Imperial College 4th Year Physics UG, 2013-14

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General RelativityLecture notes

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Toby Wiseman

October 3, 2013

1

Toby Wiseman; Huxley 507, email: t.wiseman@imperial.ac.uk

O�ce hour: 10-11 Monday, 10-11 Thursday

Rapid feedback class: 1-2 Tuesday with Andrew Hickling (please hand inwork by Monday lunchtime in the UG o�ce for marking)

During the course 8 example sheets will be handed out - one per week,starting week 3 (ie. Oct 22nd). Some questions will be for rapid feedback,some more advanced questions to develop understanding. By the conclusionof the course students will have the technical these example sheets, and theseare to be considered an essential part of the learning process.

Books

This course is not based directly on any one book. Recommended readingfor the course is;

• Sean Carroll, ”Spacetime and Geometry”(Pearson, Addison Wesley)

• James Hartle, ”Gravity: An introduction to Einstein’s General Rela-tivity”(Pearson, Addison Wesley)

• Ray D’Inverno, ”Introducing Einstein’s Relativity”(Oxford Univerity Press)

• Robert Wald ”General Relativity”(Univ. Chicago Press)

• Steven Weinberg, ”Gravitation and Cosmology”(Wiley)

I have largely tried to use the notation and conventions of Wald, which Iregard as an excellent book. In particular the book goes significantly beyondthis course, and so familiarity with it can be very useful long term. Also ofinterest in terms of Di↵erential Geometry;

• Nakahara, ”Geometry, Topology and Physics”(IOP)

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1 Classical geometry

1.1 Recap: 3-d Euclidean space

Consider 3-dimensional Euclidean space, E3, with its canonical Cartesian co-ordinates {x, y, z}.

The line element (metric) on this space is,

ds2 = dx2 + dy2 + dz2 (1)

Writing the coordinates as xi = {x, y, z} with i = 1, 2, 3, this defines themetric matrix,

ds2 = gij

dxidxj (2)

where,

gij

=

0

@1 0 00 1 00 0 1

1

A (3)

Then a vector v is a d-component object with components vi = {v1, v2, v3} ={vx, vy, vz} (in a slight bending of notation). We may compute the dot prod-uct of two vectors v and w by ‘contracting them with the metric’;

v ·w = gij

viwj = v1w1 + v2w2 + v3w3 or vxwx + vywy + vzwz (4)

For Euclidean space the norm of a vector, v, gives its length or magnitude,and is,

|v|2 = gij

vivj = vivi = (vx)2 + (vy)2 + (vz)2 (5)

although we note that for general geometries a vector doesn’t have a notionof length (although it does have a norm). We compute the angle betweentwo vectors as,

cos ✓ =v ·w|v||w| (6)

and note that the angle between two vectors retains its meaning in generalgeometries.

3

1.1.1 Spherical coordinates

We may also take spherical polar coordinates x0i0 = {r, ✓,�} on E3, where;

x = r cos ✓

y = r sin ✓ cos�

z = r sin ✓ sin� (7)

(8)

In these coordinates the same line element (metric) becomes,

ds2 = dr2 + r2d✓2 + r2 sin2 ✓d�2 (9)

with metric matrix,

ds2 = g0i

0j

0dxi

0dxj

0(10)

where,

g0i

0j

0 =

0

@1 0 00 r2 00 0 r2 sin2 ✓

1

A (11)

In di↵erent coordinates the values in gij

and g0i

0j

0 are di↵erent. However itis a very important point that this describes the same geometric quantityas before- i.e. the line element length. Quantities such as g

ij

that have co-ordinate indexes and transform from one coordinate system to another areknown as tensors. We shall define them carefully later.

Now a vector has components v0i0= {v0r, v0✓, v0�}. Like the metric, the

components of a vector in one coordinate system di↵er from those in an-other. Here the spherical components are related to the Cartesian compo-nents {vx, vy, vz} as,

0

@vx

vy

vz

1

A =

0

@cos ✓ �r sin ✓ 0

sin ✓ cos� r cos ✓ cos� �r sin ✓ sin�sin ✓ sin� r cos ✓ sin� r sin ✓ cos�

1

A ·

0

@vr

v✓

v�

1

A(12)

An important point is that; viwjgij

= v0i0w0j0g0

i

0j

0 - the length of a vectordoesn’t mind which coordinates one uses. (Ex. check this!)

4

1.1.2 Curves, tangents and straight lines

To state the very obvious, a straight line in E3 take a simple form in Cartesiancoordinates being parametrized as,

xi(�) = ai + �bi (13)

where the two vectors ai and bi have constant components (ie. independentof position xi) and give the o↵set and direction, and the variable � parame-terizes the line.

In general we may define a curve parametrically as xi = xi(�). The tangentvector v or ‘speed and direction’ to the curve is then the rate of change withrespect to the parameter, having components,

vi(�) ⌘ dxi(�)

d�(14)

For example; for the straight line above in Cartesian coordinates the tangentvector is constant with vi = bi at any point on the line.

An important point is that this is true in any coordinate system. It is anexample of a tensor equation. The values of the components of a tangentvector for a curve will di↵er depending on the coordinates - eg. Cartesian orSpherical. But the equation defining the tangent vector will be the same. Aswe will see all quantities in geometry can be phrased in terms of such tensorequations.

We say two curves that pass through the point xi

(0) are tangent if their tan-gent vectors at that point are a constant multiple of each other ie. they havethe same direction (although not necessarily the same speed).

The distance traversed along a section of a curve between parameter �0 and�1, which we denote s(�0,�1), is given by integrating the norm of the tangentvector,

s(�0,�1) =

Z�1

�0

d�

rgij

(x(�))dxi(�)

d�

dxj(�)

d�(15)

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1.2 Classical Geometry of surfaces

In order to describe interesting geometries we may take the classical approachof considering the geometry induced on a curved surface in E3. Suppose wedescribe a surface by the embedding,

z = f(x, y) (16)

for some function f at least over some range of the x, y coordinates. Thefunction f must encode the geometry of this curved surface.

For example, we may embed the upper hemisphere of a unit 2-sphere asz =

p1� (x2 + y2).

Now consider a curve that lies within this surface. Again we may para-metrically describe it by xi = xi(�), but we note it is constrained to obeyz(�) = f(x(�), y(�)), and hence,

dz

d�=

@f

@x

����y

dx(�)

d�+

@f

@y

����x

dy(�)

d�(17)

using the chain rule.

Thus the tangent vector to the curve is;

vi =

⇢dx

d�,dy

d�,@f

@x

dx(�)

d�+

@f

@y

dy(�)

d�

�(18)

1.2.1 Tangent space to a surface

Consider a tangent vector to a curve at some point. Since we have not chosenany particular curve through that point, then dx

d�

, dy

d�

are free for us to choose.Hence we may say that at any point in the surface, the tangent to a curvethrough that point takes the form,

vi =

⇢vx, vy,

@f

@xvx +

@f

@yvy�

(19)

Any vector that is tangent to a curve in this surface – a tangent vector – isof the form above.

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We say the set of all tangent vectors at a point on the surface is the tangentspace to the surface. This tangent space is a 2-dimensional ‘vector space’.

If one is given two tangent vectors to the surface at some point, ui and vi,these can be linearly combined to define a third vector wi as,

ui =

⇢ux, uy,

@f

@xux +

@f

@yuy

�

vi =

⇢vx, vy,

@f

@xvx +

@f

@yvy�

! wi =

⇢aux + bvx, auy + bvy,

@f

@x(aux + bvx) +

@f

@y(auy + bvy)

�(20)

where a, b are constants which we note is also tangent to the surface. Hencewe see that we may think of a vector in this tangent space of the surface asbeing simply given by the 2-dimensional vector {vx, vy} by the map,

{vx, vy} !⇢vx, vy,

@f

@xvx +

@f

@yvy�

(21)

and that linearly combining such 2-dimensional vectors {ux, uy} and {vx, vy}at the same point on the surface gives a third {aux + bvx, auy + bvy} tangentvector in the surface. Formally we say that the tangent space at a point isthen given by the vector space R2.

1.2.2 Induced metric on a surface

Now consider the dot product of two vectors in the surface. Denoting theseas 2-dimensional vectors {ux, uy} and {vx, vy} then their dot product in E3

is,

u · v = uxvx + uyvy +

✓@f

@xux +

@f

@yuy

◆✓@f

@xvx +

@f

@yvy◆

= uxvx 1 +

✓@f

@x

◆2!

+ uyvy 1 +

✓@f

@y

◆2!

+ (uxvy + uyvx)@f

@x

@f

@y

= (ux, uy) ·

0

@1 +

�@f

@x

�2@f

@x

@f

@y

@f

@x

@f

@y

1 +⇣

@f

@y

⌘2

1

A ·✓

vx

vy

◆(22)

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Thus thinking about two tangent vectors u,v at a point on the surface as 2-dimensional vectors ua, va, with a = 1, 2, then we learn that we may computetheir dot product,

u · v = gab

uavb (23)

where the 2⇥ 2 matrix gab

is the induced metric on the surface,

gab

=

0

@1 +

�@f

@x

�2@f

@x

@f

@y

@f

@x

@f

@y

1 +⇣

@f

@y

⌘2

1

A (24)

and is a symmetric matrix defined as a function over x, y.

Having a metric, it is natural to define its inverse. Let us define the inversematrix gab to g

ab

, so that,

gabgbc

= �ac

(25)

Comment: If the surface is embedded trivially as f =constant then the sur-

face is just a ‘flat’ plane and the induced metric is simply gab

=

✓1 00 1

◆, ie.

the 2-dimensional Euclidean metric. If the surface is a plane we may alwaysperform a rotation of the Euclidean space so that the embedding is simplyf =constant.

1.2.3 A surface is locally a plane

Consider a surface in E3 and choose any point on that surface. Without lossof generality we may translate the surface (so we don’t change its geometry)so that the coordinates of that point are x = y = 0. Perhap less obviousis that by appropriate rotation about the x and y axes we may generallyarrange that the embedding function for the surface has the property,

@f

@x=

@f

@y= 0 at x = y = 0 (26)

Let us focus on some region of the surface at this chosen point x = y = 0.We may Taylor expand the embedding function f in the neighbourhood of

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this region about x = y = 0 as,

f(x, y) = f(0, 0) +@f

@x

����0

x+@f

@y

����0

y (27)

+1

2

@2f

@2x

����0

x2 +1

2

@2f

@2y

����0

y2 +@2f

@x@y

����0

xy + . . .

= O�x2, y2, xy

�(28)

due to our chose of positioning and orientation of our surface, so that 0 =@f

@x

= @f

@y

at x = y = 0.

The implication of this is that the induced metric above for the surface nearthe chosen point x = y = 0 simply becomes,

gab

=

✓1 00 1

◆+O

�x2, y2, xy

�(29)

and thus we see the induced metric appears to be that of E2 up to quadraticcorrections in the displacement away from the chosen point.

Locally the quadratic deviation of the metric from being Euclidean describesthe fact that the surface is ‘curved’. It indicates that any quantity measuring‘curvature’ at a point should involve two derivative of the metric.

This is a rather formal way of saying that if you look at a curved surfacevery closely (relative to the curvature scale!) then it looks ‘flat’. This is acrucial observation in our development of GR later.

1.2.4 Geodesics of a surface

We define a geodesic to be a curve within the surface that has minimal length.In this sense it is the most direct or straightest path. We may parameterizethe curve as xa = xa(�) so that xi = {x1(�), x2(�), f(x1(�), x2(�))}.

We see from eqn (18) that the tangent vector at a point on the curve can beneatly written as the 2-vector va,

va =dxa(�)

d�=

⇢dx

d�,dy

d�

�(30)

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The length of the path is obtained by integrating the norm of the tangentvector which we have seen we may write as,

s =

Z�1

�0

d�pgab

vavb =

Z�1

�0

d�

rgab

(x)dxa(�)

d�

dxb(�)

d�(31)

Now an important observation is that under a reparameterization of the path,� = �(⌧), the path length is invariant. To check,

s =

Zd�

rgab

dxa(�)

d�

dxb(�)

d�

=

Zd⌧

d�

d⌧

rgab

dxa(⌧)

d⌧

d⌧

d�

dxb(⌧)

d⌧

d⌧

d�=

Zd⌧

rgab

dxa(⌧)

d⌧

dxb(⌧)

d⌧(32)

However, the length element,

L�

⌘ gab

(x)dxa(�)

d�

dxb(�)

d�

= gab

dxa(⌧)

d⌧

dxb(⌧)

d⌧

✓d⌧

d�

◆2

= L⌧

✓d⌧

d�

◆2

(33)

Thus we may always choose a parameterization of the path, by parameter�, such that L

�

= 1. Then the parameter � measures proper distance alongthe path.

Now we write the length of a path between two points x0 = x(�0) andx1 = x(�1) as,

s =

Z�1

�0

d�pL

�

(34)

and we vary this path length, keeping the points x0 and x1 fixed, so that�x(�) = 0 for � = �0 or �1. Then we find (in the sense of Euler-Lagrange),

�s =

Zd��

⇣pL

�

⌘=

Zd�

�L�

2pL

�

(35)

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and now we make the choice that our original path, before the variation, wasin a proper distance parameterization, so,

�s =1

2

Zd��L

�

(36)

Now we may evaluate this variation using Euler-Lagrange. Recall for anaction,

s =

Z�1

�0

d�L(xa,dxa

d�) (37)

then the variation �s = 0 for fixed end point data �xa(�0) = �xa(�1) = 0implies;

d

d�

@L

@(dxa

d�

)

!=

@L

@xa

(38)

Proof:

�s =

Z�1

�0

d��L(xi,dxi

d�)

=

Z�1

�0

d�

�xi

@L

@xi

+ �(dxi

d�)

@L

@(dxi

d�

)

!(39)

then recall �(dxi

d�

) = d

d�

(�xi), and so,

�s =

Z�1

�0

d�

�xi

@L

@xi

+@L

@(dxi

d�

)

d

d�(�xi)

!

=

Z�1

�0

d�

�xi

@L

@xi

� �xi

d

d�

@L

@(dxi

d�

)

!!+

"�xi

@L

@(dxi

d�

)

#�1

�0

=

Z�1

�0

d��xi

@L

@xi

� d

d�

@L

@(dxi

d�

)

!!(40)

Requiring that �s = 0 for all possible variations of path �xi then implies the‘E-L’ equations. End of proof.

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Now the equations we require come from varying the Lagrangian,

L =1

2L

�

=1

2gab

(x)dxa(�)

d�

dxb(�)

d�(41)

where we recall gab

is a function of xi (but not dx

a(�)d�

). Thus the E-L equationsgive,

d

d�

@L

@(dxc

d�

)

!=

@L

@xc

d

d�

✓gcb

dxb

d�

◆=

1

2

dxa

d�

dxb

d�

@gab

(x)

@xc

gcb

d2xb

d�2+

dxb

d�

d

d�gcb

(x) =1

2

dxa

d�

dxb

d�

@gab

(x)

@xc

gcb

d2xb

d�2+

dxb

d�

@gcb

(x)

@xa

dxa

d�=

1

2

dxa

d�

dxb

d�

@gab

@xc

(42)

Then ’raising’ the c index (ie. multiplying by the inverse matrix gdc andusing gdcg

cb

= �db

;

0 = gdcgcb

d2xb

d�2+ gdc

dxb

d�

@gcb

(x)

@xa

dxa

d�� gdc

1

2

dxa

d�

dxb

d�

@gab

@xc

= �db

d2xb

d�2+ gdc

dxa

d�

dxb

d�

✓@g

cb

(x)

@xa

� 1

2

@gab

@xc

◆

=d2xd

d�2+ gdc

dxa

d�

dxb

d�

✓@g

cb

(x)

@xa

� 1

2

@gab

@xc

◆(43)

and we finally conclude that a geodesic follows a path that obeys �s = 0 andhence the geodesic equation;

d2xd

d�2+ �d

ab

dxa

d�

dxb

d�= 0 (44)

where we have defined,

�d

ab

(x(�)) ⌘ 1

2gdc✓@g

cb

@xa

+@g

ca

@xb

� @gab

@xc

◆����x

a(�)

(45)

and recall that we picked a parameterization L(x(�)) = gab

dx

a

d�

dx

b

d�

���x

a(�)= 1.

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This is a set of coupled second order o.d.e.s in the parameter � which we maysolve for xa(�). In order to solve it we must therefore give both the initialposition xa and also the velocity dx

a

d�

at some starting value of �. Knowingthese one may then integrate the equation to deduce the curve which is givenparameterically by the solution xa(�).

An important comment here is that we must remember the solution is con-strained by the condition g

ab

dx

a

d�

dx

b

d�

= 1. This is a constraint also on ourinitial data. It is a non-trivial fact (that we will check later) that the solu-tion then preserved this condition, provided it is true for the initial data.

Comment: An important point should be noted here. Whilst we have de-rived the geodesic equation for a surface embedded in Euclidean space, wehave made no explicit reference to this embedding as we have left the metrica general matrix (rather than giving its explicit form in terms of the embed-ding function f).

Comment: Suppose our surface is a plane and so we embed it as f =constant.Then we saw before that the induced metric is just the 2-d Euclidean metricand then �c

ab

= 0, and hence,

d2xc

d�2= 0 (46)

which has general solution,

xc = ac + bc� (47)

for constant vectors ac and bc which give the starting point and direction ofthe line respectively.

Comment: Suppose that our surface is not a plane. Let us choose to considera geodesic starting at some point on the curve which we will take (w.l.o.g) tobe the point x = y = 0. As above arrange the surface so that the embeddinghas @f/@x = @f/@y = 0 at this point. Recall that then at this point themetric is just that of 2-d Euclidean space.

Now a general curve going through x = y = 0 with parameter � being zerothere can be written near this point as,

xc = bc�+ cc�2 +O(�3) (48)

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by simply Taylor expanding. The constants bc determine the tangent to thecurve. The constants cc determine the curvature of the curve at the point,with cc ⇠ the ‘inverse radii of curvatures’.

Now from above expression we see that near x = y = 0,

�c

ab

= O(x, y) (49)

and so a geodesic passing through this point will have solution,

xc = bc�+O(�3) (50)

where bc again gives the direction of the curve through the point.

Hence we see that ‘locally’ a surface is a plane and a geodesic in thatsurface is a ‘straight line’ having no curvature there. It does however deviatefrom a straight-line as O(�3). Geodesics generalize the notion of a straightline to a curved geometry, having the property that they are locally straight.

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2 ‘Modern’ (Riemannian) Geometry

In classical geometry we have examined a 2-d curved geometry by embeddingit into 3-d Euclidean space. However, in all the above we have seen thatthe notion of vectors, distances, geodesics and local flatness actually reduceto concepts that can be expressed in an ’intrinsically’ 2-d manner. Theessence of modern di↵erential geometry is to describe geometry from a purelyintrinsic perspective without every needing to resort to a higher dimensionalembedding space.

Our discussion here will try to avoid giving a formal development of dif-ferential geometry. Many of the subtleties are global ones and are not strictlyrelevant for a first course in GR.

The starting point of geometry is that any geometry locally looks like flatspace (as we have seen above for embedded curves). To that end the basicobject we employ is the space Rd (or at least subsets of it).

When we have an embedded surface, the notion of tangent vectors, met-rics that describe distance, and geodesics are clear from our understandingof Euclidean geometry. The first task is to understand what these objectsare from an intrinsic perspective. A pragmatic way to define tensors (ob-jects that include vectors and the metric) is via their transformation undera coordinate change.

2.1 Coordinate transformations and tensors fields onRd

The space Rd is topologically the same as d-dimensional Euclidean space. Anatural coordinate system onR is given by the coordinates xi = {x1, x2, . . . , xd}.There exist (infinitely many) other coordinates; for example the generaliza-tion of spherical coordinates.

The reason we are now not considering Euclidean space Ed, but rather Rd,is that we will not consider the geometry of the space to be the same (onlythe topology – and strictly speaking the di↵erential structure).

We will now consider a change of coordinates xi ! x0i0 so that x0 = x0(x) –for example one might have in mind the Cartesian and Spherical coordinates.

15

We shall restrict ourselves to ‘well behaved’ coordinate transformations -namely smooth and invertible (so that x = x(x0) exists).

Comment: We are not really allowed to transform from cartesians to sphericalcoordinates as the assumptions above break down on the axis of symmetry.

Under a change of coordinates; xi ! x0i0 = x0i0(x), we define a transformationmatrix, M, with components,

M i

0

j

⌘ @x0i0(x)

@xj

(51)

This is the Jacobian matrix for the transformations, with the Jacobian givenas J = detM.

As we know, a function transforms trivially under a coordinate transform,namely;

f(x) = f(x(x0)) eg. ; f = x2 = (x0 � 1)2 (52)

and so we may think of the function as a function of the new coordinatesx0 instead of x. We say a function is a scalar under coordinate transforma-tions. However there are other objects in geometry which do transform in aparticular way - the tensors.

A vector field vi(x) has the property that it transforms as,

vi0(x(x0)) = M i

0

j

(x)vj(x) (53)

Thus we see that even sitting at the same actual point the value of the vectorchanges.

At some point with coordinates x, the value of a vector field vi(x) is avector vi at that point. This is also called a tangent vector. (Sometimesalso referred to as a ’contra-variant vector’).

The set of all vectors at a point are the tangent vector space at the point.

16

Example: {vr, v✓} = {0, 1}, and a transform x = r cos ✓, y = r sin ✓. Then,

✓vx

vy

◆=

✓@x

@r

@x

@✓

@y

@r

@y

@✓

◆·✓

01

◆

=

✓cos ✓ �r sin ✓sin ✓ r cos ✓

◆·✓

01

◆=

✓�r sin ✓r cos ✓

◆=

✓�yx

◆(54)

Comment: A curve in the geometry is described parametrically as xi = xi(�).The tangent vector to a curve is defined as its velocity, vi = dxi(�)/d� , andtransforms correctly as a vector due to the chain rule:

vi0=

dx0i0(�)

d�=

dx0i0(x)

@xj

dxj(�)

d�= M i

0

j

dxj(�)

d�(55)

Fundamentally a vector at a point may be thought of as a tangent to curvethrough that point.

A covector field wi

has its index down and has the property that it trans-forms ’oppositely’ as,

wi

(x) = w0j

0(x(x0))M j

0

i

(x) (56)

or inverting this,

w0i

0 = wj

M j

i

0 (57)

where we have defined the components M j

i

0 to be those of the the inversematrix to M, so that,

M j

i

0 =�M�1

�j

i

0 (58)

and then,

M i

0

j

M j

k

0 = �i0

k

0 and M i

j

0M j

0

k

= �ik

(59)

where �ij

is the Kronecker delta, so that,

�ij

=

⇢1 i = j0 i 6= j

(60)

17

(ie. M ·M�1 = M�1 ·M = 1)

Due to our assumptions on the allowed coordinate transformations, the in-verse matrix M j

i

0 always exists.

At a point x the value of a covector field wi

(x) is a covector or cotangentvector w

i

at the point. (Also sometimes known as a covariant vector).

The set of all covectors at a point form the cotangent space.

Due to the property of partial derivatives,

M j

i

0 ⌘@xj(x)

@x0i0 (61)

so that we may check;

M i

0

j

M j

k

0 =@x0i0(x)

@xj

@xj(x)

@x0k0 =@x0i0

@x0k0 = �i0

k

0 (62)

Comment: given a function f(x) there is a natural covector, with componentsin the coordinates xi defined as,

wi

⌘ @f(x)

@xi

(63)

Again, due to the chain rule this transforms correctly.

wi

0 =@f(x0)

@x0i0 =@f(x)

@xj

@xj(x0)

@x0i0 =@f(x)

@xj

M j

i

0 (64)

Fundamentally covector at a point may be thought of as a di↵erentials offunction at that point.

Giving a vector or covector in one coordinate system this then defines it inany coordinate system.

18

A (q, r) tensor field has q ‘up’ indices and r ‘down’ indices, and in generalis defined by giving its components T i1i2...iq

j1j2...jrin some coordinate system xi.

Under a transform to new coordinates xi

0then the new components are,

T0i01i02...i0qj

01j

02...j

0r(x(x0)) = M

i

01i1. . .M

i

0q

iq

⇣T i1i2...iq

j1j2...jr(x)

⌘M j1

j

01. . .M jr

j

0r

(65)

At a point, the value of a tensor field gives a (q, r) tensor at the point.

The set of all (q, r) tensors at a point is the (q, r) tensor space at the point.

Comment: The metric gij

is a (0, 2) tensor. A vector is a (1, 0) tensor.

Comment: We may view a function as a (0, 0) tensor. We call this a scalarquantity as It transforms trivially; f(x) ! f(x0).

Comment: Note that the Kronecker delta �ij

is a (1, 1) tensor. It has thespecial property that it is invariant under a coordinate transform;

M i

0

i

�ij

M j

j

0 = M i

0

i

M i

j

0 = �i0

j

0 (66)

We say it is an invariant tensor.

2.2 Tensor operations

We may linearly combine two (q, r) tensors, A and B, to give a third (q, r)tensor C, defined in components in a coordinate system xi as,

C i1i2...iq

j1j2...jr⌘ f(x)Ai1i2...iq

j1j2...jr+ g(x)Bi1i2...iq

j1j2...jr(67)

for functions f, g. Note that the transformation of the components of A andB imply that the components of C correctly transform as a tensor.

Thus we see the tensor space at a point is a vector space with dimension dq+r.

We may take an outer product of a (q1, r1) tensor A and a (q2, r2) tensor Bto give a (q1 + q2, r1 + r2) tensor C defined by its components as,

Ci1...iq1j1...jq2a1...ar1b1...br2

⌘ Ai1...iq1a1...ar1

· Bj1...jq2b1...br2

(68)

19

Again we note that the components of C transform correctly as a tensor.

Finally we may contract the ath ’up’ and bth ’down’ index of a (q, r) tensorA to give a (q � 1, r � 1) tensor B, defined as,

Bi1...ia�1ia+1...iq

j1...jb�1jb+1...jr⌘ Ai1...ia�1k ia+1...iq

j1...jb�1k jb+1...jr(69)

Just as we saw Kronecker delta �ij

is a tensor, a similar calculation confirmsthis contraction yields a well defined tensor B.

Comment: Suppose we have vector fields vi and wi. We may view their dotproduct g

ij

viwj as a suitable outer product of the metric g, and these vec-tors, and then the contraction over all the indices. We are reduced to a (0, 0)tensor ie. a function. Thus this dot product is a scalar quantity.

Given a tensor we may symmetrize or antisymmetrise its upper or lowerindices. We use the notation;

T(ij) =1

2(T

ij

+ Tji

)

T[ij] =1

2(T

ij

� Tji

) (70)

With more indices we have,

T(i1i2...in) =1

n!(T

i1i2...in + sym) (71)

and likewise for antisymmetry.

Comment: Since these are linear operations, the resulting symmetrized orantisymmetrized objects are also tensors.

Important comment: We may not add tensors at di↵erent points. For ex-ample given vector fields Ai(x) and Bi(x) it is straightforward to check thatthe object,

Ci(x) = Ai(x) +Bi(x+ v) (72)

for some constant vi does not transform as a vector.

20

Is this surprising? We are used to adding vectors in Euclidean space that liveat di↵erent points. However, in our curved surface example above it would beless ‘natural’ to compare vectors at di↵erent points on the surface. In orderto compare vectors at di↵erent points we need a notion of parallel transportwhich allows us to take a vector at one position and move it somewhere elsepreserving its ‘direction’. For that we need a notion of geometry.

2.3 The metric tensor

The metric is defined to be a (0, 2) tensor gij

that is symmetric so thatgij

= gji

. At any point this is a symmetric matrix and hence has real eigen-values. The metric is required to have all its eigenvalues being strictly positiveat all points. A corollary of this is that det g

ij

> 0 everywhere, and hence gij

is invertable.

The inverse of the metric, gij, is a symmetric (2, 0) tensor, defined as theinverse of the metric at every point. This implies that in any coordinatesystem we must have,

gijgjk

= �ik

(73)

Recall this makes sense as the (1, 1) tensor �ij

is invariant - its componentsare indeed the same in any coordinate system.

The metric provides a notion of the norm of a vector vi as |v|2 = gij

vivj.

We measure the distance s along a curve by integrating the norm of thetangent vector as,

s =

Zd�

pgij

vivj (74)

where the tangent vector vi = dxi(�)/d�.

The metric also provides a notion of angle, ✓, between two vectors v, w at apoint,

|v||w| cos ✓ = viwjgij

(75)

21

Comment: The distance and angle measures above are independent of thechoice of coordinates. It is a coordinate invariant quantity. All measurablequantities in geometry must be coordinate invariant.

Comment: (At least In GR) The metric tensor completely specifies the ge-ometry. Two metrics that di↵er by a coordinate transformation describe thesame geometry.

Comment: In order to specify the geometry of our 2-surface embedded in E3

that we started with we see we should take R2 and endow it with the metrictensor,

gij

=

0

@1 +

�@f

@x

�2@f

@x

@f

@y

@f

@x

@f

@y

1 +⇣

@f

@y

⌘2

1

A (76)

or any coordinate transformation of this. However in general you cannot takean n-dimensional geometry and embed it into En+1.

2.4 Tensors and the metric

The metric gives an important new structure for tensors, namely the abilityto ‘raise’ and ‘lower’ indices. If we have a (q, r) tensor T i1...iq

j1...jr, then we may

define a (q � 1, r + 1) tensor, A, by lowering the nth index with the metricas,

Ai1...in�1in+1...iq�1

aj1...jr= g

a inTi1...iq

j1...jr(77)

Likewise we can define a new (q+1, r�1) tensor B, by raising the mth indexwith the metric as,

Bai1...iq�1

j1...jm�1jm+1...jr= ga jmT i1...iq

j1...jr(78)

New notation: We introduce the notation that we order both the upperand lower indices together. Then we can simply raise or lower an index.Writing the tensor above as T i1...iq

j1...jr then we may understand,

Ai1...in...iq�1

aj1...jr= T i1...in�1 in+1...iq

a j1...jr(79)

22

and

Bai1...iq�1

j1...jm�1jm+1...jr= T i1...iq a

j1...jm�1 jm+1...jr(80)

2.5 Geodesics and the Christo↵el symbol

Our discussion of geodesic curves follows exactly the same lines as that above.Such a curve between two points extremizes (minimises) its length. Thegeodesic equation is derived in the same manner as before – we note thatderivation was essentially intrinsic. Using the a�ne parameterization,

gab

dxa

d�

dxb

d�= 1 (81)

one finds the geodesic equation,

d2xc

d�2+ �c

ab

dxa

d�

dxb

d�= 0 (82)

where as before we have defined,

�c

ab

(x) ⌘ 1

2gcd

✓@g

db

@xa

+@g

da

@xb

� @gab

@xd

◆(83)

Note that a solution of this geodesic equation is always compatible with hav-ing g

ab

dx

a

d�

dx

b

d�

= 1.

The object �c

ab

is called the Christo↵el symbol and will play an importantrole later. However, for now we simply note that it is symmetric in itslower indices but is not a tensor as it does not transform correctly. Under atransformation we find;

�c

0

a

0b

0(x0(x)) = M c

0

c

�c

ab

Ma

a

0M b

b

0 �@2xc

0(x)

@xa@xb

Ma

a

0M b

b

0

= M c

0

c

�c

ab

Ma

a

0M b

b

0 +M c

0

c

@2xc(x0)

@x0a0@x0b0 (84)

where we explicitly see the trailing term violates the tensor transformationlaw. (See example sheet 2)

23

In fact rather than being a tensor, it is a connection, which is something thattransforms as the above. The Christo↵el symbol is known as the Levi-Civitaconnection.

The Christo↵el symbol has a symmetry in its lower 2 indices,

�c

ab

= �c

ba

(85)

One can also check that (see exercise sheet 3),

@gab

@xc

= gma

�m

bc

+ gmb

�m

ac

(86)

2.6 Riemann Normal coordinates

Consider a point at the origin xi = 0. (Note that by taking a coordinatetransform of the form x0i0 = xi + ai for some ai we can always map anyparticular point to the origin of coordinates).

Consider a coordinate transformation that locally is written near the originxi = 0 as a Taylor expansion,

xi

0= bi

0

j

xj +1

2bi

0

j

cjmn

xmxn + . . . (87)

where the bi0j

and ci0jk

are constant matrices (in the x coordinates). We

take the matrix bi0j

to be invertible with inverse bij

0 so that,

bi0

j

bjk

0 = �i0

k

0 , bij

0bj0

k

= �ik

(88)

Now it is required to be invertible so that the inverse coordinate transformexists. We take c to be symmetric in its lower indices, ci

jk

= cikj

= ci (jk)

as only this symmetric part contributes above.

Then the matrix M i

0i

takes the form,

M i

0

i

⌘ @x0i0(x)

@xj

= bi0

i

+ bi0

m

cmij

xj +O(x2) (89)

The inverse matrix M i

i

0 takes the form,

M i

i

0 = bii

0 � cimj

bmi

0xj +O(x2) (90)

24

to ensure that M i

i

0M i

0j

= �ij

+O(x2), M i

0i

M i

j

0 = �i0j

0 +O(x2).

Consider Taylor expanding the metric at the origin as,

gij

(x) = gij

|0 +@

@xk

gij

����0

xk +1

2

@2

@xk@xl

gij

����0

xkxl + . . . (91)

Then the metric transforms in the new coordinates to,

g0i

0j

0(x0(x)) = bii

0bjj

0 gij

|0 + bii

0bjj

0

✓@

@xm

gij

� gin

cnmj

� gnj

cnmi

◆����0

xm +O(x2)

Now there always exists an appropriate choice of matrix bii

0 such that,

bii

0bjj

0 gij

|0 = �i

0j

0 (92)

Furthermore we have previous determined that,

@gab

@xc

= gma

�m

bc

+ gmb

�m

ac

(93)

and we see we can set the O(x) term to zero by taking;

cmab

= �m

ab

|0 (94)

(which is consistent with the symmetry properties of c) so then,

g0i

0j

0(x0(x)) = �i

0j

0 +O(x2) (95)

As discussed in example sheet 3, this choice of cmab

is unique.

These coordinates are called Riemann normal coordinates. As earlier for thesurface we see that for a geometry we may pick any point and use a coor-dinate transform to render the metric the Euclidean metric up to quadraticorder. The quadratic deviations indicate the geometry is in fact curved, notthe Euclidean one. Thus we learn the important fact that all geometries are‘locally’ Euclidean. This will be crucial later.

25

[ Aside:Now there always exists an appropriate choice of matrix bi

i

0 such that,

bii

0bjj

0 gij

|0 = �i

0j

0 (96)

The matrix bii

0 is equal to,

b = O ·D ·O0 (97)

using matrix notation, where O is a rotation that diagonalizes the symmetricmatrix g at x = 0, so that,

OT · (g|0) ·O = ⇤ , ⇤ =

0

BBBBB@

�1 0 . . . 0

0 �2...

... �d�1 0

0 . . . 0 �d

1

CCCCCA(98)

and the matrix D is diagonal with,

D = ⇤�1/2 =

0

BBBBBB@

��1/21 0 . . . 0

0 ��1/22

...

... ��1/2d�1 0

0 . . . 0 ��1/2d

1

CCCCCCA(99)

and O0 is then an arbitrary rotation - ie. the matrices that leave �ij

invariant- which represents the freedom in choosing the bi

i

0 .

Note that the �i

are the eigenvalues of the metric matrix g and hence mustall be strictly positive, so there is not concern over taking the negative frac-tional power. (See example sheet 3 for more details). End of aside. ]

Comment: Consider taking the above Riemann normal coordinates so thatthe metric mapped from the origin point is locally Euclidean. Then theChristo↵el connection also vanishes at that point and behaves linearly aboutit,

�c

ab

= O(x) (100)

26

Hence a general geodesic passing through this point will have solution,

xc = bc�+O(�)3 (101)

and thus appears locally as a straight line - ie. no quadratic terms aboutx = 0.

2.7 Lie derivatives and symmetries

Given a vector field vi(x) we can define a flow, which is a set of curvesparameterized by a parameter � and their starting point x(0), which we writexi

(v)(�, x(0)), with the properties that,

vi(x(v)(�, x(0))) =dxi

(v)(�, x(0))

d�(102)

for any starting point x(0), and also,

xi

(0) = xi

(v)(0, x(0)) (103)

Note that a flow obeys;

xi

(1) = xi

(v)(�1, x(0)) , xi

(v)(�2, x(1)) = xi

(v)(�1 + �2, x(0)) (104)

The flow is a (dense) set of curves in space. For every point there is a uniquecurve in the flow going through it. We say that the vector field vi generatesthe flow xi

(v)(�, x0).

Suppose we have a function f(x). How does it vary as we move along a flowline? The answer is given by the Lie Derivative, defined on a function as,

Lie(v, f) ⌘ vi@f

@xi

(105)

Note that using the fact, vi = dxi

(v)/d�, then,

Lie(v, f) =@f

@xi

dxi

(v)

d�=

df(x(v)(�))

d�(106)

27

so the Lie derivative at a point p gives the rate of change of the function fat p along the flow curve going through p.

Suppose we have a function f(x) and a flow generated by vi(x) such thatLie(v, f) = df/d� = 0. We say that the flow is a symmetry of f , and thatvi(x) generates this symmetry.

Example: Consider E3 in spherical polar coordinates (r, ✓,�). Then the flowgenerated by vi = (0, 0, 1) is,

xi

(v) = (r0, ✓0,�0 + �) (107)

so it is a flow in the aximuthal direction. Suppose we have a function f =f(r, ✓) - so it is axisymmetric. Then,

Lie(v, f) =@f

@xi

dxi

(v)

d�=

@f

@�= 0 (108)

Thus vi = (0, 0, 1) generates the symmetry of this axisymmetric function.

Now suppose we want to do the same with a vector field wi(x), namely tocompute its rate of change along the flow generated by vi(x). We can usecoordinates adapted to the flow, using the flow curves to define our coor-dinates. Then we may (at least in some region) choose vi(x) = (1, 0, . . . , 0).In these coordinates we want,

Lie(v, w)i =@

@x1wi (109)

so the rate of change of wi in the direction of the flow (ie. in the x1 direction).

Let’s try to write down the tensor expression, valid in any coordinate system.We might try,

Lie(v, w)i = vj@

@xj

wi ? (110)

Whilst this obviously agrees for the particular coordinate system above, itdoes not transform correctly as a (1, 0) tensor. So this is wrong.

28

In fact the correct tensor expression is,

Lie(v, w)i ⌘ vj@

@xj

wi � wj

@

@xj

vi (111)

This fixes the problem with transformation (Ex to check!), and agrees withour expectations in the special adapted coordinate system where the secondterm vanishes.

Comments: It is linear in each argument and it obeys a ’product’ rule, for afunction f ,

Lie(v, fw)i = fLie(v, w)i + wiLie(v, f) (112)

BUT it isn’t quite like a derivative we are used to. Unlike the Lie derivativeof a function which at a point depends only on the direction of the flow vi

there, now for a vector field the Lie derivative depends on the local propertiesof the flow, so both vi and @vi/@dxj at the point.

In order to get a more usual derivative, we will need to introduce more struc-ture as we shall see later when we discuss the covariant derivative.

We may define a Lie derivative of a covector field !i

(x) along a flow vi(x).One finds that,

Lie(v,!)i

⌘ vj@

@xj

!i

+ !j

@

@xi

vj (113)

For a general (q, r) tensor the expression is;

Lie(v, T )i1i2...ipj1j2...jq

⌘ vk@k

T i1i2...ip

j1j2...jq

�T ki2...ip

j1j2...jq@k

vi1 � T i1k...ip

j1j2...jq@k

vi2 � . . .� T i1i2...k

j1j2...jq@k

vip

+T i1i2...ip

kj2jq@j1v

k + T i1i2...ip

j1k...jq@j2v

k + . . .+ T i1i2...ip

j1j2...k@jqv

k

Note the new notation: @i

= @/@xi.

29

These have the property we want, namely they are tensor expressions, andin a coordinate system adapted to the flow, so that vi = (1, 0, . . . , 0) then,

Lie(v, T )i1i2...ipj1j2...jq

=@

@x1T i1i2...ip

j1j2...jq(114)

We say that vi generates a symmetry of a tensor T i1i2...ip

j1j2...jqif Lie(v, T ) = 0.

In particular the Lie derivative of the metric gij

is;

Lie(v, g)ij

= vk@k

gij

+ gik

@j

vk + gjk

@i

vk (115)

A symmetry of the metric is called an isometry. If for some vector vi wehave Lie(v, g) = 0 we say vi generates an isometry of the metric. In fact thenvi is called a ‘Killing’ vector.

Ex. For a 2-sphere we have the metric ds2 = d✓2 + sin2 ✓d�2, and vi =(vr, v✓) = (0, 1) is an isometry of the sphere. There are others - see the ex-ample sheet!

It means that if you ’flow along in the � direction’ then the geometry remainsinvariant.

2.8 Christo↵el symbol and covariant derivatives

For convenience we now use the standard notation: @a

⌘ @

@x

a

Consider a function �(x) on our geometry. Suppose we are interested incomputing its gradient. The natural quantity to compute is, @

a

�. We haveearlier mentioned that this quantity indeed transforms as a (0, 1) tensor (seeEx Sheet 2).

However, suppose we try to do the same thing for a vector. We might try to

30

compute its gradient as, dv b

a

⌘ @a

vb. Let us see how this transforms,

dv b

0

a

0 =@vb

0

@xa

0 =@xa

@xa

0

@

@xa

✓vb@xb

0

@xb

◆

=@xa

@xa

0

@xb

0

@xb

@vb

@xa

+ vc@xa

@xa

0

@xb

0

@xa@xc

= Ma

a

0dv b

a

M b

0

b

+ v0c0Ma

a

0M c

c

0@xb

0

@xa@xc

(116)

Now the first term is precisely the transformation of a (1, 1) tensor, but thesecond term doesn’t behave nicely at all.

So our usual notion of derivative, simply using @µ

, doesn’t work as it doesn’tgive tensor (q, r + 1) quantities when you act on general (q, r) tensors. Inorder to proceed recall,

�b

0

a

0c

0 = M b

0

b

�b

ac

Ma

a

0M c

c

0 �Ma

a

0M c

c

0@2xb

0(x)

@xa@xc

(117)

so that,

v0c0�b

0

a

0c

0 = M b

0

b

(vc�b

ac

)Ma

a

0M c

c

0 � v0c0Ma

a

0M c

c

0@2xb

0(x)

@xa@xc

(118)

Hence we may define a new derivative, the covariant derivative ra

as,

ra

vb ⌘ @a

vb + �b

ac

vc (119)

and this does transform correctly as a tensor. The naughty second terms inthe transformation of dv b

0a

0 above and �c

0a

0b

0 cancel. The key point is thatneither @

a

vb nor this connection transform properly as tensors. However theircombination above exactly does (see example sheet 3).

What is this object ri

? It is the natural generalisation of the partial deriva-tive @

i

we use in Euclidean space to a curved geometry. This is best seen byrecalling that at any point p (take it to be at x = 0) we may choose RiemannNormal coordinates so that g

ij

= �ij

+ O(x2) and so �k

ij

= O(x). Thenlocally at this point,

ra

vb|0 = @a

vb|0 + �b

ac

vc|0 = @a

vb|0 (120)

31

so locally the covariant derivative is just the partial derivative.

A very useful fact is that in Riemann Normal coordinates if you see a @i

thento get a general tensor expression in general coordinates you just have toreplace @

i

! ri

.

This construction generalizes for a (q, r) tensor T i1...iqj1...jr ;

ra

T i1...iqj1...jr ⌘ @

a

T i1...iqj1...jr

+�i1ab

T bi2...iqj1...jr + �i2

ab

T i1bi3...iqj1...jr + . . .

��b

aj1T i1...iq

bj2...jr � �b

aj2T i1...iq

j1bj3...jr + . . . (121)

For convenience we define the covariant derivative on a function (a (0, 0)tensor) simply to be the partial derivative;

ri

f ⌘ @i

f (122)

In particular we find for a covector,

ra

wb

⌘ @a

wb

� �c

ab

wc

(123)

The covariant derivative has the usual ‘derivative’ properties. It obeys aproduct (Liebnitz) rule,

ra

�Ai1...

j1...Bm1...

n1...

�= Ai1...

j1...ra

Bm1...n1... + Ai1...

j1...ra

Bm1...n1... (124)

You can prove this either by direct calculation or by observing that in Rie-mann Normal coordinates at a point x = 0, then,

ra

�Ai1...

j1...Bm1...

n1...

�|0 = @0

�Ai1...

j1...Bm1...

n1...

�|0 = Ai1...

j1...@aBm1...

n1...|0 + Ai1...j1...@aB

m1...n1...|0

using the product/Leibnitz rule of partial derivatives. Now using the veryuseful fact above we know that going back to a general coordinate systemwe simply observe the answer is equation (124) when we take @

i

! ri

.

Properties of ra

:

Most importantly it vanishes on the metric;

ra

gbc

= @a

gbc

� �d

ab

gdc

� �d

ac

gbd

= 0 (125)

32

by the property of the Christo↵el symbol.

We say it is metric compatible.

Note: this is simple to see in RN coordinates (say about x = 0); then,

ra

gbc

|0 = @a

gbc

|0 = @a

�bc

|0 = 0 (126)

It also vanishes on the invariant tensor �ij

;

ra

�bc

= @a

�bc

+ �b

ad

�dc

� �d

ac

�bd

= �b

ad

�dc

� �d

ac

�bd

= �b

ac

� �b

ac

= 0 (127)

by the property of the Christo↵el symbol.

Again this is straightforward to derive in RN coordinates.

Likewise ra

gbc = 0 as we shall show. Recall gbcgcd

= �bd

so,

0 = ra

�gbcg

cd

�= g

cd

ra

gbc + gbcra

gcd

= gcd

ra

gbc (128)

and hence ra

gbc = 0.

Hence we may write ri = gijri

= ri

gij.

Comment: This covariant derivative is the unique derivative that is metriccompatible and built solely from the metric (without any other data such astorsion).

Laplacian: Consider the Laplacian on a function;

r2� = gijri

rj

� = riri

� = ri

ri� (129)

We may write this out as;

r2� = gijri

rj

� = gijri

(@j

�)

= gij�@i

@j

� �k

ij

@k

�� (130)

In fact one may also write this as,

r2� =1pg@i

�(pg)gij@

j

��

(131)

where g = det gij

. (See example sheet 4).

33

2.9 Parallel transport

The covariant derivative allows us to determine the notion of vectors beingparallel at di↵erent points.

Suppose we have a curve that joins two points x(�0) and x(�1), and thetangent to that curve is vi = dxi/d�. Then we say a vector field wi is paralleltransported along that curve if;

viri

wj

��x(�)

= 0 (132)

We say then that wi(�0) is parallel to wi(�1) with respect to the curve.

Note however this notion of parallel depends on the curve joining the twopoints. In general two curves going from x(�0) to x(�1) will not agree onwhat is parallel.

It is a special property of Euclidean space that the notion of parallel is inde-pendent of the curve. In fact as we see later, it is precisely the presence ofcurvature that means di↵erent curves give rise to di↵erent parallel transportfor di↵erent curves.

We may also think of the equation above as a way to evolve a vector fromone point, x(�0), to another at x(�1).

0 = viri

wj =dxi

d�

@wj

@xi

� �j

ik

viwk

=dwj(�)

d�� �j

ik

viwk

��x=x(�)

(133)

which we view as a first order o.d.e. in � which is integrated along the curvein �, using an initial condition, wi(�0), to obtain the value of wi(�1).

A more geometric way to understand a geodesic is a curve whose tangentvector is parallel along the curve - ie. as ‘straight as possible. For a geodesic,

viri

vj��x(�)

= 0 (134)

34

(see Ex Sheet 4). This is simply shown as;

0 = viri

vj = vi@i

vj + vi�j

ik

vk =dxi

d�

@vj

@xi

+ �j

ik

dxi

d�

dxk

d�

=dvj

d�+ �j

ik

dxi

d�

dxk

d�

=d2xj

d�2+ �j

ik

dxi

d�

dxk

d�(135)

35

3 Special Relativity

3.1 Brief review of Special Relativity

The basis of Special Relativity (SR) is the Lorentz transformation. Considertwo inertial observers with frames F, F 0 with spacetime coordinates t, x, y, zand t0, x0, y0, z0 respectively. Suppose frame F 0 is moving at a constant velocityv in the x-axis direction of frame F so that the origin spacetime point ofF and F 0 coincide ie. the point t0 = x0 = y0 = z0 = 0 is the same ast = x = y = z = 0. Then SR states that the spacetime coordinates of thetwo frames are related by a Lorentz transformation,

t0 = �⇣t� v

c2x⌘

, x0 = � (x� vt) , y0 = y , z0 = z (136)

where the gamma factor,

� =1q

1� v

2

c

2

(137)

and c is the speed of light, c = 108ms�1.

Special Relativity (SR) is based on the Minkowski space-time geometry. Itis essentially the natural extension of Euclidean space to include time. Webegin with R4 with coordinates xµ with µ = 0, 1, 2, 3 and xµ = {t, x, y, z}.Note the use of Greek indices to denote space-time coordinates. We continueto use xi = {x, y, z} with Roman indices i = 1, 2, 3 to represent only thespatial coordinates.

We now may immediately identify this Lorentz transformation as a specialtype of coordinate transform xµ ! x0µ0

= x0µ0(x) which can be written as;

0

BB@

t0

x0

y0

z0

1

CCA =

0

BB@

� �� v

c

2 0 0�� v � 0 00 0 1 00 0 0 1

1

CCA ·

0

BB@

txyz

1

CCA (138)

and similarly for boosts in the y and z directions.

36

The key point is that the Lorentz transformation preserves the length of aspace-time interval, so;

(�s)2 = �c2(�t)2 + (�x)2 + (�y)2 + (�z)2

= �c2(�t0)2 + (�x0)2 + (�y0)2 + (�z0)2 (139)

In more ‘grown-up’ language the Lorentz transformation leaves invariant theline element or spacetime interval,

ds2 = �c2dt2 + �ij

dxidxj = ⌘µ⌫

dxµdx⌫ (140)

which defines a (Lorentzian) metric,

⌘µ⌫

=

0

BB@

�c2 0 0 00 1 0 00 0 1 00 0 0 1

1

CCA (141)

Note this is not a Riemannian metric - it has a negative eigenvalue.

Relativistic units:

Recall from astrophysics/cosmology the familiar choice of measuring time inunits of years, and measuring distance in light years. In these (natural) unitsthe speed of light is simply c = 1 and becomes dimensionless.

Example: Let us choose units of minutes for time. It takes 8 minutes forlight to reach us from the sun so time T = 8 in these units. The sun is 8light minutes away so is a distance D = 8 in these units. Hence the speed oflight, c = D/T = 1.

From this point on we will work in analogous natural units where c = 1 andis dimensionless.

More conventionally one derives units of time from distance. Ex. 1 lightmeter ' 0.3⇥ 10�8sec

37

In these natural units

⌘µ⌫

=

0

BB@

�1 0 0 00 1 0 00 0 1 00 0 0 1

1

CCA (142)

so that, so that ⌘00 = �1, ⌘0i = ⌘i0 = 0 and ⌘

ij

= �ij

. This is the Minkowskimetric and gives the geometry the Minkowski spacetime.

The spacetime interval

Then we have that a space-time interval dxµ is;

• spacelike if ds2 > 0 ; ie. an infinitessimal displacement dxµ correspondsto a distance separation

p+ds2.

• timelike if ds2 < 0 ; ie. an infinitessimal displacement dxµ correspondsto a time separation

p�ds2.

• null or light like if ds2 = 0.

Note that this space-time interval is invariant under Lorentz transformationso all observers agree on which class an interval is.

Inertial observers: Recall that inertial observers just follow straight lines;

xµ(⌧) = aµ + vµ⌧ (143)

for constants aµ and vµ where ⌧ is the proper time measured by that ob-server, and vµ is their 4-velocity.

Since ⌧ is proper time, we must have ds2 = �d⌧ 2. Hence,

⌘µ⌫

vµv⌫ = ⌘µ⌫

dx

d⌧

µdx

d⌧

⌫

=

✓ds

d⌧

◆2

= �✓d⌧

d⌧

◆2

= �1 (144)

Hence the 4-velocity is timelike with unit norm, so we may write it as,

vµ =��, �vi

�(145)

38

where vi is the 3-velocity of the particle in the inertial frame of the coordi-nates xµ.

Accelerated observers follow a general curve xµ(⌧).

Then instantaneous 4-velocity is the derivative of space-time position of theobserver with respect to their proper time;

vµ =dxµ(⌧)

d⌧(146)

and as above is unit norm and timelike.

Rest frame: The rest frame of an internal observe is one where vµ =(1, 0, 0, 0) - remember for an inertial observer vµ is constant, independentof ⌧ . For an accelerated observer at some time ⌧ one may Lorentz transformto their instantaneous rest frame where vµ(⌧) = (1, 0, 0, 0) for a particularvalues of ⌧ .

Comment: The component v0 gives the time dilation of a clock carried bythe moving observer as measured by a static observer in the reference framexµ whose passes through the same space-time point.

4-momentum of the particle is pµ = mvµ with m the rest mass of the par-ticle. Recall that p0 is interpreted as the energy of the particle, E, measuredby a static observer in the reference frame xµ who meets the particle; pi arethe usual 3-momentum.

Comment: In the rest frame of the particle we have the famous pµ =(m, 0, 0, 0), so E = m (recall we have units so c = 1 - really E = mc2). Asmeasured in the reference frame xµ, a moving particle has energy E = �m,and, (putting back factors of ‘c’);

�mc2 =mc2q1� v

2

c

2

' mc2✓1 +

1

2

v2

c2+ . . .

◆= mc2 +

1

2mv2 + . . . (147)

so the energy as low speeds has a rest mass component and then the usualNewtonian kinetic energy.

39

Generally an observer at point p with 4-velocity vµ measures a particle with4-momentum pµ at point p to be;

E = �⌘µ⌫

vµp⌫ = �vµ

pµ (148)

In relativistic mechanics it is the 4-momentum that is conserved - hence bothenergy and momentum as usual.

The generalisation of Newton’s law is;

fµ =dpµ

d⌧(149)

for a 4-vector force fµ. Correspondingly the 4-vector acceleration is fµ =maµ. We see for an inertial observer there is no force as pµ is constant in ⌧ .

Comment: In the instantaneous rest frame of an accelerates particle, so thatvµ = (1, 0, 0, 0) the force 4-vector should take the form fµ = (0, f i), with f i

the usual Newton force 3-vector.

Light rays and photons

Recall that the light rays follow null straight lines;

xµ(�) = aµ + �bµ (150)

for bµ a null vector.

Photons have null 4-momentum pµ, and their energy is again given by E =�v

µ

pµ, or in the reference frame of the coordinates E = p0.

40

4 The geometry of Special Relativity

Let us for the moment ignore the fact that the Minkowski metric doesn’thave all positive eigenvalues, and try to treat it as the metric and see whatwe find. In fact we will see that many of the physical features of SR have asimply geometric description.

4.1 Lorentz transformations

The set of coordinate transformations that leave the Minkowski metric in-variant - ie. the symmetries of the Minkowski geometry - are the translationstogether with Lorentz transformations,

x0µ0= aµ

0+ bµ

0

µ

xµ (151)

with aµ0, bµ

0µ

constants and bµ0µ

2 O(1, 3) where O(1, 3) are the Lorentztransformations. These are a global transformation - all points are trans-formed in the same way.

The definition of an O(1, 3) matrix A is one that leaves invariant ⌘µ⌫

so,

⌘ = AT · ⌘ ·A (152)

Again the determinant is ±1.

In fact we will always concern ourselves only with the transformations pre-serving both spatial orientation and the direction of time - the proper or-thochronous Lorentz transformations. This is the group SO+(1, 3) and formsa subgroup of the Lorentz group O(1, 3). Note that detA = +1 for A 2SO+(1, 3).

[ We may write; O(1, 3) = SO+(1, 3) ⇥ {1, T, P, T · P}, where T it time re-versal (t, xi) ! (�t, xi), and P is parity reveral (t, xi) ! (t,�xi). ]

Now we see that Mµ

0µ

= bµ0µ

2 SO+(1, 3), and again if Mµ

0µ

2 SO+(1, 3)then its inverse is also, Mµ

µ

0 2 SO+(1, 3).

41

The metric ⌘µ⌫

transforms as,

⌘0µ

0⌫

0 = ⌘µ⌫

Mµ

µ

0M ⌫

⌫

0 (153)

and hence if Mµ

µ

0 2 SO+(1, 3) then we have ⌘µ⌫

is invariant;

⌘µ

0⌫

0 = ⌘µ⌫

Mµ

µ

0M ⌫

⌫

0 (154)

Since the above is symmetric in µ, ⌫ this constitutes 10 equations. TheSO+(1, 3) matrix has 16 components. The SO+(1, 3) Lorentz transforma-tion has 16� 10 = 6 parameters. There are 3 rotations and 3 boosts, as wediscuss later in more detail.

The Lorentz transformations together with translations in time and spacegive the Poincare transformations.

Comment: This is analogous to the Galilean group (the translations + rota-tions) that leave the Euclidean metric invariant (see Ex Sheet 4).

4.2 The geometry of Minkowski space-time

Consider the Minkowski geometry. In Minkowski coordinates this defines ametric,

gµ⌫

= ⌘µ⌫

=

0

BB@

�1 0 0 00 1 0 00 0 1 00 0 0 1

1

CCA (155)

However, in general the metric in other coordinates will look like;

gµ

0⌫

0 = Mµ

µ

0M ⌫

⌫

0⌘µ⌫

(156)

For example, in spherical coordinates (t, r, ✓,�) we have,

gµ⌫

=

0

BB@

�1 0 0 00 1 0 00 0 r2 00 0 0 r2 sin2 ✓

1

CCA (157)

42

An infinitesimal displacement dxµ in the coordinates is timelike/spacelike/nullis ds2 = g

µ⌫

dxµdx⌫ < 0, > 0 = 0 respectively. If timelike then the infinites-imal proper time is d⌧ =

p�ds2, and it space like the infinitesimal proper

length is ds.

4.2.1 Vectors in Minkowski geometry

A vector vµ at a point has norm |v|2 = gµ⌫

vµv⌫ and is timelike if |v|2 < 0,spacelike if |v|2 > 0 and null if |v|2 = 0.

Key point: Since the norm is coordinate invariant - ie. is a scalar - then allobservers using any coordinates must agree on whether a vector is timelike,spacelike or null.

In Minkowski coordinates the norm is just |v|2 = ⌘µ⌫

vµv⌫ = �(v0)2 + (vi)2.

In a coordinate system where the coordinate t = x0 is timelike, meaning, theinterval dxµ = (dt, 0, 0, 0) is timelike, then we say a timelike or null vector isfuture directed if v0 > 0. Otherwise it is past directed.

For coordinate transforms that preserve the direction of time (such as properLorentz) they also agree of future or past direction. We shall restrict to suchcoordinate transforms from now on.

As usual we define the product between vectors v, u as u ·v = gµ⌫

uµv⌫ . If thisvanishes we say the vectors are orthogonal. Note a null vector is orthogonalto itself!

4.2.2 Example: proper Lorentz transforms of vectors

Consider Minkowski coordinates, so gµ⌫

= ⌘µ⌫

. Then by a (proper) Lorentztransformation we may always find a frame so that a vector vµ that is...

• ... future timelike with norm |v|2 = �a2 has components vµ = (|a|, 0, 0, 0).This is its rest frame.

• ... spacelike with norm |v|2 = +b2 has components, vµ = (0, b, 0, 0).

43

• ... null has components vµ = (1, 1, 0, 0).

[ From a group theory point of view; the proper Lorentz transform preservesa vector’s norm, and for timelike/null vectors preserves future/past direct-edness, but otherwise has a transitive action. ]

4.2.3 Curves in Minkowski geometry

Consider a curve xµ(�). A curve is defined to be a timelike curve if the tan-gent vector vµ = dxµ/d� is everywhere timelike. Likewise we have space likeand null curves.

Comment: Note generally a curve will have indefinite character, where thetangent varies at di↵erent points between timelike, space like and null, sothese timelike, space like and null curves are rather special ones.

Spacelike curve

The proper length s along a curve is measured as,

s =

Zd�

s✓gµ⌫

dxµ

d�

dx⌫

d�

◆(158)

and is reparameterization invariant.

We may use the proper length s as a parameter, so xµ(s). This gives ana�ne parameterization as then,

+ 1 = gµ⌫

dxµ

ds

dx⌫

ds(159)

Timelike curve

The proper time ⌧ along a curve is measured as,

⌧ =

Zd�

s✓�g

µ⌫

dxµ

d�

dx⌫

d�

◆(160)

44

and is reparameterization invariant.

We may use the proper time ⌧ as a parameter for the trajectory, so xµ(⌧).This gives an a�ne parameterization as then,

� 1 = gµ⌫

dxµ

d⌧

dx⌫

d⌧(161)

Null curve

A null curve xµ(�) has gµ⌫

dx

µ

d�

dx

⌫

d�

= 0. Note that a massless curve is alwaysin an a�ne parameterization. Any reparameterization is also a�ne.

4.2.4 Geodesics

Recall geodesics are curves with tangents vµ = dxµ/d� such that in an a�neparameterization so, g

µ⌫

vµv⌫ = constant then,

vµrµ

v⌫ = 0 =) d2xµ

d�2+ �µ

↵�

dx↵

d�

dx�

d�= 0 (162)

Since |v|2 is constant we see a geodesic will either be timelike, space like ornull.

For a spacelike or timelike geodesic we may always choose the proper lengths or time ⌧ respectively as the a�ne parameter.

In Minkowski coordinates then gµ⌫

= ⌘µ⌫

so that �µ

↵�

= 0. Then thegeodesics simply obey for constant aµ, bµ,

xµ = aµ + bµ� , with ⌘µ⌫

bµb⌫ =

8<

:

�1 timelike , � = ⌧0 null+1 spacelike , � = s

(163)

Thus we see that a key physical point of SR is geometric;

Non-accelerated massive particles/observers follow timelike geodesics.

Non-accelerated massless particles follow null geodesics.

45

4.2.5 ‘Newton’s laws’

In general the relativistic ‘Newton’ laws for a massive particle with trajectoryxµ(⌧), for a�ne proper time ⌧ so vµ = dx

µ

d⌧

and |v|2 = �1, takes the form;

aµ = v↵r↵

vµ , fµ = maµ (164)

where m is the rest mass and the vectors aµ and fµ are the 4-accelerationand 4-force.

Note: If aµ = 0, then the particle follows a geodesic.

Check: In Minkowski coordinates;

aµ =dx↵

d⌧r

↵

vµ =dx↵

d⌧

@

@x↵

vµ =d2vµ

d⌧ 2(165)

as we saw earlier.

4.2.6 Isometries of Minkowski

We said earlier that the Poincare transformations are symmetries of theMinkowski metric. More precisely the Poincare transformations generatethe isometries of the geometry. In Minkowski coordinates they look like;

x0µ0= aµ

0+ bµ

0

µ

xµ (166)

with aµ0, bµ

0µ

constant and bµ0µ

2 SO+(1, 3). In other coordinate systemsthe Poincare transformations may look very complicated - for example inspherical coordinates.

We may write the three boosts and three rotations as follows;

Rotations: There are 3 rotations; about x, y and z;

M =

0

BB@

1 0 0 00 1 0 00 0 cos ✓1 sin ✓10 0 � sin ✓1 cos ✓2

1

CCA ,

0

BB@

1 0 0 00 cos ✓2 sin ✓2 00 � sin ✓2 cos ✓2 00 0 0 1

1

CCA ,

0

BB@

1 0 0 00 cos ✓3 0 sin ✓30 0 1 00 � sin ✓3 0 cos ✓3

1

CCA

46

and two more about x and y.

Boosts: There are 3 boosts; along x, y and z,

M =

0

BB@

cosh↵1 � sinh↵1 0 0� sinh↵1 cosh↵1 0 0

0 0 1 00 0 0 1

1

CCA ,

0

BB@

cosh↵2 0 � sinh↵2 00 1 0 0

� sinh↵2 0 cosh↵2 00 0 0 1

1

CCA ,

0

BB@

cosh↵3 0 0 � sinh↵3

0 1 0 00 0 1 0

� sinh↵3 0 0 cosh↵3

1

CCA

where cosh ✓ = � and sinh ✓ = � v.

[Hence this looks very close to a rotation - in fact it is a rotation if you ana-lytically continue time t to Euclidean time ⌧ as t = i⌧ and take ↵ = i✓ and✓ is a usual rotation.]

Any element of the Lorentz group can be written in terms of generators,matrices T(A) with A = 1, 2, . . . , 6, where,

M = eT(A)

✓

(A)= 1+

�T(A)✓(A)

�+

1

2!

�T(A)✓(A)

�2+ . . . (167)

and the six quantities ✓(A) parameterize the transformation, just like 3 anglesparameterize a rotation in 3-d Euclidean space.

For the Lorentz group we may write these generators conveniently as,

✓(A)T (A)µ⌫

= c↵���µ↵

⌘⌫�

� �µ�

⌘⌫↵

�(168)

for constants c↵�.The quantity c↵� is antisymmetric, so c↵� = c[↵�], and an antisymmetric 4⇥4matrix has 6 components. To agree with the previous matrices M above wehave;

c[↵�] =

0

BB@

0 �↵1 �↵2 �↵3

0 ✓2 ✓30 ✓1

0

1

CCA (169)

47

Using these generators we may simply write down the vector fields thatgenerate the isometries of Minkowski spacetime. In Minkowski coordinatesthese are,

vµ = aµ + ✓(A)T (A)µ�

x� = aµ + c↵���µ↵

⌘��

� �µ�

⌘�↵

�x� (170)

for the 4 constants aµ parameterizing translations, and the 6 constants c[↵�]

parameterizing Lorentz transforms.

Recall that isometries obey,

Lie(v, g)µ⌫

= v�@�

gµ⌫

+ gµ�

@⌫

v� + g⌫�

@µ

v� (171)

Then in Minkowski coordinates gµ⌫

= ⌘µ⌫

is constant and aµ and c↵� areconstant, so,

Lie(v, g)µ⌫

= ⌘µ�

@⌫

�c↵�

���↵

⌘��

� ���

⌘�↵

�x�

�+ ⌘

⌫�

@µ

�c↵�

���↵

⌘��

� ���

⌘�↵

�x�

�

= ⌘µ�

c↵����↵

⌘��

� ���

⌘�↵

�@⌫

(x�) + ⌘⌫�

c↵����↵

⌘��

� ���

⌘�↵

�@µ

(x�)

= c↵� (⌘µ↵

⌘��

� ⌘µ�

⌘�↵

) ��⌫

+ c↵� (⌘⌫↵

⌘��

� ⌘⌫�

⌘�↵

) ��µ

= c↵� (⌘µ↵

⌘⌫�

� ⌘µ�

⌘⌫↵

) + c↵� (⌘⌫↵

⌘µ�

� ⌘⌫�

⌘µ↵

)

= c↵� (⌘µ↵

⌘⌫�

� ⌘µ�

⌘⌫↵

)� c�↵ (⌘⌫↵

⌘µ�

� ⌘⌫�

⌘µ↵

)

= 0 (172)

which vanishes by symmetry.

48

5 Continuous matter in Minkowski spacetime

We have now understood how to think about point particle matter in Specialrelativity from a geometric point of view. Recall massive/massless

We will now think about writing down the physical laws that govern matter.We will start with electromagnetism, and then progress to fluids. An impor-tant concept we will need later is the stress tensor.

5.1 Physical laws as geometry

Since for gµ⌫

= ⌘µ⌫

then �µ

↵�

= 0 then in Minkowski coordinates a covariantderivative r

µ

= @µ

.

Note that physical laws must be tensor expressions or else they would dependon our choice of coordinates!

Given physical laws in Minkowski coordinates, we get their general form bybeing careful to write down tensor expressions and noting that generally@µ

! rµ

.

49

5.2 Electromagnetism

In Minkowski coordinates we write electromagnetism in terms of Fµ⌫

, theantisymmetric field strength where given the electric field Ei and magneticfield Bi then,

F 0i = Ei , F ij = ✏ijkBk

(173)

The Maxwell equations can be written as,

@µ

F µ⌫ = j⌫ , @[⌫F↵�] = 0 (174)

for a 4-charge current jµ where j0 is the charge density ⇢ in the frame xµ

and ji is the charge current.

Note that symmetric implies 0 = @µ

@⌫

F µ⌫ = @µ

jµ. This is the local conser-vation of charge equation;

@µ

jµ = 0 (175)

In these Minkowski coordinates we have,

@t

⇢ = �r · j (176)

Hence, the total charge in a volume V , Q =RdV ⇢ is constant;

d

dtQ = �

ZdV @

t

⇢ = �Z

dVr · j = �Z

dSn · j (177)

with n the outer unit normal to the surface element, so n · n = 1. We seethat if the current through the surface vanishes, the total charge is conserved.Hence the equation @

µ

jµ = 0 governs charge conservation. In general we seefrom the above equation that the rate of change of charge in a volume isgiven by the total current entering the volume.

Now in a general coordinate system for the Minkowski geometry, so thatgµ⌫

6= ⌘µ⌫

and rµ

6= @µ

as the tensor equations;

rµ

F µ⌫ = j⌫ , r[⌫F↵�] = 0 (178)

where Fµ⌫

is an antisymmetric tensor field, and jµ is a vector field whichobeys, r

µ

jµ = 0.

50

5.3 Newtonian equilibrium continuum mechanics

Let us first review the stress tensor in usual Newtonian mechanics.

In order to describe continuous matter, possibly deformed by some externalsurface forces, we must introduce the stress tensor. This is the symmetric(2, 0) tensor �ij that measures stresses within a material.

Consider slicing open a material at equilibrium along some surface. In orderto keep it unchanged we have to apply forces to the surface that were pro-vided by the internal forces we have removed.

Consider a surface element area �S with unit normal n. The force we mustprovide to match the internal force is �f . Then;

• �f is a normal stress if�f is parallel to n. The normal stress is definedas �

normal

⌘ |�f |�S

, and can be thought of as a ’pressure’.

• �f is a shear stress if �f is normal to n ie. tangent to the surface.The shear stress is defined as �

shear

⌘ |�f |�S

.

Cauchy: In continuum mechanics the fundamental assumption is that at anypoint in a material we may write the infinitessimal internal force vector dfacting on an infinitesimal surface with normal dS = n dS in terms of a (0, 2)stress tensor �

ij

(x) in Euclidean space;

df i = �ij(x) dSj

(179)

Conservation: Consider the internal forces acting over the surface of someclosed volume of material V . Assuming the total external force on the bodyvanishes and it is in equilibrium, then the sum of the internal forces acting ona closed subset must also vanish. Taking Cartesian coordinates, so g

ij

= �ij

and ri

= @i

then,

0 =

Z

S

df i =

Z

S

�ij(x) dSj

=

Z

V

@j

�ij(x) dV (180)

Now this must be true for any surface S, which implies,

@j

�ij(x) = 0 (181)

51

everywhere within the material.

Since �ij is a tensor, then the real expression in general coordinates (eg.spherical polar) in Euclidean space must be;

rj

�ij(x) = 0 (182)

Symmetry: Likewise the moments about any point, say xj

0, from the surfaceforces df i must also vanish for a closed surface. Hence,

0 =

Z

S

✏ijk

(xj � xj

0)dfk =

Z

S

✏ijk

(xj � xj

0)�km(x) dS

m

=

Z

V

@m

�✏ijk

(xj � xj

0)�km(x)

�dV

=

Z

V

�✏ijk

(@m

xj)�km(x) + (xj � xj

0)@m�km(x)

�dV

=

Z

V

✏ijk

�jm

�km(x) dV =

Z

V

✏ijk

�kj(x) dV (183)

where we have used the fact that @j

�ij(x) = 0. As this must be true for anyclosed surface,

✏ijk

�kj(x) = 0 =) �ij(x) = �ji(x) (184)

and hence the stress tensor is symmetric.

Note this symmetry means that �ij can be diagonalized by a coordinatetransformation, leading to the existence of the principle normal stresses.

Comment: The component �ij gives;

�ij = (force in i direction ) / ( area normal to j )

but using force = d (momentum) / d(time) then;

�ij = (momentum in i direction ) / ( time * area normal to j )= momentum flux in i direction through surface normal to j.

52

5.4 Relativistic continuum mechanics

Just as momentum and energy get combined into a 4-vector relativistic mo-mentum, this momentum flux also is combined with energy flux and energydensity to give a relativistic symmetric (2, 0) stress tensor T µ⌫ .

Definition of the stress tensor: At a point we choose the rest frame ofthe matter there (ie. its total 3-momentum is zero at that point) and then;

• T 00 = rest mass density of the matter

• T 0i = 0

• T ij = the usual stress tensor ie. momentum flux in direction of xi

through surface normal to xj.

Note that since T ij is symmetric, then T µ⌫ must be symmetric - if it is sym-metric in one frame, it must be in all frames.

In a general frame the stress tensor is;

• T 00 = energy density of the matter

• T 0i = energy flux through surface normal to xi

(or momentum density of the matter in the xi direction)

• T ij = momentum flux in direction of xi through surface normal to xj.

For an observer moving with 4-velocity vµ they observe the local 4-momentumof the matter, pµ, to be;

pµ = �v⌫

T µ⌫ (185)

In Minkowski space-time, taking the usual Minkowski coordinates xµ so thatgµ⌫

= ⌘µ⌫

we may define the total 4-momentum measured in the frame xµ ina subset V of material;

P µ =

ZdV T 0µ (186)

Hence,

d

dtP µ =

ZdV

d

dtT 0µ (187)

53

However, as the material moves, the rate of change of this momentum mustbe equal to the sum of the 4-forces acting on this volume - these are theinternal forces that act at the surface. Hence,

d

dtP µ =

ZdSniT iµ (188)

Equating these expressions and using usual Minkowski coordinates xµ so themetric is just ⌘

µ⌫

and rµ

= @µ

gives;

ZdV

d

dtT 0µ =

ZdSniT iµ =

ZdV @

i

T iµ (189)

using the divergence theorem, and hence,

0 =

ZdV

✓� d

dtT 0µ + @

i

T iµ

◆=

ZdV @

µ

T µ⌫ (190)

Now since this must be true for any volume we have,

@µ

T µ⌫ = 0 (191)

everywhere in the material.

Comment: For a state/equilibrium situation this reproduces the previousequilibrium condition @

i

T ij = 0.

We have used Minkowski coordinates. In general coordinates in the Minkowskigeometry the answer is;

rµ

T µ⌫ = 0 (192)

54

5.5 Stress tensor for electromagnetism

The stress tensor for electromagnetism is,

Tµ⌫

= Fµ↵

F ↵

⌫

� 1

4gµ⌫

F↵�

F ↵� (193)

Note that indeedrµ

T µ⌫ = 0 provided the EM equations hold (which requirescurrent conservation r

µ

jµ = 0).

In Minkowski coordinates you should find that,

T 00 ⇠ E2 +B2 , T 0i ⇠ E⇥B (194)

giving the energy density and momentum density (the Poynting vector!).

55

5.6 Perfect fluid matter

A perfect fluid is matter that is described simply by a local density, ⇢, and4-velocity, uµ (so that u2 = �1). The pressure in the fluid is isotropic - nopreferred direction - and hence is described only by a scalar function P . Thispressure is determined just in terms of the density ⇢ by the equation of state,so that P = P (⇢).Examples:

• Cold matter matter (‘Dust’) fluid has no pressure, so P = 0.

• Hot gaseous matter has pressure, so P = 13⇢.

• Radiation fluid (eg. the gas of photons) has pressure P = 14⇢.

The stress tensor is then determined as,

T µ⌫ = (⇢+ P )uµu⌫ + P⌘µ⌫ (195)

In general, uµ = (�, �vi) where the 3-velocity vi is a function of position inthe fluid. However at a point in the fluid we may move to the instantaneousrest frame so uµ = (1, 0, 0, 0) and then,

T tt = ⇢ , T ti = T it = 0 , T ij = P �ij (196)

so that T tt gives the local rest mass density, and T ij the stress (which is justnormal for a perfect fluid).

The conservation equation then determines the evolution - or equation ofmotion - of the fluid when combined with the equation of state. (See exam-ple sheet 5 for the details!)

Let us work with Minkowski coordinates so rµ

= @µ

.

Let us project our evolution equation into the direction uµ, recalling thatu2 = �1. Then,

uµ

@⌫

T µ⌫ = 0 =) u⌫@⌫

⇢+ (⇢+ P ) @⌫

u⌫ = 0 (197)

Let us now project onto an orthogonal direction nµ to the motion uµ, so thatuµn

µ

= 0. Note that nµ must be space like. Then,

nµ

@⌫

T µ⌫ = 0 =) nµ

((⇢+ P )u⌫@⌫

uµ + ⌘µ⌫@µ

P ) = 0 (198)

56

and you can show that this second equation is equivalent to;

(⇢+ P )u⌫@⌫

uµ + (⌘µ⌫ + uµu⌫) @µ

P = 0 (199)

These reduce to the usual non-relativistic Navier-Stokes equations if we canfind a coordinate frame where the fluid everywhere obeys; vi ⌧ 1 and P ⌧ ⇢.Then one finds (See example sheet 5):

@t

⇢+ @i

�⇢ vi

�= 0 , ⇢

�@t

vi + vj@j

vi�+ @iP = 0 (200)

where we recall @t

⇠ vi@i

for a fluid.

Note that for general coordinates in the Minkowski geometry we would re-place all @

µ

! rµ

.

57

6 Curved spacetime and a geometric origin

to gravity

A space-time or Lorentzian metric is a symmetric (0, 2) tensor gµ⌫

. At everypoint it is a symmetric matrix and hence has real eigenvalues. A Lorentzianmetric must everywhere have non-vanishing eigenvalues (so it is invertible)with 1 being negative and 3 positive. We say the signature is (1, 3).

Ex. An example is obviously the Minkowski metric.

Then this Lorentzian metric gµ⌫

defines the geometry of space-time in exactlythe same way a Riemannian metric g

ij

defines the geometry of space.

Our previous discussion of Riemannian geometry works in the same way forspace-time/Lorentzian geometry.

Two di↵erences with Riemannian geometry;

• Geodesics xµ(�) with a�ne � no longer minimise length; instead theystill extremize,

L =

Zd�g

µ⌫

dxµ

d�

dx⌫

d�(201)

[ One can always make a timelike curve between two points with arbi-trarily small proper time - make it move near the speed of light! ]

• The geometry is locally Minkowski (rather than locally Euclidean as inthe Riemannian case).

The second point is of key physical significance. We may choose the analog ofRiemann normal coordinates - local inertial coordinates or the local inertialframe (LIF) - so that, at some point, say x = 0,

gµ⌫

(x) = ⌘µ⌫

+O(x2) , �↵

µ⌫

= O(x) (202)

Are these coordinates unique? Almost, but given LIF coordinates at a point,we may construct other LIF coordinates by a (global) Lorentz transformation,

x0µ0= ⇤µ

0

µ

xµ (203)

58

for a constant matrix ⇤ 2 SO+(1, 3). This is the only freedom in specifyingthe LIF coordinates.

[ The proof of this follows exactly the same method as for the Riemann nor-mal coordinates in Riemannian geometry, worked out in Qu. 3 of Ex Sheet3. ]

An important consequence of this is that the tangent space at any pointagain divides into timelike (|v|2 = g

µ⌫

vµv⌫ < 0), space like (|v|2 > 0) andnull vectors (|v|2 = 0) . Again curves can be timelike, space like, null.

A timelike geodesic, parameterized by proper time ⌧ , passing through thepoint x = 0 locally takes the form,

xµ(⌧) = vµ⌧ +O(⌧ 3) (204)

where, ⌘µ⌫

vµv⌫ = �1 so we may write,

vµ = (�, �vi) (205)

as in SR. Note the absence of the quadratic terms indicate that these curvesare as ’straight’ as possible locally. Since there is a residual Lorentz invari-ance in the coordinates, for any given timelike geodesic - and hence inertialobserver - we may choose LIF coordinates such that at the point x = 0, weare in the instantaneous rest frame of the observer.

Thus we arrive at the key point in our discussion:An inhabitant of space-time locally sees the geometry to be Minkowski space-time.

Hence we may say: The physics of SR is built into the local geometry of any space-time.

Following from this all the geometry/physics of SR naturally generalises toa general curved space-time. We simply write Lorentz compatible laws in ageometric way, as tensor equations, taking,

⌘µ⌫

! gµ⌫

@µ

! rµ

(206)

and these are the laws for any space-time metric gµ⌫

, not just coordinatetransforms of Minkowski space-time.

59

Ex: An accelerated massive particle follows a timelike trajectory, aµ =v⌫r

⌫

vµ for 4-acceleration aµ. Note that in the LIF at a point on the trajec-tory,

aµ = v⌫r⌫

vµ = v⌫@⌫

vµ =dx⌫

d⌧@⌫

vµ =dvµ

d⌧(207)

Ex: Massless particles follow null geodesics.

Ex: The stress energy tensor is conserved as rµ

T µ⌫ = 0.

The scale at which the quadratic corrections to gµ⌫

= ⌘µ⌫

+ O(x2) becomeimportant is the curvature scale as we see later. Below this curvature scalethe geometry is Minkowski and the physics of SR is recovered.

60

6.1 Newtonian spacetime, gravity redshift and lightbending

This section and the following ones refer to calculations performed in ExSheet 6. Please review that question and its solutions.

Consider Minkowski space-time, in Minkowski coordinates xµ = (t, x, y, z),so g

µ⌫

= ⌘µ⌫

, and deform it a little in a specific way;

ds2 = gµ⌫

dxµdx⌫ , with gµ⌫

= ⌘µ⌫

� 2 ✏� �µ⌫

+O(✏3/2) (208)

where ✏ ⌧ 1 controls this small deformation.

The time t and spatial xi define the Newtonian frame.

Taking ✏ ! 0 will be the Newtonian limit, and we identify � with the New-tonian gravitational potential � as;

� = ✏� (209)

so,

�ij@i

@j

� = r2� = 4⇡GN

⇢ (210)

At the moment we just identify ✏� with the Newtonian potential. Later wewill use the Einstein equations to show the above equation holds.

We will think of ✏ simply as an accounting tool. We compute quantities onlyto lowest (non-trivial) order in ✏ as ✏ ! 0 and keep only that lowest term.We may make the separation of � into ✏� as,

✏ = max |�| =) �1 � +1 (211)

and requiring ✏ ⌧ 1 implies that the matter with density ⇢ spread over acharacteristic length scale R should obey,

GN

⇢R2

c2⌧ 1 (212)

(reinstating the factors of c). So this approximation is good for low densitiesspread over suitably large areas.

61

We will take the gravitating matter density ⇢ and hence � to be static so@t

� = 0 - (or at least very slowly varying).

Non-accelerated (or inertial) massive particles follow timelike geodesics. Inthe Newtonian limit, ✏ ! 0, the velocity of the particles must be small. Thuswe write the 4-velocity;

dxµ

d⌧=

�1 + ✏ f + . . . , ✏1/2 vi + . . .

�(213)

and likewise we take the acceleration to be small too,

d2xµ

d⌧ 2=

✓✏df

d⌧+ . . . , ✏ ai + . . .

◆(214)

Then in the Newtonian frame defined by xi both the velocity and accelerationare small in the Newtonian limit, ✏ ! 0, going as;

dxi

d⌧' ✏1/2 vi

d2xi

d⌧ 2' ✏ ai (215)

By calculating and working to lowest order in ✏ you find the inverse metricgµ⌫ and Christo↵el symbol are;

gµ⌫ = ⌘µ⌫ + 2✏� �µ⌫ + . . .

�i

tt

= �t

it

= �t

ti

= ✏ (@i

�) + . . .

�k

ij

= ✏ ((@k

�) �ij

� (@i

�) �jk

� (@j

�) �ik

) + . . . (216)

with other components, �t

tt

= �i

jt

= �t

ij

= 0.

The condition |dxµ

d⌧

|2 = �1 then yields,

f =1

2�ij

vivj � � (217)

in the Newtonian limit, and thus the time dilation of the particle relative tothe Newtonian frame is;

dt

d⌧= 1 + ✏

✓1

2�ij

vivj � �

◆+ . . . (218)

62

Thus we see now only the usual SR time dilation of a moving particle due toits speed, where we recall,

� ' 1 +1

2�ij

vivj + . . . (219)

but we see also that there is a gravitational time dilation due to the potential�.

If the particle is not accelerating then it follows a timelike geodesic;

d2x↵

d⌧ 2+ �↵

µ⌫

dxµ

d⌧

dx⌫

d⌧= 0 (220)

The time component of this yields in the Newtonian limit;

0 =d

d⌧

✓1

2�ij

vivj + �

◆+ . . . (221)

ie. conservation of the usual Newtonian energy, the sum of the KE andpotential energy �. The spatial components give,

ai = �(@i

�) + . . . (222)

or equivalently,

d2xi

d⌧ 2= �(@

i

�) + . . . (223)

recovering the usual Newton force law for the gravitational potential � toleading order.

We conclude that for this space-time, slow moving (wrt the Newtonian frame)inertial observers ‘feel’ a force of Newtonian gravity, and obey the usual New-tonian dynamics, of Newton’s theory with gravity potential �.

Hence: Curved space-time can account for the force of gravity!

[ Comment: We still need a theory that tells us why to pick this space-time. ]

63

6.2 Gravitational redshift

We have seen that a massive inertial particle incurs a time dilation from thegravitational potential �. Consider two particles sitting at fixed positionsxi

(1) and xi

(2) with proper times ⌧(1) and ⌧(2) respectively - hence they mustbe accelerated, eg. by a rocket. Then their 4-velocities (from above) will be

vµ(1,2) =dxµ

(1,2)

d⌧=

�1� ✏�(x(1,2)), 0, 0, 0

�(224)

so that |v|2 = �1, and hence,

dt

d⌧(1,2)' 1� ✏�(x(1,2)) (225)

Hence we have,

d⌧(1)d⌧(2)

' 1 + ✏�(x(1))� �(x(2))

c2(226)

where I am being careful to include the necessary factors of c2 if one doesnot use natural units.

Thus if particle one sends out radio pulses at a frequency !(1), then particletwo will observe the frequency to be !(2) such that,

!(2)

!(1)'

✓1 + ✏

�(x(1))� �(x(2))

c2

◆(227)

We see that if particle two is further from the gravity source, then �(x(1))��(x(2)) < 0, and the observed frequency is less than the emitted frequency.We call this gravitational red shift.

Pound-Rebka:

This e↵ect was precisely measured in Harvard in 1959 using a 20m verticalseparation of a gamma ray source (57Fe) and absorber. Due to the gravityredshift the gamma rays reaching the source were redshifted, e↵ecting theirabsorption. By vibrating the source (on a loudspeaker), a doppler shift wasadded so that now some gamma rays had the correct frequency and wereabsorbed. Thus the frequency shift due to redshift was deduced.

64

6.3 Light Bending

Light follows null geodesics in the Newtonian space-time. In the Ex Sheet6 you calculate directly by solving null geodesics the deflection that a lightray experiences as it moves past a gravity point source, o↵set by distance R.One finds for a source at xi = (0,�R, 0), with potential,

� = � GN

Mpx2 + (y +R)2 + z2

(228)

then a light ray moving initially along the x-axis follows a curve,

xµ(�) = (�,�, 0, 0) + ✏hµ(�) (229)

where hµ(�) ! 0 as � ! �1.

Solving the null geodesic equation to lowest order in ✏ determines hµ(�). Onefinds that asymptotically,

x(�) ! � , y(�) ! �✏4G

N

M

R� (230)

and defining the deflection angle �✓ by,

✏�✓ = lim�!1

tan�1

✓y(�)

x(�)

◆(231)

gives,

�✓ ' �4GN

M

c2Rradians (232)

where I have reintroduced c, if one is not working in natural units.

Working in units where c = 1, then,

GN

=6.7⇥ 10�11kg�1m3s�2

c2= 7.4⇥ 10�28kg�1m (233)

and one can calculate for the sun, radius R ⇠ 7.0 ⇥ 108m, mass M = 2.0 ⇥1030kg that the deflection angle for a ray grazing the surface is,

�✓ ' 8.5⇥ 10�6 radians = 1.8 arcsec (234)

65

This was measured in May 1919 by Eddington, by observing stars positionednear the surface of the sun in the sky which were visible due to a solar eclipse.Their positions were distorted by the light bending and Eddington and histeam claimed to measure this e↵ect.

66

6.4 Conservation laws and symmetries in curved spacetime

Consider Minkowski space-time in Minkowski coordinates. Recall a timelikeinertial particle follows the trajectory such that,

dvµ

d⌧= 0 =) pµ = mvµ = const (235)

Its 4-momentum ie. its energy and 3-momentum are constant.

However, such non-local conservation laws depend strongly on the space-timehaving isometries. In general they do not apply, and hence we only have thelocal laws.

Isometries

Recall that if a vector field vµ generates a symmetry or isometry of the metricthen,

0 = Lie(v, g)↵�

= vµ@µ

g↵�

+ g↵µ

@�

vµ + gµ�

@↵

vµ (236)

Recall from Ex. Sheet 4 that for the metric, this Lie derivative can actuallybe written in terms of the covariant derivative as;

Lie(v, g)↵�

= 2r(↵v�) (237)

A vector field generating an isometry is called a Killing vector, and satisfies,

r(↵v�) = 0 (238)

Energy conservation

Suppose we have a Killing vector field vµ which is everywhere timelike and aninertial massive particle following a timelike geodesic with a�ne parameter⌧ and tangent uµ = dxµ/d⌧ so that v2 = �1. Then as you showed in ExSheet 4;

d

d⌧(uµv

µ

) = u⌫@⌫

(uµvµ

) = u⌫r⌫

(uµvµ

)

= vµ

u⌫r⌫

uµ + uµu⌫r⌫

vµ

= 0 (239)

67

the first term vanishing as uµ is tangent to an a�nely parameterized geodesic,and the second term vanishing for the Killing vector vµ we have, r(µv⌫) = 0which implies r

µ

v⌫

= �r⌫

vµ

. Hence rµ

v⌫

is antisymmetric in its indices,but above is contracted with the symmetric uµu⌫ .

Suppose the particle has rest mass m, and thus 4-momentum pµ = muµ.Then,

d

d⌧(pµv

µ

) = 0 (240)

Hence we see that existence of the Killing vector implies that the momentumprojected in the Killing direction is conserved.

Recall that a timelike observer with 4-velocity vµ measures the particle’s en-ergy to be E = �pµv

µ

. Thus observers (not necessarily inertial ones) movingwith tangent vµ all measure the same energy for the particle as it passesthem. Thus energy of the particle is conserved as measured by these ob-servers.

In general, timelike isometries give rise to energy conservation laws, and spacelike isometries give rise to (angular) momentum conservation laws.

Example: In the Newtonian space-time, co-ordinates xµ = (t, x, y, z), withstatic potential, @

t

�, then the vector vµ = (1, 0, 0, 0) is a Killing vector.Hence we saw the Newtonian energy E = 1

2�ijvivj + � was conserved.

Example: In Newtonian space-time take a point source potential. This has aspherical symmetry, but not translations. In coordinates xµ = (t, r, ✓,�) withthe source at r = 0, then for example vµ = (0, 0, 0, 1) is an isometry. Thecorresponding conserved quantity is angular momentum. With no transla-tions, linear momentum is not conserved.

68

7 Curvature

We have seen that any geometry is locally Euclidean/Minkowski. The quan-tity that characterises the deviation away from this simple geometry is thecurvature, or more precisely the Riemann curvature tensor.

7.1 The Riemann tensor

Suppose we have a function f . Then the commutator of two covariant deriva-tives vanishes,

[r↵

,r�

]f ⌘ r↵

@�

f �r�

@↵

f

=�@↵

@�

f � �µ

↵�

@µ

f��

�@�

@↵

f � �µ

�↵

@µ

f�= 0 (241)

However for a covector field vµ

such a commutator does not generally vanish;in fact it can be written as,

[r↵

,r�

]vµ

= R ⌫

↵�µ

v⌫

(242)

where R µ

↵�⌫

is the Riemann (1, 3) tensor field.

We may explicitly evaluate the Riemann tensor in terms of the Christo↵elsymbol using;

r↵

r�

vµ

= @↵

(r�

vµ

)� �⌫

↵�

(r⌫

vµ

)� �⌫

↵µ

(r�

v⌫

)

= @↵

�@�

vµ

� ��

�µ

v�

�� �⌫

↵�

(r⌫

vµ

)� �⌫

↵µ

�@�

v⌫

� ��

�⌫

v�

�

= ��@↵

��

�µ

�v�

+ �⌫

↵µ

��

�⌫

v�

+@↵

@�

vµ

� �⌫

↵�

(r⌫

vµ

)� �⌫

�µ

@↵

v⌫

� �⌫

↵µ

@�

v⌫

= ��@↵

��

�µ

�v�

+ �⌫

↵µ

��

�⌫

v�

+ . . . (243)

where the . . . terms are symmetric in ↵ $ �.

Taking the commutator projects out these symmetric terms, and we find,

[r↵

,r�

]vµ

= (r↵

r�

�r�

r↵

) vµ

= R �

↵�µ

v�

(244)

with,

R �

↵�µ

= @�

��

↵µ

� @↵

��

�µ

+ �⌫

↵µ

��

�⌫

� �⌫

�µ

��

↵⌫

(245)

69

As we shall see, the Riemann tensor characterises the curvature of the geom-etry - ie. how it deviates from Euclidean/Minkowski space-time.

The commutator [r↵

,r�

] on other tensors is also determined by Riemann.

Using the fact that (vµ

wµ) is a scalar, we find using the Liebnitz rule,

0 = [r↵

,r�

](vµ

wµ) = vµ

[r↵

,r�

]wµ + wµ[r↵

,r�

]vµ

= vµ

[r↵

,r�

]wµ + wµR ⌫

↵�µ

v⌫

(246)

and as this is true for any vµ

and wµ, this implies, for a vector field,

[r↵

,r�

]wµ = �R µ

↵�⌫

w⌫ (247)

Then repeating this logic for the scalar (uµv⌫Aµ⌫

) one concludes (Exercise);

[r↵

,r�

]Aµ⌫

= R �

↵�µ

A�⌫

+R �

↵�⌫

Aµ�

(248)

[ Comment: Generally one finds for a tensor T µ1...µr⌫1...⌫r that,

[r↵

,r�

]T µ1...µr⌫1...⌫r = R �

↵�⌫1T µ1...µr

�⌫2...⌫r +R �

↵�⌫2T µ1...µr

⌫1�⌫3...⌫r + . . .

�R µ1↵��

T �µ2...µr⌫1...⌫r �R µ2

↵��

T µ1�µ2...µr⌫1...⌫r

]

70

7.2 Riemann in LIF coordinates:

Now recall that in LIF coordinates we have,

gµ⌫

(x) = ⌘µ⌫

+O(x2) , �↵

µ⌫

= O(x) (249)

Thus we note that at x = 0 we have �↵

µ⌫

= 0, but we note that @�

�↵

µ⌫

6= 0.

Thus in LIF coordinates at x = 0 we have;

R �

↵�µ

��x=0

= @�

��

↵µ

� @↵

��

�µ

��x=0

(250)

Comment: For Minkowski or Euclidean space where the metric is constantand �µ

↵�

= 0 everywhere then the Riemann tensor vanishes.

Generally we think of LIF coordinates as giving,

gµ⌫

(x) = ⌘µ⌫

+ Tµ⌫↵�

x↵x� +O(x3) (251)

for some constants Tµ⌫↵�

= T(µ⌫)(↵�). However, in fact one may find coordi-nates such that,

gµ⌫

(x) = ⌘µ⌫

� 1

3R

µ↵⌫�

|x=0 x

↵x� +O(x3) (252)

which in fact fixes the coordinates entirely, up to the overall Lorentz trans-formation. Hence we see the Riemann tensor is the obstruction to the metricbeing simply Minkowski.

71

7.3 Symmetries of the Riemann tensor

Firstly, by definition it has the antisymmetry property; R ⌫

↵�µ

= �R ⌫

�↵µ

Secondly, by direct calculation one can show; R ⌫

[↵�µ] = 0

[ Start with the fact;

R ⌫

[↵�µ] v⌫ = r[↵r�

vµ] (253)

Then using,

r[�vµ] = @[�vµ] � �↵

[�µ]v↵ = @[�vµ] (254)

one then finds,

r[↵r�

vµ] = r[↵@�vµ] = @[↵@�vµ] � @

⌫

v[µ�⌫

↵�] � �⌫

[↵µ@�]v⌫ = 0 (255)

and since this is true for any v we have, R ⌫

[↵�µ] = 0. ]

Thirdly, since r is metric compatible; R↵�µ⌫

= �R↵�⌫µ

[ Since r↵

gµ⌫

= 0, then,

0 = [r↵

,r�

]gµ⌫

= R �

↵�µ

g�⌫

+R �

↵�⌫

gµ�

= R↵�µ⌫

+R↵�⌫µ

(256)

as claimed. ]

A consequence of the above symmetries is that; R↵�µ⌫

= +Rµ⌫↵�

.

72

7.4 The Ricci tensor and scalar and the Einstein tensor

The natural contraction of the Riemann tensor gives the (0, 2) Ricci tensor ;

Rµ⌫

= R ↵

µ↵⌫

(257)

Due to the Riemann symmetry: Rµ↵⌫�

= R⌫�µ↵

we see the Ricci tensor issymmetric;

Rµ⌫

= R⌫µ

(258)

Taking a further trace we obtain the Ricci scalar ;

R = R µ

µ

(259)

Let us define the symmetric (0, 2) Einstein tensor in terms of the Ricci tensorand scalar;

Gµ⌫

⌘ Rµ⌫

� 1

2gµ⌫

R (260)

This has precisely the same content as Rµ⌫

(see Ex Sheet 7, Qu 1), but wehave reorganised its trace a little. For a 4-dimensional spacetime,

Rµ⌫

⌘ Gµ⌫

� 1

2gµ⌫

G (261)

where G = Gµ⌫

gµ⌫ .

73

7.5 The Bianchi identity

The Bianchi identity for the Riemann tensor is;

r[µR⇢

⌫↵]� = 0 (262)

or equivalently;

rµ

R ⇢

⌫↵�

+r⌫

R ⇢

↵µ�

+r↵

R ⇢

µ⌫�

= 0 (263)

This is straightforwardly shown in LIF coordinates at a point x = 0;

r⌫

R �

↵�µ

|x=0 = @

⌫

�@�

��

↵µ

� @↵

��

�µ

+ �⌫

↵µ

��

�⌫

� �⌫

�µ

��

↵⌫

���x=0

=�@⌫

@�

��

↵µ

� @⌫

@↵

��

�µ

���x=0

(264)

since �↵

µ⌫

|x=0 = 0, and so,

r[⌫R�

↵�]µ |x=0 =

�@[⌫@��

�

↵]µ � @[⌫@↵��

�]µ

���x=0

= 0 (265)

since @[↵@�] = 0 as partial derivatives commute.

74

Contracting the Bianchi identity:

Contracting the above Bianchi identity as,

r[µRµ

⌫↵]� = 0 (266)

yields;

0 = rµ

R µ

⌫↵�

+r⌫

R µ

↵µ�

+r↵

R µ

µ⌫�

= rµ

R µ

⌫↵�

+r⌫

R↵�

�r↵

R⌫�

(267)

Now a second contraction gives;

0 = rµ

R ⌫µ

⌫↵

+r⌫

R ⌫

↵

�r↵

R ⌫

⌫

= 2rµ

R µ

↵

�r↵

R ⌫

⌫

(268)

This identity;

rµ

R µ

↵

� 1

2r

↵

R = 0 (269)

is physically of key importance.

The Einstein tensor Gµ⌫

⌘ Rµ⌫

� 12gµ⌫R has the crucial property that it is

conserved ;

rµ

Gµ⌫ = rµ

✓Rµ⌫ � 1

2gµ⌫R

◆

= rµ

Rµ⌫ � 1

2r

⌫

R = 0 (270)

from the contracted Bianchi identity.

75

7.6 Riemann tensor and parallel transport

The commutator [rµ

,r⌫

] can be thought of as arising when one paralleltransports a vector around a closed curve.

Consider the square in the x, y plane, with corners at (x, y) = (0, 0), (0, ✏), (✏, ✏), (✏, 0).Then consider the two curves,

C(1) : (0, 0) ! (0, ✏) ! (✏, ✏)

C(2) : (0, 0) ! (✏, 0) ! (✏, ✏) (271)

Consider a vector vµ at (0, 0) which is parallel transported along the twocurves C(1,2) to yield two vectors vµ(1,2) at (✏, ✏). The di↵erence vµ(2) � vµ(1)characterises the dependence of parallel transport on the path taken. Calcu-lation (see Ex Sheet 7) shows;

vµ(2) � vµ(1) = ✏2 R µ

xy⌫

v⌫��(0,0)

+O(✏3)

Hence we see that the Riemann curvature is the obstruction to parallel trans-port being path independent.

Comment: In Euclidean or Minkowski space-time where R �

µ⌫↵

= 0, thenparallel transport is path independent as we are familiar with.

76

8 The Einstein’s equation

Recall that deforming Minkowski space to the Newtonian spacetime,

gµ⌫

= ⌘µ⌫

� 2✏�(x)�µ⌫

+O(✏2) (272)

for coordinates xµ = (t, x, y, z) yields Newton’s law of gravity for slow movingobjects in the Newtonian limit (governed by ✏ ! 0), where the Newtongravity potential � obeys the usual,

�ij@i

@j

� = 4⇡GN

⇢ (273)

for mass density ⇢. It also predicts gravitational red shift and light bend-ing which are indeed observed. However, why should we pick this particulardeformation of Minkowski? We will now discuss Einstein’s equation thatgoverns how the spacetime geometry depends on the matter present in it.

Recall that the stress energy tensor Tµ⌫

gives a characterization of the matterin a spacetime. An important point is that it is conserved r

µ

T µ⌫ = 0.Consider the stress tensor for a perfect fluid;

Tµ⌫

= (⇢+ P )uµ

u⌫

+ Pgµ⌫

(274)

where ⇢ and P are energy density and pressure, and uµ gives the local 4-velocity of the fluid, so uµu

µ

= �1.

Consider Minkowski spacetime with xµ = (t, x, y, z). Recall from Ex Sheet5 that the Newtonian fluid equations are recovered in the limit of small 3-velocity vi ⌧ 1 and P ⌧ ⇢ (ie. the dominant contribution to the local energycomes from the local rest mass energy). Then the stress tensor to leadingorder is simply,

Ttt

' ⇢ , Tti

' 0 , Tij

' 0 (275)

Now any equation governing the geometry of spacetime must recover in theNewtonian case the equation �ij@

i

@j

� = 4⇡GN

⇢ above.

Note that two derivatives of the geometry are locally related to the energydensity. This strongly suggests we should try to equate local curvature with

77

the stress tensor.

A first attempt at the Einstein equation:

We see that the stress tensor Tµ⌫

provides a natural tensor to characterize thelocal matter content, giving just ⇢ in the Newtonian limit. In 1915 Einsteinmade a first attempt;

Rµ⌫

= Tµ⌫

(276)

for some constant . This certainly has the correct flavour. Recall that theRicci tensor involves two derivatives of the metric as we require.

However there is a serious problem with this equation. Note that since Tµ⌫

is conserved we obtain,

rµRµ⌫

= 0 (277)

But by the Bianchi identity,

rµRµ⌫

=1

2r

⌫

R (278)

and hence this implies r⌫

R = 0 and so R =constant. Taking the trace ofthe equation,

R = T (279)

where T = Tµ⌫

gµ⌫ . Thus we see this Einstein equation implies the trace ofTµ⌫

=constant. In the Newtonian case above, T ' ⇢, but this is far toorestrictive.

Hence the key problem is reconciling the stress energy conservation with theBianchi identity.

The Einstein equation:

Einstein realised that stress energy conservation was a crucial ingredientthat must be consistent with the Einstein equation, rather than an additionalrestriction on it. The key observation is that the Einstein tensor;

Gµ⌫

⌘ Rµ⌫

� 1

2gµ⌫

R (280)

78

is conserved, so that,

rµGµ⌫

= 0 (281)

This is the unique conserved (0, 2) tensor involving two derivatives of themetric. Hence we conclude the Einstein equation must take the form;

Gµ⌫

= Tµ⌫

(282)

[ Comment: It may seem mysterious that the Einstein curvature tensor obeysprecisely a conservation equation. In fact the origin of the stress energy andEinstein tensor conservation is exactly the same, essentially due to the ge-ometric nature of the theory, namely that tensor equations are coordinateinvariant. ]

Now consider the Newtonian spacetime. In Ex Sheet 7, Question 4, youcomputed that (to leading order in ✏, which we then ignore);

Gtt

= 2 �ab@a

@b

�

Gti

= Git

= Gij

= 0 (283)

Further we have seen that in the Newtonian limit, the stress tensor onMinkowski spacetime takes the form,

Ttt

= ⇢

Tti

= Tit

= Tij

= 0 (284)

To leading order in ✏ we will find the same result in the Newtonian spacetime(since to leading order it is simply Minkowski). To our relief, the ti and ijcomponents of this Einstein equation are consistent, and the non-trivial ttcomponent yields;

2 �ab@a

@b

� = ⇢ (285)

Recall in Newton’s theory we expect �ab@a

@b

� = 4⇡GN

⇢. Hence we concludethat = 8⇡G

N

, and hence we finally arrive that the celebrated Einsteinequation (putting c back in);

Gµ⌫

= 8⇡GN

Tµ⌫

(286)

which links the local matter to the local curvature.

79

Comment on units:

Choosing general rather than natural units so c = 1 we would obtain;

Gµ⌫

=8⇡G

N

c4Tµ⌫

(287)

Recall that,

[GN

] = Mass�1 Length3 Time�2 (288)

In units where c = 1 we derive time units from length (or vice versa).

We may go further and choose units such that GN

= 1 or 8⇡GN

= 1. Forinstance, we may derive mass and time units from length. Such units arecommon in GR.

Indeed we may go only one step further (a step too far?) to also fix thequantum constant ~ = 1, where we recall,

[~] = Mass Length2 Time�1 (289)

For instance we first derive mass and time from length using GN

= 1, andthen choose length such that h = 1. These are called Planck units.

In SI units; GN

= 6.7⇥ 10�11kg�1m3s�2, h = 6.6⇥ 10�34kgm2s�1

In c = 1 units; GN

= 7.4⇥ 10�28kg�1m, h = 2.2⇥ 10�42kgm

In c = 1, 8⇡GN

= 1 units; ~ = 6.8⇥ 10�69m2

Hence we see 1m = 2.0⇥ 10�34 Planck lengths.

The Planck length units is the natural length/time units are the scales atwhich we expect quantum gravity to become important.

80

More remarks about Einstein’s equation:

An important point is that the Ricci tensor is a contraction of Riemann,but does not contain all the information of Riemann. Hence the Einsteinequation does not totally fix the space-time geometry.

An important example of this is the case of the vacuum Einstein equation,where T

µ⌫

= 0. Note that;

Tµ⌫

= 0 =) Gµ⌫

= 0 =) Rµ⌫

= 0 (290)

Hence space-time in vacuum must the Ricci flat. However, that does notmean that Riemann must vanish - as we shall see, the Schwarzschild blackhole is an example of this.

Furthermore to any solution of the Einstein equation one can add gravitywaves - ripples in space-time - which encode the freedom in Riemann notconstrained by the Ricci contraction.

It is also worth noting that these Einstein equations together with the mat-ter can be thought of as a dynamical system, where one gives initial data- an initial 3-spatial geometry + its momentum, together with the matterand its momentum - and then evolves in a time direction to build up the fullspacetime.

We have seen that Newtonian gravity arises from these Einstein equations.However, one can ask what other solutions they contain. We will spendthe remainder of our time considering two crucial solutions - the cosmo-logical solution (the FLRW cosmology) and the black hole space-time (theSchwarzschild metric).

81

9 The FLRW space-time and cosmology

The cosmological principle states;

There is no preferred point or direction in the universe (once we havesmoothed out small scale features)

Geometrically we take this to require spatial homogeneity (no point is special)and isotropy (no direction is special). The unique homogeneous isotropicspace-time geometry is the Friedmann-Lemaitre-Robertson-Walker space-time with metric;

ds2 = �dt2 + a(t)2hij

(x)dxidxj (291)

where xµ = (t, x, y, z), t is a time coordinate and hij

(x) is the metric on ahomogeneous isotropic 3 dimensional spatial geometry. There are 3 possibil-ities; flat space, the round 3-sphere, and 3-hyperbolic space;

hij

(x)dxidxj =

8<

:

dr2 + r2d⌦2

dr2 + sin2 rd⌦2

dr2 + sinh2 rd⌦2(292)

with d⌦2 = d✓2 + sin2 ✓d�2 the round 2-sphere metric.

In fact these can be written in the unified manner;

hij

(x)dxidxj =d⇢2

1� k⇢2+ ⇢2d⌦2 (293)

where k = +1 for the sphere, k = 0 for flat space and k = �1 for hyperbolicspace.

However, for simplicity, and since it appears to be the physically importantcase, we will focus on the flat case k = 0. Then,

hij

(x)dxidxj = dr2 + r2d⌦2 = dx2 + dy2 + dz2 (294)

so,

ds2 = �dt2 + a(t)2�dx2 + dy2 + dz2

�(295)

82

The function a(t) is called the scale factor and controls the expansion orcontraction of the spatial geometry. When we say the universe is expanding,technically we mean da/dt > 0.

A measure of this expansion is the Hubble constant H defined as,

H ⌘ 1

a

da

dt(296)

which is not actually a constant.

Conformal time ⌧

Rather than using the time coordinate t - the proper time of the timelinecurves above - the geometrically more natural coordinate is conformal time⌧ where,

ds2 = a2(⌧)��d⌧ 2 + dx2 + dy2 + dz2

�= a2(⌧)⌘

µ⌫

dxµdx⌫ (297)

for xµ = (⌧, x, y, z). Hence,

a2(⌧)d⌧ 2 = dt2 =) t =

Zd⌧ a(⌧) (298)

and we see the metric is Minkowski up to an overall scaling (a conformalfactor) given by the scale factor.

Note that in conformal time the Hubble constant is;

H =1

a2da

d⌧(299)

9.1 Geodesics of FLRW

We will now compute the geodesics of FLRW. Recall that an a�nely param-eterized geodesic, xµ(�), obeys the equation vµr

µ

v⌫ = 0 for vµ = dxµ/d�,

which can be written, d

2x

µ

d�

2 + �µ

↵�

dx

↵

d�

dx

�

d�

= 0.

Recall also that the norm of the tangent, |v|2, is conserved along the geodesic;

d

d�

�|v|2�

=dxµ

d�

@

@xµ

�g↵�

v↵v��= vµr

µ

�g↵�

v↵v��

= 2v↵vµrµ

v↵

= 0 (300)

83

since vµrµ

v↵

= 0 for the geodesic.

However, also recall that geodesics in a�ne parameterization extremize,

L =

Zd� g

µ⌫

dxµ

d�

dx⌫

d�(301)

and in practice this is often the most convenient way to compute geodesics.

We will now use this variational method to compute the geodesics of FLRW.Please see the example sheet 8 Qu 2 for calculations with geodesics using thegeodesic equation rather than the variational approach.

We write the geodesic curve in terms of an a�ne parameter � as,

xµ(�) =�⌧(�), x(�)i

�,

dxµ(�)

d�=�⌧ 0(�), x0(�)i

�(302)

where ⌧ 0(�) = d⌧/d�, so we have,

L =

Zd�L =

Zd� a(⌧(�))2

��⌧ 0(�)2 + �

ij

x0(�)ix0(�)j�

(303)

The Euler-Lagrange equation for x(�) variations is;

d

d�

✓@L

@x0(�)i

◆=

@L

@x(�)i(304)

giving,

d

d�

�2a2(⌧(�))�

ij

x0(�)j�= 0 (305)

which implies

x0(�)i =vi

a2(⌧(�))(306)

where vi are constants giving the spatial direction of the geodesic.

Anyhow we conclude that,

a2(⌧(�))x0(�)i = constant (307)

84

This is resulting from the fact that @L/@xi vanishes, and hence the metricis independent of coordinates xi resulting in the spatial isometries generatedby Killing vectors,

uµ

(1) = (0, 1, 0, 0) , uµ

(2) = (0, 0, 1, 0) , uµ

(3) = (0, 0, 0, 1) (308)

We see that as a result of this spatial translation invariance we recover themomentum conservation above, where we see the momentum of a particle inFLRW is a2(⌧)x0i.

The time variation gives;

d

d�

✓@L

@⌧ 0(�)

◆=

@L

@⌧(�)(309)

which yields,

d

d�

��2a2(⌧(�))⌧ 0(�)

�= 2a

da(⌧)

d⌧

��⌧ 0(�)2 + �

ij

x0(�)ix0(�)j�=

2

a

da(⌧)

d⌧L

Timelike geodesics

We now consider a class of timeline geodesics, those where,

x(�)i = constant =) x0(�)i = 0 =) vi = 0 (310)

These are called co-moving observers. As we shall see later, since the metricis homogeneous and isotropic so is the matter, so these observers sit at restwith respect to the matter - the move with it, hence co-moving.

Hence the timeline geodesics (in conformal time coordinates) have tangent;

dxµ

d�=

✓d⌧(�)

d�, 0, 0, 0

◆(311)

Requiring � to be the proper time, so that |dx/d�|2 = �1 then implies,

d⌧(�)

d�=

1

a(⌧(�))(312)

85

Further note that this proper time � of these observers is the same as the tcoordinate in the form,

ds2 = �dt2 + a(t)2dxidxi (313)

which recall also obeyed,

d⌧(�)

dt=

1

a(314)

Null geodesic

Note that the metric it conformal to Minkowski. In fact this generally meansthe null geodesics are the same as for Minkowski as we shall see.

We will consider the general null geodesic. We require the curve to be nullso L = 0. Then the ⌧ variation (310) gives,

d

d�

�a2(⌧(�))⌧ 0(�)

�= 0 (315)

and hence,

⌧ 0(�) =�

a2(316)

for a constant �.

However, for L = 0, then,

a2��⌧ 02 + �

ij

x0ix0j� = 0 (317)

and x0i = vi/a2. Hence we see actually, �2 = �ij

vivj, so the tangent is,

dxµ

d�=

p�ij

vivj

a2,vi

a2

!(318)

Again, one would integrate,

d⌧(�)

d�=

p�ij

vivj

a2(⌧)=) � =

1p�ij

vivj

Zd⌧a2(⌧) (319)

86

to obtain the actual curve.

Note that the set of points the curve passes through is the same as a nullgeodesic in Minkowski spacetime - ie. a 45o angle line! However the a�neparameter � evolves di↵erently along the curve.

9.2 Cosmological redshift

Consider two comoving massive particles, on a source emitting pulses of ra-diation at a constant rate and the other receiving these pulses. The radiationpulses simply travel along 45o lines in the conformal time coordinates (⌧, xi).The key point is that this remains true independent of the time of emission.

Suppose the emitter at (⌧e

, xi

e

) emits signals every small proper time interval�t

e

. Corresponding to this proper time is the conformal time interval �⌧e

;

(�te

)2 = a(⌧e

)2(�⌧e

)2 (320)

An observer receives these signals at (⌧o

, xi

o

) seeing a proper time interval�t

o

, and corresponding conformal time interval �⌧o

;

(�to

)2 = a(⌧o

)2(�⌧o

)2 (321)

Now the radiation then follows a null geodesic, a 45o line in the (⌧, xi) coor-dinates between emitter and observer. This implies the crucial result that,independent of the emission time ⌧

e

,

�⌧o

= �⌧e

(322)

Hence we conclude the proper time intervals of emitter and receiver are re-lated as;

�to

�te

=a(⌧

o

)

a(⌧e

)(323)

Now in an expanding universe then a(⌧o

) > a(⌧e

) and hence the observedtime interval �t

o

> �te

, and hence the frequency is redshifted.

87

Thus observers in an expanding universe looking at signals emitted in theirpast see them redshifted. Let us expand the scale factor around the observerstime as,

a(⌧) = a(⌧o

) + a(⌧o

)2Ho

(⌧ � ⌧o

) + . . . (324)

where Ho

> 0 for an expanding universe and is the value of Hubble constant,H, at the observer time. (Recall H = 1

a

2da

d⌧

.)

Then write ⌧0 = ⌧e

+�⌧ , and so we find,

�to

�te

=a(⌧

o

)

a(⌧e

)=

a(⌧o

)

a(⌧o

) + a(⌧o

)2Ho

(⌧e

� ⌧o

) + . . .

=1

1� a(⌧o

)Ho

(⌧o

� ⌧e

) +O((⌧e

� ⌧o

)2)

= 1 + a(⌧o

)Ho

(⌧o

� ⌧e

) +O((⌧e

� ⌧o

)2) (325)

Now since null geodesics are 45o lines in the(⌧, xi) coordinates, then,

|xo

� xe

| =q

�ij

(xi

o

� xi

e

) = ⌧o

� ⌧e

(326)

The proper distance, D(⌧), between the observer and emitter in the spatialslice at constant ⌧ is,

D(⌧) = a(⌧)|xo

� xe

| (327)

Hence,

Do

⌘ D(⌧0) = a(⌧o

)(⌧o

� ⌧e

) (328)

and we conclude that;

�to

�te

= 1 +Ho

Do

+O((⌧e

� ⌧o

)2)

(329)

We define the Cosmological Redshift Z as,

Z ⌘ �to

�te

� 1 (330)

88

so Z = 0 implies no di↵erence in clock speed, and Z > 0 means the observedsignal is redshifted.

Then we find the Hubble law ;

Z = Ho

Do

(331)

which states the redshift of any object is proportional to its distance from us.

Note: For an object in Minkowski space receding from us at velocity v ⌧ 1,then,

Z =�t

o

�te

� 1 ' v (332)

so for small redshift Z ⌧ 1, then Z gives the apparent recession velocity ofthe emitter. Thus the apparent velocity of recession of objects is proportionalto their distance form us.

Hubble and latter observations found,

Hnow

⇠ 70(km/s)/Mpc (333)

where 1Mpc ⇠ 31⇥ 1015m or roughly 3 light years.

In fact this Hubble law is true for nearby objects only where Z ⌧ 1. Ingeneral one must use the full behaviour of a(⌧) to compute the redshift -distance relation.

89

9.3 The Friedmann equation

In Question Sheet 8 you computed the Einstein tensor of the FLRW metricin conformal time coordinates to show;

G⌧⌧

= 3

✓a

a

◆2

Gij

=

✓a2 � 2 a a

a2

◆�ij

(334)

where a = da/d⌧ .

We take the matter in the universe to be a perfect fluid so that,

Tµ⌫

= (⇢+ P )uµ

u⌫

+ Pgµ⌫

(335)

Then the cosmological principle states that the density ⇢ and pressure Pshould be homogeneous and isotropic - ie. not depend on the coordinatesxi at all. Further since there is no preferred direction then the 4-velocityuµ = ( 1

a

, 0, 0, 0) so that the 3-velocity vanishes and uµuµ

= �1.

Since, uµ

= (a, 0, 0, 0), this implies,

T⌧⌧

= a2(⌧)⇢(⌧)

Tij

= a2(⌧)P (⌧)�ij

(336)

Hence from the tt component of the Einstein equation,

Gµ⌫

= 8⇡GN

Tµ⌫

(337)

we obtain the Friedmann equation;

H2 =

✓1

a2da

d⌧

◆2

=a2

a4=

8⇡GN

3⇢ (338)

In fact due to the contracted Bianchi identity we do not need to considerthe ij component provided we ensure stress energy conservation. The fluidequations are given by r

µ

T µ⌫ = 0 and following the calculation you did inexample sheet 8 one finds;

⇢+ 3a

a(⇢+ P ) = 0 (339)

90

For a perfect fluid the pressure P is determined in terms of the density ⇢ bythe equation of state parameter w;

P = w ⇢ (340)

The equation of state parameter takes the values;

w =

8<

:

0 cold matter+1

3 radiation / hot matter�1 dark energy

(341)

Note that for dark energy we have Tµ⌫

= P (x)gµ⌫

. However (see Ex Sheet 7,Qu 2) this is simply acosmological constant,

Tµ⌫

= ⇤gµ⌫

(342)

where ⇤ is a constant. The Bianchi identities imply that @µ

P = 0 so indeedP = ⇤ is just a constant.

It is simple to derive from the fluid equations (see Ex Sheet 8, Qu 1) that;

⇢+ 3(1 + w)a

a⇢ = 0 =) ⇢(⌧) =

✓a0a(⌧)

◆3(1+w)

(343)

Hence cold matter dilutes simply with the 3-volume (a3⇢ =const), whereashot matter/radiation dilutes with 4-volume (a4⇢ =const). This latter resultis because not only do the photons get diluted as the space expands, butthey also become redshifted, which further decreases their energy density.

Now substituting these into the Friedmann equation one finds,

a2

a4=

8⇡GN

3

✓a0a(⌧)

◆3(1+w)

(344)

so that,

a(⌧)32 (1+w)�2 da

d⌧=

s8⇡G

N

a3(1+w)0

3=

1

k(345)

and hence,

⌧ � ⌧0 = k0 a(⌧)32 (1+w)�1 = k0 a(⌧)

12 (1+3w) (346)

91

for constant of integration ⌧0 and appropriate constant k0. So,

a(⌧) = k00 (⌧ � ⌧0)2

(1+3w) (347)

Hence for cold matter (w = 0) we have,

a ⇠ (⌧ � ⌧0)2 (348)

and for hot matter/radiation (w = 1/3) we have,

a ⇠ (⌧ � ⌧0) (349)

Note: both hot and cold matter drive expansion for a flat universe. Also notethat for hot matter, as one goes back in time the density and temperatureincreases.

We believe that early on in the universe, for approximately the first 10000years the universe was hot. The time ⌧0 = 0 is the Big Bang !

For hot matter, a4⇢ =constant, so if a ⇠ (⌧ � ⌧0) then,

⇢ ⇠ (⌧ � ⌧0)�1/4 ! +1 as ⌧ ! ⌧0 (350)

However, note the proper time measured by a comoving observer after thebig bang is finite!

t(⌧) =

Z⌧

⌧0

d⌧a(⌧) 'Z

⌧

⌧0

d⌧k(⌧ � ⌧0) =k

2(⌧ � ⌧0)

2 (351)

(Interestingly the term ‘Big Bang’ was given by Fred Hoyle, who didn’t be-lieve in it!)

As the universe expands it cools, and at some point the matter becomes cold(around Z ⇠ 1000, a proper time ⇠ 100000 years after big bang), and thescale factor then goes as a ⇠ ⌧ 2. Note that cold matter dilutes more slowlythan hot matter.

A decade ago a big surprise came when supernovae observations actuallyindicated that a new dark energy component seems to be dominating theexpansion! For w = �1 one finds H2 ⇠ ⇤ = constant. So the expansioncontinues indefinitely - the cosmological constant doesn’t dilute!

92

10 The Schwarschild spacetime

Already in 1918 Karl Schwarschild wrote down the space-time;

ds2 = �✓1� R

S

r

◆dt2 +

✓1� R

S

r

◆�1

dr2 + r2d⌦2 (352)

where d⌦2 = d✓2 + sin ✓2d�2 is the line element on the round 2-sphere. Wewill work with these Schwarschild coordinates xµ = (t, r, ✓,�).

A calculation of the geometry of this yields;

Rµ⌫

= 0 =) Gµ⌫

= 0 (353)

and hence this is a vacuum space-time - Tµ⌫

= 0.

It was later proven by Birkho↵ (1923) that this is the unique vacuum space-time that is spherically symmetric - ie. it has the isometries of the 2-sphere.

An important point is that this space-time is static, meaning that @t

gµ⌫

= 0and hence,

vµ = (1, 0, 0, 0) (354)

is a Killing vector generating the time translation isometry. Birkho↵’s theo-rem means that spherical symmetry in vacuum implies a static metric.

Whilst Ricci vanishes, Riemann does not. An interesting geometric invariantis the Kretchmann scalar;

Rµ⌫↵�

Rµ⌫↵� =12R2

S

r6(355)

Since the Kretchmann scalar becomes infinite at r = 0, this is a physicalsingularity that cannot be removed by a coordinate transformation.

Something very funny happens at r = RS

the Schwarzschild radius. Forr > R

S

then t is a time coordinate and r is spatial. However for r < RS

infact t is spatial and r is timeline!

93

We may change radial coordinates;

r = ⇢

✓1 +

RS

4⇢

◆2

(356)

to attain,

ds2 = �

⇣1� RS

4⇢

⌘2

⇣1 + RS

4⇢

⌘2dt2 +

✓1 +

RS

4⇢

◆4 �d⇢2 + ⇢2d⌦2

�(357)

This are called isotropic coordinates. Expanding for ⇢ � RS

we find,

ds2 = �✓1� R

S

⇢+ . . .

◆dt2 +

✓1 +

RS

⇢+ . . .

◆�d⇢2 + ⇢2d⌦2

�(358)

Changing from spherical coordinates (t, ⇢, ✓,�) to Cartesian (t, x, y, z), so⇢2 = x2 + y2 + z2 we have,

ds2 =

✓⌘µ⌫

+R

S

⇢�µ⌫

+ . . .

◆dxµdx⌫

= (⌘µ⌫

� 2��µ⌫

+ . . .) dxµdx⌫ (359)

and hence asymptotically we may identify the space-time as having Newto-nian form,

2� = �RS

⇢(360)

and recall for a point mass M at ⇢ = 0 we expect,

� = �GN

⇢=) R

S

= 2GN

M (361)

Hence we conclude that the Schwarzcshild metric represents a sphericallysymmetric static space-time carrying a mass M given by R

S

as above. It isoften written as,

ds2 = �✓1� 2G

N

M

c2r

◆dt2 +

✓1� 2G

N

M

c2r

◆�1

dr2 + r2d⌦2 (362)

94

Interpretation of the Schwarzschild solution:

Interpretation 1) - Stellar exterior, valid for R > Rstar

Wemay view the Schwarschild metric as the exterior solution to any sphericalmatter distribution (static or otherwise!). For example, a star with mass Mand su�ciently low density such that its radius is > R

S

will be Schwarschildin its exterior, with some static spheric metric in the interior which is notvacuum and hence not Schwarschild.

Example: For the sun, M ' 2⇥1030kg. The Schwarschild radius RS

' 3km.Clearly the radius of the sun is much bigger being millions of km.

Example: Typical neutron stars has mass similar to the sun, but with radiusof a few kilometres. Indeed they often rotate 1000 times per second!

Interpretation 2) - Black hole, valid except in the past

This is the end state of gravitational collapse - where a star burns its fueland collapses below its Schwarzschild radius. This is a black hole space-time.The radius R

S

is called the horizon of the black hole.

Now the space-time is taken for all radii R but since it is the end state ofcollapse, one should regard it as approximating the true physical space-timeonly after the point of collapse. In particular the far past of the Schwarzschildmetric is not physical in this picture.

For this reason we should not allow any matter to be at a radius inside thehorizon with an outward velocity.

95

10.1 Timelike geodesics of Schwarzschild

We write our space-time curve in terms of functions T,R,⇥,� in terms ofan a�ne parameter �.

xµ(�) = (T (�), R(�),⇥(�),�(�)) (363)

so the tangent is given by,

dxµ(�)

d�=⇣T (�), R(�), ⇥(�), �(�)

⌘(364)

where T = dT/d� and similarly for R, ⇥, �.

The define L the norm of the tangent for the curve. For an a�ne parame-terization this will be constant; for a timeline geodesic L = �1;

L ⌘ gµ⌫

dxµ(⌧)

d⌧

dx⌫(⌧)

d⌧= �

✓1� R

S

R

◆T 2 +

✓1� R

S

R

◆�1

R2 +R2⇣⇥2 + sin2 ⇥�2

⌘

Then we obtain the geodesics by varying this Lagrangian so that L =Rd�L

is extremized.

The Euler-Lagrange equations are;

d

d�

✓@LdT

◆=

@LdT

(365)

and similarly for the other functions. In fact we need only vary T , ⇥ and �,and the remaining equation can be obtained from the condition L = �1. InEx Sheet 8 you explicitly performed these calculation to find,

@LdT

= �2

✓1� R

S

R

◆T

@Ld⇥

= 2R2⇥

@Ld�

= 2R2 sin2 ⇥� (366)

96

and,

@LdT

= 0

@Ld⇥

= 2R2⇥

@Ld�

= 0 (367)

where the former and latter vanish due to the static symmetric and azimuthalsymmetry of the metric. We expect these to yield energy and angular mo-mentum conservation.

We then obtain the variational equations;

T variation :d

d�

✓✓1� R

S

R

◆T

◆= 0

⇥ variation :d

d�

⇣R2⇥

⌘= R2 sin⇥ cos⇥�2

� variation :d

d�

⇣R2 sin2 ⇥�

⌘= 0 (368)

Consider motion in the equatorial plane ⇥ = ⇡/2. Hence ⇥ = 0 andsin⇥ = 1, cos⇥ = 0. We see this is consistent with the ⇥ equation of motion.

Note: By the spherical symmetry of the space-time we may always chooseour coordinates so that the motion lies in this plane.

Now the first and last equations above simply imply conservation of energyE, and angular momentum L, where,

E =1

2

✓1� R

S

R

◆2

T 2

L = R2 � (369)

for this equatorial plane motion. We must further supplement this with theequation L = where = �1 for a timeline geodesic (so � = ⌧ the propertime);

� 1 = �✓1� R

S

R

◆T 2 +

✓1� R

S

R

◆�1

R2 +R2�2 (370)

97

Now this may be written in terms of our conserved quantities;

� 1 = �✓1� R

S

R

◆�1

2E +

✓1� R

S

R

◆�1

R2 +1

R2L2 (371)

and rearranging this gives;

E =1

2R2 + V (R) (372)

with,

V (R) =1

2

✓1� R

S

R

◆✓1 +

L2

R2

◆(373)

Hence we see that motion is given by motion of a unit mass particle withenergy E in the potential V (R).

Properties of timeline geodesics:

The potential is then,

V (R) =1

2� R

S

2

1

R+

L2

2

1

R2� L2R

S

2

1

R3(374)

which is the same as for the Newtonian theory except for the last term!

Newtonian motion occurs for large radius R > RS

and low angular momen-tum R > L.

The extrema of this potential occur for;✓R

L� L

RS

◆2

=L2

R2S

� 3 (375)

Hence for low angular momentum L < 3RS

there are no extrema. ForL > 3R

S

there are two roots, R+ > R�, which you computed in Ex Sheet 8;

3

2R

S

< R� < 3RS

< R+ (376)

where R+ is a stable minimum giving a stable circular orbit, and R� is un-stable.

98

Hence the innermost stable circular orbit (ISCO) is R = 3RS

= 6GN

M/c2.

Note that unlike the Newtonian theory, the potential is attractive near R = 0.Anyone geodesic passing R� with R < 0 will inevitably reach R = 0.

Perihelion advance:

The classic result of the Newtonian potential is that one obtains closed el-liptical orbits. However, this is a very special property of the 1/R potential.

Let us expand the potential V (R) about the stable circular orbit R+ as,

V (R) = V (R+) +1

2(R�R+)

2 V 00(R+) +O�(R�R+)

3�

(377)

As you show in the Ex Sheet 8, one finds,

V 00(R+) =L2

R4+

✓1� 3

RS

R+

◆(378)

Hence for a nearly circular orbit (ie. |R�R+| ⌧ RS

for the duration of theorbit) we see the radius R(⌧) performs simple harmonic motion about R+

with a period, Tradial

, given by,

Tradial

=2⇡p

V 00(R+)(379)

Recall from the definition of L we have,

d�

d⌧=

L

R2' L

R2+

+O ((R�R+)) (380)

and hence for a nearly circular motion to traverse 2⇡ radians in � takes timeTangular

, where,

Tangular

'2⇡R2

+

L(381)

Hence for nearly circular orbits,

Tradial

Tangular

=1q

1� 3RSR+

' 1 +3

2

RS

R++ . . . (382)

99

Thus in the Newtonian limit we recover closed orbits - Tradial

= Tangular

.

However, there is a small advance of the perihelion each orbit, with angle ofadvance;

↵ ' 2⇡

✓Tradial

� Tangular

Tradial

◆radians/orbit =

⇡RS

R+radians/orbit (383)

For Mercury about the sun; the solar mass M� = 2.0⇥1030kg and the orbitalradius is R ' 50⇥109m. It orbits 4 times a year, so over 100 years the angleof advance;

↵|100yrs ' 43 arcsec (384)

Note; the moon in the sky subtends ⇠ 2000arcsec.

10.2 Black holes

Recall the peculiar property that r = 0 is strongly attractive (unlike for New-tonian gravity), and further more r becomes a timeline coordinate for r < R

S

.

Consider a massive particle with proper time ⌧ , and position,

xµ(⌧) = (T (⌧), R(⌧),⇥(⌧),�(⌧)) (385)

as above. Now we will not assume it is inertial (non-accelerated).

Consider the particle to be equipped with a rocket so that it may sit atconstant R,⇥,�. Then its 4-velocity of the form,

vµ =dxµ

d⌧=

0

@ 1q1� RS

R

, 0, 0, 0

1

A (386)

so that vµvµ

= �1. We may compute the 4-acceleration of this particle,

aµ = v⌫r⌫

vµ (387)

100

and find that,

aµ =

✓0,

RS

2R2, 0, 0

◆=) |a| =

paµa

µ

=R

S

2R2q1� RS

R

(388)

(see Ex Sheet 8 for the details of this calculation). Recall that this magnitudeis a physical quantity and measures the usual acceleration experienced in theinstantaneous LIF. The components of aµ do not really measure anythingphysical themselves as they depend on the coordinates.

Hence we see that the force required to hold the particle at the fixed coor-dinate location becomes infinite at the horizon r = R

S

. The horizon is the“radius of infinite attraction”!

The horizon is also a surface of infinite gravitational redshift. Consider theparticle above emits a signal at a constant frequency, and we are very far fromthe black hole (ie. at r ⌧ R

S

) observing it. Our proper time is approximatelygiven by the coordinate t. However from the 4-velocity of the particle we see,

d⌧

dt=

1q1� RS

R

(389)

An important point: Since the space-time is static the gravity redshift isjust given by the ration of our proper time t to the particles.Hence the frequency we observe !

o

is related to that emitted !e

as,

!o

!e

=

r1� R

S

R(390)

Thus for a particle approaching the horizon this redshift becomes infinite.

Falling in to a black hole:Now consider the particle to have a general (accelerated) timeline trajectoryso;

� 1 = �✓1� R

S

R

◆T 2 +

✓1� R

S

R

◆�1

R2 +R2⇣⇥2 + sin2 ⇥�2

⌘(391)

101

Note that since r is time and t is spatial inside the horizon 1� RSR

< 0. Thusfor R < R

S

we may rearrange to;

+

✓R

S

R� 1

◆�1

R2 = 1 +

✓R

S

R� 1

◆T 2 +R2

⇣⇥2 + sin2 ⇥�2

⌘� 1

and hence inside the horizon we have,

R2 � RS

R� 1 (392)

Now consider a particle that has just entered the horizon, so that R < RS

and R < 0. Thus R is decreasing, but for R < RS

we cannot have R = 0and hence R will continue to decrease! This shows its timeline nature - youcannot reverse time.

Suppose this (unfortunate) particle falls in at time ⌧i

. Then,

�r

RS

R� 1 � R =) d⌧ � �

rR

RS

�RdR

=) ⌧(R)� ⌧i

�Z

R

RS

rR

RS

�RdR

=) ⌧(R)� ⌧i

Z

RS

R

rR

RS

�RdR (393)

Hence we see the time to reach R = 0 after falling in, �⌧ = ⌧(0) � ⌧i

isbounded from above as;

�⌧ Z

RS

0

rR

RS

�RdR =

⇡RS

2(394)

Putting in the ’c’ explicitly;

�⌧ ⇡RS

2c(395)

This is approximately the time it takes light to travel a distance RS

.

Example: For a solar mass black hole, RS

⇠ 3km which implies, �⌧ ⇠ 10�5s.

102

Example: For a super massive black hole, M ⇠ 109M� so RS

⇠ 1012m whichimplies, �⌧ ⇠ 104s.

The horizon

Comment: To the particle falling into a black hole nothing special happenswhen they pass the horizon - this is simply because for them, locally space-time is Minkowski!

The metric looks singular at the horizon with gtt

! 0 and grr

! 1. Howeverthis is a coordinate singularity, and is analogous to the coordinate singularitythat occurs at the origin of polar coordinates;

ds2Euc

= dr2 + r2d⌦2 (396)

where the determinant of the metric vanishes at r = 0.

Just as for Euclidean space, we know there is no problem with r = 0 since wemay go to Cartesian coordinates where the metric is perfectly regular there,the same is true for the horizon.

We define an ‘ingoing’ time coordinate v as,

v = t+ r⇤ = t+ r +RS

ln

����r

RS

� 1

���� (397)

where we note the tortoise radial coordinate r⇤ obeys;

dr⇤

dr=

1

1� RSr

(398)

In the coordinates (v, r, ✓,�) the metric takes the ingoing Eddington-Finklestein form;

ds2 = �✓1� R

S

r

◆dv2 + 2dv dr + r2d⌦2 (399)

which is perfectly fine at r = RS

. In fact one finds the eigenvalues,

{�1, 1, R2S

, R2S

sin2 ✓} (400)

103

there. These coordinates cover all the physically interesting region of theblack hole - ie. the bit after the collapse.

The crucial point is that in these coordinates, nothing particularly dramatichappens at the horizon.

The point r = 0 is, however, a physical singularity - a mysterious point ofinfinite curvature.

104