Lecture 1, 2007-10-17. General informa-

tion

Advanced General Relativity and Cosmology

Instructor: Dr. Sergei Winitzki

Office: Philosophenweg 19, Room 014

Office hours: Thursdays 14:00-16:00

Home page: http://www.tphys.uni-heidelberg.de/~winitzki/AGRC/

Lectures: Wednesday and Friday, 11:15-13:00,

Philosophenweg 19, Seminarraum

Material shown in green was not presented due

to lack of time or for other reasons

1

Course outline

Problem of the year: inflation and dark energy

Tensor calculus explained without indices

Cosmological evolution in f(R) gravity

Birrell-Davies demystified: Basics of quantum

field theory in curved spacetime

Classical and quantum theory of cosmological

perturbations

Geometry of null surfaces, causal structure,

chronology protection, time machines in GR

Thermodynamics of horizons, entropy of black

holes, "holographic principle"

Pseudotensors demystified: Energy in asymp-

totically flat spaces; EMT of gravitational waves

Hawking-Ellis demystified: Classical singularity

theorems

2

Problem of the year

Cosmological observations yield:

- primordial inflation

- late-time inflation

Early work on inflation:

A. Starobinsky, Phys. Lett. B91, p.99, 1980

de Sitter stage in cosmological evolution

Mechanism: modification of gravity due to

quantum backreaction, for example:

S[g] =

∫

d4x√−g

(

R

16πG+ αR2

)

Inflationary spacetime: de Sitter with flat spa-

tial sections, Hubble rate H:

gµνdxµdxν = dt2 − e2Ht

(

dx2 + dy2 + dz2)

3

Problem of the year

To explain dark energy and primordial inflation

within a single model of modified gravity, and

to remain in agreement with other tests of GR.

A model of modified gravity is f(R) gravity:

S =1

16πG

∫

d4x√−g (R+ f(R)) + Smatter

Recent proposals: f(R) = c1(

1 + c2R−n)−1

How to describe consequences of f(R) gravity?

- Need equations of motion for g

- Solutions for Solar system tests

- Homogeneous cosmological solutions: have

inflation and/or dark energy?

- Primordial cosmological perturbations

There are > 50 papers on aspects of f(R)

gravity in 2006-2007

4

Deriving the EOM for f(R) gravity

Note: f(R) is nonlinear ⇒ EOM are 4th order.

Equations for gµν are complicated

Use the method of Lagrange multipliers! Then

we can reduce the Lagrangian to one that is

linear in R and then EOM will be 2nd order

Analogy with mechanics:

S =

∫

dt

[

1

2x2 + f(x, x)

]

– nonlinear in x

EOM would be 2nd order if S were linear in x

Trick: Introduce new variable y and Lagrange

multiplier λ that enforces y = x:

S =

∫

dt

[

1

2x2 + f(x, y) + λ (x− y)

]

EOM for y is f,y(x, y)− λ = 0.

Since f,yy 6= 0, we can solve for y through x, λ:

y = Y (x, λ). Also replace λx→ −λx. Finally:

S =

∫

dt

[

1

2x2 − λx+ f(x, Y (x, λ))− λY (x, λ)

]

Obtained a first-order Lagrangian with extra

d.o.f. λ and potential energy V (x, λ) = f − λf ′5

Whether one can/cannot substitute into

Lagrangians

Suppose we have L(x, x, y, y, ...); we can change

variables x, y → A(x, y), B(x, y), ...

We cannot change variables to new variables

involving derivatives, x→ A(y, y, ...)

Example: L = x2, substitute x = y + y, get

wrong EOM: y(4) − y = 0

Suppose we have L(x, x, y, y, ...) and one of the

EOM is x = f(y, y, ...) ⇒ We can eliminate x.

Suppose we have constraints: L+λF . We can

substitute F into L as long as L+λF remains.

This works even if F involves derivatives!

But: we cannot substitute EOM into L if

EOM is x = f(x, x, ...) or x = f(y, z) (change

the number of relevant initial conditions). Also

cannot substitute arbitrary relationships, e.g. spe-

cific solutions x = f(y, y, const), unless they

are enforced by constraints.

6

Practice problems:

1. Show that L = xf(x, x) can be reduced to

1st order by adding a total derivative ddtM(x, x).

2. Reduce L = f(x, x, x,...x ) to 1st order by

introducing appropriate new variables.

3. Reduce L =...x f(x, x, x) to 1st order (first

reduce to 2nd order, i.e. remove...x ).

Incorrect substitution of EOM into Lagrangian

Example: L = 12x

2 + 12xy

2

Equations of motion:

x =1

2y2

d

dt(xy) = 0

Solve y = C1x−1, substitute into EOM for x:

x =C2

1

2x

But: what if we substitute y = C1x−1 into L?

L =1

2x2 +

C21

2

1

xEquation of motion for x now is wrong:

x = − C21

2x2

7

Want to derive EOM for f(R) gravity!

Apply this technique to f(R) gravity:

S[g] =1

16πG

∫

d4x√−g (r+ f(r) + λ (R− r))

EOM for r is 1+ f ′(r)− λ = 0. Solve r = r(λ)

and substitute back:

S[g] =1

16πG

∫

d4x√−g

(

f − rf ′+ λR)

Now apply a trick: conformal transformation:

g → g ≡ e2Ωg

Choose Ω such that λR√−g becomes R

√−gplus some terms — need a formula for R!

Result: GR with scalar field Ω and metric g

“Einstein frame” vs. “Jordan frame” – these are

different variables, not reference frames

Note: coupling to matter is through g, not

through g — matter feels Ω and g

8

Tensor calculus without indices

Geometric view: manifolds, topology

Geometric operations on vector and tensor fields

Lie derivative, metric, curvature

Index-free notation: instead of

uνuµ;ν = 0 ⇒ uν∂ν

(

gαβuαuβ

)

= 0

write

∇uu = 0 ⇒ ∇ug(u,u) = 2g(∇uu,u) = 0

Practical tools for calculations without indices

Vielbein formalism; curvature and Einstein eqs.

9

Quantum field theory in curved spacetime

Key applications in cosmology:

* Prediction of the primordial fluctuations

* Dynamics of chaotic inflation

Key applications in black hole physics:

* Hawking radiation

* Black hole entropy

* Thermodynamics of horizons

Many calculations do not require “hard” QFT!

10

Causality and horizons

In GR, matter determines the causal structure

Time machines? Chronology protection?

Horizons are boundaries of causal regions

Horizons have thermodynamical properties!

Black holes have temperature and entropy

“Holographic principle”: a bound on entropy

due to quantum effects and nonlinear coupling

to gravity (not due to quantum gravity)

Can study BHs without quantum gravity!

11

Energy and conservation laws in GR

Ill-defined local energy density; “pseudotensors”

Energy in asymptotically flat spaces

Energy of matter + gravity is conserved!

Energy-momentum tensor of gravitational waves

12

Classical singularity theorems

GR predicts singularity in many cases

Limit of applicability of classical GR!

Proofs of singularity theorems have 3 parts:

* Geodesic focusing theorem

* Energy conditions – guarantee focusing

* Topological argument ad absurdum

Applications:

* Gravitational collapse of stars

* Past cosmological singularity (Big Bang)

* Future cosmological singularity (Big Crunch)

13

Lecture 2, 2007-10-19. Manifolds

Gravity in GR curved geometry, not a force

Observations: Small scales R4, large scales –?

Manifold M is a generalized “curved space”

Visualize manifolds as embedded surfaces

Examples: sphere x2 + y2 + z2 = R2.

Torus:(√

x2 + y2 −A)2

+ z2 = R2, A > R

0

z

A

Rρ

14

Local coordinates

Locally like Rn means local coordinate system

which is a map U → Rn, where U ⊂ M is a

patch; all of M must be covered by patches

Local coordinates on torus: 0 < (φ , θ) < 2π,

x = ρ cosφ, y = ρ sinφ, z = R cos θ,

ρ = A+R sin θ

Another choice of coordinates: ρ = ab−sin θ,

z = c cos θb−sin θ, for appropriately chosen a, b, c. The

coordinate change map Rn → R

n is smooth

Intrinsic description: no embedding needed!

Coordinates are scalar functions on M

Functions on manifold = functions on patches

= functions on subsets of Rn

Intrinsic definition of smoothness: smooth func-

tions on subsets of Rn + smooth coordinate

change maps

15

Tangent spaces

Vectors in physics mean velocities / directions

Visualize a vector as velocity of a point moving

within M. Path γ(t) given in local coord’s

In embedding picture: vector is in tangent

plane

E.g.: hypersurface f(x) = 0, normal vector

nj = ∂xjf(x0), tangent plane is n · (x− x0) = 0

Tangent plane→tangent space TpM at point p

Intrinsic picture: velocity means change; tan-

gent vector v = derivative operator on scalar

functions f : M → R; this derivative is with

respect to the “time” of the moving point

Dγf =d

dtf(γ(t)) =

∑

j

∂f

∂xj

∣

∣

∣

∣

x=γ(t0)

dγj

dt

∣

∣

∣

∣

∣

t=t0

The derivative operator is v ≡ ∑

jdγj

dt∂∂xj

. Write

v f instead of Dγf .

Tangent vectors v form a vector space; basis

tangent vectors are ∂xj ≡ ∂j.16

Calculations with tangent vectors

Chain rule: v [f(g)] = f ′(g)v gConsequence: components of v in a coordi-

nate system xµ are computed as v xµ, and

then v =∑

µ (v xµ) ∂xµ.

Practice problems: 1. Find components of

x∂x + y∂y in coordinates a = x+ y, b = 3xy2. In 2D, find tangent vector γ to the curve

γ(t) = x = cos t, y = 2sin t in coord’s x, y3. Given u = ∂x+x∂y, find coord’s X,Y such

that u = ∂Y .

Other properties: v (fg) = fv g + gv f ,linearity v (f + g) = v f + v g – as a 1st

derivative. All 1st derivatives are equivalent to

tangent vectors.

Tangent vectors as short curve segments:

γf ≈ f(γ(t0 + σ)− γ(t0))σ

, f(p′)−f(p) ≈ (σγ)f

Local vector field = choice of tangent vector

v|p ∈ TpM at each point p ∈ U ⊂MFlow of vector field; show that v t = 1

17

Commutator of vector fields

Consider [u,v] f = u (v f)− v (u f)This is a 1st derivative operator, thus [u,v] is

a new tangent vector

Example: [x∂x, ∂x + ∂y] has no 2nd derivatives

Proof: show that [u,v] (fg) = ... [Derivation]

Practice problem: compute commutators [a,b],

[a, c], [b, c] for

a = x∂x + y∂y + z∂z, b = x∂y − y∂x, c = ∂z

Geometric view: [a,b] is the arrow p′ → p here:

a

b

p′

p0

p

p1

p2

Approximately f(p)− f(p′) ≈ (δt)2 [a,b] f |p018

Lecture 3, 2007-10-24. Connecting vector

fields

Vector field c is connecting for v if [c,v] = 0

(these vector fields commute)

Connecting vector c 6= v points across v and

along lines of equal “time”, for some choice of

“time” across the flow

s=0

τ = 3

τ = 2

τ = 1

v

c

Note: coordinate vector fields ∂xµ always com-

mute (no discrepancy when following coord. lines)

Question: Given a vector field v, can we find

coord’s xµ s.t. v = ∂x1? (Yes. Need a basis

of connecting vector fields for v. But [c,v] = 0

is an ODE for c along flow of v.)

19

Practice problems: 1. A vector v is tangent

to a surface f(p) = const iff v f = 0. If u,v

are tangent to the surface, show that [u,v] is

also tangent to the same surface.

2. A flow is given explicitly by the equations

x, y, z 7→ x+ τy, y, z − τxCompute the vector field v that generates this

flow. Find two examples of connecting vector

fields for v.

Differential forms

1-form is a linear function chosen in each tan-

gent space (element of cotangent space T ∗pM)

Example: v 7→ v f for fixed f ; this 1-form is

denoted df , so (df) v ≡ v f . The 1-form df

is called the gradient of f

Interpretation: (df)v is the change of f over

a short curve segment represented by v

Coordinate 1-forms: dx, dy, dz are gradients

of coordinate functions x, y, z

(dx) ∂x = 1, (dx) ∂y = 0, etc.

Practice problems: 1. For ω = d(x2y3) and

v = y2∂x, compute ω v

2. Let ω a ≡ y (a x) y2 − 2a y; express ω

through dx, dy

3. Find all 1-forms ω = adx + bdy such that

ω (∂x + x∂y) = 0

4. Find all vectors v = a∂x + b∂y such that

(xdy+ ydx) v = 0

Tensor fields = tensor product of vector fields

and 1-forms

20

Calculus of n-forms

Antisymmetric tensor product of 1-forms:

(ψ ∧ ω) (a,b) = ψ(a)ω(b)−ψ(b)ω(a)

Exterior differential:

d (λω) = (dλ) ∧ ω+ λ ∧ dω

d(ψ ∧ ω) = (dψ) ∧ ω+ (−)|ψ|ψ ∧ dω

d (dω) = 0

Examples: d (xdx) = 0; d(xdy−ydx) = 2dx∧dyGeneral formulas: (θ is 1-form, ω is n-form)

(dθ) (x,y) = x θ(y)− y θ(x)− θ ([x,y])

(dω) (v1, ...,vn+1) =n+1∑

s=1

(−)s−1 vs ω(v1, ...vs...,vn+1)

+∑

1≤r<s≤n+1

(−)r+s−1ω(

[vr,vs] ,v1, ...vrvs...,vn+1

)

Insertion operator: (ιxω)(a,b, ...) = ω(x, a,b, ...)

Integration of ω over a domain U : ∫U ωStokes theorem:

∫

U dω =∫

∂U ω

Practice problem: show that dθ(x,y) does

not depend on derivatives of x,y

21

Frobenius theorem for 1-forms

1-form ω is exact iff ω = df ; then ω(v) = 0

means v is tangent to the surface f = const

Given a 1-form ω, how to find whether ω(v) =

0 is surface-forming?

Example: ω = xdy is surface-forming, but not

xdy − ydx+ dz (it’s a spiral-shaped field!)

Theorem: ω is surface-forming iff ω ∧ dω = 0

Proof: 1. If ω is surface-forming, then ∃f : ∀v,

(ω(v) = 0) ⇔ (v f = 0). Choosing a basis

df, θ1, θ2, ... in T ∗pM we find ω = λdf (else

∃v : v f 6= 0, ω(v) = 0) ⇒ ω ∧ dω = 0.

2. From ω ∧ dω = 0, need to show that

flows of ∀v : ω(v) = 0 lie within a single sur-

face. Commutator of u,v should remain within

the surface! Need to show ∀u,v : ω(u) = 0,

ω(v) = 0 ⇒ ω([u,v]) = 0. Choose a vector

t such that ω(t) = 1 (“transverse”). Com-

pute: 0 = (ω ∧ dω)(t,u,v) = ω(t)dω (u,v) =

dω(u,v) = u ω(v)− v ω(u)− ω([u,v]), thus

ω([u,v]) = 0.

Practice problem*: If F ∧ ω = 0, where F is

n-form and ω is 1-form, show that ∃ θ : F =

θ ∧ ω where θ is a suitable (n− 1)-form.

22

Lie derivative

Need derivative of vectors in direction v but...

Cannot add tangent vectors at different points!

Solution: transport vectors by the flow of v

v

a pb

p′a′

b′

Lva = limδt→0

T−1v

[

a(p′)]− a(p)

δt

LvX measures how much X differs from TvX(i.e. X transported by the flow of v)

Properties that follow from this picture:

* Lvf = v f on scalar functions f* Lva = [v, a] on vectors a

(Note: connecting vector is exactly transported.)

* Lv (X + λY ) = LvX + λLvY + (Lvλ)Y* Lv (ω a) = (Lvω) a+ω Lva on 1-forms ωNote: LvX depends on the flow of v, not only

on the value of v at a point!

Lie derivative of tensors: Lv(X⊗Y ) = (LvX)⊗Y +X ⊗LvY ; same for wedge product ω ∧ ψ

23

Lecture 4, 2007-10-26. Calculations with

Lie derivatives

LvX depends on derivatives of v as well as on

derivatives of X:

Lλva = λLva− (a λ)v 6= λLva

Lie derivative of 1-form:

(Lvω) u = v (ω u)− ω [v,u]

Compute L∂x (dy):(

L∂xdy)

v = L∂x (v y) −dy [∂x,v] = v (dy ∂x) = 0.

Lie derivative of a bilinear form: (LvA)(x,y) =

v [A(x,y)]−A([v,x] ,y)− A(x, [v,y])

Cartan homotopy formula: Lxω = ιxdω+ dιxω

Practice problems: 1. Compute L∂xdx; Lx∂x (x∂y);

Lx∂y (xydx); Lx∂y+xy∂x

(

2z2dz)

; Lx∂x (xdx ∧ dy).

2. Let A(a,b) ≡ (a x) (b y) z−(a y) (b z)x;compute Lx∂yA. (A is a 2nd rank tensor)

3. Let T (a)b = (b x) ya− (a y)xb; compute

L∂xT . (T is a 3rd rank tensor)

4. Show that Lvdλ = dLvλ for scalar λ

5. Show that Lλvω = λLvω+ (ω v)dλ

24

Important property of Lie derivative

Statement: If L∂1T = 0 for a tensor T =∑

α,β,... Tαβ...dxα⊗dxβ⊗... then the components

Tαβ... do not depend on x1

Proof: use Leibnitz rule and L∂1dxα = 0 and

L∂1∂α = 0

Interpretation: if LvT = 0 then components of

T are constant along the flow of v

Which components? – In a basis of connecting

vectors for v

The property LvT = 0 does not depend on

other basis vectors, only on v!

Note: x-component of a tensor depends on the

function x(p) and on the vector field ∂x. But

∂x depends not only on the function x(p), but

also on the choice of every other coordinate.

25

Metric

Need to compute distances along paths

Visualize using the embedding picture:

z

a

y x

b

Distance between a and b along M is

D(a,b) ≈√

(ax − bx)2 + (ay − by)2 + (az − bz)2

For infinitesimal lengths along x(τ), y(τ), z(τ):

δL = δτ

√

(

dx

dτ

)2

+

(

dy

dτ

)2

+

(

dz

dτ

)2

Use intrinsic coordinates (u, v): δL =

= δτ√

(x,uu+ x,vv)2 + ... = δτ

√

Au2 + 2Buv+ Cv2

Need to know only A,B,C to compute length!

The length in larger space gives the induced

metric. Intrinsically, metric is a quadratic form

describing the length of infinitesimal curve seg-

ments γδτ : δL ≈√

g(γδτ, γδτ)

Note: in GR the metric has signature +−−−26

Applications of metric structure

Example: if g(γ, γ) = Au2 + 2Buv + Cv2 incoords. (u, v), theng = Adu⊗du+B (du⊗ dv+ dv ⊗ du)+Cdv⊗dvShorthand notation: g = Adu2 + 2B dudv +C dv2

Practice problem: compute the metric forthe surface z = x2 + y2 in coords. x, y

* Metric maps vectors into 1-forms and back

v 7→ gv : (gv) x ≡ g(v,x)ω 7→ g−1ω : g(

(

g−1ω)

,x) ≡ ω x

* Orthonormal frame: ea; g(ea, eb) = ηab* Dual basis of 1-forms θa; θa eb = δab

Levi-Civita tensor ε ≡ θ0 ∧ θ1 ∧ θ2 ∧ θ3– independent of basis, up to orientationǫ(a,b, c,d) = the 4-volume of a parallelepiped

Components in a coordinate basis: gµν ≡ g(∂µ, ∂ν)All 4-forms are proportional to ε. Formula:

ε = |det gµν|1/2 dx0 ∧ dx1 ∧ dx2 ∧ dx3

Proof: volume spanned by ∂µ equals ε(∂0, ...)and also equals |det g(∂µ, ∂ν)|1/2(introduce transition matrix ∂α = M

βαeβ, then

the volume spanned by ∂µ equals det M)

27

Affine connection (covariant derivative)

Want to have directional derivative of vectors:

∇vx that depends only on the value v|p (Lie

derivative Lvx contains derivatives of v at p)

Problem: need to subtract vectors in different

tangent spaces; need a transport operator T :

∇vx = limσ→0

Tp′→p x|p′ − x|pσ

Motivation: consider surface embedded in flat

space; “roll” the tangent plane along the vector

v from p′ to p

Tx = x + λn

Tp′→p

n

x|ppp′

x|p′

Note: n-dimensional “rolling” is 2-dim. rota-

tion in the v,n plane – to 1st order this is

equivalent to adding a multiple of n to vectors

in larger space; to 2nd order – a multiple of v

28

Properties of the induced connection

Denote by g and by ∂ the metric and the aff. con-

nection in large space, and by g the induced

metric on the surface

Since Tx = x + λn ⇒ ∇vx = ∂vx + λ(v,x)n

Since ∇vx ∈ TpM, we have ∇vx = Proj(n)∂vx

(orthogonal projection w.r.t. g onto TpM)

Proj(n)x ≡ x− ng(n,x)

g(n,n)

Need a formula for ∇vx without embedding

Note properties: ∂uv − ∂vu = [u,v]; ∂ug = 0

Define Tors (a,b) ≡ ∇ab − ∇ba − [a,b]; ver-

ify that Tors (a,b) is dependent of derivatives;

then Tors (a,b) = Proj(n)˜Tors (a,b) = 0

Verify ∇vg = 0 for tangent vectors v:

(∇vg) (a,b) = v g(a,b)− g(∇va,b)− g(a,∇vb)

= v g(a,b)− g(∂va,b)− g(a, ∂vb) = 0

∇vx has the properties of 1st order derivative.

Verify: ∇vx is independent of derivatives of v:

∇λvx = Proj(n)∂λvx = λProj(n)∂λvx = λ∇vx

→ Express ∇vx through the intrinsic metric g!

29

Levi-Cività (LC) connection

Properties: the connection is torsion-free:

[x,y] = ∇xy −∇yx

Compatible with the metric:

∇xg(a,b) = g(∇xa,b) + g(a,∇xb)

Will now derive an explicit formula for ∇yx.

Define derivative tensor B(x)(a,b) ≡ g(∇ax,b).

Consider first the antisymmetric part of B(x):

B(x)(a,b)−B(x)(b, a) = g(∇ax,b)−g(∇bx, a) =

∇ag(x,b) − ∇bg(x,b) − g(x, [a,b]) = (dgx) (a,b).

Symmetric part: B(x)(a,b)+B(x)(b, a) = g(∇xa−Lxa,b)+g(a,∇xb−Lxb) = ∇xg(a,b)−g(Lxa,b)−g(a,Lxb) = (Lxg) (a,b).

Hence

g(∇ax,b) =1

2(d (gx) + Lxg) (a,b)

Fully explicit formula (Koszul): 2g(∇ax,b) =

a g(x,b)−b g(x,a)− g(x, [a,b])+x g(a,b)−g([x, a] ,b)− g(a, [x,b])

30

Geodesic vector fields and curves

A vector field v is geodesic if ∇vv = 0.

A curve γ(τ) is geodesic with affine parameter

τ if ∇vv = 0, where v = γ is the velocity.

Can change τ → Aτ +B if A,B = const

Normalization property: ∇vg(v,v) = 0, so we

can choose τ such that g(v,v) = 0,±1

Null geodesic: g(n,n) = 0. If observed in a ref-

erence frame with 4-velocity u, the frequency

of the photon is ν = g(u,n).

31

Killing vectors

Motivation: geometry does not change in the

direction of a symmetry

Example: g = dt2 − e2N(t)(

dx2 + dy2 + dz2)

does not change in directions x, y, zGeneralize: components of g are constant along

the flow of v

A vector field v is a Killing vector if Lvg = 0.

Example: g = f(r)dt2 − g(r)dr2 − r2dΩ2 has

Killing vectors ∂t, ∂φ (and others!)

Formula: (Lxg) (a,b) = g(∇ax,b)+ g(a,∇bx)

⇒ Killing equation:

k : g(∇ak,b) + g(a,∇bk) = 0 for ∀a,bNote: for a Killing vector k, the symmetric part

of B(k) vanishes ⇒ B(k) = 12dgk is a 2-form

Conformal Killing vector: Lkg = 2λg for some

scalar λ. ← Geometry is conformally rescaled

under the flow of k.

Example: FRW universe, g = dt2−a2(t)γ. Ex-

pect conformal Killing vector k = f(t)∂t with

some f(t). Calculation: assume a = at∂t+ax∂xsuch that ∂tat,x = 0, then (Lkg) (a,b) =

−f∂t(

a2)

γ(a,b)+2f ′atbt. Thus we must have

2f ′ = 2a′a f , hence f(t) = a(t) · const.

32

Redshift factor and gravitational potential

Metric g is stationary if there exists a timelike

Killing vector k : g(k,k) > 0, Lkg = 0.

Note: metric g is static if g(k,k) > 0, Lkg = 0,

and k is everywhere orthogonal to a surface.

Stationary observers move with 4-velocity k√g(k,k)

Consider a photon along null geodesic γ(τ); lo-

cally observed frequency (energy) is ν = g(k,γ)√g(k,k)

.

Statement: g(k, γ) = const along geodesic γ.

Proof: Let v = γ, then ∇vg(k,v) = g(∇vk,v) =12 (dgk + Lkg) (v,v) = 0.

Hence ν = ν0 [g(k,k)]−1/2. Define redshift z =√

g(k,k).

Photons are redshifted by factor z

33

Lecture 5, 2007-10-31. Geodesics extrem-

ize proper length

Statement: Geodesic curves γ(τ) extremize

proper length∫

√

g(γ, γ)dτ

Proof (for timelike curves): Deform γ(τ) by

transverse vector field t; obtain connecting field

v; compute: Lt∫

√

g(v,v)dτ =∫ g(∇tv,v)√

g(v,v)dτ =

∫ g(∇vt,v)√g(v,v)

dτ =∫ ∇v (...) dτ−∫ g(t,∇v

v√g(v,v)

)dτ .

Hence, ∇vv√g(v,v)

= 0. Reparameterize τ →∫ τ√

g(v,v)dτ to achieve g(γ, γ) = 1.

Note: for spacelike curves need√

−g(v,v); for

null curves need∫

Ng(v,v)dτ .

Universal variational principle:

δ∫

[

g(γ, γ)

N+KN

]

dτ = 0 where N ≡ N(τ)

EOM: g(γ, γ) = KN2; ∇v

(

N−1v)

= 0

For null geodesics: choose K = 0 (but N 6= 0!)

34

Killing vectors II

Compute:

(Lλkg)(a, a) = 2g(∇aλk, a) = 2 (a λ) g(k, a)+(Lkg)(a, a), so Lλkg = λLkg+dλ⊗gk+gk⊗dλor (Lλkg)αβ = λ (Lkg)αβ + λ,αk,β + λ,βk,α

Lk (λg) = (k λ) g+ λLkg

Example: FRW metric g = dt2 − a2dx2, let us

determine conformal Killing vector k = f∂t:Lf∂tg = fL∂t

(

dt2 − a2dx2)

+df⊗g∂t+g∂t⊗df =

−2faa dx2 + 2fdt⊗ dtIf f = a then Lf∂tg = 2ag.

For timelike conformal Killing vectors: redshift

z =√

g(k,k) still meaningful since ∇γg(γ,k) =12 (dgk + Lkg) (γ, γ) = 0 for null geodesics γ

Note: (Lkg) (k,k) = 2g(∇kk,k) = 2λg(k,k),

so λ =g(∇kk,k)g(k,k)

= ∇k ln z

Also compute g(∇kk,x) = 2λg(k,x)−g(∇xk,k) =

2λg(k,x)− 12∇xg(k,k), thus

∇kk = −zg−1dz+ 2λk

Practice problem: Can one always choose a

vector field v such that all covariant derivatives

vanish at a point? (∀a : ∇av = 0)

35

Redshift and gravitational potential

Consider a stationary spacetime with a Killing

vector k; redshift z =√

g(k,k)

Redshift is time-independent: ∇kz = 0 (since

∇k ln z = λ = 0 for a Killing vector)

Is k geodesic? No: ∇kk = −zg−1dz 6= 0 unless

z = const.

Stationary observers move along u = 1zk.

Acceleration of stationary observers, a = ∇uu =

−g−1dΦ where Φ is the gravitational potential.

Proof: with u = z−1k we have ∇uu = 1z∇k

kz =

1z2∇kk = −1

z g−1dz = −g−1dΦ where Φ ≡ ln z.

Example 1: Schwarzschild spacetime, k = ∂t,

g =

(

1− 2M

r

)

dt2 −(

1− 2M

r

)−1

dr2 − r2dΩ2

z =(

1− 2Mr

)1/2; Φ = 1

2 ln(

1− 2Mr

)

≈ −Mr

Example 2: FRW spacetime, k = a(t)∂t,

g = dt2 − a2dx2, Lkg = 2λg

z = a(t), λ = a(t); u = ∂t; but ∇uu = 0:

2g(∇uu,x) = (dgu + Lug) (u,x) = 0

36

Force at a distance. Surface gravity

Consider stationary spacetime; Lkg = 0

Energy transfer and conservation: zE = zg(u, γ)

is conserved in local processes

Stationary observers feel acceleration in trans-

verse direction, a = −g−1d ln z

Transferring rope tension: 1) If rope moves by

∆r and transfers energy ∆E, then F∆r = ∆E.

2) If energy ∆E is sent back by a particle, then

z∆E = const. Hence zF = const

Force needed locally to held a unit mass sta-

tionary: F0 = g−1d ln z

Force at infinity (z = 1): F∞ = zF0 = g−1dz;

|F∞| =√

g−1 (dz,dz)

Example: Schwarzschild spacetime; k = ∂t;

z =√

1− 2Mr ; dz = M

r2dr

√

1−2Mr

; g−1dz = Mr2

√

1− 2Mr ∂r;

|F∞| = Mr2

. This is still finite at horizon, r =

2M , |F∞| = 14M ≡ κ - surface gravity

Note: at horizon k is null so the formula for

∇kk = −zg−1dz does not apply.

37

Change of LC connection under conformal

transformation

Consider g = e2Ωg. What is the relationship

between ∇ and ∇?

Compute: 2g(∇•a, •) = dˆga + Lag = de2Ωga +

Lae2Ωg = e2Ω(dga+Lag)+2(dΩ)∧ˆga+2(a Ω) g,

hence g(∇xa,y) = g(∇xa,y) + (x Ω) g(a,y)−(y Ω) g(a,x) + (a Ω) g(x,y) or, without y,

∇xa = ∇xa+(x Ω) a+(a Ω)x−g(a,x)g−1 (dΩ)

Practice exercises: 1. If k is a Killing vector

for g, show that k is a conformal Killing vector

for e2Ωg, for any Ω.

2. If γ(τ) is a null geodesic for metric g, show

that a reparameterization τ → τ is possible

such that γ(τ) is a null geodesic for g = e2Ωg.

3. Let k be a Killing vector for g and γ(τ) be

a geodesic. Show that γ is transported by the

flow of k to another geodesic. (If [v,k] = 0

then Lk∇vv = 0.)

4. In a stationary spacetime, n photons are

sent each second from point a to point b. How

many photons per second arrive at b? (Express

the answer through the redshift function z.)

38

Lecture 6, 2007-11-02. Energy extraction

from horizons

Consider a stationary spacetime, Lkg = 0, with

a Killing horizon g(k,k) = 0

Lower a mass m very slowly (adiabatically) on

a strong rope from an initial point p0 to point

p1 near horizon. How much energy can be

extracted at point p0?

Force needed to hold m at p1 is F1 = mg−1 d ln z|p1

Due to redshift, force at p0 is F0 = −z1z0F1 =

−mz0 g−1 dz|p1

To describe movement, introduce vector field

s = −g−1dz√

g−1(dz,dz). After a shift ∆L along s,

energy gained at p0 is ∆E = g(F0, s∆L) =

−(

mz0

s∆L)

z. Shifting slowly from p0 to p1

yields ∆E = − ∫ p1p0mz0

dz = z0−z1z0

m

Move to horizon, z1 ≈ 0 so ∆E ≈ m, extracted

the entire energy (mc2)!

39

Energy extraction: finite rope strength

If rope has finite max. tension Fmax, then move

only up to zmin 6= 0: zminFmax ≈ z0mκ, hence

∆Emax ≈(

1− mκFmax

)

m < m.

Optimum choice of mass: mopt = 12Fmaxκ , then

extract only 50% of mc2

Note: the formula for ∆E works only for m .

mopt because of the near-horizon approxima-

tion g−1(dz,dz) ≈ κ2

Note: ∆E is independent of z0, but energy is

redshifted, so more “absolute” energy is gained

if p0 is further from the horizon

Practice problem: For the metric g = f(r)dt2−dr2

f(r)− r2

(

dθ2 + sin2 θ dφ2)

, assume f(r0) = 0,

f ′(r0) 6= 0, so that r = r0 is a horizon. Com-

pute the surface gravity κ for that horizon.

40

Expansion (divergence) of a vector field

The rate of change of volume under flow of t:

Volume spanned by t, a,b, c is ε(t, a,b, c)

Choose a,b, c connecting for t:

tε(t, a,b, c) = Ltε(t, a,b, c) = (Ltε)(t, a,b, c)Due to antisymmetry, must have Ltε = fε

where f ≡ div t is the divergence or expan-

sion of the vector field t

A better formula for computing div t:

Choose orthonormal frame ea, g(ea, eb) =

ηab; then

div t =∑

aηaag(ea,∇eat)

=∑

aηaaB(t)(ea, ea) ≡ TrxB(t)(x,x)

Proof: Consider the dual basis θa and com-

pute Ltε ≡ Lt

(

θ0 ∧ θ1 ∧ ...)

=(

Ltθ0)

∧θ1∧ ...+... Now Ltθ

0 =∑

a θaιea

(

Ltθ0)

, but only the

term with θ0 survives. So Ltε = (ιe0Ltθ0 +

...)ε. Simplify one term:(

Ltθ0)

e0 = −θ0 [t, e0] = −η00g(e0, [t, e0]) = η00g(e0,∇e0t); we

used g(e0,∇te0) = 12∇tg(e0, e0) = 0.

41

Calculations without indices

From index to index-free notation:Introduce appropriate vector & tensor objectsContract each free index with arbitrary vectorSubtract terms coming from nested derivativesTraces: lower both indices, introduce gαβ, thenreplace gαβAαβγ...x

γ... by TraA(a, a,x, ...)Substitute a⊗ a→ g−1 into A(a⊗ a,x, ...)Tricks to simplify calculations:assume zero 1st derivatives of dummy vectorsuse properties of d, L, ∇, Tr where it helps

Example: uαuβuγwγ;αβ = Xλαλβv

αvβ

Introduce vectors u,v,w and a transformation-valued function X : X(x,y)z→ Xµ

αβγxαyβzγ

Rewrite Xλαλβ = gκλXκαλβ; introduce tensor

X(a,x,y, z) := g(a, X(x,y)z)Result: g(u,∇u∇uw−∇∇uuw) = Trag(a, X(v, a)v)

Properties of the trace operation:Linear: Tra∇xF = ∇xTraF ; same for Lx, ιx, dTrag(a, a) = 4; Trag(a,x)F(a,y, ...) = F(x,y, ...);Trag(a,∇ab) = divb; Tra∇a∇aX = X.

Example: Px = x− 2ng(n,x); Tr P =?Pαα = gαβPαβ; Pαβx

αyβ = g(x, Py); substitutexαyβ → gαβ and introduce dummy vector a:Trag(a, Pa) = Trag(a, a− 2ng(n,a))= 4− 2Trag(a,n)g(a,n) = 4− 2g(n,n).

42

Curvature tensor

Define R(a,b)c = ∇a∇bc−∇b∇ac−∇[a,b]c

Verify that R(a,b)c is independent of deriva-

tives of a,b, c

Thus: R is a transformation-valued 2-form.

“Covariant” tensor: R(a,b, c,d) = g(R(a,b)c,d)

Properties: exchange a ↔ b, c ↔ d, ab ↔ cd,

Bianchi identities (derive from Jacobi identity)

Example: Show that R(a,b,x,x) = 0, then set

x = c + d and derive c↔ d antisymmetry.

Start with ∇ab−∇ba− [a,b] g(x,x) = 0;

assume ∇a,bx = [a,b] = 0; get 0 = [∇a,∇b] g(x,x)

= 2g([∇a,∇b]x,x) = 2R(a,b,x,x).

Ricci tensor: Ric (a,b) = TrxR(a,x,b,x) =

Ric (b, a)

Ricci scalar: R = TrxRic(x,x)

Einstein equation: Ric−12Rg = −8πGT , where

T is energy-momentum tensor of matter,

T =2√−g

δ

δg−1Smatter

43

Lecture 7, 2007-11-07. FRW spacetime

Metric g = dt2 − a2(t)(

dx2 + dy2 + dz2)

= dt⊗ dt− a2(t)h; h is the flat 3-dim. metric

Need the Einstein tensor, Ric− 12Rg

Compute covariant derivatives:

g(∇u∂t,v) = 12

(

L∂tg)

(u,v) = −aah(u,v), so

∇∂t∂t = 0 and ∇∂x∂t = aa∂x = ∇∂t∂x.

g∂x = −a2dx; so g(∇u∂x,v) = 12 (dg∂x)(u,v) =

(−aadt ∧ dx) (u,v); so g(∇∂y∂x, •) = 0 and

g(∇∂x∂x, ∂t) = aa, so ∇∂x∂x = aa∂t. (∂y, ∂z...)

Need Ricv ≡ R(∂t,v)∂t − 1a2

∑

s=∂x,∂y,∂z

R(s,v)s.

For v = ∂t need R(s, ∂t)s = ∇s∇∂ts−∇∂t∇ss =

a2∂t − (aa)· ∂t = −aa∂t, so Ric∂t = 3aa∂tFor v = ∂x we have Ric∂x = R(∂t, ∂x)∂t −1a2R(∂y, ∂x)∂y− 1

a2R (∂z, ∂x) ∂z = 1

a2(aa+2a2)∂x.

Therefore Ricv =

(

aa + 2a2

a2

)

v+2

(

aa −

a2

a2

)

g(v, ∂t)∂t

Ric(u,v) =

(

aa + 2a2

a2

)

g(u,v)+2

(

aa −

a2

a2

)

g(u, ∂t)g(v, ∂t

Ricci scalar: R = Trvg(Ricv,v) = 6

(

aa + a2

a2

)

Einstein tensor:

Ric− 1

2Rg = −

(

2a

a+a2

a2

)

g+2

(

a

a− a

2

a2

)

dt⊗dt

44

Scalar field cosmology

Lagrangian for scalar field:

S[φ] =∫

d4x√−g

(

1

2g−1 (dφ,dφ)− V (φ)

)

Energy-momentum tensor:

T ≡ 2√−gδS

δg−1= dφ⊗dφ−

[

1

2g−1 (dφ,dφ)− V (φ)

]

g

Cosmological solution: φ = φ(t); dφ = φdt;

T = φ2dt⊗ dt−(

φ2

2− V

)

g

Einstein equations for FRW spacetime:(

2a

a+a2

a2

)

= −8πG

(

φ2

2− V

)

2

(

a

a− a

2

a2

)

= −8πGφ2

Hence: Friedmann equation

(

a

a

)2

=8πG

3

(

φ2

2+ V

)

≡ 8πG

3ρ(φ, φ)

EOM for φ: φ+ 3Hφ+ V ′(φ) = 0, H ≡ aa

Expanding universe: a > 0, then H =√

8πG3 ρ(φ, φ),

then assume φ ≡ v > 0, so v = v(φ), getdv(φ)dφ = f(φ, v) ← autonomous dynamics

45

Attractors and the slow-roll approximation

Consider the auxiliary variable u(φ) defined bydφdt ≡ −u(φ)

d√V

dφ ≡ −u(φ)F(φ). Then we have

φ =(

u′F + uF ′)

uF ; the EOM for φ becomes

2 + uu′F√V

+ u2 F′

√V

= u

√

√

√

√24πG

(

1 +1

2u2F

2

V

)

If V (φ) satisfies the slow-roll conditions,

Fu√V≪ 1,

F ′u2√V≪ 1,⇒ V ′√

24πG≪ V,

V ′′

24πG≪ V,

the EOM becomes u ≈ u0 ≡ 1√6πG

= const

— the slow-roll approximation is φ ≈ −(√V)′

√6πG

Note: need to verify u20F

2 ≪ V , u20F′ ≪√V .

Generically, there exists a single solution with

u0(φ) → const at φ → ∞. This u0(φ) is ap-

proximated by the slow-roll formula. This is

an attractor under the slow-roll assumptions:

u(φ) = u0(φ) + δu,

d

dφδu = C(u, φ)δu, C > 0

Further approximations: use perturbative ex-

pansion, u = u0(φ)+δu(φ); or substitute u0(φ)

into the l.h.s. of the equation, u(φ) =F(u,u′,φ)√

24πG

46

Lecture 8, 2007-11-09. Conformal trans-

formation of curvature

How do Ric and R change under g → g = e2Ωg?

LC connection: ∇xy = ∇xy + Γ(x,y) where

Γ(x,y) ≡ g(w,y)x + g(w,x)y − g(x,y)w,

where w ≡ g−1dΩ, g(w,x) ≡ x Ω

Ric(a, c) = Trbg(ˆR(a,b)c,b) = Trbg(

ˆR(a,b)c,b)

For calculation of ˆR(a,b)c, assume that ∇a,∇b

of a,b, c all vanish: ˆR(a,b)c = ∇a∇bc−∇b∇ac =

∇a (∇bc + Γ(b, c))+Γ(a,∇bc+Γ(b, c)−[...b↔ a...]

= R(a,b)c+∇aΓ(b, c)+Γ(a,Γ(b, c))−[...b↔ a...]

Need to express ∇xw; note that g(∇•w, •) =12 (dgw + Lwg) but dgw = ddΩ = 0; define a

symmetric bilinear form HΩ(x,y) ≡ g(∇xw,y) ≡ιy∇xdΩ. (In the index notation, this is Ω;αβ)

∇aΓ(b, c) = ∇a (g(w,b)c + g(w, c)b− g(b, c)w)

= HΩ(a,b)c+HΩ(a, c)b−g(b, c)∇aw; antisym-

metrize [b↔ a] and take g(...,b): HΩ(a, c)g(b,b)−HΩ(b, c)g(a,b)−g(b, c)HΩ(a,b)+g(a, c)HΩ(b,b).

Now take trace:

Trb(1st term) = (N − 2)HΩ(a, c) + g(a, c)Ω

47

Conformal transformation of Ricci tensor

2nd term: let’s write 〈xy〉 instead of g(x,y),

then Γ(a,Γ(b, c)) = 〈wa〉Γ(b, c) + 〈wΓ(b, c)〉 a−〈aΓ(b, c)〉w = 〈wa〉 (〈wc〉b + 〈wb〉 c− 〈bc〉w)+

(〈wb〉 〈wc〉+〈wc〉 〈wb〉−〈ww〉 〈bc〉)a−(〈ab〉 〈wc〉+〈ac〉 〈wb〉−〈aw〉 〈bc〉)w. Antisymmetrizing [b↔ a],

only terms 5,6,8 remain: Γ(a,Γ(b, c))−Γ(b,Γ(a, c))

= 〈wc〉 (〈wb〉 a−〈wa〉b)+〈ww〉 (−〈bc〉 a+〈ac〉b)+

(−〈ac〉 〈wb〉+〈bc〉 〈aw〉)w. Compute g(b, ...) =

〈wc〉 (〈wb〉 〈ab〉−〈wa〉 〈bb〉)+〈ww〉 (−〈bc〉 〈ab〉+〈ac〉 〈bb〉)+(−〈ac〉 〈wb〉+〈bc〉 〈aw〉) 〈wb〉. Trace:

Trb(2nd term) = (N − 2) (〈ac〉 〈ww〉 − 〈aw〉 〈cw〉)Finally, Ric = Ric + (N − 2) (HΩ + 〈ww〉 g −dΩ⊗ dΩ) + (Ω) g

Ricci scalar: R = TrxRic(x,x) = e−2ΩTrxRic(x,x)⇒

R = e−2Ω [R+ 2(N − 1)Ω + (N − 1) (N − 2) 〈ww〉]Note that 〈ww〉 ≡ g(w,w) = g−1(dΩ,dΩ).

Einstein tensor: Ric−12Rg = Ric−1

2Rg+(N − 2) [HΩ−dΩ⊗ dΩ−

(

Ω + 12 (N − 3) 〈ww〉

)

g ]

Einstein tensor does not change in 2D!

A conformal transformation can make R = 0

in 2D - any 2D space is conformally flat.

48

Equations of motion for f(R) gravity

After introducing Lagrange multiplier λ:

S[g] + Sm =∫

d4x√

−g(

λR+ U(λ)

16πG+ Lm

)

,

U(λ) ≡ f(r)− rf ′(r)∣

∣

∣

r=r(λ), λ = 1 + f ′(r)

Now conformal transformation: g ≡ e2Ωg; note:√−g = e4Ω√−g and R = e−2Ω(

R+ 6Ω + 6 |dΩ|2)

Note: “” is with respect to g now!

Choose Ω such that λR√−g equals R

√−g plus

some terms:√−g

(

λR+U(λ)16πG + Lm

)

= e4Ω√−g(

116πG

[

λe−2Ω(

R+ 6Ω + 6 |dΩ|2)

+ U(λ)]

+ Lm

)

.

Set λ = e−2Ω and omit Ω (total divergence):

S =∫

d4x√−g

(

R+ 6 |dΩ|2 + e4ΩU(e−2Ω)

16πG+ e4ΩLm

)

Standard action for Einstein gravity (but with

weird matter coupling)← “Einstein frame”, i.e. the

variables g,Ω with 2nd order EOM

The variables g are “Jordan frame” (have 4th

order EOM)

49

Lecture 9, 2007-11-14. Cosmology with

f(R) gravity

Define canonical scalar field: φ ≡ Ω√

34πG ≡

Ωs

Assume f(R) = 12αR

2; neglect Lm; then λ =

1+αR, U(λ) = −(λ−1)2

2α , so (in Einstein frame)

S =∫

d4x√−g

(

R

16πG+

1

2|dφ|2 − V (φ)

)

,

V (φ) =1

32πGα(exp [2sφ]− 1)2 =

(

e2sφ − 1)2

24s2α.

Trick: Write EOM for φ through x(φ) ≡ EkinEpot

,

so dφdt ≡ −

√

2x(φ)V (φ) (assumed φ < 0, α > 0!)

dx

dφ+V ′

V(1 + x)− 6s

√

x (1 + x) = 0

Set V (φ) ≈ V0e2nsφ (we will have n = 2), then

x′ = 2s√x+ 1

(

3√x− n

√x+ 1

)

. Assume x →x0 at φ → ∞: x0 = n2

9−n2; φ ≈ −√

2x0V (φ); so

φ(t) ≈ − 1ns ln

(t−t0)n2√

12α(

9−n2)

Stable attractor : x = x0 + δx, then δx′ ≈ Cδx,C = s

n

(

9− n2)

> 0 as long as n < 3.

50

Do we have inflation?

Note: slow-roll conditions are not valid,

V ′

V

1√48πG

=2

36≪ 1!

Compute

H(t) ≈ s√

2 (1 + x0)V (φ) = 3n−2 (t− t0)−1

a(t) = exp

∫

H(t)dt = (t− t0)3n−2

Conformal transformation:

g = e2Ωg =const

(t− t0)2/n(

dt2 − a2(t)dx2)

Define physical time: want g = dt2− a2(t)dx2;

so define t = (t− t0)1−1/n, then

a(t) = (t− t0)3n−2−n−1

= t3−n

n(n−1)

With n = 2 we have radiation-dominated era,

a ∝ t1/2

Note: inflation will be with φ < 0 and φ > 0!

(Slow roll approximation is then valid)

Practice problem: Determine (approximately)

the potential V (φ) for f(R) gravity with f(R) =

R− α2R

2+βR−n, n > 0, in the regimes of large

R and small R

51

Conditions for viable f(R) theory

If f ′(R) < 0 then wrong sign of the action in

Einstein frame:∫ −R

16πG, so graviton is a ghost

In Einstein frame, have scalar field with

V (φ) =1

16πG

rf ′ − ff ′2

∣

∣

∣

∣

∣

f ′+1=exp(−2sφ)

If f ′′(R) < 0 at Solar system scales then

M2φ

M2Pl

≡ 16πGd2V (φ)

dφ2

=1

3f ′′+Rf ′ − 4 (R+ f)

(1 + f ′)2≈ 1

3f ′′< 0,

which gives tachyonic instability (small-scale)

At high R we must have f(R)≪ R at observed

scales (MφL≫ 1); f ′′ → 0

Hence f(R) must be negative, monotonically

growing in R

Practice exercise: Derive the formula for M2φ .

52

Lecture 10, 2007-11-16. Matter-dominated

era

The full action in Einstein frame is

S =

∫

d4x√−g

[

R

16πG+|dφ|2

2− V (φ) + Lme

4sφ

]

where s ≡√

4πG3 , V = 1

24αs2

(

e2sφ − 1)2

Mock-up Lagrangian for matter:

Lm(ψ, ψ) = pm, ρm = ψLm,ψ − Lm,

and we assume pm = wmρm (nondynamical

matter with wm = const) in Jordan frame

Also assume homogeneous φ = φ(t), ψ = ψ(t)

Evolution equations for φ, ψ:

d

dt

[

φ√−g

]

= −√−gV,φ + 4s√−ge4sφpm

d

dt

[

e4sφLm,ψ

√−g]

= −√−ge4sφLm,ψ

With√−g = a3(t) we get

ψLm,ψψ + ψLm,ψψ + 3HLm,ψ + 4sφLm,ψ − Lm,ψ = 0;

d

dtρm =

(

ψLm,ψψ + ψLm,ψψ − Lm,ψ

)

ψ

53

Finally, using pm = wmρm, we get

d

dtρm = −

(

3H + 4sφ)

ρm (1 + wm) ;

φ = −3Hφ− V,φ + 4se4sφwmρm

Friedmann equation (Einstein frame):

a2

a2≡ H2(t) =

8πG

3

(

ρφ + e4sφρm)

= 2s2[

φ2

2+ V (φ) + e4sφρm

]

Obtained a closed system for φ, ρm, H

How to analyze attractor behavior?

1. Use energy conservation equation

d

dtρφ = −3H

(

ρφ + pφ)

+ 4se4sφφwmpm

2. Use the energy quotient

q(t) ≡ e4sφρm

e4sφρm + ρφ, 0 ≤ q ≤ 1

3. Assume that q(t) → const and wφ → const

on the attractor

4. Derive closed equations for q(t), φ(t)

dq

dt= −3Hq

[

(

wm − wφ)

(1− q) +4s

3wm

φ

H

]

(Complete analysis of dynamics is in Amendola

et al., gr-qc/0612180 — using Jordan frame)

54

Attractor for dust-dominated era (wm = 0)

Integrate the equations for ρm, ρφ:

ρm = ρ(0)m a−3(1+wm)e−4s(1+wmφ)

ρφ = ρ(0)φ a−3(1+wφ)

where we assumed pφ = wφρφ with wφ ≈ const

Now assume wm = 0, derive equation for q(t):

q = 3Hq (1− q)wφFixed point if wφ → 0 or q → 1 (not if q → 0)

If q → 1 then need wφ > 0 so that q > 0.

Since wφ ≥ 0 then at late times ρm(t)≫ ρφ(t)

or at most ρm(t) ≈ C0ρφ(t), hence

H = s√

2

√

ρφ + ρme4sφ ≈ C1a

−3/2

and then a(t) ∝ t2/3, H(t) ≈ 23t−1 in any case.

So ρφ = ρ(0)φ t−2(1+wφ) and φ = C2t

−1−wφ.If wφ > 0 then φ → const at late times, but

this contradicts EOM for φ with V,φ 6= 0 (or

else we must require a fine-tuned value of φ)

55

Dust tracker solution (wφ = 0)

Consider wφ → 0, q → q0 6= 1 (the value q0 is

determined by the EOM for φ) then

H2 =

(

2

3t−1

)2

= 2s2(

ρφ + e4sφρm)

,

φ(t) =

√

2 (1− q0)3s

ln t, esφ = t

√2(1−q0)

3 ≡ tν, ν > 0

Transforming back to Jordan frame:

g = t2ν(

dt2 − a2dx2)

≡ dτ2 − a2(τ)dx2

τ = t1+ν, a(τ) = τ

(

23+ν

)

/(1+ν) 6= τ2/3

To have a dust-dominated era with a(τ) ∝ τ2/3we need ν ≪ 1

(This is a constraint on V (φ) after q0 is deter-

mined)

(See Amendola et al. for precise conditions)

56

Lecture 11, 2007-11-21. Introduction to

perturbation theory for gravity

What is the metric resulting from a small per-

turbation in the matter source?

• Compute Ric(x,y) for g = g+ h to 1st order

in h (treating h as a small perturbation)

Change in the LC connection is a tensor,

Γ(x,y) ≡ ∇xy −∇xy = Γ(y,x)

g(Γ(x,y), z) = g(∇xy, z)−g(∇xy, z)−h(∇xy, z) =12

(

dhy + Lyh)

(x, z)−h(∇xy, z) = 12 (∇xh) (y, z)−

12 (∇zh) (x,y) + 1

2

(∇yh)

(x, z).

Since we are computing to 1st order, we may

set g(Γ, ·) ≈ g(Γ, ·) and disregard ΓΓ.

Assuming ∇x,y x,y, z = 0, write the Riemann

tensor ˆR for g through R for g:

ˆR(x,y)z = ∇x∇yz− ∇y∇xz ≈ R(x,y)z

+∇xΓ(y, z)−∇yΓ(x, z)

57

Compute the Ricci tensor: to 1st order,

Ric(x, z) = TryR(x,y, z,y) ≈ Tryg( ˆR(x,y)z,y)

Covariant Riemann tensor: to first order,

R(x,y, z,y) ≈ R(x,y, z,y)+g(∇xΓ(y, z)−∇yΓ(x, z),y)

= R(x,y, z,y)+∇xg(Γ(y, z),y)−∇yg(Γ(x, z),y)

Change in the Ricci tensor: Ric(x, z) ≈ Ric(x, z)+

∇xTryg(Γ(y, z),y)−Try∇yg(Γ(x, z),y)= Ric(x, z)+12∇xTry(∇yh(y, z) +∇zh(y,y)−∇yh(y, z))

−12Try∇y(∇xh(y, z) + ∇zh(x,y) − ∇yh(x, z))=

Ric(x, z)+12∇x∇zTrh+1

2h(x, z)−12∇x (divh) (z)−

12∇z (divh) (x), where (divh) (x) ≡ Try(∇yh)(y,x).

Practice problem: Show that div (λg) = dλ

By taking trace of the Einstein equation, find

R = 8πGTrT , hence Ric = −8πG(

T − 12gTrT

)

Assume that a background spacetime (Ric, T)

is given and we have a small perturbation δT

of the EMT. We need to solve the equation

1

2∇x∇zTrh+

1

2h(x, z)− 1

2∇x (divh) (z)

−1

2∇z (divh) (x) = −8πG(δT − 1

2gTrδT)

to first order in δT . Complicated!

58

The 3+1/SVT decomposition

Questions: 1. What is the metric gµν gener-ated by a small δTµν in flat spacetime?2. What is the metric g generated by a smallδTµν superimposed onto the cosmological Tµνin FRW spacetime (g = dt2 − a2dx2)?

To solve the difficult tensor equation, use tricks:1. Compute time components separately fromspatial components; h =

h00, h0j, hij

2. Use scalar-vector-tensor (SVT) decompo-sition for h and T

Helmholz decomposition: ω = dλ+ν, divν = 0(we define divν ≡ div (g −1ν) for 1-forms)Determine λ: divω = div dλ = ∆λ, then λ =1∆divω if the BVP (λ → 0 at |x| → ∞) has aunique solution (true in Euclidean signature!)

Motivation for the 3+1 split:- background will depend on t via a(t), can useFourier transform only in 3D but not in time- cannot use SVT decomposition with Lorentzianmetric signature (no unique solution to BVP)!

Motivation for the SVT decomposition:- equations for S,V,T sectors are decoupled- different physical sources contribute to S,V,Tin δT (usually have only scalar sources)

59

Plan:

1. Perform 3+1/SVT decomposition

2. Compute Ric in flat background

3. Resolve gauge issues using 3+1/SVT

4. Solve Einstein equation, determine h

5. dt2 − a2dx2 = a2(η)(

dη2 − dx2)

; make con-

formal transformation to FRW background with

fixed a(t) to compute Ric for perturbed FRW

3+1/SVT decomposition of vectors

Decompose 4-vectors as u = Ae0 + v; e0 ≡ ∂t;g(e0,v) = 0; u ≡

A, vj

. Further, decompose

v as gradient + divergence-free, vj = B,j+Cj,

Cj,j = 0. In 3-dim. Fourier space: v(k) =

iBk+C, k ·C = 0. So B := 1iv·kk·k, C := v− iBk.

In real space: B = 1∆divv; C = v − g−1dB.

Note:(

1∆f

)

(x) = − ∫ f(x,y,z)d3x4π|x−x| , where x ∈ R3

Remarks on the SVT decomposition:

1. The decomposition is nonlocal in space

2. Only defined when divv → 0 sufficiently

quickly at spatial infinity (at least ∝ |x|−2)

3. Defined also without Fourier transform!

4. Useful only because perturbation equations

are algebraic in Fourier space

60

3+1/SVT decomposition of (0,2) tensors

Consider a symmetric tensor Tij in 3-dimensional

Euclidean space with metric δij

Algebraic decomposition: trace + traceless part:

Tij = Aδij + T(1)ij , T

(1)aa = 0, A := 1

3Taa

Fourier decomposition: project along tensors

made out of k

Component parallel to kikj − 13k

2δij:

T(1)ij =

(

kikj − 13k

2δij)

B + T(2)ij , T

(2)ij kikj = 0,

T(2)aa = 0, B := 3

2

kikjT(1)ij

k4

Component parallel to ki & orthogonal to kj:

T(2)ij = i

(

kiCj + kjCi)

+T(3)ij , T

(3)ij kj = 0, T

(3)aa =

0, Caka = 0, Cj := 1i

kaT(2)aj

k2; relabel Dij ≡ T (3)

ij

Final decomposition in real space: Tij = Aδij−(

∂i∂j − 13δij∆

)

B + Ci,j + Cj,i +Dij

Remarks:

1. Tij must decay at least as |x|−2 at infinity

2. Terms proportional to δij can be gathered

because we cannot have B,ij = Aδij (check!)

61

Properties of SVT decomposition

SVT decomposition can be applied to arbitrary

(Euclidean) tensors. It is useful because:

1. Tensor equations split into simpler eqs.

2. Eqs. for S/V/T components decouple

3. Often get algebraic eqs. in Fourier space

Theorem 1: If Aδij+B,ij+Ci,j+Cj,i+Dij = 0

and A,B,Ci, Dij → 0 at least as |x|−2 at infinity,

then each of A,B,Ci, Dij = 0 everywhere.

(Not true in spaces with Lorentzian signature!)

Theorem 2: EOM for perturbations of a ten-

sor X in the presence of (several) background

tensors will decouple after using a complete

decomposition of X with respect to projections

on the background tensors and SVT. (In our

case, the background tensors are k and δij)

Lemma: If a linear function of k is invariant

under 3D rotations, it is equal to zero.

62

Lecture 12, 2007-11-23. Why 4D/SVT

does not work; gauge issues

Try avoiding 3+1 decomposition (only SVT):

hµν = Agµν +B;µν + Cµ;ν + Cν;µ +Dµν

where divC = 0, divD = 0, TrD = 0. (Index-

free notation is more bulky now!)

Gauge transformation: h→ h+δh, where δh ≡Lug; so δh(x,y) = g(∇xu,y) + g(∇yu,x); in

index notation: δhµν = uµ;ν+uν;µ; decompose

uµ = b,µ+cµ with div c ≡ cµ;µ = 0; then δhµν =

2b;µν + cµ;ν + cν;µ. This removes B, C terms!

Compute: Trh = 4A; divh = dA; so δRicµν =

A;µν+ 12gµνA+ 1

2Dµν; Einstein equation will

set A = ..., Dµν = ... Note: Dµν has 5 d.o.f.

It appears that we have 5 d.o.f. in Dµν, but

this is wrong: the gauge has not been fixed

(have 3 free solutions of u = 0, divu = 0).

But these “gauge” solutions are not explicit.

• 3+1/SVT will make the d.o.f. explicit. Also

we will avoid having to solve u = 0 in FRW

background!

63

Perturbations in flat spacetime

Use ordinary derivative instead of ∇Use 3+1/SVT decomposition for h (notation

from V. F. Mukhanov, Physical Foundations

of Cosmology, chapter 7):

h00 = 2Φ; h0j = B,j + Sj; Sj,j = 0

hij = 2Ψδij + 2E,ij + Fi,j + Fj,i + tij

Gauge transformations: hµν → hµν+δhµν where

δhµν = uµ,ν + uν,µ. Apply 3+1/SVT to uµ:uµ →

a; b,j + cj

, cj,j = 0. Now compute:

δh00 = 2a,0; δh0j = a0,j + b,0j + cj,0;

δhij = 2b,ij + ci,j + cj,i

By the SVT theorem 1, we have

δΦ = a, δB = a+ b, δSj = cj,

δΨ = 0, δE = b, δFj = cj, δtij = 0

Hence we can set E = B = 0, Fj = 0 (this is

called longitudinal or Newtonian gauge)

Remarks: 1. Scalar parts of u contribute to

scalar parts of h, vector parts of u to vector

parts of h (cf. SVT theorem 2)

2. The source Tµν will, in general, change

under a gauge transformation as Tµν → Tµν +

LuTµν, but in flat space Tµν = 0. Will have to

take care with FRW background!

64

Einstein equations in almost flat space

Compute Ricµν to first order in Newtonian gauge:

h00 = 2Φ; h0j = Sj; Sj,j = 0

hij = 2Ψδij + tij, taa = 0, tij,j = 0

Need to compute

Ricµν =(Trh),µν + hµν − ∂µ (divh)ν − ∂ν (divh)µ

2Trh = 2Φ− 6Ψ

(divh)0 = gµνh0µ,ν = h00,0 − h0j,j = 2Φ

(divh)j = hj0,0 − hji,i = Sj − 2Ψ,j

Hence (in the SVT decomposition):

Ric00 = −3Ψ−∆Φ, Ric0j = −2Ψ,j −1

2∆Sj

Ricij = (Φ−Ψ),ij +1

2tij + δijΨ− 1

2

(

Sj,i + Si,j)

R = Ric00 −Ricjj = −6Ψ− 2∆Φ + 4∆Ψ

Need to decompose the EMT in 3+1/SVT:

T0j = α,j + βj; βj,j = 0

Tij = µδij + λ,ij + σi,j + σj,i + qij

σj,j = 0, qjj = 0, qij

,j = 0

65

Explicit solutions

Explicit expressions for 3+1/SVT components:

α =1

∆T0i,i λ =

1

∆

[

3

2

1

∆Tik,ik −

1

2Tii

]

µ =1

2

[

Tii −1

∆Tik,ik

]

σj =1

∆

[

Tji,i −(

1

∆Tik,ik

)

,j

]

Einstein equations in 3+1 decomposition are

−2∆Ψ = −8πGT00

−2Ψ,j −1

2∆Sj = −8πG

(

α,j + βj)

(Φ−Ψ),ij +1

2tij −

1

2

(

Sj,i + Si,j)

−[

2Ψ + ∆(Φ−Ψ)]

δij = −8πG(µδij + λ,ij+ σi,j + σj,i + qij)

Explicit solution using SVT theorem 1:

Ψ =1

∆4πGT00, Φ−Ψ = −8πGλ,

Sj =1

∆16πGβj, tij =

1

16πGqij

Remarks: 1. Other equations are consequencesof Tij

,j = 0, do not need to solve them!2. Φ,Ψ, Sj are constrained and not free d.o.f.3. A free wave solution can be added to tij(this gives 2 vacuum d.o.f.)4. For Tµν = (ρ+ p)uµuν−pgµν we have α = 0,λ = 0, µ = p, σj = βj = 0, qij = 0, T00 = ρ+ p

66

Lecture 13, 2007-11-28. Perturbations of

gravity and matter in FRW spacetime

(see Mukhanov 2005, chapters 7 and 8)

Perturbations due to a single, ideal comoving

fluid (if one matter component dominates and

if vector perturbations are absent, e.g. scalar

field/dust/radiation but no cosmic strings):

Tµν = (ρ+ p)uµuν − pgµνIn flat spacetime it follows that Φ = Ψ and

Sj = 0, so the perturbed metric is simply

gµν =

(

1 + 2Φ 00 −δij + 2Φδij + tij

)

, uµ =

1000

To pass to FRW spacetime, note that any

FRW spacetime is conformally flat:

g = dt2 − a2(t)dx2 = a2(t)[

dt2 − dx2]

Note: t is the conformal time, t is the proper

time, t =∫ dta(t)

Example: de Sitter, a(t) = eHt, t = −H−1e−Ht,a(t) = |Ht|−1

Note that −∞ < t < 0 for de Sitter; t → 0 is

“late times”

In matter-dominated FRW, a(t) ∝ tp where

usually 0 < p < 1 and so 0 < t <∞67

Ricci tensor for perturbed metric in FRW

Ansatz for metric perturbations: g = a2g, so

gµν = a2(t)

(

1 + 2Φ 00 −δij (1− 2Φ) + tij

)

Now need to compute the Einstein tensor of g.

Use linear approximation in Φ, tij. Note: a(t)

is fixed and has no perturbations!

Use the conformal transformation formula with

g = e2Ωg, Ω ≡ ln a(t):

Ein = Ein + 2HΩ − 2dΩ⊗ dΩ− (2Ω + |w|2)gwhere w = g−1dΩ. Need the Hessian: HΩ(x,y) =

(∇.∇.Ω)(x,y) = (∇xdΩ) (y) = g(∇xw,y).

Compute: dΩ = (∂t ln a)dt, denote b(t) ≡∂t ln a = ∂ta; then dΩ = bdt; w ≈ (1− 2Φ) b∂t;

|w|2 ≈ (1− 2Φ) b2 to first order in Φ.

Covariant derivative ∇ is computed w.r.t. g

and contains perturbations:

HΩ(x,y) = g(∇xw,y) = b (1− 2Φ) g(∇x∂t,y)+

g(∂t,y)∇xb (1− 2Φ)

g(∇.∂t, ·) = 12

(

dg∂t + L∂tg)

= dΦ ∧ dt+ 12g

So HΩ(x,y) = −[(x Φ) (y t)+(y Φ) (x t)]b+b2g(x,y)+(yt)(xt)b; and Ω = TrxHΩ(x,x) =

−4bΦ + (1− 2Φ) b because Trxh(x,x) = −4Φ

68

Use 3+1/SVT decomposition to computeHΩ and the Einstein tensor:

dΩ = bdt; HΩ00 = b − bΦ; HΩ0j = −bΦ,j;HΩ ij = bΦδij + 1

2btij; w0 = (1− 2Φ) b; |w|2 =

(1− 2Φ) b2

use Ein = Ein+2HΩ−2dΩ⊗dΩ−(2Ω+|w|2)gEin00 ≡ Ein (∂t, ∂t) = −2∆Φ− 3b2 + 6bΦ

Ein0j ≡ Ein(

∂t, ∂xj)

= −2(

Φ + bΦ)

,j

Einij = 12tij+ btij−

(

b2 + 2b)

tij+[(

b2 + 2b)

−2Φ− 4

(

b2 + 2b)

Φ− 6bΦ]δij

Einstein equations:

Ein(0) + Ein(1) = −8πG(

T (0) + T (1))

Background Friedmann equ.

b2 =8πG

3T

(0)00 ; b2 + 2b = −8πGµ(0)

Perturbations in 00, 0j, ij:

∆Φ− 3bΦ = 4πGT(1)00

Φ + bΦ = 4πGα(1)

tij + 2btij − 2(

b2 + 2b)

tij = 4πGq(1)ij

In usual scenarios q(1)ij = 0, so only scalar per-

turbations are sourced

69

Lecture 14, 2007-11-30. Perturbations of

gravity and scalar field in FRW spacetime

Cosmology with scalar field:

S[g, f ] =

∫

d4x√−g

[

R

16πG+

1

2|dφ|2 − V (φ)

]

Perturbations in scalar field: φ = f0(t)+f(t,x)

Metric perturbations (only scalar):

g = a2(t)

(

1 + 2Φ 00 − (1− 2Φ) δij

)

Energy-momentum tensor of φ:

Tµν = φ,µφ,ν − gµνp, p ≡ 12 |dφ|

2 − V (φ)

p =1− 2Φ

a2

(

12f

20 + f0f

)

− V (f0)− fV ′(f0)

T00 = f20 + 2f0f − a2 (1 + 2Φ) p

=(

12f

20 + a2V0

)

+(

f0f + a2V ′0f + 2a2V0Φ)

T0j = f0f,j

Tij = (1− 2Φ) a2pδij =(

12f

20 − a2V0

)

δij + ...

Einstein equations in 3+1/SVT:

(background) 3b2 = 8πG(

12f

20 + a2V0

)

∆Φ− 3bΦ = 4πG(

f0f + a2V ′0f + 2a2V0Φ)

Φ + bΦ = 4πGf0f

70

Equations of motion for the field φ = f0(t)+f :

∂µ(√−ggµνφ,ν

)

+√−gV ′(φ) = 0

Compute:√−g = a4 (1− 2Φ);√−gg0νφ,ν = a4 (1− 2Φ) a−2 (1− 2Φ)

(

f0 + f)

=

a2(

f0 + f − 4Φf0)

;√−ggjνφ,ν = −a2 (1− 2Φ) (1− 2Φ) f,j = −a2f,j;V ′(φ) = V ′0+fV ′′0 ; so EOM for φ is a−2[a2(f0+

f −4Φf0)]·−∆f + a2

[

(1− 2Φ)V ′0 + V ′′0 f]

= 0.

Background EOM: a−2[

a2f0]·

+ a2V ′0 = 0

Perturbation EOM:

a−2[

a2f − 4Φa2f0]·−∆f−2Φa2V ′0+a2V ′′0 f = 0

Collect terms with Φ using background EOM:

a−2(

a2f)·− 4Φf0−∆f +2Φa2V ′0 + a2V ′′0 f = 0

Change variables Φ = a−1h, f = a−1x in the

00 and 0j Einstein equations; use background

EOM (3b2Φ = ...; a2V ′0f = ...) and simplify:

∆h− 3bh = 4πG[

−f20h+ f0x− 3bf0x− f0x

]

h = 4πGf0x

Make Fourier transform in (flat) 3-space:[

f20 −

k2

4πG

]

h = f0x− f0x;h

4πG= f0x

EOM for perturbations (f = ...) is a conse-

quence of this 1st order system

The Mukhanov-Sasaki variable

Would like to have a closed 2nd order equation

For quantization, expect x to be fundamental

(occurs as 12x

2 + ... in canonical field action)

Look for a combination v = x+B(t)h such that

v satisfies an equation of the form v+k2v = Cv

Substitute x = v −Bh into 2nd Einstein equ.:

h

4πG= f0 (v −Bh)

Expect relationships of the form

v = A1h+A2h, k2h = A3v+A4v+ 0 · hbecause then we will have

k2v = A1 (A3v+A4v) +A2 (A3v+A4v)·

Compute k2h from 1st Einstein equ.:[

f20 −

k2

4πG

]

h = f0(

v − Bh−Bh)

− f0 (v −Bh)

then substitute h = 4πGf0 (v −Bh) and find

− k2

4πGh = f0v −

(

4πGBf20 + f0

)

v

+ h(

−f0B + 4πGf20B

2 + f0B − f20

)

71

Now we need to cancel the last term:

f0B = f20

(

4πGB2 − 1)

+ f0B

This is a Riccati equation; convert to linear:

B = −f0Q

Q; Q = 4πGf2

0Q;

A suitable solution is Q = a−1 due to the back-ground equation 4πGf2

0 = b2− b (found by sub-tracting b2 + 2b = −8πGp from 3b2 = ...); so

B =f0b, v = x+

f0bh; v =

f0bh+

1

4πGf0h

− k2

4πGh = f0v −

(

4πG

bf30 + f0

)

v

Equation for v: define z ≡ af0b−1, then

v =z−1

4πG

(

zh

f0

)·, − k

2h

4πG= f0z

(

v

z

)·

−k2v =z−1

4πG

(

−k2h zf0

)·= z−1

(

f0z2

f0

(

v

z

)·)·

= v − zzv

Expressing x through v:

x = v − f0bh; h = 4πGf0z

1

∆

(

v

z

)·

Action for v that reproduces the EOM:

S = 12

∫

d4x

(

v2 − v,jv,j +z

zv2)

Qualitative considerations for solutions

Solution v(t) for k2 ≫ z/z is an oscillator ∝ eiktwith constant amplitudeThe function z(t) grows quickly as ≈ a(t), soeventually every mode k will cross the horizonSolution for k2 ≪ z/z has two branches, onegrowing as z, the other decayingThe growing mode means v ∝ z, so field per-turbation is x ∝ z/a = H−1∂tf0 ≈ const butactually slightly growing towards the end of

inflation. (Figure from Mukhanov 2005)

∝ φ/H

f

ttc tf

f0

∝ 1a

Note: describing the evolution of perturba-tions in non-φ dominated epochs requires adifferent calculation! (We used the fact thatthe background is φ-dominated.)

72

Lecture 15, 2007-12-07. Quantization of

free fields in FRW spacetime

(See also my lecture notes of 2006)

Consider a scalar field in a fixed FRW back-

ground spacetime:

S =

∫ √−gd4x[

1

2gαβ (∂αφ)

(

∂βφ)

− 1

2m2φ2

]

Use conformal time: g = a2(t)[

dt2 − dx2]

S =1

2

∫

d3x dt a4

a−2[

φ2 − (∇jφ)2]

−m2φ2

Change variables: χ ≡ a(t)φ, then

a2φ2 = χ2−2χχa

a+χ2

(

a

a

)2

= χ2+χ2a

a−[

χ2a

a

].

So the action is

S =1

2

∫

d3x dt(

χ2 − (∇jχ)2 −m2eff(t)χ2

)

m2eff(t) ≡ m2a2(t)− a(t)

a(t)

Scalar field φ(t,x) in FRW spacetime is equiva-

lent to scalar field χ(t,x) with a time-dependent

mass meff(t) in flat Minkowski spacetime

Note: the action for Mukhanov-Sasaki variable

v(t,x) is of this form with m = 0 and using z(t)

instead of a(t)

73

Quantization of free field with m(t)

Mode expansion: Fourier transform in space

(not in time)

χ (x, t) =

∫

d3k

(2π)3/2χk(t)e

ik·x, (χk)∗ = χ−k

Then a term such as∫

d3x(

∇jχ)2

becomes

∫

d3x(

∇jχ)2

=

∫

d3xd3k d3k′

(2π)3

(

ei(k+k′)·xi2k · k′χkχk′)

=

∫

d3k k2χkχ−k =

∫

d3k |χk|2 k2

So the action for χk(t) becomes

S[χk(t)] =

∫

d3k dt

(

1

2|χk|2 −

1

2|χk|2

[

k2 +m2eff(t)

]

)

Every Reχk, Imχk is a separate harmonic os-

cillator with time-dependent frequency

ω(t) ≡√

k2 +m2eff(t)

It is more convenient to consider complex-valued

variables χk

Equation of motion:

χk + ω2k(t)χk = 0

74

Quantization of time-dependent oscillators

Action for a unit-mass oscillator:

S =

∫

dt

[

1

2q2 − 1

2ω2(t)q2

]

Equation of motion: q+ ω2(t)q = 0

General solution (complex-valued) is in 2-dim.

space of solutions spanned by v(t), v∗(t):q(t) = a+v(t) + a−v∗(t); v+ ω2(t)v = 0

Quantization replaces q(t)→ q(t) and a± → a±

Standard quantization requires commutation

relations [q, p] = i where p = q, hence

[a+v+ a−v∗, a+v+ a−v∗]= [a−, a+](v∗v − vv∗) = i

A mode function is a solution v(t) such that

1 =v∗v − vv∗

i≡ 2Im

(

v∗v)

Note: v(t) must be complex-valued; have one-

parametric family of admissible mode functions

(can set v(t0)/v(t0) = λ arbitrarily if Imλ > 0)

Note: v(t) 6= 0 for all t because Im (v∗v) is a

Wronskian and is constant in time

For a given v(t) we have a− = 1i [v(t)q(t)− v(t)p(t)],

so the definition of a± depends on v(t)

75

Fock space: a− |0〉 = 0; |n〉 = 1√n!

(

a+)n |0〉,

depends on the choice of v(t)

Hamiltonian H = 12p

2 + 12ω

2q2 is expressed as

H(t) =|v|2+ω2|v|2

2

(

2a+a−+ 1)

+v2+ω2v2

2 a+a++

v∗2+ω2v∗22 a−a− [exercise]

Note: in time-independent cases, ω(t) = ω0,

one chooses v(t) = 1√2eiω0t, so H is diagonal in

the Fock basis: H |n〉 = En |n〉.

In general, |n〉 are not eigenstates of H(t), ex-

cept possibly at one t = t0 if we choose v(t)

such that v(t0) = iω(t0)v(t0). The state |0〉 is

then the instantaneous vacuum at t0.

Statement 1: It is impossible to chose v(t)

such that H(t) is diagonal at all times, i.e. v2+

ω2v2 = 0 for all t.

Proof: v = ei∫

ωdt does not satisfy EOM.

Statement 2: The instantaneous vacuum |t0〉minimizes 〈0| H(t) |0〉 over all possible |0〉.Proof: Consider |λ0〉 defined through v(t) such

that v/v = λ at t = t0. Compute 〈λ0| H(t0) |λ0〉 =|v0|2+ω2

0|v0|2

2 =|λ2|+ω2

02(λ−λ∗). If λ = A+ iB, then the

minimum is at A = 0, B = ω0. So v0 = iω0v0.

76

Lecture 16, 2007-12-12. Quantized free

scalar field in FRW

Decompose the Fourier modes as

χk(η) = a−kv∗k(t) + a+−k

vk(t), a+k

=(

a−k

)∗

Apply this to field mode expansion,

χ(x, t) =

∫

d3k

(2π)3/2

[

a−kv∗k(t) + a+−k

vk(t)]

eik·x

=∫

d3k

(2π)3/2

[

a−kv∗k(t)e

ik·x + a+kvk(t)e

−ik·x]

The canonical commutation relation is

[χ(x1, t), ˙χ(x2, t)] = iδ(x1 − x2)

Choose vk(t) for every k (same for equal |k|)

a−k

=vkχk − vk ˙χk

i; [a−

k, a+

k′ ] = δ(

k− k′)

Once vk(t) are chosen, define vacuum: a−k|0〉 =

0, ∀k. Excited states are interpreted as parti-

cles because 〈kn| H |kn〉 = 〈0| H |0〉+ nωk

Note: no instantaneous minimum energy pos-

sible if ω2k < 0. If so, can have e.g. 〈kn| H |kn〉 <

〈0| H |0〉. No particle interpretation possible!

77

Bogoliubov transformations

Consider an oscillator with “in-out” transition:

PSfrag replacements

ω(t)ω2

ω1

t1 t2t

Natural definitions of minimum-energy vacuum:

v1(t)|t<t1 =1√2eiω1t, v2(t)|t>t2 =

1√2eiω2t

Since

v1, v∗1

is a basis, we must have

v2(t) = αv1(t) + βv∗1(t)

(Bogoliubov transformation/coefficients)

Condition Im(

v2v∗2

)

= 12 gives |α|2 − |β|2 = 1

Express a±2 through a±1 , get a−2 = αa−1 − βa+1

The state∣

∣

∣t10⟩

is not vacuum at t = t2 but

has particles. Compute mean particle number:⟨

t10∣

∣

∣ a+2 a−2

∣

∣

∣t10⟩

= |β|2

Numerical computation of β requires knowing

v1(t), v1(t), v2(t), v2(t) at any one moment t0:

β =1

i[v1v2 − v2v1]t=t0

78

Number density for quantum particles

Bogolyubov transformation: uk(t) = αkvk(t)+

βkv∗k(t), and |αk|2 − |βk|2 = 1

Fourier expansions must agree:

eik·x(

v∗k(t)a+k

+ vk(t)a−k

)

= eik·x(

u∗k(t)b+k

+ uk(t)b−k

)

Hence b−k

= αka−k− βka+−k

, b+k

= α∗ka+k− β∗ka

−−k

Mean k-particle number in the ak-vacuum state:

〈a0| b+k b−k|a0〉 = 〈a0| (α∗ka

+k−β∗ka

−−k

)(αka−k−βka+−k

) |a0〉= |βk|2 〈a0| a−−k

a+−k|a0〉 = |βk|2 δ(3D)(0). The

divergent factor δ(3D)(0) represents 3-volume

of the space. Therefore the number density of

k-particles is nk = |βk|2.

Need to calculate the coefficients βk:

βk =ukvk − ukvk

i

Need to know the mode functions uk(t0), vk(t0)

at any one time t0

79

Particle production by gravity in FRW space-time

Toy model: a(t) = const for t < t1 and t > t2,nonstationary regime for t ∈ [t1, t2]

Mode functions are solutions of

vk +

(

k2 +m2a2 − aa

)

vk ≡ vk + ω2k(t)vk = 0

Natural definitions of vacuum for t < t1 andt > t2 are the positive-frequency solutions:

vk =1

√

2ω(1)k

eiω(1)k t for t < t1

uk =1

√

2ω(2)k

eiω(2)k t for t > t2

ω(1,2)k ≡ ωk(t1,2) =

√

k2 +m2a2(t1,2) = const

Initial state is vacuum ⇒ choose vk as modefunctions ⇒ final state has particle density

|βk|2 = |ukvk − ukvk|2 =1

2ω(2)k

∣

∣

∣

∣

vk(t2)− vk(t2)iω(2)k

∣

∣

∣

∣

2

Need to compute vk(t2). If we use WKB,

vk(t2) ≈1

√

2ωk(t2)exp

[

i∫ t2

t1ωk(t)dt

]

then |βk|2 ≈ 0, so this precision is insufficient!

80

Improved WKB approximation: “instanta-

neous squeezed state”

Consider Bogolyubov coefficients αk(t), βk(t)relating instantaneous vacua at t and at t1:

αk(t) = −v∗k − iωkv

∗k

i√ωk

, βk(t) =vk − iωkvk

i√ωk

Define “instantaneous squeezing parameter”

Z(t) ≡ −βk(t)α∗k(t)

=vk − iωkvkvk + iωkvk

Since |βk|2 =|Z|2

1−|Z|2 ≪ 1, we expect Z(t)≪ 1

Equation for Z(t) is

Z + 2iωkZ =ωk2ωk

(

1− Z2)

, Z(t1) = 0

First approximation (neglect Z2):

Z(t) ≈∫ t

t1

ωk(t′)

2ωk(t′)

exp

[

−2i∫ t

t′ωk(t

′′)dt′′]

dt′

Adiabatic estimate of Z(t): for t ∈ [t1, t2] weintegrate by parts and use ωk(t1) = 0,

iZ(t) ≈ exp

[

−2i∫ t

t1ωk(t

′′)dt′′]

×(

ωk4iω2

k

− 1

2iωk

d

dt

(

ωk4iω2

k

)

+ ...

)

81

This is an adiabatic series (the “small param-

eter” is the slowness of ωk, i.e. ωk ≪ ω2k etc.)

Note: this series is divergent! At t > t2 this

series gives Z(t2) ≈ 0 because Z(t2) is smaller

than the precision of this series.

Asymptotic estimate of |βk|2 ≈ |Z(t2)|2

Note: for high-energy particles (k2 ≫ m2a2,

k ≫∣

∣

∣

aa

∣

∣

∣) the effective frequency ωk(t) changes

slowly (is adiabatic)

Assuming t1 → −∞, t2 → +∞, ωk(t) analytic:

Change variable θ(t) ≡ ∫ tt1 ωk(t)dt, then

Z(θ) ≈ e−2iθ∫ θ

−∞ωk2ω2

k

e2iθ′dθ′

If t = t∗is a zero of ωk(t) of n-th order, then

ωk(t) ≈ ω(0)k (t− t∗)n, so we have near t ≈ t∗

that θ ≈ θ∗ +ω

(0)k

n+1 (t− t∗)n+1. The functionωk2ω2

k

≈ n

2ω(0)k (t−t∗)n+1

= n2(n+1)

1θ−θ∗ .For Imθ∗ > 0,

we get a contribution n2(n+1)

e−2Imθ∗

If δt is the characteristic timescale, Imθ∗ ∼ωkδt ≈ kδt, so |βk|2 ∝ e−4kδt is exponentially

small at high energy (ultrarelativistic particles)

Lecture 17, 2007-12-14. De Sitter space,

Bunch-Davies vacuum

De Sitter spacetime: a(t) = − (Ht)−1, so the

effective frequency is ωk(t) = k2+(

m2H−2 − 2)

t−2

Consider m≪ H (relevant in cosmology); then

vk +

(

k2 − 2

t2

)

vk = 0

has the general solution

vk = Ak

(

1 +i

kt

)

eikt +Bk

(

1− i

kt

)

e−ikt

Normalization yields |Ak|2 − |Bk|2 = 12k

How to choose Ak, Bk?

Consider instantaneous vacuum at early time

|t| ≫ k−1, then vk(t) ≈ 1√2keikt is a natural

choice. Interpretation: short-wave modes do

not feel the curvature, behave as in flat space.

Bunch-Davies vacuum state: for every k, choose

vk/vk → iωk at t→ −∞ (Minkowski vacuum at

early times). This yields Ak = 1√2k

, Bk = 0

82

Amplitude of quantum fluctuations

Fluctuations averaged on spatial scales L:

χL(t) ≡∫

d3x χ(t,x)WL(x),

where WL(x) is the window function such that

WL(x) ≪ 1 for |x| & L and∫

d3xWL(x) = 1,

for example WL(x) = 1

(2π)3/2L3exp

[

− |x|2

2L2

]

The quantum expectation value of χLχL is

〈0| χL(t)χL(t) |0〉 =∫

d3k

(2π)3|vk(t)|2 e−|k|

2L2

≈∫ L−1

0

k2dk

2π2|vk(t)|2 ∼

1

2π2|vk(t)|2 k3

∣

∣

∣

∣

k=L−1

This is called the power spectrum δχ2L(k) of

fluctuations in χ on scales L ∼ k−1

Power spectrum of free field in flat space:

vk =ei√k2+m2t

√2(k2 +m2)1/4

, δχ2L(k) =

1

2π2

k3√

k2 +m2

Power spectrum in de Sitter space (BD):

δφL(t) =k3/2 |vk(t)|a(t)π

√2

=H

2π

√

t2k2 + 1

At late times (t→ 0), δφL ≈ H2π = const!

83

Primordial fluctuations from inflation

Plan: 1. Quantize the MS variable v. 2. De-termine vacuum state |0〉. 3. Determine power

spectrum of Φ at the end of inflation.

vk +

(

k2 − zz

)

vk = 0, z(t) ≡ af0b

BD vacuum state for v: mode functions

vk(t) =1√2keikt, t→ −∞

To determine time tc(k) of horizon crossing,

need to have approximate expression for z(t)

Use slow roll approximation, conformal time:

b2 =a2

a=

8πG

3

(

1

2f20 + a2V (f0)

)

f0 + 2bf0 + a2V ′(f0) = 0

Slow-roll parameters: (of order 10−2)

ε =1

16πG

V ′2

V 2, η =

1

8πG

(

V

V

′′− 1

2

V ′2

V 2

)

Slow-roll solution: (note conformal time)

f0 ≈ −(√

V)′

√6πG

a = −a√

2

3V (f0)ε

z(t) ≈ − a√ε√

4πG;

z

z≈ 2b2 ≈ 16πG

3a2V (f0)

84

Approximately treat ε as a constant for now,and estimate a(t) ∝ t−1, z(t) ∝ t−1, z/z ≈ 2t−2

Horizon crossing: ktc(k) ≈√

2Approximate vk(t) after horizon crossing:

vk(t) ≈ Akz(t)[

1 + k2C(t)]

, Ak unknown

Find leading correction of order k2:

−1 ≈ C + 2z

zC = z−2

(

Cz2).

C(t) =

∫ tz−2(t′)

∫ t′z2(t′′)dt′′ ≈ −1

2t2 +O(t3)

vk(t) ≈ Akz(t)[

1− 1

2k2t2

]

, kt≪ 1

To determine Ak, match at horizon crossing:

Akz(tc(k)) ≈1√2keiktc(k), Ak ≈

1√2kz(tc(k))

Substitute vk into the formula for Φ:

Φ =1

a4πGf0

1

∆z

(

v

z

).

Φk(t) ≈ Aka(t)tε(t)√

8πG

3V (f0(t))

≈ 4πGε(t)

3√

kε(tc(k))

a(t)t

a(tc(k))

√

V (f0(t))

≈ 4πGε(t)

3√

kε(tc(k))

√2

k

√

V (f0(t))

85

Power spectrum

δΦk(t) =k3/2

π√

2|Φk(t)| =

4Gε(t)

3√

ε(tc(k))

√

V (f0(t))

Note: at the end of inflation ε(t) ≈ 1, so δΦk ∝ε−1/2(tc(k))

Towards the end of inflation ε grows, so the

spectrum is “red tilted”

86

Lecture 18, 2007-12-19. Null surfaces,

horizons

A smooth hypersurface (a submanifold of

codimension 1) S ⊂M is a set of points p : f(p) = 0where f is a smooth function such that df 6= 0

on S.

Normal vector n ≡ g−1df ; can redefine f →λf with λ 6= 0 on S, then n→ λn.

Tangent vectors: t f = 0, g(n, t) = 0

Note: 1. A null vector cannot be orthogonal to

a timelike vector. 2. If g(n,n) = g(l, l) = 0 and

g(n, l) = 0 then n = λl. 3. A timelike vector

cannot be orthogonal to a timelike vector.

Hypersurface is timelike / spacelike / null

iff the normal vector n is everywhere space-

like / timelike / null. A spacelike hypersurface

has only spacelike tangent vectors. A timelike

hypersurface has timelike, null, and spacelike

tangent vectors. A null hypersurface has 1

null and 2 spacelike basis tangent vectors (no

timelike!). For a null hypersurface, the normal

vector lies within the hypersurface!

Let us call hypersurfaces just simply surfaces.

87

Examples of null surfaces

Example 1. In Minkowski spacetime, consider

f(p) = 0, f(p) ≡ t−√

x2 + y2 + z2 ≡ t− rthen the normal vector is null,

n ≡ g−1df = ∂t +1

r(x∂x + y∂y + z∂z)

This surface is a lightcone focusing through

the origin.

Example 2. In Schwarzschild spacetime,

g =

(

1− 2m

r

)

dt2 −(

1− 2m

r

)−1

dr2 − r2dΩ2,

the surface r = 2m (the horizon H) is a null

surface.

Proof: Tangent space to r = 2m is spanned

by ∂φ, ∂θ, ∂t. The vector ∂t is null at r = 2m

(but not at r 6= 2m). Also, ∂t is normal to ∂θand ∂φ. Hence, ∂t is normal to TH everywhere

on H. Hence, H is a null surface.

Definition: a null function f(p) is such that

n ≡ g−1df is null everywhere.

88

Geodesic generators of null surfaces

Null surfaces are “made of” geodesic lines called

generators. These are the flow lines of the

normal vector n ≡ g−1df . Note: n is the

affine normal to the surface. (Other normals,

n′ = λn with λ 6= const, are not geodesic.)

Statement: If f is a null function then the

normal vector field n ≡ g−1df is geodesic.

Lemma (stronger!): If a vector n is integrable

(n = g−1dλ for some function λ) and normal-

ized (∇xg(n,n) = 0 for ∀x), then n is geodesic.

Proof: If n is integrable then dgn = 0. Let

us apply the zero 2-form dgn to two vectors

(n,x) where x is arbitrary:

0 = dgn(n,x) = ∇ng(n,x)−∇xg(n,n)− g(n, [n,x])= g(∇nn,x)− g(∇xn,n) = g(∇nn,x).

Interpretation of null functions:

Surface f(p) = 0 is (locally) a lightcone emit-

ted by some event at one moment. If we con-

sider a moving source, then f(p) is the retarded

time function, i.e. time at source when the

lightray that crosses p was emitted.

89

Causal properties

A piecewise smooth curve is causal if its tan-

gent vector is everywhere timelike or null.

A piecewise smooth surface is made of pieces

of smooth surfaces. (It is not smooth on sub-

manifolds of smaller dimension.)

Spacelike surfaces contain only spacelike curves;

null surfaces contain null and spacelike curves.

Surfaces of different kinds are useful for differ-

ent things:

1. Spacelike surfaces: for setting initial condi-

tions. A spacelike surface S is (locally) Cauchy

if every causal curve (in a neighborhood of S)intersects S exactly once.

2. Timelike surfaces are worldsheets of 2-

dimensional objects, e.g. boundaries of imagi-

nary boxes where some system is enclosed.

3. Null surfaces are boundaries of causally con-

nected regions.

The future influence domain I+(F) is the

set of points connected to F by future-directed

causal curves starting at F. (Similarly, the past

influence domain I−.)90

Causal boundaries are null surfaces

Statement: The boundary ∂I+(F) of the fu-

ture influence domain of any set F is a null

surface (piecewise, and away from F).

Proof: Consider a point p ∈ ∂I+(F) where

the normal vector n exists and is not null, and

also p 6∈ F. Go to a local Lorentz frame where

∂I+(F) is locally a plane orthogonal to n.

“Condition B” holds: Points infinitesimally

close to ∂I+(F) on one side are in I+(F) but

point infinitesimally close to ∂I+(F) on the

other side are not in I+(F).

If n is timelike, then a curve going along n is

causal and violates condition B.

If n is spacelike, then there exists a timelike

vector t within ∂I+(F). Then t + λn is time-

like for very small λ; a causal curve along t+λn

crosses ∂I+(F) and violates condition B.

Note: statement is not true if p ∈ F or if n is

undefined (a “corner” of a surface)

Definition: Event horizon H is the boundary

of the past influence domain of the asymptotic

far future (can be timelike or null future)

91

Stationary (Killing) horizons

A null surface H is a stationary (Killing) hori-zon for a Killing vector field k if k is normal toH : g(k,x) = 0 for every tangent vector x tothe surface. (Note: k must be hypersurface-orthogonal to admit a horizon)

Properties:1. Killing vector k is null on H; ∇kk = κk2. Surface gravity κ is constant along flowlines of k

3. (without proof) The set of points wherek = 0 is a 2-sphere B (bifurcation 2-surface)4. Surface gravity κ is constant on B5. Surface gravity κ changes sign across B

Derivations:

1. Note: k is proportional to the affine nor-mal n to H, so k is itself null on H. If t istangent to H, we must have 0 = 1

2∇tg(k,k) =g(∇tk,k) = −g(∇kk, t) for every t such thatg(k, t) = 0, hence ∇kk = λk on H. Recallthat z ≡

√

g(k,k), g−1dz = −∇kkz . Assume

g(k,k) > 0 in some domain bounded by H.Surface gravity (defined as force at infinity) is|κ| =

∣

∣

∣g−1dz∣

∣

∣H = limp→H|∇kk|z = |λ|.

Let us define κ (not only |κ|!) by the formula∇kk ≡ κk on H. (Signed surface gravity.)

92

Lecture 19, 2007-12-21. Zero-th law of

black hole thermodynamics

2. Need formula: ∇x∇yk−∇∇xyk = R(x,k)y

(derivation postponed), then ∇k∇kk = κ∇kk

and so 0 = ∇k (∇kk− κk) = −k∇kκ. Hence

κ = const along k (unless k = 0)

Similarly, ∇k

(

κ2)

= limp→H∇kg(∇kk,∇kk)

g(k,k)= 0

3. Assume κ 6= 0, introduce an affine null

normal n to H, write k = µn. By definition,

∇nn = 0, hence ∇kk = µn∇nµ = κµn, so

∇nµ = κ = const along a flow line (orbit) of

n, as long as µ 6= 0. Hence there is a point on

the orbit where µ changes sign. At that point,

k = 0, changes direction (bifurcation point).

Define B as the set of points where µ = 0.

4. Derive another formula for κ. Since (by

the Frobenius theorem) gk ∧ dgk = 0, a useful

identity is found from a double trace with dgk:

0 = Trx,yιxιy (gk ∧ dgk) ιxιy (dgk) = ...

But ιxιkdgk = 2g(∇kk,x), so ιkdgk = 2κgk.

Also 14Trx,ydgk (x,y)dgk (x,y) = Trxg(∇xk,∇xk) =

Trx(∇xg(k,∇xk)−g(k,∇x∇xk)) = 12z2−Ric(k,k).

93

Now ιyιx (gk ∧ dgk) = g(k,x)ιydgk−g(k,y)ιxdgk+gkdgk(x,y), then we compute 0 = 1

4Trx,yιyιx(gk ∧ dgk) ιyιxdgk = gk(|∇•k|2 + 2κ2), where

we denoted |∇•k|2 ≡ Trxg(∇xk,∇xk). Hence

κ2 = −1

2Trxg(∇xk,∇xk) = −1

4z2− 1

2Ric(k,k)

5. For any vector t tangent to B, we have

∇t∇xk = ∇∇txk + R(t,k)x, hence −∇tκ

2 =

Trxg(∇t∇xk,∇xk) = TrxR(t,k,x,∇xk) = 0 since

k = 0 on B.

Derivation of the formula for ∇a∇bk

g(∇a∇bk, c) = g(∇∇abk, c) +R(b, c, a,k)

Assume that all covariant derivatives of vec-

tors a,b, c vanish. We are trying to reverse the

order of vectors a,b by using the Killing prop-

erty g(∇ak,b) = −g(∇bk, a) and the definitionof R(). Write g(∇a∇bk, c) = ∇ag(∇bk, c) =

−∇ag(∇ck,b) = −g(∇a∇ck,b) since ∇a b, c =

0. Replace g(∇a∇ck,b) = g(∇c∇ak,b)+R(a, c,k,b)

and so g(∇a∇bk, c) = R(c, a,k,b)−g(∇c∇ak,b).

Now repeat this for the permutations (abkc)→(cakb)→ (bcka) and find g(∇a∇bk, c) = R(c, a,k,b)−R(b, c,k, a) + R(a,b,k, c) − g(∇a∇bk, c). Now

use the first Bianchi identity to express R(b, c, a,k) =

−R(a,b, c,k)−R(c, a,b,k). Hence we get

2g(∇a∇bk, c) = 2R(b, c, a,k)

Vector fields orthogonal to a single 3-surface

Statement: If a vector field v is either geodesic

or Killing, and is orthogonal to a single 3-

surface∑

(either spacelike or timelike but not

null) then it is everywhere hypersurface-orthogonal.

Proof: We have ∇vg(v,v) = Lvg(v,v) = 0 in

either case. Hence, g(v,v) is constant along

orbits of v. We may rescale v → u = λv

such that g(u,u) = const everywhere. (If v

is Killing, we will not need to rescale; see be-

low.) Now it is sufficient to show that u is

hypersurface-orthogonal everywhere.

Construct a family of surfaces Σ(τ) by trans-

porting Σ along u, where τ is the affine pa-

rameter, u τ = 1. Now prove that tangent

vectors to Σ(τ) remain orthogonal to u.

Choose a connecting vector c ∈ TpΣ such that

[c,u] = 0 everywhere; by construction g(c,u)Σ =

0. It remains to prove that Lug(c,u) = 0:

Lug(c,u) = (Lug) (c,u) + g(Luc,u) + g(c,Luu)

= (Lug) (c,u) = g(∇cu,u) + g(∇uu, c).

This vanishes either when Lug = 0 (Killing

case, set λ ≡ 1) or when g(u,u) = const and

∇uu = 0 (geodesic case).

94

Lecture 20, 2008-01-09. Near-horizon ge-

ometry for Killing horizons

Assumptions: a metric with hypersurface-orthogonal

Killing vector, and a Killing horizon H with

surface gravity κ =∣

∣

∣g−1dz∣

∣

∣ 6= 0, where z ≡√

g(k,k) is the redshift function (defined only

at that side of the horizon where k is timelike)

Want to investigate near-horizon properties.

Use t as time (k = ∂t), due to hypersurface-

orthogonality have g = z2dt2−(spatial metric).

Use z as a spatial coordinate; introduce spatial

coordinates

y1, y2

within surfaces z = const

(at least locally b/c dz 6= 0 on H)

Metric is written as

g = z2dt2−g(∂z, ∂z)dz2−∑

A,B

yAyBγAB(z, y1, y2)

To compute g(∂z, ∂z), consider the inverse met-

ric; the coefficient of g−1 at ∂z⊗∂z is g−1(dz,dz) =∣

∣

∣g−1dz∣

∣

∣

2 ≡ κ2f(z, yA) where f = 1 on H. Hence

g = z2dt2 − f−1κ−2dz2 − γ(...)Note: g is degenerate on H (det g = 0 on H).

95

Local Rindler frame

Coordinate z is undefined beyond the horizon.

Trick: Define ζ ≡ z2 and consider ζ < 0

g = ζdt2 − dζ2

4ζκ2f(ζ)− γ(...)

Near horizon, f ≈ 1, introduce coordinates

T,X via κT ≡ ±√

|ζ| sinhκt, κX ≡ ±√

|ζ| coshκtThen g ≈ dT2−dX2−γ(...). (2D flat spacetime×2D

space)

Rindler spacetime:

T

X

z = 0z =const

ζ > 0ζ > 0

ζ < 0

ζ < 0

z < 0

96

Removing coordinate singularity

Metric g = ζdt2 −(

4κ2ζf(ζ))−1

dζ2 − γ, Killing

vector k = ∂tSingularity det g = 0 at ζ = 0 is not physical

(det g is a coordinate-dependent quantity!)

Introduce (Painlevé) infalling coordinates: τ ≡t + A(ζ), where A(ζ) is chosen so that the

metric is nonsingular on H in coords.

τ, ζ, yA

:

g = ζ(

dτ −A′dζ)2 − γ

= ζdτ2 − 2A′ζdτdζ −(

1

4ζκ2f− ζA′2

)

dζ2 − γ(...)

Infalling means that ∂τ points inwards: g(∂τ , ∂ζ) >

0; this holds if A′(ζ) > 0.

Can choose A(ζ) such that A′ζ 6= 0 and that1

4ζκ2f−ζA′2 is finite as ζ → 0. For example, can

have 14ζκ2f

−ζA′2 → 1 on H. (Need A′(ζ) ≈ 12κζ

near H in any case.)

Examples: spherically symmetric spacetimes

g = Q(r)dt2 − dr2

Q(r)− r2

(

dθ2 + sin2 θdφ2)

Zeros of Q(r) are horizons. Reissner-Nordström:

Q(r) = 1 − 2mr + q2

r2; Schwarzschild-de Sitter:

Q(r) = 1− 2mr −H2r2

97

Computing surface gravity

Exercise: verify ∇kk = κk directly in Painlevé

coordinates

τ, ζ, yB

using the formula g(∇kk,x) =12 (dgk) (k,x).

Note: surface gravity depends on the normal-

ization of the Killing vector k! (This needs to

be fixed ad hoc, say g(k,k)→ 1 at infinity)

For metric of the form

g = P(x)dt2 − dx2

Q(x)− γ(x, yB)

where P(x0) = Q(x0) = 0, while P ′(x0) > 0,

Q′(x0) > 0. (Assuming k = ∂t in the given

coordinate system!)

Need to transform the metric to g = ζdt2 −(

4κ2ζf(ζ))−1

dζ2−γ where ζ = 0 at x = x0 and

f(0) = 1: change coordinates to ζ = P(x),

then

Q(x) ≈ Q′(x0)P ′(x0)

ζ, g = ζdt2 − P ′(x0)Q′(x0)ζ

dζ2

P ′2(x0)

and hence κ = 12

√

P ′(x0)Q′(x0).

98

Unruh effect

Rindler spacetime (2D): g = (1 + ax)2 dt2 −dx2; Killing vector k = ∂t; Killing horizon x =

−a−1; redshift z = 1+ ax (for x > −a−1 only!)

Proper acceleration of stationary observers:

a = −g−1d ln z =a

1 + ax∂x

Observer at x = 0 has constant acceleration a

Interpretation: horizon plane x = −a−1 due to

constant acceleration. Surface gravity κ = a

To be shown: quantum field in Minkowski vac-

uum appears to have thermal spectrum of par-

ticles, as measured by accelerated detectors

Unruh temperature: T = a2π in Planck units

(Analytic continuation arguments: 1. Euclidean

metric a2x2dt2E + dx2 is nonsingular at origin

only if tE is periodic with period 2π/a. 2.

Fields are functions of sin atE and must be pe-

riodic in tE. – Inconclusive!)

99

Lecture 21, 2008-01-11. Hawking radia-

tion. First law for Killing horizons

Metric g = ζdt2 −(

4κ2ζf(ζ))−1

dζ2 − γConsider lightrays in the t− ζ plane near H:

4ζ2κ2f(ζ)t2 = ζ2, t = ±∫

dζ

2κζ√

f(ζ)

Statement: Any null curve in 2D spacetime

is a null geodesic (lightray).

Proof: g(∇nn,n) = 0 but there is only one

vector orthogornal to n, so ∇nn = λn and

hence ∇µn (µn) = 0 for some µ.

Define the tortoise coordinate r ≡ ∫ dζ

2κζ√f(ζ)

,

then lightrays are t ± r = const; 2D metric is

conformally flat, g = ζ(

dt2 − dr2)

.

Consider scalar field (any coupling/potential):

S [g, φ] =

∫

dt dζ d2y

√γ

2κ√f

[

1

2ζφ2 − 2κ2fζ

(

φ,ζ)2

−1

2|∇⊥φ|2 − V (φ,R)

]

=

∫ √γd2y

∫

dt dr

[

1

2φ2 − 1

2φ2,r + (...) ζ

]

Coupling/potential is unimportant at ζ → 0!

Essentially, free field in 2D Rindler spacetime.

100

Unruh effect

Quantization of free field in Rindler spacetime:use two coordinate systems, t, r and T,XTortoise coordinate: r = a−1 ln (1 + ax)

Minkowski frame: aT = ear sinh at, aX = ear cosh at

g = dT2 − dX2 = e2ar(

dt2 − dr2)

Mode functions: uk = 1√2ωk

eiωkt, ωk = |k|Mode expansions for quantization:

φ(T,X) =

∫

dk√2π

1√

2 |k|

[

e−i|k|T+ikX a−k + ei|k|T−ikX a+k

]

φ(t, r) =∫

dk√2π

1√

2 |k|

[

e−i|k|t+ikrb−k + ei|k|t−ikrb+k

]

Use lightcone coordinates: u = t−r, v = t+r,U = T − X, V = T + X and rewrite modeexpansions using ωk as integration variable:

φ(U, V ) =∫ ∞

0

dΩ√2π

1√2Ω

[

e−iΩU a−Ω + eiΩU a+Ω

+e−iΩV a−−Ω + eiΩV a+−Ω

]

φ(u, v) =∫ ∞

0

dω√2π

1√2ω

[

e−iωub−ω + eiωub+ω

+e−iωvb−−ω + eiωvb+−ω]

Note: both expansions are for the same oper-ator φ, so we can express b±±ω through a±±Ω

101

Bogoliubov transformation

Relationship between u, v and U, V :

aU = −e−au, aV = eav

Hence the u-dependent part of φ is equal to

the U-dependent part: (same for v and V )∫ ∞

0

dω√2π

1√2ω

[

e−iωub−ω + eiωub+ω]

=

∫ ∞

0

dΩ√2π

1√2Ω

[

e−iΩU a−Ω + eiΩU a+Ω

]

Compute Fourier transform in u of both sides:

∫ +∞

−∞du√2πeiωu

∫ ∞

0

dω′√2π

1√2ω′

[

e−iω′ub−ω′+ eiω

′ub+ω′]

=1

√

2 |ω|

b−ω , ω > 0

b+|ω|, ω < 0

∫ +∞

−∞du√2πeiωu

∫ ∞

0

dΩ√2π

1√2Ω

[

e−iΩU a−Ω + eiΩU a+Ω

]

=

∫ +∞

−∞du

2π

∫ ∞

0

dΩ√2Ω

[

eiωu−iΩU a−Ω + eiωu+iΩU a+Ω

]

≡∫ ∞

0

dΩ√2Ω

[

F(Ω, ω)a−Ω + F(−Ω, ω)a+Ω

]

,

where we defined

F(Ω, ω) ≡∫ +∞

−∞du

2πexp

[

iωu+ iΩ

ae−au

]

102

Then the Bogoliubov transformation is

b−ω =

∫ +∞

0dΩ

[

αΩωa−Ω + βΩωa

+Ω

]

,

where (for ω > 0,Ω > 0)

αΩω =

√

ω

ΩF(Ω, ω), βΩω =

√

ω

ΩF(−Ω, ω)

Need to compute 〈Ω0| b+ω b−ω |Ω0〉 = ∫+∞0 dΩ |βΩω|2

where |Ω0〉 is the vacuum state annihilated bya−Ω (vacuum in the Minkowski frame T,X)

Trick 1: The function F(Ω, ω) satisfies

F(Ω, ω) = F(−Ω, ω) expπω

a, F ∗(Ω, ω) = F(−Ω,−ω)

Derivation: Shift the integration contour intothe line u = −iπa−1+ t with t ∈ R, then e−au =−e−at and a factor eπω/a appears.Trick 2: Canonical commutation condition,[

b−ω , b+ω′]

= δ(ω − ω′), entails

δ(ω − ω′) =

∫ +∞

−∞dΩ

[

αΩωα∗Ωω′ − βΩωβ∗Ωω′

]

=

∫ +∞

−∞dΩ

√ωω′

ΩF(−Ω, ω)F ∗(−Ω, ω′) exp

π

a

(

ω+ ω′)

The mean particle density is computed by set-ting ω = ω′ and discarding the δ(0) factor:

〈nω〉 =(

exp

[

2πω

a

]

+ 1

)−1

=

(

exp

[

E(k)

T

]

+ 1

)−1

First law for Killing horizons: δE = TδS

where δE is the energy flow through H, T = κ2π

is the temperature, S = 14GA is the “entropy of

the horizon”

Consider an almost Killing horizon (∇kk ≈ κk

on H) and a small amount of matter (Tµν)

that falls into the horizon very slowly.

“Area of the horizon” is the 2-area of the space-

like cross-section of H at t = const

Consider spacelike connecting vectors s1, s2 for

k; they remain approximately orthogonal to k

since Lkg(k, s) = (Lkg) (k, s) ≈ 0

Introduce a null vector l such that g(k, l) = 1,

then ∇kl ≈ −κl near HThe rate of change of 2-area under flow of k:

2-area A = 4-volume spanned by k, l, s1, s2,so A = ε(k, l, s1, s2) and

Lkε(k, l, s1, s2) = (div k)A+ ε(k,Lkl, s1, s2)

The last term approximately vanishes since

Lkl has almost no component parallel to l:

g(k,∇kl−∇lk) ≈ g(k,−κl + κl) = 0. Hence

div k ≈ ∇n lnA

103

Assume that the horizon is everywhere smooth

(this excludes e.g. collision of black holes!)

Then H is (locally) entirely covered by the flow

lines of k. Let t be the flow parameter for k.

Denote θ(t) ≡ div k along flow lines of k. Then

the total change in A is δA =∫+∞−∞ dt

∮

d2A∇n lnA =∫+∞−∞ dt

∮

d2Aθ(t)Compute ∂tθ(t) = ∇kTrxg(∇xk,x) = Ric(k,k)+div∇kk ≈ Ric(k,k) + κθ(t) since ∇kκ is small.

Then θ(t) = − ∫∞t Ric(k,k)eκ(t−t′)dt′ and so

∫ +∞

−∞θ(t)dt = −1

κ

∫ +∞

−∞Ric(k,k)dt′

=8πG

κ

∫ +∞

−∞T(k,k)dt′

The quantity δE ≡ ∮

d2A∫+∞−∞ dt′T(k,k) is the

total flux of energy through the horizon, so

δE =κ

2π

δA

4G= TδS

Remarks: 1. The null energy condition guar-

antees T(k,k) ≥ 0 and thus δS ≥ 0. But this

is only a weak “2nd law” b/c we assumed very

small variations of smooth Killing horizon.

2. Introducing a rotational Killing vector ∂φsuch that k = ∂t + Ω∂φ we get

δE−ΩδJ = TδS, δJ ≡∮

d2A∫ +∞

−∞dt′T(∂φ, ∂φ)

Lecture 22, 2008-01-16. Raychaudhuri equa-tion for timelike geodesics

Motivation: Want to describe the distortion ofa spacelike 3-volume by the Lie flow of a time-like geodesic field, compared with the paralleltransport (comoving flow)

v

γ

ca τ=0

b

τ

a(τ)

b(τ)

c(τ)

Let v be a timelike geodesic field. Let γ(τ) beone flow line. Let c be a connecting vector forv; values of c on γ are c(τ). Let cP (τ) be theresult of parallel transport of c(0) along γ.

104

Distortion of transverse connecting basis

Some useful properties:

Statement 1: A connecting vector c stays

orthogonal to v if v is normalized and geodesic

(or Killing)

Proof:∇vg(v, c) = g(v,∇cv) = 0;Lkg(k, c) = 0

Statement 2: (Distortion of connecting ba-

sis) a) The connecting vector c satisfies ∇vc =

Bv(c) where Bv is the derivative tensor of v;

g(Bv(c),x) ≡ g(∇cv,x). b) The transforma-

tion operator Z(τ) defined by c(τ) = Z(τ)cp(τ)

satisfies Z(0) = 1 and ddτ Z(τ) = Bv(τ)Z(τ)

Proof: g(∇vc,x) = g(∇cv,x) ≡ B(v)(c,x);(

ddτ Z

)

cP = ∇vZcP = ∇cv = Bvc = BvZcP for

any (parallelly transported) basis vector cP .

Statement 3: Transverse 3-volume satisfiesd

dτV = Lvε(v, a,b, c) = (divv)V = V TrBv

Statement 4: Derivative tensor is transverse,

B(v)(v, ·) = B(v)(·,v) = 0

Proof: g(∇xv,v) = 0, g(∇vv,x) = 0

So the tensor B(v) is effectively 3-dimensional.

105

Orthogonal decomposition of Bv for time-

like v

Consider B(v) as a bilinear form in 3-dim. space

with reduced metric hµν ≡ gµν − vµvνNote: divv = Trxg(∇xv,x) = Trxh(∇xv,x);

Trxh(x,x) = 3; h(v, ·) = 0

Decompose B(v) into pure trace, traceless sym-

metric, and antisymmetric parts (w.r.t. h):

B(v) =1

3hdivv + σ(v) + r(v)

where σ(v) is shear and r(v) = 12dgv is rotation

of v. Note: σ(v) and r(v) are transverse to v.

Statement 5: Rotation vanishes for hypersurface-

orthogonal, non-null geodesic v.

Proof: Let ω ≡ gv; by assumption ω ∧ dω = 0

and ω(v) 6= 0. Compute 0 = ιv (ω ∧ dω) =

ω(v)dω − ω ∧ ιvdω = ω(v)dω − ω ∧ ιv2r(v) =

ω(v)dω, hence dω = 0.

Counterexample for null v: set v = y (∂t + ∂x)

in Minkowski spacetime; then v is null, hypersurface-

orthogonal, and geodesic, but r(v) 6= 0.

Practice exercise: check the counterexample

106

Raychaudhuri equation

Compute ddτdivv. (Note: all 1st derivatives of

x vanish under trace, but [v,x] 6= 0.)

∇vTrxg(∇xv,x) = TrxR(v,x,v,x)

+ Trxg(∇x∇vv,x) + Trxg(∇[v,x]v,x)

= Ric(v,v)−Trxg(BvBvx,x)

Now simplify for timelike geodesics v using

the 3-dim. decomposition. Since Tr() is ascalar product in the space of matrices, we

have Tr(AB) = 0 if AT = A and BT = −B.

Traceless matrices are orthogonal to h. Hence

d

dτdivv = Ric(v,v)−1

3(divv)2−Trσvσv−Trrvrv

Comments: shear changes shape but not vol-

ume; rotation does not change shape; Tr canbe computed with h instead of g here; h is

negative-definite.

Focusing conditions

Statement 6: Trσvσv ≥ 0 and Trrvrv ≤ 0 forany timelike v.

Proof: If a matrix A satisfies AT = ±A then

±TrAa = ±Trxh(AAx,x) = Trxh(Ax, Ax) ≤ 0

Strong energy condition: Ric(v,v) ≤ 0 for alltimelike v (attractive nature of gravity)

107

Focusing theorem

Statement 7: a) For a hypersurface-orthogonal

timelike geodesic field v, assuming strong en-

ergy condition, we have

d

dτdivv ≤ −1

3(divv)2

b) If divv = −θ0 < 0 at τ = 0 then divv = −∞at finite time τ = τ0, where τ0 ≤ 3θ−1

0 .

Proof: a) Follows from Statements 5 and 6.

b) Consider auxiliary variable x(τ) ≡ (divv)−1,

then dxdτ ≥

13 and x(0) = −θ−1

0 . It follows that

x(τ0) = 0 at τ0 < 3θ−10 .

Interpretation: The volume of connecting ba-

sis shrinks to zero. So geodesics that are in-

finitesimally nearby focus onto γ(τ) at τ = τ0.

This is congruence-dependent, not just a prop-

erty of spacetime.

Statement 8: Volume vanishes as V ∝ (τ − τ0)nnear τ = τ0, where we may have n = 1,2,3

Proof: 3-volume element V (τ) is proportional

to det Z(τ) while Z satisfies ddτ Z = BvZ; hence

ddτ det Z = (divv) det Z. Since d

dτ det Z is al-

ways finite, we must have det Z = 0 at τ = τ0where divv = −∞. Since all components of Z

are regular at τ = τ0, the determinant vanishes

as polynomial of τ − τ0 of degree at most 3.

108

Lecture 23, 2008-01-18. Rauchaudhuri

equation for null geodesics

Orthogonal decomposition of Bn requires pro-

jecting onto the 2-dim. spacelike subspace or-

thogonal to n. Choose a null connecting vec-

tor l such that g(n, l) = 1,[n, l] = 0 everywhere.

Properties: ∇ng(n, l) = 0; g(∇ln, l) = 0.

Projection onto n⊥ is the operator

Pnx = x− ng(l,x)− lg(n,x)

2-dim. reduced metric h(2)µν = gµν − lµnν − nµlν

(Note: l is defined up to l → l + λn which

affects Pn but not the final results. l is the

lightray going towards inside of the horizon.)

Consider Bn as a bilinear form in 2-dim. space

with reduced metric h(2)

Note: divn = Trxg(∇xn,x) = Trxh(2)(∇xn,x);

Trxh(2)(x,x) = 2; h(2)(n, ·) = h(2)(l, ·) = 0

Decompose Bn into components w.r.t. the pro-

jector Pn: Bn(x,y) = Bn(Pnx, Pny)+Bn(Pnx, l)g(n,y)+

Bn(l, Pny)g(n,x)+Bn(l, l)g(n,x)g(n,y). Denote

for brevity Bn(Pnx, Pny) ≡ B⊥n (x,y). Compute

trace: Trxg(BnBnx,x) = Trx,yBn(x,y)Bn(y,x) =

Trx,yB⊥n (x,y)B⊥n (y,x) because all other com-

ponents generate Pnn = 0 or g(n,n) = 0.

109

Now decompose B⊥n (x,y) according to sym-

metry and trace properties:

B⊥n = 12h

(2)divn + σ⊥(n) + r⊥(n)

σ⊥n ≡ 12[B

⊥n (x,y) +B⊥n (y,x)]− 1

2h(2)(x,y)divn

r⊥n ≡ 12[B

⊥n (x,y)−B⊥n (y,x)]

Raychaudhuri equation:

d

dτdivn = Ric(n,n)−(divn)2

2−Tr

(

σ⊥n σ⊥n

)

−Tr(

r⊥n r⊥n

)

Statement 1: r⊥n = 0 if n is hypersurface-

orthogonal. (Even if rn does not vanish.)

Proof: The tensor r⊥n is equivalently defined

by r⊥n (x,y) = 12dgn(Pnx, Pny). This vanishes

(by construction) if either x or y is not in n⊥.It remains to consider x and y both in n⊥,so that Pnx = x and Pny = y. Then ιxgn =

ιygn = 0 and (by the Frobenius condition) 0 =

ιxιy (gn ∧ dgn) = gnιxιydgn = 2gnr⊥(n)

(x,y). So

r⊥(n)

(x,y) = 0.

Null energy condition (NEC): Ric(n,n) ≤ 0 for

any null n. [Equivalently, T(n,n) ≥ 0]

Focusing theorem: If NEC holds and divn <0 then focusing occurs within finite interval of

affine parameter τProof: Define x ≡ (divn)−1; then dx

dτ ≥12 and

x(0) = −θ−10 , so x(τ0) = 0 where τ0 < 2θ−1

0

Geodesic deviation equation

If c is a connecting vector to a geodesic fieldv then c satisfies the 2nd order equation:

∇v∇vc = R(v, c)v

Choose v as the field of all the geodesics emit-ted from a point p. Solve geodesic deviationequation along a single curve γ. Obtain c(τ)such that c(0) = 0, ∇vc(0) ≡ c 6= 0.The vector c(τ) is a Jacobi field for γ(τ).Point q ∈ γ is conjugate to p ∈ γ if ∃c(τ) notidentically zero, solving GDE, c = 0 at p and q.Interpretation: an infinitesimally nearby geodesiccurve emitted at p will intersect γ again at q.

Define the operator A(τ) as the map fromc(τ = 0) to c(τ) in a parallelly propagated

basis. Then ¨A = GA, where Gx ≡ R(v,x)v.Initial conditions: A(0) = 1− Pv, ˙A(0) = Pv

Conjugate points and focusing

1. A vector field emitted at p is always hypersurface-orthogonal (proof below)2. If divv < 0 at some point p then focus-ing is guaranteed (assuming energy conditions)within a finite distance δτ from p3. How to compute divv for this field: divv =ddτ ln det A. So det A→ 0 when divv→ −∞

110

Lecture 24, 2008-01-23. Extremal prop-

erties of geodesics

To understand the extremal properties it is im-

portant to consider a particular kind of vector

field: namely geodesics emitted from a point.

This field is hypersurface-orthogonal (see be-

low) and hence we may use the bounds on

focusing from the Raychaudhuri equation. Fo-

cusing will control the extremal properties.

Statement 2: A geodesic vector field v emit-

ted at a single point p is hypersurface-orthogonal.

Proof: Let ω ≡ gv; we will show that ω∧dω =

0. a) Suppose that v is timelike. Since dω =

2rv while rv is transverse to v, the only nonzero

possibility is ω∧dω(v, a,b) = ω(v)2rv(a,b) where

a,b ∈ v⊥. However, 2rv(a,b) = g(∇av,b) −g(∇bv, a). Let us choose a basis of vector

fields a,b, c connecting to v; by construc-

tion these fields vanish at p. So rv(a,b)|p =

0. Now compute the derivative of 2rv(a,b)

along v. We find 2∇vrv(a,b) = ∇vg(∇va,b)−∇vg(∇vb,a) = R(v, a,v,b) − R(v,b,v, a) = 0.

Hence rv(a,b) = 0 everywhere.

111

b) Suppose that v is null. Choose a null con-

necting vector l such that g(v, l) = 1. The only

nonzero possibility is ω ∧ dω(l, a,b) = 2rv(a,b)

where a,b ∈ v⊥. Then the same argument in-

volving a connecting basis shows that rv(a,b) =

0 everywhere.

Statement 3: The transformation operator

G is symmetric, GT = G, with respect to the

metric g

Proof: Compute g((G− GT )a,b) = g(Ga,b)−g(a, Gb) = R(v, a,v,b) − R(v,b,v, a) = 0, so

G− GT = 0

Conjugate points γ(0) and γ(τ0) correspond to

focusing of infinitesimally nearby geodesics:

τ = 0

γ(τ)

τ = τ0c(τ)

Second variation of action in mechanics

Consider mechanical system with action

S[x] =

∫ T

0

[

1

2x2 − V (x)

]

dt

Is the action minimized by a trajectory x(τ)

that solves x+ V ′(x) = 0?

Compute first and second variation of S:

δS =

∫ T

0

[

−x− V ′(x)]

δx(t)dt

δ2S = −∫ T

0

[

δx+ V ′′(x)δx]

δx(t)dt

If there exists a solution for “geodesic devi-

ation” such that δx(0) = 0, δx(0) 6= 0, and

δx(t0) = 0 for some t0 > 0, call t0 a conju-

gate point to t = 0. Suppose no conjugate

points exist until t = T . Then the solution

A(t) of A+ V ′′(x)A = 0, A(0) = 0, A(0) = 1

is nonzero for 0 < t ≤ T . For any δx(t) define

s(t) ≡ δx(t)/A(t) and substitute into δ2S:

δ2S = −∫ T

0

[

As+ 2As]

Asdt

=∫ T

0

[

−(

A2ss).

+A2s2]

dt

=

∫ T

0A2s2dt > 0 if s(t) 6= 0 somewhere

Then S has a minimum on x(t).

112

No minimum of action with conjugate points

Statement: If a conjugate point exists at t0 <

T , then S[x] has no minimum at x(t).

Proof: We will present explicitly a δx(t) such

that δ2S < 0. By assumption, a solution c(t)

of c + V ′′(x)c = 0 exists such that c(t0) =

0, c(t0) 6= 0. Choose δx(t) = c(t)θ(t0 − t) −c(t0)ε(t), where ε(t) is very small and positive

and only nonzero in a small neighborhood of

t = t0. Then δx+ V ′′(x)δx ≈ −c(t0)δ(t− t0) +

O(ε) near t− t0, so

δ2S =∫ T

0

[

δx+ V ′′(x)δx]

δx(t)dt

= −∫ T

0

[

c2(t0)ε(t)δ(t− t0) +O(ε2)]

dt

= −c2(t0)ε(t0) +O(ε2) < 0

for sufficiently small ε. (Note: sign of ε is

chosen to “cut the corner.”)

T

c(t)

0

tδx(t)

t0113

Second variation of geodesic length

Consider a timelike curve γ perturbed by a vec-

tor field t. Define Lie-propagated vector field

v such that Ltv = 0. Assume that t = 0 at

endpoints τ1, τ2. Proper length of the curve

γ(τ ;σ) displaced by parameter σ is

L[γ(τ ; σ)] =

∫ τ2

τ1

√

g(v,v)dτ

Compute first and second derivatives w.r.t. σ:

d

dσL[γ;σ] = −

∫ τ2

τ1g

(

t,∇vv

N

)

dτ, N ≡√

g(v,v)

d2

dσ2L[γ;σ] = −

∫ τ2

τ1

[

g

(

∇tt,∇vv

N

)

+ g

(

t,∇t∇vv

N

)]

dτ

For a geodesic curve, ∇vvN = 0, and we can

set N = 1 to simplify calculations. Moreover:

g(t,∇t∇vv) = R(t,v,v, t) + g(t,∇v∇vt)

Hence the 2nd variation evaluated on γ is

d2

dσ2L[γ;σ] =

∫ τ2

τ1g(

t, Gt−∇v∇vt)

dτ

If this is negative for any field t(τ) then L[γ]is locally maximized by the curve γ

Note: if γ has conjugate points τ1, τ2 with a

Jacobi field c(τ), then choosing t = c yields

the absence of maximum! (See next theorem)

114

Theorem of geodesic extremum:

a) If a timelike geodesic curve γ(τ) has no

conjugate point to τ = 0 within an interval

[0, T ], then γ is a local maximum of L[γ] for

curves between γ(0) and γ(T).

b) If τ = τ0 < T is a conjugate point to τ =

0 on γ(τ) then there exists a longer timelike

curve between γ(0) and γ(T).

Proof: a) Consider the operator A(τ) for γ.If there are no conjugate points for γ(0) within

τ ∈ [0, T ], then A(τ) is nonsingular, so A−1(τ)exists for τ ∈ [0, T ]. We will now show that

d2

dσ2L[γ; σ] =

∫ T

0g(

t, Gt−∇v∇vt)

dτ < 0

for all nonzero t(τ) ∈ v⊥ that vanish at τ = 0

and τ = T . Define s(τ) ≡ A−1(τ)t(τ), denote

∇v under the integral by the overdot, and ex-

press the integral through s(τ):∫ T

0g(

As, GAs− (As)··)

dτ = −∫ T

0g(

As,2 ˙As + As)

dτ

Now we add a total derivative ∇vg(As, As) =

g( ˙As + As, As) + g(As, ˙As + As) and obtain

d2

dσ2L[γ; σ] =

∫ T

0g(

As, As)

dτ

+

∫ T

0

[

g(

˙As, As)

− g(

As, ˙As)]

dτ

115

Rewrite g( ˙As, As)−g(As, ˙As) = g(

s, ( ˙AT A− AT ˙A)s)

and examine this Wronskian-like combination:˙AT A− AT ˙A = 0 at τ = 0, while ¨AT = AT GT =

AT G. Hence(

˙AT A− AT ˙A).

= 0 and finally

d2

dσ2L[γ;σ] =

∫ T

0g(

As, As)

dτ < 0

since A is nondegenerate (has no zero eigen-

vectors) and As is always spacelike, while s 6= 0

at least for some τ .

b) If there exists a conjugate point at τ =

τ0 < T , then ∃ Jacobi field c(τ) 6= 0 such that

c(0) = c(τ0) = 0. Using c(τ) we will explicitly

construct a field t(τ) such that d2

dσ2L[γ(τ ;σ)] >0 when evaluated on the geodesic γ. The

idea is to “cut the corner”: set t(τ) ≈ c(τ)for 0 < τ < τ0 and t ≈ 0 for τ > τ0. Explicitly,

t = c(τ)θ(τ0 − τ) − εq(τ)b, where b = const,

g(b, c(τ0)) = −1, and ε > 0, q(τ) > 0. Then

Gt− t(τ) = c(τ0)δ(τ − τ0) +O(ε)

and finally we obtain

d2

dσ2L[γ;σ] =

∫ T

0g(t, Gt− t)dτ

=

∫ T

0g(−εq(τ)b, c(τ0)δ(τ − τ0))dτ

= εq(τ0) +O(ε2) > 0

Lecture 25, 2008-01-25. Extremal prop-

erties of null geodesics

Since null curves cannot be obtained by ex-

tremizing∫

√

g(v,v)dτ , we will use an alterna-

tive variational principle:

L[γ] =

∫ T

0

g(γ, γ)

Ndτ,

where N(τ) is a Lagrange multiplier. Define

the vector field v as the Lie transport of γ by

a perturbation field t, then LtN = 0, Ltv = 0.

First variation w.r.t. parameter σ:

d

dσL[γ; σ] = Lt

∫

g(v,v)

Ndτ =

∫

2g(∇tv,v)

Ndτ

=

∫

2g(∇vt,v

N)dτ =

∫

d

dτ(...)dτ −

∫

2g(t,∇vv

N)dτ

= −2

∫

g(t,∇vv

N)dτ

Second variation (using ∇vvN = 0 on shell):

d2

dσ2L[γ;σ] = −2

∫

g(t,∇t∇vv

N)dτ

= −2

∫ [

R(t,v,v

N, t) + g(t,∇v/N∇v/Nt)

]

dτ

= 2

∫

g(t, G(τ)t− t)Ndτ,

where G(τ)t ≡ R( vN , t)

vN ; overdot means d

d(Nτ).

116

Restrict N = 1 everywhere, so have ∇vg(v, t) =

g(v,∇vt) = g(v,∇tv) = 12∇tg(v,v) = 0 and

since t = 0 at endpoints, must have g(v, t) = 0

everywhere.

So it suffices to consider t ∈ v⊥.Also, t = λv gives no variation since Gv = 0

and g(t, t) ∝ g(v,v) = 0. So it suffices to con-

sider t ∈ v⊥ up to multiples of v.

Introduce a null vector u such that g(u,v) = 1

and fix the projector Pvx ≡ x − vg(x,u) =

ug(x,v). Then it suffices to consider t ∈ (u,v)⊥.Define A(τ) by the equation GA − ¨A = 0,

A(0) = 1 − Pv, ˙A(0) = Pv. Then focus point

τ = τ0 exists iff a solution A(τ) exists with

det A(τ0) = 0.

Now the same argument involving a solution

A(τ) and t(τ) ≡ A(τ)s(τ) shows that

d2

dσ2L =

∫

g(As, As)dτ < 0

since As is in (u,v)⊥ and thus everywhere space-

like. Similarly, if a focus point exists then there

exist a deformation t(τ) such that d2L/dσ2 >0. This indicates that the deformed curve is

at least somewhere timelike. An explicit ge-

ometric consideration indicates that the curve

can be deformed to everywhere timelike.

Deforming the null curve into timelike

1. Near the focus point τ0 we have a solu-

tion c(τ) such that c ∈ (u,v)⊥, c(τ0) = 0,

c(τ0) 6= 0. Consider the curve γ that follows

v + c until τ = τ0 and then v. The vector γ is

piecewise null.

2. Consider the tangent vector w at τ =

τ0− δτ . The vectors v and w are both future-

directed and null, so g(v,w) > 0, g(v,v) =

g(w,w) = 0. Hence b ≡ v − w is spacelike:

g(b,b) = g(v − w,v − w) = −2g(w,v) < 0.

Also g(b,w) > 0, g(b,v) < 0.

3. The deviation vector t(τ) can be chosen as

t = εq(τ)b where ε > 0, q(τ) > 0, signq(τ) =

sign (τ0 − τ) near τ = τ0, for example q(τ) =

1− |τ − τ0|. Then q(τ)g(γ,b) > 0 for all τ .

4. The tangent vector to the deformed curve

is ˙γ = γ + t = γ + εq(τ)b.

5. We verify that ˙γ is everywhere timelike:

g(γ + t, γ + t) = 2εq(τ)g(γ, b) +O(ε2) ≥ 0

117

t

vv

ww

118

Extremal properties for geodesics emitted

from a surface

Consider a spacelike surface Σ and a point p 6∈Σ. What is the shortest distance from p to Σ?

It is the geodesic γ emitted normally from Σ

towards p unless there exists a conjugate point

on γ before p.

Example: let Σ be the half-hyperboloid given

by t2 − x2 − y2 − z2 = 1, t < 0 in Minkowski

spacetime. The normal vector field is n =

t∂t+x∂x+y∂y+z∂z. A straight line (geodesic)

at point t, x, y, z towards the origin is x =

−t,−x,−y,−z and so it is orthogonal to Σ

because x ‖ n.

Hence the origin is the focus of geodesics emit-

ted normally from Σ.

Consequence: There is no (timelike) curve of

maximal proper length connecting 1,0,0,0and Σ

119

The easiest singularity theorem

If Σ is a compact Cauchy surface with (time-

like) normal vector n, such that divn < 0 ev-

erywhere on Σ, and if the strong energy condi-

tion holds to the future of Σ, then no timelike

geodesic to the future from Σ is complete.

Proof: 1. Consider the timelike geodesic field

n emitted normally from Σ. Since Σ is com-

pact, divn has an upper bound such that divn <−θ0 everywhere on Σ with some θ0 > 0. Since

n is hypersurface-orthogonal, by focusing the-

orem every geodesic has a conjugate point to

Σ at τ < τ0 ≡ 3θ−10 .

2. By extremal length theorem, no geodesic

maximizes proper length longer than τ0.3. Since Σ is compact and Cauchy, for any

point p to the future of Σ there is a nonempty

set of timelike or null curves intersecting Σ and

p. These curves intersect Σ at a compact set,

hence there is a point p′ ∈ Σ and a curve γ be-

tween p′ and p that maximizes proper length

between p and Σ. Then the curve γ would

have to be a geodesic emitted normally from

Σ. But such geodesic cannot maximize proper

length if it is longer than τ0. Therefore, no

points p can exist such that there is a timelike

curve from Σ to p with proper length > τ0.

120

Lecture 26, 2008-01-30. Area theorem

2nd law of horizon thermodynamics: The

combined area of event horizons does not de-

crease with time. (Assuming some conditions!)

Event horizon H = boundary of the set Bwhere signals do not escape to infinity.

Statement 1: (a) Event horizon H is locally

a null surface. (b) Null generators of H may

enter H from the past (at “corners”) but can-

not exit H to the future. (Assuming no naked

singularity directly on H, i.e. that the space-

time is asymptotically predictable.)

Proof: (a) There are no timelike curves cross-

ing H from interior outwards. Hence H cannot

contain timelike tangent vectors but must con-

tain a null tangent vector. B must be on the

“future side” from H. (b) Let γ be a null gen-

erator of H; then ∃γ′ that is arbitrarily close

to γ but never enters B, for instance, γ′ couldbe chosen as γ shifted a little bit towards the

past. Suppose p is a point where the null gen-

erator γ leaves H. By assumption p is not a

singularity, so we may continue γ past p; then γ

enters B\H. However, arbitrarily nearby curves

γ′ never enter B at all; contradiction.

121

Statement 2: If the NEC holds and if there

are no naked singularities, then divn ≥ 0 for

affine generators n of the horizon.

Proof: Suppose divn < 0 at some p ∈ H;

then divn < 0 also in a neighborhood of p.Consider the null geodesic γ emitted from p.By focusing theorem, the null generators n will

focus within a finite interval of affine parame-

ter. After focusing, some point q on γ will be

reachable from p by a timelike curve. However,

a timelike curve from p must be entirely within

B (or else p is visible from infinity). Hence, γleaves H since a point q on γ is not in H. How-

ever, Statement 1 says that γ cannot leave H;

contradiction. Area theorem (Hawking): Let Σ1 and Σ2

be two Cauchy surfaces (Σ2 later than Σ1).

Let H be the event horizon. Then the cross-

section (2D) area of H∩Σ2 is not smaller than

the area of H∩Σ1.

Proof: The generators of H emitted from

H ∩ Σ1 cannot leave H and thus all intersect

Σ2, generating a portion of H∩Σ2. The area

of that portion cannot be smaller than the area

of H∩Σ1 since divn ≥ 0. Other disconnected

pieces of H∩Σ2 might have appeared, making

the area of H∩Σ2 even larger. Remark: Since divn ≥ 0 everywhere, the area

of each connected piece of H is non-decreasing.

122

Singularity in collapsing universe

The existence of a singularity = the existence

of a geodesic that cannot be continued beyond

a finite parameter range ∆τ .

Note: If a piecewise timelike or null curve be-

tween points p, q has proper length L, then

there exists a timelike geodesic between p, q

of length at least L. Same for curves between

a point p and a surface Σ.

Statement 3: If Σ is a spacelike Cauchy su-

face with normal vector field n such that divn ≤−θ0 < 0 everywhere, and if the strong energy

condition holds, then every timelike geodesic

from Σ has proper time ∆τ ≤ 3θ−10 .

Proof: Select any point p to the future of Σ

and consider the set of all causal (timelike or

null) geodesic curves leading to p. All these

curves intersect Σ only once (Cauchy prop-

erty of Σ). The set of tangent vectors at p is

compact (it includes the past timelike and null

directions), hence the set of intersection points

with Σ is a compact subset of Σ. Hence, there

123

exists the longest (by proper length) geodesic

curve γ(p)max between Σ and p. Hence, γ

(p)max

is also the longest piecewise causal curve be-

tween Σ and p.

The curve γ(p)max must be timelike (since it has

proper length 6= 0). By the focusing theorem,

every timelike geodesic orthogonal to Σ will

have a focus within ∆τ ≤ 3θ−10 . Hence, no

geodesic longer than 3θ−10 maximizes proper

length. Hence, γ(p)max is not longer than 3θ−1

0 ,

for any point p to the future of Σ.

Singularity in closed universe

Closed universe = containing a compact space-

like 3-surface Σ without boundary (but not

necessarily a Cauchy surface!)

Cauchy horizon of Σ is ∂I+(Σ) and is locally

a null surface such that I+(Σ) is on the future

side of ∂I+(Σ). Null generators of ∂I+(Σ)

cannot leave it (same reason as before).

Statement 4: If a compact spacelike 3-surface

Σ without boundary, with the normal vector

n such that divn < 0 on Σ, and if the SEC

holds, and if the spacetime has no closed time-

like curves (strongly causal), then there is at

least one incomplete timelike geodesic emitted

from Σ.

Proof: Since Σ is compact, divn < −θ0 < 0

for some θ0 everywhere. By the focusing theo-

rem and extremal theorem, no geodesic emit-

ted from Σ that is longer than 3θ−10 maxi-

mizes length. Let us show that, for every point

p ∈ I+(Σ), the set of causal geodesics be-

tween Σ and p is compact. The intersection

124

I−(p) ∩Σ is a subset of a compact set Σ and

is thus bounded. It remains to show that it is

closed. Any sequence in I−(p)∩Σ corresponds

to a sequence of tangent vectors in the past

lightcone at p and thus has an accumulation

point, a tangent vectorv. If the geodesic γ

emitted from p in the direction v does not in-

tersect Σ, then a neighborhood around γ also

does not intersect Σ; this is impossible.

Hence for each p there exists a timelike geodesic

γ(p)max of maximum length, among all geodesics.

But no other causal curve between Σ and p

can be longer than 3θ−10 ; this is so because

any causal curve can be deformed to a smooth

causal curve of no smaller proper length, and

then, if it is not a geodesic, deformed further

to a curve of longer proper length. Hence,

a nongeodesic curve cannot maximize proper

length. The only remaining possibility is that

there are (nongeodesic) causal curves of ar-

bitrarily large proper length. However, this is

impossible because the set of all causal curves

between a compact Σ and p is compact (in the

C0 topology in the space of curves). Proper

length is upper semicontinuous as a functional

of piecewise differentiable curves, and hence

there exist a causal curve of finite maximum

length between Σ and p. This curve must be

a geodesic and thus is shorter than 3θ−10 .

Hence, no p ∈ I+(Σ) can be further than 3θ−10

from Σ. , so I+(Σ) ⊂ Σ ×[

0,3θ−10

]

. Hence

I+(Σ) is compact.

Now consider the Cauchy horizon ∂I+(Σ): it

is either compact or noncompact. If it is com-

pact, then it contains null generators that wind

around infinitely many times. Then ∃ closed

timelike curve. Hence ∂I+(Σ) is noncompact.

But I+(Σ) is compact; contradiction.

Remark: Hawking & Ellis give a longer proof

without requiring the no-CTC condition.

Lecture 27, 2008-02-01. Singular collapse

There are two families of null geodesics emit-ted normally from a spacelike 2-surface; “in-

wards” and “outwards”.Trapped 2-surface = both “inwards” and “out-wards” future null geodesics emitted normally

to the 2-surface have negative divergence.

Trapped surface theorem (Penrose): If a

time-orientable spacetime M has a noncom-pact Cauchy surface C and a compact, space-

like trapped 2-surface S, then there exists anincomplete null geodesic to the future of S.Proof: Consider ∂I+(S): it is generated bynull geodesics n emitted normally from S. SinceS is compact, divn < −θ0 < 0 on S. Suppose

that every null geodesic has length ∆τ ≥ 2θ−10

from S. By the focusing theorem and the ex-

tremal theorem for null geodesics, every pointp on a null geodesic after ∆τ = 2θ−1

0 will beinside I+(S). Hence, ∂I+(S) ⊂ S ×

[

0,2θ−10

]

and so ∂I+(S) is compact.Since M is time-orientable, ∃v : g(v,v) = 1.

Each orbit of v intersects ∂I+(S) and also Cexactly once. Hence orbits of v map 1-into-1 ∂I+(S) → C. So the image of ∂I+(S) is

a compact subset of C, hence its boundary isnonempty, but ∂∂I+(S) = ∅. Contradiction.

125

Topology change ⇒ closed timelike curves

Theorem (Geroch): If the spacetime domain

M between two compact, spacelike, bound-

less surfaces Σ1, Σ2 is compact and admits

no closed timelike curves, then Σ1∼= Σ2 (have

the same topology) and M ∼= Σ1 × [0,1].Proof: Choose a timelike, nonvanishing vec-

tor field v in M. Orbits of v from Σ1 either

end on Σ2 or never reach Σ2. If an orbit never

reaches Σ2, it winds around itself and can be

deformed to a CTC. Hence, all orbits reach

Σ2 in finite parameter time and provide a 1-

to-1 map Σ1 → Σ2. Parameter time can be

rescaled to [0,1].

Theorem (Hawking): If there is a tube-like

domain T connecting spacelike 3-surfaces Σ1

and Σ2 of different topology, and if ∂T con-

sists of a timelike surface and of Σ1,2, then

there is a closed timelike curve in T .

Proof: One can choose a nonvanishing time-

like vector field v in T such that v is tangent to

the timelike side boundary of T . If each orbit

of v from Σ1 ends on Σ2, then one would have

a 1-to-1 map Σ1 → Σ2, but they have different

topology. So there are some orbits that wind

around in T and never reach Σ2. These orbits

can be deformed into CTCs.

126

Holographic principle

Generalized 2nd law of thermodynamics:δS ≥ 0

where S =combined entropy of matter + 14 of

combined area of black hole horizons.

Susskind process: A system can be made into

a black hole with entropy ≤ πR2 by adding

more energy to it.

Spacelike entropy bound: If a system is bounded

by a sphere of radius R then its entropy is not

larger than πR2.

GSL = A black hole is the highest-entropy ob-

ject possible in a given spherical volume.

This suggests that the fundamental degrees of

freedom scale as area rather than as volume;

hence the name “holography”.

127

Spacelike bound is not generally covariant and/or

not generally valid.

Difficulties with generalizing the spacelike bound:

1. Not clear what is inside and what is outside

a sphere if the universe is closed.

2. In an expanding universe, obviously one

can find an arbitrarily large volume and the en-

tropy density is constant on large scales, hence

S ∝ A3/2

3. In a collapsing star, we may enclose it in

arbitrarily small (infalling) sphere and violate

the bound.

4. Choose a spacelike 3-surface (“surface of

constant time”) such that the boundary of the

system is a 2-surface of vanishing 2-area, then

area is arbitrarily small while entropy “enclosed

in it” remains finite. (The entropy bound, as

formulated, is noncovariant!)

These pitfalls are in some sense “kinematical”

since they do not involve a violation of the

Susskind process. Need a covariant formula-

tion of the entropy bound that avoids these

pitfalls.

128

Covariant entropy bound (Bousso)

A lightsheet L for a 2-surface Σ is a null sur-

face made by emitting null geodesics normally

to Σ such that divn ≤ 0 everywhere. (L needs

to be cut if the nonconvergence condition is

violated somewhere.)

Entropy on a lightsheet is the integral over

L of the 3-form dual to the entropy current

s. (Note: integration in the null direction is

independent of the choice of the affine param-

eter!)

Covariant entropy bound: The total entropy

on a lightsheet is smaller than 14 of the area of

the initial surface Σ.

Interpretation: This is the entropy visible from

Σ. This avoids the counterexamples above!

Spacelike projection theorem (Bousso): If

a system is enclosed in Σ and the lightsheet

L emitted from Σ has Σ as its only boundary,

then the entire entropy of the system is not

larger than the entropy on the worldsheet.

Proof: All matter within Σ will pass through

L; the entropy will not decrease until that time.

129

Lecture 28, 2008-02-06. Inflation is not

eternal to the past

Inflation = accelerated expansion of spacetime.

Example: de Sitter spacetime, g = dt2−e2Htd~x2

Expansion of spacetime = ∃ timelike geodesic

vector field v such that divv > 0 everywhere

throughout a spacetime domain between two

Cauchy surfaces. For de Sitter spacetime: v =

∂t. Compute divv = Trxg(∇xv,x) using or-

thonormal basis

v, e−Ht∂j

: since g(∂j, ∂j) =

−e2Ht, we get divv = g(∇vv,v)−∑3j=1 e

−2Ht

×g(∇∂jv, ∂j) = −3e−2Ht12v g(∂j, ∂j) = 3H.

Hence, in a general spacetime we estimate the

local Hubble rate as H ≡ 13divv ≈ −g(n,∇nv)

for any spacelike, normalized n (g(n,n) = −1)

Theorem (Borde & Vilenkin): If a spacetime

is everywhere expanding, i.e. there exists a

timelike geodesic field v such that −g(n,∇nv) ≥H0 > 0 for any spacelike vector n (assuming

g(v,v) = 1, g(n,n) = −1, g(v,n) = 0) every-

where to the past of a Cauchy surface Σ, then

there exists a timelike geodesic that is incom-

plete to the past of Σ.

130

Proof: Visually, assume that spacetime is filled

with particles move along worldlines that are

orbits of v. The incomplete timelike geodesic

will be constructed explicitly. It will be the

worldline of an observer not at rest with re-

spect to the comoving particles. One can choose

a geodesic vector field u such that g(u,u) = 1

and g(v,u) > 1 everywhere. Denote γ ≡ g(v,u)

(this is the gamma factor of special relativity,

γ = (1− ~vrel)−1/2, where ~vrel is the relative ve-

locity of observer w.r.t. comoving particles).

Then u = αv+βn for some constants α, β and

spacelike vector n orthogonal to v. Explicitly,

projecting u onto v⊥, we get

n =u− vg(u,v)

|u− vg(u,v)| =u− γv√

γ2 − 1≡ αu + βv.

The observer can measure ∇uγ by observing

relative velocities of comoving particles. Then

the observer estimates the local expansion fac-

tor as

H = −g(∇nv,n) = −g(∇αu+βvv, αu + βv)

= −α2g(∇uv,u) = −α2∇ug(v,u) = − ∇uγ

γ2 − 1.

Replace ∇uγ = ∂τγ and integrate H along the

observer’s worldline:∫ τ2

τ1Hdτ = −

∫ τ2

τ2

γ

γ2 − 1dt =

1

2lnγ + 1

γ − 1

∣

∣

∣

∣

∣

τ2

τ1

<1

2lnγ(τ2) + 1

γ(τ1)− 1

since γ > 1 and ln(...) > 0. On the other hand,∫ τ2τ1Hdτ ≥ H2(τ2 − τ1). Hence, there is a lower

bound on the value of τ1:

τ1 > τ2 −1

2H0lnγ(τ2) + 1

γ(τ2)− 1≡ τ1min.

Hence, the worldline u cannot be extended

to the past beyond τ1min and is incomplete.

(Note: orbits of v are complete, but all other

geodesics are incomplete.)

Conclusion: Either there exists a singularity

to the past, or inflation is not past-eternal.

Remarks: 1. The theorem is purely kinemat-

ical - no Einstein equations or energy condi-

tions are used.

2. In de Sitter space, either one uses global

coordinates where the universe is initially con-

tracting, or the worldline u reaches the “edge”

where the geodesic field v is not expanding any

more.

Superluminal travel

“Warp drive” (Alcubierre 1994):

g = dt2 − (dx− vsf(rs)dt)2 − dy2 − dz2,

vs(t) ≡dxs

dt, rs ≡

√

(x− xs(t))2 + y2 + z2,

f(r) ≈

1, |r| < R

0, |r| > R

The worldline γ ≡ τ, xs(τ),0,0 is timelike be-

cause dτ = dt on γ (no time dilation!).

Properties: 1. γ is always a geodesic. Proof:

Let u ≡ γ = ∂t + vs(t)∂x. Compute gu = dt,

hence g(∇uu,x) = 12(Lug)(u,x) = g(∇xu,u) =

0 since g(u,u) ≡ 1.

2. The function xs(t) is completely arbitrary,

which allows superluminal travel.

3. All energy conditions are violated. [Omitted

calculation: T(∂t, ∂t) ∝ −v2s f ′2]Negative energy is concentrated in the “skin”

of the bubble.

131

Superluminal travel needs negative energyK. D. Olum, Phys.Rev.Lett.81, p.3567, 1998; gr-qc/9805003

Generic superluminal travel: Assume that there

exists a null geodesic that travels further (in a

fixed, externally defined time) than any neigh-

bor null geodesic.

Formalize this construction. Let lightrays be

emitted orthogonally from an extrinsically flat

(made of spacelike geodesics), spacelike 2-surface

ΣA. The “externally defined time” is a space-

like 3-surface C intersected by the lightrays.

Let the “furthest-reaching” lightray be between

A ∈ ΣA and B ∈ C. Hence, there exists an

extrinsically flat 2-surface ΣB ⊂ C such that

B ∈ ΣB but no other point p ∈ ΣB can be

reached by any lightrays from ΣA. Then call

the path γ between A and B superluminal.

Theorem (K. Olum): The NEC is violated on

a superluminal path γ if Ric(γ, γ) < 0 at some

point on γ.

Proof: Consider a congruence n of geodesics

emitted normally from ΣA. There are no con-

jugate points to ΣA on γ (or else B is timelike-

reachable from A). Hence, there is no focus-

ing, and divn remains finite everywhere on γ.

132

Initially, divn = 0 on ΣA because there exists a

basis of (spacelike) geodesic vector fields ci ∈TΣA and we have g(n, ci) = 0 and g(∇cin, ci) =

∇cig(n, ci)− g(n,∇cici) = 0.

Raychaudhuri’s equation gives

∇ndivn = Ric(n,n)−Tr(σ⊥n σ⊥n )− 1

2(divn)2

Since Ric(n,n) < 0 somewhere on γ, we must

have divn < 0 at B.

Now we show that divn ≥ 0 at B. Define a

basis of connecting fields ci for n, and consider

ci at point B. The fields ci are not geodesic

at B; the orbits of ci must leave ΣB at B

towards the future of B, or else points near

B are in I+(ΣA). Hence, the deviation ∇cicimust point to the future, so g(n,∇cici) ≥ 0.

Also, γ is normal to ΣB at B (or else it can be

deformed to reach other points near B).

Also, g(ci,n) = 0 everywhere because ∇ng(ci,n) =

0 while g(ci,n) = 0 everywhere on ΣA.

To compute divn at B, consider g(∇cin, ci) =

−g(∇cici,n) at B. Since g(∇cici,n) ≥ 0, we

find divn =∑

i

g(∇cin,ci)

g(ci,ci)≥ 0. Contradiction.

Remarks: 1. The NEC is violated directly on

the path of the signal.

2. Singularities do not help prevent this.

133

Lecture 29, 2008-02-08. Conservation laws.

Komar’s global energy integral

Consider a stationary spacetime with a Killing

vector k. The derivative tensor Bk = 12dgk is

a 2-form, hence its divergence is a divergence-

free 1-form (conserved current):

divBk = Trxg(∇x∇xk, ·) = g(k, ·)[Why is it conserved? divBk = −∗d∗Bk, hence

(∗d∗)(∗d∗)Bk = 0.]

More generally: any rank 2 tensor ω satisfies

ωαβ;[µν]

= Rλβµνωαλ+Rλαµνω

λβ, hence a 2-form

ω satisfies ωαβ;αβ = 2Trx,yRic(x,y)ω(x,y) = 0.

The “conserved charge” Q is obtained by inte-

grating the flux of the conserved current q ≡g−1divBk = k through a 3-surface C:

Q ≡∫

Cg(q,n)d3V =

∫

Cg(k,n)d3V

Since k is a total divergence, rewrite this as

Q =

∫

Σ=∂C

(

∇αkβ)

nαvβd2Σ

Can write using forms:∫

C d ∗Bk =∫

∂C (∗Bk)

The conserved charge Q is an integral over an

infinitely large 2-surface enclosing the system.

134

Example: Schwarzschild spacetime, k = ∂t.

The redshift function is z = z(r) =√

1− 2Gmr .

The metric is g = z2dt2 − z−2dr2 − r2dΩ2

Integrate∫

ΣBk(v,n)d2Σ, where Σ is a sphere

of large radius R. Vectors normal to the sphere

are v ≡ z−1k, n = z∂r. Need to compute

Bk(v,n) = g(∇vk,n) = g(∇z−1kk, z∂r). We

know that the proper acceleration of station-

ary observers is ∇kk = −12g−1dz2. Hence

g(∇kk, ∂r) = −1

2∂rz

2 = −Gmr2

Integrating over the 2-sphere of radius R:

Q =

∫

Σ

(

−mr2

)

d2Σ = −4πGm.

This is the mass of the black hole, times (−4πG).

Now we show that this is not a coincidence:

the total energy (including gravity) of a sta-

tionary spacetime is E = − 14πGQ

135

Interpretation of Q as the energy

Statement: In the Newtonian limit, Q = −4πE,

where E is the total mass of the matter sources.

Proof: The property ∇x∇yk = R(x,k)y yields

g(k,x) = Tryg(∇y∇yk,x) = TryR(y,k,y,x) =

Ric(k,x) and hence Q =∫

CRic(k,n)d3V .

“Newtonian limit” = particles move along or-

bits of k, normal to the equal-time surface C;the energy-momentum tensor of matter sources

is Tµν = ρkµkν, where g(k,k) ≈ 1 and ρ is the

distribution of mass density.

By Einstein equations: Ric(k,n) = −8πG(T(k,n)−12(TrT)g(k,n)) ≈ −4πGρg(k,n) since TrT ≈ ρ.Hence Q =

∫

C(−4πG)ρd3V = −4πGE.

Remarks: 1. In the non-Newtonian case, we

get a combined energy of gravity and matter.

2. The energy (or Komar mass) is time-

independent.

3. Can define the total angular momentum

similarly, using the azimuthal Killing vector ∂φ4. If only have approximate Killing vectors at

infinity, can use the formula with the surface

integral and obtain conserved quantities

136

Conservation laws and energy-momentum

pseudotensors

Killing vectors are special and do not always

exist. But there is always invariance under dif-

feomorphisms.

An arbitrary diffeomorphism is generated by a

vector field v. The variation of the metric is

δg = Lvg. The variation of the action gives

δS =

∫

([Einµν + 8πGTµν]δgµν + qα;α)

√−gd4x

where Ein = Ric − 12Rg is the Einstein ten-

sor, and q is the vector field such that qα;α =

gαβδRαβ. (Explicitly, δRic = LvRic and q =

g−1divdgv.) Since the above is an identity,

while δgµν = −vµ;ν − vν;µ, so we have (using

Bianchi identity and conservation of Tµν) that

[Einµν+8πGTµν]δgµν = −2 ([Einµν + 8πGTµν]v

µ);ν

and so

(qν − 2[Einµν + 8πGTµν]vµ);ν = 0.

On-shell (Ein + 8πGT = 0) we have qα;α = 0,

i.e. a conserved current. Up to a total deriva-

tive, one can write

qν = 2[Einµν + 8πGTµν]vµ +

(

vλUλ[µν]

),µ

137

One can add an exact 1-form to qα such that

the result does not contain 2nd derivatives of

g (but contains derivatives of vα):

qν = qν +(

vβUβ[µν]

)

,µ, Uβ

[µν] = ...

Explicitly, one will have

qν = 8πGTµνvµ + τµνv

ν + σµνλvν;λ

for some (non-tensor) quantities τ and σ.

A “conserved density” is obtained as√−gqα

such that(√−gqα),α =

√−gqα;α = 0 on-shell.

Choose 4 different vectors v0,v1,v2,v3 and

obtain a “pseudotensor”. For example, in a co-

ordinate system xα choose v0 = 1,0,0,0,v1 = 0,1,0,0 etc., i.e. v

µα = δ

µα (noncovariant

definition!). Then one obtains four conserved

currents q0, ..., q3. Since the numerically cho-

sen components of vα are constant, we have,

for example,

[q0]ν = 8πGT0ν + τ0ν

and so on; hence

(8πGTµν + τµν),µ = 0

so 18πGτµν is the “energy-momentum pseudoten-

sor for gravity” such that the combined EMT

of gravity and matter is (non-covariantly) con-

served.

In this way, choosing different vectors vα and

different coordinate systems xα, various “energy-

momentum pseudotensors” can be constructed.

These pseudotensors are not tensors for the

following reasons:

1. They are defined through vectors vα given

by numerical components in a fixed coordinate

system.

2. They are collections of four conserved vec-

tors, rather than single conserved rank 2 ob-

jects.