Lecture Notes for General Relativity Fall...

Lecture Notes for General Relativity

Fall 2010

Lecturer: Professor Lam HuiTranscriber: Alexander Chen

December 9, 2010

Contents

1 Lecture 1 41.1 Basic Information . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2 Newtonian Gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 General Relativity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.4 Applications of General Relativity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.5 Special Relativity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.6 Some Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 Lecture 2 62.1 Transformations the Preserves Distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.2 A Note about Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.3 Lorentz Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.4 Integration of One-form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

3 Lecture 3 83.1 Review of Last Lecture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83.2 Tensor Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

4 Lecture 4 104.1 Some note about Λ Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104.2 More Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104.3 Particle Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104.4 Electrodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

5 Lecture 5 125.1 Continue Electrodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125.2 Fluid Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1

6 Lecture 6 156.1 Einstein’s Happiest Thought . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156.2 Equivalence Principles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166.3 Some Remarks on Metric . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

7 Lecture 7 187.1 The Geodesic Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187.2 Altenative Derivation Using Action Principle . . . . . . . . . . . . . . . . . . . . . . . . . . 19

8 Lecture 8 208.1 An Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

9 Lecture 9 229.1 Newtonian Limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229.2 Gravitational Redshift . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239.3 More Formalism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

10 Lecture 10 2510.1 Another Look at the Geodesic Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2510.2 Digression on Parallel Transport . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

11 Lecture 11 2711.1 Principle of General Covariance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2711.2 Riemann Curvature Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

12 Lecture 12 3012.1 Alternative Way of Looking at the Riemann Tensor . . . . . . . . . . . . . . . . . . . . . . . 3012.2 The Einstein’s Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

13 Lecture 13 3313.1 Return to Einstein’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

14 Lecture 14 3514.1 Linear Perturbation Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3514.2 Newtonian Limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

15 Lecture 15 3715.1 Summary of Last Lecture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3715.2 Study of the Newtonian Metric . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

16 Lecture 16 4016.1 Gravitational Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

17 Lecture 17 4317.1 Particles in Gravitational Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4317.2 Solution to the Nonlinear Einstein Equations . . . . . . . . . . . . . . . . . . . . . . . . . . 44

2

18 Lecture 18 4618.1 The Schwartzschild Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4618.2 The Orbit of Mercury . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

19 Lecture 19 5019.1 Perihelion Precession of Mercury . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5019.2 Gravitational Redshift . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

20 Lecture 20 5320.1 Going Across the Horizon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

21 Lecture 21 5621.1 Penrose Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

22 Lecture 22 5922.1 Metric for a Homogeneous, Isotropic and Expanding Universe . . . . . . . . . . . . . . . . . 59

23 Lecture 23 62

24 Lecture 24 66

25 Lecture 25 70

26 Lecture 26 7226.1 Time Travel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

3

General Relativity Lecture 1

1 Lecture 1

1.1 Basic Information

Course Instructor: Lam HuiOffice Hour: Tue 4:30 - 5:30 pmTextbook: Gravitation and Cosmology by Steven Weiberg & Spacetime and Geometry by Sean Carroll

1.2 Newtonian Gravity

The general theory of relativity is a theory of gravity. We want to contrast it to Newtonian gravity, whichis characterized by the following two aspects:

1. The equation for the gravitational potential:

∇2ϕ = 4πGρ (1.1)

where ϕ is the gravitational potential, G the Newton constant, and ρ the matter density.2. The equation of motion for a test particle in the gravitational field:

F = −m∇ϕ = ma (1.2)

where an important assumption is made that the gravitational mass is equal to the inertial mass.

1.3 General Relativity

The structure of the theory of general relativity is similar to the Newtonian one. We also have two aspects:1. The field equation:

Gµν = 8πGTµν (1.3)

which is the relation between the metric (describing the space-time geometry) and the energy and momen-tum content in the space-time.

2. The equation of motion of a test particle, which is the geodesic equation.

1.4 Applications of General Relativity

There are many applications of general relativity, for example:

1. Gravitational lensing

2. Cosmology

3. Black holes

4. Gravitational waves

5. Hawking Radiation

6. Time-machines

4


1.5 Special Relativity

We shall begin our discussion of general relativity by reviewing special relativity, and develope the languagewe use in discussing the general theory. First we consider 2D Euclidean space. Let us denote the distancebetween two arbitrary points as ∆s. The distance can be calculated by Pythagoras theorem:

∆s2 = ∆x2 + ∆y2 (1.4)

If we rotate the coordinates, the same expression holds for distance. Distance between two points isinvariant under rotation.

Now we come to 3+1 dimensional space-time. Here the analog of a coordinate transformation likerotation is a change of reference frame. The points in the space-time are not physical points but ratherevents. The space-time distance between two arbitrary points is:

∆s2 = −(c∆t)2 + ∆x2 + ∆y2 + ∆z2 (1.5)

which is unchanged under a “rotation” in the 3+1 dimensional space-time. From now on we will take c = 1for notational brevity.

Let us look for the physical meaning of the space-time distance ∆s. Let us define ∆τ2 = −∆s2.Suppose we can find a frame of reference such that |∆x|2 = 0, i.e. the observer finds the events A and Bto happen at the same place, then ∆τ is just the proper time measured in this rest frame. Such a framecan be found when ∆τ2 > 0, or ∆s2 < 0. Such separation in space-time is called time-like.

Suppose the opposite. If we have ∆τ2 < 0 and ∆s2 > 0 then we can find a frame in which A and Bhappens at the same time. Such separation is called space-like because now ∆s is the physical distancemeasured in this frame.

What if ∆τ2 = ∆s2 = 0? Then we say that the separation is null or light-like, because it can only bealong the trajectory of a light beam.

1.6 Some Notations

We will denote a vector in 3+1 space-time as xµ where index µ runs from 0 to 3. We set x0 = t. We willdenote a vector in 3D Euclidean space as xi where the index runs from 1 to 3. We require xi = (x, y, z).

We write the space-time distance in Einstein’s notation:

∆s2 = ηµν∆xµ∆xν (1.6)

where summation is assumed at repeated upper and lower indices. The quantity ηµν is a 4× 4 matrix with16 entries in general. But here η = diag(−1, 1, 1, 1). This is called the Minkowski metric.

To Be Continued. . .

5


2 Lecture 2

2.1 Transformations the Preserves Distance

Now we ask the question: what coordinate transformation will preserve the distance between two arbitrarypoints? First we consider the case of R3, the 3 dimensional Euclidean space. Remember we defined thedistance in this space to be δs2 = δijδx

iδxj . If we have a coordinate transformation x → x′, we want topreserve this distance:

δs2 = δijδxiδxj = δijδx

′iδx′j (2.1)

so what transformations are allowed?

• Translation. Obviously translation will not modify the relative coordinates of any two points. ∆xi =∆x′i

• Rotation. To see this, let’s imagine a transformation x′i = Rijxj where R is a 3 × 3 matrix. We

demand that:δij∆x

′i∆x′j = δijRijR

jl∆x

k∆xl = δkl∆xk∆xl (2.2)

i.e. we require that RTR = I, which means that R ∈ O(3). Here we just need the part which isconnected to identity, therefore we take R ∈ SO(3).

Now we pass to Minkowski space, R3,1. Similarly we demand:

δs2 = ηµνδxµδxν = ηµνδx

′µδx′ν (2.3)

We write the transformation as x′µ = Λµνxν , and ask questions about Λ.

• Translations are still allowed.

• We now need the transformation to satisfy ηµνΛµσΛνρ = ησρ, i.e. ΛT ηΛ = η. This is the generalizationof the defining relation for orthogonal group in Euclidean space. We say that Λ ∈ SO(3, 1). Againwe have specialized to the part connected to identity.

Note that in our argument we have assumed that the coordinate transformation is linear. This isbecause we require the distance function to remain second order, which is in general not the case if thetransformation is nonlinear.

2.2 A Note about Notation

When we write Aµν = B αµ Cαν then in matrix form it is just A = BC. However if we write Aµν = Bα

µCαν ,

then the corresponding matrix form is A = BTC. We contract neighboring indices in matrix multiplicationand a transpose is needed if non-neighboring indices need to be contracted.

6


2.3 Lorentz Transformations

Let us do a little counting first. Because the matrix Λ is a 4×4 matrix, it has 16 entries. However we havea matrix equation ΛT ηΛ = η which constrains the value of Λ. Because η is a symmetric matrix, there are10 independent constraint equations. Therefore there are 16− 10 = 6 degrees of freedom.

Among those 6 degrees of freedom, 3 are just rotations in 3D space. Now we want to show that therest 3 are Lorentz boosts in the 3 directions. Consider the transformation in x and t direction (discard theother 2 coordinates first): (

Λ00 Λ0

1

Λ10 Λ1

1

)T (−1 00 1

)(Λ0

0 Λ01

Λ10 Λ1

1

)=

(−1 00 1

)(2.4)

Writing out the equations for individual components, we get the equations:−(Λ0

0)2 + (Λ10)2 = −1

−Λ00Λ0

1 + Λ10Λ1

1 = 0−(Λ0

1)2 + (Λ11)2 = 1

Solving these equations, we can get:

Λ =

γ −vγ 0 0−vγ γ 0 0

0 0 1 00 0 0 1

=

cosh ξ − sinh ξ 0 0− sinh ξ cosh ξ 0 0

0 0 1 00 0 0 1

(2.5)

where γ =1√

1− v2is the well-known γ factor in special relativity, and the ξ, the rapidity, is defined by

cosh ξ = γ and tanh ξ = v.

2.4 Integration of One-form

Suppose we have a particle propagating in the spacetime and traces out a time-like worldline from A toB. We want to calculate the total ∆s from A to B that was traversed by the particle, i.e. we should writethe differential form of the distance ds2 = ηµνdx

µdxν and integrate it with respect to a parameter, say t:

∆s =

∫ B

A

√ηµν

dxµ

dt

dxν

dtdt =

∫ B

A

√−1 + |v|2dt (2.6)

However for a time-like worldline the above integration usually yields an imaginary number, becausethe speed is less than 1. In this case we should instead integrate the proper time dτ :

∆τ =

∫ B

A

√−ηµν

dxµ

dt

dxν

dt=

∫ B

A

√1− |v|2dt (2.7)

which makes a lot more sense for a usual particle.

7


3 Lecture 3

3.1 Review of Last Lecture

Recall from last time, we denote a Lorentz transformation as x′µ = Λµαxα. We showed last time that theLorentz transformations are those that preserve length in the Minkowski space R3,1. Which means:

ds2 = ηµνdxµdxν = ηαβdx

′αdx′β (3.1)

And the equality obeyed by the Lorentz transformation matrices is:

ηαβ = ηµνΛµαΛνβ, η = ΛT ηΛ (3.2)

the former is in component form and the latter is in matrix form.

3.2 Tensor Notations

Some definitions first:Vector: is an object which transforms like V ′µ = ΛµαV α. For example, dxµ is a vector, the 4-velocity

uµ = dxµ

dτ is also a vector because dτ is a scalar under Lorentz transformations.1-Form: Recall ds2 = ηµνdx

µdxν . We can group the first two terms on RHS and call it dxµ = ηµνxν .

This is called a 1-form. More accurately, a 1-form is an object that contracts with a vector to form ascalar. Now we can write ds2 = dxµdx

µ.As an example, let’s consider 1-forms in R3,1. The expression is xµ = ηµνx

ν , and we can evaluate thecomponents:

dx0 = −dx0, dx1 = dx1, dx2 = dx2, dx3 = dx3 (3.3)

In the previous definiton we say that the index of xν is lowered by the metrix ηµν . We can similarlyget a vector from a 1-form by applying the inverse of the metric:

dxν = ηνµdxµ (3.4)

note that ηµν are components of η−1, instead of η itself. Somewhat confusing notation, but it makes sensebecause ηµνηνα = δµα. The rule of thumb is: use metric to lower indices and use its inverse to raise indices.

Let’s consider the object Λ νµ = ηνβηµαΛαβ, obtained by raising and lowering appropriate indices of the

Lorentz transformation matrix. Written in matrix form, we call this object Λ, and it is the inverse of thetransposed Lorentz transformation, ΛT Λ = 1. Written in components form, this means ΛµαΛ ν

µ = δ να .

Now we are finally in a position to define a 1-form like we defined the vector. Let’s consider howa 1-form transforms under Lorentz transformations. Recall that dxµ = ηµνdx

ν , and we know how dxν

transforms. We require that dx′µ = ηµνdx′ν in the new frame. Working out the mathematics:

dx′µ = ηµνΛναdxα = ηµνΛναη

αβdxβ = Λ βµ dxβ (3.5)

1-Form: is an object which transforms like dx′µ = Λ νµ dxν . This definition is similar to our definition

of a vector. This definition is good because if we take any vector vµ and any 1-form uν and contract them,the result s = vµuµ is a scalar:

s′ = v′µu′µ = ΛµνvνΛ ρ

µ uρ = δ ρν v

νuρ = vνuν = s (3.6)

8


This is the main reason to introduce these cumbersome notations.

Question: Is∂ϕ

∂xµa vector or 1-form? We need to look at its transformation properties:

∂ϕ

∂x′µ=∂xα

∂x′µ∂ϕ

∂xα(3.7)

Let’s consider the coordinate transformation x′µ = Λµνxν , so we know∂xν

∂x′µ= Λ ν

µ . Therefore the

directional derivative is a 1-form, so we write it as ∂µϕ.

9


4 Lecture 4

4.1 Some note about Λ Matrices

Remember Λµν is the components of the Lorentz transformation matrix Λ, and we have x′µ = Λµνxν . Thetransformation for 1-forms is Λ ν

µ which we denote Λ in matrix form. Remember we said ΛT Λ = 1. Butthe transpose is very confusing. It is better just use index notation throughout. The transpose is onlynecessary because we are contracting the first indices from both matrices.

4.2 More Notations

Recall:

• A vector transforms as V ′µ = ΛµνV ν

• A 1-form transforms as W ′µ = Λ νµ Wν

Last time we found that ∂ϕ/∂xµ is a 1-form, which we denote ∂µϕ.Now let’s define some more interesting objects. Given two positive integers, we can form a (m,n) tensor

whose components are T(m upper indices)(n lower indices) . We will be interested, just like the case in vectors and 1-forms, in

the transformation properties of these tensors. The metric tensor gµν is a (0, 2) tensor.The transformation of a tensor, say, of the form (2, 1), then it looks like Tµνρ. It transforms like:

T ′µνρ = ΛµαΛνβΛ γρ Tαβγ (4.1)

Basically we can play around with the metric and inverse metric tensors to raise and lower indiceswhatever we want.

4.3 Particle Mechanics

We define the four-velocity of a particle to be:

Uµ =dxµ

dτ=

(dt

dτ,dxi

dτ

)=(γ, γvi

)(4.2)

where vi is the actual velocity in the 3-space. The 4-velocity rotates nicer in Lorentz transformations,whereas the 3-velocity will get mixed up with the time component.

We can similarly define 4-momentum of a particle as:

Pµ = mUµ =(γm, γmvi

) nonrelativistic limit−−−−−−−−−−−−→(m+

1

2mv2,mvi

)(4.3)

where m is the rest mass of the particle. Note that the 0-component of the 4-momentum is the energy ofthe particle E = P 0 = γm.

Let’s construct a Lorentz invariant out of the 4-momentum by contracting it with itself:

PµPµ = m2UµUµ = m2

(dxµ

dτ

dxµdτ

)= −m2 (4.4)

If we translate the equation a little bit, and expand the summation, we can see that:

10


E2 = p2 +m2 (4.5)

which is the famous formula from special relativity. Note that the rest mass is frame-independent, thereforea Lorentz invariant. This is ok for a massive particle, like the electron, but for a massless particle the 4-velocity is not well defined this way. For a massless particle, τ is no longer a good parametrization of theworldline, as it is zero everywhere. We then use a parameter λ called the affine parameter, and defineUµ = dxµ/dλ. The magnitude of the 4-velocity is just zero because:

UµUµ =dxµ

dλ

dxµdλ

= 0 (4.6)

If we define the momentum as Pµ = Uµ then we also have PµPµ = 0, and E2 = p2.

4.4 Electrodynamics

Recall the Maxwell equations:

Sourceful

∇ ·E = ρ

∇×B− ∂E∂t = J

(4.7)

Sourceless

∇×E + ∂B

∂t = 0

∇ ·B = 0(4.8)

Claim: ρ and J form a 4-vector Jµ (we will not prove). We have an equation expressing chargeconservation:

∂µJµ =

∂ρ

∂t+∇ · J = 0 (4.9)

which is just the continuity equation which we are familiar with in classical electrodynamics. Note thatwritten in this form, the equation is manifestly independent of frames, i.e. Lorentz invariant. Now we wantto consider electromagnetic field. Let us look at the first Maxwell equation: ∂iE

i = ρ. Because we knowthat ρ is part of a 1-index object, and the differential operator is a 1-form, so we expect the electrodynamicfield to have 2 indices, i.e. a rank-2 tensor. However, a rank-2 tensor has 16 degrees of freedom which isway too many if we only want electromagnetic field, which has only 6 degrees of freedom. A simple solution(and correct) is to require the rank-2 tensor to be antisymmetric, which reduces the degree of freedom to6. Now we have enough motivation to define the electromagnetic tensor:

Fµν =

0 E1 E2 E3

−E1 0 B3 −B2

−E2 −B3 0 B1

−E3 B2 −B1 0

(4.10)

Check that Fµν transforms correctly as rank-2 tensor, and that the two sourceful Maxwell equationswill become:

∂µFµν = Jν (4.11)

11


5 Lecture 5

5.1 Continue Electrodynamics

Remember last time we introduced the Maxwell equations:

Sourceful

∇ ·E = ρ

∇×B− ∂E

∂t= J

(5.1)

Sourceless

∇×E +∂B

∂t= 0

∇ ·B = 0(5.2)

and we said that the sourceful equations can be greatly simplified by the introduction of the field strengthtensor:

∂µFµν = Jν (5.3)

The advantage of writing equations in this form is that we can easily know what are the values ofcorresponding physical variables in another frame (just do Lorentz transformations), and that the form ofequations in different frames remains the same.

In order to write down the other two Maxwell equations we need to define another tensor called thedual of the field strength tensor:

Fµν =1

2εµναβFαβ (5.4)

The εµναβ is called the Lavi-Civita symbol in 4 dimensions, and is defined by the following

εµναβ =

−1 , if (µναβ) is an even permutation of (0123)

1 , if (µναβ) is an odd permutation of (0123)

0 , otherwise

(5.5)

Note that εµναβ = −εµναβ . And both objects are rank 4 tensors under Lorentz transformations (checkedin homework).

With the aid of the dual tensor, we can simplify the last two Maxwell equations into

∂µFµν = 0 (5.6)

To get a flavor of what exactly is the dual field tensor, let’s evaluate one entry, say F 01

F 01 =1

2

(ε0123F23 + ε0132F32

)= −F23 (5.7)

If we carry out all the calculations we can find the dual tensor to have the following value:

Fµν =

0 −B1 −B2 −B3

B1 0 E3 −E2

B2 −E3 0 E1

B3 E2 −E1 0

(5.8)

12


The familiar Lorentz force law for a charged particle moving inside a electromagnetic field can besimplified with the introduction of field tensors, too:

dPµ

dτ= qgνσF

µν dxσ

dτ(5.9)

with a little bit of algebra we can see this gives the usual form of Lorentz force law. Note that the zerothcomponent of this equation gives the rate of change of energy of the charged particle.

5.2 Fluid Dynamics

Suppose we have a bunch of particles, and we want to describe the physics of this collection. We havesome variables of particular interest. First is the number density n, defined as the number of particles perunit volume in the rest frame of the collection. We can also introduce a collective velocity of the particlesas Uµ. Using these two variables we can introduce the number current (or number flux, as the dimensionis number per area per time):

Nµ = nUµ (5.10)

In particular N0 = γn gives the number density in the boosted frame. Very similar to the idea of 4-currentin Electrodynamics.

How about energy density? The energy density can’t be a scalar, because it is not invariant underLorentz transform. When it transforms under a boost, it gets two γ factors, therefore we expect it to bepart of a rank 2 tensor. This object is usually denoted Tµν and is called the energy-momentum tensor.

Let’s consider an example. Consider a cloud of dust, i.e. a collection of non-interacting massiveparticles. The energy-momentum tensor of such a system is:

Tµν = PµNν = mnUµUν (5.11)

The coefficient mn is nothing other than the restframe energy density (if we think of mass as energy). Ifwe consider T 00 = mnγ2 is just the energy density in lab frame. How about other components? ConsiderT 0i:

T 0i = P 0N i = mnγ2vi (5.12)

which has the meaning of energy flux in lab frame. Similarly T i0 = P iN0 which can be interpreted as themomentum density. Because by definiton T is symmetric, the energy flux equals the momentum density.Finally let’s look at T ij :

T ij = P iN j (5.13)

which has the dimension of momentum per unit area per unit time, which can be called “pressure”. Notethat this is a more general form of pressure, but it is not isotropic, so we usually call it stess. Because ofthis, the energy-momentum tensor is also sometimes called the stress tensor.

Let’s now study the energy-momentum tensor of a perfect fluid. In local fluid rest frame, the tensor is

Tµν = diag(ρ, P, P, P ) (5.14)

where ρ is the energy density at the local coordinate and P is the physical pressure. The dust exampleabove is a special case of a perfect fluid, where local pressure is zero.

The above energy-momentum tensor is written in the local rest frame. We want to find the expressionin some other frame. Recall that dust has Tµν = ρUµUν , but we want one which is more general than

13


this. Let’s make a guess Tµν = (ρ+ P )UµUν , but it doesn’t work. To correct this we add a term to givethe right P dependence:

Tµν = (ρ+ P )UµUν + Pηµν (5.15)

Because this is a tensor by definiton, and it is right in the local rest frame, so we can conclude that this isthe right definition for all frames. It is instructional to work out the explicit components of this tensor:

Tµν =

((ρ+ P )γ2 − P (ρ+ P )γ2vi

(ρ+ P )γ2vj (ρ+ P )γ2vivj + Pδij

)(5.16)

In general Tµν is symmetricTµν = T νµ (5.17)

This is true by our definition above, but there is physical reason behing it. Ref. Schutz Chapter 4.5.In electrodynamics we have charge conservation. Here we have energy-momentum conservation con-

straints on the energy-momentum tensor:∂µT

µν = 0 (5.18)

We want to plug the above tensor for perfect fluid into this equation. First we take the nonrelativisticlimit γ ≈ 1 and P ρ. The resulting equations are:

∂ρ

∂t+ ∂i(ρv

j) = 0 (5.19)

∂

∂t(ρvj) + ∂i

(ρvivj + Pδij

)= 0 (5.20)

=⇒ ∂vj

∂t+ vi∂iv

j = −∂jP

ρ(5.21)

14


6 Lecture 6

Starting from this lecture we will be talking about General Relativity.

6.1 Einstein’s Happiest Thought

Galileo observed that the inertial mass is equal to the gravitational mass, i.e. different objects fall at thesame acceleration under gravitation. The inertial mass is the mass which shows up in Newton’s secondlaw, which relates force with acceleration. The gravitational mass, however, is of totally different origin.It is the coefficient of the gravitational field which shows up in Newton’s law of gravitation:

F = GMm

r2(6.1)

These two quantities of totally different origin turns out to be the same, so all bodies fall at the samerate under gravity. Now what is the implication of this assertion?

1. Suppose somebody is trapped in a free falling elevator. He feels weightless because he falls with theelevator at the same rate, i.e. as if there is no gravity.

2. Suppose somebody is trapped in a elevator. There is no gravitational field around, but the elevator isbeing accelerated upwards. Because there is force on your feet, it feels like that there is gravitationalforce acting on you. This means that an accelerating frame acts as though this is gravity.

Einstein pushed this thought to an extreme, saying that there is absolutely no difference between anaccelerating frame and gravitation. Before we examine his thought, let’s firstly contrast gravity with E&M.Suppose we have 2 test particles with masses m1, m2, and charges q1, q2. Say the particles are on thesurface of a planet of mass M . The motion of the two particles in the gravitational field are the same:

d2r1

dt2=d2r2

dt2=GM

R2(6.2)

Now if we go to a frame r′ = r +GMt2/2R2, then in that frame we have d2r′/dt2 = 0 for both particles.This is the frame where gravitational force does not exist. Note that in this local frame all objects do notfeel gravity. The crucial thing is that in one transformation we can eliminate gravity from all test particles.

How about E&M? The motion of the test particles on a planet with charge Q (suppose such planetexists), then their equations of motion are:

d2r1

dt2= − Q

R2

q1

m1,

d2r2

dt2= − Q

R2

q2

m2(6.3)

But the charge-to-mass ratio is not a universal constant in the universe, so different things fall in differentrate, and we can’t find one single frame where electric forces go away.

As shown in the above example, we want to consider general coordinate transformations for gravity,i.e. x′ = x′(x), which may be highly nonlinear, unlike the Lorentz transformations. This seems necessaryto describe gravity in corresponding frames. Also we can only transform away gravity locally, i.e. there isin general no global transformation to eliminate gravity in all space. This local gravity-free frame is calleda local inertial frame.

15


6.2 Equivalence Principles

We are now ready to introduce the equivalence principle. There are several versions of this principle:

1. (Weak Equivalence Principle) At every space-time point in some gravitational field, it is possible tochoose a locally inertial coordinate system where test particles move as if there is no gravity.

2. (Einstein Equivalence Principle) At every space-time point in some gravitational field, it is possibleto choose a locally inertial coordinate system where the laws of physics takes the same form as inan unaccelerated frame in the absence of gravity (i.e. special relativity). (This is much more generalthan the above, because it covers all laws of physics, including E&M, QM, etc.)

3. Strong Equivalence Principle The statement that inertial mass equals gravitational mass includes allpossible contribution to mass, including gravitational binding energy itself.

The last statement is very hard to test, as in usual objects the gravitational binding energy is extremelyweak. The only objects in which gravitational contribution to mass is significant are black holes and neutronstars.

In order to consider general coordinate transformations, we need to go beyond Minkowski metric, andconsider

ds2 = gµν(x)dxµdxν (6.4)

where gµν is in general dependent on spacetime coordinate and gµν 6= ηµν . gµν will be our gravitationalfield, which is defined to be symmetric, as the antisymmetric part does not have any physical meaning. Aswe talked about in the first lecture, we will consider two kinds of questions:

1. How does mass source gµν?

2. Given gµν , how do test particles move?

The second question is easier to answer, since we can always use the equivalence principle to locallyeliminate the gravitational force, so that particles will only move in a straight line. But the questionbecomes how to cleverly choose such a local inertial frame.

6.3 Some Remarks on Metric

Remember we defined ds2 = gµνdxµdxν to be the invariant interval under any coordinate transformation.

That is to say, the meric transforms as:

g′µν =∂xα

∂x′µ∂xβ

∂x′νgαβ (6.5)

In special relativity the partial derivatives are just Λ matrices, because the transformation was linear. Nowwe consider more general transformations. Note that in special relativity the Minkowski metric is invariantunder Lorentz transformations, this is just a special case of the above transformation law.

It is useful to remember some simple examples which builds our intruition. For example we have thecartisian coordinates in R2:

ds2 = dx2 + dy2, gµν = diag(1, 1) (6.6)

We can also consider R2 in a totally different frame, i.e. the polar coordinates, where

ds2 = dr2 + r2dθ2, g′µν = diag(1, r2) (6.7)

16


We have the transformation law from cartesian coordinates to polar coordinates:

g′mn = gij∂xi

∂x′m∂xj

∂x′n(6.8)

Check that this transformation works, yielding the explicit coordinates we wrote down above.Another example is S2, which is embedded in R3. Consider first the inherited metric from R3 which is

flat. By the relation x2 + y2 + z2 = R2, we have

z = ±√R2 − x2 − y2, dz = ∓

(xdx√

R2 − x2 − y2+

ydy√R2 − x2 − y2

)(6.9)

so we can get the metric to be:

ds2 =R2 − y2

R2 − x2 − y2dx2 +

R2 − x2

R2 − x2 − y2dy2 +

2xy

R2 − x2 − y2dxdy (6.10)

We can also use spherical coordinates x = R sin r

R cos θ

y = R sin rR sin θ

z = R cos rR

(6.11)

and we can obtainds2 = dr2 +R2 sin2 r

Rdθ2 (6.12)

with the range of variables 0 ≤ θ < 2π and 0 ≤ r ≤ πR. This metric looks very like the polar coordinatesfor R2, but with different topology.

17


7 Lecture 7

7.1 The Geodesic Equation

The this lecture and next, we will attempt to replace the Newton’s second law by a generalized equationwhich describes the motion of a particle under gravitational field, which is just the geodesic equation. Theessence of the geodesic equation is that we don’t think of gravity as a “force” on the particle, but theparticle just falls freely in a curved space-time.

There are two ways that we can derive the geodesic equation, first is from the equivalence principle,and the next is from the principle of least action. We will do the former in this lecture.

Remember the equivalence principle which says that there always exists a local inertial coordinatesystem in which there is no gravity, i.e. space-time is flat

ds2 = gµνdxµdxν = ηµνdξ

µdξν (7.1)

where ξµ is the local coordinates where the space-time is flat. The statement that the particle feels noforce translates into

d2ξ

dτ2= 0 (7.2)

where τ is the affine parameter and dτ2 = ds2. This equation is written in the local inertial frame, so it isnatural to ask what does it look like in another frame? Let’s do a coordinate transformation:

d

dτ

dξα

dτ=

d

dτ

[∂ξα

∂xµdxµ

dτ

]= 0 (7.3)

∂ξα

∂xµd2xµ

dτ2+

∂2ξα

∂xν∂xµdxν

dτ

dxν

dτ= 0 (7.4)

We multiply the second equation by a term ∂xλ/∂ξα, we can write it in a new and illuminating way

d2xλ

dτ2+ Γλµν

dxµ

dτ

dxν

dτ= 0 (7.5)

where

Γλµν =∂xλ

∂ξα∂2ξα

∂xµ∂xν(7.6)

is called the Christoffel symbol. From the definition it is evident that the µν indices are symmetric becausepartial derivatives are symmetric. However it is not a tensor although it has three indices, and we willreturn to this issue later. The equation (7.5) is called the geodesic equation, and it explicitly involves thegeometry of spacetime. Now we want to write the Christoffel symbol in terms of the metric gµν .

Remember we changed the coordinates from the original space-time to a flat inertial one, so

gµν = ηαβ∂ξα

∂xµ∂ξβ

∂xν(7.7)

we want to evaluate the derivative of metric, and identify terms in the Christoffel symbol

∂λgµν = ηαβ

(∂2ξα

∂ξλ∂xµ∂ξβ

∂xν+∂ξα

∂xµ∂2ξβ

∂xλ∂xν

)= ηαβ

(∂ξα

∂xγΓγλµ

∂ξβ

∂xν+∂ξα

∂xµ∂ξβ

∂xγΓγλν

)= gγνΓγλµ + gµγΓγλν

(7.8)

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Now we will massage the equation a little bit and create some different copies of it by changing aroundindicies, and add them up to isolate one copy of Γ:

Γλµν =1

2gλρ (∂µgρν + ∂νgρµ − ∂ρgµν) (7.9)

7.2 Altenative Derivation Using Action Principle

Remember the proper time defined by d2τ = −d2s is a manifestly invariant quantity under coordinatetransformation. If we want to construct an invariant action it is a good starting point. So let us define theaction to be

S =

∫dτ =

∫ √−gµν

dxµ

dλ

dxν

dλdλ (7.10)

where λ is an affine parameter of the motion of the particle. This action makes sense because a particleshould minimize the proper time it travels when it feels no force. We carry out the variation as follows (weuse dot for derivatives with respect to λ)

δS =

∫dλδ

√gµν xµxν =

∫dτδ(gµν x

µxν)

2√gµν xµxν

=

∫dλ

1

2√gµν xµxν

(∂σgµν xµxνδxσ + gµνδx

µxν + gµν xµδxν)

=

∫dλ

1

2√gµν xµxν

(∂σgµν xµxνδxσ − ∂σgµν xσxνδxµ − ∂σgµν xσxµδxν − 2gµν x

µδxν)

=

∫dλ

1

2√gµν xµxν

(∂σgµν xµxν − ∂µgσν xµxν − ∂νgµσxν xµ − 2gµσx

µ)δxσ

(7.11)

Because the variation is arbitrary, the term in the bracket must vanish, and we have our equation ofmotion:

d2xρ

dλ2+ Γρµν

dxµ

dλ

dxν

dλ= 0 (7.12)

where Γ is the same symbol defined above. It is the exactly the same equation we got above.

19


8 Lecture 8

Note that in our definition of a local inertial frame, we left out one condition, which is that the derivativeof the metric should also vanish, i.e.

gµν = ηµν , Γαµν = 0 (8.1)

8.1 An Example

Let us we write down a metric which describes a worm hole:

ds2 = −dt2 + dr2 + (b2 + r2)(dθ2 + sin2 θdφ2) (8.2)

This looks like a connected hole in the embeding diagram. b is the radius of the throat of the hole. Weneed to calculate the Christoffel symbols for this metric. The metric is just

gµν =

−1 0 0 00 1 0 00 0 (b2 + r2) 00 0 0 (b2 + r2) sin2 θ

(8.3)

So the inverse metric is just gµν = diag

(−1, 1,

1

b2 + r2,

1

(b2 + r2) sin2 θ

). Let’s calculate all the nonzero

Christoffel symbols and they are:

Γrθθ = −r, Γrφφ = −r sin2 θ, Γθrθ =r

b2 + r2(8.4)

Γθφφ = − sin θ cos θ, Γφrφ =r

b2 + r2, Γφφθ = cot θ (8.5)

We can plug in these Christoffel symbols into the geodesic equation and get the equation for components:

xt = t = 0 (8.6)

r + Γrθθθ2 + Γrφφφ

2 = r − rθ2 − r sin2 θφ2 = 0 (8.7)

θ + 2Γθrθθr + Γθφφ = θ +2r

b2 + r2rθ − sin θ cos θφ2 = 0 (8.8)

φ+2r

b2 + r2rφ+ 2 cot θθφ = 0 (8.9)

Consider an initial value problem where initially r = R, r = v, θ = φ = 0, and let the particle have aradial free-fall into the wormhole. The dot on the quantities are derivatives with respect to τ . The initialfour-velocity is just

Uµ =dxµ

dτ=(√

1 + v2, v, 0, 0)

(8.10)

In our geodesic equations if θ and φ start at zero, then θ = 0 and φ = 0 and these two quantities willnot change at all. Therefore the radial equation is simple, too: r = 0. The particle essentially moves withconstant speed on the r direction, and the (proper) time it takes to travel from r = R to r = −R is just

∆τ =2R

v(8.11)

20


Note we have some conserved quantities for our metric. Obviously the metric does not explicitly dependon time t and angle φ, however there is another symmetry. We will discuss this later by introducing Killingvectors. Let’s look at the Lagrangian again with the affine parameter equal to proper time, and L = dτ

dτ = 1.Writing out the Euler-Lagrangian equation we have:

d

dτ

(∂L∂xµ

)− ∂L∂xµ

= 0 (8.12)

We can look at the µ = 0 component, i.e. time component, first. Working out the equation we getimmediately t = 0. Carrying out all the calculations we can get all the equations of motion shown above.Note that ∂L/∂φ = 0, therefore if we define ∂L/∂φ = πφ, then by equation of motion we have πφ = 0. Sowe have a conserved quantity. This is a special case of the Noether Theorem.

21


9 Lecture 9

9.1 Newtonian Limit

Remember we had the geodesic equation

d2xµ

dτ2+ Γµαβ

dxα

dτ

dxβ

dτ= 0 (9.1)

We want to recover the Newton’s equation in the limit of weak field:

gµν = ηµν + hµν , |hµν | 1, ∂0hµν ∼ 0 (9.2)

Another assumption we make is that the motion is nonrelativistic, i.e. |dx/dτ | 1. Under this assumptionthe only contribution to the acceleration is just

d2xµ

dτ2+ Γµ00

dt

dτ

dt

dτ= 0 (9.3)

We need to evaluate the Christoffel symbol

Γµ00 =1

2gµλ [2∂0gλ0 − ∂λg00] = −1

2ηµλ∂λh00 (9.4)

We ignored the time derivatives because everything moves slowly and these terms are subdominate.Now we can write out the µ = i components of the geodesic equation, discarding the terms with v

which are subdominate and taking dt/dτ ≈ 1

d2xi

dτ2=

1

2

∂h00

∂xi(9.5)

This looks like the Newton’s law of motion. We identify

h00 = −2φ, g00 = −(1 + 2φ) (9.6)

where φ is identified as the gravitational potential. Note that |φ| 1 in most circumstances:

• On Earth’s surface φ ∼ 10−9

• On Sun’s surface φ ∼ 10−6

• On the surface of a White dwarf φ ∼ 10−4

• On the surface of a proton φ ∼ 10−29

• Near black holes we have φ ∼ O(1)

22


9.2 Gravitational Redshift

We have seen that the metric in the weak field limit is

ds2 = g00dt2 + 2g0idx

idt+ gijdxidxj (9.7)

and we have g00 = −(1+2φ) and ∂0g00 = 0. Under assumption of nonrelativistic speed, the only importantcontribution is from the first term.

Now let’s consider a wave propagating from an observer at point 1 to another observer at point 2.Suppose two successive wave peaks passes through point 1 and then passes through point 2. We choosethe coordinate separation to be ∆t1 = ∆t2, because the metric is static. The time differences recordedby the two observers are determined by the proper time they record in their frame. We can calculate theproper time from observer 1 as (under the nonrelativistic assumption)

τ1 =

∫ √−g00

dt

dλ

dt

dλdλ =

√−g00(1)∆t1 (9.8)

Similarly we have τ2 =√g00(2)∆t2. So we can calculate the ratio of the proper times

τ2

τ1=

√1 + 2φ1

1 + 2φ2∼ 1 + (φ2 − φ1) + o(φ2) (9.9)

Because we know that the frequency of the photon ν ∝ 1/τ , so we know that when

• φ1 > φ2 then ν1 > ν2 so we have redshift

• φ1 < φ2 then ν1 < ν2 so we have blueshift

9.3 More Formalism

Remember we could raise and lower indices in Special Relativity using the inverse metric and the metric.But now in GR we need to replace ηµν with gµν and the inverse metric no longer has the same componentsas the metric.

Remember also that in SR we defined the transformation property for vectors and 1-forms as

V ′µ = ΛµνVν , W ′µ = Λ ν

µ Wν (9.10)

where Λ are Lorentz transformations. Now in GR we need to replace Λµν = ∂x′µ/∂xν . The tensorstransform the same way, just replace the Lorentz transforms with derivatives.

Note that in SR ∂V µ/∂xν is a tensor, but this in general is not true in GR:

∂V ′µ

∂x′ν=∂xγ

∂x′ν∂

∂xγ

(∂x′µ

∂xβV β

)=∂xγ

∂x′ν∂x′µ

∂xβ∂V β

∂xγ+∂xγ

∂x′ν∂2x′µ

∂xβ∂xγV β

(9.11)

It is the second term that destroys the tensorial transformation property. We can rewrite the second termas

∂xγ

∂x′ν∂2x′µ

∂xλ∂xγ=∂xγ

∂x′ν∂x′µ

∂xβ∂2x′α

∂xλ∂xγ∂xβ

∂x′α=∂xγ

∂x′ν∂x′µ

∂xβΓβλγ (9.12)

23


This motivates the definition of a covariant derivative, which converts a tensor into another tensor

∇αV β = ∂αVβ + ΓβαγV

γ (9.13)

Let’s try to work out the covariant derivative of various objects:

• The covariant derivative of a scalar is just the derivative of that scalar ∇αφ = ∂αφ.

• ∇αωβ = ∂αωβ − Γµαβωµ.

• ∇ρ(Tµνλ) = ∂ρTµνλ + ΓνραT

µαλ + ΓµραT

ανλ − ΓαρλT

µνα

The way we defined the Christoffel symbol satisfies the metric compatibility condition which says that

∇αgµβ = 0 (9.14)

If we define the connection as not metric compatible, then there is extra degrees of freedom, and we canactually convert them into scalar fields.

Let’s check some consistency. First we need that ∇α(V βWβ) = ∂α(V βWβ). This can be checkeddirectly:

∇α(V βWβ) = (∇αV β)Wβ + V β(∇αWβ) (9.15)

We can check that the plus and minus terms with Christoffel symbols exactly cancel to yield only thederivative term. Another thing to check is that ∇2 yields the correct Laplacian in different coordinatesystems. For example in 2D we have

∇2φ =∂2φ

∂x2+∂2φ

∂y2=

1

r

∂

∂r

(r∂φ

∂r

)+

1

r2

∂2

∂φ2(9.16)

This is checked in the homework.Finally we calculate the divergence of V µ

∇µV µ = ∂µVµ + ΓµµαV

α

= ∂µVµ +

1√|g|∂α√|g|V α

=1√|g|∂µ(√|g|V µ)

(9.17)

The derivation involve using determinant trick.

24


10 Lecture 10

Remember the definition of a geodesic is the curve on which the integral∫dτ is extremized. Also remember

that the principle of equivalence tells us around any space-time point, we can always choose a coordinatesuch that a particle under no other force moves in a straight line with constant velocity. This frame iscalled local inertial frame, and is characterized by the condition

gµν(x) = ηµν , ∂αgµν(x) = 0 (10.1)

at that point x. Note that the second order derivatives of gµν do not vanish in the local inertial frame, sowe have curvature even in the inertial frame, but as long as we don’t go too far then the curvature doesn’treally matter as it is a second order effect.

These two principles both lead to the geodesic equation

d2xµ

dτ2+ Γµαβ

dxα

dτ

dxβ

dτ= 0 (10.2)

which is true in general coordinates. In locally coordinates this translates into

d2xµ

dτ2= 0 (10.3)

which is just a motion at constant velocity.Note that the Christoffel symbol is not a tensor, although it has three indices and looks like one.

We worked out in the homework the transformation property of this connection. However the geodesicequation is a tensor equation because the second derivative of xµ is not a tensor either. However, the partswhich do not behave nicely under transformation from these quantities cancel so that the sum transformsnicely.

10.1 Another Look at the Geodesic Equation

We want to write the geodesic equation in a new way to show manifestly that it is a good tensor equation.Let’s first consider the 4-velocity Uµ = dxµ/dt which is a good tensor under coordinate transformations.It obeys the transformation law

dx′µ

dτ=∂x′µ

∂xνdxν

dτ(10.4)

However xµ is not a good tensor because it doesn’t transform as above, whereas dxµ is a good tensor.Now recall that we defined the covariant derivative

∇αUµ = ∂αUµ + ΓµαβU

β (10.5)

which is a good tensor regardless of general coordinates, i.e.

∇′αU ′µ =∂xβ

∂x′α∂x′µ

∂xν∇βUν (10.6)

The geodesic equation can be readily written as:

Uα∇αUµ =dxα

dτ

(∂αdxµ

dτ+ Γµαβ

dxβ

dτ

)=d2xµ

dτ2+ Γµαβ

dxα

dτ

dxβ

dτ= 0 (10.7)

So in this tensor form we know manifestly that the geodesic equation is a good tensor equation, and it hasthe same form in all general coordinates.

25


10.2 Digression on Parallel Transport

Firstly let us define the concept of parallel transport. Consider a path x(λ) parametrized by some parameterλ. We want to define the transport of a vetor V µ along this path, and define a derivative D

DλVµ such that

• D

DλV µ =

dV µ

dλin a local inertial frame.

• D

DλV µ is a good tensor.

First note that dV µ/dλ is not a good tensor because

dV ′µ

dλ=

d

dλ

(∂x′µ

∂xνV ν

)6= ∂x′µ

∂xνd

dλV ν (10.8)

Now we claim that the definitionDV µ

Dλ=dxα

dλ∇αV µ (10.9)

satisfies the two conditions above. This is easy to check because for the first requirement, in local inertialframe the covariant derivative just reduces to ordinary derivative. And for second requirement the covariantderivative and the velocity vector are good tensors, so the product is a tensor. Now that we have defined aderivative, and want to know what does it mean. We claim that parallel transport means that DV µ/Dλ = 0along the path of interest x(λ). The intuitive picture is a vector being transported along the curve x(λ)while keeping its length and orientation with respect to the path.

Consider the 2-sphere S2. The path we are interested is part of the equator, i.e. part of a great circle.Then parallel transport of a vector along this path is just moving it while keeping the angle with thetangent vector of the path. Now consider the geodesic equation Uα∇αUµ = 0 again, it can be written as

DUµ

Dλ= 0 (10.10)

along the geodesic x(λ). The geodesic is the path on which the 4-velocity is parallel-transported.Let’s look at the 2-sphere again. If it is an Euclidean space then two parallel lines do not intersect.

But on the sphere two “parallel straight lines” can be defined as two geodesics from two points at whichtheir tangent vectors are both perpendicular to the geodesic connecting theme. But these kind of parallellines eventually meet, so this space is a curved space.

Another relation between curvature and parallel transport is that in flat space if you parallel transporta vector along a loop, it will come back to itself, while in a curved space after parallel transport along aclosed loop the vector comes back at an angle.

26


11 Lecture 11

Remember our formulation of the geodesic equation, we can either write down the explicit form with theChristoffel symbol:

d2xµ

dτ2+ Γµαβ

dxα

dτ

dxβ

dτ= 0 (11.1)

or can write the covariant derivative of velocity equals zero:

D

DτUµ = 0 (11.2)

where Uµ is the 4-velocity. For massless particle like photon the worldline is null, so we can’t parametrizethe path using proper time. We need to choose some other λ as the affine parameter. Usually for masslessparticles, however, it is usually more convenient to use the equation gµνdx

µdxν = 0 which is often strongerthan the geodesic equation.

11.1 Principle of General Covariance

The principle of general covariance states the following. An equation that describes some physical law inthe presence of gravity satisfies the following 2 properties:

• The equation is generally covariant, i.e. it takes the same form in all coordinate systems.

• The same equation holds in the absence of gravity, i.e. it agrees with special relativity if gµν = ηµνand Γµαβ = 0.

Note that this is just a restatement of the equivalence principle. But this provides a machinery to easilywrite down equations that are automatically covariant and including gravity. Note that we don’t have towrite the equation in manifestly covariant form, just like the usual geodesic equation, but it is useful andpossible to write the same equation in general tensor form, which is in general simpler and makes life easier.

Let’s consider some examples. Firstly we can derive the geodesic equation using this principle. Considera massive particle and Uµ = dxµ/dτ . In special relativity the equation of motion for a free particle is

dUµ

dτ= 0 =

dxα

dτ

dUµ

dxα= Uα∂αU

µ (11.3)

Note in the final part the derivative is just usual derivative. Now using the above principle, we just replacethe usual derivative by covariant derivative, so we get the equation

Uα∇αUµ = 0 (11.4)

This is a good tensor equation, and it is just our geodesic equation. The principle boils down to “replaceordinary derivatives with covariant derivatives”.

Let’s try this for E&M. Remember the Maxwell equations with source was written as

∂µFµν = Jν (11.5)

We just replace the ordinary derivative with covariant derivative

∇µFµν = Jν (11.6)

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There is also the source-free Maxwell equations

∂µFµν = 0 (11.7)

Now it is tempting to just replace the ordinary derivative with a covariant derivative. But there is onecaveat. When we defined the dual of the electromagnetic field tensor as Fµν = 1

2εµναβ , we left an important

part. We actually need to define it as

Fµν =1

2εµναβ , εµναβ =

1√|g|εµναβ (11.8)

In this way we can write the covariant version of the equation to be

∇µFµν = 0 (11.9)

Finally let’s do fluid mechanics. The conservation of energy-momentum equation can be directlygeneralized as

∂µTµν = 0 −→ ∇µTµν = 0 (11.10)

Let’s look at this equation in more detail. Write out the component form

∇µTµν = ∂µTµν + ΓµµβT

βν + ΓνµβTµβ = 0 (11.11)

This can be interpreted as follows. Formerly if we want to change the energy density, we need the fluid toflow out or into the volume element, which relates the time change of energy density with the divergenceof fluid momentum. Now the extra terms mean that another way to change energy density is to movein a direction where the gravity is different. The charge conservation equation also gets “covariantized”,because the current density involves the volume element, which is related to the space-time measure.

11.2 Riemann Curvature Tensor

We now want to introduce a tensor to characterize whether your space-time is flat or not. By the equivalenceprinciple, at any point we can choose a coordinate frame where the metric is Minkowski and first derivativevanishes, but in general we can’t touch the second derivative. So we anticipate that the curvature is relatedto the second derivatives of the metric.

Remember if we parallelly transport a vector on the surface of a 2-sphere over a complete loop, thevector will not return to itself. We would use this property and consider an infinitesimal loop and askunder what condition does the vector return to itself. Suppose we have a family of curves of constant x1,and another family of curves of constant x2. The loop we choose is the loop enclosed by the curves x1 = aand x1 = a + δa, x2 = b and x2 = b + δb. Let’s consider first the segment on x2 = b. The path can becharacterized as

x2(λ) = b,dx2

dλ= 0,

dx1

dλ6= 0 (11.12)

The parallel transport of vector V along this path means

dxµ

dλ∇µV α = 0 (11.13)

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But remember that the second component of velocity is 0, so we can write the above equation as

dx1

dλ∇1V

α = 0, ∇1Vα = 0, ∂1V

α = −Γα1µVµ (11.14)

So we can write the change of vector component as (A,B,C,D clockwise)

V α(B) = V α(A)−∫ a+δa

aΓα1µV

µ|x2=bdx1 (11.15)

Similarly we can work out the change of the component in all four segments, and get

V α(Aafter)− V α(A) =

∫Γα2µV

µ|x1=adx1 −

∫Γα2µV

µ|x1=a+δadx1

+

∫Γα1µV

µ|x2=b+δbdx1 −

∫Γα1µV

µ|x2=bdx1

= δaδb[−∂1

(Γα2µV

µ)

+ ∂2

(Γα1µV

µ)] (11.16)

We can expand the derivatives and use the relations ∂iVα = −ΓαiµV

µ to get

δV α = δaδb[− (∂1Γα2δ) + Γα2µΓµ1δ + (∂2Γα1δ)− Γα1µΓµ2δ

]V δ (11.17)

This is a special case of a more general identity

δV λ =[−∂µΓλβν + ∂νΓλβµ − ΓλµηΓ

ηβν + ΓλνηΓ

ηβµ

]V βδxµδxν (11.18)

And the terms inside the square brackets are called the Riemann curvature tensor [∗] = −Rαδλσ. Notethere is a minus sign in the convention. As promised, this quantity is related to the second derivative ofthe metric.

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12 Lecture 12

Recall we defined the Riemann tensor last time

Rλβµν = ∂µΓλβν − ∂νΓλβµ + ΓλµηΓηβν − ΓλνηΓ

ηβµ (12.1)

And we found that if we go around a loop then the change in the vector is

δV λ = −RλβµνV βδxµδxν (12.2)

(There is a problem in this equation, i.e. the Riemann tensor is antisymmetric in µν indices. We will fixthis next time.)

There are some properties for this quantity:

• It is a tensor. This is proved by mathematicians and we won’t show this here.

• Rλβµν = 0 everywhere is equivalent to the spacetime being flat.

• Define Rαβµν = gαλRλβµν . This tensor has following symmetry in indices:

Rαβµν = −Rαβνµ, Rαβµν = −Rβαµν , Rαβµν = Rµναβ (12.3)

• The following identity is trueRαβµν +Rαµνβ +Rανβµ = 0 (12.4)

It is conventional to define the Ricci tensor as

Rβν = Rλβλν = gαµRαβµν (12.5)

which has the symmetry property Rαβ = Rβα.We can also contract all the indices and define

R = gαβRαβ = Rαα (12.6)

and this quantity is called the Ricci scalar.

12.1 Alternative Way of Looking at the Riemann Tensor

In a local inertial coordinates, we have Γ’s vanishing, so the curvature looks like

Rλβµν = ∂µΓλβν − ∂νΓλβµ (12.7)

Let’s consider the quantity ∇α∇βV µ. Let’s do it now

∇α(∇βV µ) = ∂α(∇βV µ) + terms with Γ = ∂α∂βVµ + (∂αΓµβν)V ν (12.8)

We can also calculate the same thing with α and β reversed

∇β∇αV µ = ∂β∂αVµ + (∂βΓµαν)V ν (12.9)

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So we get something like

(∇α∇β −∇β∇α)V µ = (∂αΓµβν − ∂βΓµαν)V ν = RµναβVν (12.10)

Note that this equation is true in local inertial coordinates, and it is a valid tensor equation, so it holds inall frames. This is actually an equivalent way to moving the vector in a loop.

It is good to introduce the following, the so called Bianchi identity

∇λRαβµν +∇µRαβνλ +∇νRαβλµ = 0 (12.11)

One useful corollary of this identity is to contract this identity with gαµ and we get

∇λRβν +∇µRµβνλ −∇νRβλ = 0 (12.12)

We can contract this expression again with gβν

∇νR−∇µRµν −∇µRµν = 0 (12.13)

Therefore we know that

∇µR = 2∇νRνµ , ∇µ[Rµν −

1

2gµνR

]= 0 (12.14)

It is conventional to call the term in the bracket as Gµν which is called the Einstein tensor. The Bianchiidentity implies that ∇µGµν = 0.

12.2 The Einstein’s Equations

So far, we have done two things. Given a metric, we know how particles move in this field. We have alsounderstood how to rewrite special relativistic laws in GR form. That left out an important question whichis—What determines the metric?

Let’s do some educated guess. We have on our hands some clues from before:

1. In E&M we know that ∂µFµν = −Jν is the Maxwell’s equation with source. This relates the second

derivative of some field to a source. We will similarly look for some equation with two derivatives onthe metric, and relating it to some source which should be associated with mass and energy. Howeverthe Maxwell equation is linear, but we probably don’t want that, because gravitational field itself cansource the gravity, which is a nonlinear effect. We also want to write down a good tensor equationwhich is valid in any frame.

2. In appropriate limit we should recover Newton’s theory of gravity, which says

∇2φ = 4πGρ (12.15)

The Newtonian limit is when v 1, ρ P , and φ 1, i.e. the motion is nonrelativistic, and thegravitational field is weak.

For the second part, we need to relate the gravitational potential to the metric. We have already donethis in lecture 9, and got

g00 = −(1 + 2φ) (12.16)

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So the Einstein’s equation, whatever it is, should give the following equation in Newtonian limit

∇2

(−1

2g00

)= 4πGρ (12.17)

Now let’s write down the Einstein’s equation. We want the energy-momentum tensor Tµν on the righthand side, and something with 2 indices and second derivatives on metric on the left hand side. Einstein’sfirst guess is

Rµν = ATµν (12.18)

where A is some constant. But a problem is that we want energy-momentum conservation ∇µTµν = 0,while Rµν does not have this property in general. But remember Gµν has exactly this property, so we canguess

Gµν = Rµν −1

2gµνR = ATµν (12.19)

so that energy-momentum conservation is enforced automatically. Now we need to fix the constant A bygoing back to the Newtonian limit. Recall that for perfect fluid we have

Tµν = (ρ+ P )UµUν + Pgµν (12.20)

In Newtonian limit, P ρ and Uµ = (1, vi) and Uµ = (−1, vi) so Tµν has only one nonzero componentT00 = ρ. So in this limit the only nontrivial component of the Einstein’s equation is the 00 component.

Let’s look at the ij-th component of Gµν first, which should vanish, so we get

Rij −1

2gij(R

00 +Rii) = 0 =⇒ Rii + 3R00 = 0 (12.21)

We will show next time that A = 8πG.

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13 Lecture 13

There is some correction to what we covered in Lecture 11. We derived the curvature tensor using paralleltransport around a loop consisting of lines with x1 = a, x1 = a+ δa and x2 = b, x2 = b+ δb. We derivedthat

δV α = −δaδbV βRαβ12 (13.1)

The correct way to generalize is δα→ δaλ and δβ → δβδ and 1→ λ, 2→ δ. So the correct generalizationis

δV α = −δaλδbδV βRαβλδ (13.2)

so now δa and δb are some infinitesimal vectors.

13.1 Return to Einstein’s Equation

Recall the form of the Einstein’s Equation

Gµν = ATµν (13.3)

where A is a constant to be determined. There are some features about this equation

• The equation has two derivatives on gµν which is analogous to Maxwell equations. The deep reasonfor the equation to contain no higher derivatives is that higher derivatives will usually introduceghost perturbations with negative kinetic energy, which is not desirable in any sensible theory.

• This equation is consistent with energy-momentum conservation ∇µTµν = 0. This is a result of theBianchi identity. This is analogous to charge conservation in E&M.

Let’s fix A by matching Newtonian limit where gµν = ηµν + hµν , v 1 and P ρ. Remember in thislimit the only nonvanishing element of Tµν is the component T00 = ρ, where we discarded terms of orderO(ρv, ρh).

Now we return to the Einstein Equation and take its trace

gµν(Rµν −

1

2gµνR

)= R− 2R = AgµνTµν = ATµµ = AT (13.4)

So we can rewrite the equation as

Rµν = A

(Tµν −

1

2gµνT

)(13.5)

In our Newtonian limit the left hand side of the equation becomes

A(Tµν −1

2gµνT ) = −1

2Aρ δµν (13.6)

so now we only need to work out R00 and equate it to the 00 component of the above quantity and compareit to the Newton’s equation. We will find that R00 ∼ −h00

2 +O(h2), where = −∂2t +∇2. So we can just

plug into the Newton’s equation and read off A = 8πG. So we can write the complete Einstein’s equation

Gµν = Rµν −1

2gµνR = 8πGTµν (13.7)

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How do we add the speed of light c back into the equation? The left hand side has dimension [L]−2.The left hand side has dimension [Gmc2/L3], which is actually c4/[L]2. So the equation with dimensionfulc would be

Gµν =8πG

c4Tµν (13.8)

Because of the symmetry of indices of both Gµν and Tµν , we only have 10 independent equations.Because Bianchi identities further give 4 constraint equations, so we only have 6 degrees of freedom inthe field gµν . But gµν has 10 independent components, so there are 4 degrees of diffeomorphism freedom,which corresponds to the freedom to choose different coordinates for any given physical space. We can alsoview it in another way. We can choose some coordinate transformation x′µ(xµ) for the same space-timemanifold, so we need to subtract 4 degrees of freedom from the 10 independent components. This verymuch resembles the gauge symmetry in Electrodynamics. So in general relativity we also need to “choosea gauge” in the very beginning before we start to solve the Einstein Equations.

How do we choose a gauge (frame)? Recall in Electrodynamics we introduced the vector potentialwhich has a gauge degree of freedom A −→ A + ∇ϕ. A common gauge is the Coulomb gauge where∇·A = 0. We can always choose such a gauge because we only need to find a ϕ which solves the Poisson’sequation ∇2ϕ = −∇ ·A. Note that a gauge condition is usually not gauge invariant, otherwise it wouldnot act as a gauge condition.

In gravity let’s give an example of a gauge, which is called the harmonic/Lorenz/de Donder gauge:

gµνΓλµν = Γλ = 0 (13.9)

Note that this is not a tensor equation, i.e. it is not gauge invariant. We would like to show that we canalways choose this gauge. Recall the transformation of the Christoffel symbol

Γ′λµν =∂x′λ

∂xα∂xβ

∂x′µ∂xγ

∂x′νΓαβγ +

∂x′λ

∂xα∂2xα

∂x′µ∂x′ν(13.10)

If we contract the above quantity with g′µν , so we get

Γ′λ =∂x′λ

∂xγΓγ − gαβ ∂2x′λ

∂xα∂xβ(13.11)

We can solve the above equation equal to zero for x′µ as a function of xµ. That is just a second order PDEand its solution exists. So we can always choose a new coordinate such that Γ′λ = 0.

Next time we will impose the harmonic gauge on the Einstein’s Equations in the weak field limit, andsolve for all the components of the metric.

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14 Lecture 14

14.1 Linear Perturbation Theory

Today we want to solve the Einstein Equation completely to linear order. Remember the equation lookslike

Gµν = 8πGTµν , or Rµν = 8πG

(Tµν −

1

2gµνT

)(14.1)

To linear order means to first order in metric fluctuation hµν = gµν − ηµν .First step is to write down all the affine connections

Γµαβ =1

2gµν (∂αgνβ + ∂βgνα − ∂νgαβ) (14.2)

Because the derivatives of ηµν vanish, so we can take the gµν in front to be ηµν to first order in h, becausethe terms in the bracket are already of order h. So to first order we have

Γµαβ ∼1

2

(∂αh

µβ + ∂βh

µα − ∂µhαβ

)(14.3)

Note that here all the indices are raised using the Minkowski metric ηµν .After the affine connection we need to compute the Riemann tensor

Rλβµν = ∂µΓλβν − ∂νΓλβµ + ΓλµηΓηβν − ΓλνηΓ

ηβµ (14.4)

Note that we can throw the later two terms away because they are at order O(h2). We will need to carryout the calculation in the homework. After computing this tensor we need to contract indices to get theRicci tensor

Rµν = gαβRµανβ = Rαµαν = −1

2

[hµν + ∂µ∂νh− ∂µ∂βhβν − ∂ν∂βh

βµ

](14.5)

where h = ηµνhµν is the trace of hµν . If we take a trace again we can get the Ricci scalar

R = ηµνRµν = − [h− ∂µ∂νhµν ] (14.6)

We will verify these expressions in the homework. After knowing the Ricci tensor and scalar we can writedown the Einstein tensor

Gµν = Rµν −1

2gµνR = −1

2

[hµν − ηµνh+ ∂µ∂νh+ ηµν∂α∂βh

αβ − ∂µ∂βhβν − ∂ν∂βhβµ

](14.7)

Before proceeding any further, we want to verify a very important identity which is the Bianchi identity.We only want to verify it to first order in h, so the covariant derivative goes to ordinary derivative:

∇µGµν = ∂µGµν = −1

2

[∂µhµν −∂νh+ ∂νh+ ∂ν∂α∂βh

αβ −∂βhβν − ∂ν∂

µ∂βhβµ

]= 0 (14.8)

So it is reassuring to know that the Bianchi identity is satisfied.Remember we still have a lot of freedom. We want to choose a gauge, so let’s choose the harmonic

gauge (which is the only one we introduced so far):

gµνΓλµν = 0 (14.9)

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To first order in h we have1

2(2∂νh

µν − ∂µh) = 0 (14.10)

So our gauge condition to first order in h is just ∂µh = 2∂νhµν . With this condition we can simplify the

Ricci tensor

Rµν = −1

2

[hµν + ∂µ∂νh−

1

2∂µ∂νh−

1

2∂ν∂µh

]= −1

2hµν (14.11)

So we can now write the Einstein Equation using this very much simplified vershion of Ricci tensor

hµν = −16πG

(Tµν −

1

2ηµνT

)(14.12)

The reason we use ηµν in the right hand side is that because T is sourcing the h field, so it should havethe same order as h, so to first order we can just write ηµν . Note that this equation is exactly a waveequation with a source, so already this predicts gravitational waves. If there is no source then this is justthe homogeneous wave equation with plane wave solutions satisfying kµk

µ = 0, so it travels at the speedof light.

14.2 Newtonian Limit

We have done a lot of approximation in the left hand side of the Einstein Equations. Now we want to takethe Newtonian limit where Tµν ∼ diag (ρ, 0, 0, 0). So the right hand side is

Tµν −1

2ηµνT =

1

2ρδµν (14.13)

So the equations translate into

h00 = −8πGρ, h0i = 0, hij = −8πGρδij (14.14)

The first equation says that we can identify h00 = −2φ where φ is just the usual gravitational potential.The second equation is just the homogeneous wave equation, so it tells us that there can always be wavesin the background. We can just ignore these terms and choose boundary conditions such that there is nowaves. The third equation looks exactly the same as the first one, so we know that hij = −2φδij .

Now that we have the solution, we can write down the metric in a very simple way

ds2 = −(1 + 2φ)dt2 + (1− 2φ)(dx2 + dy2 + dz2

)(14.15)

where∇2φ = 4πGρ. This is the full solution of the Einstein Equations in linear order of metric perturbation.Now that we have the metric we can work out how particles move in this background. We will do this

in the next time.

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15 Lecture 15

15.1 Summary of Last Lecture

Remember we wrote the Einstein Equation in the form

Rµν = 8πG

(Tµν −

1

2gµνT

)(15.1)

which, in the weak field limit, we have

hµν = −16πG

(Tµν −

1

2ηµνT

)(15.2)

where hµν is the metric perturbation hµν = gµν − ηµν which is assumed to be small. Note that we used theharmonic gauge condition in deriving the above equation from the general one. We argued that the gaugecan always be chosen. The harmonic gauge condition is

∂µ(hµν −

1

2ηµνh

)= 0 (15.3)

The equation was simplified further by assuming Newtonian limit where Tµν ≈ diag (ρ, 0, 0, 0). We thensolved the metric completely in this limit. Refer to lecture 14 for details. Remember on top of that solutionwe can always add the solution of the homogeneous equation hµν = 0 which corresponds to gravitationalwave solutions. For now we focus on solutions without gravitational waves, which amounts to setting theboundary condition ∂0 ∼ 0. With this boundary condition the metric turned out to be

ds2 = −(1 + 2φ)dt2 + (1− 2φ)δijdxidxj (15.4)

15.2 Study of the Newtonian Metric

Now we try to study the trajectory of test particles in the above metric. Let’s write down the geodesicequation first

d2xµ

dτ2+ Γµαβ

dxα

dτ

dxβ

dτ= 0 (15.5)

Note we use the form for massive particles where the proper time can be taken as the parameter of thecurve. For a slowly moving particle we have

dxµ

dτ∼ (1, 0, 0, 0) +O(v, hµν) (15.6)

If we study the µ = i parts of the geodesic equation, which can be approximated to be

d2xi

dτ2+ Γi00 ∼ 0 (15.7)

Remember that Γi00 = ∂iφ, so in this limit we recover the Newton’s law of motion

d2xi

dt2= −∂iφ (15.8)

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xα

θ

M

bb

cos θ

Figure 15.1: Deflection of Trajectory

Let’s consider a test particle passing beside the sun, and because of the gravitation, its trajectory willbe deflected by a small angle α

We can get the deflection angle α by integrating the acceleration on the perpendicular direction. Theacceleration on the perpendicular direction is just

dv⊥dt

=d2x⊥dt2

=GM

(b/ cos θ)2cos θ (15.9)

while the velocity on the parallel direction is

v‖ =d(b tan θ)

dt=

b

cos2 θ

dθ

dt(15.10)

Combining the equations we get a nice differential equation

dv⊥dθ

=dv⊥dt

dt

dθ=GM

bv‖cos θ (15.11)

We can integrate this equation from −π/2 to π/2 and get

v⊥(t→∞) =2GM

bv‖, α =

v⊥(t→∞)

v‖=

2GM

bv2‖

(15.12)

Now we want to work out the same thing for a massless particle, most notably the photon. We writethe geodesic equation as

d2xµ

dλ2+ Γµαβ

dxα

dλ

dxβ

dλ= 0 (15.13)

Remember for photons we can’t assume that dxµ/dλ ∼ (1, 0, 0, 0) because photons are relativistic. Weneed to work out more components than for massive particles. First for µ = 0 the affine connections are

Γ000 = 0, Γ0

0i = ∂iφ, Γ0ij = 0 (15.14)

Note that we only keep up to first order in φ. We will then work out the µ = i parts

Γi00 = ∂iφ, Γi0j = 0, Γijk = −∂jφδik − ∂kφδij + ∂iφδjk (15.15)

In order to solve for the deflection angle, we again write down the unperturbed path first, and calculatethe deviation from this path and integrate it. The unperturbed path is just the solution of the equation

d2xµ

dλ2= 0 (15.16)

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We can choose t(λ) = λ, which can always be done because this is essentially just a parametrization. Wecan write xi = niλ where ni is along the velocity of the unperturbed path. We can write the perturbedequation as

0 =d2

dλ2(x+ δx) + Γi00

d

dλ(t+ δt)

d

dλ(t+ δt) + Γijk

d

dλ(xj + δxj)

d

dλ(xk + δxk)

=d2

dλ2δxi + ∂iφ+

[−∂jφδik − ∂kφδij + ∂iφδjk

]njnk

(15.17)

We can replace the λ in the last line by t which only introduces higher order corrections which we are notinterested in. So in the end we have

d2δxi

dt2= −2

[∂i − ninj∂j

]φ (15.18)

Note that ni · d2δxidt2

= 0, so the change of the trajectory is perpendicular to the direction of velocity. Thenwe get the change in the perpendicular direction to be

d2δxi⊥dt2

= −2∂⊥i φ (15.19)

The extra factor of 2 results in the difference from the Newtonian result, and now we get α = 4GM/bc2.

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16 Lecture 16

Remember last time we have worked out the deflection of light around a massive object. Today we aregoing to discuss the homogeneous solutions, i.e. the gravitational wave solutions of the linearized EinsteinEquations.

16.1 Gravitational Waves

Remember our linearized Einstein Equations

hµν = −16πG

(Tµν −

1

2ηµνT

)(16.1)

where we have used the harmonic gauge condition ∂µ(hµν − 1

2ηµνh)

= 0.Let’s define hµν = hµν − 1

2ηµνh. Then we can write the Einstein Equations in a nicer way

hµν = −16πGTµν (16.2)

This form is more convenient for us.We can write down the formal solution to the equation (16.2) by using Green’s function

hµν(x) =

∫d4y (−16πGTµν) (y)G(x− y) (16.3)

where the Green’s function satisfies xG(x− y) = δ4(x− y). We explicitly write out the Green’s functionfor the wave operator

G(x− y) = − 1

4π|x− y|δ(|x− y| − |x0 − y0|

)θ(x0 − y0) (16.4)

The step function θ(x0 − y0) is added to enforce causality. This form of the Green’s function is called theretarded Green’s function. So we have the solution

hµν(x, t) =

∫d3y

4G

|x− y|Tµν(tR,y) (16.5)

where the retarded time tR = t− |x− y|. We can in principle work out the Tµν of a binary star and workout how it sources the gravitational wave. Ref. Carroll Chapter 7 or Schutz Chapter 9.

We will consider the propagation of gravitational waves in vacuum, where Tµν = 0. That is to say, weconsider

hµν = 0 (16.6)

of course with the gauge condition ∂µhµν = 0. Remember in 1 + 1 dimensions the solution of the equationf(x, t) = 0 is

f = fR(x− t) + fL(x+ t) (16.7)

and we have left travelling modes and right travelling modes. We can use the familiar method of Fourieranalysis, expanding f(x, t) =

∑k A(k)eikαx

α, we can get

f = A(k)eikαxα(−kµkµ) = 0 (16.8)

40


which just says that kµkµ = 0 so the energy is equal to the magnitude of the wave number. Let’s studythe wave with k0 = k1, i.e. the wave vector looks like

kµ = k0(1, 1, 0, 0) (16.9)

in 3 + 1 dimensions, which is just a special Fourier mode. The corresponding metric perturbation is

hµν = Aµνeikαxα (16.10)

where kµ satisfies the above constraint. The harmonic condition translates into kµAµν = 0. This gives 4independent equations. Originally Aµν is symmetric because the metric is symmetric, so there are only 6independent components for Aµν . Actually if we work out the matrix it looks like

Aµν =

A00 −A00 A02 A03

−A00 A00 −A02 −A03

A02 −A02 A22 A23

A03 −A03 A23 A33

(16.11)

There are indeed 6 free parameters. However we claim that there are actually only 2 degrees of freedomwhich propagates, just like the case of photons.

Let’s have a little detour first. We want to know how coordinates transformations change the metric.Remember under coordinate transformations we have

g′µν(x′) = gαβ(x)∂xα

∂x′µ∂xβ

∂x′ν(16.12)

Let’s consider a small coordinate change x′µ = xµ − ξµ where ξµ is small. So we can write

g′µν(x′) = gαβ(x′ + ξ)∂(x′α + ξα)

∂x′µ∂(x′β + ξβ)

∂x′ν

=[gαβ(x′) + (∂γgαβ)ξγ

] [δαµ + ∂′µξ

α] [δβν + ∂′νξ

β]

= gµν(x′) + (∂γgµν)ξγ + gαν∂µξα + gαµ∂νξ

α + o(ξ2)

(16.13)

Therefore we can write down the change in gµν as

g′µν − gµν = δgµν = gµα∇νξα + gνα∇µξα = ∇νξµ +∇µξν (16.14)

The covariant derivatives show up because of the term (∂γgµν)ξγ =(

Γλγµgλν + Γλγνgλµ

)ξγ . If we apply

this to our perturbed metric then we just get

δhµν = ∂νξµ + ∂µξν , δhµν = ∂νξµ + ∂µξν − ηµν∂αξα (16.15)

This is the analogue of our gauge transformation in electrodynamics. Now let’s consider our harmonicgauge condition more carefully. We want to have the gauge condition ∂µhµν = 0. So given any hold

µν we canalways satisfy this condition by doing a transformation satisfying

∂µ(∂νξµ + ∂µξν − ηµν∂αξα) = ξν = −∂µholdµν (16.16)

41


We just solve this equation in ξν to find the coordinate transformation to the harmonic gauge. Againwe have a wave equation, we can always add the homogeneous solutions to the solution and the solutionremains a solution. This introduces 4 degrees of freedom and we want to remove them. So our freedom inAµν is further reduced by 4 and we only have 2 free parameters in Aµν , corresponding to two polarizationsof the gravitational wave. These 4 removed degrees correspond to artifacts of coordinate transformation.

The homogeneous solutions to equation (16.16) can be Fourier expanded. Suppose we have ξν =Bνe

ikαxα then we can always change our Aµν by this amount

Aµν = Aoldµν + ikνBµ + ikµBν − ηµνikαBα (16.17)

while not screwing up the harmonic gauge condition. Note if kµ = k0 (1, 1, 0, 0) we can simplify and writethe matrix as

−2ik0B0 + i(k0B0 + k0B1) i(−k0B1 + k0B0) −ik0B2 −ik0B3

i(−k0B1 + k0B0) 2ik0B0 − i(k0B0 + k0B1) ik0B2 ik0B3

−ik0B2 ik0B2 −i(k0B0 + k0B1) 0−ik0B3 ik0B3 0 −i(k0B0 + k0B1)

(16.18)

We can exploit our freedom and cancel out as many terms in Aµν as possible, and finally we only have A23

and A22 = −A33 non-zero. The two polarizations of gravitational waves are usually called + and ×. Wewrite

hµν =(h(+)e(+)

µν + h(×)e(×)µν

)eikαx

α(16.19)

where we have

e(+)µν =

0 0

0

(1 00 −1

) , e(×)µν =

0 0

0

(0 11 0

) (16.20)

The first polarization distorts space-time in y and z directions like . The second polarization

distorts the space like .

42


17 Lecture 17

Recall that the gravitational waves are solutions of the equation

hµν = −16πGTµν (17.1)

This is the linearized Einstein Equations in the harmonic gauge. Last time we looked at the homogeneoussolutions of the equation and found that 4 out of the 6 apparent degrees of freedom are artificial and thereare only two different polarizations. We showed that making use of residual coordinate freedom to imposethat hαα = 0 and h0α = 0. In literature these conditions are collectively called the transverse-tracelessgauge. This gauge is mainly used in studying the propagation of gravitational waves.

Remember last time we introduced the two polarizations of the gravitational wave e(+)µν and e

(×)µν

e(+)µν =

0 0

0

(1 00 −1

) , e(×)µν =

0 0

0

(0 11 0

) (17.2)

We want to study the physical interpretation of these waves, i.e. how do particles respond to these waves.

17.1 Particles in Gravitational Waves

We need to solve the geodesic equation which describes the trajectory of a particle in the gravitationalwave field. Again we work in the Newtonian limit where Uµ = (1, 0, 0, 0) + O(v). The only interestingcomponent of the geodesic equation for us is just

d2xi

dτ2+ Γi00 = 0 (17.3)

Remember in the linear perturbation we have

Γi00 =1

2ηiγ (2∂0hγ0 − ∂γg00) (17.4)

But plugging in the h expression for both polarizations we find these Christoffel symbols vanish! So a testparticle does not respond the gravitational waves!

The problem in the above analysis is that gravitational waves are just disturbances in the geometry ofthe spacetime. In terms of coordinates, the particles will not move at all. The sensible question to ask iswhat is the physical separation of two points under the gravitational wave perturbations.

Let’s consider two particles sitting on the y-axis. Their physical separation is just

ds2 = gµνdxµdxν = gyydy

2 (17.5)

Let’s first consider the (+) component of the gravitational wave. If the wave is given by hµν = h+e(+)µν eikµx

µ,

then we can work out the separation of the particles to be

dsy =√

1 + hyydy ≈(

1 +1

2h+e

ikµxµ)dy (17.6)

43


Now suppose our particles are sitting on the z-axis then we should use the gzz component of the metric,and note that there is a sign difference from gyy, so we have

dsz =

(1− 1

2h+e

ikµxµ)dz (17.7)

So the effect of the wave is to distort the space, squeezing the space in z direction and stretching in ydirection at some point, and stretch z and squeeze y at some other point. It can be shown that (×)polarization does similar things, except at rotated axises.

17.2 Solution to the Nonlinear Einstein Equations

Now we come back the the Einstein Equations with its full content. It is in the nonlinearity that thingsbecome interesting. However it is impossible to solve it in general, so we want to impose symmetries.Recall the equations are

Rµν = 8πG

(Tµν −

1

2gµνT

)(17.8)

The assumptions we want are

• Spherical symmetry

• Time independent (static)

It turns out for Einstein gravity in vacuum the first assumption is enough to guarantee the second.This statement is called the Birkhoff Theorem. We will focus on vacuum solutions which are solutions tothe equation Rµν = 0.

Let’s first make sense of our assumptions. If we choose a set of coordinates t, x1, x2, x3, the staticassumption can well be taken to mean that gµν does not depend on time t. We write down an ansatz

ds2 = −F (r)dt2 + 2E(r)dtx · dx +D(r) (x · dx)2 + C(r)dx · dx (17.9)

Remember r = |x|. The spatial part of this metric is explicitly invariant under O(3). So this ansatzindeed satisfies our two assumptions. To simplify the problem we can make coordinate transformations tomake most of the coefficients go away. Let’s first switch to spherical coordinates where

x1 = r sin θ cosφ

x2 = r sin θ sinφ

x3 = r cos θ

(17.10)

Note that in this coordinate we have

dx · dx = dr2 + r2dθ2 + r2 sin2 θdφ2 (17.11)

x · x = r2 (17.12)

x · dx = rdr (17.13)

So the metric now looks like

ds2 = −F (r) dt2 + 2E(r)r dtdr + r2D(r) dr2 + C(r)[dr2 + r2dθ2 + r2 sin2 θdφ2

](17.14)

44


Now we define t′ = t+ f(r) and find f to get rid of the mixed term dtdr. We have

dt = dt′ − df

drdr (17.15)

Plug this expression to our previous expression for the metric we have

ds2 = −F(dt′ − df

drdr

)2

+ 2Er dr

(dt′ − df

drdr

)+ r2Ddr2 + C

[dr2 + r2dθ2 + r2 sin2 θdφ2

](17.16)

Obviously we need to choose f such that

2df

drF + 2Er = 0 (17.17)

So the rest of the metric looks like

ds2 = −F (r)dt′2 +G(r)dr2 + C(r)[dr2 + r2dθ2 + r2 sin2 θdφ2

](17.18)

We can make a further substitution such that r′2 = C(r)r2, to make C(r) go away. So finally we get(dropping all the primes)

ds2 = −B(r)dt2 +A(r)dr2 + r2(dθ2 + sin2 θdφ2

)(17.19)

This is the most general time-independent and spherically symmetric metric we can write down. We willplug this ansatz into the Einstein Equations and solve it next time.

45


18 Lecture 18

18.1 The Schwartzschild Solution

Recall from last time that we want to solve the fully nonlinear Einstein Equations in the vacuum withspherical symmetry. Birkhoff theorem implies that spherical symmetry also guarantees time-independenceof the metric. Remember last time we wrote down the most general spherically symmetric ansatz

ds2 = −B(r)dt2 +A(r)dr2 + r2(dθ2 + sin2 θdφ2

)(18.1)

And we will use this minimal form as a starting point.Note that this metric is diagonal. The components are

gtt = −B, grr = A, gθθ = r2, gφφ = r2 sin2 θ (18.2)

We first need to figure out the inverse, which is easy because we have a diagonal metric

gtt = − 1

B, grr =

1

A, gθθ =

1

r2, gφφ =

1

r2 sin2 θ(18.3)

The next steps are standard, we first compute the affine connections, and then the Riemann tensor,and then contract to get the Ricci tensor. The Einstein Equations we want to solve in vacuum is just

Rµν = 0 (18.4)

If we follow the steps we will eventually get the Ricci tensor

Rrr = −B′′

2B+

1

4

B′

B

(A′

A+B′

B

)+

1

r

A′

A(18.5)

Rθθ = 1− r

2A

(−A

′

A+B′

B

)− 1

A(18.6)

Rφφ = sin2 θRθθ (18.7)

Rtt =B′′

2A− 1

4

B′

A

(A′

A+B′

B

)+

1

r

B′

A(18.8)

where all the primes are with respect to r, and that Rµν = 0 for µ 6= ν. It is a little difficult to solve theseequations to get an ODE for A or B. Now observe that the first and last equations have a lot in common,let’s take a linear combination of them to get

RrrA

+RttB

=1

r

A′

A2+

1

r

B′

AB= 0 (18.9)

This equation reveals a very nice relation between A and B:

A′

A= −B

′

B(18.10)

Integrating this equation we get

logA = − logB + constant , or A ∝ 1

B(18.11)

46


Let’s fix the proportionality constant between A and B as follows. Imagine we are extremely far awayfrom the star, then the metric should reduce to the Minkowski metric. So in the limit r → ∞ we haveA = B = 1. This fixes the proportionality constant to be 1. A = 1/B.

Now we only need to compute one of them from one of the differential equations above. Let’s considerRθθ component and replace A with 1/B. We have

Rθθ = 1− rB

2

(2B′

B

)−B = 1− rB′ −B = 0 (18.12)

This is a pretty simple differential equation. We write it in a more explicit form

d(rB)

dr= r

dB

dr+B = 1 (18.13)

Integrating this expression, we just get

B(r) = 1 +C

r(18.14)

where C is some constant we want to determine. We are going to fix the constant by resorting to theweak-field limit which should reproduce the Newtonian gravitational potential. In the weak-field limit wejust have

B(r) = 1 + 2φ(r), where φ = −GMr

(18.15)

So the constant is just C = −2GM . The full metric is therefore

ds2 = −(

1− 2GM

r

)dt2 +

(1− 2GM

r

)−1

dr2 + r2(dθ2 + sin2 θ dφ2

)(18.16)

This is the celebrated Schwartzschild metric. Note that this metric is fully nonlinear, so it is applicableeven when the field strength is of order unity.

Let’s look at the metric in a more detail. Suppose we have two points 1 and 2 as shown in figure 18.1.

12

Figure 18.1: Physical Separation of Two Points

The physical separation of the two points is just the integration

s12 =

∫ r2

r1

dr√1− 2GM/r

(18.17)

And the ratio of the area of the two spheres is

A2

A1=r2

2

r21

(18.18)

47


Note when r1 = 2GM , the distance s12 is actually divergent! It takes you forever to travel from point 1 topoint 2. But the ratio of the areas of the two spheres is always a finite number. This is a feature of curvedspace, where radial distance and angular distance are decoupled.

This kind of behavior is reminiscent of the 2-sphere, where when you go from one pole to the otheryour circle has constant angular span but has variable circumference. In Schwartzschild space-time we justhave one more complication that the distance could be divergent. The place of divergence r = 2GM iscalled the Schwartzschild radius.

The Schwartzschild radius must be there for any celestial object. But why don’t we observe this kindof divergence in space-time on our earth? The answer is that this metric is only applicable outside theobject, but for the earch or even the sun the radius is actually way smaller than the radius of the objectitself. To compute the radius we need to insert c back to the expression and find

Rs =2GM

c2(18.19)

which is about 3 km for the sun. So for most astronomical objects this number is so fantasitically smallthat we don’t need to consider it. Only for dense objects like a neutron star or a black hole do we find thisradius larger than the radius of the object.

18.2 The Orbit of Mercury

Historically GR triumphs because of another famous result, which is the prediction of the perihelionprecession of the Mercury orbit. The Schwartzschild solution correctly predicts the amount of precessionof the orbit, so we are going to carry out this calculation. To contrast with our calculation of the deflectionof photon, where we only used the Newtonian limit, here we need to carry the expansion of the metric tonext order.

The geodesic equation isdUα

dλ+ ΓαµνU

µUν = 0 (18.20)

where Uα = dxα/dλ and we can take λ = τ for a massive particle. The equation can also be written morecompactly as

Uµ∇µUα = 0 (18.21)

where we have used a metric to lower the index of U inside derivative. This is the form more useful to us.We can write it out as

Uµ(∂µUα − ΓγµαUγ

)= 0 (18.22)

Note thatdUαdλ

= ΓγµαUµUγ =

1

2gγη (∂µgηα + ∂αgηµ − ∂ηgµα)UµUγ (18.23)

The first and third term in the bracket cancel with each other because of symmetry between two U ’s. Sothe geodesic equation can be simplified to be

dUαdλ

=1

2(∂αgηµ)UηUµ (18.24)

So whenever the metric does not depend on some coordinate, the velocity 1-form on the direction of thatcoordinate is conserved. In this case of Schwartzschild metric, we know Ut and Uφ are constants. We willcall these

− E = Ut, L = Uφ (18.25)

48


So the corresponding vector components are

U t = gttUt =E

1− 2GM/r, Uφ = gφφUφ =

L

r2 sin2 θ(18.26)

If we think about the physics of the problem, we will solve it in one plane, namely the equatorial planewhere θ = π/2. So actually Uφ = L/r2.

Now let’s consider other components. First recall that

dτ2 = −ds2 =

(1− 2GM

r

)dt2 − dt2

1− 2GM/r− r2dφ2 (18.27)

where we have set θ = π/2 by hand. We can directly use this equation to get an equation of motion for amassive particle

1 =

(1− 2GM

r

)(dt

dτ

)2

−(

1− 2GM

r

)−1(drdτ

)2

− r2

(dφ

dτ

)2

(18.28)

If we substitute what we know about dt/dτ and dφ/dτ we can get a simple equation for dr/dτ :(dr

dτ

)2

= E2 −(

1− 2GM

r

)(1 +

L2

r2

)(18.29)

At perihelion or aphelion, this derivative dr/dτ will be zero, and we can solve for the two points to get themaximum and minimum distances of Mercury from the sun. But we are interested in dr/dφ instead. Thatis not difficult as we have already known dr/dτ and dφ/dτ , we can compute by

dr

dφ=dr

dτ

/dφdτ

(18.30)

which we will do the next time.

49


19 Lecture 19

19.1 Perihelion Precession of Mercury

Recall last time we argued that two components of the velocity are constant in the Schwartzschild metric

Ut = −E =⇒ U t =E

1− 2GM/r, Uφ = L =⇒ Uφ =

L

r2(19.1)

And we have the radial velocity(dr

dτ

)2

= E2 −(

1− 2GM

r

)(1 +

L2

r2

)= (E2 − 1) +

2GM

r− L2

r2+

2GML2

r3

(19.2)

We can call the terms in powers of 1/r the effective potential and this equation looks very much like energyconservation. The effective potential is

Veff = −2GM

r+L2

r2− 2GML2

r3(19.3)

If we plot the potential it looks like

r

V

Figure 19.1: Effective Potential

Recall the Newtonian result we should have

1

2r2 +

1

2

L2

r2− GM

r= EN (19.4)

So the radial velocity should satisfy

r2 = 2EN +2GM

r− L2

r2(19.5)

The term proportional to L2 is the centrifugal energy. The GR correction is essentially the third term.Now to get precession we need the derivative dr/dφ(

dr

dφ

)2

=

(dr

dτ

)2/(dφdτ

)2

=r4

L2

(E2 − 1 +

2GM

r− L2

r2+

2GML2

r3

)(19.6)

50


We use a trick from our mechanics class, substituting u = 1/r. Then(du

dφ

)2

=E2 − 1

L2+

2GM

L2u− u2 + 2GMu3 (19.7)

We anticipate that the correction from the last term is very small so we integrate the Newtonian resultfirst and treat the last term as a perturbation. Complete the square for u in the right hand side we have(

du

dφ

)2

= −(u− GM

L2

)2

+E2 − 1

L2+G2M2

L4(19.8)

Substitute y = u−GM/L2 we can solve this equation to get

y =

√E2 − 1 +G2M2/L2

L2cos (φ+ φ0) =

1

r− GM

L2(19.9)

This means that in Newtonian gravity we have a closed orbit because this function has period 2π. Nowwe add back the GR correction and get(

dy

dφ

)2

= −y2 +E2 − 1 +G2M2/L2

L2+ 2GM

(y +

GM

L2

)3

(19.10)

Let’s contemplate what is the physical meaning of y. It is actually the deviation from the circular orbit.So as long as the orbit is not far away from circular we can expand the cubic binomial and keep terms upto y2, and get(

dy

dφ

)2

= −y2

(1− 6G2M2

L2

)+ 6y

G3M3

L4+

[E2 − 1 +G2M2/L2

L2+

2G4M4

L6

](19.11)

This differential equation can be solved using the same technique as above, completing the square and shiftthe variable and integrate directly. Schematically we have(

dy

dφ

)2

= −ay2 + by + c = −a(y − b

2a

)2

+ c+b2

4a(19.12)

If we substitute z = y − b/2a we can solve the equation and get

z =

√c+

b2

4acos(√aφ+ φ0

)(19.13)

Remember√a =

√1− 6G2M2/L2 and this is a small deviation from 1, as the gravitational field from the

sun is of order 10−6, so we can expand it as

√a ∼ 1− 3

G2M2

L2(19.14)

So the period of the orbit is

T =2π√a∼ 2π

(1 +

3G2M2

L2

)(19.15)

so the precession per orbit is just

∆T ∼ 6πG2M2

L2=

6πGM

r(19.16)

If we plug in numbers we find that the precession of perihelion for mercury is about 0.43′′ per year, thatis 43′′ per century. Note that this is just the correction from GR and the total precession is about 5600′′

per century, but the most contribution is from the other planets.

51


19.2 Gravitational Redshift

Now that we have the Schwartzschild metric

ds2 = −(

1− 2GM

r

)dt2 +

(1− 2GM

r

)−1


)(19.17)

we can study the gravitational redshift in greater detail. This is to study the motion of photons in radialdirection, so the part of the metric that concerns us is just

ds2 = −(

1− 2GM

r

)dt2 +

(1− 2GM

r

)−1

dr2 = 0 (19.18)

It is zero because photons travel on a null geodesic. Consider a source sending continuous signal from asmaller radius r1 to a larger radius r2. We consider two adjacent crests A and B on the wave and let ∆t1be the time difference between emission of A and B, and ∆t2 be the time difference between reception ofA and B at point 2.

r1 r2

A B

Figure 19.2: Emission and Absorption of Photon

The time of travel for the two crests should be equal because the metric is static, so we know that∫ t2

t1

dt =

∫ t2+∆t2

t1+∆t1

dt (19.19)

So ∆t1 = ∆t2. Now we want to convert all these coordinate time to physical time ∆τ1 and ∆τ2. Now weshould have

∆τ1 =

√1− 2GM

r1∆t1 (19.20)

And also

∆τ2 =

√1− 2GM

r2∆t2 =

√1− 2GM/r2

1− 2GM/r1∆τ2 (19.21)

So we know that gravitational redshift is just

λobs

λem=

∆τ2

∆τ1=

√1− 2GM/r2

1− 2GM/r1(19.22)

Note this result diverges at r1 = 2GM . So if we emit some photon at the Schwartzschild radius then thephoton sent out is infinitely redshifted.

52


20 Lecture 20

20.1 Going Across the Horizon

Recall that there is some strange thing for the Schwartzschild metric at the Schwartzschild radius. Peopleafter a long time realized that the Schwartzschild radius is not a singularity. The curvature tensor doesnot blow up at the Schwartzschild radius, so we can do a coordinate transform to make this “singularity”go away, and go across the horizon without problem.

Recall the Schwartzschild metric

ds2 = −(

1− 2GM

r

)dt2 +

(1− 2GM

r

)−1


)(20.1)

We argued that the angular part does not play significant role in our analysis. Another thing to note isthat across the horizon the signs of time and radial component switch, so inside the horizon time and spacechange into each other.

We want to study the lightcone structure of this metric. From the lightcone condition we can deduce

dt

dr= ± 1

1− 2GM/r(20.2)

Note as r → ∞ this slope goes to ±1. So the space looks like a flat space. For r → 2GM the slope goesto ±∞, so the lightcone becomes infinitely narrow. We want to change into a coordinate where this doesnot happen. We manipulate the above equation to get

dt =r dr

r − 2GM=

(1 +

2GM

r − 2GM

)dr (20.3)

We can readily integrate this expression to get

t = r + 2GM (log (r − 2GM)− log 2GM) = r + 2GM log( r

2GM− 1)

(20.4)

Now we define a new spacial coordinate r∗ which equals the right hand side of the above equation. Now ifwe use this then all the null rays have slope ±1. The metric now looks like

ds2 =

(1− 2GM

r

)[−dt2 +

dr2

(1− 2GM/r)2

]+ r2dΩ2 =

(1− 2GM

r

)(−dt2 + dr2

∗)

+ r2dΩ2 (20.5)

Now it is better because the metric does not blow up at any place. Plus at any place the lightcone hasslope ±1. But now where is the Schwartzschild radius? If we plug r = 2GM into the expression for r∗,then it corresponds to r∗ → −∞. So we have effectively pushed the horizon to −∞.

Now we want to define some new coordinate system which is called lightcone coordinates. We define

v = t+ r∗, u = t− r∗ (20.6)

These are called lightcone coordinates, because if v = C where C is a constant, then r∗ = −t + C whichis essentially an incoming particle at the speed of light. This coordinate is essentially the frame of anobserver travelling at the speed of light. Now we have

− dt2 + dr2∗ = −dv du (20.7)

53


Now we want to bring our horizon back from infinity by yet another coordinate transformation. Wedefine

u = −4GMe−u/4GM , v = 4GMev/4GM (20.8)

It is reasonable because for v → −∞ the new variable v → 0. We essentially inverse the infinities. Nowwe can plug in the expression for r∗ in terms of r and get

v = 4GMe(t+r)/4GM( r

2GM− 1)1/2

u = −4GMe(−t+r)/4GM( r

2GM− 1)1/2 (20.9)

Note that this expression, the horizon is now at u = v = 0. Now we ask again in this new coordinatewhat does the metric look like

ds2 =

(2GM

r− 1

)dv du+ r2dΩ2

= −2GM

re−r/2GMdv du+ r2dΩ2

(20.10)

where we have used

dv du =dv

dv

du

dudv du = e−r∗/2GMdv du = e−r/2GM

( r

2GM− 1)−1

dv du (20.11)

Now in our metric nothing blows up and nothing irregular. But the form is not nice because time andspace are mixed into the coordinates. Now let’s define

v = T +R, u = T −R (20.12)

and we can write −dv du = −dT 2 + dR2. We write the metric as

ds2 =2GM

rer/2GM (−dT 2 + dR2) + r2dΩ2 (20.13)

This is called the Kruskal metric. Note that the lightcone again has slope ±1. The coordinate T is indeedtime-like because of the signature in the metric. The metric has all the nice regularity structure for all Tand R. Now we want to recover our original t and r from this new set of coordinates

T =1

2(v + u) = 4GMer/4GM

( r

2GM− 1)1/2

sinht

4GM(20.14)

R =1

2(v − u) = 4GMer/4GM

( r

2GM− 1)1/2

cosht

4GM(20.15)

So far in our expression everything only makes sense outside the Schwartzschild radius. We need aform where we can go across the horizon. Now consider the quantity

T 2 −R2 = (4GM)2(

1− r

2GM

)er/2GM (20.16)

In addition let’s consider their ratio

T/R = tanh

(t

4GM

)(20.17)

54


R

T r = 2GM, T = R, t→∞

r = 2GM, T = −R, t→ −∞

r = C

t = C

r = 0

Figure 20.1: Relation between T and R

Note the first expression only depends on r and the second only depends on t. So for any r and t we canmap into a pair of T and R, even for r beyond the horizon. We can plot a relation between R and T

Now let’s see where is the horizon. The horizon r = 2GM is where T = ±R. Curves of constant rare hyperbolae in the horizontal direction, while curves of constant t are lines passing through the origion.The curve t → ∞ corresponds to T = R and t → −∞ corresponds to T = −R. So the curves of horizonand infinite times coincide!

The trajectory of a particle falling into the horizon looks like the blue line inside the graph. But asyou go across the horizon, there is no turing back, because the horizon is parallel to the lightcone at anypoint. The best you can do is delay your inevitable fate of meeting the singularity.

55


21 Lecture 21

21.1 Penrose Diagrams

Last lecture we did some coordinate transformations. Now we want to take one more coordinate transfor-mation to put everything into a compact region in space-time. Recall the steps we have taken from theSchwartzschild metric

1. Find r∗ so that we have normal lightcone everywhere in the spacetime, so that the space looks flatexcept for a conformal factor. We have

ds2 =

(1− 2GM

r

)(−dt2 + dr2

∗)

(21.1)

2. We switch to lightcone coordinates (u, v) = t± r∗ and get

ds2 =

(1− 2GM

r

)(−dv du) (21.2)

3. We defined the “exponentiated coordinate” so that the coordinate covers the whole space and get

ds2 = −2GM

re−r/2GMdu dv (21.3)

4. Finally we switched from lightcone coordinate back to “pseudo flat” spacetime

ds2 = −2GM

re−r/2GM

(−dT 2 + dR2

)(21.4)

Now we take an extra step from the third

v = 4GM tan v, u = 4GM tan u (21.5)

And now the range of u and v are from −π/2 to π/2. We have essentially mapped the infinite space intoa compact space. Again we carry out the final step and define

v = T + R, u = T − R (21.6)

What does the space-time look like?Remember we have

uv = (4GM)2er/2GM(

1− r

2GM

)(21.7)

We want to translate this equation into u and v and get a relation of these quantities to r. We get

tan u tan v = er/2GM(

1− r

2GM

)(21.8)

The singularity corresponds to the left hand side equal to 1. The line corresponding to singularity is drawnon the diagram as red line. This is just u + v = ±π/2, so that tan u = cot v. We claim that the regionbetween the two lines is physical.

56


R

T

vu

π/2

π/2

i+

i−

i0

J +

J −

Figure 21.1: Penrose Diagram

Now where is the horizon? Remember the horizon is parallel to a null line. We want to have

tan u tan v = 0 (21.9)

which means either u or v is zero. The horizon is drawn on the diagram as green lines.Now where is infinity? We want to find the place

tan u tan v = −∞ (21.10)

This corresponds to the blue lines on the diagram. We focus on the diamond on the right side. The linesof constant r can be draw as orange lines. The point i+ is very strange. It is the intersection of the horizonand the infinity. But this is just an artifact of our coordinate choice.

Now what about surfaces of constant t? We need to work out the relation between u, v and t

v + u

v − u= tanh

(t

4GM

)(21.11)

57


Remember in Kruskal coordinates we have the lines of infinite time coincide with the horizon. Same thinghappens here. The lines of constant t are draw in the diagram in purple.

Let’s examine the causal structure of the diagram. The point i+ is the place where every time-liketrajectory end up, and the point i− is the place everything begins. The point i0 is the spatial infinity,because that is the place if you do a spacial translation at a constant time, this is the farthest point youcan get to. The edges J + and J − are called the future and past null infinity, because these are the placesthe null lines end up.

Let’s try to draw the Penrose diagram for the Minkowski space. We essentially carry out all theprocedure as above. We are going to suppress y and z coordinates and write

ds2 = dt2 + dx2 (21.12)

We introduce a new coordinate

t = eξ sinh η, x = eξ cosh η (21.13)

and the metric becomes

ds2 = −(eξ sinh ηdξ + eξ cosh ηdη

)2+(eξ cosh ηdξ + eξ sinh ηdη

)2= e2ξ(−dη2 + dξ2) (21.14)

This is analogous to the first step in our way to Kruskal metric and Penrose diagrams. Now note thatt2 − x2 = −e2ξ, so a line of constant ξ will be a hyperbola. Plus t/x = tanh η, and lines of constant η willbe rays from the origin. So this coordinate is very close to the Kruskal coordinate which we drew in thelast lecture.

The line of ξ = −∞ looks like a horizon in Kruskal coordinate. This is an artifact of our cooridatechoice again. This set of coordinates is actually the set of observers of constant acceleration, where ξ isactually proportional to the acceleration.

The Penrose diagram for Minkowski space is almost identical to the diagram for Schwartzschild space.The crucial difference is that if an observer goes around the origin in Schwartzschild it can get a massof the black hole which characterizes the horizon. However in Minkowski space there is no singularity orblack hole, so the mass will go to zero.

58


22 Lecture 22

22.1 Metric for a Homogeneous, Isotropic and Expanding Universe

Let’s consider a homogeneous, isotropic and possibly a time-dependent universe. By homogeneous we meanthat the universe has more or less the same density at all area. To be precise, if we average the matterdensity, or the density of galaxies, over a very large volume, this fluctuation of this average will tend tozero when the volume of averaging becomes large enough. Suppose we have a sphere of radius R centeredat some point in the universe. Let n be the number of galaxies within this sphere. We can do this countingin another sphere of radius R. Let n be the average n when we move the sphere around all the universeand we can evaluate the variance

〈(n− n)2〉 = 〈n2〉 − n2 (22.1)

We can define a new quantity

δ2R =

〈(n− n)2〉n2

(22.2)

which is called the fractional variance depending on R. Observation shows that this quantity drops whenR becomes larger like shown in the graph

logR

log δ2R

1

10−2

10 Mpc

100 Mpc

Figure 22.1: Dependence of δR on R

Note the size of the observable universe is about 3000 Mpc. The distance between Milkyway and M31galaxy is about 700 Mpc. The universe is isotropic in the sense that when we look around the sky we seeroughly the same density and background microwave radiation. Our measurements are pretty precise inthis aspect. Note that isotropy and homogeneity are not the same and we can imagine a universe withhomogeneity but not isotropy.

Hubble’s law tells us that the universe is not static. It was Hubble’s observation that the knownspectral lines like Hydrogen lines are redshifted from the distant galaxies, and this tells us that thosegalaxies are moving away from us. So we will try to write down a metric that is homogeneous, isotropicand time-dependent which describes our universe.

Let’s consider a first ansatz

ds2 = −dt2 + a2(t)(dx2 + dy2 + dz2

)(22.3)

where a(t) is called the scale factor and if it increases with time then we have an expanding universe. Thismetric is automatically homogeneous and isotropic, because obviously x, y and z are equivalent in thismetric, and this does not depend on the origin of the coordinates.

59


This metric is not the most general isotropic and homogeneous expanding universe we can write down,but we will study this for a moment first. It is usually written in a spherically symmetric way

ds2 = −dt2 + a2(t)[dχ2 + χ2

(dθ2 + sin2 θdφ2

)](22.4)

Note in this metric even when the coordinate separation between two points is unchanged with time,the physical separation increases with time because the metric is enlarging. It is not the particles aremoving, but the notion of spacetime distance are changing. We call these coordinate separation comovingdistance which does not change with time, and we call the physical separation the proper distance.

The metric we wrote down in equation (22.3) is a flat metric, because the spatial metric in the bracketis just like a Euclidean space. Now we can also write down the spatial metric as that of a 3-sphere, thiswill be another possibility of a homogeneous and isotropic metric. The 3-sphere is defined as the surfacein R4 by the equation

R2 = x2 + y2 + z2 + w2 (22.5)

We can parametrize the sphere just as we did the 2-sphere:w = R cosβ

z = R sinβ cos θ

x = R sinβ sin θ cosφ

y = R sinβ sin θ sinφ

(22.6)

Then the metric will become

ds2 = R2(dβ2 + sin2 β dθ2 + sin2 β sin2 θ dφ2

)(22.7)

Now if we define χ = Rβ to contrast with equation (22.4) we get

ds2 = −dt2 + a2(t)[dχ2 +R2 sin2

(χR

) (dθ2 + sin2 θ dφ2

)](22.8)

This is called the closed universe, because the universe looks like a sphere which is closed and compact.The final possiblity of a homogeneous and isotropic universe is

ds2 = −dt2 + a2(t)[dχ2 +R2 sinh2

(χR


)](22.9)

and this is called the open universe. One can obtain this spatial metric by considering the submanifold ofR4 defined by the equation R2 = x2 + y2 + z2 − w2. Some people call it the hyperbolic space because theequation seems to define a hyperbola. The three possibilities combined is called Friedmann–Robertson–Walker (FRW) metric and is the most general homogeneous and isotropic metric.

These metrics are continuously deformable to each other, for example when R → ∞ in the closeduniverse, we recover the flat universe. We can get the open universe from the closed one by analyticcontinuation R→ −iR. So the three kinds of metric are equivalent under certain limits.

Let’s write the metric in a form that is often found in textbooks. Ref. Carroll Chapter 8. Consider theclosed universe metric, let’s define a new variable

r = R sin(χR

), dr = cos

(χR

)dχ (22.10)

60


If we use r to replace χ then we get

ds2 = −dt2 + a2(t)

[dr2

1−Kr2+ r2

(dθ2 + sin2 θ dφ2

)](22.11)

where K is the curvature of the space and K = 1/R2 for closed universe, K = 0 for flat universe andK = −1/R2 for the open universe. This is an elegant way of summarizing, but the more explicit forms weintroduced earlier will be more useful for calculations.

The next logical step will be to plug the metric into the Einstein’s Equations and solve for the evolutionof a(t). Unlike in the Schwartzschild case we need to plug in some Tµν or we will get something uninteresting.The Tµν we would like to choose is the perfect fluid one


A theorem by Weinberg says that for a homogeneous and isotropic universe all the energy-momentumtensors take the form of a perfect fluid.

61


23 Lecture 23

Recall last time we introduced the FRW metric

ds2 = −dt2 + a2(t)[dχ2 +R2 sin2

(χR


)](23.1)

This is the closed universe we wrote down. We need to replace sin2 χ/R by χ2/R2 for flat universe orreplace sin by sinh for open universe. We also wrote it as

ds2 = −dt2 + a2(t)

[dr2

1−Kr2+ r2

(dθ2 + sin2 θ dφ2

)](23.2)

where K takes ±1/R2 or 0 depending on whether you are describing close, open or flat universe. Notethat in a closed universe r < R. When r ≥ R the metric is not defined. This is indeed a closed universe.

The scale of the factor a is arbitrary. We usually set a = 1 today by convention. Today we want tostudy what is the dynamics of a(t). We want to first derive the Hubble’s Law using heuristic arguments.Consider the physical separation of two points separated by distance ∆x, it will be

dphys = a(t)∆x (23.3)

Now assume we have a triangle with vertices A, B and C. The triangle will change its physical size dueto expansion, but the its shape remains the same because the expansion of the universe is isotropic. Thetriangle at t1 must be similar to that at t2. The velocity of A with respect to B is just

vA = dAB = a∆xAB (23.4)

similarly we knowvC = dCB = a∆xCB (23.5)

Now because ∆xAB = dAB/a, we have derived the Hubble law

dAB =a

adAB (23.6)

So the rate of an object moving away from us is proportional to its distance from us. The assumption ofisotropic expansion automatically gives you Hubble’s Law. Note that this law should break down in largerscale because then we need to define precisely what time are we talking about.

We can also have a heuristic derivation of the redshift due to expansion. Think of a photon emittedfrom some source being sent to some observer. Because the universe expands during the time the photontravels between the two points, we would expect that the wavelength of the photon to be stretched due toexpansion, so we expect

λsrc

λobs=a(tsrc)

a(tobs)(23.7)

This is heuristic, but pretty accurate. We will make it precise later.Now we will try to plug the metric into the Einstein’s Equations and solve for the evolution of a. The

form of metric we will use is equation (23.2). We will write down directly the Ricci tensor

Rtt = −3a

a(23.8)

Rrr =aa+ 2a2 + 2K

1−Kr2(23.9)

Rθθ = r2(aa+ 2a2 + 2K

)(23.10)

Rφφ = r2(aa+ 2a2 + 2K

)sin2 θ (23.11)

62


And the Ricci scalar is

R = gµνRµν = 6

[a

a+

(a

a

)2

+K

a2

](23.12)

We also need the energy-momentum tensor, which we use the perfect fluid one


We use a static fluid where Uµ = (1, 0, 0, 0) and Uµ = (−1, 0, 0, 0).Now the tt component of the Einstein’s equation reads

− 3a

a+ 3

[a

a+

(a

a

)2

+K

a2

]= 8πGρ (23.14)

which simplifies to (a

a

)2

=8πGρ

3− K

a2(23.15)

The equation in this form is called the Friedmann equation. A heuristic derivation of this equation can isjust to consider the energy conservation of a particle of unit mass at the surface of a fictitious sphere

1

2a2 −

G(

4π3 a

3ρ)

a= −K

2(23.16)

This reproduces exact the same equation as (23.15).Now we consider the rr component of Einstein’s Equation

1

1−K2

[aa+ 2a2 + 2K − 1

26

[a

a+

(a

a

)2

+K

a2

]]= 8πGP

a2

1−Kr2(23.17)

After simplification we have

− 2a

a− a2

a2− K

a2= 8πGP (23.18)

Now we add the Friedmann equation to this equation, and we due to cancellation we have something verysimple left

− 3a

a= 4πG (ρ+ 3P ) (23.19)

Note that a < 0 if ρ + 3P > 0 whereas a > 0 if ρ + 3P < 0. The second case is actually our universebecause it seems its expansion is actually accelerating.

Now we want to multiply Friedmann equation (23.15) by a2 and take a time derivative. The idea is toget rid of K. We will get

d

dta2 =

d

dt

(8πGρ

3a2

)=⇒ 2aa =

8πG

3

[dρ

dτa2 + 2ρaa

](23.20)

Now we use equation (23.18) to get rid of a and after simplification we get

− a

a(ρ+ 3P ) =

dρ

dt+ 2ρ

a

a=⇒ dρ

dt+ 3

a

a(ρ+ P ) = 0 (23.21)

63


We can manipulate this equation into

d(ρa3)

a3dt= −P da3

a3dt=⇒ d(ρa3) = −Pda3 (23.22)

This looks just like the first law of thermodynamics dE = −PdV . This equation combined with Friedmannequation will give you the evolution of the universe. In the homework we will show that this equationactually comes from the conservation equation ∇µTµν = 0. Now we have two equations but three variables,i.e. a, ρ and P . We need one more equation which is the equation of state relating ρ to P . The equation ofstate tells us what kind of matter are we thinking about. We will give three kinds of examples of equationof state.

The first example is matter. The equation of state for normal matter is just

P = 0 (23.23)

This means that we think of all non-relativistic matter as moving slowly such that P ρ. We also call itdust. The second example is radiation. In this case we have

P =1

3ρ (23.24)

This comes from black-body radiation. We can also obtain this result from comparing the Tµν of electro-magnetic field and that of a perfect fluid. The final example is cosmological constant, i.e. vacuum energy.In that case we have

P = −ρ (23.25)

Let’s say something more for the cosmological constant. The energy-momentum tensor for it is just

Tµν = −ρgµν (23.26)

The cosmological constant is due to the vacuum processes going on everywhere.Let’s do some simple examples. Assume K = 0 the Friedmann equation becomes(

a

a

)2

=8πGρ

3(23.27)

Consider matter P = 0 we have d(ρa3) = 0 and we have ρ ∝ 1/a3. Plug this into the Friedmann equationand solve for a we get

a3/2 ∝ t+ constant (23.28)

If we take the constant to be zero, corresponding to a = 0 at t = 0 at big bang, then we have the simple

a ∝ t2/3 (23.29)

So the universe expands but is decelerating. Then we can find

a

a=

2

3t(23.30)

and we can find the age of the universe by solving for t. Although this is only one universe, the result isfairly general and only off by a factor of about 2.

64


Now we consider radiation in K = 0 universe. We have

d(ρa3) = −1

3ρda3 =⇒ ρ ∝ 1

a4(23.31)

And following the above method we can geta ∝ t1/2 (23.32)

Note in both cases the universe expands forever, but decelerates as it evolves.

65


24 Lecture 24

Last time we introduced the equations of state for matter and radiation

Pmat = 0, Prad =1

3ρrad (24.1)

We would like to make these choices more plausible. First we consider matter, using a box of balls floatingaround and the number density of the balls is n. On the x direction the balls have average momentum pxand velocity v. The pressure on the wall at x direction is just

P =Force

Area∼ npxv (24.2)

Of course a more rigorous way to derive the pressure is to use Boltzmann distribution and do it thestatistical way. But here heuristic way is enough. Now the density is

ρ =Energy

Volume= n

√p2 +m2 (24.3)

Now in nonrelativistic limit we have v 1 and p2 m2, so we know P ρ. For ultrarelativistic particlesv → 1 and p2 m2, so we will get P ∼ ρ. This makes the above equations of state much more plausible.

Now we think about the cosmological constant. We asserted that P = −ρ. Now if we plug into theequation, we get

d(ρa3) = ρda3 (24.4)

This says that ρ is a constant regardless of the expansion of universe. Recall the simple harmonic oscillatorthe energy level is

En =

(n+

1

2

)~ω (24.5)

Here E0 = ~ω/2 is the zero-point energy, which is there even when there is no excitation at all. We thinkof quantum field theory the same way and this zero-point energy also appears. This is an energy which isthere even when there is absolutely nothing, certainly not the expansion of the universe. This is a goodcandidate of this kind of weird material. Now let’s plug this into the Friedmann equation and get(

a

a

)2

=8πGρ

3− K

a2(24.6)

Now for simplicity we set K = 0 for flat universe, then the result is

a ∝ eHt (24.7)

So when there is only vacuum energy, the universe expands exponentially! The acceleration a/a = H2

is definitely larger than zero. This is one way to drive the accelerated expansion of the universe. Thisuniverse is called the de Sitter space with metric

ds2 = −dt2 + e2Ht(dx2 + dy2 + dz2

)(24.8)

and we will come back to this metric later.

66


Now we have two kinds of universes. Both expands but one accelerates and the other decelerates.Suppose we consider the case K > 0 which corresponds to closed universe and for simplicity we considermatter where ρ ∝ 1/a3, the equation is (

a

a

)2

=8πG

3a3− K

a2(24.9)

Now in very early times a→ 0 and the first term is dominant and there seems to be no curvature. Howeverif a→∞ the second term is dominant, but then because K > 0 the right hand side will be negative, whichis absurd. This tells us that there is a maximum for a when the two terms cancel. After that a will notincrease but come back and collapse. Another way to see this is the Newtonian way where we can write

1

2a2 −

G(ρ4π

3 a3)

a= −K

2(24.10)

So when K > 0 there will not be an escape velocity and a must come back. The last point when a comesback to 0 is called the big crunch.

What is our universe then? Our universe has all three kinds of matter, we need to write the Friedmannequation as

H2 =

(a

a

)2

=8πG

3(ρm + ργ + ρΛ)− K

a2(24.11)

Divide both sides by H2, we write it as

1 =8πGρm

3H2+

8πGργ3H2

+8πGρΛ

3H2− K

a2H2= Ωm + Ωγ + ΩΛ + ΩK (24.12)

Today observation tells us that Ωm ∼ 0.3, Ωγ ∼ 10−4, ΩK ∼ 0.7 and ΩK ∼ 0. The last one is accuratewithin 5%. So our universe has all three things except curvature. Now we can write the Hubble constantlike (“0” in subscript and superscript means today)

H2 = H20

[Ω0m

a3+

Ω0γ

a4+ Ω0

Λ +Ω0K

a2

](24.13)

to account for the evolution of various components with respect to time, or the expansion factor. We canplot the evolution of various components of the universe

From the diagram we would expect that at very very early universe the dominant thing is radiation,although is portion today is very small. Because ργ ∝ T 4, at the beginning of the universe the temperaturewill be extremely high as T ∼ 1/a. So we call that the hot big bang.

We now want to rigorously derive the cosmological redshift. We are interested in the trajectory of anull geodesic in the FRW metric, so we have

− dτ2 = 0 = −dt2 + a2dχ2 (24.14)

To integrate the equation we only need to integrate the equation∫ t2

t1

dt

a(t)=

∫ 2

1dχ (24.15)

67


log a

log ρ

log ρΛ

log ρm

log ργ

Figure 24.1: The Evolution of Various Components in the Universe

Now I’m sending the signal from point 1 to point 2, and suppose I emit another signal at a later time weshould have ∫ t2+∆t2

t1+∆t1

dt

a(t)=

∫ t2

t1

dt

a(t)(24.16)

because the coordinate separation of the two points is unchanged regardless of cosmological expansion.Subtracting the integral we have

∆t2a(t2)

=∆t1a(t1)

=⇒ ν1

ν2=

∆t2∆t1

=a(t2)

a(t1)(24.17)

So we actually have λ ∝ a when the universe is expanding. We define

λobs

λemit=

1

aemit= 1 + z (24.18)

where z is the redshift factor. For example, if we say some galaxy has redshift z = 3, then it means thataemit = 1/(1 + z) = 1/4, so the light is emitted when the universe is a quarter of its current size. Now weare equipped to derive the Hubble law from first principles.

We want to first define the notion of distance. We define the comoving radial distance which is just∫dχ =

∫ tobs

temit

dt

a(t)=

∫da

Ha2=

∫ z

0

dz′

H(z′)(24.19)

We can plug in the expression for H in terms of a or z to compute this distance

H(z) = H0

√Ω0m(1 + z)3 + Ω0

γ(1 + z)4 + Ω0Λ + Ω0

K(1 + z)2 (24.20)

We also have another distance which is called Luminosity distance. For this we need to assume astandard candle, which has a known standard luminosity L in the rest frame of that lightsource. We knowthat standard candle send out certain energy flux, so we know

Observed Flux =Energy

Time ·Area=

L

4πr2(χ)(1 + z)2(24.21)

68


Now in the denominator we have r2(χ) because of angular integral. One factor of (1 + z) comes from theredshift of photon, which decreases its energy, and another factor comes from the stretch of unit time againdue to expansion. We define the luminosity distance

dL = r(χ)(1 + z) (24.22)

If we have a bunch of standard candles at different distances, we can plot the distance against z and weexpect to recover Hubble law. If we measure it very precisely and go to very high redshift, then the curvedepend on the functional form of H depending on z, so in that way we can fit the parameters and find thecomposition of the universe today.

69


25 Lecture 25

Remember last time we introduced the luminosity distance and the comoving distance. Ref. the equationsin the last lecture note. Recall we have the following definition

ΩK = − K

a2H2, Ω0

K = − K

H20

(25.1)

In the limit z 1 a good approximation for the comoving distance is

χ(z) ∼ z

H0(25.2)

and the luminosity distance isdL ∼ r(χ) ∼ χ (25.3)

The last ∼ applies when χ is small, so that the curvature is not very important. If we take the equationand put back the speed of light, we get

dL ∼cz

H0(25.4)

which is just Hubble law. However this linear relationship holds only for small z and it is dangerous toextrapolate it to large z. For large z it must deviate from a straight line. H0 today is approximately

H0 ≈ 70 km s−1 Mpc−1 (25.5)

If we divide the speed of light by this number we get

c

H0∼ 103 Mpc (25.6)

This is approximately the size of the observable universe.Consider the following problem. If we send out a photon at big bang, travelling at the radial direction.

Its position at some later time is called the horizon. Because photons travel on null geodesic, the comovingdistance it travels is ∫ t

0

dt′

a(t′)=

∫dχ (25.7)

To translate this into a physical distance we just multiply it by the scale factor at the time interested, weget

d = a(t)

∫ t

0

dt′

a(t′)(25.8)

For radiation dominated flat universe we know a ∝ t1/2 the physical horizon is

d = t1/2∫ t

0

dt′

t′1/2= 2t = H−1 ∝ a2 (25.9)

For matter dominated flat universe we know a ∝ t2/3 and we have H = 2/3t, the horizon is

d = t2/3∫ t

0

dt′

t′2/3= 3t = 2H−1 ∝ a3/2 (25.10)

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In both cases we have the horizon d ∼ H−1, and we call the quantity H−1 the Hubble radius. And inboth cases the horizon grows faster than a. These two observations tells us some problem. If we pick twopoints in the universe today. We know at present that the universe is fairly homogeneous, and the physicalseparation between the two points goes proportional to a. So if today the two points are within horizon,at some earlier point in time they become out of the horizon. If they were outside the horizon, they can’thave communicated so that they equilibriate and get to the same density and temperature. This is calledthe horizon problem.

The way to evade this problem is to break the relation d ∼ H−1. We can’t change the fact that Hubbleradius goes faster than a, so we can only change this. Let’s consider a cosmological constant dominatedflat universe where a ∝ eHt. The horizon is

d = a(t)

∫ t

0

dt

a(t′)= H−1eHt −H−1 (25.11)

For Ht > 1 we have d ∼ H−1eHt H−1 so the horizon is much larger than the Hubble radius. IfH−1 ∼ 10−36 s then if we enforce this kind of exponential amplification for a period of t ∼ 60 × 10−36 sthen the horizon will be e60 larger than the Hubble radius, and this period is called inflation, which solvesthe horizon problem.

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26 Lecture 26

26.1 Time Travel

First we need to talk about an object called cosmic strings. Suppose we have a scalar field with twocomponents

φ = (φ1, φ2) (26.1)

We have a “Mexican Hat” potential V (φ) which is like the Higgs potential.

φ1

V (φ)

φ2

Figure 26.1: Mexican Hat Potential

When the field goes around the minima of this potential, and there is something in the middle, so itpicks up a non-trivial factor. This is sometimes called a topological defect, which is robust under deformand perturbations. If the defect extends in one direction then it is called a cosmic string. We take it to beinfinitely thin. The energy-momentum tensor is

Tµν = δ(x)δ(y)µ

1 0 0 00 0 0 00 0 0 00 0 0 −1

(26.2)

where µ is the mass per unit length. Note the cosmoic string has negative pressure just like the cosmologicalconstant, but it is confined on one dimension. We are not going to justify this form of energy-momentumtensor. If we plug this into the Einstein equation we will find the space to be almost flat, with expression

ds2 = −dt2 + dz2 + dr2 + (1− 4πGµ)2 r2dφ2 (26.3)

This is just a constant correction to the flat cylindrical Minkowski coordinate. In fact if we do a transfor-mation of coordinate φ′ = (1− 4πGµ)φ then the metric looks exactly flat

ds2 = −dt2 + dz2 + dr2 + r2dφ′2 (26.4)

But there is one point. Originally we have φ ∈ [0, 2π). But now with the new variable we have φ′ ∈[0, 2π(1− 4πGµ)). The difference is called the deficit angle

∆φ = 8πGµ = 2α (26.5)

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This is something like a flat space with a wedge cut out and glued into a cone. Note this is purely spatial,nothing about time yet.

Now let’s consider what happens in the x-y plane. We can draw a diagram

B Ad

w0w0 y0 y0

x0 x0

α

Figure 26.2: Light Signal in Presence of Cosmic String

We have from geometry the following relation

w20 = (d+ y0 cosα)2 + (x0 − y0 sinα)2 (26.6)

This is the way photon travels from A to B that crosses the wedge. We want to minimize it by varying y0

dw20

dy0= 0 = d cosα− x0 sinα+ y0, y0 = x0 sinα− d cosα (26.7)

Now we want to compare w0 with x0

w20 − x2

0 = d2 + 2dy0 cosα− 2x0y0 sinα+ y20 (26.8)

Now if we plug in the y0 which takes w20 to be minimum, we get

w20 − x2

0 = d2 − y20 (26.9)

So w0 is smaller than x0 when d is smaller than y0. Now this is only the trajectory of light. But the factthat the path length is shorter tells us that it is possible to achieve this for speeds less than that of light.Because w0 < x0 we can always find some v < 1 such that

w0

v< x0 (26.10)

Now if we have a rocket that travels at v and going along the trajectory w0 we can go from A and arriveat B even before photon arrives.

Now let event 1 to be the event when the rocket takes off A. The space-time coordinate of the event is

E1 =

(−w0

v, x0, 0, 0

)(26.11)

The event 2 of the rocket arriving at B has space-time coordinate

E2 =(w0

v,−x0, 0, 0

)(26.12)

Consider the space-time separation between E1 and E2, which is just ds2. We get

∆s2 = −4w20

v2+ 4x2

0 (26.13)

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Now because we know w0/v < x0 we find that the space-time separation is larger than 0, so this separationis spacelike! So we have managed to send a rocket that is slower than the speed of light across a space-likeseparation. This is essentially a short-cut to connect two space-like separated points, which we thoughtwas impossible.

The meaning of a space-like separation is that the order of events can be arbitrary. We can find areference frame such that the two events happen at the same time! What’s more, we can even boost intoa frame that the rocket lands at B before it even departs A. However, this is not quite a time machine.Let’s put another cosmic string adjacent to the original one with the wedge in the other direction. We cannow have a complete loop, i.e. sending a rocket from A to B and come back to A to the same event. Ifwe let the upper string move to the right and the lower string move to the left, then essentially we have atime machine.

For a complete treatment refer to Gott 1991, Phys. Rev. L. ,66, P1126.

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