484 Advanced Electrodynamics and Gravitygratus/484/484.pdfH Flanders, Differential forms with...

484 Advanced Electrodynamics and Gravity

Jonathan Gratus

Physics Department

Lent 2019

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Motivation

Our goal is

Introduce differential geometry.

Introduce coordinate free notation.

Introduce covariant electrodynamics.

Further develop ideas for general relativity.

These have application to many ideas of physics:

Covariant electrodynamics: Cosmology, QED

String theory

Quantum Gravity.

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Example coordinate free electrodynamics.Maxwell’s equations

dF = 0 and d ? F = ?J

and the Lorentz force equationm∇VV = qiVF where g(V,V) = −1

where with respect to a coordinate system (t, x1, x2, x3)

F = dt ∧ E + ?(dt ∧ B) and V = γ∂

∂t+

3∑i=1

γvi ∂

∂xi

For this we need to understand a large quantity of new notation and theunderlying ideas they encapsulate.

A manifold and a coordinate system (t, x1, x2, x3)A vector field: V , and thinking of vectors as differential operatorsA metric g and metric dual •Forms: F is a 2-form, E and B are 1-formsThe wedge product ∧The internal contraction iV•The Hodge dual ?The exterior derivative dThe covariant derivative ∇

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Books

Some books which you can use on the course:

H Flanders, Differential forms with application to the physical sciences,Dover Publications

CW Misner, KS Thorne, JA Wheeler, Gravitation, W.H.Freeman & Co Ltd

IM Benn, RW Tucker, Introduction to Spinors and Geometry withApplications in Physics

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Some idiosyncratic notation

In this course I will use three notations which are not universal. In fact theyare mainly used by me.The goal is to remove the ambiguity associated with parentheses.

My Usual Commentnotation notation

Evaluation of scalars ata point

f |p f (p) Both are common nota-tion.

Vectors acting onscalars

V〈f 〉 V(f ) andVf

My unique notation

1-forms combiningwith vectors

α : V α(V) The notation does existelsewhere but it is rare.

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Contents

1 Basic geometric ideas

2 Black Holes

3 Orthonormal frames and co-frames

4 Forms

5 Physical Applications (Knowledge hooks)

6 Tensors

7 Connections

8 Einstein’s equation and Conservation Laws

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1 Basic geometric ideas1.1 Scalar Fields1.2 Coordinated functions1.3 Manifolds1.4 The summation convention recap1.5 1–forms and the differential d on scalar fields1.6 Vector Fields1.7 Combining a 1–form (field) and a vector field1.8 The tensor product1.9 The metric1.10 Curves1.11 Lie bracket of vector fields and the Lie Derivative1.12 Different notions of linearity

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Scalar fields on Rn

A Scalar field on Rn is simply a real valued function f : Rn → R.For example f : R2 → R,f (x, y) = (x2 + y2 + 1/10)−1 + ((x− 1)2 + y2 + 1/11)−1.[1]If p = (2, 3) ∈ R2 what is f (p)We also write f |p instead of f (p).On spacetime R4 coordinated by (t, x, y, z), examples include entropyS(t, x, y, z), temperature T(t, x, y, z) and pressure p(t, x, y, z).In two dimensions R2 coordinated by (x, y) we may represent scalar fields as3 dimensional functions or contour plots.

−1.5

−1.0

−0.5

0.02.0

2.5

y1.50.51.0

5.0

f

0.51.00.0

7.5

x −0.5 1.5

10.0

−1.0

0.5

−0.5

x

0.0−0.5 0.5

0.0

0.25

0.75

−0.25

1.5

y

1.0

−0.75

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Addition and multiplication of scalar fields

The set of all scalar fields on Rn is denoted ΓΛ0Rn

Let f and h be two scalar fields.

The sum of f and h is simply the sum at each point. That is

(f + h)|p = f |p + h|p

The product of f and h is simply the product at each point. That is

(fh)|p = f |p h|p

We need these in order to define vector fields.

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Coordinate functions

Let Rn have coordinates (x1, . . . , xn), then each

xa : Rn → R ; xa|(p1,...,pn) = pa

is an (indexed) scalar field for each a = 1, . . . , n.

For example [2]x2|(5,3,−2) =? .

[3]If p = (2, 4,−3, 7) ∈ R4, what is xa|p for a = 1, . . . , 4?

The functions xa : Rn → R are called the coordinate functions.

Observe that each point p ∈ Rn is uniquely determined by the values of thefunctions (x1|p, . . . , xn|p).

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Alternative coordinate functionsGiven Rn and a set of functions ya : Rn → R. Now for any set of numbers(ξ1, . . . , ξn) assume there is a unique point p ∈ Rn such that ya|p = ξa. Thenthe set of functions (y1, . . . , yn) is an alternative coordinate system.For example. Let

y1|(x1,x2) =√

(x1)2 + (x2)2 and

y2|(x1,x2) = arctan(x2/x1),

0 ≤ y2|(x1,x2) ≤ π if x2 ≥ 0π < y2|(x1,x2) < 2π if x2 < 0

Then (x1, x2) represent Cartesian coordinates for R2 and (y1, y2) representplane polar coordinates. Usually we write (y1, y2) = (r, θ).[4]Draw picture[5]What are the range of y1 and y2?We may also think of (y1, . . . , yn) as Rn.Then we have a new set of functions xa : Rn → Rn, xa|(y1,...,yn).For example

x1 = y1 cos(y2) and x2 = y1 sin(y2)11/165

Subsets of Rn

There is no requirement that (x1, . . . , xn) have all possible values of Rn. Wecan simply have a subset. Which we write S ⊂ Rn

For example in the polar (r, θ) coordinate system, we have

S =

(r, θ)∣∣ r > 0 and 0 < θ < 2π

⊂ R2

For technical reasons, we require open subsets of Rn. These are sets withoutboundaries. For example 0 < θ < 2π and not 0 ≤ θ ≤ 2π.

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Manifold example: The Sphere

Consider the sphere(x1, x2, x3) ∈ R3

∣∣∣ (x1)2 + (x2)2 + (x3)2 = 1

There are several possible coordinate systems. It is a two dimensionalmanifold. So every coordinate system has two coordinate functions.

One coordinate system (the spherical coordinate system) is given by(θ, φ) ∈ R2

∣∣∣ 0 < θ < π and 0 < φ < 2π

x1 = sin θ cosφ , x2 = sin θ sinφ and x3 = cos θ

Observe that the north and south poles are not covered by this coordinatesystem.Also since φ 6= 0 and φ 6= 2π then there is a line of longitude that is excluded.

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Manifold example: The Sphere

Another coordinate system is given by(y1, y2) ∈ R2

∣∣∣(y1)2 + (y2)2 < 1

x1 = y1 , x2 = y2 and x3 =√

1− (y1)2 − (y2)2

This coordinate system covers only the north hemisphere of the sphere.

Given the two coordinate systems, this gives rise to two maps mappingbetween (y1, y2) and (θ, φ).

y1 = sin θ cosφ , y2 = sin θ sinφ

and

θ = arcsin(√

(y1)2 + (y2)2), φ = arctan(y2/y1)

Using these two coordinate systems we can cover more of the sphere.

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A Manifold:

Very sketchy definition.

It is a set M such that there exists coordinate systems.

A coordinate system, is a subset S ⊂ M and a set of functions (x1, . . . , xn),xa : S→ R such that each point p ∈ S is uniquely determined by the numbers(x1|p, . . . , xn|p).

In general no coordinate system covers the entire manifold. This is the reasonwe need several coordinate systems.

However for Minkowski spacetime there does exist a single coordinate system(t, x, y, z) ∈ R4 covering the entire manifold.Alternative coordinate systems are given by rotations and Lorentz boosts.

Given two coordinate systems (x1, . . . , xn) and (y1, . . . , yn). This will giverise to two sets of functions

ya = ya(x1, . . . , xn) and xa = xa(y1, . . . , yn)

The set of all scalar fields on M is denoted ΓΛ0M.15/165

The summation convention recapIndices are dummy indices or external (also known as free) indices and are

either high or low.∂

∂xa is considered to be a low a index.An index which occurs once each term is an external index.All terms must have exactly the same external indices and in the same position.We can always replace an external index with another unused index.[6]What does Va = Ua + Wa mean for a = 1, 2, 3?[7]Why is it the same as Vb = Ub + Wb?An index which occurs twice in the same term, one high and one low is adummy index, and is (implicitly) summed over all its possible values.We can replace any dummy index with another unused index.[8]What does Va = Aa

bUb mean?[9]Why is it the same as Va = Aa

cUc?If an equation does not conform to these rules, for example having 3 copies ofthe same index, it is not a valid equation.Sometimes the summation convention cannot be used. [10]For examplewhen using eigenvalue expressions or determinants. In this case we mustwrite down sums explicitly. We should also write (no implicit sum).

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1–form fields and the differential d on scalar fieldsI have decided this year to have an unconventional order:Defining 1–forms before vector fields.Consider the differential operator d which acts on scalar fields to give a“1–form” field.We will use the differential operator to define 1-froms. This is circularhowever there is a formal non circular definition, (see below).The set of 1–form fields is writen ΓΛ1M.For each scalar field f there is a corresponding 1–form df ∈ ΓΛ1M.Given two 1–forms α, β ∈ ΓΛ1M and a scalar field f ∈ ΓΛ0M then

α+ β ∈ ΓΛ1M and f α ∈ ΓΛ1M

Thus we can add together two 1-forms and multiply a 1-form by a scalar field.All 1–forms fields can be constructed by d of a scalar, addition andmultiplying by scalars. I.e.

ΓΛ1M =

k∑

µ=1

fµ dhµ

∣∣∣∣∣ f1, . . . , fk, h1, . . . , hk ∈ ΓΛ0M

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The differential d on scalar fields

d is defined simply via the following criteria:d : ΓΛ0M → ΓΛ1M, i.e. it maps scalar fields (0-forms) to (1–forms).d is ‘+’-linear

d(f + h) = df + dh

If λ ∈ R is a constant then dλ = 0and d is Libnitz

d(fh) = h df + f dh

The set of 1–forms α ∈ ΓΛ1M such that there is a scalar field f ∈ ΓΛ0M withα = df are call exact.

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Visualising exact 1–forms

x

y

A 1–form in 2 dimensions.The hypersurface are1–dimensional.

A 1–form in 3 dimensions.The hypersurface are 2–dimensional.

An exact 1–form is simply the contour lines or hypersurfaces.That is they have dimension one less that the space in which they live.They do not have boundaries.They also have a transverse arrow.This was the direction of increasing of the original field by now representsimply an orientation.[11]Where must the 1–form vanish?

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Integrating an exact 1–form

Let [a, b] ∈ R be the intervalτ ∈ R, a ≤ τ ≤ b and let C : [a, b]→ M bea curve.

Given an exact 1–form α = df then we canintegrate α along the curve C : [a, b]→ M∫

Cα =

∫C

df = f (b)− f (a)

This means simply counting the number ofcontour lines of α crossed by C.[12]If C is a loop, what is

∮C α?

An exact 1–form corresponds to aconservative field.

Ca

bα = df

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Addition of 1–forms

In 2 dimension, adding the two 1–forms, red and blue to give the 1–formgreen.

In the left picture, the blue 1–form is increasing when going up and the red isincreasing when going right. Therefore the result green form increases goingup and right.

Orientation is important when we add forms.

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The differential d: Function of functions

Let h1, . . . , hk ∈ ΓΛ0M be a set of k scalar fields and let f : Rk → R be asmooth function. Then f (h1, . . . , hk) ∈ ΓΛ0M is a scalar field and

d(f (h1, . . . , hk)

)=

k∑µ=0

∂f∂hµ

dhµ

We can show this is true: [13]See additional handout.

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The differential d in coordiantes

Given a coordinate system (x1, . . . , xn) then we can consider the scalar fieldf ∈ ΓΛ0M as a function f (x1, . . . , xn).

df =∂f∂xa dxa

Given two coordinate systems (x1, . . . , xn) and (y1, . . . , yn) thenya = ya(x1, . . . , xn) hence

dya =∂ya

∂xb dxb

[14]If α = αadxa is exact, i.e. α = df , what is a criteria for the αa ?[15]In 3-dim with coordinates (x, y, z) calculate df .[16]Compare df and∇f .Also look at the pictures of both.Why do we need the metric?

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Arbitrary 1–forms and thermodynamicsIn a coordinate system (x1, . . . , xn) then all α ∈ ΓΛ1M can be writen[17](show)

α = αa dxa

Thus the set

dx1, . . . , dxn

form a basis of 1–forms.[18]Let α = αa dxa = αa dya show the relationship between the componentsαa and αa.1–forms are also known as covector fields.In Phys411 these were written αa.I.e. in index notation αa is a 1–form, whereas in coordinate free notation, αa

are scalars which are the components of α.You have seen the df symbol representing “A small increase in f ”. Forexample in thermodynamics.[19]Explain dU = T dS− P dV[20]Explain d/U = T dS− P dV?

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Non exact forms

A non exact 1–form can be thought of as hypersurfaces with boundaries.They can also be thought of as little patches of hypersurface.

y

x

The non exact 1–form y dx in two dimensions.Each dash is a bit of a hypersurface. Howeverthey each have a boundary.

Integrating

[21]On R2, coordinates (x, y), calculate∮

Cx dy where C joins the points (0, 0),

(2, 0), (2, 2), (0, 2), (0, 0) along straight segments.

These are non conservative fields.

An integral depends on the path taken.

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Non exact 1-forms

The left is an integrable 1-form in three dimensions. Two1-form-submanifolds (green) each with a scalar field, represented by contourlines (red). The exterior differential of this 1-form are the closed red contourlines.

The right is a non exact, non integrable 1–form in three dimensions. This oneis more complicated. Going round the centre (red line) may not land you inthe same place.

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Vector Fields

Intuitively we think of a vector field as an arrow at each point in our manifold

We have seen vector fields in 3 dimensions. For example the electric,magnetic or gravitational field.

The challenge is: How do we specify the direction of a vector?

If we have a coordinate system, (x1, . . . , xn), then we can simply specify thenumbers (V1, . . . ,Vn), called the components of V .

Since V is a field, the quantities Va depend on position. I.e. Va : S→ R. Thisis an (indexed) scalar field so that at p ∈ M the value of Va is Va|p.Thus the components of V|p are (V1|p, . . . ,Vn|p).

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Vector Fields

There is no preferred coordinate system. Given the components of V withrespect to one coordinate system, what are the components of V with respectto another coordinate system?

Let V have components (V1, . . . ,Vn) with respect to (x1, . . . , xn) andcomponents (V1, . . . , Vn) with respect to (y1, . . . , yn) then

Va = Vb∂ya

∂xb and Vb = Va∂xb

∂ya

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Vector Fields: The partial derivative

Let f ∈ ΓΛ0M be a scalar field and (x1, . . . , xn) be a coordinate system.

Let a ∈ 1, . . . , n. Consider the partial derivative∂f∂xa .

It means the differential in the direction of xa keeping all the other derivativesconstant.

0.0

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−0.25

−0.5

0.75

−0.5

0.0

1.50.5

0.5

1.0

0.25

x2

x1

-

6∂∂x2

∂

∂x1

Thus∂

∂xa may be thought of as vector in the direction of increasing xa.

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Vector Fields: The partial derivativeIn two dimensions V〈f 〉 = V1 ∂f

∂x1 + V2 ∂f∂x2

0.0

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0.75

−0.5

0.0

1.50.5

0.5

1.0

0.25

x2

x1

-

6∂∂x2

∂

∂x1

V

Here x1|p = 0.4, x2|p = −0.1,V1|p = 1 and V2|p = 2.

Think of f (x, y) as a mountainous terrain then V〈f 〉 the gradient of f is thedirection of V .I.e. look in the direction V , then V〈f 〉 is the slope at your feet, times |V|.Note the use of angle brackets for vectors acting on scalar fields. V〈f 〉. This isan idiosyncratic notation of mine. The text book notation is simply V(f ) orVf . But I believe this is more likely to lead to misinterpretation.

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Vector Fields: A differentialLet V have coefficients (V1, . . . ,Vn) with respect to (x1, . . . , xn) andcoefficients (V1, . . . , Vn) with respect to (y1, . . . , yn) then

Va = Vb∂ya

∂xb and Vb = Va∂xb

∂ya

Let us write

V = Va ∂

∂xa

[22]Show that

V = Va ∂

∂ya

Thus writing the vector V as a differential automatically encodes the changeof coordinates.Can we define a vector field without reference to a coordinate system?

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Vector Fields: A differential

WithV = Va ∂

∂xa (1)

Let f ∈ ΓΛ0M be a scalar field.

V〈f 〉 = Va ∂f∂xa ∈ ΓΛ0M (2)

is another scalar field.

Let f , h ∈ ΓΛ0M be two scalar fields. Then [23] show

V〈f + h〉 = V〈f 〉+ V〈h〉 (3)

and [24]Leibniz rule

V〈f h〉 = f V〈h〉+ h V〈f 〉 (4)

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Vector Fields: A differentialA vector field V is defined as a function which takes a scalar field f ∈ ΓΛ0Mand gives a new scalar field V〈f 〉 ∈ ΓΛ0M. I.e.

V : ΓΛ0M → ΓΛ0M , V : f 7→ V〈f 〉

such that it is ‘+’-linear

V〈f + h〉 = V〈f 〉+ V〈h〉

and Leibniz ruleV〈f h〉 = f V〈h〉+ h V〈f 〉

[25]Given (1) show

Va = V〈xa〉

The set of all vector fields on M is written ΓTM.In “classical tensor analysis”, Va is a vector field. In “modern differentialgeometry” V is a vector field and Va is a component of V .

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Vector Fields: The AlgebraAction on scalar fields: f , h ∈ ΓΛ0M and V,U ∈ ΓTM

V〈f + h〉 = V〈f 〉+ V〈h〉 and V〈f h〉 = h V〈f 〉+ f V〈h〉

Addition of two vector fields gives a new vector field.

V + U ∈ ΓTM where (U + V)〈f 〉 = U〈f 〉+ V〈f 〉

Multiplication of a vector field by a scalar field gives a new vector field.

fV ∈ ΓTM where (fV)〈h〉 = f V〈h〉

Proof the fV is a vector: For f , h, k ∈ ΓΛ0M

(f V)〈h + k〉 = f V〈h + k〉 = f V〈h〉+ f V〈k〉 = (fV)〈h〉+ (fV)〈k〉

and

(f V)〈h k〉 = f V〈h k〉 = f k V〈h〉+ f h V〈k〉 = k (fV)〈h〉+ h (fV)〈k〉

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Vector Fields: Basis and NotationGiven a coordinate system (x1, . . . , xn) then

∂

∂x1 , . . . ,∂

∂xn

is a basis for vector the fields since every vector field may be written

V = Va ∂

∂xa .

If the coordinate system has different letters, for example (r, θ) then we write

∂r =∂

∂rand ∂θ =

∂

∂θ

If the coordinate system (x1, . . . , xn) is fixed then we often write

∂a =∂

∂xa

This notation does not work if we have two coordinate systems: e.g.(x1, . . . , xn) and (y1, . . . , yn). In this case we would have to write ∂a and ∂ya .

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Vector Fields: Example

Consider the change of coordinates from R2 Cartesian coordinates (x, y) andplane polar coordinates (r, θ)

x = r cos θ and y = r sin θ

[26]Show that

∂x = cos θ ∂r −sin θ

r∂θ and ∂y = sin θ ∂r +

cos θ

r∂θ

and

∂r =x√

x2 + y2∂x +

y√x2 + y2

∂y and ∂θ = x ∂y − y ∂x

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Vectors at a point

pVp

We have introduced a vector field. This is a vector at each point in on M.A vector at a point consists of two pieces of information: A base point p ∈ Mand a vector Vp at p.Given a coordinate system (x1, ..., xn) on M. [27]How many numbersdescribe Vp?The set of all vectors at points is written TM. [28]What dimension is TM?A point in spacetime is an event, a vector in spacetime is a point in TM. Itcorresponds to an event with a 4-velocity at that point.[29]Why can we not take a vector at 1 point and convert it to a vector field?

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Combining a 1–form (field) and a vector field

Given a 1–form (field) α ∈ ΓΛ1M and a vector field V ∈ ΓTM, then there is anatural contraction which gives a scalar field α : V ∈ ΓΛ0M.[30]Draw picture

This is ‘+’-linear and ‘f’-linear so that

(α+ β) : V = α : V + β : V , α : (U + V) = α : V + α : U

and (fα) : V = α : (f V) = f (α : V)

and

(h df ) : V = h V〈f 〉

[31]Calculate (αa dxa) : (Vb∂b).

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The tensor productGiven two 1–form fields α, β ∈ ΓΛ1M we can create a tensor α⊗ β.We call α⊗ β a tensor of type (form,form).Given two tensors of type (form,form) we can add them together to create anew tensor of type (form,form).

α1 ⊗ β1 + α2 ⊗ β2

In fact we can add an arbitrary number of such tensors.We call ⊗ the tensor product.It has the following properties:‘+’-linearity:

(α1 + α2)⊗ β = α1 ⊗ β + α2 ⊗ βand α⊗ (β1 + β2) = α⊗ β1 + α⊗ β2

‘f ’-linearity:(fα)⊗ β = α⊗ (fβ) = f α⊗ β

for α, α1, α2, β, β1, β2 ∈ ΓΛ1M and f ∈ ΓΛ0M.In general there is no symmetry property.

α⊗ β 6= β ⊗ α39/165

The tensor productGiven a coordinate system (x1, . . . , xn) [32]show that any tensor of type(form,form) may be written

S = Sab dxa ⊗ dxb

A tensor of type (form,form) may act on two vectors

(α⊗ β)(U,V) = (α : U) (β : V)

The first form acts on the first vector, the second form on the second vector.This extends by linearity:[33]What is

(α1 ⊗ β1 + α2 ⊗ β2)(U,V)

[34]Show that a general tensor S of type (form,form) is‘+’-linearity and‘f’-linearity in both arguments.

S(U + V,W) = S(U,W) + S(V,W) and S(f U,W) = f S(U,W)

S(U,W + V) = S(U,W) + S(U,V) and S(U, f W) = f S(U,W)

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The tensor product

[35]What is

(Sabdxa ⊗ dxb)(U,V)

[36]Show that any map T : ΓΛTM × ΓΛTM which takes two vector fields andgives a scalar field, and also is ‘+’-linear and ‘f’-linear in both its arguments isa tensor of type (form,form)

We will see later that we can tensor together any collections of vectors andforms so that V ⊗ α⊗ U is of type (vector,form,vector).

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The metric

The metric is the the most important structure that one can add to a manifold.

It is the generalisation of the “dot” product.

A metric is a (invertable) tensor of type (form,form). So it takes two vectorfields and gives a scalar field.

We usually assume there is a single metric g.

In a coordinate system we can write g = gab dxa ⊗ dxb. Invertable means thatgab is invertable as a matrix.

The signature of the metric is the number of positive eigenvalue and thenumber of negative eigenvalues as a list.

General relativity is the study of 4-dimensional manifolds with signature(−,+,+,+).

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The power of the metric

The metric is necessary for so many important properties of the objets weconsider.

Length of vectors.Angle between vectors.Distance between points.It gives one a volume element so that you can integrate scaler fields.It alows you to convert between vectors and 1-forms.It gives the density of p-formsIn general relativity it gives whether a vector or curve is a physical velocity orworldline for a particle or a lightray.The metric automatically give a conection so you can differentiate vectors. Youcan find geodesics. You can find the acceloration of a particle.It gives the curvature of the manifold. For GR this gives the gravity andEinstiens field equations.

Without a metric your manifold is made of rubber, infinitely strechable.(Differential toplogy)

With a metric your manifold is made of concrete.

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Manifolds without metrics.

If a metric is so useful why consider manifolds without one.Not all manifolds in physics posess a (prefered) metric. For example

TM and Λ1M, even when M does have a metric.6-dimensional phase space.7-dimensional phase-time space.Configuration space.States space in thermodynamics.Manifolds used for solving differential equations.

In other cases it may be useful to consider manifolds with several metrics.In tensformation optics one can consider spacetime with two metrics.In calculus of variation one considers metrics which vary. This leads to thestress-energy tensor,

Thus it is (sometimes) very important to know how objects depend on themetric.

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The metric, viewing the metric

The unitspheres in flatRiemannianspace.

The unitspheres incurvedRiemannianspace.

The metric is a tensor of type (form,form), so it takes two vectors and gives ascalar. We can write

g(U,V) ∈ ΓΛ0M where U,V ∈ ΓTM

The best way to visulise the metric is to draw the set of all vectors with unitlength. I.e. V at a point p such that g(V,V) = ±1.

Clearly for R3 with the standard metric the set of unit vectors forms a sphere.

If the metric has signature with all +’s, then we are considering Riemanniangeometry. In this case the metric is represented by unit ellipsoids.

If the manifold is curved then the ellipsoids are different at different points inspace.

45/165

The unit hyperboloids in spacetime.tim

e

spacespace

(a) The two-sheet (upperand lower) hyperboloid:g(V,V) = −1.Timelike vectors.

spacetime

space

(b) The double cone:g(V,V) = −0.Lightlike vectors.

spacetime

space

(c) The one-sheethyperboloid:g(V,V) = 1.Spacelike vectors.

In spacetime the signature is (−,+,+,+).This means g(V,V) can be positive, negative or zero.The pictures are for three dimensional spacetime, with signature (−,+,+).For curved spacetime, the hyperboloids are different at different points.For spacetime it is usual to index the coordinate starting at 0 giving(x0, x1, x2, x3), with ∂0 timelike and ∂1, ∂2, ∂3 spacelike.Some textbooks and Phys411 use the opposite signature, i.e. (+,−,−,−).The geometry is equivalent. 46/165

The metric for the plane in Cartesian and polar coordinatesA metric g is a tensor of type (form,form) such that:

It is symmetric g(U,V) = g(V,U).It is non degenerate.

In a coordinate system g = gab dxa ⊗ dxb

where gab = gba and det(gba) 6= 0.Thus

g(U,V) = gabUa Vb

This is akin to PHYS411 notation

ds2 = gabdxa dxb

The metric for the plane in Cartesian coordeinates (x, y) is simply

g = dx⊗ dx + dy⊗ dy[37]Show that in plane polar coordinates it given by

g = dr ⊗ dr + r2 dθ ⊗ dθ[38]Since ∂θ = x ∂y − y ∂x show that g(∂θ, ∂θ) = r2 in both coordinatesystems.

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Minkowski spacetimeThe simplest spacetime is Minkowski spacetime.A special coordinate system is given by (x0, x1, x2, x3). It is usual to set x0 = tand call it the laboratory time.The metric for Minkowski spacetime is given by

g = η where η = −dt ⊗ dt + dx1 ⊗ dx1 + dx2 ⊗ dx2 + dx3 ⊗ dx3

[39]What are the values of ηab?Special relativity is simply the study of Minkowski spacetime.The study of spacetimes which are not Minkowski spacetime is generalrelativity. These consider the effects of gravity.Let v ∈ R3 with the standard dot product.

V ∈ ΓTM , V = γ∂t +

3∑i=1

γvi∂i where γ = (1− v · v)−1/2

[40]Show that g(V,V) = −1.48/165

Length of vectors in Riemannian geometry.

0.51

1.52

Using the metric ellipsoid to calculatethe length of a vector. The unitellipsoid is scaled to various sizes. Thevector has length 2.

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Length of vectors in spacetime geometry.

1

2tim

e

1 2space

The diagonal (green) lines are the path oflight rays. Thus the diagonal vector (darkgreen) is a lightlike vector.

• The left and right (red) curves are thehyperboloid for measuring the length ofspatial vectors. (In two dimensions thesetwo curves are disjoint. However in threeor more dimensions these two linesconnect to create the one-sheethyperboloid.) The right-pointing (black)vector is spatial and has length 2.

• The upper and lower (blue) lines are for measuring the length of timelikevectors. The upward (orange) line is timelike and has length 1 in the forwarddirection. The downward (purple) line is timelike and has length 2 in thebackward direction.

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Angles: Riemannian geometry

In Riemannian geometry, the cosine ofthe angle between to vectors U,V is

g(U,V)

(g(U,U)g(V,V))1/2

Using the metric ellipsoid to measurethe angle between two vectors. The(red) lines are orthogonal to one of thevectors (black). The length ` is thecosine of the angle between the twovectors.

`

51/165

Angles: Spacetime geometry

In spacetime geometry the quantity

g(U,V)

(g(U,U)g(V,V))1/2

where U and V are timelike, givesthe gamma factor and the timedilation.

Using the metric hyperboloid tomeasure the angle between twovectors. The (red) lines areorthogonal to one of the vectors(black). The length γ is theγ-factor required to boost one ofthe vectors to the other.

time

space

γ

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Metric dual

Given a vector field V ∈ ΓTM then the metric dual is written

V ∈ ΓΛ1M , V : U = g(V,U)

[41]Taking the metric dual V 7→ V is ‘+’-linear and ‘f’-linear, i.e.˜(U + V) = U + V and (fV) = f U

For every 1–form field α∈ΓΛ1M there is a unique V ∈ ΓTM such that α = V .(This is due to the fact that g is nondegenerate.)We therefore say that V is the metric dual of α and write V = α

The metric dual maps 1–forms to vectors and vectors to 1–forms. Furthermore

˜V = V and ˜α = α

Alternative “musical” notation is α# = α and V[ = V .

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Metric dual: Pictorially

The metric dual of the 1–form (red) isthe vector (blue).

Note that the length of the vector 2,corresponds to the density of the1–form.

time

space

The metric dual of the 1–form (red) isthe vector (blue).

Note that the length of the vector 2.5,corresponds to the density of the1–form.

The opposite direction is because thevector is timelike and we use signature(−,+,+,+).

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Raising and lowering indices

[42]Show

V = Vagab dxb for V = Va∂a

The inverse of gab is given by gab. That is

gabgbc = δac

[43]Show that

α = αagab∂b for α = αadxa

[44]Calculate

dxc and ∂c

55/165

Dual of metricThe dual metric g is a tensor of type (vector,vector). I.e. it takes two 1–formsand gives a scalar field

g : ΓΛ1M × ΓΛ1M → ΓΛ0M

It is defined byg(α, β) = g

(α, β

)Combining the various duals we have for V ∈ ΓTM, α ∈ ΓΛ1M [45](show)

α : V = g(α,V) = g(α, V) = V : α

[46]Show that

gab = g(dxa, dxb)

We can therefore write

g = gab∂a ⊗ ∂b

[47]Expand g(α, β).56/165

The dual metric and the denisty of 1-forms

The value g(α, α) gives the density squared of the 1–form α.

Pictorially using the metric to measurethe magnitude of a 1-form in 2dimensions. Simply count the numberof form-submanifold inside the ellipseand divide by 2. In this case the1-form has magnitude 3.5

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Tangent to a curve

A curve in M is a map C : R→ M. The parameter on the curve is usuallywrite τ . We will consider only non intersecting curves.

At each point p = C(τ) on the curve, there is a vector C(τ) which representsthe tangent to the curve C.

We emphasise that C is not a vector field over the whole of M, only over thosepoints which are in the image of C.

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Tangent to a curve

We can define C(τ) via its actions on scalar fields. Observe that τ 7→ f (C(τ))is a function R→ R. Thus we can differentiate it

C(τ)〈f 〉 =d(f |C(τ)

)dτ

(5)

We can also use a coordinate system (x1, . . . , xn). Let Ca(τ) = xa(C(τ)).Thus the curve is represented by a set of n functions (C1(τ), . . . ,Cn(τ)).Then

C(τ) =dCa

dτ∂a

∣∣∣C(τ)

(6)

[48]Show that (5) and (6) are equivalent.

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Timelike, spacelike and lightlike curves

In spacetime the sign of g(C, C) characterises the nature of the curve C.C is timelike if g

(C(τ), C(τ)

)< 0 for all τ .

C is spacelike if g(C(τ), C(τ)

)> 0 for all τ .

C is lightlike if g(C(τ), C(τ)

)= 0 for all τ .

Timelike curves represent the motion of point particles in spacetime.

Lightlike curves represent the motion of particles moving at the speed of lightin spacetime.

No particle can travel in a spacelike curve.

There is no mathematical reason the C cannot pass from timelike, thoughlightlike to spacelike. However for physically reasonable curves they retaintheir nature.

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Integral Curves

Let V ∈ ΓTM be nonvanishing. A curve C : R→ M is an integral curve of Vif at each point p = C(τ) on C, the tangent C(τ) is equal to V evaluated at p.That is

C(τ) = V|C(τ)

Given p ∈ M there always exists a unique integral curve of V that passesthrough p.

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Integral Curves and ODEsConsider C(τ) = V|C(τ) in coordinates

Ca(τ) = Va|C(τ)

where the set Va : Rn → R are known functions.This can be written

dC1

dτ= V1(C1(τ), . . . ,Cn(τ))

...dCn

dτ= Vn(C1(τ), . . . ,Cn(τ))

This is a general time independent system of ordinary differential equations.Thus all ODEs systems are simply the search for integral curves and via versa.[49]How do we geometrise time dependent ODEs?[50]How do we geometrise higher order ODEs?

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Lie bracket of two vector fieldsGiven two vector fields U,V ∈ ΓTM we can define a new vector field the Liebracket of U and V

[U,V] ∈ ΓTM , [U,V]〈f 〉 = U⟨V〈f 〉

⟩− V

⟨U〈f 〉

⟩This is the formal definition. It is necessary to show that [U,V] is indeed avector field. This requires showing that [U,V]〈f 〉 is [51]‘+’-linear in f :

[U,V]〈f + h〉 = [U,V]〈f 〉+ [U,V]〈h〉

[52] And Leibniz in f

[U,V]〈fh〉 = f [U,V]〈h〉+ h[U,V]〈f 〉

Given a coordinate system (x1, . . . , xn) and U = Ua∂a and V = Va∂a[53]Show that

[U,V] =(

Ua∂Vb

∂xa − Va∂Ub

∂xa

)∂b

This is a formal definition, but what does [U,V] mean?63/165

Lie bracket of coordinate vector fields

If U and V are coordinate vector fields. I.e. U = ∂a and V = ∂b. Then

[U,V] = 0[54]Why?

The Lie bracket is then a measure of how well two vector fields can becoordinate vector fields.It poses the question “Does the black square close?”

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Lie derivative of scalars and vectorsThe Lie derivative of scalars is given by

LU : ΓΛ0M → ΓΛ0M LUf = U〈f 〉and of vectors

LU : ΓTM → ΓTM LUV = [U,V]

This satisfies ‘+’-linearity and Leibniz

LU(V + W) = LUV + LUW and LU(fV) = LU〈f 〉V + fLU(V)

Let M be a manifold with coordinates (t, x1, . . . , xn) and let U = ∂t[55]show

that

L∂t

(Va∂a

)=∂Va

∂t∂a

where t = x0 and the sum is over a = 0, . . . n.We see that the Lie derivative with respect to a coordinates vector field issimply partial derivative of the components.However LfU(V) = −V〈f 〉U + f [U,V] so LU also “differentiates U”.LU is not ‘f’-linear in U.

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Lie derivative of 1–forms and tensors

The Lie derivative is Leibniz across contraction. That is for a vector field Wand a 1–forms field α,

LV(α : W) = (LVα) : W + α : (LVW)

This is the definition of LVα.

The Lie derivative is Leibniz across a tensor product. That is for tensors S,Tthe

LV(S⊗ T) = (LVS)⊗ T + S⊗ (LVT)

[56] Given a tensor T of type (form,form) so that it act on two vectors, then

LV(T(W,Y)

)= (LVT)(W,Y) + T(LVW,Y) + T(W,LVY)

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Different notions of linearity and LeibnizIn order to discuss linearity we must have a map which takes one field toanother field. The three fields we have encountered so far are scalar fieldsΓΛ0M, vector fields ΓTM and 1–form fields ΓΛ1M.We have encountered three different concepts of linearity.‘+’-linearity: If we add two arguments do the results add?E.g. V : ΓΛ0M → ΓΛ0M is ‘+’-linear since V〈f + h〉 = V〈f 〉+ V〈h〉.‘R’-linearity: If we multiply the argument by a real number λ ∈ R does theresult scale?E.g. V : ΓΛ0M→ΓΛ0M is ‘R’-linear since V(λf ) = λV〈f 〉 where λ∈R.‘f ’-linearity: If we multiply the argument by a scalar field f ∈ ΓΛ0M does theresult scale?E.g. g : ΓTM × ΓTM → ΓΛ0M is ‘f ’-linear since g(fV,U) = f g(V,U) wheref ∈ ΓΛ0M.(Clearly ‘f ’-linearity implies ‘R‘-linearity.)If a map is not f -linear it may satisfy some kind of Leibniz rule.E.g. V : ΓΛ0M → ΓΛ0M is not ‘f ’-linear since V〈f h〉 6= fV〈h〉. However itdoes satisfy a Leibniz rule: V(f h) = fV〈h〉+ h V〈f 〉

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Linearity and Leibniz

Vector acting on a scalar ‘+’-linear V〈f + h〉 = V〈f 〉+ V〈h〉V : ΓΛ0M → ΓΛ0M ‘R’-linear V〈λf 〉 = λV〈f 〉

Leibniz V〈f h〉 = fV〈h〉+ h V〈f 〉Metric dual V ‘+’-linear V + W = V + WV 7→ V ‘R’-linear λV = λV: ΓTM → ΓΛ0M ‘f ’-linear fV = f VLie derivative ‘+’-linear LU(V + W) = LU(V) + LU(W)LU : ΓTM → ΓΛ0M ‘R’-linear LU(λV) = λLU(V)

Leibniz LU(fV) = fLU(V) + LU(f )V

A complete list of linearities of all the operators we deal with is on a handout.

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‘f’-Linearity and tensors

pV|p = U|pV is redU is blueU − V is yellow

‘f’-linearOperations which are ‘+’-linear and ‘f’-linear consist of adding and multiplyingcomponents.These can be defined even when we only know the argument at one point.For example if we only know U|p at the point p ∈ M we can still calculate(α : U)|p. [57]Show.These operations are called tensorial. We will define a tensor later.

LeibnizOperations which are Leibniz involve differentiating components.These require knowledge of arguments in a region.For example to evaluate LU(V) at the point p if we need to know V in a regionabout p.

This will become important when we look at connections.69/165

Lie derivatives and the force equation for particles.Given a point particles. The motion of the particle is described by theworldline, a curve C : R→ M where M is spacetime.The 4-velocity of the particle is given by its derivative C(τ) ∈ TM.If τ is the particle’s proper time then g

(C(τ), C(τ)

)= −1.

The dynamics of the particle is given by

Rest mass× 4-Acceleration = 4-Force

where the 4-acceleration is the derivative of 4-velocity (in the direction of C).However we have no method yet of defining the derivative of 4-velocity.We cannot use the Lie-derivative since LU(V) is not ‘f’-linear in U. Thismeans that to define LC(C) at a point C(τ) requires knowledge of C(τ) in aregion about C(τ). But we only know C(τ) in a line about C(τ).Besides LUU = 0.We will see later that the correct acceleration is given by the connection

4-acceleration = ∇CC

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2 Black Holes2.1 Killing vectors, stationary and static spacetimes2.2 Kruskal-Szekeres coordinates for Schwarzschild Black hole2.3 Penrose-Carter diagrams2.4 The Reissner-Nordström black hole2.5 The Kerr-Newman black hole


4 Forms


6 Tensors

7 Connections

8 Einstein’s equation and Conservation Laws71/165

Lie derivative of the metric

Recall a spacetime is determined by its metric g : ΓTM × ΓTM → ΓΛ0M.The Lie derivative LV(g) of a metric g with respect to a vector field V isdefined so that

LV(g) is a ‘+’-linear, LV(g) : ΓTM × ΓTM → ΓΛ0M.It is Leibniz.

In fact the Lie derivative on anything is defined so that is preserves thestructure of the object and that it is Leibniz.

Leibniz gives us

LV(g(X,Y)) = (LVg)(X,Y) + g(LVX,Y) + g(X,LVY)

I.e

(LVg)(X,Y) = V〈g(X,Y)〉 − g([V,X],Y)− g(X, [V,Y])

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Killing vectors

A vector field K ∈ ΓTM is Killing if

LKg = 0

I.e.

K〈g(X,Y)〉 − g([K,X],Y)− g(X, [K,Y]) = 0

Given a coordinate system (x1, . . . , xn). [58]Show that K = ∂1 is killing ifand only if

∂gab

∂x1 = 0

Killing vectors correspond to symmetries.

We will see that Killing vectors are required for conservation laws.

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Killing vectors of Minkowski spacetimeMinkowski spacetime has 10 killing vectors. If (x0, . . . x3) are the standardcoordinate these are given by

1 temporal translation:

∂0

This give rise to the conservation of energy.3 spacial translation:

∂1 , ∂2 , ∂3

These give rise to the conservation of momentum.3 rotations:

x1∂2 − x2∂1 , x2∂3 − x3∂2 , x3∂1 − x1∂3

These give rise to the conservation of angular momentum.3 boosts

x0∂1 + x1∂0 , x0∂2 + x2∂0 , x0∂3 + x3∂0

What conservation do these give rise to?74/165

Asymptotically flat spacetimeTo define flatness we really need the concept of curvature which we see later.However we know that Minkowski spacetime is flat.A spacetime is Asymptotically flat if “as we go towards spacelike infinity themetric approximates Minkowski”.If this is possible we can write

g = habdxa ⊗ dxb + ηabdxa ⊗ dxb

where

hab(x)→ 0 as r →∞where r2 = (x1)2 + (x2)2 + (x3)2.Thus we can think of this as the following 2-dimensional picture.

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Stationary and static spacetimesA spacetime is stationary spacetime if it

Has a killing vectorIs asymptotically flat.The killing vector is timelike in the asymptotically flat region.

If a spacetime is stationary it can be written in coordinates (t, x1, x2, x3)

g = −dt ⊗ dt + ωidt ⊗ dxi + ωidxi ⊗ dt + gijdxi ⊗ dxj

with sum i, j = 1, 2, 3, where ∂ωi∂t = 0 and ∂gij

∂t = 0 and where ωi → 0 andgij → δij as (x1)2 + (x2)2 + (x3)2 →∞.This is true in the region where ∂t is timelike.The functions ωi represent the rotation of the spacetime.A spacetime is static if

It is stationary.It has zero rotation, i.e. ωi = 0

I.e.

g = −dt ⊗ dt + gijdxi ⊗ dxj

and the vector ∂t is orthogonal to the surfaces of constant t.76/165

Examples of static spacetimesMinkowski spacetime. In spherical coordinates this can be written

g = −dt ⊗ dt + dr ⊗ dr + r2dθ ⊗ dθ + r2 sin2 θdφ⊗ dφ

Schwarzschild spacetime (A static uncharged black hole)

g = −(1− 2mr )dt ⊗ dt + (1− 2m

r )−1dr ⊗ dr + r2dθ ⊗ dθ + r2 sin2 θdφ⊗ dφ

Observe that in the asymptotic limit r →∞, it becomes Minkowski. (As everwe have chosen units so that G = 1, c = 1 and ε0 = 1.)Here m represents the “mass” of the black hole. That is in the asymptoticlimit, slow moving test particles behave as though there was a Newtonianmass m at the origin.∂t is a killing vector field. [59]Is it always timelike?Reissner-Nordström spacetime

g = −(1− 2mr + q2

4πr2 )dt ⊗ dt + (1− 2mr + q2

4πr2 )−1dr ⊗ dr

+ r2dθ ⊗ dθ + r2 sin2 θdφ⊗ dφ

Here m is the mass and q is the charge of the black hole.77/165

Example of stationary but not static spacetimesThe Kerr-Newman spacetime in coordinates (t, r, θ, φ)

g = −(dr ⊗ dr

∆+dθ ⊗ dθ

)ρ2 +

(dt − α sin2 θ dφ

)⊗(dt − α sin2 θ dφ

)∆

ρ2

−((r2+α2)dφ− α dt

)⊗((r2+α2)dφ− α dt

)sin2 θ

ρ2

where α =Jm, ρ2 = r2 + α2 cos2 θ , ∆ = r2 − 2mr + α2 +

( q4π

)2

This is a charged rotating black hole. m is the mass, q is the charge and J isthe angular momentum.The Kerr spacetime is given by setting q = 0 and represents an unchargedrotating black hole.The only stationary spacetime which axially symmetric and a solution to thevacuum Einstein’s equations is the Kerr spacetime. A special case is theSchwarzschild spacetime.The “no hair” theorem states that Black hole is completely described by itsmass m, charge q and angular momentum J.This is true classically. But is it true quantum mechanically?

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Non stationary spacetimes

An example of a non stationary spacetime isFriedmann-Lemaître-Robertson-Walker metric in spherical polars (t, r, θ, φ)is given by

g = −dt ⊗ dt + a(t)2(

(1− kr2)−1dr ⊗ dr + r2dθ ⊗ dθ + sin2 θdφ⊗ dφ)

where a(t) is a function representing the expansion or contraction of theuniverse.

This is used in cosmological models.

Observe that ∂t is not a Killing vector.

The constant k has the value −1, 0 or 1 depending on whether the universehas negative, zero or positive curvature.

The equation for a(t) is given by the Einstein equation. This depends of thechoice of matter (given in the stress-energy-momentum tensor) and thecosmological constant.

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Kruskal-Szekeres coordinates for Schwarzschild Black holeRecall that a Schwarzschild Black hole metric is

g = −(1− 2mr )dt ⊗ dt + (1− 2m

r )−1dr ⊗ dr + r2Ω

where Ω = dθ ⊗ dθ + sin2 θdφ⊗ dφ is the metric in the sphere.The Kruskal-Szekeres coordinates are given by (T,R, θ, φ)

T = ( r2m − 1)1/2er/4m sinh(t/4m) and R = ( r

2m − 1)1/2er/4m cosh(t/4m)

ThendT =

er/4m

8m2

(2m( r

2m− 1)1/2

cosh( t

4m

)dt +

sinh( t

4m

)r

( r2m − 1)1/2 dr

)dR =

er/4m

8m2

(2m( r

2m− 1)1/2

sinh( t

4m

)dt +

cosh( t

4m

)r

( r2m − 1)1/2 dr

)Thus32mer/2m

r(−dT ⊗ dT + dR⊗ dR) = −(1− 2m

r )dt ⊗ dt + (1− 2mr )−1dr ⊗ dr

henceg =

32me−r/2m

r(−dT ⊗ dT + dR⊗ dR) + r2Ω

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Kruskal-Szekeres coordinates for Schwarzschild Black holeT = ( r

2m − 1)1/2er/4m sinh(t/4m) and R = ( r2m − 1)1/2er/4m cosh(t/4m)

where r is given by implicitly solving( r2m − 1

)er/2m = T2 − R2

Inside the event horizon

T = ( r2m − 1)1/2er/4m cosh(t/4m) and R = ( r

2m − 1)1/2er/4m sinh(t/4m)

g =32me−r/2m

r(−dT ⊗ dT + dR⊗ dR) + r2Ω

The event horizon r = 2m is given by

T2 − R2 = 0 i.e. T = ±R

Observe that the event horizon is no longer a coordinate singularity.From g we see that the vector ∂T ± ∂R are lightlike.The singularity r = 0 is the curve

T2 − R2 = 1

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Kruskal-Szekeres coordinates for Schwarzschild Black hole

R

T

This it the maximal extended Schwarzschild Black hole.Lightlike linear are at 45.Timelike likes are more vertical than 45.Spacelike lines are more horizontal that 45.

If exterior 1 is our universe. There is another universe exterior 2. You cannottravel to exterior 2 from exterior 1.There is also a white hole.

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Penrose-Carter diagrams: Minkowski spacetimeFor the two dimensional Minkowski spacetime

g = −dt ⊗ dt + dr ⊗ dr

Let a new coordinate system (T,R) be given by

tan(T + R) = t + r and tan(T − R) = t − r

Thus the metric becomes

g = sec2(R + T) sec2(T − R)(−dT ⊗ dT + dR⊗ dR)

This is conformally equivalent to

gconf = −dT ⊗ dT + dR⊗ dR

i.e. gconf = λ2g where λ ∈ ΓΛ0M, λ > 0.Although conformally equivalent metrics do not preserve length, they dopreserve angles.

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Penrose-Carter diagrams: Minkowski spacetimeObserve that g is only defined inside the square

|T + R| < π2 and |T − R| < π

2

Lightlike worldlines are at 45. Timelike more vertical. Spacelike morehorizontal.

Note that if T + R = π2 then t − r =∞. Thus this square represents the

boundary of spacetime.The edges of the square are lightlike infinity. They are reached by light rays.[60]Show a timelike curve which reaches the lightlike infinityMost timelike curves go from the bottom corner to the to corner. 84/165

Penrose-Carter diagrams: Schwarzschild black holesSingularity

Our Universeoutside ofevent horizon

Another Universe

Singularity

Inside ofblack hole

Even

t Horizon

This is the equivalent diagram for Schwarzschild black holes withKruskal-Szekeres coordinates.We have crunched up the lightlike infinities.Lightlike curves and the event horizon are at 45.[61]Describe the blue, red and yellow curvesNear the singularity the line of constant r are spacelike. Hence we can drawthe singularity as a horizontal line.

85/165

Einstein-Rosen Bridge

Consider the region ofSchwarzschild spacetime witht = 0. Then

g = (1− 2mr )−1dr ⊗ dr + r2Ω

Now set

r = ρ(

1 +m2ρ

)2so that

( drdρ

)2 1(1− 2m/r)

= (1 + m/2ρ)4

Hence

g = (1 + m/2ρ)4(dρ⊗ dρ+ ρ2(dθ ⊗ dθ + sin2 θdφ⊗ dφ))

The upper paraboloid is the region r > 2m and is asymptotically flat. Thelower paraboloid is the region r < 2m and is also asymptotically flat.The minimum of r is at r = 2m and ρ = m/2. This corresponds to the throatof the Einstein-Rosen bridge.This is a spacelike worm-hole and it is not possible to cross it.

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The Reissner-Nordström black hole

Recall the Reissner-Nordström metric.

g = −(1− 2mr + q2

4πr2 )dt ⊗ dt + (1− 2mr + q2

4πr2 )−1dr ⊗ dr

+ r2dθ ⊗ dθ + r2 sin2 θdφ⊗ dφ

This has two event horizons at

r+ = m + (m2 − q2/4π)1/2 and r− = m− (m2 − q2/4π)1/2

What does the the Penrose-Carter diagram look like for a Reissner-Nordströmblack hole?

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The Reissner-Nordströmblack hole

Sin

gularity


Another Universe

Inside ofblack hole

Out

er E

vent H

orizon

Another UniverseAnother Universe

Inner Event Horizon

Sin

gularity

Sin

gula

rity

Sin

gularity

Inside ofblack hole

Inside ofblack hole

Inside ofblack hole

88/165

The Kerr-Newman black hole

The Kerr-Newman spacetime in coordinates (t, r, θ, φ)

g = −(dr ⊗ dr

∆+dθ ⊗ dθ

)ρ2 +

(dt − α sin2 θ dφ

)⊗(dt − α sin2 θ dφ

)∆

ρ2

−((r2+α2)dφ− α dt

)⊗((r2+α2)dφ− α dt

)sin2 θ

ρ2

where α =Jm, ρ2 = r2 + α2 cos2 θ , ∆ = r2 − 2mr + α2 +

( q4π

)2

The singularity occurs when ρ = 0, i.e. when r = 0 and θ = π/2.This is a ring singularity, which can be avoided.

We can pass by the ring into another universe, which contains a naked ringsingularity. In this r < 0.

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The Kerr-Newmanblack hole

Sin

gularity


Another Universe

Inside ofblack hole

Out

er E

vent H

orizon

Another UniverseAnother Universe

Inner Event Horizon

Sin

gularity

Inside ofblack hole

Inside ofblack hole

Inside ofblack hole

Ass

ym

toti

cally fl

atre

gion

con

tanin

gnak

ed r

ing

singu

lari

ty

90/165

Acausal regionConsider q = 0 and the curve C : R→ M with

t(C(τ)) = 0 , r(C(τ)) = r0 , θ(C(τ)) = π/2 and φ(C(τ)) = τ

ThenC(τ) = ∂φ

and this curve is periodic with C(τ) = C(τ + 2π).Set q = 0, [62]Show

g(∂φ, ∂φ

)=

1ρ2

(α2(r2 − 2mr + α2)− (r2+α2)2

)Assume r0 is small then

g(∂φ, ∂φ

)≈ −2mr0

α2

ρ2 < 0

Near the ring singularity there is a torus which contains closed timelikecurves!!

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The Kerr-Newman black hole: The Ergoregion

There is a region outside the event horizon where ∂t is spacelike. All timelikecurves must rotate.

Erg

ospher

e

Inside outerevent horizon

From above the surfaces of lightlikeemanating from each point looks like.

Penrose suggested this was a way of extracting energy from a rotating blackhole. You throw stuff into the ergosphere and when it emerges it must have agreater velocity.

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2 Black Holes


4 Forms


6 Tensors

7 Connections


93/165

Orthonormal frames and co-framesOrthonormal frames are a very useful tool for simplifying calculationsinvolving the metric, such as the metric dual and the Hodge dual (see below).On an n-dimensional Manifold a frame and co-frames are set of vector fieldsX1, . . . ,Xn, Xa ∈ ΓTM and 1-form fields

e1, . . . , en

, ea ∈ ΓΛ1M such

that

g(Xa,Xb) = ηab , g(ea, eb) = ηab and ea : Xb = δab

whereηab = ηab =

±1 if a = b0 if a 6= b

Recall the number of +1 versus −1 is the signature. E.g.2 dimensional Riemannian geometry (+,+), so gab = δab

n dimensional Riemannian geometry (+, . . . ,+), so gab = δab

Spacetime (−,+,+,+), so gab = ηab

Henceg = ηabea ⊗ eb =

n∑a=1

(±1)ea ⊗ ea

[63]Show that

Xa = ηabeb , ea = ηabXb , U = (ea : U) Xa and α = (α : Xa) ea94/165

Orthonormal framesIf the metric with respect to a coordinate system is diagonal (i.e.g(∂a, ∂b

)= 0 if a 6= b) then we can find an orthonormal frame by simply

rescaling the coordinate vectors.[64]For the Schwarzschild metric coordinates (t, r, θ, φ) which is diagonal and

g = −(1− 2mr )dt ⊗ dt + (1− 2m

r )−1dr ⊗ dr + r2dθ ⊗ dθ + r2 sin2 θdφ⊗ dφ

find an orthonormal co-frames and orthonormal frame.Many operations such as metric duals and Hodge duals are much easier usingthe orthonormal frame.Other operations that taking the exterior derivative are easier using acoordinate frame.You must therefore be able to switch between frames.[65]Using the frame given above for the Schwarzschild metric, express thevector

U = a∂r + b∂θ

in terms of the orthonormal frame.95/165

Example of metric duals I

For a coordinate system where the metric is not diagonal, then you need todiagonalise the metric first. This is the same as diagonalising a symmetricmatrix.

For the manifold with coordinates (u, v, z) and metric

g = 2du⊗ dv + 2dv⊗ du + a2dz⊗ dz

[66]Find an orthonormal frame and write down the metric[67]Find

du , dv and dz

[68]Write down the dual metric in both coordinates and orthonormal frame

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4 Forms4.1 Visualising Forms4.2 2-forms4.3 The electromagnetic 2-form field4.4 Higher degree forms4.5 Exterior derivative d on higher forms4.6 Integration, Stokes theorem and conserved currents4.7 Internal Contraction4.8 f-Linearity, Leibniz, Graded Leibniz4.9 Hodge dual4.10 Lie derivatives of forms

97/165

Visualising forms

In 3-dimensions, the wedge productof a 1–form (vertical, blue) with a1–form (horizontal, red) to give a2-form (green).

In 3-dimensions, the wedge product ofthree 1–forms to give a 3–form (green).The first vertical (blue) followed by thesecond facing reader (brown) then thethird horizontal (red).

Given two 1–forms one can combine them together to create a 2–form.Given three 1–forms one can create a 3–form.The result is given by taking the intersection. This is called wedging.The orientation is given by concatenating the orientations together.Thus in n–dimensions a 1–form in an (n−1) dimensional hypersurface. A2–form is an (n−2) dimensional hypersurface.An (n−1)–form consists of lines and an n–form consists of dots.

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2-formsThe set of all 2-form fields is written ΓΛ2MGiven two 1-forms α, β ∈ ΓΛ1M we can create a 2-form by wedging themtogether with the wedge product

α ∧ β ∈ ΓΛ2M

We can add together 2-forms

α ∧ β + ζ ∧ ξ ∈ ΓΛ2M where α, β, ζ, ξ ∈ ΓΛ1M

We can multiply a 2-form by a scalar field.

fα ∧ β ∈ ΓΛ2M where α, β ∈ ΓΛ1M and f ∈ ΓΛ0M

The wedge product of two 1-forms has the following properties:‘+‘-linear in both arguments:(α1 + α2) ∧ β = α1 ∧ β + α2 ∧ β and α ∧ (β1 + β2) = α ∧ β1 + α ∧ β2

‘f ‘-linear in both arguments:(fα) ∧ β = fα ∧ β and α ∧ (fβ) = fα ∧ β

It is antisymmetricα ∧ β = −β ∧ α

So α ∧ α = 0.99/165

Basis for 2-formsRecall that given a coordinate system (x1, . . . , xn) the basis for 1-forms isgiven by

dx1, . . . , dxn

.

A basis for 2-forms isdxa ∧ dxb

∣∣∣ 1 ≤ a ≤ n and a < b ≤ n

For example if n = 4 then the basis isdx1 ∧ dx2, dx1 ∧ dx3, dx1 ∧ dx4, dx2 ∧ dx3, dx2 ∧ dx4, dx3 ∧ dx4

A general 2-form α ∈ ΓΛ2M may be written

α =

n∑a=1

n∑b=a+1

αabdxa ∧ dxb where αab ∈ ΓΛ0M

(The summation convention is not used.)[69]If ξ = ξadxa ∈ ΓΛ1M and β = βadxa ∈ ΓΛ1M what is ξ ∧ β?[70]In 3 dimensions compare with ξ × β. (Why do we need Hodge dual?)

100/165

The problem with 12

Given α ∈ ΓΛ2M we can write this using the basis

α =

n∑a=1

n∑b=a+1

αabdxa ∧ dxb (7)

We can also write this using the summation convention

α = αabdxa ∧ dxb where we set αba = −αab (8)

However (7) and (8) are different sums.Setting n = 3 then (7) is

α = α12dx1 ∧ dx2 + α13dx1 ∧ dx3 + α23dx2 ∧ dx3

whereas (8) isα = α11dx1 ∧ dx1 + α12dx1 ∧ dx2 + α13dx1 ∧ dx3

+ α21dx2 ∧ dx1 + α22dx2 ∧ dx2 + α23dx2 ∧ dx3

+ α31dx3 ∧ dx1 + α32dx3 ∧ dx2 + α33dx3 ∧ dx3

= 2α12dx1 ∧ dx2 + 2α13dx1 ∧ dx3 + 2α23dx2 ∧ dx3

Hence we must set αab = 12αab.

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The problem with 12

There are therefore three ways of expressing α ∈ ΓΛ2M in terms of its basis.

α =

n∑a=1

n∑b=a+1

αabdxa ∧ dxb

α = αabdxa ∧ dxb

α = 12αabdxa ∧ dxb (9)

where we have set αba = −αab.I will always use (9).Different authors use different notations.I am sometimes inconsistent and this gives me problems.The difference is the factor 1

2 .If higher p-forms the factor is 1

p! .

102/165

The electromagnetic 2-form fieldWe can combine the electric field E and the magnetic field B into a single2-form field.Here we will present it for Minkowski space with coordinates (t, x1, x2, x3).We represent the electric and magnetic fields as 1-forms fields instead ofvector fields. I.e. E,B ∈ ΓΛ1M.We can always convert E and B into vector fields by taking the metric dual Eand B.E and B are spacial 1-forms, that is

g(E, ∂∂t

)= E : ∂∂t = 0 and g

(B, ∂∂t

)= B : ∂∂t = 0

We set

F ∈ ΓΛ2M , F = dt ∧ E − B1dx2 ∧ dx3 − B2dx3 ∧ dx1 − B3dx1 ∧ dx2

We will see after we have defined the Hodge dual that there is a more elegantway of writing F which is also valid on general spacetimes.The electromagnetic 2-form is more fundamental than electric and magneticfields, since E and B depend on the choice of the velocity of the observer.This is consistent with Lorentz transformations of t, xi and E,B.

103/165

The electromagnetic 2-form field: Change of basisWe now have sufficient technology to rewrite the electric and magnetic fieldsin a boosted frame.Let (t, x, y, z) and (t′, x′, y, z) be two coordinate systems related via Lorentzboost.

t′ = γ(t − vx) , x′ = γ(x− vt) , y′ = y , z′ = z

where v ∈ R and γ2(1− v2) = 1.Let

F = dt ∧ E − B1dy ∧ dz− B2dz ∧ dx− B3dx ∧ dy

= dt′ ∧ E′ − B′1dy′ ∧ dz′ − B′2dz′ ∧ dx′ − B′3dx′ ∧ dy′

where

E′ = E′1dx′ + E′2dy′ + E′3dz′ and B′ = B′1dx′ + B′2dy′ + B′3dz′

then

E = E′1dx + (γE′2 + B′3γv)dy + (γE′3 − B′2γv)dz and

B = B′1dx + (γB′2 − E′3γv)dy + (γB′3 + E′2γv)dz

104/165

The electromagnetic 2-form field: Change of basisThe proof is included here (in dark orange). It is particularly ugly. We willdemonstrate a much prettier proof once we have defined

The Hodge dual.The “(3 + 1)” split of the Hodge dual.The exterior derivative.The “(3 + 1)” split of the exterior derivative.

I will then reproduce all the proofs in dark orange in 1 line.

F = dt′ ∧ E′ − B′1dy′ ∧ dz′ − B′2dz′ ∧ dx′ − B′3dx′ ∧ dy′

= (γdt − γvdx) ∧ (E′1(γdx− γvdt) + E′2dy + E′3dz)

− B′1dy ∧ dz− B′2dz ∧ (γdx− γvdt)− B′3(γdx− γvdt) ∧ dy

= E′1dt ∧ dx + dt ∧ dy(γE′2 + B′3γv) + dt ∧ dz(γE′3 − B′2γv)

− B′1dy ∧ dz− dz ∧ dx(γB′2 + E′3γv)− dt ∧ dz(γB′3 − E′2γv)

= dt ∧ E − B1dy ∧ dz− B2dz ∧ dx− B3dx ∧ dy

105/165

Higher degree forms

We have seen 0-forms (scalar fields), 1-forms and 2-forms.

We now define p-forms for all p = 3, 4, 5, . . ..

The set of all p-forms fields is denoted ΓΛpM.

We create a p-form by wedging together p 1-forms

α1 ∧ α2 ∧ · · · ∧ αp ∈ ΓΛpM where α1, . . . , αp ∈ ΓΛ1M

We also add together p-form to give a new p-form

α, β ∈ ΓΛpM then α+ β ∈ ΓΛpM

We can multiply a p-form by a scalar field.

fα ∈ ΓΛpM where α ∈ ΓΛpM and f ∈ ΓΛ0M

The degree of a p-form is p. That is if α ∈ ΓΛpM then deg(α) = p.

106/165

Wedge productGiven a p-form α ∈ ΓΛpM and a q-form β ∈ ΓΛqM then we can wedge thesetogether or form

α ∧ β ∈ ΓΛp+qM

Thus the deg(α ∧ β) = deg(α) + deg(β).Wedging is ‘+’ linear in both arguments

(α1 + α2) ∧ β = α1 ∧ β + α2 ∧ β and α ∧ (β1 + β2) = α ∧ β1 + α ∧ β2

‘f ‘-linear in all its arguments:(fα) ∧ β = fα ∧ β and α ∧ (fβ) = fα ∧ β

Associative(α ∧ β) ∧ ζ = α ∧ (β ∧ ζ)

So we can write α ∧ β ∧ ζ without having to state the order in which they arewedged.

f ∧ α = f α For f ∈ ΓΛ0M and α ∈ ΓΛpM

Graded commutative. Given α ∈ ΓΛpM and β ∈ ΓΛqM then [71]show.

α ∧ β = (−1)pqβ ∧ α107/165

Basis of p-formsGiven a coordinate system (x1, . . . , xn) then a basis for ΓΛpM is given by

dxi1 ∧ dxi2 ∧ · · · ∧ dxip∣∣∣ i1 < i2 < · · · < ip

The number of basis forms, and hence the dimension of ΓΛpM is given by

dim(ΓΛpM) =n!

p!(n− p)!

For p = n then dim(ΓΛpM) = 1 and the unique basis element is

dx1 ∧ · · · ∧ dxn

There are no p-forms when p > n.A general p-form is written

α =

n∑i1=1

n∑i2=i1+1

· · ·n∑

ip=ip−1+1

αi1···ipdxi1 ∧ dxi2 ∧ · · · ∧ dxip

α = 1p!αi1···ipdxi1 ∧ dxi2 ∧ · · · ∧ dxip

The notation gets pretty hairy here. Fortunately we will never deal with n > 4.108/165

Exterior derivative: dThe exterior derivative d maps p-forms into (p + 1)-forms.

d : ΓΛpM → ΓΛp+1M

It obeys the following 4 rules.Usual definition for 0-forms

df : V = V〈f 〉 for f ∈ ΓΛ0M and V ∈ ΓTM

‘+’-linearity:

d(α+ β) = dα+ dβ

Graded Leibniz:

d(α ∧ β) = dα ∧ β + (−1)pα ∧ dβ where deg(α) = p

And

d2 = 0

[72] Let dim(M) = 3 with coordinates (x, y, z) and α = f dx + h dy andβ = f dx ∧ dz where f , h ∈ ΓΛ0M. Calculate dα and dβ.

109/165

The electromagnetic and the electromagnetic potential

Given

F ∈ ΓΛ2M , F = dt ∧ E − B1dx2 ∧ dx3 − B2dx3 ∧ dx1 − B3dx1 ∧ dx2

then

dF = 0

[73]is equivalent to the two source free Maxwell’s equations:

∇ · B = 0 and ∇× E +∂B∂t

= 0

110/165

Electromagnetic potential

Since d2 = 0 then if we set F = dA then automatically dF = 0.

A ∈ ΓΛ1M is the electromagnetic potential.

In Minkowski space

A = φdt +

3∑i=1

Aidxi

where φ is the electric potential and A is the magnetic potential

E = −∇φ+∂A∂t

and B = ∇× A

[74]Show F = dA.

A is not unique. We can replace A with A + dψ for any ψ ∈ ΓΛ0M. [75]Why?

We can always find A locally.

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Bases and cohomology

For a general p-form

α = 1p!αi1···ipdxi1 ∧ dxi2 ∧ · · · ∧ dxip

dα = 1p!dαi1···ip ∧ dxi1 ∧ dxi2 ∧ · · · ∧ dxip

= 1p!

∂αi1···ip∂xj dxj ∧ dxi1 ∧ dxi2 ∧ · · · ∧ dxip

since αi1···ip is a 0-form. [76]Show

Given α with dα = 0 the existence of β such that α = dβ is call deRhamcohomology. It is to do with the global nature of manifolds. E.g. circlesversus lines and sphere, tori and planes etc.[77]Think about dθ on circle.

112/165

Exterior derivative (boundary) of non exact forms andStokes’ theorem.

−

+

−+

−

+

−

+

−+

−

+

−

+

In 2–dimensions, the non exact twisted 1–form is given by the black segments.It is twisted to that the arrows lie inside along the segments.The exterior derivative of the 1–form is a 2–form: the red and blue dots.The orientation of the exterior derivative is also twisted: the + and − signs.Stokes’ theorem states that the integral of the exterior derivative of a forminside a domain equals the integral of original for on the boundary.In this case the number of dots in the green domain (-7) equals the number ofblack curves crossing the boundary (also -7).

113/165

Integration and Stokes theorem

Let M be a manifold with dim M = n. Given a form α ∈ ΓΛnM of maximaldegree.We can define integration using a coordinate system∫

Mα =

∫M

fdx1 ∧ · · · ∧ dxn =

∫· · ·∫

f (x1, . . . , xn)dx1 · · · dxn

where α = fdx1 ∧ · · · ∧ dxn.

Let ∂M be the boundary of M. (I will not give a formal definition of this.)The dimension of ∂M is n− 1.

Let β ∈ ΓΛn−1M. Stokes theorem states∫M

dβ =

∫∂Mβ

Stokes theorem is the generalisation of both the 3-dimensional Stokes theoremand the divergence theorem [78]and the fundamental theorem of calculus.

114/165

Conserved currentsIn spacetime a current J ∈ ΓΛ3M is conserved if dJ = 0.From Maxwell’s equations the electric current is conserved. [79]Why?

Σ1

Σ2

S U

Integrating J over the 4-volume U gives

0 =

∫U

dJ =

∫∂UJ = −

∫Σ1

J +

∫Σ2

J +

∫SJ (10)

Where the boundary ∂U of U is split into three 3-spaces: Σ1, Σ2 are twospacelike bounded hypersurfaces (with opposite orientation) and S a timelikehypersurface.

115/165

Conserved currents

Let Σ1 represents a 3-volume V at a time t = t1 and Σ2 the same volume attime t2, whereas S is the boundary of the 3-volume × time, i.e. betweent1 < t < t2 then (10) states

Q2 = Q1 −∫

SJ

where Q1 =∫

Σ1J is the charge in volume V at time t = t1. Likewise for Q2

and∫

S J is the current passing out of the volume V .

I.e.The charge Q2 at time t2 equals the charge Q1 plus the charge flowing

into the 3-volume V .

116/165

The Aharonov-Bohm Effect and Maxwell’s equations

The Aharonov-Bohm effect considers passingelectron (waves) around a solenoid.

The electron waves interfere on the other side.The interference pattern is effected by themagnetic field inside the solenoid.

Thus although the electrons do not experenceany B field, they are still effected in a non localway!

⊗ ⊗ ⊗ ⊗ ⊗⊗

⊗

⊗

⊗

⊗

⊗

⊗

⊗

B 6= 0

e−e−

B = 0

The mathematics is quite easy with C = ∂S as the curve bounded by theelectron paths (in black)

Phase = exp(

i∮

CA)

= exp(

i∫

SdA)

= exp(

i∫

SF)6= 0

BUT: A is not a physical field and the electrons do not see F.[80]Explain how the Aharonov-Bohm effect leads to dF = 0.

117/165

Internal ContractionThe internal contraction takes a vector field V ∈ ΓTM and a p-formα ∈ ΓΛpM field and gives a (p−1)-form field iVα ∈ ΓΛp−1M.I.e.

i : ΓTM × ΓΛpM → ΓΛp−1M , (V, α) 7→ iVα

It is ‘+’-linear in both arguments

i(U+V)α = iUα+ iVα and iV(α+ β) = iVα+ iVβ

It is ‘f ’-linear in both arguments

i(fV)α = f iVα and iV(fα) = f iVα

On 0-form it is simply zero. iV f = 0 where f ∈ ΓΛ0M.On 1-forms it is simply the action of a 1-form and a vector to give a scalar.

iVα = α : V

Given α ∈ ΓΛpM and β ∈ ΓΛqM then on the wedge product we have

iV(α ∧ β) = (iVα) ∧ β + (−1)pα ∧ (iVβ)

This relationship is known as the graded Leibniz rule.118/165

Internal Contraction[81]Show that if αi ∈ ΓΛ1M then

iV(α1 ∧ · · · ∧ αp) = α1 : V α2 ∧ · · · ∧ αp

− α2 : V α1 ∧ α3 ∧ · · · ∧ αp

+ . . .

+ (−1)p−1αp : V α1 ∧ · · · ∧ αp−1

=

p∑r=1

(−1)r−1αr : V α1 ∧ · · · ∧ αr−1 ∧ αr+1 ∧ · · · ∧ αp

In particular iUiV(α ∧ β) = (α : V)(β : U)− (α : U)(β : V)Let

F ∈ ΓΛ2M, F = dt ∧ E + B1dx2 ∧ dx3 + B2dx3 ∧ dx1 + B3dx1 ∧ dx2

andV ∈ ΓTM , V = γ

∂

∂t+

3∑i=1

γvi ∂

∂xi where γ2(1− v · v) = +1

[82]Show that iVF = γ(E · v)∂

∂t+

3∑i=1

γ(E + v× B)i ∂

∂xi

119/165

An incredibly useful formula

The formula for dα where α ∈ ΓΛ1M may be written

iV iUdα = U〈α : V〉 − V〈α : U〉 − α : [U,V]

(See Worksheet)

Also useful

iUiV = −iV iU

120/165

‘f’-Linearity, Leibniz and Graded LeibnizRecall a map is tensorial if it is ‘+’-linear and ‘f’-linear. For example themetric dual

˜(U + V) = U + V and (f U) = f U

A map may be ‘+’-linear and Leibniz. For example the Lie-derivative

LU(V + W) = LUV + LUW and LU(fV) = LU(f )V + f LUW

The last possibility is that it is ‘+’-linear and graded Leibniz.Only objects acting on forms can be graded Leibniz. Otherwise there is noconcept of degree.For example the exterior derivative

d(α+ β) = dα+ dβ and d(α ∧ β) = dα ∧ β + (−1)degαα ∧ dβ

and the interior derivative

iU(α+ β) = iUα+ iUβ and iU(α ∧ β) = iUα ∧ β + (−1)degαα ∧ iUβ

121/165

Hodge dual

The Hodge dual takes a p-form α ∈ ΓΛpM and gives a (n− p)-form?α ∈ ΓΛn−pM

I.e.? : ΓΛpM → ΓΛn−pM , α 7→ ?α

It is ‘+’-linear?(α+ β) = ?α+ ?β

It is ‘f ’-linear?(fα) = f ? α

We define the Hodge dual by induction. We first define ?1 and then define?(α ∧ ζ) where ζ is a 1-form.

122/165

Hodge dual, ?1

If dim M = n then ?1 ∈ ΓΛnM is a form with the highest degree.

We define ?1 using the orthonormal co-frame (e1, . . . , en)

?1 = e1 ∧ · · · ∧ en

?1 is uniquely defined up to a sign ±1.This is known as an orientation on M.Compare left handed and right handed coordinate system on R3. It isimportant, (and a bit of a nightmare) to get the right orientation, otherwisethings like the magnetic field change sign.

In terms of a coordinate basis (x1, . . . , xn) for M,

?1 =√| det(gab)|dx1 ∧ · · · ∧ dxn

where gab = g(∂∂xa ,

∂∂xb

)123/165

Hodge dual, ?(α ∧ ζ)

We now define ?α inductively via

?(α ∧ ζ) = iζ? α

We know what ?α is if deg(α) = 0. [83]It is

Thus we know what ?α is if deg(α) = 1. [84]It is

Thus we know what ?α is if deg(α) = 2. [85]It is

. . .

[86]In a spacetime with orthonormal co-frame (e0, . . . , e3) and signature(−,+,+,+) what are ?e0, ?e1, ?(e0 ∧ e2)?

125/165

Three more identitiesStar pivot. If α, β ∈ ΓΛqM then

α ∧ ?β = β ∧ ?α

The Hodge dual applied twice on a q-form gives the original q-form times ±1.If α is a q-form then

? ? α = s(−1)(n−q)qα

where dim M = n, and s = ±1 is given by

s =det(gab)

|det(gab)|= (−1)number of negative eigenvalues of g

?(iVα) = (−1)q+1V ∧ ?α for α ∈ ΓΛqM

All three of these identities can be proved inductively. Though they may be alittle non trivial.

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Lie derivative of formsThe Lie derivative is already defined on 0-forms and vectors.We extend the Lie derivative to forms by stating that it takes a p-form andgives a p-form, that is

LU : ΓΛpM → ΓΛpM

and it is ‘+’-linear and Leibniz.Thus Lie derivative on 1-forms is given by

LU(α : V) = (LUα) : V + α : (LUV) for α ∈ ΓΛ1M

And the Lie derivative of higher forms is given by induction

LU(α ∧ β) = (LUα) ∧ β + α ∧ LUβ for α ∈ ΓΛ1M, β ∈ ΓΛpM

We can now show Cartan’s identity (Worksheet)

LU = diU + iUd

[87]Show dLUα = LUdα.If K is Killing then

LK ? α = ?LKα127/165


2 Black Holes


4 Forms

5 Physical Applications (Knowledge hooks)5.1 Vector calculus in 3 dimensions5.2 Maxwell’s equations

6 Tensors

7 Connections


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Vector calculus in 3 dimensionsLet M be R3 with the Flat Euclidean metric gab = δab, a, b = 1, 2, 3 withrespect to coordinates (x1, x2, x3).We have the following results: (Here we treat the classical vectors as 1-formson M, α, β ∈ ΓΛ1M)

?(α ∧ β) = α× β for α, β ∈ ΓΛ1M

df = ∇f for f ∈ ΓΛ0M

?dα = ∇× α for α ∈ ΓΛ1M

?d ? α = ∇ · α for α ∈ ΓΛ1M

?d ? df = ∇2f for f ∈ ΓΛ0M

[88]Show.129/165

Maxwell’s equations[89]In Minkowski spacetime M show

F = dt ∧ E − B1dx2 ∧ dx3 − B2dx3 ∧ dx1 − B3dx1 ∧ dx2

= dt ∧ E + ?(dt ∧ B)Let ρ ∈ ΓΛ0M and J ∈ ΓΛ1M be spacial, i.e. satisfy J( ∂∂t ) = 0.[90]Show that (3 + 1 splits handout)

d ? F = ?J where J = ρdt + J

is equivalent to Maxwell’s two source equations for the vacuum:(ε0 = µ0 = 1).

∇ · E = ρ and ∇× B +∂E∂t

= J

We now have Maxwell’s equations written in geometric notationdF = 0 and d ? F = ?J

These are Maxwell’s equations for any spacetime. Thus we can examine howelectromagnetic fields are affected by gravity.[91]Where is the metric?[92]How is gravity affected by electromagnetic fields?

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Some solutions to Maxwell’s equations(Worksheet) in Minkowski space the pane wave solution to Maxwell’sequations with J = 0 is given by

F = Aeiω(t−z)(dt ∧ dx + dx ∧ dz)

where A is a constant.In Reissner-Nordström spacetime the metric is given with signature(−,+,+,+) and orthonormal co-frame

e0 = (1− 2mr + q2

4πr2 )1/2dt , e1 = (1− 2mr + q2

4πr2 )−1/2dr ,

e2 = rdθ , e3 = r sin θdφ

[93]Show that

A =qr

dt

is potential for Maxwell’s equations. I.e.

d ? dA = 0

for r > 0.131/165

Electromagnetic field with respect to a general observer

A general (accelerating) observer is given by a curve C : R→ M, such thatg(C, C) = −1.

A family of observers is given by a vector field V ∈ ΓTM such thatg(V,V) = −1.

With respect to a V we define the electric and magnetic fields as

E = iVF and B = −iV ? F

[94]Show that

F = E ∧ V + ?(B ∧ V)

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2 Black Holes


4 Forms


6 Tensors6.1 Tensors and tensor product6.2 Tensors as maps

7 Connections


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Tensors and tensor productI shall write the collection of all tensor fields Γ

⊗M. (Not standard)

A vector field V ∈ ΓTM and a 1-form field α ∈ ΓΛ1M are both tensor fields.We create new tensor fields by taking the tensor product of two tensor fieldsThus if A,B ∈ Γ

⊗M then

A⊗ B ∈ Γ⊗

M

The type of a tensor is the list

(vector or form, vector or form, . . . )

which describes how the tensor was constructed.For example if α, β ∈ ΓΛ1M and V ∈ ΓTM then

α⊗ β is of type (form,form)α⊗ V is of type (form,vector)V ⊗ β is of type (vector,form)α⊗ V ⊗ β is of [95]type?

[96]What type is a the metric g and the co-metric g?134/165

Tensors and tensor productTwo tensor of the same type may be added. The results has the same type.For example with α, β ∈ ΓΛ1M and U,V ∈ ΓTM then

α⊗ U + β ⊗ V

is a tensor of type (form,vector).We cannot add α⊗ U + V ⊗ β.The tensor product is ‘+’-linear. Given tensors A,B,C ∈ Γ

⊗M, where A and

B are of the same type then

(A + B)⊗ C = A⊗ C + B⊗ C and C ⊗ (A + B) = C ⊗ A + C ⊗ B

The tensor product is ‘f’-linear. Given tensors A,B ∈ Γ⊗

M and f ∈ ΓΛ0Mthen

(f A)⊗ B = A⊗ (f B) = f (A⊗ B)

The tensor product is associative. Given tensors A,B,C ∈ Γ⊗

M then

(A⊗ B)⊗ C = A⊗ (B⊗ C) = A⊗ B⊗ C

There is no concept of commutativity, anti-commutativity orgraded-commutativity for tensors.

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Tensors and tensor productFor example, let M have coordinates (x, y, z) and

A = ydx⊗ dy⊗ ∂

∂y+ 4xdy⊗ dz⊗ ∂

∂z

Let M have coordinates (x1, . . . , xn) a general tensor of type(vector,form,vector) is

Sab

c ∂

∂xa ⊗ dxb ⊗ ∂

∂xc

A tensor in index notation is simply a multi-index object. Each upper indexrefers to a vector, each lower index to a form.My definition of the type of a tensor is not standard. Normally they simplyrefer a tensor of type (r, s) where r is the number of vectors in the list and sthe number of 1-forms. However this does not distinguish correctly differenttypes. For example α⊗ V and V ⊗ α are both type (1, 1).Observe that in index notation, it is not possible to distinguish between thedifferent type (r, s) tensors. That is αaUb = Ubαa.

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Tensors as mapsA tensor act as a function whose arguments are the opposite of its type. Thatis a tensor of type (vector,form,vector) will have as its arguments (A form, Avector, A form).As a list each form will act on each vector. The result is always a scalar.For example let α, β ∈ ΓΛ1M and U,V ∈ ΓTM then

(α⊗ U)(V, β) = (α : V) (β : U)

Tensors are ‘+’-linear and ‘f’-linear in all their arguments. Foe example ifα, β ∈ ΓΛ1M, U,V ∈ ΓTM, f ∈ ΓΛ0M and S is a tensors of type(vector,form) then

S(α+ β,U) = S(α,U) + S(β,U) , S(fα,U) = f S(α,U)

and

S(α,U + V) = S(α,U) + S(α,V) , S(α, f U) = f S(α,U)

Tensors are described as multi-linear.137/165

Tensors as maps

For example

T = ydx⊗ dy⊗ ∂

∂y+ 4xdy⊗ dz⊗ ∂

∂z, U = z

∂

∂x+ sin(y)

∂

∂y+ ez ∂

∂z,

V =∂

∂x+ cos(y)

∂

∂y+ x

∂

∂zand α = dz + x3dx

[97]what is T(V,U, α)?

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Reinterpreting tensorsGiven a tensor of type which starts with form,

(vector,something,something,. . . )

We can view it as a map which takes the remaining arguments and gives avector.For example consider a tensor S of type (vector,form,form). Then we canview the map

Sm : ΓTM × ΓTM → ΓTM , Sm(V,W) = S(ea,V,W)Xa

This is precisely what happens with the metric dual. I.e. α = g(ea, α)Xa.The reverse is true, given a map with is ‘+’-linear and ‘f’-linear in all itsarguments and takes form and vector fields and gives a form or vector we canview this as a tensor.For example the curvature tensor

R(α,W,U,V) = α : (R(U,V)W)

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7 Connections7.1 Introduction7.2 Connection and Covariant Differentiation7.3 Leibniz and the Christoffel symbols7.4 Levi-Civita connection7.5 Geodesics and the force equation7.6 Curvature7.7 The geodesic deviation equation7.8 The connection and Killing vectors

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Differentiating vectors

Recall the definitions of the derivative of a function f : R→ R

f ′(x) = limε→0

( f (x + ε)− f (x)

ε

)Observe that we have to take the difference between the value of the functionf at two different points: x + ε and x.

This is OK for the function f since its values are simply real number and weknow how to add (and subtract) real numbers.

If we want to differentiate a vector field V ∈ ΓTM we must take the subtractvectors at different points.

However there is no general way of adding vectors at two different point.

This is due to the curvature (non flatness) of a general manifold.

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Curvature of the sphere

Consider transporting one of the “down” vectors towards the North pole.

The vectors remain on the sphere.

Observe that the resulting vector depends on the path taken by the vector.

Thus there is no general way of adding vectors at two different point.

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Requirements for differentiation

The differential of a scalar field f ∈ ΓΛ0M, in the direction of V is given byV〈f 〉.Therefore to define the derivative, we need two fields:

The field we want to differentiate.The vector field giving the direction we want to differentiate in.

As we mentioned before we cannot use the Lie derivative to define the4-acceleration of a point particle.

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Connection and covariant differentiationLet us define a connection∇UV where V ∈ ΓTM is the vector field we wantto differentiate and U ∈ ΓTM is the direction we want to differentiate in. Theresult∇UV ∈ ΓTM is also a vector field.That is∇ takes two vector fields and gives a third vector field.

∇ : ΓTM × ΓTM → ΓTM , (U,V) 7→ ∇UV

We say that∇UV is the covariant derivative of V in the direction U.Since we are differentiating V we want ∇UV to be ‘+’-linear in V

∇U(V + W) = ∇UV +∇UW

and Leibniz in V∇U(f V) = U〈f 〉 V + f∇UV

Since we are differentiating in the direction of U we want∇UV to be‘+’-linear in U

∇U+WV = ∇UV +∇WV

and ‘f ’-linear in U∇(fU)V = f∇UV

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Leibniz and the Christoffel symbolsFor scalar fields f ∈ ΓΛ0M, we define

∇Uf = U〈f 〉

Thus the Leibniz rule becomes

∇U(f V) =(∇U f

)V + f∇UV

Given a coordinate system (x1, . . . , xn) let

Γabc = dxa :

(∇∂b∂c

)These are called the Christoffel symbolsThese are sufficient information to define the connection since given twovector fields U = Ua∂a and V = Va∂a then [98]show

∇UV = U〈Va〉 ∂a + ΓabcUbVc∂a

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Levi-Civita connection

A connection is metric compatible if for all U,V,W ∈ ΓTM

U〈g(V,W)〉 = ∇U

(g(V,W)

)= g(∇UV,W) + g(V,∇UW)

A connection is torsion free if

∇UV −∇VU = [U,V]

[99]Show that torsion free implies

Γabc = Γa

cb

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Levi-Civita connection

Given a manifold M with a metric g, there is a unique connection on M whichis both metric compatible and torsion free.This is called the Levi-Civita connection. The Christoffel symbols are givenby [100](show)

Γabc = 1

2 gad(∂gdb

∂xc +∂gdc

∂xb −∂gbc

∂xd

)We will always use the Levi-Civita connection. But see worksheet.

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Vector fields along a curve

Given a curve C : R→ M then we can define a vector field along C.

This is the map V : R→ TM, such that V(τ) is a vector with base point C(τ).

For example the derivative C is a vector field along C.

If M is spacetime and C(τ) is the world-line of a particle, then C(τ) is the4-velocity of the particle.

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Vector fields along a curveIn general V(τ) with not be tangent to C.

For example the acceleration vector.If we have a family of curves, then V(τ) can represent the divergence ofneighbouring curves.

This is crucial when looking at the geodesic deviation, and measuring gravity.149/165

Differentiating along a curveGiven a curve C : R→ M and a vector field along C, V : R→ TM then wecan define the covariant derivative of V in the direction C.

∇CV

This is possible although V is only defined along the curve C.

To see this look at the formula in terms of Christoffel

∇CV = C〈Va〉 ∂a + ΓabcVbCc∂a

Now (ΓabcVbCc)|C(τ) only requires information about Va at C(τ) and V is

defined there.

And C〈Va〉 = dVa(τ)dτ which is well defined.

Thus

∇CV =dVa(τ)

dτ∂a + Γa

bcVbCc∂a

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Geodesics and the force equationThe 4-acceleration of a world line C : R→ M is given by ∇CC. This is also avector field along C.[101]Show that the acceleration is always orthogonal to the velocity.

g(∇CC, C) = 0

The geodesic equation is given by∇CC = 0

We can write the geodesic equation as [102](show)

d2Ca

dτ 2 + Γabc

dCb

dτdCc

dτ= 0

Given a force field W ∈ ΓTM, where g(W, C) = 0, the force equation for aparticle of mass m is given by

m∇CC = W

In Minkowski spacetime, with Cartesian coordinates this becomes [103](show)

d(γv)

dτ=

3∑i=1

W i∂i anddγdτ

= W0

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Length of a curves and geodesics.

Recall that in R2 the length of a curve C : R→ R2, C(τ) = (C1(τ),C2(τ))between τ0 ≤ τ ≤ τ1 is given by

Length =

∫ τ1

τ0

√(C1)2

+(C2)2dτ =

∫ τ1

τ0

√CaCbδab =

∫ τ1

τ0

√g(C, C)dτ

Assume we have a spacetime M with a metric g with Lorentzian signature. Ingeneral g depends on position.

For timelike curves g(C, C) < 0, therefore we set

T =

∫ τ1

τ0

√−g(C, C)dτ (11)

T represents the proper time between C(τ0) and C(τ1) of the particleundergoing the motion given by the curve C.

If we minimise the length of a curve we obtain the geodesic equation.[104](show: Handout) .

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CurvatureThe Curvature of a connection is given by the function that takes three vectorfields and gives a new vector field

R : ΓTM × ΓTM × ΓTM → ΓTM ,

R(X,Y)U = ∇X∇YU −∇Y∇XU −∇[X,Y]U

It is tensorial, i.e. ‘+’-linear and ‘f’-linear in all its arguments. [105](Show)If Minkowski spacetime or any flat manifold R = 0. [106](Show)Clearly

R(X,Y)U + R(Y,X)U = 0

Assuming∇ is metric compatible and torsion free, the curvature also satisfiesthe Bianchi identities

R(X,Y)Z + R(Z,X)Y + R(Y,Z)X = 0

g(R(X,Y)Z,W) + g(R(X,Y)W,Z) = 0

and

∇U(R(V,W)X

)+∇V

(R(W,U)X

)+∇W

(R(U,V)X

)= 0

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The geodesic deviation equation

The curvature of spacetime and hence gravity is measured by the GeodesicDeviation equation.

Let C : R→ M be a geodesic and V : R→ TM is a vector field along C,which represents the deviation of neighbouring geodesics to C then V satisfiesthe geodesic deviation equation.

∇C∇CV = R(C,V)C

This is the equation which allows us to physically test whether we are in agravitational field or not.

It involves observing you neighbouring particles. Does the distance betweenthen change.

Consider two neighbouring masses in a falling lift.

What is the difference between weight and gravity?

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The geodesic deviation equation

To demonstrate this let Cσ(τ) be a family of geodesics. We can view this as aribbon so that C : R2 → M, (τ, σ)→ C(τ, σ) = Cσ(τ).For f ∈ ΓΛ0M set

C(τ, σ)(f ) =∂f (C(τ, σ))

∂τand C′(τ, σ)(f ) =

∂f (C(τ, σ))

∂σ

[107]Now show

[C,C′] = 0

[108]and

∇C∇CC′ = R(C,C′)C

Setting σ = 0 and C′(τ, σ) = V(τ) we have the geodesics deviation equation.

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Covariant differentiating formsGiven a p-form α ∈ ΓΛqM and a vector field V ∈ ΓTM we can differentiate αin the direction of V to give a p-form∇Vα.The resulting form always has the same degree as original form.The rule with covariant differentiate is always use Leibniz.For 0-forms, we define∇V f = V〈f 〉.For 1-forms

∇V(α : U) = (∇Vα) : U + α :∇VU

For higher forms∇V(α ∧ β) = (∇Vα) ∧ β + α ∧ (∇Vβ)

[109]Show that ∇V(αadxa) = V〈αa〉 dxa − ΓabcαaVbdxc

[110]Show that the connection commutes with the metric dual and the hodgedual, is Leibniz with respect to internal contraction.

∇UV = ∇UV , ∇U(?α) = ?(∇Uα) , ∇U(iVα) = i(∇UV)α+ iV(∇Uα)

[111]Show thatdα = ea ∧∇Xaα

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Covariant derivatives of a tensor

The covariant derivative is Leibniz across a tensor product. That is for tensorsS,T the

∇V(S⊗ T) = (∇VS)⊗ T + S⊗ (∇VT)

If we view tensors as maps then we and rewrite this in terms the covariantderivative of its arguments. For example for a tensor field S of type(form,vector) [112](show)

(∇US)(V, α) = U〈S(V, α)〉 − S(∇UV, α)− S(V,∇Uα)

Hence∇Ug = 0 if an only if g is metric compatible, i.e. for all V,W ∈ ΓTM

U〈g(V,W)〉 = g(∇UV,W) + g(V,∇UW)

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The connection and Killing vectors

Recall that a vector field K ∈ ΓTM is Killing if LKg = 0, i.e.

LK(g(U,V)

)= g(LKU,V) + g(U,LKV)

[113]Show that this is equivalent to

g(∇UK,V) + g(∇VK,U) = 0

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Connection 1-forms (*)A connection may be expressed in terms of the Christoffel symbolsΓa

bc = dxa :∇∂a∂b.Alternatively if the orthonormal frame X1, . . . ,Xn and co-frame e1, . . . , en areused then we define the connection 1-forms

ωab ∈ ΓΛ1M , ωa

b : U = ea : (∇UXb)

In which case with U = UaXa, V = VaXa and α = αa ea

∇UV = U〈Va〉Xa + (ωab : U) Vb Xa

and ∇U(αaea) = U〈αa〉 ea − (ωab : U)αaeb

The ‘f’-linearity of∇ guarantees that ωab is a 1-form.

If ∇ is metric compatible [114]show that

ωab + ωba = 0

where ωab = ωcbηca.

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2 Black Holes


4 Forms


6 Tensors

7 Connections

8 Einstein’s equation and Conservation Laws8.1 The Einstein tensor8.2 The vacuum Einstein equations8.3 The stress-energy-momentum tensor

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The Einstein tensorThe Ricci tensor is a type (form,form) tensor and is given by

Ric(U,V) = ea : (R(U,Xa)V)

From the Bianchi identities this is symmetric

Ric(U,V) = Ric(V,U)

The Ricci scalar is given by

R = Ric(Xa,Xa)

where Xa = ηabXb.The Einstein tensor is a type (form,form) tensor is given by

Ein = Ric− 12Rg

The Einstein tensor is symmetric and divergenceless.

(∇XaEin)(Xa,U) = 0 for all U ∈ ΓTM

This follows directly from the Bianchi identities. I.e. from the symmetries ofthe curvature tensor R.

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The vacuum Einstein equations

The vacuum Einstein equations are

Ein = 0

[115]Show this is equivalent to Ric = 0.

How do you show Schwarzschild is a solution to the vacuum Einsteinequations?

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The stress-energy-momentum tensor

The source for gravity in the Einstein equations is thestress-energy-momentum tensor.

Ein = 8πT

This is a tensor field T ∈ Γ⊗

M of type (vector,vector).

It is symmetric

T(U,V) = T(V,U)

The (3+1) split of the stress-energy-momentum tensor gives the stress, energyand momentum.Given an orthonormal frame Xa with X0 timelike and representing the field ofobservers, then

T(X0,X0) is the energy.T(X0,Xµ) with µ = 1, 2, 3 is the 3-momentum.T(Xµ,Xν) with µ, ν = 1, 2, 3 the components of the stress tensor.

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The stress-energy-momentum tensor: ExamplesSome common stress-energy-momentum tensors:Dust is given by a vector field U ∈ ΓTM and the density the scalar field ρ

T = ρU ⊗ U

[116]Show that for a dust, U satisfies the auto-parallel equation.For a fluid with density ρ and pressure p

T = (ρ− p)U ⊗ U + pg

The stress-energy-momentum tensor for electromagnetism is

T = iXaF ⊗ iXaF − 12 ? (F ∧ ?F)g

[117]Show that this is equivalent to

τV =12

(iVF ∧ ?F − F ∧ iV ? F

)This is the stress-energy-momentum tensor that is used Einstein’s equation forthe Reissner-Nordström and Kerr-Neumann spacetimes

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Conservation laws

From Einstein’s equations the stress-energy-momentum tensor must also bedivergenceless, i.e.

(∇XaT)(Xa,U) = 0 for all U ∈ ΓTM

The fact that the stress-energy-momentum tensor is divergenceless is oftencall a conservation law. However in order to obtain a closed current we need aKilling vector.

If K is a Killing vector then [118]show

dτK = 0

where the stress 3-forms are given by

τU ∈ ΓΛ3M , (?τU) : V = T(V,U)

Hence if there is a Killing vector, then we have a conserved quantity.

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