XITS Math [version=setB,StylisticSet=1]XITS Math

1

Cool stuff in General Relativity

Afiq Hatta

January 13, 2020

Contents

1 Starting off with non-relativistic particles 41.1 Exploring different Lagrangians . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 Special relativity 5

3 Introducing Differential Geometry for General Relativity 73.1 Tangent vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83.2 Treating tangent vectors as derivatives of curves on the manifold . . . . . . . . . . . 103.3 Vector fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103.4 Integral curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113.5 Differentiating vector fields with respect to other vector fields . . . . . . . . . . . . . 13

3.5.1 Push-forwards and Pull-backs . . . . . . . . . . . . . . . . . . . . . . . . . . . 133.5.2 Components for Push-forwards and Pull-backs . . . . . . . . . . . . . . . . . 133.5.3 Introducing the Lie derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

4 Tensors 154.1 A note about dual spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

4.1.1 Vector and Covector spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164.2 Taking the Lie Derivative of Covectors . . . . . . . . . . . . . . . . . . . . . . . . . . 174.3 Tensor fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194.4 Differential forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

4.4.1 Wedge products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214.4.2 Properties of wedge products . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

4.5 The exterior derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224.5.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

4.6 Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254.6.1 Integrating over submanifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . 254.6.2 Stokes’ theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

5 Introducing Riemannian Geometry 285.1 The metric . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

5.1.1 Riemannian Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285.1.2 Riemannian Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295.1.3 The Joys of a metric . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

5.2 Connections and Curvature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

2

5.3 The Levi-Civita Connection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345.3.1 The Divergence Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

5.4 Parallel Transport . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365.4.1 Geodesics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

5.5 Normal coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365.5.1 Constructing Normal Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . 38

5.6 Path Deviations and Curvature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385.7 Derived tensors from the Riemann Tensor . . . . . . . . . . . . . . . . . . . . . . . . 405.8 Connection 1-forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

6 The Einstein Equations 436.0.1 The Einstein-Hilbert Action . . . . . . . . . . . . . . . . . . . . . . . . . . . . 436.0.2 Dimensional analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 456.0.3 The Cosmological Constant . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

6.1 Diffeomorphisms Revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 456.2 Some simple solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

6.2.1 De Sitter Spacetime . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 466.2.2 Anti de-Sitter Space-time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

6.3 Symmetries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 506.3.1 More examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

6.4 Conserved quantities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 526.5 Asymptotics of Spacetime . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

6.5.1 Penrose Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 536.5.2 4 dimensional Minkowski space . . . . . . . . . . . . . . . . . . . . . . . . . . 556.5.3 de Sitter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

7 Coupling to matter 577.1 Energy Conservation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

7.1.1 A foray into conservation in Electromagnetism . . . . . . . . . . . . . . . . . 59

8 When Gravity is weak 628.1 Linearised theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

8.1.1 The Fierz-Pauli Action . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 638.1.2 Gauge Symmetries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

8.2 The Newtonian Limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 658.3 Gravitational Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 658.4 Measuring gravitational waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 668.5 Making Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 688.6 Power Radiated . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 708.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

8.7.1 Gauge symmetries in linearised theory . . . . . . . . . . . . . . . . . . . . . . 738.7.2 Gravitational wave solutions and polarisations . . . . . . . . . . . . . . . . . . 738.7.3 Gravitational waves from a source . . . . . . . . . . . . . . . . . . . . . . . . 73

9 Example Sheet 1 749.1 Question 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 759.2 Question 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 769.3 Question 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

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9.4 Question 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 789.5 Question 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 809.6 Question 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 819.7 Question 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 839.8 Question 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 859.9 Question 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 869.10 Question 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

10 Example Sheet 2 88

11 Example Sheet 4 8911.1 Question 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8911.2 Question 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8911.3 Question 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

11.3.1 First part . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9211.3.2 Second Part . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

11.4 Question 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9511.4.1 Deriving the Linearised Einstein Hilbert action . . . . . . . . . . . . . . . . . 95

11.5 Question 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

12 Question 6 98

13 Problem Solving Tips 10013.1 Being careful about how we expand things . . . . . . . . . . . . . . . . . . . . . . . . 10013.2 Tricks in the Levi-Civita connection . . . . . . . . . . . . . . . . . . . . . . . . . . . 10013.3 Integrating over volume forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10013.4 Showing that magnitude is constant . . . . . . . . . . . . . . . . . . . . . . . . . . . 10013.5 Symmetries of the Riemann Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10013.6 General techniques . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

1 Starting off with non-relativistic particles

Deriving the Euler Lagrange equations. We perturb our action

S =

∫ t2

t1

dtL

If we perturb our action slightly,

S[xi + δxi] =

∫ t2

t1

dtL(xi + δxi, xi + δxi)

= S[xi] +

∫ t2

t1

dt

(∂L

∂xiδxi +

∂L

∂xiδxi)

= S[xi] +

∫ t2

t1

dt δxi(∂L

∂xi− d

dt

(∂L

∂xi

))This implies that the integrand goes to zero.

4

1.1 Exploring different Lagrangians

Consider the Lagrangian in Euclidean coordinates

L =1

2m(x2 + y2 + z2)

The E-L equations imply that x = 0. So we have a constant velocity. Now in different coordi-nates,

L =1

2gij(x)xixj

This is a metric. Our distance in these general coordinates between xi → xi + δxi is now

ds2 = gijdxidxj

Some gij do not come from R3, and these spaces are curved. This means there is no smoothmap back into R3. Our equations of motion that comes from the Euler Lagrange equations are thegeodesic equations.

Observe that

d

dt

(∂L

∂xi

)=

d

dtgikx

k

= gikxk + xkxl∂lgik

And, differentiating the lagrangian with ∂i, we get

∂L

∂xi=

1

2∂igklx

kxl

Substiting this into the EL equations, this reads

gikxk + xkxl∂lgik −

1

2∂igklx

kxj = 0

Now, the second term is symmetric in k, l, so we can split this term in two. We also multiply bythe inverse metric to cancel out this annoying factor of g that we have in front of everything. Thisgives us the final expression that

xi +1

2gil (∂kglj + ∂jglk − ∂lgjk) xj xk = 0

2 Special relativity

We’ll now put time and space on the same ’footing’ per se, and talk about special relativity. Inspecial relativity, instead of time being its own separate variable, we have that dynamic eventstake place in 4 spacetime coordinates denoted xµ = (t, x, y, z), where we now use greek indices todenote four components µ = 0, 1, 2, 3. Now, we wish to construct an action and extremize this path,but since our t variable is already taken, we need to parametrise paths in spacetime by a differentparameter. We’ll call this parameter σ, and show that there’s a natural choice for this, somethingcalled ’proper time’, later.

5

We define our metric, the Minkowski metric, on this spacetime to be ηµν = diag(−1,+1,+1,+1).Thus, distances in Minkwoski spacetime are denoted as

ds2 = ηµνdxµdxν = −dt2 + dx2 + dy2 + dz2

We have names for different events based on their infinitesimal distance. Since our metric is nolonger positive definite, we have that events can have a distance of any sign.

• If ds2 < 0, events are called timelike.

• If ds2 = 0, events are called null.

• If ds2 > 0, events are called spacelike.

Our action, then, should look like (now with the use of an alternate parameter σ to parametrizeour paths)

S[xµ(σ)] =

∫ σ2

σ1

√−ds2

Now, we can parametrise the integrand with sigma to get

S[xµ(σ)] = m

∫ σ2

σ1

dσ

√−ηµν

dxµ

dσ

dxν

dσ

In this case, our Lagrangian L = m√−ηµν dx

µ

dσdxν

dσ . Now, before we begin analysing what thisequation gives us, there are two symmetries we’d like to take note of. One of our symmetriesis invariance under Lorentz transformations. This means, if we boost our frame with a Lorentztransformation xµ → Λµνxν , one can easily verify, using the condition that

ΛαµΛβνηαβ = ηµν

that the Lagrangian remains invariant under this. One can also verify that this action is invariantunder reparametrisation of the curve via a new function σ′ = σ′(σ).

Using the chain rule, we reparametrise by rewriting the action as

S = m

∫dσ

dσ′dσ′

√−ηµν

(dσ′

dσ

)2 dxµ

dσ′dxν

dσ′=

∫dσ′√−ηµν

dxµ

dσ′dxν

dσ′

In this case, we’re just applying the chain rule but factoring out the the dσ′

dσ term. But this isexactly the same as what we had before. Thus, we have reparametrisation invariance.In analogywith classical mechanics, we compute the conjugate momentum

pµ =∂L

∂xµ

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3 Introducing Differential Geometry for General Relativity

Our main mathematical objects of interest in general relativity are manifolds. Manifolds are topo-logical spaces which, at every point, has a neighbourhood which is homeomorphic to a subset ofRn, where we call n the dimension of the manifold. In plain English, manifolds are spaces in which,locally at a point, look like a flat plane. This can be made more rigourous by the creation of maps,which we call ’charts’, that take an open set around a point (a neighbourhood), and mapping thisto a subset of Rn.

Precisely, for each p ∈ M, there exists a map φ : O → U ⊂ Rn, where p ∈ O ⊂ M, and O isan open set of M defined by the topology. Think of φ as a set of local coordinates, assigning acoordinate system to p. We will write φ(p) = (x1, . . . xn) in this regard.

This map must be a ’homeomorphism’, which is a continuous, invertible map with a continuousinverse. In this sense, our idea of assigning local coordinates to a point in M becomes even moreclear.

We can define different charts to different regions, but we need to ensure that they’re we’ll behavedon their intersections. Suppose we had two charts and two open sets defined on our manifold, andlooked at how we transfer from one chart to another. For charts to be compatible, we require thatthe map

φα φ−1β : φβ(Oα ∩ Oβ)→ φα(Oα ∩ Oβ)

is also smooth (infinitely differentiable).

A collection of these maps (charts) which cover the manifold is called an atlas, and the maps takingone coordinate system to another (φα φ−1

β ), are called transition functions.

Some examples of manifolds include

• Rn and all subsets of Rn are n-dimensional manifolds, where the identity map serves as asufficent chart.

• S1, S2 are manifolds, with modified versions of polar coordinates patched together forming achart (as we’ll see in the case of S1.

Let’s start simple and try to construct a chart for S1. Our normal intuition would be to use a singlechart S1 → [0, 2π), which indeed covers S1 but doesn’t satisfy the condition that the target set isan open subset of R. This yields problems in terms of differentiation functions at the point 0 ∈ R,because the interval is closed there, not open. One way to remedy this is to define two coordinatecharts then path them together to form an atlas. Our first open set will be the set of points on thecircle which exclude the rightmost point on the diameter, a set denoted by O1, and our second openset is the whole sphere excluding the leftmost point. We’ll denote this O2.

We assign the following charts which are inline with this geometry

φ1 : O1 → θ1 ∈ (0, 2π)

φ2 : O2 → θ2 ∈ (−π, π)

It’s easy to verify that if we take a point on the manifold, our transition matrix reads that

θ2 = φ2(φ−11 (θ1)) =

θ1, θ1 ∈ (0, π)

θ1 − 2π, θ1 ∈ (π, 2π)

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Now that we have coordinate charts, we can do things that we usually do on functions described inRn, like differentiate. Furthermore, we can define maps between manifolds (which don’t necessarilyhave the same dimension), where smoothness is defined via smoothness on coordinate charts. Theseare called diffeomorphisms. A function

f :M→N

is a diffeomorphism if the corresponding map between RdimM and RdimN is smooth:

ψ f φ−1 : U1 → U2

for all coordinate charts φ : O1 → U1 and ψ : O2 → U2 defined on the manifolds M and Nrespectively.

3.1 Tangent vectors

Throughout our whole lives, we’ve been thinking of a ’vector’ as a way to denote some positionin space. However, this idea of a vector is only really unique to the manifold Rn. A much moreuniversal concept of a vector is the idea of ’velocity’, the idea of movement and direction at a givenpoint. A tangent vector is a ’derivative’ form at a given point in the manifold. This means that wedefine it to obey properties that one might expect in our usual notion of a derivative for functionsin R. We denote a vector at a point p ∈ M as Xp. This means that a vector is simply a mapXp : C∞ → R, which satisfies

• Linearity:Xp(αf + βg) = αXp(f) + βXp(g), ∀f, g ∈ C∞(M), α, β ∈ R

• Xp(f) = 0 for constant functions on the manifold.

• Much like the product rule in differentiation, tangent vectors should also obey the Leibnizrule where

Xp(fg) = f(p)Xp(g) + g(p)Xp(f)

Remember that with the Leibniz rule, the functions which are not differentiated are evaluatedat p! This is useful for our theorem afterwards.

This next proof is about showing that tangent vectors can be built from differential operators inthe n dimensions of the manifold. We will now show that all tangent vectors Xp have the propertythat they can be written out as

Xp = Xµ ∂

∂xµ

∣∣∣∣p

What we’re saying here is that ∂µ at the point p ∈M forms a basis for the space of tangent vectorsat a point. To do this, take your favourite arbitrary function f :M→ R. Since this is defined onthe manifold, to make our lives easier we’ll define F = f φ−1 : Rn → R, which we know how todifferentiate. The first thing we’ll show is that we can locally move from F (x(p)) → F (x(q)) bydoing something like a Taylor expansion:

F (x(q)) = F (x(p)) + (xµ(q)− xµ(p))Fµ(x(p))

Here, we’re fixing p ∈ M and Fµ is some collection of n functions. One can easily verify that Fcan be written in this way by precisely doing a Taylor expansion then factorising out the factors of

8

(xµ(q) − xµ(p)). We can find an explicit expression for Fµ(x(p)) by differentiating both sides andthen evaluating at x(p). We have that

∂F

∂xν

∣∣∣∣x(p)

= δµνFµ + (xµ(p)− xµ(p))∂Fµ∂xν

∣∣∣∣x(p)

= Fν

The second term goes to zero since we’re evaluating at x(p), and our delta function comes fromdifferentiating a coordinate element. Our initial F (x(p)) term goes to zero since it was just aconstant. Recalling that φ−1 xµ(p) = p, we can just rewrite this whole thing as

f(q) = f(p) + (xµ(q)− xµ(p))fµ(p)

where in this case we’ve defined that fµ(p) = Fµ φ−1. However, we can figure out what this isexplicitly

fµ(p) = Fµ φ(p) = Fµ(x(p)) =∂F (x(p))

∂xµ=f φ−1(x(p))

∂xµ:=

∂f

∂xµ

∣∣∣∣p

Now, its a matter of applying our tangent vector to our previous equation, recalling that Xp(k) = 0for constant k, and that all functions are evaluated at the point p. We have that, upon applicationof the Leibniz rule

Xp(f(q)) = Xp(f(p)) +Xp(xµ(q)− xµ(p))fµ(p) + (xµ(p)− xµ(p))Xp(fµ(p))

= Xp(xµ(p))fµ(p)

= Xµfµ(p)

= Xµ ∂f

∂xµ

∣∣∣∣p

In the first line we’ve replaced q with p in the last term since Leibniz rule forces evaluation atp. We’ve declared Xµ = Xp(x

µ) as our components. Since f was arbitrary, we have now writtenthat

Xp = Xµ ∂

∂xµ

To show that ∂µ forms a basis for all tangent vectors, since we’ve already shown that they spanthe space we need to show they’re linearly independent. Suppose that

0 = Xµ∂µ

Then, this implies that if we take f = xν , then 0 = Xν for any value of the index ν we take. So,we have linear independence.

Tangent vectors should be basis invariant objects

A tangent vector is a physical thing. However, so far we’ve expressed it in terms of the basis objects∂µ which are chart dependent.So, suppose we use a different chart which is denoted by coordinatesxµ. This means that our new tangent vector needs to satisfy the condition that

Xp = Xµ ∂

∂xµ

∣∣∣∣p

= Xµ ∂

∂xµ

∣∣∣∣p

9

This relation allows us to appropriately relate the components Xµ to that of Xµ, in what is calleda contravariant transformation. Using the chain rule, we have that

Xµ ∂

∂xµ

∣∣∣∣p

= Xµ ∂xν

∂xµ

∣∣∣∣φ(p)

∂

∂xν

∣∣∣∣p

Notice that when differentiating a coordinate chart with respect to another, we’re evaluating atthe coordinate chart of the point. This is why we subscript with φ(p) in the terms. Comparingcoefficients, we have that

Xν = Xµ ∂xν

∂xµ

∣∣∣∣φ(p)

3.2 Treating tangent vectors as derivatives of curves on the manifold

We present a different way to think of tangent vectors, which is viewing them as ’differential oper-ators along curves’. Consider a smooth curve along our manifold, which we can parametrise fromon an open onterval I = (0, 1), and define the starting point of this curve at p ∈M;

λ : (0, 1)→M, λ(0) = p

We now ask the question, how do we differentiate along this thing? To do this, we’ll have toapply coordinate charts so that we can make sense of differentiation. So, suppose we would like todifferentiate a function f along this manifold. We apply our chart φ to λ to get a new functionφ λ : R→ Rn, which we’ll suggestively write as xµ(t). In addition, to be able to differentiate f ina sensible way we also construct the function F = f φ−1 . Thus, differentiating a function alonga curve xµ(t) = φ λ(t) should look like

d

dtf(t) =

d

dt

(F φ−1 φ λ(t)

)=

d

dt

(F φ−1 xµ(t)

)∣∣∣∣t=0

=dxµ

dt

∂F φ−1

∂xµ

∣∣∣∣φ(p)

=dxµ

dt

∣∣∣∣t=0

∂f

∂xµ

∣∣∣∣p

= Xµ∂µ(f)

= Xp(f)

Thus, differentiating along a curve gives rise to a tangent vector acting on f .

3.3 Vector fields

Thus far we’ve defined tangent spaces at only a specific point in the manifold, but we’d like to knowhow we can extend this notion more generally. A vector field X is an object which takes a function,and then assigns it avector at any given point p ∈M. So, we’re taking

X : C∞(M)→ C∞(M), f 7→ X(f)

10

X(f) is a function on the manifold which takes a point, and then differentiates f according to thetangent vector at that point.

X(f)(p) = Xp(f), X = Xµ∂µ

In this case, Xµ is a smooth function which takes points on the manifold to the components of thetangent vector Xµ

p at p. We call the space of vector fields X as X (M). So, since X(f) is now also asmooth function on the manifold, we can apply another vector field Y to it, for example. However,is the object XY a vector field on it’s own? The answer is no, because vector fields also have toobey the Liebniz identity at any given point, ie the condition that

X(fg) = fX(g) + gX(f)

However, the object XY does not obey this condition since

XY (fg) = X(fY (g) + gY (f))

= X(f)Y (g) + fXY (g) +X(g)Y (f) + gXY (f)

6= gXY (f) + fXY (g)

We do get from this however, thatXY − Y X := [X,Y ]

does obey the Leibniz condition, because it removes the non-Leibniz cross terms from our differen-tiation. The commutator acts on a function f by

X(Y (f))− Y (X(f)) = [X,Y ]f

One can check that the components of the new vector field [X,Y ]µ are given by

[X,Y ]µ = Xν ∂Yµ

∂xµ− Y ν ∂X

µ

∂xν

The commutator obeys the Liebniz rule, where

[X, [Y,Z]] + [Y, [Z,Z]] + [Z, [X,Y ]] = 0

3.4 Integral curves

We’ll now do an interesting diversion to discuss flows on a manifold. Think of a body of watermoving smoothly on a surface, where each point on the manifold moves to a different point aftersome amount of time. This is a what a flow is. More specifically, it is a smooth map σt :M→Mon the manifold (which makes it a diffeomorphism), where t is our ’time’ parameter that we weretalking about. As such, these flow maps actually form an abelian group, where

• σ0 = IM. So, after time t = 0 has passed, nothing has moved so we have that this is theidentity map.

• If we compose the same flow after two intervals in time, we should get the same flow whenwe’ve let the sum of those times pass over. So,

σs σt = σs+t

11

If we take a flow at a given point p ∈M, we can define a curve on the manifold by setting:

γ(t) : R→ (M), γ(t) = σt(p)

where without loss of generality we have γ(0) = p. Since this is a curve, we can define it’s associatedcurve in Rn space with a given coordinate chart, and hence associate with it a tangent vector Xp.We can also work backwards. From a given vector field, we can solve the differential equation

Xµ(x(t)) =dxµ(t)

dt

with the initial condition that xµ(0) = ψ(p) at some point in the manifold, and have that thisdefines a unique curve. The set of curves then together form a flow. Thus, we’ve seen a one to onecorrespondence between vector fields and flows.

12

3.5 Differentiating vector fields with respect to other vector fields

3.5.1 Push-forwards and Pull-backs

A sensible question to now as is that, since we have these smooth vector fields, how do we differentiatea vector field with respect to another one? For example, if we have X,Y ∈ M, what constitutesthe notion of a change in X with respect ot Y . The notion of derivatives on manifolds is difficultbecause we can’t compare tangent spaces at different points in the manifold, for example Tp(M) andTq(M) are tangent spaces at different points, and we could define different charts for each space,hence we have some degrees of freedom (and our derivative wouldn’t make sense). To make senseof comparing different tangent spaces, we need to create way to compare the same functions, buton different manifolds. These are called push forwards and pull backs.

Let’s start by defining a smooth map between manifolds φ : M → N (φ is not a chart here). We’renot assuming thatM and N are even the same dimension here, and so we can’t assume φ−1 doesn’teven exists.

Suppose we have a function f : N → R. How can we define a new function based on f that makessense, which goes from M → R? We define the pull back of a function f , denoted (ψ∗f) : M → Ras

(ψ∗f)(p) = f(ψ(p)), p ∈M

So, we’ve converted this thing nicely.

Our next question then is how, from a vector field Y ∈ X (M), can we make a new vector field inX ∈ X (M)? We can, and this is called the push-forward of a vector field, denoted φ∗Y ∈ X (N).We define that object as the vector field which takes

(φ∗Y )(f) = Y (φ∗f)

This makes sense because φ∗f ∈ C∞(M), so applying Y makes sense. Now, to show φ∗Y ∈ X (N),we should verify that

φ∗Y : C∞(N)→ C∞(N), f 7→ C∞

Well, this object philosophically maps

f 7→ φ∗Y (f) = Y (φ∗f)

But the object on the left hand side is a vector field ready to be turned into a tangent vector whenwe assign it to a point on the manifold:

p 7→ Yp(φ∗f)

Hence this object agrees with our definition. The fact that we have a map φ : M → N and arepushing the vector field from X (M) to X (N) is the reason why we call this new mapping a pushforward.

3.5.2 Components for Push-forwards and Pull-backs

Now, since ψ∗Y is a vector field, it’s now in our interest to find out about what the components arefor this thing. We want to find that the components (ψ∗Y )ν such that

ψ∗Y = (ψ∗Y )ν∂ν

13

We can work first by assigning coordinates to φ(x), which we denote by yα(x) = φ(x), x ∈M,α = 1, . . . dim(N). If we write out our vector field Y as Y = Y µ∂µ, then our push-forward mapin summation convention looks like

(φ∗Y )f = Y µ∂f(y(x))

∂xµ= Y µ ∂y

α

∂xµ∂f

∂yα

In the second equality, we’ve applied the chain rule. Remember, y pertains to coordinates in themanifold N , so on our push-forward, we have that our new components on the manifold N , we havethat

(φ∗Y )α = Y µ ∂yα

∂xµ

3.5.3 Introducing the Lie derivative

In what we’ve just presented, some objects are naturally pulled back and some are naturally pushedforward. However, things become when our map between manifolds is a diffeomorphism and henceinvertible; which means we can pull back and push forward with whatever objects we want. We canuse this idea to differentiate vector fields now. Recall that if we’ve given a vector field, X ∈ X (M),we can define a flow map σt : M → M This flow map diffeomorphism allows us to push vectorsalong flow lines in the manifolds, from the tangent spaces

Tp(M)→ Tσt(p)(M)

This is called the Lie derivative LX , a derivative which is induced by our flow map generated by X.For functions. we have that

LXf = limt→0

f(σt(x))− f(x)

t=df(σt(x))

dt

∣∣∣∣t=0

However, the effect of doing this is exactly the same as if we were to apply the vector field X to thefunction:

df

dxµdxµ

dt

∣∣∣∣t=0

= Xµ ∂f

∂xµ= X(f)

Thus, a Lie derivative specialised to the case of functions just gives us LX(f) = X(f). Now, thequestion is about how we can do this differentiation on vector fields. What we need to do is to ’flow’vectors at a point back in time to where they originally started, and look at this difference.

(LXY )p = limt→0

(σ∗−t(Y )p − Yp)t

Let’s try and compute the most basic thing first, the Lie derivative of a basis element of the tangentspace ∂µ:

σ∗−t∂µ = (σ∗−t∂µ)ν∂ν

Let’s try and figure out what (σ∗−t∂µ)ν is. Because of the fact that the diffeomorphism is inducedby the vector field X, we have that

(σ∗−t(x))ν = xν − tXν + . . .

14

Thus our components of a push-forward of an arbitrary vector field are given by

(σ∗−tY )ν = Y σ ∂(σ−t(x))ν

∂xσ, in our case Y σ = δσµ

Substituting the expressions above with one another gives us that

(σ−t(x)∂µ)ν = δνµ − t∂Xν

∂xµ+

Contracting this with ∂ν , and then subtracting off ∂µ, we have that

LX∂µ = −∂Xν

∂xν∂ν

We require that a Lie derivative obeys the Leibniz rule, so we have that applying on a general vectorfield Y ,

LX(Y ) = LX(Y µ∂µ)

= LX(Y µ)∂µ + Y µLX(∂ν)

= Xν ∂Yµ

∂xν∂µ − Y µ∂X

ν

∂xµ∂ν

We realise that this however are just components of the commutator! So

LX(Y ) = [X,Y ]

4 Tensors

4.1 A note about dual spaces

In linear algebra, if we have a vector space which we call V , then we can define a natural objectwhich we call it’s dual space, denoted by V ∗. The dual space is the space of linear functions whichtakes V → R:

V ∗ = f | f : V → R, f is linear

Now it may seem from first glance that the space of all functions is a lot larger than our originalvector space, so it’s counter intuitiveto call it the ’dual’. However, we can prove that these vectorspaces are isomorphic. Suppose that eµ is a basis of V . Then we pick what we call a dual basisof V ∗, by choosing the

B(V ∗) = fµ | fµ(eν) = δµν

One can show that this set indeed forms a basis of V ∗.

Theorem.The above basis forms a basis of V ∗.

Proof. First we need to show that the linear maps above, span our space. This means we need tobe able to write any linear map, say ω, as a linear combination

ω =∑

ωµfµ, ωµ ∈ F

15

To do this, we appeal to the fact that if two linear maps agree on the vector space’s basis, then theyagree. So, let the values that ω takes on the basis be ωµ = ω(eµ). Then, taking

Ω =∑

ωµfµ

We find that Ω also satisifes Ω(eµ) = ωµ. Thus, the maps are the same. Hence, ω can be writtenas the span of our dual basis vectors. To show that these basis vectors are linearly independent, weassume that there exists a non trivial sum such that they add to the zero map.

0 =∑

λµfµ

If we apply this map to an arbitrary basis vector ei, then we get

0 =∑

λµfµ(ei) = λi

for arbitrary i. Hence, we must have that all λi are zero. Thus the basis vectors are independent.

Now, assuming that our original vector space V had finite dimension, the way we’ve defined thebasis of V ∗ means that we had the same number of basis elements. This means that V and V ∗ havethe same dimesion. One can prove that vector spaces with the same dimension are isomorphic, sowe have that

V ' V ∗

Think of a dual space as a ’flip’ of a vector space. We can identify the dual of a dual space as theoriginal space itself, so that

(V ∗)∗ = V

This is because given an object in the dual space ω, we can define a natural map from V : V ∗ → Rgiven by

V (ω) = ω (V ) ∈ R

4.1.1 Vector and Covector spaces

Now, since we’ve identified tangent spaces as vector spaces, we can proceed to construct its dual.If we have a tangent space Tp(M) with a basis eµ, our natural dual basis is given by

B(T ∗p (M)) = fµ | fµ(eν) = δµν

The corresponding dual space is denoted as T ∗p (M), and is known as the cotangent vector space.For brevity, elements in this space are called covectors. In this basis, the elements fµ have theeffect of ’picking’ out components of a vector V = V µeµ.

fν(V ) = fν(V µeµ) = V µfν(eµ) = V ν

There’s a different way to pick elements of this dual space in a smooth way. They’re chosen bypicking elements of a set called the set of ’one forms’. We denote the set of one forms, with an index1 as Λ1(M). We can construct elements form this set by taking elements from C∞(M). Supposethat we have an f ∈ C∞(M), then the corresponding one-form is a map

df : Tp(M)→ R V 7→ V (f)

16

From one the set of one forms, we then have an obvious way to get the dual basis. The dual basisis obtained by just taking the coordinate element of the manifold, so that our one form is

dxν : Tp(M)→ R

This satisfies the property of a dual basis, since

dxν(eµ) =∂xν

∂xµ= δνµ

With this convention, we can check that V (f) is what its supposed to be by observing that

df(X) =∂f

∂xµdxµ(Xν∂ν) = Xν ∂f

∂xν= X(f)

So we recover what we expect by setting this as a basis. Now, we should check whether a changein coordinates leaves our properties invariant.

Suppose we change our basis from xµ → zµ(x), then, we know that our basis vector transformslike

∂

∂xµ=∂xµ

∂xνdxν

We guess that our basis of one forms should transform as

dxµ =∂xµ

∂xνdxν

This ensures that, when we contract a transformed basis one form with a transformed basis vector,that

dxµ∂

∂xν=∂xµ

∂xρdxρ

∂xσ

∂xν∂

∂xσ

=∂xµ

∂xσ∂xσ

∂xνdxρ

(∂

∂dxσ

)=∂xµ

∂xρ∂xρ

∂xν= δµν

It’s no coincidence that this looks like a Jacobian!

Now, as with basis elements in our vector space, we need to determine how these objects transformunder a change of basis. *Need to finish this section on basis transformations for covectors*

4.2 Taking the Lie Derivative of Covectors

We would like to repeat what we did for vectors, and take derivatives of covectors. To do this, weneed to define pull-backs for covectors. Suppose we had a covector ω living in the tangent spaceof some manifold N , in T ∗p (N). We can then define the pull back of this vector field based on firstpushing forward the vector field X. Thus, if φ : M → N , then we define the pullback φ∗ω as thecovector field in Tp(M) as

(φ∗ω)(X) = ω(φ∗X)

What information can we glean from this? Well, we can try to figure out what the components ofφ∗ω are. If we let yα to be coordinates on N , then we expand this covector as

ω = ωµdyµ

17

Also recall that the components of a pushed forward vector field are

(φ∗X)µ = Xν ∂yµ

∂xν

Now, if we take the equation(φ∗ω) = ω(φ∗X)

Then, expanding out in terms of components, we have that

(φ∗ω)µdxµ(Xνeν) = wµdyµ(φ∗X)ν∂

∂yν

Remember, we have to expand in the correct coordinates. The object φ∗ω lives in the space M sowe expand in the dxµ basis. On the other hand we have that ω originally lives in N so we expandin dyµ. Substituting our expression for our push forward vector field, and we get that

(φ∗ω)µXµ = ων

∂yν

∂xµXµ

This step requires a bit of explanation. After we substitute in the components for the pushedforward vector field, we then use the fact that on both manifolds, our basis vectors and our basiscovectors contract to give a delta function

dxµ(

∂

∂xν

)= δµν , dyµ

(∂

∂yν

)= δµν

This implies that the components of our pulled back vector field are

(φ∗ω)µ = ων∂yν

∂xµ

As in the case of vectors, we can also make rigourous the definition of a Lie derivative with respectto a covector field. This is denoted LXω, where X is our underlying vector field we’re differentiatingwith. If our vector field X imposes a flow map which we label as σt, then our corresponding Liederivative is derivative is defined as

LXω = limt→0

(σ∗t ω)− ωt

There’s an important point to be made here. In our previous definition of a Lie derivative for avector field, we took the inverse diffeomorphism σ−t. But in this case, we need to take t positivesince we’re doing a pull-back instead of a pushforward, like we did with vector fields.

Let’s go slow and try to compute the components of this derivative. Recall that for a flow map, wehave that infinitesimally,

yν = xν + tXν , =⇒ δνµ + tdXν

dxµ

Thus for a general covector field, our components for the pull back are

(σtω)µ = ωµ + tωνdXν

dxµ

Hence, the components of a basis element under this flow becomes

(σ∗t dxν) = dxν + tdxµ

dXν

dxµ

18

So, taking the limit, we have that our components of our Lie derivative are given by

limt→0

σ∗t dxν − dxν

t= dxµ

dXν

dxµ

Now, as before, we impose the Liebniz property of Lie derivatives, and expand out a general covector.Hence, we have that

LX(ωµdxµ) = ωµLdxµ + dxµLX(ωµ)

= ωµdxν dX

ν

dxµ+ dxνXµdων

dxµ

This implies that our components of our Lie derivative can be written nicely as

LX(ω)µ = (Xν∂νων + ων∂µXν)

4.3 Tensor fields

Now, we can combine both maps from tangent and cotangent spaces to create tensors. A tensor ofrank (r, s) is a multilinear map from

T : T ∗p (M)× . . . T ∗p (M)× Tp(M)× . . . Tp(M)→ R

where we have r copies of our cotangent field and s copies of our tangent field. We define the totalrank of this multilinear map as r + s. Since a cotangent vector is a map from vectors to the reals,this is a rank (0, 1) tensor. Also, a tangent vector has rank (1, 0) since it’s a map from the cotangentspace. A tensor field is the smooth assignment of a rank (r, s) tensor to a point on the manifoldp ∈M . We can write the components of a tensor object by writing down a basis, then sticking thisinto the object. If eν was a basis of Tp(M), and fν was the basis of the dual space, then thetensor has components

Tµ1µ2...µrν1ν2...νs = T (fµ1 , fµ2 , . . . , fµr , eν1 , . . . eνs)

For example, a (2, 1) tensor acts as

T (ω, ε;X) = T (ωµfµ, ενf

ν , Xρeρ)

We have that the covectors ω, ε ∈ Λ1(M), and X ∈ X (M). The object above is then equal bymultilinearity to

= ωµενXρTµνρ

Under a change of coordinates, we have that

eν = Aµνeµ, Aµν =∂xµ

∂xν

Similarly, we have that for covectors we transform as

fρ = Bρσf

σ, Bρσ

∂xρ

∂xσ

19

Thus, a rank (2, 1) tensor transforms as

Tµνρ = BµσB

ντA

λρT

στλ

There are a number of operations which we can perform on tensors. We can add or subtract tensors.We can also take the tensor product. If S has rank (p, q), and T has rank (r, s), then we can constrictT ⊗ S, which has rank (p+ t, q + s).

S ⊗ T (ω1, . . . ωp, ν1, . . . νr, X1, . . . Xq, Y1, . . . Ys) = S(ω1, . . . , ωp, X1, . . . , Xq)T (ν1, . . . , νr, Y1, . . . Ys)

Our components of this are

(S ⊗ T )µ1...µpν1...νr

ρ1...ρlσ1...σs = Sµ1...µp

ρ1...ρqTν1...νq

σ1...σs

We can also define a contraction. We can turn a (r, s tensor into an (r− 1, s− 1) tensor. If we haveT a (2, 1) tensor, then we can define a

S(ω) = T (ω, fµ, eµ)

The sum over µ is basis independent. This has components

Sµ = Tµνν

This is different from (S′)µ = T νµν We can also symmetrise and anti symmetrise. Given a (0, 2)tensor, we can define

S(X,Y ) =1

2(T (X,Y ) + T (Y,X)), A(X,Y ) =

1

2(T (X,Y )− T (Y,X))

This has components which we write as

T(µν) :=1

2(Tµν + Tνµ )

T[µν] :=1

2(Tµν − Tµν )

We can also symmetrise or anti symmetrise over multiple indices. So

Tµ(νρσ) =1

3!(Tµνρσ + 5 perms )

We can also anti symmetrise by multiplying by the sign of permutations.

Tµ[νρσ] =1

3!(Tµνρσ + sgn(perm) for 5 perms )

20

4.4 Differential forms

Differential forms are totally antisymmetric (0, p) tensors, and are denoted Λp(M). 0 -forms arefunctions. If dim(M) = n, then p-forms have n choose p components by anti-symmetry. n-formsare called top-forms.

4.4.1 Wedge products

Given a ω ∈ Λp(M) and ε ∈ Λq(M), we can form a (p+ q) form by taking the tensor product andantisymmetrising. This is the wedge product. Our components are given by

(ω ∧ ε)µ1...µpν1...νq =(p+ q)!

p!q!ω[µ1...µpεν1...νq ]

An intuitive way to thing about this is that we are simply just adding anti-symmetric combinationsof forms, without dividing (other than to make up for the previous anti-symmetry). So, we havethat, for example, when we wedge product the forms dx1 with dx2, that

dx1 ∧ dx2 = dx1 ⊗ dx2 − dx2 ⊗ dx1

In terms of components one can check that, for example, for one forms we have that

(ω ∧ ε)µν = ωµεν − ωνεµ

We can iteratively wedge the basis of forms dxµ together to find that

dx1 ∧ . . . dxn =∑σ∈Sn

ε(σ)dxσ(1) ⊗ · · · ⊗ dxσ(n)

To show this, we use an example. Note that the components of dx1 ∧ dx2 are

dx1 ∧ dx2 = 2δ1[µδ

2ν]dx

µdxν

Now, this means that wedging this with dx3 gives components

(dx1 ∧ dx2) ∧ dx3 =3!

22δ1

[[µδ2ν]δ

3ρ]dx

µdxνdxρ

But this is the sum of all permutations multiplied by the sign, since a set of antisymmetrised indicesnested in a bigger set it the original set.

4.4.2 Properties of wedge products

Our antisymmetry property of forms gives it properties we might expect. One of these is thatswitching a p and q form picks up a sign: we have that

ω ∧ ε = (−1)pqε ∧ ω

In general, for an odd form we have that

ω ∧ ω = 0

21

For the manifold M = R3, with ω, ε ∈ Λ1(M), we have that

(ω ∧ ε) = (ω1dx1 + ω2dx

2 + ω3dx3) ∧ (ε1dx

1 + ε2dx2 + ε3dx

3)

expanding this thing, we have that

ω ∧ ε = (ω1ε2 − ε2ω1)dx1 ∧ dx2

+ (ω2ε3 − ω3ε2)dx2 ∧ dx3

+ (ω3ε1 − ω1ε3)dx3 ∧ dx1

These are the components of the cross product. The cross product is really just a wedge productbetween forms. In a coordinate basis, we write that

ω =1

p!wµ1...µpdx

µ1 ∧ · · · ∧ dxµp , ω = wµ1...µpdxµ1 ⊗ · · · ⊗ dxµp

This is useful because writing out forms in terms of wedge products as their basis turns out to makecalculations a lot easier.

4.5 The exterior derivative

Notice that given a function f , we can construct a 1-form

df =∂f

∂xµ

In general, there exists a map d : Λp(M)→ λp+1(M). this is the exterior derivative. In coordinates,we have that

dw =1

p!

∂ωµ1...µp∂xν

dxν ∧ · · · ∧ dxµp

One should view this as a generalised curl of some vector field. In components, we have that

(dω)µ1...µp+1 = (p+ 1)∂[µ1ωµ2...µp+1]

Let’s try to gain an intuition about why these two definitions are equivalent. If we contract ourcomponent definition with the tensor product dx1 ⊗ · · · ⊗ dxp+1, then we are summing over

dω =1

p!

∑σ∈Sn

ε(σ)∂σ(µ1)wσ(µ2)...σ(µp+1)dxµ1 ⊗ · · · ⊗ dxµp+1

However we can transfer our permutations to permutations on tensor product, but by definitionthis would just be the wedge product on our basis one-forms.

By antisymmetry, we have a very significant identity that

d(dω) = 0

We write this as d2 = 0. To show this, the easiest way is not to use our definition of dω incomponents but rather to use our definition in terms of wedge product basis vectors. Let’s thinkcarefully about how the exterior derivative acts on some on p form but with ’components’ in our

22

wedge product basisdx1 ∧ · · · ∧ dxp

. In our wedge product basis, our components are 1

p!ωµ1...µpsince

ω =1

p!ωµ1...µp

We have that under the exterior derivative, we are mapping

d :1

p!ωµ1...µpdx

µ1 ∧ · · · ∧ dxµp 7→ 1

p!∂νωµ1...µpdx

ν ∧ dxµ1 ∧ · · · ∧ dxµp

So, in our new fancy wedge product basis, we are mapping

wµ1...µp 7→ ∂νωµ1...µp

Hence, we have that, in our wedge product basis, our components of d(dω) are

d(dω) =1

p!∂ρ∂νwµ1...µpdx

ρ ∧ dxν ∧ dxµ1 ∧ · · · ∧ dxµp = 0

since we have symmetry of mixed partial derivatives in ρ, ν contracted with the wedge productwhich is antisymmetric in those indices.

It’s also simple to show that

• d(ω ∧ ε) = dω ∧ ε+ (−1)pω ∧ dε

• For pull backs, d(φ∗ω) = φ∗(dω)

• LX(dω) = d(LXω

A p-form is closed if dω = 0 everywhere. A p form is exact if ω = dε everywhere for some ε. Wehave that

d2 = 0 =⇒ exact =⇒ closed

Poincare’s lemma states that on Rn, or locally onM, exact implies closed.

23

4.5.1 Examples

Consider a one form ω = ωµ(x)dxµ. Using our formula for the exterior derivative:

(dω)µν = ∂µων − ∂νωµ

Or, in terms of our form basis,

dω =1

2(∂µων − ∂νωµ)dxµ ∧ dxν

In three dimensions,

dω = (∂1ω2 − ∂2ω1)dx1dx2 + (∂2ω3 − ∂3ω2)dx2 ∧ dx3 + (∂3ω1 − ∂1ω3)dx3 ∧ dx1

These are the components of ∇ × ω. The exterior derivative of a 1 form is a 2 form, but 2 formshave just three components in R3 by anti symmetry ( with the components shown there). So wethink of it has another vector field, if we identify the basis vectors

dxi ∧ dxj

with components

in R3!. However, getting another ’vector field’ by doing an exterior derivative on the same type ofobject we had before is not the case in general.

Consider B ∈ Λ2(R3). So, we have a 2-from in a 3 dimensional manifold. Let’s label the componentsout explicitly here.

B = B1(x)dx2 ∧ dx3 +B2(x)dx3dx1 +B3(x)dx1 ∧ dx2

Before we do any explicit calculation, we know that the exterior derivative pushes this up to a 3-form, which only has one component in a three dimensional manifold. We compute the exteriorderivative explicitly by differentiating each component and then adding on the wedge.

dB = ∂1B1dx1 ∧ dx2 ∧ dx3 + ∂2B2dx

2 ∧ dx3 ∧ dx1 + ∂3B3dx3 ∧ dx1 ∧ dx2

= (∂1B1 + ∂2B2 + ∂3B3) dx1 ∧ dx2 ∧ dx3

In the last step, we permuted the indices cyclically so that we don’t have a sign change. Note thatwe get our components of a grad operator acting on B!

For our final example, we take something from electromagnetism. The gauge field, or perhaps morecommonly known as our vector potential A ∈ Λ1(R4), can be written out as a one-form in R4.

If we expand this as a one form, we can write

A = Aµdxµ

What happens when we take the exterior derivative of this thing? We get that

dA = ∂νAµdxν ∧ dxµ

=1

2∂[µAµ]dx

ν ∧ dxµ

=1

2(∂νAµ − ∂µAν) dxν ∧ dxµ

=1

2Fµνdx

µdxν

24

We identify here that Fµν = ∂µAν−∂νAµ is our electromagnetic field strength tensor! We also havethat since dF = d2A = 0, we get for free what one may recognise as the Bianchi identities.

We can also introduce gauge transformations which act on our electromagnetic vector potential.These act as

A→ A+ dα =⇒ F → d(Adα) = dA invariant

where we treat α ∈ Λ0(M) = C∞(M) We also get Maxwell’s equations for free!

F = dA =⇒ dF = d2A = 0

There’s are two of Maxwell’s equations.

4.6 Integration

On a manifold, we integrate functionsf :M→ R

with the help of a special kind of a special kind of top form. The kind of form we need is calleda volume form or orientation. This is a nowhere vanishing top form. Locally, it can be writtenas

v = v(x)dx1 ∧ · · · ∧ dxn, v(x) 6= 0!

For some manifolds, globally we may not be able to glue volume forms together over the wholemanifold. If such a form exists, the manifold is said to be orientable. Not all manifolds areorientable, for example the Mobius strip. This says that v(x) must change direction and hence bezero. Or, RPn. Given a volume form, we can integrate any function f : M → R overM. In chartO ⊂M, we define ∫

Ofv =

∫Udx1 . . . dxnf(x)v(x)

Now, this tells us how to integrate a patch. Then, summing over patches gives us the whole integral.v(x) can be thought of as our measure - ’ the volume of some part of the manifold’. There is freedomin our choice of volume form here, we could’ve chosen lots of different volume forms which satisfyour condition above.

4.6.1 Integrating over submanifolds

We haven’t defined how to integrate, say a function over a p-form on an n dimensional manifold,where p < n (since so far our definition of integration has only pertained to top forms.

So to do things like this, we need to find a way to ’shift down’ into a lower dimensional subspaceand do things there. This is why we define the concept of a submanifold.

A new manifold Σ of dimension k < N is called a submanifold ofM if there exists an injective mapφ : Σ → M such that φ∗ : Tp(Σ) → Tp(M), the pullback of φ, is also injective. We require thecondition of injectivity so that our submanifold doesn’t intersect itself when we embedded it in ourlarger manifold. The first condition is so that there are no crossings. The second condition is thereso that we have no cusps in our tangent space.

We’re now fully equipped to integrate over some portion of a submanifold ofM. You can think ofthis as a ’surface’ or ’line’ of some sort embedded in our manifold. We can integrate any ω ∈ Λk(M)

25

over Σ by first identifying it with the embedded portion of Σ inM, and then pulling it back intoΣ itself. Now, we’re in a p dimensional space, and we know how to integrate here since φ∗ω is nowa top-form ∫

φ(Σ)ω =

∫Σφ∗ω

Let’s do an example where we integrate say over a line embedded in a bigger manifold. We definean injective map σ which takes our line C into our manifold.

σ : C →M

defines a non intersecting curve in M. Then, forA ∈ Λ1(M), we have, integrating over our embeddingof our line inM, ∫

σ(C)A =

∫Cσ∗A =

∫dτAµ(x)

dxµ

dτ

The last equality comes from the fact that the components of a one-form pulled back transformsas

(σ∗A)ν = Aµdxµ

dtν

where in this case, since we’re pulling back to a one dimensional manifold, we only have ν = 0(which we write for brevity as τ).

4.6.2 Stokes’ theorem

How does this tie in to the bigger picture? What use does Stoke’s theorem have in general relativity?How do we prove Stoke’s theorem?

Definition.(Boundaries). So far we’ve only considered manifolds which are smooth. However, wecan ’chop off’ a portion of our manifold to make a slightly different map that what we are used to,a new map

φα : Oα → Uα ⊂1

2Rn = (x1, . . . xn), | x1 ≥ 0

Our boundary of our manifold is the set of points onM which are mapped to (0, x2, . . . xn). Bound-aries are n− 1 dimensional manifolds of our original manifold which has dimension n

Theorem.Stokes’ Theorem. Let M be a manifold with boundary. This is a manifold which juststops and gets cutoff somewhere. If we call the manifoldM, we call the boundary ∂M. If we takeω ∈ Λn−1(M). Our claim is that ∫

Mdω =

∫∂M

ω

This Stokes’ theorem. It’s presented in a more general form than what we’re used to, but we’ll seein the examples that this way of expressing this gives us both Stokes’ theorem in three dimensionsas well as Green’s theorem.

Example.Stokes’ theorem in one dimension. TakeM as the interval I with x ∈ [a, b]. ω(x) is afunction and

dω =dω

dx· dx

26

Figure 1: Here we have a manifold with boundary.

We have that ∫Mdω =

∫ b

a

dω

dxdx

∫∂M

ω = ω(b)− ω(a)

In one dimension, we’ve recovered integration by parts on a line.

Example.Stokes’ theorem in two dimensions. In the second case, we have that

M ⊂ R2, ω ∈ Λ1(M)

This recovers ∫Mdω =

∫M

(∂ω2

∂x1− ∂ω1

∂x2

)dx1 ∧ dx2

By Stokes theorem this is ∫∂M

ω =

∫∂M

ω1dx1 + ω2dx

2

This equality is Green’s theorem in a plane.

Example.Stokes’ theorem in three dimensions. Finally, takeM⊂ R3 and ω ∈ Λ2(M)∫Mdω =

∫dx1dx2dx3(∂1ω1 + ∂2ω2 + ∂3ω3)∫

∂Mω =

∫∂M

ω1dx2dx3 + ω2dx

3dx1 + ω3dx1dx2

Equating the two expressions above gives us our usual notion of Stokes’ theorem in three dimensions,which is disguised as Gauss’ divergence theorem.

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5 Introducing Riemannian Geometry

5.1 The metric

Definition.(The metric tensor). We’ll now do introduce a tensor object which turns our tangentspace into an inner product space. We do this by introducing an object called a metric, whichintuitively has been a way in which we define the notion of ’distance’ in a space. A metric g is a(0, 2) tensor that is

• symmetric g(X,Y ) = g(Y,X)

• non-degenerate: g(X,Y )p = 0, ∀Yp ∈ Tp(M) =⇒ Xp = 0

Notice that our condition for non-degeneracy is not the same as having a point where g(X,X)p = 0as we shall soon see. In a coordinate basis, g = gµνdx

µ⊗ dxν . The components are obtained by ourstandard way of subbing in basis vectors into our tensor.

gµν = g

(∂

∂xµ,∂

∂xν

)We often write this as a line element which we call

ds2 = gµνdxµdxν

This is something that perhaps we’re more familiar with. Since our metric is non-degenerate, itis a theorem in linear algebra that we can diagonalise this thing, and furthermore we have nozero eigenvalues. If we diagonalise gµν , it has positive and negative elements (non are zero). Thenumber of negative elements is called the signature of the metric. There’s a theorem in linearalgebra (Sylvester’s law of inertia) which says that the signature is invariant, which means that itmakes sense to talk about signatures in a well defined sense.

5.1.1 Riemannian Manifolds

A Riemannian manifold is a manifold with metric with signature all positive. For example, Euclideanspace in Rn endowed with the usual Pythagorean metric.

g = dx1 ⊗ dx1 + · · ·+ dxn ⊗ dxn

A metric gives us a way to measure the length of a vector X ∈ X (M). Since our signature ispositive, we have that g(X,X) is a positive number, and hence we can take a square root to definea norm.

|X| =√g(X,X)

We can also find the angle between vectors, where

g(X,Y ) = |X||Y | cos θ

It also gives us a way to measure the distance between two points, p, q. Along the curve

σ : [a, b]→M, σ(a) = p, σ(b) = q

28

our distance is given by the integral of the metric at that point where X is the tangent to thecurve’

s =

∫ b

adt√g(X,X) |σ(t)

where at each point X is our tangent to the curve. If our curve has the coordinates xµ(t), then ourdistance is

s =

∫ b

adt

√gµν(x)

dxµ

dt

dxν

dt

Note that this notion of distance still makes sense since and is well defined since it’s easy to checkthat s is invariant under re-parametrisations of our curve.

5.1.2 Riemannian Geometry

A Lorentzian manifold is a manifold equipped with a metric of signature (−+ +...). For example,Minkowski space is Rn but our metric is

η = −dx0 ⊗ dx0 + dx1 ⊗ dx1 + · · ·+ dxn−1 ⊗ dxn−1

with componentsηµν = diag(−1, 1 . . . , 1)

This is slightly different to our Riemannian manifold case since now we can have vectors withnegative or zero length. We classify vectors Xp ∈ Tp(M) as

g(Xp, Xp) =

< 0 timelike= 0 null> 0 spacelike

At each point p ∈ M, we draw null tangent vectors called lightcones, and as we’ll soon see, thisregion outlines our area of possible causality.

A curve is called timelike if it’s tangent vector at every point is timelike. We can see this in thefigure where we have two light-cones for future and past time. In this case, we can measure thedistance between two points.

τ =

∫ b

adt

√−gµν

dxµ

dt

dxν

dt

This object τ is called the proper time between two points. Philosophically, this is a parameterwhich is invariant in all frames. If we were to reparametrise this curve, our definition of τ remainsinvariant.

5.1.3 The Joys of a metric

Claim.Metrics induce a natural isomorphism from vectors to 1-forms. The metric gives a natural(basis independent) isomorphism

g : Tp(M)→ T ∗p (M)

29

Figure 2: Here we show timelike vectors with negative norm!

Given X ∈ X (M), we can construct g(X, ·) ∈ Λ1(M). If X = Xµ∂µ, our corresponding one formis

gµνXµdxν := Xνdx

ν

In this formula, we’ve written the index on X downstairs! The metric provides a natural isomor-phism between our vector space and our one-forms. Hence, this metric allows us to raise and lowerindices , which means that our metric switches the mathematical space we are working in. Loweringan index is really the statement that there’s a natural isomorphism. Because g is non-degenerate,there’s an inverse

gµνgνρ = δµρ

This defines a rank (2, 0) tensor g = gµν∂µ ⊗ ∂ν , and we can use this to raise indices. We have

Xµ = gµνXν

Claim.Metrics induce volume forms to integrate with. There’s something else that the metric givesus. We also get a natural volume form. On a Riemannian manifold, our volume form is defined tobe

v =√detgµνdx

1 ∧ · · · ∧ dxn

We write g = detgµν On a Lorentzian manifold, v =√−gdx0 ∧ · · · ∧ dxn−1 This is independent of

coordinates. In new coordinates,

dxµ = Aµν , Aµν∂xµ

∂xν

We see how they change.

dx1 ∧ · · · ∧ dxn = A1µ1 . . . A

nµndx

µ1 ∧ · · · ∧ dxµn

=∑

perms π

A1π(1) . . . A

nπ(n)dx

1 ∧ · · · ∧ dxn

= det(A)dx1 ∧ · · · ∧ dxn

30

If we have that detA > 0 , coordinate change preserves orientation. Meanwhile,

gµν =∂xρ

∂xµ∂xσ

∂xνgρσ

= (A−1)ρµ(A−1)σν gρσ

Hence,detgµν = (detA−1)2detgρσ

Thus we have thatv =

√|g|dx1 ∧ · · · ∧ dxn

in components, we have that

v =1

n!vµ1...µndx

µ1 ∧ · · · ∧ dxµn

where our components are given byvµ1...µnεµ1...µn

we can integrate functions as ∫Mfv =

∫Mdnx

√|g|f(x)

The metric provides a map from ω ∈ Λp(M) to (∗ω) ∈ Λn−p(M) defined by

(∗ω)µ1...µn−p =1

p!

√|g!|εµ1...µn−pν1...νpων1...νp

This object is called Hodge dual. We can check that

∗(∗ω) = ±(−1)p(n−p)ω

with + used in with a Riemannian metric, and − used with a Lorentzian metric. We can thendefine an inner product on forms. Given ω, η ∈ Λp(M), let

〈η, ω〉 =

∫Mη ∧ ∗ω

The integrand is a top form so this is okay. This allows us to introduce a new object. If we have ap-form ω ∈ Λp(M), and a p-1 form α ∈ Λp−1(M), then

〈dα, ω〉 =⟨α, d†ω

⟩when d† : Λp(M)→ Λp−1(M), is

d† = ±(−1)np+n−1 ∗ d∗where again our ± signs depend on whether we have a Riemannian or Lorentzian metric. To showthis, on a closed manifold Stokes’ theorem implies that

0 =

∫Md(α ∧ ∗ω) =

∫Mdα ∧ ∗ω + (−1)p−1α ∧ d ∗ ω

But the term on the right is just

= 〈dα, ω〉+ (−1)p−1sign 〈α, ∗d ∗ ω〉

When we fix our sign, we get the result. There’s actually a close relationship between forms indifferential Geometry and fermionic antisymmetric fields in quantum field theory.

31

5.2 Connections and Curvature

What’s the point of a connection? This is going to be our final way of differentiation as opposed tothe things we have already written down. A connection is a map∇ : X (M)×X (M)→ X (M)

We write this as ∇(X,Y ) = ∇XY . The purpose of doing this is to make it look more like differen-tiation. Here, we call ∇X the covariant derivative, and it satisfies

• Linearity in the second argument ∇X(Y + Z) = ∇XY +∇XZ

• Linearity in the first argument ∇fX+gY Z = f∇XZ + g∇Y Z∀f, g ∈ C∞(M)

• Leibniz ∇X(fY ) = f∇XY + (∇XF )Y , with f a function.

• In the above, we have that ∇Xf = X(f), agrees with usual differentiation.

Suppose we have a basis of vector fields eµ. We write

∇eρeν = Γµρνeµ

This expression is a vector field because we know the derivative spits out a vector field. We usethe notation ∇eµ = ∇µ, to make the connection look like a partial derivative. Then, applying theLeibniz rule we can do a derivative on a general vector to give

∇XY = ∇X(Y µeµ) +X(Y µ)eµ + Y µ∇Xeµ= Xνeν(Y µ)eµ + Y µXν∇νeµ= Xν(eνY

µ + ΓµνρYρ)eµ

Because the vector sits out front we can write

∇XY = Xν∇νY

with∇νY = (eν(Y µ) + ΓµνρY

ρ)eµ

or, we define we equivalently define

(∇νY )µ := ∇νY µ = eν(Y µ) + ΓµνρYρ

Comparing this to the Lie derivative however, we have that LX depends on X and ∂X, so we can’twrite " LX = XµLµ ". If we take a coordinate basis for the vector fields eµ = ∂µ, then washave that

∇νY µ = ∂νYµ + ΓµνρY

ρ

In terms of notation, we can replace differentiation with punctuation. So, we have that

∇νY µ := Y µ;ν := Y µ

,ν + ΓµνρYρ

The connection is not a tensor! Consider a change of basis

eν = Aµνeµ, with Aµν =

∂xµ

∂xν

We have that∇eρ eνΓµρν eµ = ∇Aσρeσ(Aλνeλ) = Aσρ∇σ(Aλνeλ)

32

This simplifies further to give

= Aσρ(AλνΓτσλeτ + eλ∂σA

λν

= Aσρ(AλνΓτσλ + ∂σA

τλ)eτ , eτ = (A−1)µτ eµ

This implies thatΓµρν = (A−1)µτA

σρA

λνΓτσλ + (A−1)µτA

σρ∂σA

τν

So we have that an extra term is added on. We can also use the connection to differentiate othertensors. We simply ask that it obeys the Leibniz rule. For example, ω ∈ Λ1(M), Y ∈ X (M)

X(ω(Y )) = ∇X(ω(Y )) = (∇Xω)(Y ) + ω(∇XY )

Thus, rearranging the terms we have that

∇Xω(Y ) = X(ω(Y ))− ω(∇XY )

So, in terms of coordinates,

Xµ(∇µων)Y ν = Xµ∂µ(ωνYν)− ωνXµ(∂µY

ν + ΓνµρYρ) = Xµ(∂µωρ − Γνµρων)Y ρ

Hence, we have that∇µωρ = ∂µωρ − Γνµρων

Given a connection, we can construct two tensors.

1. Torsion is a rank (1, 2) tensor

T (ω;X,Y ) = ω(∇XY −∇YX − [X,Y ])

where we have that ω ∈ Λ1(M), and X,Y ∈ X (M). We can also think of T as a map fromX (M)×X (M)→ X (M), with

T (X,Y ) = ∇XY −∇YX − [X,Y ]

2. Our second quantity that we can create is called curvature. This is a rank (1, 3) tensor

R(ω,X, Y, Z) = ω(∇X∇Y Z −∇Y∇XZ −∇[X,Y ]Z

This is called the Riemann tensor. We can also think of it as a map from X (M)×X (M) toa differential operator which acts on X (M).

R(X,Y ) = ∇X∇Y −∇Y∇X −∇[X,Y ]

To show that these objects are tensors, we just need to check linearity in all the arguments. Forexample,

T (ω, fX, Y ) = ω(∇fXY −∇Y (fX)− [fX, Y ])

= ω(f∇XY − f∇YX − Y (f)X − (f [X,Y ]− Y (f)X))

= fω(∇XY −∇YX − [X,Y ])

= fT (ω,X, Y )

33

Linearity is inherited from the fact that our covariant derivative is linear when you add. In acoordinate basis eµ = ∂µ, and our dual basis of one-forms fµ = dxµ, we have that thetorsion in our components is

T ρµν = T (fρ, eµ, eν)

= fρ(∇µeν −∇νeµ − [eµ, eν ])

= Γρµν − Γρνµ‘

A connection with Γρµν = Γρνµ has T ρµν = 0 and is said to be torsion free. In addition, our curvaturetensor has components

Rσρµν = R(fσ; eµ, eν , eρ)

= fσ(∇µ∇νeρ −∇ν∇µeρ −∇[eµ,eν ]eρ

= fσ(∇µ(Γλνρeλ)−∇ν(Γλµρeλ))

= ∂µΓσνρ − ∂νΓσµρ + ΓλνρΓσµλ − ΓλµρΓ

σνλ

Clearly, we have an antisymmetry property here, where

Rσρµν = −Rσρνµ

5.3 The Levi-Civita Connection

The fundamental theorem of Riemannian Geometry is that there exists a unique, torsion-free con-nection with the property obeying

∇Xg = 0,∀X ∈ X (M)

To prove this, suppose that this object exists. Then,

X(g(Y,Z)) = ∇X [g(Y,Z)]

= ∇Xg(Y,Z) + g(∇XY,Z) + g(Y,∇X , Z)

= g(∇XY,Z) + g(Y,∇XZ)

The fact that our torsion vanishes implies that

∇XY −∇YX = [X,Y ]

Hence, our equation on the left hand of our blackboard reads

X(g(Y,Z)) = g(∇YX,Z) + g(∇XZ, Y ) + g([X,Y ], Z)

Now we cycle X,Y, Z, where we find that

Y (g(X,Z)) = g(∇Z , Y,X) + g(∇YX,Z) + g([Y, Z], X)

Z(g(X,Y )) = g(∇XZ, Y ) + g(∇ZY,X) + g([Z,X], Y )

34

If add the first two equations and then subtract by the third one we get that

g(∇YX,Z) =1

2

[Xg(Y,Z) + Y g(X,Z)− Zg(X,Y )

− g([X,Y ], Z)− g([Y,Z], X) + g([Z,X], Y )

In a coordinate basis, we have that eµ = ∂µ, we have that

g(∇νeµ, eρ) = Γλνµgλρ =1

2(∂µgνρ + ∂νgµρ − ∂ρgµν)

Where we have thatΓλµν =

1

2gλρ(∂µgνρ + ∂νgµρ − ∂ρgµν)

This is the Levi-Civita connection, and the Γλµν are called the Christoffel symbols. We still need toshow that it transforms as a connection, which we leave as an exercise.

5.3.1 The Divergence Theorem

Consider a manifold M with metric g, with boundary ∂M , and let nµ be an outward pointingvector orthogonal to ∂M. Then, for any Xµ, our claim is that∫

Mdnx√g∇µXµ =

∫∂M

dn−1x√γnµX

µ

On a Lorentzian manifold, this also holds with √g →√−g and this also holds provided ∂M is

purely timelike or purely spacelike.

First, we need a lemma, that Γµµν = 1√g∂ν√g. To prove this, we have that, writing out the definitions,

thatΓµµν =

1

2gµρ∂νgµρ =

1

2tr(g−1∂ν g

)But from this we have that

. . . =1

2tr(∂ν log g)

=1

2∂ν log det g

=1

2

1

det g∂ν det g

=1√g∂ν√g

35

5.4 Parallel Transport

Now that we have a connection ∇, we will show that we can use this object to make vectors ’travel’along the surface of a manifold. This means that our connection has given us a map from the vectorspace at a point p ∈M, to another point q ∈M. To set up this map, we take our favourite vectorfield X and then generate integral curves. At a specific point p ∈ M, this specifies a unique curveC. This curve is given by

Xµ |C=dxµ(τ)

dτ

A vector field X (M) is said to be parallel transported along this curve if at every point, we havethat the covariant derivative of Y with respect to X is zero.

∇XY = 0, at all p ∈ C

From this object we have an equation which specifies the components Y µ at each point on the curve.Our initial condition here is what Y µ(τ = 0) is. The above condition reads

Xµ(∂µY

ν + ΓνµρYρ)

=dY ν(τ)

dτ+XµY ρΓνµρ = 0

5.4.1 Geodesics

From our idea of parallel transport which we introduced above, we shall talk about a very specialtype of curve called a geodesic. Suppose we have a curve C. Now, this curve C has a tangentvector X(τ) at every point, where τ is some parameter we have used to construct the curve. If thetangent vector X is parallel transported along this curve, that is, we have that

∇XX = 0, for all p ∈ C

then, we call this object a geodesic. Our equation for our tangent vector component Xµ is

dXν

dτ+XµXρΓνµρ = 0

However, we can go a bit further here since we’re on an integral curve of Xµ.

Xµ =dxµ(τ)

dτ=⇒ d2xµ(τ)

dτ2+dxν

dt

dxρ

dτΓµνρ = 0

This specifies a unique geodesic given a set of initial conditions on xµ(0) and xµ(0), and is determinedfrom our connection on our manifold.

5.5 Normal coordinates

Working with metrics is difficult, because they can be verbose and involved. What if, there wassome way to change our basis such that we can work with a simple (diagonal) and flat metric at agiven point. Lucky for us, there is a way! This is called switching to normal coordinates. In thisbit we’ll show that at a point p ∈M, we can always choose coordinates xµ such that

gµν = δµν , gµν,ρ = 0

36

In this context our tensor δ can be either the standard Euclidean or Minkowski metric dependingon whether we’re on a Riemannian or Lorentzian manifold.

To show this, we count degrees of freedom (which in my opinion I think is a bit of a weird technique).Suppose we’re in a coordinate system xµ at a point p. We can, without loss of generality, setxµ(p) = 0. Our aim then is to find a change of basis which yields a new coordinate system xµ(without loss of generality we also have that xµ(p) = 0, such that in this basis, the above conditionson the metric are satisfied.

This means that our change of basis satisfies

gµν =∂xα

∂xµ∂xβ

∂xνgαβ

Now, we employ a trick. We Taylor expand xα in terms of xµ about the point p, which gives us theexpansion

xµ =∂xµ

∂xν

∣∣∣∣x=0

xν +xρxν

2

∂2x2

∂xρxν

∣∣∣∣x=0

When we differentiate this Taylor expansion we get that,

∂xµ

∂xν=∂xµ

∂xν

∣∣∣∣x=0

Now, when we substitute this into our change of basis formula and evaluate at the point p = 0, wefind that we need to solve the system

δµν =

(∂xα

∂xµ

∣∣∣∣x=0

∂xβ

∂xν

∣∣∣∣x=0

gαβ(p)

)This equation looks like

I = AGAT

where A represents our change of basis, and G our metric. From our change of basis, we are freeto choose n2 different components for each element. Now, since this equation is symmetric upontaking a transpose, we only have 1

2n(n + 1) constraints. Thus, it’s possible to choose a change ofbasis such that this works. Now, to second order, we have to expand further. We find that

gµν = xγ(∂2xα

∂xµxγ

∣∣∣∣x=0

∂xβ

∂xν+

∂2xβ

∂xγxν

∣∣∣∣x=0

∂xα

∂xµgαβ

)Our condition for our metric to be flat means we need to differentiate this to get

gµν,ρ =

(∂2xα

∂xµxρ

∣∣∣∣x=0

∂xβ

∂xν

∣∣∣∣x=0

+∂2xβ

∂xµxρ

∣∣∣∣x=0

∂xα

∂xµ

∣∣∣∣x=0

)From this equation, we need to make sure we have enough components to specify our change ofbasis. Now, let’s do some counting of our available degrees of freedom. gµν is symmetric and hencehas 1

2n(n− 1) free components. The addition of our derivative gives us a further n free componentsfor each element of g. Thus, we have 1

2n2(n − 1) components. Furthermore, our second derivative

term ∂2xα/∂xµ∂xρ, due to symmetry of mixed partials, has 12n(n− 1) for each index α. Thus, we

have in total also 12n

2(n− 1) free components to choose from. This is enough to specify our changeof basis!

37

5.5.1 Constructing Normal Coordinates

This section is going to be a short one. We basically will pull a cat of the bag and construct a setof normal coordinates with a map. The tangent space associated at a point on a manifold has acanonical map to recover points on the manifold. Consider the exponential map

Exp : Tp(M)→M

This map is given by picking a vector, then following the geodesic curve there for one unit ofparameter distance τ = 1. This means we will arrive at some point q. Now, that this point q , wedefine our coordinates to be

xµ(q) = Xµ

where Xµ was the vector we had originally chosen.

5.6 Path Deviations and Curvature

When we parallel transport a vector on a manifold along a curve, how the vector ends up dependson the specific path taken. Specifically, on an infinitesimal ’square’ path, this change in the vectoris given by the Riemann curvature tensor. To illustrate this, we start by generating two curves bysome vector fields X,Y , chosen such that the vector fields commute, hence [X,Y ] = 0. We thenmove an infinitesimal amount in these two directions in a rectangle like fashion.

Let’s take the path via q to r. The first thing we will do is to Taylor expand Z from p to q using τas a parameter. This is a technique we shall repeatedly use, so take note!

Zµq = Zµp + δτdZµ

dτ

∣∣∣∣p

+1

2(δτ)2 d

2Zµ

dτ2

∣∣∣∣p

+ . . .

Now, by assumption of how we set up the problem, we have that Z is parallel transported alongthis integral curve of X, so obeys this equation

dZµ

dτ= −ZρXνΓµνρ

If we evaluate this derivative at p, it vanishes since Γµνρ(p) = 0 since we’re working in normalcoordinates! If we differentiate the above term to find our second order term, then applying theproduct rule, we only have one term which survives when we evaluate at p since non-derivatives ofΓ vanish.

d2Zµ

dτ2

∣∣∣∣p

= −ZρXν dΓµρνdτ

= −ZρXνXσΓµνρ,σ

Going into the last equality, we used the chain rule. So, we now have an expression for the Taylorexpansion

Zµq = Zµp −1

2

(XνXσZρΓµρν,σ

)p

(dτ)2 + . . .

The next step to do is to go from q to r, repeating the same process. We expand Zµr about Zµq .Since we are Taylor expanding along the integral curve Y which is generated by the parameter λ,

38

we have that

Zµr = Zµq + δλdzµ

dλ

∣∣∣∣p

+1

2(δλ)2 d

2Zµ

dλ2

∣∣∣∣p

= Zµq − δλ(Y νZρΓµρν)q −1

2(δλ)2(Y νY σZρΓµρν,σ)q + . . .

Now don’t despair. You can see we’re in a bit of a bind here since we have to evaluate the connectionΓ at q. Thus, we can’t use normal coordinates to make this term disappear. We can however, do thenext best thing and Taylor expand this term about p, along the path of the integral curve generatedby X

(Γµρβ)q = (Γµρβ)p + δτdΓµρβdτ

∣∣∣∣∣p

+ · · · = −(Y νZρXσΓµρν,σ)dτ

Note that we used the chain rule to rewrite d/dτ . Now, from normal coordinates the first term herevanishes. And, since we’re substituting this into an expression which is first order in δλ anyway, weneed only keep the derivative term. By, the same logic, expanding out our second order derivativeterm and then Taylor expanding Γ around p gives us

d2Z

dλ2

∣∣∣∣q

= −(Y νY σZρΓµρν,σ)p

The upshot of this is that, we can now write the whole thing, from p to q, as

Zµr = Zµp −1

2(Γµρν,σ)p

[XνXσZρdτ2 + 2Y νZρXσdτdλ+ Y νY σZρdλ2

]Now, if we were to go the other way and go through s, we get exactly the same result but with X,Yswitched to reflect the fact that we’re changing the order of vectors we’re travelling with, and alsoa switch in λ and τ to reflect the change in the order of parametrisation. This means that only themixed term is changed.

Zµr = Zµp −1

2(Γµρν,σ)p

[XνXσZρdτ2 + 2XνZρY σdτdλ+ Y νY σZρdλ2

]Now, subtracting one from the other gives us our Riemann curvature tensor in normal coordinates,and hence since both sides are tensors, this holds in all coordinate systems.

∆Zµr = Zµr − (Z ′r)µ

=(Γµρν,σ − Γµρσ,ν

)(Y νZρXσ)pdλdτ

= (RµρνσYνZρXσ)pdλdτ

Going into the last line, we’ve once again used the fact that since our expression is a tensor innormal coordinates, this holds in all coordinate systems.

39

5.7 Derived tensors from the Riemann Tensor

From our Riemann tensor, we can easily build new tensors by contracting over some indices. In thisbit, we’ll define some really important tensors given by our curvature.

Definition.(Ricci Tensor) Our Ricci tensor is given by contracting our first index with our thirdcomponent of the Riemann tensor.

Rµν = Rρµρν

This obeys the symmetry property that Rµν = Rνµ.

Definition.(Ricci Scalar) From this object, we can then get our Ricci scalar which is given by

R = gµνRµν

Applying our Bianchi identity to the Riemann tensor, we have that

∇µRµν =1

2∇νR

This can be shown by considering the Bianchi idenity

∇[ρRαβ]γδ = 0, =⇒ ∇ρRαβγδ +∇αRβργδ +∇βRραγδ = 0

Now, raising and lowering indices, and making use of the symmetries of the Riemann tensor shouldmake the above result pop out. In this spirit, we define the Einstein tensor

Gµν = Rµν −1

2Rgµν

This obeys the rule that the covariant derivative is

∇µGµν = 0

5.8 Connection 1-forms

We now will introduce a handy technology to compute our Riemann tensor, which will make iteasier to do instead of computing all the Christoffel components. If we had a basis eµ = ∂µ,we can always introduce a different basis which is a linear sum of the coordinate induced basis. Wecall this basis

ea = e µa ∂µ

The upshot of this is that on a Riemannian manifold, we can pick a basis such that

g(ea, eb) = gµνeνa e

νb = δab

We call the components e µa are called vielbeins. We can raise and lower indices using gµν and a, b

using δab The basis of one formsθa

obey our standard dual basis relation

θa(eb) = δab

They are θa = eaµdxµ, with eaµe

µb = δab . This property satisfies eaµe ν

a = δ νµ . Our metric is

g = gµνdxµ ⊗ dxν = δabθ

a ⊗ θb

40

This means thatgµν = eaµe

bνδab

An example would be to consider the metric (Schwarzchild metric)

ds2 = −f(r)2dt2 + f(r)−2dr2 + r2(dθ2 + sin2 θdφ2)

This is ds2 = ηabθa ⊗ θb, with non-coordinate 1 forms

θ0 = fdt, θ1 = f−1dr, θ2 = rdθ, θ3 = r sin θdφ

In the basis ea, the components of our connection are defined as (not to be confused with ourusual connection) that

∇ec eb := Γabcea

We define the connection 1-form to be ωab = Γabcθc, this is called the spin connection.

We have the first Cartan structure equation.

dθa + ωab ∧ θb = 0

In addition, we have for the Levi-Civita connection, that our Cartan structure components ωab =−ωba. In the Vielbein basis, we set

Rabcd = R(θa; ec, ed, eb)

with Rabcd = −Rabdc. We define the curvature 2 form to be

Rab =1

2Rabcdθ

c ∧ θd

This gives our second structure equation.

Rab = dωab + ωac ∧ ωcb

We should think about what information is compressed and what is not.

Let’s go back to our example. We compute dθa, which gives us

dθ0 = f ′dr ∧ dtdθ1 = f ′dr ∧ dr = 0

dθ2 = dr ∧ dθdθ3 = sin θdr ∧ dφ− r cos dθ ∧ dφ

Now, we use our Cartan structure equation to get

dθ0 = 0ω0b ∧ θb =⇒ ω0

1 = f ′fdt = f ′θ0

But, we use our anti-symmetry and the Minkowski metric to get that

w01 = −ω01 = ω10 = ω1

0

41

To check the consistency of this system, we check that our equation for

dθ1 = 0

Proceeding like this, we find that

ω01 = ω1

0 = f ′θ0

ω21 = −ω1

2 =f

rθ2

ω31 = −ω1

3 =f

rθ3

ω32 = −ω2

3 =cot θ

rθ3

Now the curvature tensor

R01 = dω0

1 + ω0c ∧ ωc1

= f ′dθ0 + f ′′dr ∧ θ0

= ((f ′)2 + f ′′f)dr ∧ dt= −((f ′)2 + f ′′f)θ0 ∧ θ1

=⇒ R0101 = (ff ′′ + f ′′f)

We can convert back by usingRµνρσ = eaµe

βν ecρe

dσRabcd

This means that Rtrtr = ff ′′ + (f ′)2.

42

6 The Einstein Equations

Space time is a manifold M equipped with a Lorentzian metric g. We view our metric as somematrix which varies from point to point on the manifold. The right way to write things down hereis to write down an action principle, then vary the actions to get our equations of motion.

6.0.1 The Einstein-Hilbert Action

The dynamics is governed by the Einstein-Hilbert action. There’s not very much we can write here.The only object we have to play with is our metric g, and from this we have our natural volumeform we can use. The only one we have is the square root of our determinant of our metric. Ouraction must precede

S =

∫d4√−g

Now, what scalar function can we put here? The only thing we can do is to pull out our Ricciscalar, so that

S =

∫d4x√−gR

Note that schematically, R ∼ ∂Γ + ΓΓ, and Γ ∼ g−1∂g. Thus, R is a function of two derivativesof our metric. Note, this means that the only connection we could cook up is the Levi-Civitaconnection. This is the simplest thing we can do! Now, to derive the equations of motion, we needto vary this field. We take the metric and push it wish a small change

gµν → gµν + δgµν

Now, from this we can also deduce how the inverse metric changes with this infinitesimal change.

gρµgµν = δνρ =⇒ δgρµg

µν + gρµδgµν = 0

This means we can read off our change δgµν as

δgµν = −gµρgνσδgρσ

When we vary the whole thing, the trick is to write out the Ricci scalar in terms of the Ricci tensorcontracted with the metric. Then, we apply the product rule when looking at a small variation.When we do the small variation, we have to remember to hit the colume form as well.

δS =

∫d4x

[δ(√−g)gµνRµν +

√−g (δgµνRµν + gµνδRµν)

]We need to figure out what δ

√−g is. To do this, we prove a claim that

Claim.We claim thatδ√−g = −1

2

√−ggµνδgµν

Proof. We use the fact that log detA = tr logA, which implies that

1

detAδ(detA) = tr = tr(δ logA) = tr

(A−1δA

)= tr

(A−1δA

)43

Now, susbtituting in our metric g as the matrix A in our identity above, this finally implies that

δ√−g =

1

2

1√−g

(−g)gµνδgµν

=1

2

√−ggµνδgµν

Now, we compute the variation of our Ricci tensor.

Claim.We have that our variation is

δRµν = ∇ρδΓρµν −∇νδΓρµρwith

δΓρµν =1

2gρσ(∇µδgσν +∇νδgσµ −∇σδgµν)

Proof. First note that importantly, Γµρν is not a tensor. However, the difference δΓµρν is a tensorsince its the difference between two connections. This means that it is a well defined action to takethe covariant derivative of the objects In normal coordinates, at some point

δΓρµν =1

2gρσ (∂µδgσν + ∂νgσµ − ∂σδgµν)

=1

2gρσ (∇µδgσν +∇νδgσµ −∇σδgµν)

Since we’re in normal coordinates, we can now replace our partial derivatives ∂ with covariantderivatives ∇. This gives us a tensorial relation, and hence this holds true in all coordinate frames.We can play the same game with our Riemann tensor,

Rσρµν = ∂µΓσνρ − ∂νΓσµρ

This implies that our small change is given by

δRσρµν = ∂µδΓσµρ − ∂νδΓσµρ

= ∇µδΓσνρ −∇νδΓσµρ

Since Γ = 0 in normal coordinates.

Theorem.(Vacuum Einstein Equations)

Now we have that, after all the dust has settled, that

δS =

∫d4x√−g[Rµν −

1

2Rgµν

]δgµν +∇µXµ

Where we’ve writtenXµ = gρνδΓµρν − gµνδΓρνρ

When we impose the condition of stationarity for some arbitrary change in the metric, we then havea condition for our integrand. This yields the vacuum Einstein equations.

δS = 0,∀gµν =⇒ Gµν = Rµν −1

2Rgµν = 0

Multiplying this quantity by gµν , we have that R = 0, which implies the vacuum Einstein equations

Rµν = 0

44

6.0.2 Dimensional analysis

We can use dimensional analysis to fill out the constants here. Our action S has dimensionML2T−1,Our metric is dimensionless, and R has units of L−2. This implies that our full action is

S =c3

16πG

∫d4x√−gR

Our Planck mass isM2pl = ~c

8πG and this is approximatelyMpl ∝ 1018GeV . Work with units withc = 1, and ~ = 1. Throughout the next of the section, we shall work in terms of this Planckmass.

6.0.3 The Cosmological Constant

We could add a further term to the action

S =1

2Mpl

∫d4x√−g(R− 2Λ)

We will study these. We have the equation

Rµν −1

2Rgµν = −λgµν =⇒ R = 4Λ =⇒ Rµν = Λgµν

6.1 Diffeomorphisms Revisited

Our metric has 124 · 5 = 10 components. But two metrics related by xµ → xµ(x) are physically

equivalent. This means that actually we only have 10 − 4 = 6 degrees of freedom. The change ofcoordinates can be viewed as a diffeomorphism

φ :M→M

such ’diffeos’ are the ’gauge symmetry’ of general relativity. Consider diffeomorphisms which takethe form

xµ → xµ(x) = xµ + δxµ

Moreover, we can view this as a diffeomorphism generated by a vector field Xµ = δxµ. Themetric transforms as gµν(x)→ gµν(x) with gµν(x) = ∂xρ

∂xµ∂xσ

∂xν gρσ(x). From our definition that thesecoordinates are generated by a vector field, this is equal to

. . . = (δ ρµ + ∂µX

ρ)(δ σν + ∂νX

σ)(gρσ(x) +Xλ∂λgρσ(x))

= gµν(x) +Xλ∂λgµν(x) + gµρ(x)∂νXρ + gνρ∂µX

ρ

This means that our infinitesimal change is

δgµν = Xλ∂λgµν + gµρ∂νXρ + gνρ∂µX

ρ

= (LXg)µν

45

Alternatively, we can write this as

δgµν = ∂µXν + ∂νXµ +Xρ (∂ρgµν − ∂µgρν − ∂νgµρ)= ∂µXν + ∂νXµ + 2XρgρσT

σµν

= ∇µXν +∇νXµ

Now, let’s look back at the action

σS =

∫d4√−gGµνδgµν

If we restrict to these changes of coordinates, the above

· · · = 2

∫d4x√−gGµν∇µXν = 0∀X = −2

∫d4x√−g(∇µGµν)Xν

this is because changing coordinates is a gauge symmetry. This means that our Einstein tensorobeys

∇µGµν = 0

This means that diffeomorphisms are a symmetry implies the Bianchi identity. So, not all of theGµν are independent. We have found four conditions, which means that the Einstein equationsGµν is really only 6 equations, which is the right number of components to determine our metricgµν .

6.2 Some simple solutions

Let’s look at what happens when we have a vanishing cosmological constant with Λ = 0. We needto solve Rµν = 0. It’s tempting to write gµν = 0, but this isn’t allowed because the metric requiresan inverse! This makes gravity different from other field theories, we need to have the constraintthat non of the eigenvalues are zero. This may suggest that the metric is not a fundamental field.There are lots of similarities to this an fluid mechanics. The simplest solution is Minkowski spacetime.

ds2 = −dt2 + d~x2

6.2.1 De Sitter Spacetime

In the case, Λ > 0, we can look for solutions to Rµν = Λgµν which have some spherical symmetry.We look for solutions of the form

ds2 = −f(r)2dt2 + f(r)−2dr2 + r2(dθ2 + sin2 θdψ2)

We can compute our Ricci tensor to find

Rtt = −f4Rrr = f3

(f ′′ +

2f ′

r+ (f ′)2f

), Rψψ = sin2 θRθθ = (1− f2 − 2ff ′r) sin2 θ

46

Now, with this ansatz we can solve for f(r) by setting Rµν = Λgµν , and then solving the equationby substituting in specific indices. By setting µ, ν = tt, rr we get that

f ′′ +2f ′

r+

(f ′)2

f= −Λ

f

Similarly, by setting µ, ν = θθ, ψψ, we have

1− 2ff ′r − f2 = Λr2

This equation is solved by

f(r) =

√1− r2

R2, R2 =

3

Λ

This means that our resulting, final metric that we get takes the form

ds2 = −(

1− r2

R2

)2

dt2 +

(1− r2

R2

)−2

dr2 + r2dΩ22

To save ink, we’ve written r2(dθ2 + sin2 θdφ2) as r2dΩ22, which is the radius squared, multiplied by

the familiar metric on a unit 2-sphere. In particular, if we have a set of three Cartesian coordinatessuch that

x2 + y2 + z2 = r

then since this parametrises a 2-sphere of radius r, our resulting metric in spherical coordinatesis

d2x+ dy2 + dz2 = dr2 + r2dΩ22

This will save a lot of time later in our calculations which involve different ways to parametrise thismetric.

This is de Sitter space time (or alternatively called the static patch of d S ). Note that we havea valid range from r ∈ [0, R], but the metric appears to be singular at r = R. Throughout thissection, we will check whether this singularity comes from merely a poor choice of coordinates, oris a genuine physical space-time singularity. Recall, it’s totally okay for this to be a coordinatesingularity (in fact, as we shall see, it is) because when we define coordinates, we only define themin terms of coordinate patches. We can look at geodesics. We have the action, where σ representsproper time

S =

∫dσ[−f(r)2t2 + f(r)−2r2 + r2

(θ2 + sin2 θφ2

)]Here, there are two conserved quantities. Nothing depends on φ, so we have that l = 1

2∂Ldφ

=

r2 sin2 θφ. The other conserved quantity we get is the energy of the particle with

E = −1

2

∂L

∂t= f(r)2t

For a massive particle, we also require that the trajectory is timelike. Since σ represents propertime, this means that the Lagrangian itself is equal to

−f2t2 + f−2r2 + r2(θ2 + sin2 θφ2) = −1

Look for geodesics with θ = π2 and θ2. This gives

r2 + Veff (r) = E2

47

with

Veff(r) = (1 +l2

r2)(1− r2

R2)

For l = 0, we have thatr(σ) = R

√E2 − 1 sinh

( σR

)This hits r = R in finite σ. Meanwhile

dt

dσ= E

(1− r2

R2

)−1

Solutions to this have t → ∞ as r → R. To see this, suppose r(σ∗) = R and expand σ = σ∗ − ε.We find that

dt

dε' −α

ε

this of course means that t ∼ −α log(εR

), and t→∞ as ‘ε→ 0. so, it takes a particle finite proper

time σ but infinite coordinate time t.

In fact, the de Sitter space time can be embedded in 5 dimensional Minkwoski space, R1,4. whichis

ds2 = −(dX0)2 +4∑i=1

(dXi)2

Our surface which we are embedding is

−(X0)2 +

4∑i=1

(Xi)2 = R2

This can be viewed as a hyperbola in Minkowski space. The flat metric is inherited onto the surface.We will show that this is true. We let r2 =

∑3i=1(Xi)2 and X0 =

√R2 − r2 sinh

(tR

), and also that

X4 =√R2 − r2 cosh

(tR

). This means that, we have

−(X0)2 + (X4)2 = R2 − r2

Now just substitute and plug those in. We can check that of you compute dX0 and dX4, and pluginto the metric in 5 dimensional Minkowski space, we’ll recover the de Sitter metric. Note that weabove is a really weird parametrisation! We only singled out X0 and X4. Moreover, X4 only runsfrom −∞ to ∞, but in our parametrisation we only have that X4 > 0. These coordinates are notparticularly symmetric, and moreover cover only X4 ≥ 0 . A better choice of coordinates is

X0 = R sinh (τ/R) , and Xi = cosh (τ/R) · yi

where we have a normalisation constraint on y, such that∑4

i=1(yi)2 = 1. Another small calculation,we substitute this into our 5d Minkowski metric. We have that

ds2 = −dτ2 +R2 cosh2 (τ/R) dΩ23

where dΩ23 is a metric on S3. This is a hard calculation, but we’ve swept this under the rug and put

these into the 3 sphere metric. We can see that these are a better description of de Sitter space timebecause they describe the whole space, and has no more coordinate singularities. These are globalcoordinates. This metric also solves the Einstein equations with the same cosmological constantsince it’s just a change of coordinates from our original solution. So now we have two metric whichdescribes the same space. There’s a natural cosmological interpretation here. We have a 3-sphere,which initially shrinks, then expands. This corresponds to a contracting / expanding universe.

48

6.2.2 Anti de-Sitter Space-time

The next solution we will look at is for Λ < 0. Again, looking for solutions

ds2 = −f(r)2dt2 + f(r)−2dr2 + r2dΩ22

We find that in this case,

f(r) =

√1 +

r2

R2, R2 = − 3

ΛThis is called anti-de Sitter space-time (AdS). There is no coordinate singularity here! This time,massive geodesics obey the rule r2 + Veff = E2. This potential energy is what it was before butthere’s now a plus sign instead of a minus sign

Veff(r) =

(1 +

l2

r2

)(1 +

r2

R2

)What’s happening is that this kind of looks like a gravitational potential well in which we’re stuckin. In particular, massive particles seemed to be confined to the centre of Anti de-Sitter space. Theorigin is special here.

Now let’s look at massless particles. Massless particles follow null geodesics. This means that in thederivation of our geodesics we had a constrant which was −1, but we can change this to zero.

−f2t2 + f−2r2 + r2(θ2 + sin2 θφ2

)= 0

This means that at the point θ = π2 , θ = 0, we have that our potential obeys r2 + Veff (r) = E2,

and our null geodesic obeys

V null =l2

2r2

(1 +

r2

R2

)We introduce new coordinates, r = R sinh ρ. Plugging this into our de-Sitter metric, the sinh ischosen so the radial coordinate factors out. Thus, we get

ds2 = − cosh2 ρdt2 +R2dρ2 +R2 sinh2 ρ(dθ2 + sin2 θdφ2)

The null geodesic equation is

Rr = ± E

cosh ρ=⇒ R sinh ρ = E(σ − σ0)

Massless particles hit ρ → ∞ as δ → ∞. However, E = cosh2 ρt =⇒ R sinh ρ = R tan (t/R) =E(σ − σ0). SO, t → πR

2 as σ → ∞. Massless particles reach infinity of AdS in finite coordinatetime. AdS can be viewed as a hyperboloid in R2,3, as

−(X0)2 − (X4)2 +

3∑i=1

(Xi)2 = R2

Now, let

X0 = R cosh ρ sint

R

X4 = R cosh ρ cost

R

Xi = Ryi sinh ρ,∑

(yi)2 = 1

49

and we hence recover the AdS metric. There is one last set of coordinates which we can exam-ine.

Xi =r

Rxi, i = 0, 1, 2

X4 −X3 = rX4 +X3 =R2

r+

r

R2ηijx

ixj

The metric we get out of this is

ds2 = R2dr2

r2+r2

R2ηijdx

idxj

These don’t cover all of AdS. This is called the Poincare patch.

6.3 Symmetries

Let’s look at some symmetries of the metric. The first thing we’d like to do is explain what asymmetry actually is. Think about a 2 dimensional sphere. This has the symmetry group SO(3),since we can rotate it in any axis. Now think of a rugby ball. This has SO(2) symmetry since wecan only rotate it about one axis. The correct way to think about these things is to consider flows.Consider a 1-parameter family of diffeomorphisms σt :M→M. Recall that this is associated to avector field

Kµ =dxµ

dt

where dxµ

dt are tangent to the flow lines. We’re going to call this a symmetry if we start at any point,flow along, then our destination point looks the same.

This flow is an isometry if the metric looks the same at each point along the flow, in otherwords

LKg = 0 ⇐⇒ ∇µKν +∇νKµ = 0

We can show that the second expression is equivalent to the first by recalling the expression fora Lie derivative, and then working in normal coordinates to get the covariant expression out. Incomponents, our Lie derivative is

(LXg)µν = Kα∂αgµν + gαν∂µKα + gαµ∂νK

α

In normal coordinates, our first term disappears since the metric is flat, and we can convert thepartial derivatives to covariant derivatives, which gives us the second expression. This equation iscalled the ’Killing equation’ and any vector K which obeys this is called the Killing vector. This isthe equation which we need to solve. These objects describe the symmetries of the metric.

Note, commuting the Lie derivatives is handy because

LXLY − LY LX = L[X,Y ]

You can show this easily in the case of the Lie derivative acting on either a function or a vector field(the first is trivial, for a vector field just apply Jacobi). From this, we start to get a Lie algebrastructure that emerges for the group of continuous symmetries of the metric, (which is exactly whatwe expect from continuous symmetries).

50

Example.Minkowski space With Minkwoski space in our metric, the Killing equation implies

∂µkν + ∂νkµ = 0

In full generality, we have that

kµ = cµ + ωµνxν , ωµν = −ωνµ

We have that cµ represent our translations, and ωµν represent boosts or rotations depending onwhat indices we’re choosing. We can define Killing vectors

Pµ =∂

∂xµ, and Mµν = ηµρx

ρ ∂

∂xν− ηνρxρ

∂

∂xµ

Now, we find that [Pµ, Pν ] , and that

[Mµν , Pσ] = −ηµσPν + ησνPµ

in addition,[Mµν ,Mρσ] = ηµσMνρ + ηνρMµσ − ηµρMνσ − ηνσMµρ

These are the commutation relations of the Poincare group.

6.3.1 More examples

The isometries of dS and AdS are inherited from the 5 dimensional embedding. de-Sitter space timehas isometry group SO(1, 4), and Anti de-Sitter space has isometry group SO(2, 3). Both groupshave dimension 10, same as SO(5), and the same is the Poincare group as well. Minkowski andde-Sitter space are equally as symmetric. In 5d, the Killing vectors are

MAB = ηACXC ∂

∂XB− ηBCXC ∂

∂XA

In this case, A runs from A = 0, 1, 2, 3, 4, and we have that the metric are given by

η = (−,+,+,+,+)

η = (−,−,+,+,+)

The flows induced by MAB map the embedding hyperboloid to itself. These are isometries of ( A) dS. Let’s look at an example in de Sitter space in static patch coordinates. If the metric gµν(x),does not depend on some coordinate y, then K = ∂

∂y is a Killing vector since L∂yg =∂gµν∂y = 0. So,

for the static path, we expect ∂∂t to be a Killing vector.

We had

X0 =√R2 − r2 sinh

(t

R

)X4 =

√R2 − r2 cosh

(t

R

)

51

Look at∂

∂t=∂XA

∂t

∂

∂XA=

1

R

(X4 ∂

∂X0+X0 ∂

∂X4

)It’s interesting to note that timelike Killing vectors such that gµνkµkν < 0 are used to define energy.Minkowski and AdS have such objects. de-Sitter space has such an object in the static patch, butnot globally. For example,

K = X4 ∂

∂X0+X0 ∂

∂X4

Now, the first term increases X0 then X4 > 0, and decreases X0 when X4 < 0. The Killing vector ispositive and timelike only in the static patch. Elsewhere, it is spacelike. Energy is a subtle conceptin dS.

6.4 Conserved quantities

Consider a particle moving on a geodesic xµ(τ) in a spacetime with Killing vector Kµ. Then, wehave that

Q = Kµdxµ

dτis conserved

To see this, differentiate to find that

dQ

dτ= ∂νkµ

dxν

dτ

dxµ

dτ+ kµ

d2x

dτ2

= ∂νkµdkν

dτ

dxµ

dτ− kµΓµρσ

dxρ

dτ

dxσ

dτ

= ∇νkµdxν

dτ

dxµ

dτ

We can also see this from the action

S =

∫dτgµν x

µxν

Consider δxµ(τ) = kµ(x). We have

δS =

∫dτ ∂ρgµν

dxµ

dτ

dxν

dτ+ 2gµν

dxµ

dτ

dkν

dτ

Now, we use the fact that

gµνdkν

dτ=dkµdτ− dgµρ

dτkρ

= (∂νkµ − ∂νgµρkρ)dxν

dτ

Substituting this into the action

δS =

∫dτ2∇µKν

dxµ

dτ

dxν

dτ

This implies that σS = 0 iff ∇(µkν) = 0, the Killing equation.

52

6.5 Asymptotics of Spacetime

Given a spacetimeM, with metric gµν(x), we consider a conformal transformation

gµν(x) = Ω2(x)gµν(x)

where Ω(x) is smooth, and non zero. These metrics don’t necessarily have the same symmetries -they’re different metrics that in general describe different space times. They do have one thingon common, however, which is important. A vector which is null in the gµν spacetime is also nullin the gµν space time. This means that they have the same causal structure.

gµνXµXν = 0 ⇐⇒ gµνX

µXν = 0

Since 0 is the dividing line between positve and negative, null / spacelike / timelike vectors in gµνmap to null / spacelike / timelike vectors in gµν . Conformal transformations are something thatcrop up everywhere in physics.

6.5.1 Penrose Diagrams

The idea is to use conformal transformations to bring the infinity of space time a ’little bit closer’,in such a way that is becomes simple to visualise what infinity looks like. There is some fancytechnical way to do these diagrams, but we’ll just do examples to get the feel.

In Minkowski space in 2 dimensions, R1,1, the metric ds2 = −dt2 +dx2 (the standard metric). We’lldo two successive coordinate transformations. We introduce lightcone coordinates u = t − x, andv = t+ x. It’s simple to see that in these coordinates,

ds2 = −dudv

We have to be very careful about the range in which coordinates move. We have that u, v ∈(−∞,∞), since t, x are in the same range. Now, the idea is to come up with a second coordinatetransform which takes infinity to a finite number. We can choose any function we like which has aninfinite range in an finite domain. The simplest and most obvious choice we can make is to use thetan function. We can now map this to a finite range,

u = tan u, v = tan v

The upshot of this is that now the range of our variables u , v is now finite, with u, ˜v;w ∈ (−π2 ,

π2 ).

With differentiation of one forms, we have that for the u coordinate for example, that

du = sec2 udu

In these coordinates, the metric is

ds2 = − 1

cos2 u cos2 vdudv

Our form for the metric now is promising, because we have something of the form Ω2dudv, whichmeans we have cooked up something conformally equivalent to our lightcone coordinates in the firstplace, but has a finite range! Now we do a conformal map to get rid of this scaling factor by simplymultiplying it by the reciprocal. Consider the metric

ds2 = cos2 u cos2 vds2 = −dudv

53

Thus, we get something that looks like the Minkowski metric but now has a limited range. We writethis range to include the closure of the set (which acts as infinity) so that u, v ∈ [−π

2 ,+π2 ]. Adding

the points ±π2 that used to be ±∞ is called conformal compactification.

We’ll now try to draw a pictorial representation of this spacetime. The rule of thumb in generalrelativity is that light-cone coordinates should be drawn at 45 degrees. We make no exception inPenrose diagrams, so we draw the coordinates in a diamond shape in the range (−π

2 ,π2 ). In this set

of coordinates, time is vertical.

+π2

+π2

−π2

−π2

u v

Figure 3: A Penrose diagram of conformally compactified Minkowski spacetime in two dimensions

This is the Penrose diagram. Don’t trust distances on these diagrams, but trust the causal struc-ture. What’s the form of a general geodesic in this space? We derive the geodesic from theLagrangian

L = −t2 + x2

Then, we solve the geodesic equations to get the timelike geodesics

xµ(τ) =(Kτ +A,±

√K2 − 1τ +B

)Now, when we substitute this into our new lightcone coordinates, we find that all geodesics, in termsof the coordinates (u, v), tend to (π2 ,

π2 ) if we go far enough into the future. Similarly, we get to

(−π2 ,−

π2 ) if we go far enough into the past.

We can draw various geodesics on this diagram, in particular timelike geodesics with constant x,and spacelike geodesics and with constant t (of course, there are other geodesics but these are theones well draw in figure 5. These are geodesics in our original choice of metric.

All timelike geodesics start at the point i−, [−π2 ,−

π2 ], and end at i+, [π2 ,

π2 ]. These are called past /

future timelike infinity. Meanwhile, all spacelike geodesics start and end at two points i0, [−π2 ,+

π2 ]

or [+π2 ,−

π2 ]. These are called spacelike infinity.

All null curves start at J− "scri-minus" and end at J+ "scri-plus". These null curves would berepresented by diagonal lines on the Penrose diagram. These all called past and future null infin-ity.

There are some things we can read off from a Penrose diagram. They tell us basic things about thespacetime. For example, any two points on the spacetime have a common future and a commonpast.

54

i+

i−

i0 i0

J−J−

J+J+

Figure 4: The Penrose diagram of two dimensional Minkowski space with some geodesics drawn on

6.5.2 4 dimensional Minkowski space

Let’s now do the same thing for R1,3. We follow the same routine - get a metric, and wrangle itinto a form which is finite so we can start drawing diagrams. We do something similar. The metricis best written in polar coordinates

ds2 = −dt2 + dr2 + r2dΩ22

We, as before, do a change of coordinates into light-cone coordinates (but in t and r ) so that

u = t− r = tan u

v = t+ r = tan v

After switching to these light-cone coordinates, we then do the same coordinate transform u =tan u, v = tan v to make our range finite. We get that

ds2 = −dudv +1

4(u− v)2dΩ2

2

=1

4 cos2 u cos2 v

(−4dudv + sin2(u− v)dΩ2

2

)The only tricky thing to verify here is the prefactor in front of our angular component of our metric.This can be done however by just expanding out the sin term with double angle formulae thendividing it. Finally, we then do a conformal transformation by multiplying this whole thing by cos2

This is still 4 dimensional, so to present this in two dimensions something has to be removed, weignore the spherical part. Unlike in Minkwoski space, we also require that r ≥ 0 which implies thatv ≥ u, which means that

−π2≤ u ≤ v ≤ π

2

We hence drop the S2 and draw the Penrose diagram which appears to be sliced in two to reflectthis condition.

Note that the left hand line on the diagram is not a boundary of space time - it is merely whereu = v =⇒ r = 0, and S2 shrinks to zero here. So, for a null geodesic like the path of a photon,when it hits this left hand boundary representing r = 0, it bounces back off in the other directionto head to J+.

55

i+

i−

i0

J−

J+

Figure 5: The Penrose diagram of two dimensional Minkowski space with some geodesics drawn on

6.5.3 de Sitter

In global coordinates, de Sitter space is represented by

ds2 = −dτ2 +R2 cosh2( τR

)dΩ2

3

We introduce conformal time, which makes the metric have a factor which sits out front. Conformaltime is given by

dη

dτ=

1

R cosh (τ/R)=⇒ cos η =

1

cosh (τ/R)

where η ∈ (−π2 ,

π2 ). Plugging this in, we get the following metric

ds2 =R2

cos2 η

(−dη2 + dΩ2

3

)We write out the 3 sphere metric as dχ2 + sin2 χdΩ2

2. de Sitter is conformal to

ds2 = −dη2 + dχ2 + sin2 χdΩ22

Now we can draw our Penrose diagram.

We see that the boundary of de Sitter is spacelike. No matter how long you wait, you cannot seethe whole space, nor can you influence the whole space. This is given by a diagonal line. We cancheck that the static patch coordinates map to the intersection of the event horizon and the particlehorizon.

56

7 Coupling to matter

In this section, we’ll be looking at how matter ’backreacts’ with the metric. Matter itself changesthe dynamics of our physical system.

We’ll start simple and consider fields first. With a field in space-time, like we do in quantum fieldtheory, we associate an action to the field. This action as a kinetic part, and a potential. We firstconsider the scalar field

S =

∫d4x (−ηµν∂µφ∂νη + V (φ))

We call this thing a matter field because it dictates the dynamics of matter. In this case we’re inMinkowski space-time. Now, the natural way to generalise this would be η with our new metric g(which may depend on our position on the manifold now), and also include our canonical measure√−g in the integrand. In addition, to make sure we’re dealing with manifestly tensorial objects, we

should change our partial derivatives to covariant derivatives.

In the end, we get a form for the scalar field which is

Sscalar =

∫d4x√−g (−gµν∇µφ∇νφ+ V (φ))

The fact that we’re now not just considering the Minkowski metric means that we can include thingsin this matter field which might depend on the metric. For example, we may decide to include theeffect of the Ricci scalar in our action.

7.1 Energy Conservation

In this section, we’ll talk about the subtleties of current, energy and momentum conservation incurved space-time. What we’ll see is that while everything is fine and dandy in flat space-time,there’s something that goes wrong when we try to construct conserved charges that come from thecovariantly conserved energy-momentum tensor in curved space-time. Loosely speaking,

∇µTµν = 0 does not imply a conserved Pµ

In flat space, we have the familiar currents and energy momentum tensor which obey

∂µJµ = 0, ∂µT

µν = 0

Now, our conserved current could come from say, the symmetry of a phase rotation in a complexscalar field (using Noether’s theorem). We already know that in quantum field theory our energymomentum tensor comes from symmetries of translating space-time.

From these conservation laws, we can extrapolate conserved charges. These conserved objects areobtained by integrating the first around some spatial region which we call Σ. Σ is like some slice ofspace in space-time which we usually take to be quite large.

Q(Σ) =

∫Σd3xJ0, and Pµ(Σ) =

∫Σd3xT 0µ

Now, what exactly are the objects Q and P? Well, we know that J0 is a number, so Q(Σ) must bea number. Now, what is it a function of? Since we integrating out the spatial degrees of freedomof J0, it’s a function of time. Hence, we have that

Q(Σ) : R→ C, for any Σ

57

where Σ is some spatial slice in flat space-time. In addition, we have that Pµ is also just a func-tion

Claim.Provided that now current flows out of our space-time cylinder, the quantity Q(Σ) is con-served. Specifically, we have that between two points in time, say t1 and t2, we require that thechange in Q(Σ) between these two points remain the same. To understand conservation here, weneed to bound this in a region of space-time. What we do is that we smear our spatial region acrosscylinder in space-time, then assert that no current flows in or out of this cylinder.

Proof. We first need to define the volume which we’re integrating the conserved current over. Inthis case, we take a cylinder through space-time.

Σ

V

t1

t

Σ

Σ

t2

Figure 6: Here our volume in four space is just a cylinder.

From our conservation law, we assert that ∂µJµ = 0 here. This means that our integral over thewhole volume is zero. Thus,

0 =

∫d4x ∂µJ

µ

=

∫ t2

t1

dt

∫Σd3x ∂tJ

0 +

∫Bd3x ∂iJ

i

=

∫ t2

t1

dt ∂t

(∫Σd3xJ0

)+

∫Bd3x∂iJ

i

= [Q(Σ)](t2)− [Q(Σ)] (t1) +

∫Bd3x∂iJ

i

We recognize the expression in brackets in the first term as the definition Q. We denote B asthe boundary of this space-time cylinder. Note that this boundary is time-like (although it’s notentirely clear why this is important right now). Now, crucially, we impose the condition that on theboundary B, we have that the current vanishes, so J i = 0. This then enforces charge conservation.

We can play this same game with the conserved charge Pµ(Σ) =∫

Σ d3xT 0µ. We recognise that

from the conservation law that ∂νT νµ = 0. As before, just integrate over 4 space in the requiredregion.

0 =

∫dt

∫d3x∂0T

0µ +

∫Bd3x ∂iT

iµ = Pµ(Σ)(t2)− Pµ(Σ)(t1) + 0

Here, we imposed the condition that T iµ is zero on B. This does the trick, and we have yet anotherconserved charge.

58

This was all well and good in flat space time, but we want to generalise this to see how things workin curved spacetime. We have that instead, our quantities of interest obey covariant conservationlaws.

∇µJµ = 0, ∇µTµν = 0

In this case, we can safely say that the charge associated with the conserved current Jµ is conserved.This can be shown by recalling the divergence theorem for curved space-time with n = 4. We havethat

0 =

∫Vd4x√−g∇µJµ =

∫∂Vd3x√|γ|nµJµ

In this case, we take our closed volume V to be the space-time enclosed in the cylinder withboundary

∂V = Σ1 ∪ Σ2 ∪B

We have to be a little bit more careful here. In particular, we need to define two different surfacesΣ1 and Σ2 since there’s no canonical way to map our these surfaces throughout time. Now, the

Σ2

Σ1

BV

Figure 7: We now deal with the case when we have curved space-time

trick here is to partition this boundary integral up into the slices

0 =

∫Σ1

d3x√|γ|nµJµ +

∫Σ2

d3x√|γ|nµJµ +

∫Bd3x

√|γ|nµJµ

Again, if we impose that on the boundary, we have that Jµ = 0, then we recover charge conservation.We then have charge conservation

Q(Σ1) = Q(Σ2)

where Q is defined analogously as before, where we have Q(Σ) =∫

Σ d3x√|γ|nµJµ.

7.1.1 A foray into conservation in Electromagnetism

Example.(Charge conservation in electromagnetism) Let’s look at the case of conserved quantitiesin electromagnetism. The first thing we start with is the electromagnetic tensor F , which is atwo-form. If we want to create an action to integrate over this, the most natural thing we can dois to wedge product these two things together to get a four form, which we can integrate in fourdimensional space.

Our resulting action then looks like

S top = −1

2

∫F ∧ F

59

How do we make sense of this object in terms of physical fields which we observe? In components,this is

1

2

(1

4

)∫FµνFαβ dx

µ ∧ dxν ∧ dxα ∧ dxβ

Now, which terms in this expansion survive? Since we’re anti-symmetrising over all of the indicesµ, ν, α, β this means that any terms in which FµνFαβ contain common values like F01F01 go to zero.So, we have to consider terms that are like F01F23 and permutations of these. It’s not hard toconvince yourself that, adding all of these permutations together cancels our prefactor of 1

8 and wehave that

S top =

∫F ∧ F =

∫dx1dx2dx3dx4 ~E · ~B

Now, to incorporate this with what we know about GR, we can try couple this action with a metricby placing in the Hodge star. This gives us something slightly different

S... =

∫F ∧ ?F

We can also try to compute this object explicitly. (We’ll finish this another time)

Now, something goes wrong when we try to compute the associated charge from the covariantconservation law ∇µTµν = 0. Integrating over 4-space, we have that

0 =

∫Vd4x∇µTµν

Now, is there a way to turn this integral into a surface term which we can take to zero? Forthe divergence theorem, we could have written ∇µJµ as a total integral. This is achieved bywriting

∇µJµ = ∂µJµ + ΓµµρJ

ρ

as a total integral. Recall, this can be done

The same argument does not work for Tµν . Now we have

0 =

∫Vd4x∇µTµν

Now, we have a bit of an issue here. There’s a hanging index ν, so we can’t apply our usualdivergence theorem. In fact, there is no divergence theorem for this! Instead, we have

√−g∇µTµν = 0 ⇐⇒ ∂µ

(√−gTµν

)= −√−gΓνµρT

µρ

This comes from two occurences of the Christoffel symbols given by doing the covariant derivativeson Tµν . This looks like a driving force, which roughly speaking means that T 0µ is not conserved.This extra term can be viewed as energy seeping into the gravitational field itself, since matterbackreacts with our field!

Suppose our spacetime has a Killing vector K. Define

JµT = KνTµν

Now, this quantity is conserved as well! So, by the Killing equation and covariant conservation, wehave that

∇µJµT = (∇µKν)Tµν + kν∇µTµν = 0

60

Now, we can repeat the same steps which we did before, and define the conserved charge

QT (Σ) =

∫Σd3x

√|γ|nµJµT

We have multiple interpretations of this. If gµνkµkν < 0 in other words timelike, this can beinterpreted as energy. What if there isn’t a Killing vector is space-time? Can we make sense ofthe total energy which is possessed by both by the matter field and the gravitational field? Theshort answer : no. There is no diffeomorphism invariant local energy density for the gravitationalfield. This is statement we can prove! However, we should stop here since these types of things areunphysical as non-diffeomorphism invariant things are unphysical. There is however, one statementwe can make. Precise statements can only be made asymptotically. For example, J+ in Minkowskispace. This is called the Bondi energy. We want conserved energy since it allows us to solveequations!

61

8 When Gravity is weak

We will solve the Einstein equations perturbatively, working in what we call ’almost inertial coor-dinates’. Almost inertial coordinates are coordinates which look like Minkowski space, but addinga small term hµν on the end which we assume to be small

gµν = ηµν + hµν with hµν 1

Adding a small field onto Minkowski space like this has an important interpretation - we’re nowthinking of gravity as a field hµν that acts on the stage of Minkowski space. In particular, we thinkof gravity as a spin 2 field ηµν field hµν propagating in Minkowski space. I won’t go into the detailsof why hµν is thought of as a spin 2 field.

In particular, since we’re considering Minkowski space as the main metric here, we will raise andlower indices using ηµν rather than gµν . For example, we have that when we apply the Minkowksimetric twice to raise the indices on hµν , we get

hµν = ηµρηνσhρσ

Just as in normal field theory, we would hope that the field hµν would be invariant under sometransformations. In particular, we impose the condition that the theory that is Lorentz invariantunder transformations xµ → Λµνxν , and

hµν = ΛµρΛνσh

ρσ(Λ−1x)

8.1 Linearised theory

We work to leading order on hµν . We have that the inverse metric is given by perturbating by thenegative, so we have that

gµν = ηµν − hµν

We can find that this is indeed the correct inverse. Just verify that gµρgρν = δµν :

gµρgρν = (ηµρ − hµρ) (ηρν + hρν)

= ηµρηρν − hµρηρν + hρνηµρ +O(h2)

= δµν − hµν + hµν +O(h2)

= δµν +O(h2)

Be aware that when we’re saying ’inverse’, we actually mean ’inverse to first order in h’ in linearisedtheory. From this, we can compute the Christoffel components as well, to first order in h.

Definition.(Christoffel components in linearised theory)

Γσνρ =1

2ησλ (∂νhλρ + ∂ρhνλ − ∂λhνρ)

We can calculate our Riemann tensor as well in linearised theory. Schematically, we have thatR ∼ ∂Γ− ∂Γ + ΓΓ− ΓΓ . Now, since we’ve shown in the above that Γ is a h term, we know that

62

ΓΓ ∼ O(h2), so we can ignore these terms. If we calculate just the derivative terms in linearisedtheory, we then get

Rσρµν = ∂µΓσνρ − ∂νΓσµρ + ΓλνρΓσµλ − ΓλµρΓ

σνλ

' ∂µΓσνρ − ∂νΓσµρ

=1

2ησλ (∂µ∂σhνλ − ∂µ∂λhνρ − ∂ν∂ρhµλ + ∂ν∂λhµρ)

When we contract our indices to get the Ricci tensor, we get that

Rµν =1

2(∂ρ∂µhνρ + ∂ρ∂νhµρ −hµν − ∂µ∂νh)

R = ∂µ∂νhµν −h

where we’ve define the Laplacian operator = ∂µ∂µ. Also, we have that h = h µ

µ = ηµνhµν .Finally, we have that our Einstein-Tensor given by Gµν = Rµν − 1

2gµνR. This can also be expandedlinearly in terms of the above. Just substitute in our expressions for Rµν and R and you’ll findthat

Gµν =1

2[∂ρ∂µhνρ + ∂ρ∂νhµρ − hµν − ∂µ∂νh− (∂ρ∂σhρσ − h)hµν ]

Like the other expressions, we find that the Einstein tensor is a term in h. Because the Einsteintensor obeys the Bianchi identity

∇µGµν = 0

since Gµν is a term proportional to h and its derivatives, we can forget about the extra Christoffelsymbol here. This means we only need to include the first partial derivative, and he that that Gµνobeys the linearised Bianchi identity ∂µGµν = 0.

8.1.1 The Fierz-Pauli Action

To look at symmetries, we can reformulate the above in terms of action principles. The Einsteinequation Gµν = 8πGTµν follows from

SFP =

∫d4x

1

8πG

[−1

4∂ρhµν∂

ρhµν +1

2∂ρhµν∂

νhρµ +1

4∂µh∂

µh− 1

2∂νh

µν∂µh

]+ hµνT

µν

This arises from SEH by expanding to order h2. We can vary this action first without including thematter term to get the vacuum equations. Varying the action, we find that relabelling indices andintegrating by parts. Our variation is given first by

δSFP =1

8πG

∫d4x [−1

2∂ρhµν∂

ρδhµν +1

2∂ρhµν∂

ν (δhρµ) +1

2∂ρδhµν∂

νhρµ+

1

2∂µ (δh) ∂µh− 1

2∂νδh

µν∂µh−1

2∂νh

µν∂µδh]

Now, the trick here is to expand h to first order so that h = hµνηµν . Varying h gives δh = ηµνδhµν .

Relabelling the indices and integrating by parts gives

δSFP =1

8πG

∫d4x

1

2(∂ρ∂ρhµν) δhµν +

1

2∂νhµρ∂

ρ (δhµν) +1

2∂ρ (δhµν) ∂νhρµ+

1

2∂ρ (δhµν) ηνµδ

ρh− 1

2∂νδh

µν∂µh−1

2∂ρh

σρ∂σδhµνηµν

63

After simplifying this even further, we can factor out the Einstein tensor in linearised theory fromthis action. This is given by

δSFP =1

8πG

∫d4x

[1

2∂ρ∂

ρhµν − ∂ρ∂νhρµ −1

2(∂ρ∂ρh) ηµν +

1

2∂ν∂µh+

1

2∂ρ∂σh

ρσηµν

]We recognize this however as the action

δSFP =1

8πG

∫d4x [−Gµνδhµν ]

8.1.2 Gauge Symmetries

The action above has a lot of nice symmetries we can work with, in particular, gauge symmetries.Much like in electrodynamics, we can explore the kinds of transformations on our manifold whichleave this action invariant.

One transformation we can do is on the coordinates of the manifold. Under an infitesimal diffeo-morphism xµ 7→ xµ + ξµ(x), we induce a Lie derivative on the metric generated by the vector fieldξ.

δgµν = (Lξg)µν = ∇µξξ +∇νξµWe view this as a gauge transformation of hµν which, to leading order is

hµν → hµν + ∂µξν + ∂νξµ

Note that we only have partial derivatives here since we are assuming that not only h, but ξ is alsosmall and linear in h. This means that the terms ∼ Γξ are order h2, and we can ignore them.

We can check that Rρνσ and SFP are both invariant under gauge transformations. The first thingwe’re going to do to make life easy is to pick a gauge and use gauge symmetry. The sensible thingto do is to pick a symmetry called the de Dander gauge. This choice of gauge obeys

∂µhµν −1

2∂νh = 0

The interesting thing is that this is analogous to the Lorentz gauge in electromagnetism. Thisis analogous to the Lorentz gauge ∂µAµ = 0 in the equations of motion. An aside, in the fullnon-linear theory, the generalisation of the de-Donder gauge is something quite elegant! It’s thecondition that

gµνΓρµν = 0

This is not atensor eqaution. This is ok - this is a gauge choice and this is coorddinate depenedent.We also have four equations to solve - and this recovers our linearised gauge which we presentedearlier.

In de-Donder gauge, the linearised Einstein equations become

hµν −1

2hµνh = −16πGTµν

Given this gauge, we’re now in a nice position to define a new object which we call

hµν = hµν −1

2ηµνh

64

This is easy to invert. We have that h = ηµνhµν

= −h. And so, our inverse expressing h in termsof h. h is easy to solve! Our equation is just given by

hµν = −16πGTµν

8.2 The Newtonian Limit

Consider stationary matter with T00 = ρ(~x). This doesn’t change in time. Thus, we can just writethe wave operator

= −∂2t +∇2

and we look for solutions with ∂∂t = 0. Thus, the equation’s we’re trying to solve just look like

∇2h00 = −16πGρ(~x), and ∇2h0i = ∇2hij = 0

One of our familir solutions to this is just the Newtonian gravitational potential! These has thesolution h0i = hij = 0. And, we have that h00 = −4Φ. Where, we have that Φ is the Newtoniangravitational potential

∇2Φ = 4πGρ(~x)

This looks a lot like Maxwell’s. This is because our extra ν index enforces gauge symmetry. Whatdoes this do at a classical level? Gauge symmetry allows us to have evolution without ambiguity.So we have that this thing gives h00 = −2Φ, and that hij = −2Φδij , h0i = 0. This means that wehave our solution for a metric

ds2 = −(1 + 2Φ)dt2 + (1− 2Φ)d~x2

This is the metric for a Newtonian gravitational potential. Now, suppose that our point mass Mhas the gravitational potential,

Φ = −GMr

The metric then agrees with the Taylor expansion of the Schwarzchild metric! By the way, there’sa famous factor of 2 buried in this. We can argue in general grounds that what sits beside Φ needsto be a 2 in fornt of dt2. But classically, we have no information about what we can say in front ofd~x2. However, the Einstein equations predict the extra factor of −2Φ in front of d~x2, which givesus the offset required to agree with experiment.

8.3 Gravitational Waves

Gravitational waves are a big deal! From our Einstein momentum tensor, we have wave solutionsin GR obeying the equation

hµν = 0

The solution is hµν = Re(Hµνe

iκρxρ). Our matrix Hµν is complex, symmetric and tells us our

polarisation matrix. This object solves the wave equation providing kµkµ = 0. This tells us thatgravitational waves travel at the speed of light. There’s one thing we missed here. To derive theabove equation, we had to make sure that we were in the de Donder gauge! Thus, we have to imposethe condition that ∂µh

µν= 0, provided that kµHµν = 0 ! In terms of electromagnetism this gives

65

us restrictions on the polarisation of our wave. We can make further gauge transformations thatleave us in the de Donder gauge.

hµν → hµν + ∂µξν + ∂νξµ

This means thathµν → hµν + ∂µξν∂νξµ − ∂ρξρηµν

This leaves us in the de Donder gauge provided that

µ = 0

We can take, for example, that ξµ = λµeiκρxρ . This shifts the polarisation vector Hµν → Hµν +

i (κµλν + kνλµ − kρλρηµν). We can choose λµ such that

H0µ = 0 and Hµµ = 0

This is the transverse and traceless gauge. This has the advantage that hµν = hµν . Let’s calculatethe number of possible polarizations that we could get from this. We initially have a 4 by 4 symmetricmatrix. Then, we imposed the de Donder gauge, then we have the residual gauge transformations.This means

number of polarizations = 10− 4− 4 = 2

This is cool because this is the number of polarizations of light. This is a coincidence that thenumber of polarizations of a graviton matches that of light. As an example, consider a wave in thez-direction. We choose the wave vector kµ = (ω, 0, 0, ω). The de-Donder gauge tells us that

kµHµν = 0 =⇒ H0ν +H3ν = 0

Then in the transverse and traceless gauge, we write that

Hµν =

0 0 0 00 H+ HX 00 HX −H+ 00 0 0 0

We have some questions. How do we make gravitational waves in the first place? We’ll cover thatlater. Another reasonable question to ask is how to measure gravitational waves?

8.4 Measuring gravitational waves

A single particle is not enough to detect the waves. We need to consider a family of particles, andin particular a family of geodesics. Consider the family of geodesics xµ (τ, s) where τ is our affineparameter and s is denotes geodesics. We have that

4 - velocity uµ =∂xµ

∂τ

∣∣∣∣s

and also our displacement across geodesics.

Sµ =∂xµ

∂s

∣∣∣∣τ

66

We take particles in flat space uµ = (1, 0, 0, 0). We use the geodesic deviation equation

d2S

sτ2= Rµρσνu

ρuσsν

We work to leading order in h, and R is linear in h. Since we’re working to linear order, we leave ufixed. We also replace τ with t, the coordinate time in Minkowski space since the difference will besomething of order h . Thus, we have the equation

d2Sµ

dt2= Rµ00νS

ν

Now, using our previous result for the linearised Riemann tensor, we get that

d2Sµ

dt2=

1

2

d2hµνdt2

Sν

So now what happens? For a wave in the z direction, we see that from the form of Hµν is thecomponents, we have that

d2S0

dt2=d2S3

dt2= 0

So, all the action happens in the (x, y) plane. We’ll focus on the place z = 0. We have twopolarisations to look at.

In the H+ polarisation, we set HX = 0, which implies that we have

dS1

dt2= −ω

2

2H+e

iωtS1

d2S2

dt2=ω2

2H+e

iωtS2

The solution to this is

S1(t) = S1(0)

[1 +

1

2H+e

iωt + . . .

]S2(t) = S2(0)

[1− 1

2H+e

iωt

]

These are the geodesic equations. Taking the real part, we get that the circle is squished in and outas an ellipse. Sµ is the displacment to neighbouring geodesics. When the move in at x1, the moveout at x2.

Now, we can play a similar game for HX polarisation. We have that H+ = 0 implies

d2S1

dt2= −ω

2

2HXe

iωtS2

d2S2

dt2= −ω

2

2HXe

iωtS1

67

We now have the solutions

S1(t) = S1(0) +1

2S2(0)HXe

iωt + . . .

S2(t) = S2(0) +1

2S1(0)HXe

iωt + . . .

We define S± = S1 ± S2. It’s the same as before, but rotated by 45!

Gravitational wave detectors have two perpendicular arms. As a wave passes, the change in lengthis

L = L(

1± H+

2

)=⇒ δL

L=H+

2

We’ll see shortly that astrophysical sources give H+ ∼ 10−21. This is fine for linearised theory. Forarms of length L ∼ 3km, we need a change in the length of arm δL = 10−18m. This is small! Anaside: everything we’ve done is in the linearised model, which is fine. There is however a class ofexact gravitational wave solutions, called Brinkmann metrics. This metric is

ds2 = −dudv + dxadxa +Hab(u)xaxbdu2

We introduced lightcone coordinates here where u = t−z and v = t+z We’re indexing over a = 1, 2.Hab(u) is arbitrary with Ha

a = 0. This solves the Einstein equations.

8.5 Making Waves

In this section we’ll look at how to make grasvitational waves. We generate electromagnetic wavesby shaking charges, and we generate gravitational waves by shaking mass. We want to solve theequation

hµν = −16πGTµν

The right hand side must be small since we’re working in linearlised theory. We also have a restrictionthat Tµν must be non-slowly moving. Let’s look at how this works. Suppose we have a region ofspace, in the diagram below. (Insert here). We can solve this using Green’s functions, by puttingthe Green’s function on T , and integrating over this.

hµν = 4G

∫Σd3x′

Tµν (~x′, tret)

|~x− ~x′|

where we define the retarded time tret = t− |~x− ~x′|. Recall we derive the Green’s function by firstfourier transforming, then solving the Helmholtz equation. Alternatively, in QFT, we can Fouriertransform in 4-space and then choose the retarded time contour to find our integral. ONe comment- this equation we derived while in de Donder gauge, so we should make sure the solution obeys thede-Donder gauge condition.

∂µTµν = 0

We can check this. This is analogous to electromagentism where we impose that the current obeysthe Lorentz gauge. Now, this is the general solution, and we want to know what this solution lookslike a long way away. We Taylor expand! For r d, we have that

|~x− ~x′| ' r − ~x · ~x′

r+ . . .

68

If we take the reciprocal of this we get that

1

|~x− ~x′|' 1

r+~x · ~x′

r3+ . . .

Since ~x′ is buried in tret , we need to Taylor expand this out as well.

Tµν(~x′, tret

)= Tµν

(~x′, t− r

)+ Tµν

(~x′, t− r

) ~x · ~x′r

+ . . .

At leading order, we have that our field is

hµν '4G

r

∫Σd3x′ Tµν

(~x′, t− r

)If we look at the specific components, we have that

h00 '4GE

rwith E =

∫Σd3x′ T00

(~x′, t− r

)Similarly we have

hoi ' −4GPir

with Pi = −∫

Σd3x′T0i

(~x′, t− r

)Note, these equations are okay for the inspiral part of the motion, but not okay for the ’merger’ and’ringdown’ part of the star merger. The interesting physics part of the solution comes from

hij '4G

r

∫Σd3x′Tij

(~x′, t− r

)Now, to manipulate this we write

T ij = ∂k(T ijxj

)−(∂kT

ik)xj

But, using the fact that the energy momentum tensor is conserved, we gave that the above

· · · = ∂k

(T ikxj

)+ ∂0T

0ixj

where we used the identity ∂µTµν = 0. Anti-symmetrising, we have that

T 0(ixj) =1

2∂k

(T 0kxixj

)− 1

2

(∂kT

0k)xixj

=1

2∂k

(T 0kxixj

)+

1

2∂0T

00xixj

Now, we can put this into the integral for the expression hij , and we can drop any spatial derivativesbecause we can treat them as boundary terms. We put this into

∫Σ dx

′. Hence, we’re only left withtime derivatives in our object we and we get that

hij '2G

rIij (t− r)

where Iij =∫d3x′ T 00(~x′, t)x′ix

′j is defined as the quadrapole of the energy distribution in Σ. If

we continue this expansion, we have multiple pole expansions after this. Note that, we could nowuse ∂µh

µν= 0 to find corrections to h00 and h0i using hij . Also, if we shake matter at some

characteristic frequency ω, then we’re going to create waves which have a frequency of roughly ωas well. Because T ij has two indices, we have that we get a dquadrapole instead of the dipole as inelectromagnetism. We couldn’t have a dipole in the expansion, since momentum is conserved andcan’t shake backwards and forwards.

69

Example.(An example Binary system) Consider two objects, each with massM in a circular orbitrather than in an elliptic orbit, which will be in the (x, y) plane at distance R. If we assumeNewtonian gravity, we get that the expression for frequency is

ω2 =2GM

R3

Viewed as point particles, we get that the energy momentum tensor is

T 00 (~x, t) = Mδ(z)

[δ

(x− 1

2R cosωt

)δ

(y − 1

2R sinωt

)+ δ

(x+

1

2R cosωt

)δ

(y +

1

2R sinωt

)]To compute Iij (t) to find

Iij =MR2

4

1 + cosωt sinωt 0sin 2ωt 1− cosωt 0

0 0 0

where we’ve used the double angle formula to help simplify out the terms. This tells us that

hij =−2GMR2ω2

r

cos 2ωt ret sin 2ωtret 0sin 2ωtret − cos 2ωt ret 0

0 0 0

So we get circularly polarised gravitational waves which are travelling in the z direction. Using thefact that from Newtonian gravity, ω2 = 2GM

R2 , we get that

|hij | ∼G2M2

Rr

There’s something surprising. When we are looking at light through a telescope, the dimness goesas 1

R

2 here, it’s 1R . This is becasue we’re not measuring intesity but strains. This meanas that if we

double our intensity, we can see 8 times as more since we double our range and have an additionalvolume factor 8. To get a large |hij |, we need compact objects nearby. But, the most compactobjects are black holes and the closest they can possibly come if we have two black holes is theSchwarzchild radius, RS = 2GM . As they approach, we have that |hij | ∼ GM

r . A black hole ofa few solar masses RS ∼ 10km in the Andromeda galaxy (r ∼ 1018km ) gives us the strength ofthe gravitational wave to be |hij | ' 10−17. Observed galaxies are even farther away, with LIGOdetecting strengths at 10−21. !

8.6 Power Radiated

How much energy is emitted in gravitational waves? For electromagnetism, we compute the power

P =

∫S2

d2SiT0i

The T 0i component is called the Poynting vector. For gravitational waves, we would need to definean energy momentum tensor tµν for the gravitational field which, in the linearised theory, shouldobey

∂µtµν = 0

70

That’s the goal - we need to come up with something like a conserved energy momentum tensor,then integrate over the Poynting vector. Sadly, there’s no such object which is gauge invariant. Ormore precisely, there is no such gauge invariant object. The thing before depends on the choice ofcoordinates. There is a way to proceed - we have to work in Minkowski space. We look at J+,and look at the energy radiated at this point. A correct treatment studies energy which hit J+ inMinkowski space, and it turns out that one can define gauge invariant things at this infinity. Thisis something called Bondi energy.

Instead, we’re going to do things in a hand-wavy way. We’re gonna be looking for just order ofmagnitude estimates, and just be sloppy in this section.

We have the Fierz Pauli action. As constructed, this looks similar to all other actions in QFT withthe field hµν , as a theory in flat space, and compute the Noether current for translations. We cantreat it as any other action, and in particular compute the Noether current. However, what we getis not gauge invariant or symmetric. This is the same in Maxwell theory, but in Maxwell theory wecan add extra terms to make it symmetry and gauge invariant. However, in the Fierz-Pauli actionthere’s nothing we can do to make it gauge invariant.

In TT gauge, with h = 0 and ∂µhµν = 0 , the Fierz-Pauli action is

SFP = − 1

8πG

∫d4x

1

4∂ρhµν∂

ρhµν

Pretend that hµν are a bunch of scalar fields. The energy density is

t00 ∼ 1

G˙hµν ˙hµν

For wave solutions, the∇hµν ·∇hµν term contributes the same since we can sub in the wave equation.Previously, we had the solutions written in terms of our field

hij =2G

rIij

This is not in TT gauge, If we put it in this form, we get

hij ∼G

rQij

with Qij = Iij − 13Ikkδij . This suggests that the energy density in gravitational waves far from the

source is given by the time derivative of h, with leads to three time derivatives on Q, so that

t00 ∼ G

r2

...Qij

Integrated over a sphere, this suggests that the power emitted is

P ∼ G...Q

2ij

It turns out that this approximation is correct. The correct result is P = G5

...Q

2ij . This is from the

treatment at J+ using the Bondi energy. There was an assumption in this solution that things werestationary. The problem is that at every single step, the calculation we did relies on our choice ofcoordinates. How does it make any sense that we’re getting the right answer? This is probably dueto dimensional analysis. If we follow Bob Wald’s book, they take the Einstein equations and expand

71

them to second order. Then, we carry the second order piece to the otherside, and wrangle this intothe form of tµν . As we average over space and time, the lack of gauge invariance is less severe, sincethe terms are suppressed by larger volumes. This leads to something that we call almost gaugeinvariant. This is dubious. However, if we average over all of spacetime, we get a method thatrecovers the Bondi energy.

There’s something called the Holst Taylor binary. This is a couple of neutron stars orbiting eachother, and one of them is a pulsar. This means that measured over many years, you can measurethe frequency of the orbit. The frequency decreases by 10 micro seconds every year, and this resultagrees with the calculation we’ve done here.

The error is that if we did things this way, we would get a bunch of other stuff.

Let’s put some numbers in this and see what we’re going to get.

Example.Take a binary system ω2R ∼ GMR2 . The quadrupole is proportional to Q ∼ MR2 by

dimensional analysis. This means that the third time derivative is given by...Q ∼ ω3MR2

This means that our power is given by

P ∼ G...Q2 ∼ G4M5

R5

It turns out that in regular dimensions, this is correct. But, it turns out that the Schwarzchildradius, RS = 2GM

c2. We get that our power constructed is

P =

(RSR

)5

Lplanck

with Lplanck = c5

G which is roughly equivalent 3.6×1052Js−1. This is an enormous amount of energyemitted per second! The sun emits L ' 10−26L planck . All the stars in the observable universe emitL ∼ 10−5Lplanck . Yet, when two black holes collide, they have R ∼ RS and so L ∼ Planck . Thisis quite astonishing - they are astonishingly violent events! You might think, surely gravitationalwaves are emitted by some less violent events. Two objects with different masses M1 M2 has apower which is emitted

P ∼ G4M31M

22

R5

Now we can start calculating the gravitational waves emitted by other things. In the solar system,we can calculate the gravitational waves emitted by Jupiter, whereMJ ∼ 10−3M, and R ' 108km.If we plug this is, we get that

P ∼ 10−44Lplanck = 10−18L

Example.(Waving around arms) Suppose that you wave your arms wildly. Q ∼ 1kgm2 and...Q ∼ 1kgm2s−3. The formula for the power emitted is

P ∼ G...Q2/c5 ∼ 10−52Js−1

How small is this number? A single graviton has E = ~ω, so if ω = 1s−1 =⇒ E ' 10−34J . Toemit a single graviton, you should wave your arms for T ∼ 1018s ' 10billion years .

72

8.7 Summary

8.7.1 Gauge symmetries in linearised theory

We can significantly reduce the complexity of the equations in linearised theory by fixing thegauge.

• An infinitesimal diffeomorphism on the metric induces the following transform on the pertur-bation as well as the trace removed form

hµν → hµν + ∂µξν + ∂νξµ, hµν → hµν + ∂µξν + ∂νξµ − ηµν∂ρξρ

• With this transformation we can impose that we’re working in the de Donder gauge:

∂µhµν = 0

• This gauge gives us the forced wave equation to solve

hµν = −16πTµν

8.7.2 Gravitational wave solutions and polarisations

8.7.3 Gravitational waves from a source

We’ll summarise the procedure of finding hµν from the energy-momentum tensor Tµν (~x, t).

• We need to solve the forced wave equation

hµν = −16πTµν , = −∂2t + ∂2

x

• We import the same result from electromagnetism that applying a retarded Green’s functiongives us the desired result

hµν

=

∫d3x′

Tµν (~x′, t− |~x− ~x′|)|~x− ~x′|

where we define tret = t− |~x− ~x′|.

• Since ∂µTµν = 0, the de Donder gauge condition is also obeyed with ∂µhµν

= 0. (I shouldprobably work on proving this sometime).

• The quadrapole moment Iij and obeys∫d3x′Tij

(~x′, t

)=

1

2Iij (t) , Iij(t) =

∫d3x′T 00

(~x′, t

)xixj

• We employ the approximation |~x−~x′| = r−~x′ · ~x to approximate the stress-energy tensor andthe metric perturbation as

Tµν(t− |~x′ − ~x|, ~x′

)' Tµν

(t− r + ~x · ~x′, ~x′

)' Tµν

(t− r, ~x′

)+ ~x · ~x′∂0T

0ν(t− r, ~x′

)hij (t, ~x) =

4

r

∫d3~x′ Tij

(t− r, ~x′

)We assumed that the source is moving slowly so the second term with the derivative withrespect to time disappears.

73

9 Example Sheet 1

To: João Melo. From: Afiq Hatta

Questions 5, 7 (and the rest)

74

9.1 Question 1

If we’re given compoonents of a vector field and want to solve for its integral curve, then we needto solve the equation

dxµ(t)

dt

∣∣∣∣φ(p)

= Xµ(xν(t))|φ(p)

So for the first integral curve, we need to solve the system

dx

dt= y

dy

dt= −x

This is made a lot easier by writing out the system in polar coordinates (which is indeed a differentcart for the manifold R2, and writing (x, y) = (r cos θ, r sin θ) with the chain rule gives us

r cos θ − rθ sin θ = r sin θ

r sin θ + rθ cos θ = −r cos θ

If we multiply the first equation by sin θ and the second equation by cos θ, and then subtract thefirst equation from the second equation, we’ve eliminated the r term. We’re left wwith

θ = −1 =⇒ θ = −t+ C

for some constant C. Substituting in θ = −1 in our first equation gives the condition that r =0 =⇒ r = R for R constant. Hence our integral curves are merely circles of arbitrary radius aboutthe origin.

For our second vector fieldXµ = (x− y, x+ y)

we proceed exactly as before, with polar coordinates. One finds instead that θ = 1, and hence thatr = r. Thus, we have that

θ = t+A, r = Bet

for arbitrary constants A,B. These curves are spirals.

75

9.2 Question 2

We’re given that the map H : Tp(M)→ T ∗p (M) is a linear map. So, since H is linear,

H(X,αY + βZ) = H(αY + βZ)(X)

= (αH(Y ) + βH(Z))(X)

= αH(Y )(X) + βH(Z)(X)

= αH(X,Y ) + βH(X,Z)

Thus, H is linear in the second argument. The only fact that we’ve used here is that H is a linearmap. For the linearity in the first argument, we use the fact that H(Y ) ∈ T ∗p (M), which meansthat it’s a linear map. So

H(αX + βZ, Y ) = H(Y )(αX + βZ) = αH(Y )X + βH(Y )(Z) = αH(X,Y ) + βH(Z, Y )

Thus our map is linear in the first argument. Note that

H : Tp(M)× Tp(M)→ R

and since the map is multilinear, we have a rank (0, 2) tensor.

Similarly, if we had a linear mapG : Tp(M)→ Tp(M)

we could then define a new map

G : T ∗p (M)× Tp(M)→ R, G(ω,X) = ω(G(X))

which is also bilinear in both arguments, and hence is a rank (1, 1) tensor. If G is the identity map,then our induced function

δ : T ∗p (M)× Tp(M)→ R, δ(ω,X) = ω(X)

is indeed our standard Kronecker delta function. If we set ω = xµ, X = eν , then δµν = ∂µ(xµ) =

δµν .

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9.3 Question 3

In this question, we show that only symmetric (antisymmetric) parts of a tensor are ’conserved’whe contracted with symmetric (antisymmetric) tensors over the same indices. If Sµν is symmetric,then

V (µν)Sµν =1

2(V µν + V νµ)Sµν

=1

2V µνSµν +

1

2V νµSµν

=1

2V µνSµν +

1

2V νµSνµ

=1

2V µνSµν +

1

2V µνSµν

= V µνSµν

Going into the second last line we’ve just relabelled over summed indices. Going into the thirdline we’ve used the fact that S is a symmetric tensor.The case for when we contract V µν for anantisymmetric tensor is entirely similar.

77

9.4 Question 5

We show that our components Fµν transform appropriately under a change of coordinates. This isdone with the chain rule.

Fµν → F ′µν

=∂2f

∂x′µ∂x′ν

=∂

∂x′µ

(∂xρ

∂x′ν∂f

∂xρ

)=∂xσ

∂x′µ∂

∂xσ

(∂xρ

∂x′ν∂f

∂xρ

)=∂xσ

∂x′µ∂2xρ

∂xσ∂x′ν∂f

∂xρ+∂xσ

∂x′µ∂xρ

∂x′ν∂2f

∂xρ∂xσ

=∂xσ

∂x′µ∂xρ

∂x′ν∂2f

∂xσxρ

=∂xσ

∂x′µ∂xρ

∂x′νFρσ

There’s a reason why we’ve taken the first term to zero going into the fifth line. Since df = 0 at p,then for an arbitrary vector A in any basis, we have that at p ∈M,

df(A) = A(f) = Aµ∂µ(f) = 0, =⇒ ∂µ(f) = 0 at p, ∀µ = 1, . . . D

So this term goes to zero, since we only have a single derivative acting on f . * Thus, the Hessianobeys the tensor transformation law. Since our components transform in the two lower indices witha change of coordinates, this object is basis invariant and hence is a rank (0, 2) tensor.

Since this is a rank (0, 2) tensor, our coordinate independent way of expressing this object wouldbe

F : Tp(M)× Tp(M)→ R

This specific representation isF (V,W ) = VW (f)

We can show this by expanding with coordinates.

VW (f) = V µ∂µ(W ν∂µf)

= (∂µWν)(∂µVν)f + V µW ν∂µ∂νf

= (V µ∂µWν)∂νf +W νV µ∂µ∂νf

= W νV µ∂µ∂νf

= W νV µFµν

Here we’ve used the fact that df = 0, which implies that for an arbitrary set of components Zµ, wehave that Zµ∂µf = 0. In the above, we identify this as Zν = V µ∂µW

ν , and hence the first term inthe third line goes to zero.

This implies that Fµν are indeed the components of F . Multi linearity in both arguments is justinherited from the linearity of V,W as vector fields.

78

* A different argument would be that one can note that the first term is of the form

(Gµ′ν′)ρ∂ρf

Where we can view Gµ′ν′ as D2 separate vectors indexed by µ′ and ν ′. Thus, since df = 0, thisterm goes to zero. (I like this way since it’s manifestly a bit more basis invariant!)

79

9.5 Question 6

This question explores how the determinant of a metric transfors under coordinate transformations.For this question, we denote the determinant of a change of basis as the Jacobian:

J = det

(∂xρ

∂x′ν

)Hence, when we do a coordinate transform, since det(AB) = det(A)det(B), we have that

g′ = det(g′µν)

= det

(∂xσ

∂x′µ∂xρ

∂x′νgρσ

)= det(g)J−2

This is because the expression in the determinant is the inverse of what we’ve defined the Jacobianto be.

80

9.6 Question 7

Lie derivative of 1-form

Using the Leibniz rule for our Lie derivative, we consider the Lie derivative for LX(ωY );

LX(ω(Y )) = ω(LXY ) + (LXω)(Y )

This expression is basis independent. Now, observe that ω(Y ) is a function in C∞(M). Thus,the Lie derivative for this term is just given by X(ω(Y )). We also know that LX(Y ) = [X,Y ].Thus,

Xµ∂µ(ωνYν) = (LXω)νY

ν + ων [X,Y ]ν

YνXµ∂µω

ν + ωνXµ∂µY

ν = (LXω)νYν + ωνX

µ∂µYµ − ωνY µ∂µX

ν

Up to index relabelling of the dummy indices, the last term of the LHS and the second term of theRHS are the same, so they cancel out. Moving the negative term on the LHS to the right hand sideand relabelling gives

Yν(Xµ∂νων + ωµ∂νX

µ) = (LXω)νYν

However, since Y was arbitrary we can just read off the basis independent components here.

(Xµ∂νων + ωµ∂νX

µ) = (LXω)ν

Lie derivative for a 2-tensor

We play exactly the same game with the rank (0, 2) tensor as well.

LX(g(V,W )) = (LXg)(V,W ) + g(LXV,W ) + g(V,LXW )

In components, and multiplying out with the product rule, this term is

Xν(∂µgαβ)V αW β +XµgαβWβ∂µV

α +XµgαβVα∂µW

β =

= (LXg)αβVαW β + gαβ[X,V ]α∂µW

β + gαβVα[X,W ]β

The right hand side is just equal to, expanding the commutators in terms of components,

= (LXg)αβVαW β + gαβX

ν∂νVαW β − gαβV ν∂νX

αW β + gαβVαXν∂νW

β − gαβV αW β∂νXβ

Now up to index relabelling µ and ν, the second and fourth terms of this equation cancel out withthe second and third terms on the LHS of our first equation. Thus, we’re left with

Xµ∂µgαβVαW β + gαβV

ν(∂νXα)W β + gαβV

αW ν(∂νXβ) = (LXg)αβV

αW β

Now, as before, relabelling ν, α in the second term and ν, β in the third term recovers the expressionin the question (after factorising out V,W ).

81

Last part

The last part of the question is just a matter of substituting in definitions.

(ιXdω)µ = Xν(dω)νµ

= Xν2∂[νωµ]

= Xν∂νωµ −Xν∂µων

Also, we have

d(ιXω)µ = ∂µ(ιXω)

= ∂µ(Xνων)

= ων∂µXν +Xν∂µων

Adding these terms together gives

(ιXdω)µ + d(ιXω)µ = Xν∂νωµ −Xν∂µων + ων∂µXν +Xν∂µων = Xν∂νωµ + ωνX

ν

since we have cancellation with the second and last term.

82

9.7 Question 8

The components of the exterior derivative of a p− form consists of the antisymmetrisation of p+ 1indices. Suppose that m ω is a p− form. Then

(dω)µ1µ2...µp+1 = (p+ 1)∂[µ1ωµ2...µp+1]

Thus, we can expand the components of the exterior derivative of this object as

(d(dω))µ1...µp+1 = (p+ 2)(p+ 1)∂[µ1∂[µ2ωµ3...µp+2]]

Now, we have a tricky thing to deal with here. We have an antisymmetrisation nested inside of anantisymmetrisation. We claim that nesting an antisymmetrisation inside an antisymmetrisation isjust the larger antisymmetrisation:

X[µ1[µ2,...µp]] = X[µ1µ2...µp]

We can prove this by expanding out an antisymmetrisation based on just the µ1 index first.

X[µ1...µp] =1

p!

( ∑σ∈Sp−1

ε(σ)Xµ1µσ(2)...µσ(p)

−∑

σ∈Sp−1

ε(σ)Xµσ(2)µ1µσ(3)...µσ(p)

...

+ (−1)p+1∑

σ∈Sp+1

ε(σ)Xµσ(2)µσ(3)...µσ(p)µ1

)So, when we nest antisymmetrisations, we have terms in the sum that look like

X[µ1[µ2...µp]] =∑

similar sum as above but of

1

p!

∑σ∈Sp−1

ε(σ)Xµ1[µσ(2)...σ(p)]

But, expanding out the definition, we have that this term is just equal to

=1

p!

1

(p− 1)!

∑σ′∈Sp−1

∑σ∈Sp−1

ε(σ′)ε(σ)Xµ1µσ′σ(2)...µσ′σ(p)

But, we can compose each pair of permutations and write σ′′ = σ′σ, and since the sign operator forpermutations is a homomorphism, we can write that ε(σ)ε(σ′) = ε(σ′′). But, we have to be carefulto make sure to count twice. Hence the term above is

1

p!

1

(p− 1)!

∑σ′

∑σ

ε(σ′′)Xµ1µσ′′(2)µσ′′(3)...µσ′′(p)

Now, we can relabel the σ index as σ′′, and so we’re just summing over and extra σ′. This gives

1

p!

1

(p− 1)!

∑σ′

∑σ′′

ε(σ′′)Xµ1µσ′′(2)µσ′′(3)...µσ′′(p) =1

p!

∑σ′′

Xµ1µσ′′(2)µσ′′(3)...µσ′′(p)

83

But this just removes the effect of an antisymmetric tensor! Hence, given a set of indices, wehave

[[µ1µ2 . . . µp]] = [µ1µ2 . . . µp]

So, the nested indices have no effect. Thus we have that

d(dω)µ1...µp+2 = (p+ 2)(p+ 1)∂[µ1∂[µ2ωµ3...µp+2]] = (p+ 2)(p+ 1)∂[µ1∂µ2ωµ3...µp+2] = 0

By antisymmetry of mixed partial derivatives.

Now, we’d like to show a ’product rule for the exterior derivatives and one forms.

d(ω ∧ ε) = dω ∧+(−1)pω ∧ dε

The right hand side in components, by definition is

d(ω ∧ ε)γµ1...µpν1...νq =(p+ q + 1)(p+ q)!

p!q!

(∂[γωµ1...µpεν1...νq ] + ω[γµ1...µp−1

∂µpεν1...νq ])

Let’s see what we’ve done here. We used the product rule to expand out the derivatives, but whendoing this we need to preserve our order of our indices, which is why we kept it in this form.

Now, we reorder the indices γ, µ1, . . . µp. We do the procedure

(γ, µ1, µ2, . . . , µp)→ (−1)(µ1, γ, µ2, . . . µp)→ · · · → (−1)p(µ1, µ2 . . . µp, γ)

So, we’ve picked up a factor of (−1)p. Thus, when we stick in an extra set of antisymmetric indices(which doesn’t change things as we showed earlier), we get that our expression above is equalto

(p+ q + 1)!

p!q!∂[γωµ1,...µpεν1...νq ] + (−1)p

(p+ q + 1)!

p!q!ω[µ1,...µp∂γεν1,...νq ]

Now, we substitute our expression for an exterior derivative. The above expression is equal to

=(p+ q + 1)!

p!q!

1

(p+ 1)dω[γµ1...µpεν1...νq ] + (−1)p

(p+ q + 1)!

p!q!

1

(p+ 1)!ω[µ1...µpdεν1...νq ]

=(p+ q + 1)!

(p+ 1)!q!(dω)[γµ1...µpεν1...νq ] + (−1)p

(p+ q + 1)!

p!(q + 1)!µ[µ1...µpdεγν1...νq ]

But these are explicitly the components of what we are looking for. Hence, we’ve shown that

d(ω ∧ ε) = dω ∧ ε+ (−1)pω ∧ dε

Finally, we wish to show that a pull back of the one form ω , denoted as ψ∗ω from the manifold Mto N commutes with our exterior derivative. In other words, we wish to show that

d(ψ∗ω) = ψ∗(dω)

We do this by expanding the components explicitly first. We have that

d(ψ∗ω)νµ1...µp = ∂[ν(ψ∗ω)µ1...µp]

=∂

∂x[ν

∂yα1

∂xµ1. . .

∂yαp

∂xµp]ωα1...αp

84

We were careful here to ensure that, since ψ∗ω lives in the manifoldM , we need to differentiate withrespect to the coordinates xα. Now, here we used the fact that for a general p-form, our componentschange like

(ψ∗ω)µ1...µp =∂yα1

∂xµ1. . .

∂yαp

∂xµpωα1...αp

Since we have one differential as ∂∂xν , we can use the chain rule to expand this term out, giving that

the above expression is equal to

∂

∂yβ∂yβ

∂x[ν

∂yα1

∂xµ1. . .

∂yαp

∂xµpωα1...αp] =

∂yβ

∂x[ν

∂yα1

∂xµ1. . .

∂yαp

∂xµp]

∂

∂yβωα1...αp

Now, this step deserves some explanation. By symmetry of mixed partial derivatives, we’re allowedto commute the ∂

∂yβterm past everything. Because even though the product rule dictates that this

has to differentiate each term in this big product, the first terms are derivatives, so by symmetry ofmixed partial derivatives inside an antisymmetric tensor, all these extra terms go to zero.

Finally, due to our ability to relabel dummy indices, one can show that for a vector contraction ofthe form

Xµ1ν1 . . . X

µnνnY[µ1...µn] = Xµ1

[ν1. . . Xµ1

µn]Yµ1...µn

This means that indeed, we can shift the antisymmetric terms to the right most indices, giving

d(ψ∗ω)νµ1...µp =∂yβ

∂xν∂yα1

∂xµ1. . .

∂yαp

∂xµp∂

∂y[βωα1...αp]

But indeed, these are the components of the pulled back one form

ψ∗d(ω)

So we’re done!

9.8 Question 9

This question shows the advantage of coming up with a tensorial definition of objects first, tosimplify calculations for components. In the case when p = 1, we set the basis X1 = eµ, X2 = eν .Then, our definition in tensorial form gives

(dω)µν = dω(eµ, eν) = eµ(ω(eν))− eν(ω(eµ))− ω([eµ, eν ])

Now note that eµ, eν aren’t indexed components per say, they’re just our choice of basis vector. Theupshot of doing this is that by symmetry of mixed partial derivatives, we have

[eµ, eν ] =∂2

∂xµ∂xν− ∂2

∂xνxµ= 0

Hence, the last term vanishes and thus

(dω)µν = ∂µ(ων)− ∂ν(ωµ)

In the above line we’ve used the fact that ω(eµ) = ωµ , which can be shown by expanding ω intoits components and covector basis. Once again, when p = 3, we can still use this trick of using

85

the vector basis eµ, to forget about the commutator terms in the definition. This is because ourdefinition of dω contains terms like

ω([eµ, eν ], eα)

but since the commutator vanishes and ω is multilinear, ω(0, eα) = 0. Thus, the only terms thatare preserved from the definition is that

dω(eµ, eν , eρ) = eµω(eν , eρ)− eνω(eρ, eµ) + eρω(eµ, eν)

Using the fact that the basis vectors are derivative terms this becomes

(dω)µνρ = ∂µωνρ − ∂νωρµ + ∂ρωµν

This is consistent with our definition that

(dω)µνρ = 3∂[µωνρ] =1

2

∑anti symmetric perms

∂µωνρ

Note the seemingly extraneous factor of two here, but this cancels our since ω is a two form andtherefore we count twice the number of permutations.

9.9 Question 10

For now we’ll just show that dσ1 = σ2 ∧ σ3. Let’s calculate the right hand side explicitly, we havethat

σ2 ∧ σ3 = (cosψdθ + sinψ sin θdφ) ∧ (dψ + cos θdφ)

= cosψdθ ∧ dψ + sinψ sin θdφ ∧ dψ + cosψ cos θdθdφ+ cosψ cos θdθ ∧ dφ+ sinψ sin θ cos θdφ ∧ dφ= cosψdθ ∧ dψ + sinψ sin θdφ ∧ dψ + cosψ cos θdθdφ+ cosψ cos θdθ ∧ dφ

When we do an exterior derivative in 3 dimensions on a one form, we get that in our wedge productbasis

dω = (∂1ω2 − ∂2ω1)dx1 ∧ dx2

+ (∂2ω3 − ∂3ω2)dx2 ∧ dx3

+ (∂3ω1 − ∂1ω3)dx3 ∧ dx1

Now, if we identify dx1 = dθ, dx2 = dψ, dx3 = dφ, then our first component to calculate is

(∂θ(σ1)ψ − ∂ψ(σ1)θdθ ∧ dψ) = cosψdθ ∧ dψ

Similarly, we find that

(∂2(σ1)3 − ∂3(σ1)2)dx2 ∧ dx3 = sinψ sin θdφ ∧ dψ(∂3(σ1)1 − ∂1(σ1)3)dx3 ∧ dx1 = cosψ cos θdθ ∧ dφ

86

9.10 Question 11

The point of this question is to show that a basis which is coordinate induced is equivalent to it’scommutator vanishing. Showing one way is straightforward, we have that

[eµ, eν ] =∂2

∂xνxµ− ∂2

∂xµxν= 0

This is by the symmetry of mixed partial derivatives.

Now we go the other way. From the condition that

[eµ, eν ] = γρµν eρ

We expand this out

[e ρµ

∂

∂xρ, e λν

∂

∂xλ] = γρµν e

λρ

∂

∂xλ

Writing this out explicitly and cancelling cross terms give

e ρµ

∂e λν

∂xσ∂

∂xλ− e λ

ν

∂e σµ

∂xσ∂

∂xσ= γσµν e

λσ

∂

∂xλ

Upon relabelling dummy indices (for example replacing σ → λ in the second term), we end up withthe expression

e σµ

∂e λν

∂xσ∂

∂xλ− e σ

ν

∂e λµ

∂xσ∂

∂xλ= γσµν e

λσ

∂

∂xλ

But, we can just factor out our partials ∂∂xλ

to get our required expression. Now, we appeal to thefact that

e ρµ f

νρ = δ ν

µ

Differentiating both sides, we have that

fνρ∂e ρ

µ

∂xγ+ e ρ

µ

∂fνρ∂xσ

= 0

Hence,

fνρ∂e ρ

µ

∂xσ= e ρ

µ

∂fνρ∂xσ

e τν f

νρ

∂e ρµ

∂xσ= −e τ

ν eρµ

∂fνρ∂xσ

∂e τµ

∂xσ= −e τ

ν eρµ

∂fνρ∂xσ

Substituting this into the above,

−e σµ e λ

α e βν

∂fαβ∂xσ

+ e σν e λ

α e βµ

∂fαβ∂xσ

= γαµν eλα

We can cancel out the e λα . We get the

−e σµ e β

ν

∂fαβ∂xσ

+ e αν e β

µ

∂fαβ∂xσ

= γαµν

87

Contraction with fµλ fνσ , gives the result,

∂fρσ∂xλ

−∂fρλ∂xσ

= −γρµν fµλ f

µσ

However, this implies that each of fµ is closed since if we have [eµ, eν ] = 0, then γ = 0 for allindices. So, we get that dfµ = 0 for all µ by the above formula. Hence, the Poincare lemma statesthat we can write

fµν = ∂νηµ

Hence, ηµ are a set of functions. Our condition that

δ νµ = e α

µ fνα = e αµ ∂αη

ν = eµ(ην)

Hence, relabelling ην = xν gives us a set of coordinates given that ην are independent. However,we know this is the case since if we have a linear sum∑

µ

λµηµ = 0

contracting with e gives each coefficient 0. Hence, we have that n of these are linearly independent.Thus, the collection ηi is a map from M → Rn is injective, and since we have n of these maps,they span (by the Steinitz exchange lemma) they also span. Hence, we have a homeomorphism,and thus

ηiis a set of coordinates. Thus, the corresponding eν are a set of coordinate induced

basis vectors.

10 Example Sheet 2

88

11 Example Sheet 4

11.1 Question 1

We want to solve the linearised equation

∇2h00 = −16πMδ (~x)

Using the standard Green’s function technique, the solution to this equation is

h00 = −4M

r

This means that we haveh00 = −2M

r, hii = −2M

r

Our gravitational field is hence

ds2 = −(

1 +2M

r

)dt2 +

(1− 2M

r

)d~x · d~x

Now, where is this valid? Well in comparison to the Scharwzchild metric, the spatial perturbationto the metric is the leading order expansion for

(1− 2GM

r

)−1, thus we should require that r 2GM

Our expansion is valid when |h| 1, so all we need from this is that h00 1, for example. Thisgives the condition that r GM as well.

11.2 Question 2

The linearised Einstein equations are

hµν = −16πGTµν , hµν = hµν −1

2ηµνh

We choose to solve for hµν first because it’s easier. We then obtain hµν from hµν by just invertingthe above relation. This inverse is

hµν = hµν −1

2ηµνh

We are given the following time independent stress-energy tensor to solve for, which is

Tµν = µδ (x) δ (y) diag (1, 0, 0,−1)

We proceed by solving component by component. Since our stress energy tensor is time-independentwe can reduce the wave operator to just the Laplacian ∇2. For h00, we wish to solve

∇2h00 = −16πGµδ (x) δ (y)

For now, let’s work in cylindrical polar coordinates and isolate to the case where our solution for hand therefore h relies solely on r. We hence rewrite our equation as

∇2h00 (r) = −16πGµδ (r = 0)

89

Now, if you already know the Green’s function for this kind of problem then you can just solvethis thing right here. In case you don’t, we have that the radial part of the Laplacian gives us theequation

1

r

d

dr

(rdh00

dr

)= −16πGµδ (r = 0)

For the case r > 0, when the right hand side is zero, our solution is

h00 = A log

(r

r0

)We need to fix the constant A consistently with the delta function contribution on the right handside of the equation. Integrating both sides in cylindrical polars, to some radius R, and then takingthe limit as R→ 0, we have that

2πA

∫ R

0dr r

1

r

d

dr

(rdf

dr

)= −16πGµ

2πRd log R

r0

dR= −16πGµ

A = −8Gµ

Thus h00 = −8Gµ log(rr0

)Additionally, we have that h33 = 8Gµ log

(rr0

). All of the other compo-

nents of hij can be set to zero, since this is a valid solution in the homogeneous case. We thus haveh = 16Gµ log

(rr0

). Inverting with the formula above, to get h, the only non-zero contributions are

h11 = h22 = −8µG log(rr0

)Adding on the perturbation, our metric is thus

ds2 = −dt2 + (1− λ)(dx2 + dy2

)+ dz2

In polar coordinates, the dx2 +dy2 is just dr2 +r2dφ2. This means that our resulting metric is

ds2 = −dt2 + dz2 + (1− λ)(dr2 + r2dφ2

)Using the substitution

(1− λ2

)r2 = (1− 8µG) r2, our angular component of the metric reads

(1− 8µG) dr2. To first order, we have that

r =(1− λ)

12

(1− 8µG)12

r

' r (1 + 4µG log r) (1 + 4µG) to first order in 8µG

= (1 + 4µG− 4µG log r) r

Differentiating, this givesdr = (1− 4µG log r) dr

Squaring, we get that this is

dr2 =

(1− λ

2

)2

dr2 ' (1− λ) dr2

90

Substituting this in, we get that our metric under this change of coordinates is

ds2 = −dt2 + dz2 + dr2 + (1− 8µG) r2dφ2

Scaling the angular coordinate as φ =√

1− 8µGφ, this gives our metric as

ds2 = −dt2 + dz2 + dr2 + r2dφ2

This looks like Minkowski space time after all the first order approximations, but our original metricwasn’t. Intuitively, to get a double image, our geodesics follow a path as follows.

y

x

Object

null geodesic

Figure 8: Double image appearing

Here’s an efficient way to derive the Green’s function.

∂2x + ∂2

yG = −1∫B1

d2x∂2G = −1∫S1

rdφ (∂G) · ~n =

∫S1

dφ∂rG

(∂r

∂~v

)· ~n

= 2πr∂rG

If we integrate through, we get that

−h00 = h33 = λ = 8Gµ log

(r

r0

)

There’s a slight caveat in this question. if we rescale φ as before, this means that our periodicitychanges, we have a slightly smaller period. So, two light rays get pinched when we take straightlines.

91

11.3 Question 3

To first approximation, since our rotating sphere is moving slowly, our energy-momentum tensor isTµν ∼ ρuµuν , where we take all contributions of O

(Ω2)as approximately zero.

Tµν ' ρ

1 −Ωy Ωx 0−Ωy 0 0 0Ωx 0 0 00 0 0 0

11.3.1 First part

We first solve for our h00 component, which is the solution to the equation

∇2h00 = −4M

R2δ (r −R)

We switch to spherical polar coordinates and assume axisymmetric solutions, and assume thath00 = h00 (r). In spherical polar coordinates, the form of the Laplacian means that we have tosolve

1

r2

∂

∂r

(r2∂h00

∂r

)= −4M

R2δ (r −R)

In the regions r < R and r > R, we want to solve

1

r2

∂

∂r2

(r2∂h00

∂r

)= 0

To ensure we don’t have a singular solution at the origin and that the solution decays at infinity,this means that

h00 =

CR r < RCr r > R

where C is a constant to be determined. To determine this constant, we integrate the above equationbetween the bounds R+ ε and R− ε for small ε. In other words, we wish to calculate

4π

∫ R+ε

R−εdr r2

[1

r2

∂

∂r

(r2∂h00

∂r

)]= −4M

R2(4π)

∫ R+ε

R−εdr r2δ (r −R)

Hence we have that [R2∂h00

∂r

]R+

R−

= −4M =⇒ C = 4M

since h00 is constant taking the limit from below. Hence

h00 =

4MR r < R

4Mr r > R

We invert this to find h00. Since h00 is the only diagonal component of hµν , we have that h = −h00

to first order. This means that our solution for h00 is given by

h00 = h00 −1

2η00h00 =

2MR r < R

2Mr r > R

92

Our metric is thusds2 = −

(1− 2M

r

)dt2 +

(1 +

2M

r

)(dr2 + r2dφ2

)This is the form of Newtonian gravity!

11.3.2 Second Part

We proceed with the same strategy we used before. Making use of the spherical Laplcian and theansatz that H = f(r) sin θeiφ, we want to solve the following equation in the regions r > R andr < R: Substituting in this ansatz with the Laplacian in spherical coordinates, we have that wewant to solve the following equation in the two regions (with the correct boundary conditions)

sin2 θ

(∂

∂r

(r2∂f

∂r

))eiφ − eiθ sin θf + eiφf sin θ

(cos2− sin2 θ

)= 0

We make the appropriate cancellations and make use of standard trigonometric identities

sin2 θ

(∂

∂r

(r2∂f

∂r

))− f + f

(cos2 θ − sin2 θ

)= 0

∂

∂r

(r2∂f

∂r

)− 2f = 0

We then make the ansatz that f = Arα for some power exponent α. The solutions substituting inthis ansatz, we find that the two solutions are α = 1,−2. Due to regularity conditions, these are theexponents for the regions r < R and r > R respectively. Imposing that the function is continuousat r = R, we find that

f (r) =

Ar r < RAR3

r2r > R

Now all that’s left to do is to determine the constant A. We do this by integrating the equation overthe troublesome coordinate over a small volume around radius R. We only care about the radialpart of the Laplacian since continuity takes every other term to zero. Thus, we integrate over

sin θeiφ∂

∂r

(r2∂f

∂r

)= eiφ sin θ

4Ω

R2Mδ (r −R)

We cancel off the functions of φ and θ , and remembering to include the measure dr r2, integratingthe radial coordinate [

R2 ∂f

∂R

]R+

R−

= −i4ΩRM

Now, the crucial observation here is that we substitute in the limits from above and below accordingto the different forms of f (r). This means that

AR2 [1− (−2)] = −i4ΩRM

Thus, we have that A = −43ΩRM . Substituting in our value of A and comparing the real and

imaginary parts, we recover our formula for h0i = h0i as

h0i =

ω (y,−x, 0) r < RωR3

r3(y,−x, 0) r > R

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This means that our new metric is

ds2 =

(1− 2M

R

)dt2 +

(1 +

2M

R

)(dx2 + dy2 + dz2

)+ 2ωydxdt− 2ωxdydt

Our corresponding Lagrangian is

L = −(

1− 2M

r

)t2 +

(1 +

2M

r

)(x2 + y2 + z2

)+ 2ωyxt− 2ωxyt

Let’s look at what our geodesic equation for x is. Making use of the Euler-Lagrange equations, wehave that for the x coordinate

x

(1 +

2M

r

)+ xr(−2M

r2) + ωyt+ ωyt = −ωyt

Now, to first order we have that t ' 1. The second term on the left hand side above is second orderin |x| and ωyt is as well. So, we’re left with the equation

x(1 +2M

r)+ = −2ωy

where this time we’re differentiating with respect to t since it’s the same as differentiating withrespect to λ to first order. Similarly, we have that

y(1 +2M

r)+ = 2ωx

Taking the(

2Mr

)perturbation as small to the other side, we recover the Coriolis force.

From the above, we can just integrate out the angular parts straight way to arrive at the differentialequation

d

dr

(r2 d

drh00

)= −4MGδ (r −R)

Another way to do the last part would be to observe that since the gravitational potential insidethe shell is zero. Hence, we’re left with just the kinetic part of the Lagrangian which is

L = ~v · ~v + 2h0i · ~v

We have that h0i = ~r × ω. This means that our Lagrangian is

~v · ~v + 2 (ω × ~v) · ~r

Using the Euler-Lagrange equations

d

dt

(∂L∂d~v

)= 2

d

dt~v + 2~v × ω

∂L∂~r

= 2 (ω × ~r)

d

dt~v = 2ω × ~r

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11.4 Question 4

Our second order contribution from our Christoffel component is

Γµνρ = −1

2hµσ(hσν,ρ + hσρ,ν − hνρ,σ)

Our Ricci tensor is given by

R(2)µν [h] = ∂ρΓ

ρνµ − ∂νΓρρµ + ΓαµνΓραρ − ΓαµρΓ

ραν

To keep things simple, we look at contributions by type, and see if they match with what we’regiven. Specifically, terms of the schematic h∂∂h come from the ∂Γ−∂Γ term. After working thoughthe algebra, this term is given by

R(2)µν [h∂∂h] = −1

2hρσhσν,µρ +

1

2hρσhνµ,ρσ +

1

2hσρhρσ,µν −

1

2hρµ,σν

This agrees with the form shown. Next, we need to calculate the only other type of term there is inthe expansion which is of the form ∂h∂h schematically. The easiest way it seems to go about doingthis is by first calculating the term ∂ρΓ

ρνµ + ΓαµνΓραρ and then proceed to anti-symmetrise over the

indices ρ, ν on the bottom.

The contribution comes from

∂ρΓρνµ + ΓαµνΓραρ = −1

2hρσ,ρ (hσν,µ + hσµ,ν − hνµ,σ) +

1

4ηασηρβ (hµσ,ν + hνσ,µ − hµν,σ) (hαβ,ρ + hρβ,α − hαρ,β)

This term here is quite gnarly to deal with. We get three terms from the first term in the sum, anda further nine terms from the two first order h brackets multiplied together.

11.4.1 Deriving the Linearised Einstein Hilbert action

Recall that in the Lagrangian formulation our Einstein-Hilbert action can be written as

SEH =

∫d4x√−gR

In this question, we have to be slightly careful. We need to include terms of both order O (h) andorder O

(h2)in our Rµν term, since gµν = ηµν − hµν . We first have to find out what

√−g is. To

first order, using the fact that the derivative of the determinant of a matrix is the trace, we havethat

g = −1− h =⇒ (−1− h)12 ' −1− h

2So the total term we have to consider is

SEH =

∫d4x

(1− h

2

)(ηµν − hµν)Rµν

Our first order contribution to the Ricci tensor is

R(1)µν [h] = ∂ρ∂(µhν)ρ −

1

2∂ρ∂ρhµν −

1

2∂µ∂νh

To make things clearer, we go term by term. The first term that’s easiest to check out is the 14∂ρh∂

ρhterm in the integrand. We can selectively pick out terms from each bracket which will get us whatwe want. This is a technique we can try.

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11.5 Question 5

Let’s do some counting here. In our gauge transformation, we have 4 degrees of freedom which arechosen from λ. But, we have that it looks like we have five conditions from the conditions

H0µ = H = 0

What happened? Well, we need to impose the de Donder gauge condition so that

−H00k0 +H0iki = 0

This removes one constraint, namely, if H0i = 0, then since k0 = 0 then we get that H00 = 0 forfree!

This question is about solving the right equations to force us in the ’transverse-traceless’ gauge, orTT gauge. Since we’re solving a wave equation in two indices, we have a polarisation matrix Hµν .All we’re doing in this question is showing that when we move to a different gauge,

Hµν → Hµν + i (kµXν + kνXµ − ηµνkρXρ)

we can solve the right equations to get that

H0µ = Hµµ = 0

It’s not always obvious a priori that we have enough degrees of freedom in our gauge transformationsto be able to solve things like this, so the whole point of the question is to indeed check that this ispossible.

First observe that due to the condition kµkµ = 0, we have that all non-trivial solutions to the waveequation must include k0 6= 0.

Suppose that initiallyH0µ = f0µ

Imposing our transverse condition means that we require

f0µ = −i (kµX0 + k0Xµ − η0µkρXρ)

If we set µ = 0, carrying the i to the other side gives us the equation

if00 = k0X0 + kiXi

In addition, imposing our traceless condition means that if we set Hµµ = ρ, then we get the

condition−iρ

2= kρX

ρ = −k0X0 + kiX

i

Combining this with the condition that H00 = 0 above, we find that we can solve for X0 bysetting

X0 =i

2k0

(f00 +

ρ

2

)The other transverse conditions are

H0i = f0i = −i (k0Xi + kiX0)

96

However, since we’ve already solved for X0, we can just invert to get

if0i − kiX0

k0= Xi

These are all the gauge conditions we need. However, we still need to check that these solutions areconsistent with one another. To do this, the simplest thing to do is to take our wave vector to liein the z direction without loss off generality, so k = ω (1, 0, 0, 1). This yields a system of equationswe have to solve as well as check for consistency.

if00 = ω (X0 +X3)

− iρ2

= ω(−X0 +X3)

if01 = ωX1

if02 = ωX2

if03 = ω (X0 +X3)

We have explicit solutions for X1 and X2, but we need to check for consistency for the X0 and X3

components. Solving for the first two equations, we have that

X0 =1

2ωi(f00 +

ρ

2

), X3 =

1

2ωi(f00 −

ρ

2

)The only thing we need to check is that now, f01 = f03. However, this is okay since our deDonder gauge condition kµHµν = 0 for ν = 0 implies that f00 = f03. Thus, we have a consistentsolution.

Question: is it always valid to take the wave vector in the z direction without loss of generality?

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12 Question 6

Assuming we have slow moving bodies, the four momenta of an object is given by the vector withunity in the zeroth component as well as the Newtonian velocity.

Be careful when contracting indices. For example if we want to compute...I kk, make sure you

calculate Ikk first in the definition of Qij first.

For each particle, the associated energy-momentum tensor is the mass of each particle along with adelta function for the coordinate.

T 00 = m1δ (z) δ (x− r1 cosωt) δ (y − r1 sinωt) +m2δ (z) δ (x+ r2 cosωt) δ (y + r2 sinωt)

Using standard double angle formulae we have that

Iij =

(m1r

21 +m2r

22

2

)1 + cosωt 2 sinωt 0sin 2ωt 1− cos 2ωt 0

0 0 0

Using this, our quadrapole moment is given by

Qij (t) = Iij −1

3Ikkδij =

(m1r

21 +m2r

22

2

)13 + cos 2ωt sin 2ωt 0

sin 2ωt 13 − cos 2ωt 0

0 0 −23

Differentiating this with respect to time 3 times, and then using the expression for reduced massµUR = m1r1 = m2r2, R = r1 + r2, we have that

...Qij = 4ω3µR2

sin 2ωt cos 2ωt 0cos 2ωt − sin 2ωt 0

0 0 0

Our expression for power is given by the expression

P =1

5

...Qij

...Qij

Recalling a factor of 2 that comes from summing the triogonometric terms in the matrix, we havethat

P =32

5ω6µ2R4

Substituting our values of ω and µ, this gives our value of P as

P =32

5

G4m21m

22 (m1 +m2)

R5

For the second part of the question, we want to solve dEdt = −P. Using the chain rule for our

expression for E, we have that

RGm1m2

2R2= −32

5

G4m21m

22 (m1 +m2)

R5

Simplifying, we have thatdR

dt= −64G3m1m2 (m1 +m2)

5R3

98

Substituting in our initial condition that R = R0 at t = 0, then we get that

R4

4=R4

0

4− 64

5G3m1m2 (m1 +m2)

To extract the chirp mass from experimentally determining ω and ω, we use the following facts(note that I’ve emitted constants here). We have from the definition that

ω ∝ (m1 +m2)12

R32

Differentiating ω2 with respect to time, and substituting our expression for R above, we findthat

ω ∝ (m1 +m2)32 m1m2R

− 112

Now, just sub out R in favour of ω in to get

ω = m1m2 (m1 +m2)−13 ω

113

This means that we get an experimentally measured value of m1m2

(m1+m2)13. Just take this to the power

of 35 to get the chirp mass.

99

13 Problem Solving Tips

13.1 Being careful about how we expand things

We need to be careful about trying to apply the covariant derivative to Christoffel symbols. Forexample, writing out

∇νΓνρµ

doesn’t make sense since Γ is not a tensor! Γ doesn’t transform as a function. Thus, we need toexpand out things the right way. Let’s try to expand out the term

∇[µ∇ν]f

We need to work out from in, because we know that ∇νf is a covector.

13.2 Tricks in the Levi-Civita connection

In the Levi-Civita connection, we have that raising indexes

13.3 Integrating over volume forms

Always integrate ’as is’ - we don’t have to add the metric if it’s not there. When trying to calculatethe hodge star, write things out in terms of the Vielbein basis.

The minus sign when taking our hodge star comes from our θ0 has the Minkowski metric.

(Weird thought -

13.4 Showing that magnitude is constant

It’s a lot easier to show things like the angle between two vectors being constant

Sometimes we need to avoid coordinate singularities, a work around is to rotate the manifold.

13.5 Symmetries of the Riemann Tensor

We have 4 main symmetries of the Riemann Tensor to worry about. Antisymmetry in the 1 and 2indices, anti-symmetry in the 3 and 4 indices, Symmetry in swtiching, and Bianchi identity.

We do the first steps by writing things out in terms of an anti-symmetric basis. Then, adding onthe Bianchi identity is a matter of counting the extra independent constraints we can get.

In 2D, this means that Gµν = 0, so no matter exists.

13.6 General techniques

Looking at the number of components can help simplify things greatly.

100