing horizons and surface gravity

Definition. A null hypersurface N is a Killing horizon if there exists a KVF ξa defined in aneighbourhood of N such that ξa is normal to N .

Theorem 19. In a stationary, analytic, asymptotically flat, vacuum black hole solution, H+ is aKilling horizon.

For example, in Kerr r = r+ is a Killing horizon with normal KVF ξa = ka + ΩHma.

It can be shown that if ξa is not parallel to ka, then (M, g) is stationary and axisymmetric. Wenormalise so that k2 → −1 at ∞, and write ξa = ka + ΩHm

a, where ma is the generator ofaxisymmetry. Since ξ · ξ|N = 0, we have that dξaξa is orthogonal to N . Therefore there exists a κsuch that

∇a(ξbξb)∣∣∣N

= −2κξa.

κ is called the surface gravity of N . Since ξa is a KVF, we have ∇a(ξbξb) = 2ξb∇aξb = −2ξb∇bξa,so we can write the above as

ξb∇bξa∣∣∣N

= κξa.

Let na be tangent to the affinely parametrised generators of N . Then ξa = fna on N and we have

ξb∇bξa = fnb∇b(fna) = f2 nb∇na︸ ︷︷ ︸=0

+fnanb∇bf = ξaξb∂b log |f |,

so κ = ξa∂a log |f |.

Example. Consider a Reissner-Nordstrom black hole.

ds2 = −∆r2 dv2 + 2 dv dr + r2 dΩ2

where ∆ = (r− r+)(r− r−), r± = M ±√M2 − e2. We have k = ∂

∂v and ka|r=r± = (dr)a. Thuska is orthogonal to the null hypersurfaces r = r±, so these are Killing horizons. We have

d(kbkb) = d(−∇r2

)=(−∇

′

r2 + 2∇r3

)dr

=⇒ d(kbkb)∣∣∣r=r±

= −(r± − r∓r2±

)ka

and therefore the surface gravities at r± are κ = r±−r∓2r2±

.If we set e = 0 we obtain Schwarzschild with r+ = 2M , r− = 0 and κ = 1

4M .If we set e = M , we obtain extreme RN with r+ = r− = M and κ = 0.

Exercise: in the Kruskal spacetime, show that H± are Killing horizons N± of ka with surfacegravities ± 1

4M . This is an example of a bifurcate Killing horizon. N± are both Killing with respectto ξa, and they are orthogonal, so on B = N+ ∩N−, ξ must vanish. Any Xa tangent to B mustalso be tangent to both of N±, so Xa is spacelike, and B is a spacelike 2-surface.

– 50 –

9 Black Hole Mechanics 9.2 Interpretation of the surface gravity

9.2 Interpretation of the surface gravityThe main reason that κ is important is because the Hawking temperature (see later) is ~κ

2π . Howeverthere is another interpretation. In a static, asymptotically flat spacetime, κ is the force per unitrest mass at infinity needed to hold a particle at rest on the horizon.

9.3 Zeroth law of BH mechanicsLecture 20

29/02/16 Lemma 10. Consider a null geodesic congruence containing the generators of a Killing horizon N .Then θ = σ = ω = 0 on N .

Proof. We already have ω = 0 since the generators are hypersurface orthogonal.Let ξa be a KVF normal to N , and Ua be tangent to the affinely parameterised generators of N .We can write ξa = hUa on N . Thus Ua = h−1ξa + fV a for some V a and f that vanishes on M.Then we have

Bab = ∇bUa = (∂bh−1)ξa + h−1∇bξa + (∂bf)Va + f∇bVa.

Evaluating on N , using Killing’s equation, and symmetrising, we obtain

B(ab)

∣∣∣N

=(ξ(a∂b)h

−1 + V(a)∂b)f)N.

ξa, ∂af are parallel to Ua on N , so after projecting onto T⊥, we get

B(ab)

∣∣∣N

= 0,

so θ and σ vanish.

Theorem 20 (zeroth law). κ is constant on the future event horizon of a stationary black holespacetime obeying the dominant energy condition.

Proof. We have that H+ is a Killing horizon with respect to some KVF ξa. From the above lemma,θ = ω = σ = 0 along generators of H+, so by Raychaudhuri’s equation we have

0 = Rabξaξb∣∣∣H+

= 8πTabξaξb∣∣∣H+

,

using the fact that ξ is null in the second equality. Thus J · ξ|H+ = 0 where Ja = −Tabξb. Since ξais future-directed and causal, the DEC implies that so is Ja (unless Ja = 0). Hence we must haveJa parallel to ξa on H+. Thus

0 = ξ[aJb]

∣∣∣H+

= − ξ[aTb]cξc∣∣∣H+

= − 18π ξ[aRb]cξ

c∣∣∣H+

.

It can be shown that this implies0 = 1

8πξ[a∂b]κ,

so ∂aκ is proportional to ξa, so t · ∂κ = 0 for any vector field ta tangent to H+, and therefore κ isconstant on H+ (assuming H+ is connected).

– 51 –

9 Black Hole Mechanics 9.4 First law of BH mechanics

9.4 First law of BH mechanicsFor a Kerr black hole, if we make small perturbations to M and a, then it is possible to show that

κ

8πδA = δM − ΩHδJ. (∗)

In fact this holds under more more general perturbations. Take a spatial hypersurface Σ withΣ ⊥ ka near to i0. Make a small perturbation to the initial data on Σ, satisfying the constraintequations to 1st order. Then (∗) is satisfied, where δA is the change in area of the bifurcationsurface A, δM is the change in ADM mass, ∂J is the change in ADM angular momentum (definedanalogously to ADM mass). (For Einstein-Maxwell theory, we get an extra term ΦHδQ on theRHS, where ΦH is the potential difference between the eveny horizon and infinity.)There is another version of this law. Suppose we start with a Kerr spacetime and add an energy-momentum tensor Tab = O(ε). Then we can define Jc = −T ab kb and La = T abm

b. We find that∇aJa and ∇aLa are O(ε2), so these are conserved to first order.Let N be the portion of H+ to the future of B, and consider the case where some matter crossesN . The energy crossing N is

δM = −∫N∗J

and the angular momentum isδJ = −

∫N∗J.

Pick coordinates (y1, y2) on B, and let λ be the affine parameter distance along generators of Nfrom B and r the affine parameter along null geodesics orthogonal to N . The metric takes theform ds2

∣∣∣N

= 2 dr dλ + hij dyi dyj and hence the volume form is η|N =√h dλ ∧ dr ∧ dy1 ∧ dy2.

The orientation of N to use in evaluating the integrals is the one used in Stokes’ theorem, viewingN as the boundary of the region r > 0. This is dλ ∧ dy1 ∧ dy2. On N we then have

(∗J)λ12 =√hJr =

√hJλ =

√hU · J

where U = ∂∂λ . Hence

δM = −∫N

dλ d2y√hU · J

andδJ = −

∫N

dλ d2y√hU · L.

To order ε, we can replace h by g in these expressions. Thus we can use the fact that N is a Killinghorizon with KVF ξa = ka + ΩHm

a. Hence ξ|N = fUa, and

ξ · d log |f | = κ =⇒ U · df = κ =⇒ ∂f

∂λ= κ.

Thus f = κλ+ f0(y). Since ξa|B = 0, we have f0 = 0, so xia|N = κλUa. Hence

∂M =∫N

dλd2y√hTabU

akb

=∫N

dλd2y√hTabU

a(ξb − ΩHmb)

= κ

∫N

dλ d2y λ√hTabU

aU b︸ ︷︷ ︸= 1

8πRabUaUb

−ΩH

∫N

dλ d2y√hTabU

amb︸ ︷︷ ︸=−δJ

– 52 –

9 Black Hole Mechanics 9.5 Second law of BH mechanics

=⇒ δM − ΩHδJ = κ

8π

∫N

dλd2y√hλRabU

aU b.

Using Raychaudhuri’s equationdθdλ = . . .︸︷︷︸

O(ε2)

−RabUaU b

we have

δM − ΩHδJ = − κ

8π

∫d2y

∫ ∞0

dλ√hλ

dθdλ

= − κ

8π

∫d2y

[√hλθ]∞0 −

∫ ∞0

dλ(√h+ λ

d√h

dλ

)θ

Recall that d√h

dλ = θ√h = O(ε), so the λd

√h

dλ θ term is O(ε2) and can be ignored. We assume thatthe spacetime settles down to some stationary final state, so that

√h→ some finite limit as λ→∞.

Then ∫ ∞0

√hθ dλ =

∫ ∞0

d√h

dλ dλ = [√h]∞0 = δ

√h

is finite. Thus θ must be o(1/λ), so the boundary term [√hλθ]∞0 = 0 also. Finally we have

δM − ΩHδJ = κ

8π

∫Bδ√H d2y = κ

8πδA,

where δA is the change in the area of the horizon between early and late times.

9.5 Second law of BH mechanicsLecture 21

02/03/16 Theorem 21 (second law). Let (M, g) be a strongly asymptotically predictable spacetime that obeysEinstein’s equations and the NEC. Let U ∈ M be a globally hyperbolic region with J−(I+) ⊂ U ,Σ1,Σ2 be spacelike Cauchy surfaces for U such that Σ2 ⊂ J+(Σ1), and Hi = Σi ∩ H+. Thenarea(H2) ≥ area(H1).

Σ1

Σ2

H1

H2

Proof. Assume that the inextendible generators ofH+ are future-complete (we sayH is non-singular.Suppose θ < 0 at some p ∈ H+. By continuity, θ < 0 at some q ∈ H+ in the near future of p alonga geodesic γ, and therefore there is a point r conjugate to p along γ. Therefore we can deform γinto a timelike curve from p to r, but H+ is achronal, so this is a contradiction, and thus θ ≥ 0 onH+.Let p ∈ H1. The generator of H+ through p cannot leave H+, since generators don’t have futureendpoints, so it must intersect H2 (as Σ2 is a Cauchy surface). This defines a map φ : H1 → H2. Wehave area(H2) > area(φ(H1)) since φ(H1) ⊂ H2. Since θ ≥ 0, we have area(φ(H1)) ≥ area(H1).

– 53 –

10 QFT in Curved Spacetime

Example. In the spherical collapse of a star into a black hole, the radius of H+ can onlyincrease.

Example. Consider the merger of 2 black holes of areas A1 and A2 that are initiallySchwarzschild and well-seperated, and settle down to a Schwarzschild final state with area A3The second law tells us that A3 ≥ A1 +A2. In particular this means

16πM23 ≥ 16π(M2

1 +M22 )

and so we have a lower bound on the mass of the final black hole M3 ≥√M2

1 +M22 . The

energy radiated in the merger is M1 +M2 −M3, so we have an upper bound on the efficiencyof the merger

M1 +M2 −M3M1 +M2

≤ 1−

√M2

1 +M22

M1 +M2≤ 1− 1√

2.

Consider initial data (Σ, hab,Kab) that is asymptotically flat with a trapped surface behind anapparent horizon of area Aapp, and let Ei be the ADM energy of this data. Weak cosmic censorshipsays this should settle down to a stationary black hole at late time, and the uniqueness theoremssay it should be a Kerr black hole; let the mass of this black hole be Mf and its angular momentumbe Jf . By the second law, Aapp ≤ Ai, and we have

Ai ≤ AKerr = 8π(M2f +

√M4f − J2

f

)≤ 16πM2

f ≤ 16πE2i ,

where the last inequality holds because energy can only be carried away (by gravitational waves) inthis process. Hence we have the Penrose inequality:

Ei ≥

√Aapp16π

10 QFT in Curved SpacetimeConsider a black hole in its rest frame i.e. with ADM energy E = M . The first law of black holemechanics

dE = κ

8π dA+ ΩH dJ

looks a lot like the first law of thermodynamics

dE = T dS + µ dJ .

This suggests a correspondence

thermodynamics ↔ BH mechanicsT ↔ λκ

S ↔ A8πλ

µ ↔ ΩH

for some constant λ. The zeroth laws say that T and κ are constant in equilibrium. The secondlaws say that S and A cannot decrease.

– 54 –

10 QFT in Curved Spacetime 10.1 Free scalar field

This seems to suggest that black holes are thermal objects, but classical general relativity forbidsthe radiation that we would exect such objects to have. If however we take quantum mechanicsinto account, we will see that black holes do radiate, with temperature TH = ~κ

2π (so λ = ~2π ).

10.1 Free scalar fieldWe assume that the spacetime (M, g) is globally hyperbolic, with metric given by

ds2 = −N dt2 + hij(dXi +N i dt)(dXj +N j dt).

Let Σt denote the Cauchy surfaces given by t constant, and na be their normal. We havena = −N(dt)a and

√−g = N

√h. The action for the free scalar field is

S =∫M

dT d3x√−g

(−1

2gab∂aφ∂bφ−

12m

2φ2).

Varying φ gives the Klein-Gordon equation

gab∇a∇bφ−m2φ = 0,

and its conjugate momentum is given by

π(x) = δS

δ(∂tφ(x)) = −√−ggtµ∂µφ

= −N√h(dt)νgνµ∂µφ

=√hnµ∂µφ.

To quantise, we promote φ(x) and π(x) to operators and impose the canonical commutationrelations

[φ(t,x), π(t,x′)] = iδ3(x− x′),

[φ(t,x), φ(t,x′)] = [π(t,x), π(t,x′)] = 0.

In the classical picture we define S to be the set of complex solutions of the Klein-Gordon equation.Note that these solutions are specified uniquely by φ, ∂φ on Σ0. Let α, β ∈ S. We define a bilinearform on S by

(α, β) = −∫

Σ0d3x√hnaj

a(α, β) where j(α, β) = −i(α dβ − β dα).

Note that ∇aja = −i(α∇2β − β∇2α) = −im2(αβ − βα) = 0, so the integral is independent of thechoice of Cauchy surface.This bilinear form has the following properties:

• It is Hermitian: (α, β) = (β, α).

• It is non-degenerate: (α, β) = 0∀β =⇒ α = 0.

• (α, β) = −(β, α).

Note that it is not positive definite (as for example the last property above implies (α, α) = −(α, α)).In Minkowski space, (·, ·) is positive definite for Sp = positive frequency solutions. A basis forSp, labelled by momentum p, is given by

ψp = 1(2π)

32 (2p0)

12eip·x,

– 55 –

10 QFT in Curved Spacetime 10.2 Bogoliubov transformations

where p0 =√

p2 +m2. If k = ∂∂t , then Lkψp = −ip0ψp, so its eigenvalue is negative imaginary,

which is what we mean by positive frequency. Similarly, ψp has negative frequency, and we denotethe space of these solutions by Sp. Note that Sp and Sp are orthogonal, and we have S = Sp ⊕ Sp.In a curved spacetime, there is no a priori concept of positive frequency. We will make a choice ofdecomposition such that the following holds:

S = Sp︸︷︷︸positive frequency

(·,·)>0

⊕

negative frequency(·,·)<0︷︸︸︷Sp

In a general spacetime there are many ways to do this.Lecture 2204/03/16 We define positive frequency creation and annihilation operators for a mode f ∈ Sp by a(f) = (f, φ)

and a†(f) = −(f, φ) (this is consistent because φ = φ†). For example, in Minkowski space f = ψpfor some p, and a(f) = ap. When we promote these to operators, the canonical commutationrelations imply

[a(f), a(g)†] = (f, g) and [a(f), a(g)] = [a(f)†, a(g)†] = 0.

We define a vacuum state |0〉 by a(f) |0〉 = 0 for all f ∈ Sp, normalised so that 〈0|0〉 = 1. Given abasis ψi for Sp, we write ai = a(ψi). An N -particle state is then given by

a†i1a†i2. . . a†iN |0〉 .

The Hilbert space of the theory is spanned by |0〉, 1-particle states, 2-particle states, and so on.This space has positive definite inner product, for example∥∥∥a†(f) |0〉

∥∥∥2= 〈0|a(f)a†(f)|0〉

= 〈0|[a(f), a†(f)]|0〉= (f, f) 〈0|0〉= (f, f) > 0.

Now suppose we choose a different decomposition S = S′p ⊕ S′p. Let f ′ ∈ S′p and write f ′ = f + g,where f, g ∈ Sp. Then a(f ′) = (f, φ) + (g, φ) = a(f)− a(g)†, and we have a(f ′) |0〉 6= 0, i.e. |0〉 isnot the vacuum with respect to S′p. Thus we can see that the concept of a particle is ambiguous ingeneral when considering QFT in a curved spacetime.Consider a stationary spacetime with future-directed timelike KVF ka. This gives a map Lk : S → Sthat is antiHermitian with respect to (·, ·), so its eigenvalues are imaginary. We write Lku = −iωuand define positive frequency by ω > 0, which implies (u, u) > 0. In this case there is a preferreddefinition of particles, as we can just write Sp = positive frequency modes.

10.2 Bogoliubov transformationsLet ψi be an orthonormal basis for Sp, so that

(ψi, ψj) = δij and (ψi, ψj) = 0,

and define ai = a(ψi). We cam decompose any state φ in terms of this basis

φ =∑i

(aiψi + a†iψi

),

– 56 –

10 QFT in Curved Spacetime 10.3 Particle production in a non-stationary spacetime

and the commutation relations become

[ai, aj ] = δij , [ai, aj ] = [a†i , a†j ] = 0.

Now let S′p be another choice, with orthonormal basis ψ′i. The Bogoliubov transformation definesthis change of basis by two matrices A and B:

ψ′i =∑j

(Aijψj +Bijψj

)ψ′j =

∑j

(Bijψj +Aijψj

)Let a′i = a(ψ′i). Exercise: show that

a′i =∑j

(Aijaj −Bija†j

).

Using the commutation relations above, we can then deduce∑k

(AikAjk −BikBjk

)= δij i.e. AA† −BB† = 1

and ∑k

(AikBjk −BikAjk) = 0 i.e. ABᵀ −BAᵀ = 0.

10.3 Particle production in a non-stationary spacetimeConsider a globally hyperbolic spacetime (M, g) that is stationary only at late and early times,and write M =M+ ∪M0 ∪M−, where M+ is late times and stationary, and M− is early timesand stationary.

time

M−

M0

M+

stationary

stationary

Since (M±, g) are stationary, there exist for each of thempreferred choices of S±p . Since the spacetime is globallyhyperbolic, we can extend these to two choices for theentire spacetime.Let u±i be orthonormal bases for S±p , and write a±i =a(u±i ). As before, we can write

u†i =∑j

(Aiju

−j +Biju

−j

)and

a†i =∑j

(Aija

−j −Bija

−j†).

Assume there are no particles in M−. More concretely, we define the vacua with respect to S±p bya±i |0±〉 = 0∀ i. Also assume that no particles are present at early times, so that the state is |0−〉.

– 57 –

10 QFT in Curved Spacetime 10.4 Rindler spacetime

The number operator for the ith late time mode is N+i = a+

i†a+i , and its expectation value is

〈0−|N+i |0−〉 = 〈0−|a+

i†a+i |0−〉

=∑jk

〈0−|a−k (−Bjk)(−Bij)a−j†|0−〉

=∑jk

BijBij 〈0−|a−k a−j†|0−〉︸ ︷︷ ︸

=δjk

=∑j

BijBij = (BB†)ii.

Therefore the total expected number of particles is tr(BB†

)= tr

(B†B

)(and this is zero iff B = 0

i.e. iff S+p = S−p ).

10.4 Rindler spacetimeConsider spacetime geometry near the horizon fo a Schwarzschild black hole. We define a newcoordinate x by

r = 2M + x2

8Mand expand near x = 0. The metric becomes

ds2 = −κ2x2 dt2 + dx2 + (2M)2 dΩ2 + . . .

where κ = 14M is the surface gravity. Ignoring the subleading terms and fixing the spherical

coordinates, we obtain Rindler spacetime (x > 0):

ds2 = −κ2x2 dt2 + dx2

Let U = −xe−κt > 0 and V = xeκt > 0. Then we can write the metric as

ds2 = −dU dV = −dT 2 + dX2

where U = T −X and V = T +X. With these coordinates we can draw a Kruskal diagram:

Rindlerk

k

k

k

III

IIIIV

VU

U = 0 ∪ V = 0 is a bifurcate Killing horizon of k = ∂∂t , with surface gravity ±κ.

– 58 –

10 QFT in Curved Spacetime 10.5 Wave equation in Schwarzschild solution

We have k = κ(V ∂∂V − U

∂∂U

). Consider a “Rindler observer” on an orbit of k (lines of constant x).

The proper acceleration of the observer is given by

Aa = 1x

(dx)a =⇒ |A| = 1x.

We use k to define positive frequency (this differs from the usual Minkowskian definition that uses∂∂T ).Lecture 23

07/03/16 Consider a massless scalar field. The Klein-Gordon equation reads(− ∂2

∂T 2 + ∂2

∂X2

)φ =

(−κ2x2 ∂

2

∂t2+ ∂2

∂x2

)φ = 0.

A basis of positive frequency Minkowski modes is given by

up(T,X) = cpe−i(ωT−px) where σ = |p|.

Alternatively, a positive frequency Rindler eigenmode satisfies

Lkφ = −iσφ

for some σ > 0. From this, we deduce φ ∝ e−iσt, and solving the K-G equation gives φ ∝ e−iσtxiP ,where P = ±σ

κ . Hence a basis of positive frequency Rindler modes is given by

uRp (T,X) = cpe−i(σt−p log x) where σ = κ|p|.

Let Np be the number operator for a Rindler observer, and |0〉 the Minkowski vacuum. It can beshown that

〈0|Np|0〉 = 1e

2πσκ − 1

.

A Rindler observer has 4-velocity 1κx

∂∂t = A

κ∂∂t , so it measures frequency σ = A

κ σ, and we noticethat the number of particles follows a Planck spectrum at the Unruh temperature

TU = A

2π .

10.5 Wave equation in Schwarzschild solutionDecompose the field in terms of spherical harmonics:

φ =∞∑l=0

l∑m=−l

1rϕlm(t, r)Ylm(θ, φ)

The (massless) K-G equation ∇a∇aφ = 0 implies[∂2

∂t2− ∂2

∂r2∗

+ Vl(r∗)]ϕlm = 0

whereVl(r∗) =

(1− 2M

r

)(l(l + 1)r2 + 2M

r3

).

– 59 –

10 QFT in Curved Spacetime 10.6 Hawking radiation

Vl

r∗

I+H+

Consider a field with arbitrary initial data at t = t0. For late and early times (large |t|), thepotential vanishes, so the field will decompose into outgoing and ingoing parts:

ϕlm ≈ f±(u) + g±(v) for t→ ±∞

f± and g± are localised wavepackets, i.e. they vanish for |u| → ∞ and |v| → ∞ respectively. Thuswe have ϕlm|I+ = f+(u) and ϕlm|H+ = g+(v), and ϕlm is uniquely determined by its behaviour onI+ ∪H+.We define out and down modes as solutions vanishing on H+ and I+ respectively. We can writeany solution ϕlm as an out modes + down modes, and the inner product between the two vanishes.Similarly, define in and up modes as solutions vanishing on H− and I− respectively. Any solutioncan be written as in modes + down modes, and again their inner products vanish.

H+

H− I−

I+

out

H+

H− I−

I+

up

Consider φ ∝ e−iωt, φωlm = 1re−iωtRωlm(r)Ylm(θ, φ), ω > 0. We say that φ is “positive frequency”

if we can write it as a superposition of φωlm. If we substitute ϕlm = e−iωtRωlm into the waveequation, we obtain [

− ∂2

∂r2∗

+ Vl(r∗)]Rωlm = ω2Rωlm.

Since the potential vanishes for |r∗| → ∞, we have asymptotically Rωlm ∼ e±iωr∗ . Therefore

φωlm ∝ e−iω(t∓r∗) =e−iωu outgoinge−iωv ingoing

.

10.6 Hawking radiationConsider a collapsing star. This is time dependent, so we expect some particle production.Surprisingly, this particle production persists to late times.Since H− = ∅, there are no up modes. Let fi be a basis for in modes with positive frequencynear I−, pi be a basis for out modes with positive frequency near I+, and qi, qi be a basis ofdown modes. Thus we have two different bases for S: fi, fi and pi, pi, qi, qi. We assume thatboth bases are orthonormal.

– 60 –

10 QFT in Curved Spacetime 10.6 Hawking radiation

I+

I−

H+

Let ai = (fi, φ) and bi = (pi, φ) be annihilatorsfor the in and out modes respectively. We canexpand

pi =∑j

(Aijfj +Bijfj

),

and as before we obtain

bi =∑j

(Aijaj −Bija†j

).

Assume that the state is the “in” vacuum |0〉,i.e. ai |0〉 = 0 for all i. The expected number ofparticles in the ith out mode is

〈0|b†ibi|0〉 = (BB†)ii.Lecture 24

09/03/16 In order to calculate this, we will need to determine the Bologiubov coefficients Bij .

u

pi

2π/ωi

We choose the pi such that pi|I+ are wavepackets with fre-quencies near ωi > 0, and fi such that fi|I− are wavepack-ets with the save v-dependence as the u-dependence ofpi|I+ .For the moment, just consider Kruskal spacetime, andimagine propagating the wavepacket pi backwards in time.Part of the wavepacket would be reflected to I−, andpart would be transmitted to H−. Thus we can writepi = p

(1)i + p

(2)i , where p(1)

i is an in mode and p(2)i is an

up mode.

H+

H− I−

I+

pi

p(2)i p

(1)i

The reflection coefficient is Ri =√

(p(1)i , p

(1)i ), and the transmission coefficient is Ti =

√(p(2)i , p

(2)i ).

Since (pi, pi) = 1, we have R2i + T 2

i = 1 (since the in and up modes are orthogonal). Time reversalsymmetry implies that Ri is the fraction of fi reflected to I+, and Ti is the fraction of f i crossingH+.

– 61 –

10 QFT in Curved Spacetime 10.6 Hawking radiation

p(1)i

p(2)i

pi

Now return to the collapsing matter spacetime. We are interested in pi localised at late retardedtime. Because of this p(1)

i is localised at late advanced time , so scattering happens outside thscollapsing matter, and p

(1)i is the same as for the Kruskal spacetime. p(1)

i does not experience thenon-stationary geometry and so is positive frequency on I−. As above, (p(1)

i , p(2)i ) = R2

i . Thus inthe Bogoliubov transformation

p(1,2)i =

∑j

(A

(1,2)ij fj +B

(1,2)ij fj

),

we have B(ij1) = 0. Let Aij = A

(1)ij +A

(2)ij and Bij = B

(2)ij . Since (p(1)

i , p(2)i ) = 0 and (pi, pi) = 1, we

have (p(2)i , pi(2)) = 1−R2

i = T 2i .

Note that H+ is at u =∞, so for a wavepacket on I+ at late retarded times there are infinitelymany oscillations close to H+, and hence an observer crossing the horizon measures a very largefrequency.

v = 0

v0 < 0

– 62 –

10 QFT in Curved Spacetime 10.6 Hawking radiation

As there is a large frequency, we can use the approximation of geometric optics, and set

φ(x) = A(x) = eiλS(x),

where λ 1 and S(x) is the phase. ∇2φ = 0 implies (∇S)2 = 0, so S = constant are nullhypersurfaces.Consider then the null geodesic congruence containing the generators of H+ and surfaces of constantS. As previously, let U be the tangent vector field to the geodesics in the congruence, and let N bea null vector field such that N · U = −1 and U ·∇Na = 0. Write Sa = αUa + Sa + βNa whereαUa + Sa is the part orthogonal to Ua and tangent to H+, and betaNa points off of H+ towardsgenerators of surfaces S = constant.

H+

γ

γ′

Na Ua

−εNa

Let γ be a generator of H+, and γ′ a generatorof a constant S surface close to H+. By spher-ical symmetry, we can choose N θ = Nφ = 0.Outside the matter, ∂

∂V is tangent to the affinelyparametrised generators ofH+, so we can chooseUa =

(∂∂V

)a. N2 = 0, U · N = −1 then give

N = C ∂∂U , where C > 0 is a constant given by

gUV .γ has U = 0, so γ′ has U = −εC for some smallε (since we are considering late time modes andare outside matter). We have u = − 1

κ log(−U),and therefore γ′ is a null geodesic with

u = −1κ

log(εC).

Let F (u) denote the phase of the wavepacket pi on I+. Then the phase everywhere on γ′ is

S = F

(−1κ

log(εC)).

Near I−, γ and γ′ are ingoing radial null geodesics, so Ua ∝(∂∂u

)ain (u, v) coordinates. Near I−,

the metric isds2 ≈ −dudv + 1

4(u− v) dΩ2 .

Spherical symmetry, N2 = 0, U ·N = −1 together imply that N |I− = D−1 ∂∂v , where D > 0 is a

constant. From this we have that γ′ intersects I− at v = εD−1, and thus on I− at small v the phaseis

S = F

(−1κ

log(CDv)).

Hence we have

p(2)i

∣∣∣I−≈

0 v > 0A(v) exp

[iF(− 1κ log(CDv)

)]small v < 0,

where the amplitude A(v) is some smooth function of v.Lecture 2511/03/16 At the expense of some rigour, we will now drop the wavepacket treatment and assume that we are

dealing with plane waves, so that F (u) = −iωiu. We will write pω instead of pi, so that

p(2)ω

∣∣∣I−≈

0 v > 0Aω(v) exp

[iωκ log(CDv)

]small v < 0.

– 63 –

10 QFT in Curved Spacetime 10.6 Hawking radiation

Similarly, use a basis of in modes labelled by their frequencies

fσ|I− = (2πNσ)−1e−iσv, σ > 0.

Writing p(2)ω in terms of the fσ is just the same as finding its Fourier transform. We have

p(2)ω (σ) = Aω

∫ 0

−∞dv eiσv exp

[iω

κlog(−CDv)

].

where we have approximated Aω is constant since p(2)ω is non-zero for a small portion of v. The

inverse is

p(2)ω (v) =

∫ ∞−∞

dσ2π e

−iσvp(2)ω (σ)

=∫ ∞

0dσNσp

(2)ω (σ)fσ(v)︸ ︷︷ ︸

positive frequency

+∫ ∞

0dσNσp

(2)ω (−σ)fσ(v)︸ ︷︷ ︸

negative frequency

.

Hence we can write down some Bogoliubov coefficients:

A(2)ωσ = Nσp

(2)ω (σ), Bωσ = Nσp

(2)ω (−σ) (ω, σ > 0)

In order to evaluate the Fourier transform we will complete the contour of integration in the complexplane. We choose the branch of log to be the one given by log |z|+ i arg z, where arg z ∈

(−π

2 ,3π2

).

Consider p(2)ω (−σ) for σ > 0. We close the contour in the v-plane in the following way:

R→∞

By Jordan’s lemma, the contribution of the large semi-circular part of the contour vanishes. Hence

p(2)ω (−σ) = −Aω

∫ ∞0

dv e−iσv exp[iω

κlog(−CDv)

]= −Aω

∫ ∞0

dv e−iσv exp[iω

κ(log(CDv) + iπ)

]= −Aωe−

πωκ

∫ 0

−∞dv eiσv exp

[iω

κlog(−CDv)

](v → −v)

= −e−πωκ p(2)

ω (σ).

Therefore |Bωσ| = e−πωκ |A(2)

ωσ |. Returning to wavepackets, the corresponding result is

|Bij | = e−πωiκ |A(2)

ij |.

– 64 –

10 QFT in Curved Spacetime 10.7 Black hole thermodynamics

Note thatT 2i = (p(2)

i , p(2)i ) =

∑j

(|A(2)

ij |2 − |Bij |2

)=(e

2πωiκ − 1

)∑j

|Bij |2,

and∑j |Bij |2 = (BB†)ii = 〈0|b†ibi|0〉, so we have

〈0|b†ibi|0〉 = Γie

2πωiκ − 1

,

where Γi = T 2i is the absorption cross-section for the mode fi. This is exactly a blackbody spectrum

at the black hole’s Hawking temperature

TH = κ

2π .

Note that TH ≈ 6× 10−8 ×MM K, and the CMB is at a temperature of 2.7 K, so the effect of this

Hawking radiation is cosmologically negligible.

10.7 Black hole thermodynamicsHawking radiation implies that a black hole is a thermodynamic object at temperature TH . Thezeroth law of BH mechanics is just the application of the zeroth law of thermodynamics to theblack hole. Writing the first law as

dE = TH dSBH + ωH dJ ,

we see that the entropy associated with the black hole is

SBH = A

4 = c3A

4G~ .

Note that SBH can decrease quantum mechanically. We therefore make a generalisation of thesecond law to saying that SBH + Smatter is non-decreasing (this holds in Hawking radiation).Note that if most of the energy in the universe was in the form of black holes, the total entropy inthe universe would be far higher than it currently is – the universe seems to be in a special lowentropy state.Statistical mechanics gives that the number of quantum microstates of the black hole as N ∼ expSBH .A replication of this result is a major goal of quantum gravity research.

10.8 Black hole evaporationThe energy of Hawking radiation must come from the black hole itself. Stefan’s law for energy lossfrom a blackbody gives

dEdt ≈ −αAT

4

where α is a dimensionless constant, and we approximate γi by treating the black hole as a perfectlyabsorbing sphere of area A. Plugging in E = M , A ∝M2 and T = TH ∝ 1

M gives

dMdt ∝ −

1M2 .

Hence we have M → 0 in a finite time τ ∼M3 ∼ 1071( MM )3 seconds.

– 65 –

10 QFT in Curved Spacetime 10.8 Black hole evaporation

Black hole evaporation gives rise to the so-called information paradox. Consider the followingsequence of events. A star is initially in a pure quantum state. It collapses to a black hole, andthen evaporates. The final state is completely composed of thermal radiation which must be amixed state. So we have gone from a pure state to a mixed state, but this is impossible in quantummechanics!Information about the initial state is lost, and the only information that remains is the energy,angular momentum and charge of the black hole.

Fin

– 66 –