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General Relativity
Satyam Ladva
Newton's gravity
•

Remember gravitational force equation and N2 law and how the acceleration is obtained
...


Determined by use of two same mass and same material balls in a pivot setup
...
e
...
Observer in freefall, Lab in a space rocket orbitting Earth
vs
...


•

Combined with Newton's gravity, which acts instantaneously, Einstein's GR is formed
...
GR
•

On a Space-time diagram, an inertial observer follows a straight line - As demonstrated in special
relativity
...


•

e
...
A visual description is of a planet orbitting a star
...
e
...


Note:
•

A Perihelion advance states that an elliptical orbit is not truly elliptical due to slight shifts in
the orbit
...

On very small scale, the Geometry in GR is equal to that of the Geometry in SR but as the scale
becomes larger, curved spacetime geometry dominates
...
If you take a very
small surface area of a curved surface, this increment is deemed at i
...
A very small section of
Curved spacetime gives rise to Minkowski metric
...
A small increment of a curved line
gives rise to an approximate straight line
...


1

2

Secondterm:
Thirdterm:
Equation:

GN

Tµν

represents Newton's Gravitational constant
...


At appropriate scale and deviation e
...
Light deviation, Gravitational Redshift etc
...


Classical Geometry
Euclidean Space: Recap
• E3

is the notation used to dene 3-dimensions in Euclidean Space

•
•

Line element/metric:

•

The distance P1 to P2 =

xi = {x, y, z}

The Canonical Co-ordinates:

where i, in 3-dimensional space, = 1, 2, 3
...




•

The Metric matrix of

ds2

is represented by

ds2 = gij dxi dxj

where

gij

1
=  0
0

0
1
0


0
0  =
1

diag(1, 1, 1)

•

In matrix notation:



gij = 


g11
g21
g31

g12
g22
g32

g13
g23
g33


...


...


...



...


...


...




gnm




...


3

Vectors
vi

v i = (v 1 , v 2 , v 3 ) or (v x , v y , v z )
...
g
...
w = v x wx + v y wy + v z wz = gij v i wj = Scalar

metric matrix form:

SuxNotation:

where

See ACP Einstein Sux notation revision notes for further examples
...
e
...


This is possible since the transpose of

v

is obtainable if

via a

vi

is

=
replaced in front of




• (v 1 , v 2 , v 3 ,
...


g11
g21
g31

g12
g22
g32

g13
g23
g33


...


...


...



...


...


...


gnm












w1
w2
w3

...


...


Physical signicance of vectors
•

The dot product of the same vector gives the length of the vector, squared i
...


(length of v)
•

2

v
...


In general, the dot product gives the norm of

v

• Vectors live at a point, but in the space
...
e
...
A tangent to Earth's surface does not give a length but it is classied as a
vector
...


•

θ at a point between two vectors v and w can be calculated by using the equation:
v
...
This is a general result i
...
It is true in all spaces, not just Euclidean
...
i
...

Changing from a set of cartesian co-ordinates

• x = rcosθ, y = rsinθcosφ, z = rsinθsinφ

xi

to a set of spherical co-ordinates

xi

in

E3
...


•

Taking the derivative of the three transitions and substituting it into the line metric equation
produces:

ds2 = dr2 + r2 dθ2 + r2 sin2 θdφ2

which is not valid in an co-ordinate system
...


•

Vectors in spherical co-ordinates, with co-ordinates related to

i

gi j dx dx

j

can be written as:

ds2 = v i =

is a geometrically invariant equation i
...


•

vi ,

gi j = diag(1, r2 , r2 sin2 φ)
...
w = v i wj gij = v i w j gi j

Curves: Tangents and Straight Lines
•

In

E3 ,

where

a straight line in canonical/cartesian co-ordinates can be written as:

λ

parameterises the point on the line and

i

a

and

b

i

xi (λ) = ai + λbi

constants which give the oset and

direction
...


•

i

v (λ) =

•

λ1 ↔ λ2 is given by integrating the norm of the tangent
dxj
dλ where gij = gij (x(λ))
...


The distance along a curve between
vector:

•

xi (λ) whilst a tangent vector to a curve is dened as
dxi (λ)
which is an equation that is true in general
...
e
...

dλ

A Curve is dened parametrically as

S(λ1 , λ2 ) =

To calculate

S,

´ λ2
λ1

i

dλ gij dx
dλ

break up the curve into very small straight lines
...
e
...


•

Metrics and vectors change form in dierent co-ordinate systems however, scalar dot products
are invariant under co-ordinate system changes
...
e
...
For
any vector, one can nd an innite class of curves whose tangent vector is that vector
...
e
...


•

The tensor equationv

•

The distance

∂xi ∂x i
∂x i ∂λ

•

S =

i

´ λ1
λ0

=

dxi
dλ is expressed in metric form as
i

dλ gij dx
dλ

xµ xν gµν = x µ x ν gµ ν


...


dx i
∂x i i
dλ = ∂xi v where the second partial
derivative, on the RHS, is a 2-index matrix which transforms from one set of co-ordinate systems

Taking the inverse of this chain rule result gives

vi =

onto the other
...
Consider

a curve lying IN the embedded surface, described parametrically as

f (x(λ), y(λ))
•

z = f (x, y)

is an embedded function which encodes the geometry of the curve
...


As a consequence of the embedding

z(λ) −→

dz
dλ

=

∂f
dx
∂x |y dλ

+

∂f
dy
∂y |x dλ , due to the rule of partial

dierentiation
...


vi =

dxi
dλ

dx dy
dx
= ( dλ , dλ , ∂f dλ +
∂x

6

•

The rectangle represents a 2D plane which is tangent to the surface
...


•

A tangent space at a point = Set of all vectors at a point
...
This is possible if no curve is dened through

the point
...


•

Given two tangent vectors

ui

and

vi

at the surface, of the form:

 ui = (ux , uy , ∂f ux + ∂f uy ) ↔ (ux , uy )
∂x
∂y
 v i = (v x , v y , ∂f v x + ∂f v y ) ↔ (v x , v y )
∂x
∂y
•

A linear combination gives a third tangent vector to the surface

wi = aui + bv i

where

a

and

b

are constants
...


This implies that the tangent space at a point is given by the vector space

Notation:

v a = (v x , v y )

where

by the map

∂f y
∂y v )

Linearly combining the vectors
tangent vector in the surface

•

∂f y
∂y v )
...


wa = aua + bv a

7

•

In

E3

vector addition allowed
...
This vector addition property is only allowed
in Monkowski spacetime and not a general property
...
v = ux v x + uy v y + uz v z where the third terms can be expressed as a combination of the rst
∂f x
∂f y ∂f x
∂f y
z z
two: u v = (
∂x u + ∂y u )( ∂x v + ∂y v )
...
v = ux v x (1 + ( ∂f )2 ) + uy v y (1 + ( ∂f )2 ) + (ux v x +
∂x
∂y

uy v y ) ∂f ∂f
...
which can be re-written in the form of a metric:

(Euclidean) i j

gij

u v = gab ua v b
...
v = (ux uy )

1 + ( ∂f )2
∂x
∂f ∂f
∂x ∂y

g ab

The inverse metric

•

If the surface is embedded such that
induced metric

is dened by the equation:

gab = I

i
...


vx
vy

E 3
...
gij
is the 3 × 3 indentity matrix diag(1, 1, 1)
...


the surface i
...
The

which proves that

(g ab )−1 g ab = I
...


•

If the laws of GR were taken and the eects of curvature of spacetime removed, the laws of SR
would be the outcome
...
It can also be translated such that the embedding function of
the surface has the property

•

x=y=0

What these equations mean is that in

and

E3 ,

∂f
∂x

=

∂f
∂y

= 0
...
Essentially,
rotating from

x → x = M
...


=

8

•

x = y = 0, the function f (x, y) can be Taylor expanded to become f (x, y) =
2
2
∂2f
1
+
+ 1 ∂ f x2 + 2 ∂ f y 2 + ∂x∂y xy
...


= f (0, 0) + O(x2 , y 2 xy) +
...
i
...
All surfaces are locally

at if you look close enough, relative to the curvature of the object
...


Geodesics (In the embedded surface z = f (x, y))
•

In the context of geometry, where no time is involved, Geodesics are minimum length curves
or the most direct path between 2 points
...
g
...


•

This minimum length of curve

x(λ)

is what is required in order to reduce the work done by the

object in real-life situations
...


•

Can parameterise the curve

•

S is obtained by integrating the norm of the tangent vector
´ λ1
a
b
dxi dxj
Euc
S = λ0 dλ gij dλ dλ = λ0 dλ gab (x) dx dx where the rst form of S is dened for a 3-D
dλ dλ
object whereas the second form is an intrinsic concept
...


•

The curve length

S is a reparameterisation invariant
...


This results

b
a
gab dx dx
...


•

An ane parameterisation is where the square root expression in

S=

the equation becoming

•

If a curve

x(τ )

λ0

dλ
dτ

=

√

f

λ

This implies that

resulting in

S=

´ λ1

´ τ1
τ0

dτ

S

equals to one, resulting in

only measures distance along the curve
...


a new ane parameterisation is

√

a
b
dτ
dλ L where Lλ = gab dx dx = Lτ ( dλ )2
...


is such that the square root term

computed where

•

´ λ1

S =

results in a 1st order perturbation i
...
Any variation from this minimum path results in the curve
lengthening

x(λ) → x(λ) + δx(λ)
...


•

´λ
√
√
√
dλδ( Lλ ) = λ01 dλ 2δLλ where δ( Lλ ) is simply a small change in
λ0
Lλ
the Length element from one curve to the other i
...
Lλ → Lλ + δLλ
...


measures the proper distance along the path
...


•

This results in

L=
•

´ λ1
λ0

dλL
...


where there is a minimum length curve between two

points on the surface is known as a Geodesic
...


This

x1 and x2
∂f ∂f
∂x ∂y
is a
+ ( ∂f )2
∂y

making

1

∂x ∂y

metric that denes length in arbitrary space-time
...
A geodesic curve is dened such that dS = 0 for all dx (λ) whereby the
endpoints x1 and x2 are xed
...
λ

is a measure of the distance

λ
...


where you

Invoking the

ds

which is an extremal
...


a

L(xa , dx ) =
dλ

in an extremal curve:

´ λ2

dλ

vice-versa for the RHS
...

dλ

•

dxb
dλ
...
The choice of index isn't relevent as long as it hasn't been used
dλ
dxb dλ
a
in the expression prior e
...
Can't use x again
...
This gives
− 1 ∂gab ) = Γd (x) is the Christofell symbol
ab
2 ∂xc

by multiplying by

d x
1 ∂gab dx dx
dc ∂gac
dc ∂gac
dλ2 + g ( ∂xb − 2 ∂xc ) dλ dλ = 0 where g ( ∂xb
and denes the geodesics, notion of straight lines
...
For non-ane λ, you get another term which can be reduced to the geodesic equation
∂xa
by choosing a suitable parameterisation
...

dxa
a
To solve this, information about x (λ) and
dλ are required at some λ
...
Can use symmetries (conserved quantities along the curve) to simplify the solving
procedure
...
Γd = 0
ab
x (λ) = a + b λ and (b1 )2 + (b2 )2 = 1
Consider a plane

a

a

therefore the geodesics are curves

a

f =constant, gab = diag(1, 1) and Γa = 0
bc
a
a
a + b λ where aa and ba are cosntants and ba bb gab = 1
...
This rresults in
varies linearly with x and y
...
Rotate this surface so that
gab = diag(1, 1)+O(x2 , y 2 , xy) and Γc = O(x, y) →It
ab

This results in a geodesic equation of the form
through

x=y=0

a

a

3

λ = 0, x = b λ + O(λ )

at

which results in

d2 xa
dλ2 = O(x, y) where, if the geodesic passes
→ This equation is forbidden by the geodesic

equation
...
, xd )
...
, x d )
i
i
and related to x by x
= x i (xj ) e
...
Cartesian and Spherical coordinates
...


Consider

• x → x = x (x)

is a coord transformation and it is usually best to restrict the transformation to

those that are invertible
...
e
...


Example: Spherical to cartesian transform possible but not vice versa due to the origin of the
spherical coords breaking down
...
e
...


This is known as single point failure, a problem not addressed in this lecture
...

=

M
=

such that

M i (x) ≡
j

∂x i
∂xj

J is dened
=
∂xi (x )
due to partial
∂x j

where the Jacobian matrix

The inverse of the Jacobian transform matrix,

Mi =
j

derivatives
...


It is usual

to call a function a coordinate scalar
...


•

v i (x) which has 1 upper index
...
Including the x dependency gives v (x ) = M j v |x=x(x )
...
v
=

i

=

y = rsinθ
...


Expression
...

dx k

Mi Mj =
j
k

∂x i ∂xj
∂xj ∂x k

v

This means that

i


...

i
wi = wj M j
...


It works the same as vectors, w
...
t functions of x
...
Can use the

11

•

In

E3 ,

spherical and cartesian coordinates are the natural choice due to symmetry arguments

whereas, in

Rd ,

there are no natural choice
...


•

∂f
f (x), there is a covector associated to its dierential df = ∂xi dxi where the
∂f
covector is wi ≡
∂xi →Tensor equation
...


•

A (q,r) tensor eld has q upper indices and r lower indices
...
i

1 2
Tj1 j2
...

r

It is a generalisation of

vectors and covectors
...
iq
j1 j2
...
i

j
j
Mj 1
...
jq (x) Mi1
...
i

jr
j1
1
q
• f (x) = Tj1
...
wiq(q) )(v(1)
...
, q

1

and r vectors


...
, r
...




Examples: Metric



Examples: The metric

v i is a (1,0) tensor
...
The Riemann metric Rijl is a (1,3) tensor and the Einstein metric Gij is a (0,2)
tensor
...
Vector

gij

in a 3D vector calculus and the dot product

u
...


The indices have been contracted, reducing the dot prduct to a function
...

m
n
i
j

The primed condition:

This is a metric

transform where the rst bracket is the inverse jacobian whilst the second and third are the
Jacobian matrices
...

a b
f (x ) = gab u v = f (x)
...


j
j
i
Mii δj Mj = Mii Mj
...
The tensor transform is

1 i =j
0 i =j

=

δ

i
j

=


...
To be invariant, a metric

must have 1 up and 1 down indices, at least
...

i1
...
and Bj1
...
(x) =
j1
...

f (x)Ai1
...
where this is also transforms as a tensor
...


•

(q,r) tensors

•

Can take the outer product of a (q1 ,r1 ) tensor

i1
...
Cj
...
i
iq

...
jq1 Bjr1 +1
...
ia−1 mia+1
...
jb−1 mjb+1
...


The contraction of the

i1
...
i
Bj1
...
jq
r−1 r+1
r

•

=

i
...
jq1
1
r

=

− 1,r − 1)

tensor:

T(ij) = 1 (Tij + Tji ) and
2
1
T[ij] = 2 (Tij − Tji ) where the square bracket indicates anti symmetry
...

Given a tensor, anti-symmeterise some indices by using the notation

12

•

1
gji = gji = g(ij)
...
in ) = n! (Ti1
...
in )
...


• Tabc
•

is the tensor and you want to symmetrise a and c but not b,

Can't add tensors at dierent points i
...


T(a|b|c) = 1 (Tabc + Tcba )
...
When you

transform this, you have to multiply by the Jacobian matrix which lives at
multiply by the Jacobian which lives at

x

x+v

for B but you

for A
...


The Metric Tensor
•

The metric is a (0,2) tensor eld

gij (x)

which is symmetric at any point and so its components

forms a symmetric matrix
...


•

In Riemannian geometry (where there is no time dependence) require the real eigenvalues of the
matrix at every point to be greater than 0
...


i
g ij (x) is symmetric and dened as g ij gjk = δk
...


•

The metric determines Riemannian geometry by



dening the line element
and

i

x + ∆x

i


...


 |v| = gij v v where the norm on the vectors is dened as |v|
 The angle θ between v i and ui so that |v||u|cosθ = gij v i uj at a point
...
e
...
In

E3 ,
•

i j

cartesian

diag(1, 1, 1) =spherical diag(1, r2 , r2 sin2 θ)

A 2D surface in

E 3 embedded as z = f (x, y) has induced metric gij =

1 + ( ∂f )2
∂x

∂f ∂f
∂x ∂y
The set of geometries is the set of metrics module coord transformations but for
d+1
embed a generic geometry in E

...

1 + ( ∂f )2
∂y
> z , you can't

Tensors and the metric
•

Can raise and lower indices of a tensor by using a metric
...


tensor:

i
...

gin a Tji1
...
jr

in+1
...
No information is available to deduce
the top and bottom indice interaction
...
iq
j1
...


T

i1
...
iq
a
j1
...
ir
Aij1ij2
...
iq
j1
...
The minimum curve path obeys this equation
...

Christofell symbol:
2

•

c

a

b

The Christofell symbol is a metric with indices but it is not a tensor
...

ab
ba
(ab)

connection which is symmetric in its lower indices
...


•

If you try to transform the Christofell symbol from
2

This can be checked by

gma Γm + gmb Γm
...
However, by
a
b
∂xa ∂xb
linearly combining connections the second term will dissapear resulting in a tensor equation
otherwise the Christofell symbol can be 0 in one coordinate system but not 0 in another
...


Riemann Normal (RN) coordinates
•

P look at Euclidean as possible
...
e Take a
gij (x) doesn't look like Euclidean space unless

The idea is to make a space at a point
such that

P

is at

i

x = 0
...


•

Taylor expand

•

2

∂
1
gij (x) = gij |0 + ( ∂xk gij |0 )xk + 2 ( ∂x∂ ∂xl gij |0 )xk xl +
...
ai = 0 when you choose x1 = 0 where x =0 and the
a's, b's and c's are constants
...


This can be viewed as a matrix where, for the

inverse coord transformation to exist, inverse of this matrix has to exist
...
e
...

j

gi j = M i M j gij
i
j
+ bm cimn xn + O(x2 )
j

This leads to

i
full expression
...


= bij + bij cm xi + O(x2 )
ij

where substitute of M's to give the

RN coords
•

Choose

bij

such that

bii bjj gij |0 = δi j

and, by using the Christofell symbol derivative expression

combining the two expressions gives

gi j = δi j

•

i j |x=0 = δi j
at a point, it behaves as a metric
...
can choose cm = −Γm |0 where
ab
ab
+ O(x ) = δi j + O(x 2 ) since x = bx
...
e
...

i

and

gij |0 = g0 so that bT g0 b = I
...


is symmetric so

∃

When the Christofell symbol vanishes

an orthogonal matrix
...


OT g0 O = Λ = diag(λ1 , λ2 ,
...


=

(ODO )T g0 (ODO ) = O T DT (OT g0 O)DO = O T IO

matrix then:

== =

= == =

=

=

= == = =

=

=

for any

O

=

such

O T O = I
...


and det =

•

are positive for the metric

=

so

= ==

λs

the equation for diagonalising

=

D T ΛD = I
•

where all of the

=

In spacetime, this freedom is dened as the Lorentz transform whereby SR is embedded in GR
...
Given a function φ(x), the dierential is associated to a covector
can dierentiate a sunction to obtain a tensor ∂a φ
...
Tab ≡ ∂a v b
...



...


∂2x c
Ma Mb
...


Multiplying the Christofell equation by

∂a v b +

vc

gives

The covariant derivative

0
•

b
∂xa ∂
( ∂x
∂x a ∂xa ∂xb

Therefore,

∂ x

...


a of vector v gives
resulting in the metric looking Euclidean
...
In RN coords at x = 0,
b
b
k
a v |x=0 = ∂a v |x=0 as Γ ij |0 = 0 and gij |0 = δij
...


•

1
Γbaj2 T ibj
...
iq
j1
...
iq
i1
bi2
...

i1
...
jr +Γ ab T j1 +Γ ab T j1 −Γ aj1 T bj2 −

where all, except the rst term, vanish in RN coords
...


The only way this will transform is if

∂i →

v i → ∂i v i
i
...
It
is like transforming from a curved space to at Euclidean space
...


In 3D vector calculus, have

equivalent to

∂i v i = 0

this is equivalent to


...


15



Example 2: Conservation of the stress tensor (denes momentum of momentum and angular
momentum) in cartesian Euclidean space gives
in a general geometry

•

g

ij

δ ij ∂i Tjk = 0
...


Notation: A function can be considered a (0,0) tensor, thus, in all geometeries

if

= ∂i f

Properties of the covariance derivative
1
...


i1
...

m1
...

a (A
n1
...
) = A
j1
...
B
i1
...
)B m1
...
To prove this expression, can explicitly write out the entire covariant
n
derivative expression, in terms of the Christofell symbol (which will take too long) or can

(a) Take the outer product of two tensors to give:

convert it to RN coords (to perform the calculation) and convert it back into general coords
...
1
...
1
...
1
...
1
...

gives A
(∂a B m1
...
) + (∂a Ai1
...
)B m1
...
|x=0
...

j

(b) By considering a point (say
plying the Liebnitz rule

convert back to general coords by the principle of covariance
...
e
...


(d) The problem with this method is it only works for 1 derivative
...

2
...
i
...


= ∂a gbc − Γd gdc − Γd gbd = 0
...

(b) By considering a point (x

0
...


= ∂a gbc |x=0 = ∂a (δbc +O(x2 ))|x=0 =
varies linearly, must evaluate at x = 0
...


b
b
δc = gc = g bd gdc
...


This means that

)gcd + g bc ( a gcd ) = 0
...

(e)

Γ commutes with the metric therefore, i commuted with the metric
...
) = k gij (
...
) = j (g ij
...


The Laplacian
•

It acts on a function such that:
Because

•

φ

2

φ = g ij

i

jφ

=

iφ

=

i

i

φ

is a function (0,0) tensor, its derivative is a (0,1) tensor thus,

This means that the Laplacian in a general geometry is:
where

i

2

(Due to commutation)
...


φ = g ij (∂i ∂j φ−Γk ∂k φ) =
ij

√
1
√ ∂i ( gg ij ∂j φ)
g

g = detgij
...


•

Suppose 2 points
on the points as

x1 = x(λ1 ) and x2 = x(λ2 ) on a curve x(λ)
...
The derivative is the tangent vector to the curve
...


•

In Euclidean space, if two vectors are parallel wrt a curve connecting points

x1

and

x2

then they

are parallel for any curve i
...
Straight line parallel
...

ik
ik
dλ
dλ

Parallel transporting a vector

iv

j

(x) = 0 =

dxi
j
dλ (∂i v

A geodesic (an extremal length curve between 2 points) is geometrically dened as a curve
whose tangent is parallel transported along itself
...
e
...

ui

j
iu

i

u ui = 1

=

dxi
j
dλ (∂i u

+ Γj uk ) =
ik

duj
dλ

i

+ Γj dx
ik dλ

is preserved along the geodesic

i

u

dxk
dλ
j

iu

=

d2 x
dλ2

i

+ Γj dx
ik dλ

dxk
dλ

= 0
...


d
i
along the curve to get
dλ (u ui ) = 0
d
i
i
j
To prove the above equation is true let
i (u uj )
...

Start by dierentiating the object wrt

λ

This implies that, for ane parameterisation, geodesic is the most parallel curve
...


This means

that if you parallel transport a tangent vector along a curve, it has to be pointing in the direction
of the curve but it can still change its length
...


The black grid represents the x-coord system whilst the red one represents the
From this

•

x → x = x (x)

x

coord system
...


In the Active transport, x the coords but move the points and drag the tensor elds/functions
with them
...
The latter implies that
P is the same as that at P : f (x(p )) = f (x(p))

of the function in the x-coord at

the value

17

•

No new information for either passive or active transform but, for active,

x

Example:

For passive,

and for

f

and

f

are functions

only
...


f (x) = x2 and x = x + 1
...
e
...
e
...

tensor: Tj
...

Tji1
...

j1
...

j1
...


j
...

) = Mi1
...
Tji1
...


x = x − v(x) + O( 2 )

where

is innitesimal

18

•

A function transforms as

f (x) → f (x)
f (x ) = f (x + v(x ) + O( 2 ))

•

Taylor expand this function about

f (x ) = f (x)

A tensor actively transforms by the Lie derivative:

O( 2 )

1
...
(x) → T

∂
∂x i

f + O( 2 )

i1
...
(x)

to give an active

1
...
+ Lie(v, T )i1
...


where:

 Lie(v, f ) = v i ∂i f for a function
 Lie(v, u)j = v i ∂i uj − ui ∂i v j for a vector ui
•

using the above equation to give

x to give f (x ) = f (x ) + v i
∂
f (x) → f (x) ≈ f (x) + v i ∂xi f + O( 2 )

transformation

•

where

where the second term arises from the Jacobian

1
...

Tji1
...
= (v k ∂k Tji1
...

i1
...
∂k v i1 −
...
∂j1 v k +
...
= 0
j1
...

If



vi

generates a dieomorphism which is a

that

∂
1
...

Lie(v, T )i1
...
= ∂x1 Tji1
...

2
3
d
function of x , x ,
...
, 0) so
i1
...

1

Example: Choose coords
In these coordinates

v

is

is a

discrete steps
...
If v generates
v i is dened as the Killing vector
...




2

ds2 = dθ2 +sin2 θdφ2
...
A unit sphere (S in group theory notation) has
Let

i

θ

φ

v = (v , v ) = (0, 1)

where the coordinate system is adapted to

isometry
...
This isometry, in particular, is independent of
sphere is rotated in the

φ

direction
...


Lorentz transform
•

and

A spacetime event in






t = γ(t −

F

v
c2 x)

x = γ(x − vt)
y =y

and

γ = (1 −

where the origin is shared thus:

and the same for the primed frame
...
So, what would happen if these spacetime
coordinates coincided with a dierent inertial frame? For an experiment measured by the human
eye, a ray-tracing eect diagram would be needed to account for light arriving from dierent
locations
...


•

The Lorentz transform

Λµ
ν

can be used to give:

x µ = Λµ xµ
ν

where

xµ = (t, x, y, z) and x0 =⇒ t
...
It is also NOT a general
function and, if changing the frame, a collection of non-linear terms arise
...
g
...
g
...


and to determine units of time, you have to

1
3×108

s

Applying the relativistic conditions to the Minkowski metric gives a metric of the form:

diag(−1, 1, 1, 1)
...


•

As for vector calculus in

Ed ,

points in Minkowski spacetime (in Minkowski coordinates) can be

thought of as vectors - more so 4-vectors of the form

•

v 0 ⇒ t
...

µ

The spacetime interval

dxµ

is spacelike if

gives the Proper distance between

•

where

The line element,
is given by

•

v µ = (v 0 , v i )

The interval is timelike if
and

µ

µ

x + dx

ds2 < 0

xµ

ds2 > 0 where ds2 = ηµν dxµ dxν
...


In this situation,

ds

has no meaning since, between

xµ

and

and, in this sitution,

, the Proper time is given by

√

−ds2
...
(Draw light cone diagram)

Timelike if

v µ v ν ηµν < 0
>0

Spaclike if
Lightlike if

=0

As a result of the 4-vector product and the variables being invariant, the partioning is also
invariant
...


Inertial trajectories
•

Non-accelerated massive passive particles follow straight lines deduced by the equation:

µ

µ

a +v τ



where

−dxµ dxν η

µν

dxµ = v µ dτ , the tangent to the spacetime curve is v µ
...


The norm of a spacetime parameter is

+1
•

is the proper time of the particle
...


for time
...


dxµ
µ
dτ = v (τ ) gives the instantaneous (For
an accelerating frame of reference, there is no rest frame unless you take a point at an
µ ν
instantaneous point in time) 4-velocity
...




Since

τ

is proper time:

dxµ dxν ηµν = −dτ 2

(Norm of curve equation)
...


Rest frame - Analogous to RN coords and how they work
v µ = diag(1, 0, 0, 0)

•

A frame where

•

As dened previously, an accelerated particle does not have a frame of rest but, since
can choose a point

P

v µ (τ )|P = (1, 0, 0, 0)

For an accelerating particle, since

0

v =

dx0
dτ

v µ = v µ (τ ),

where the particle travels through and, at this point, the particle is deemed

to be in instantaneous rest
...
e
...


v µ v ν ηµν = −1

at point

can write

P
...


•

pµ = (p0 , pi ) where p0 ⇒ E (Energy) which is dened as the energy of
µ
measured in frame x
...

•

At rest,

j

This is just dened as conservation of momentum
...
For slightly
4
E = p = γm ≈ mc + 1 mv 2 +
...

2
c

which results in the equation

considered slow in general),

0

2

faster motions (Those

21

Newton's laws
•

In 4-vector notation, Newton's law is expressed as
particle and

fµ

fµ =

dpµ
dτ where

τ

is the proper time of the

is the 4-force
...


•

Newton's law can also be expressed through 4-acceleration:

•

v µ v ν ηµν = −1, it constrains the 4-velocity such that v µ f ν ηµν = 0
...


•

In the rest frame:

•

Example: Alice (Observer) has 4-velocity

Since

v µ = (1, 0)

and

f µ = (0, f i )
...


relative to the lab frame, and Bob (Particle) has

µ

p
...
In Alice's rest frame,
µ
v = (1, 0, 0, 0) thus the energy Alice observes is expressed as: EAlice = −ηµν pµ v −ν = −ηµ0 pµ
0
0
- In this situation, µ = 0 is the only point where η is non-vanishing thus EAlice = −η00 p = p
since η00 = −1
...
The latter condition implies that b
µ
µ
2
µ
ν
is null/light like and, consequently, dx = b dλ is a light like interval i
...
ds = ηµν dx dx = 0
...
e
...
These transforms
take the form:

x µ = aµ +bµ xµ
µ

where

aµ

and

bµ
µ

are constants (The rst term is the translation

whilst the second represents the transformations) and are such that

•

An

O(1, 3)

matrix

gation
...
e
...
This statement implies that

expression is termed Lorentz groups
...


invariant under conju-

η = AT η A
=

= ==

where this

22



O(3) are matrices such that I = AT I A where I = diag(1,
...


c
...


This relationship means

=

•

The coordinate transformation chosen will be as such, so that the direction of time and the
orientation of space are preserved by restricting to the proper Lorentz group dened as


•

The

O(1, 3) = ±1

The determinant of

SO

and

O

whilst det

SO+ (1, 3)
...


groups are related by the equation:

O(1, 3) = SO+ (1, 3) × {1, P, T, P T }

where a

product with either of the terms in the curvy brackets can be taken
...


connected to the identity
...

µ

=

+

• SO (1, 3)

= ==

has 6 parameters and, in order to specify each element,

T

equations, in total, present since

η

=

η = A ηA

=
is symmetric ∴LHS and RHS are 4

×


...


=
matrix, it has 16 components and 16-10=6 thus 6 parameters
...

Thus, since

•

A

The Poincare transformations:

x µ = aµ +bµ xµ
µ

is the set of translations plus the proper Lorentz

transformations
...
e
...


•

The Poincare transforms are the full set of isometries for Minkowski spacetime/metric which, for
Riemannian space, include just rotations and translations
...


gµ ν = M µ M ν gµν
ν
µ

•

Under a coordinate transformation, the metric changes as usual:

•

Any Coordinate system can be used to describe Minkowski geometry, not just Monkowski coords
...


Spacetime interval
•

ds2 = gµν dxµ dxν is a geometric
2
(i)2
statement, true in any coord system
...
This line element
µ
determines that the character of an interval, dx , is spacetime
...


2

•

The line element




•

ds2

is dened as:

√
ds2 > 0 where the proper distance is ds2
√
2
−ds2
Timelike: ds < 0 where the proper time is
2
lightlike: ds = 0 where it is null
...

that gµν transforms as a tensor
...


•

A vector is dened as a direction living at a spacetime point, just as in usual geometry
...


Consequently, 2 vectors are said to be orthogonal if

A null vector is only

orthogonal to itself
...
For

•

µ

where

0

v >0

dxµ = (1, 0, 0, 0)

is also timelike
...


The future and past directed denitions are invariant for coord transformations that preserve

x

being time directed and the space being orientated
...




Example: Consider Minkowski geometry in Minkowski coords where a vector

2

given at

|v| = a
...
In special relativity, this is only possible

a point where it is future directed timelike
be achieved in order to obtain

v µ is

2

for an instantaneous rest frame
...


is future directed null, a proper LT will give

v µ = (1, 1, 0, 0)
...




2
|v| > 0 which gives proper length of the form
dxµ dxν
µ v ν which gives an ane parameter so xµ (λ) and g
gµν v
µν dλ λ = +1
...

µ

2

|v| = 0 where ane parameter has gµν dx
dλ

dxν
λ

=

0
...


vµ

µv
transported along the curve
...
The geodesics for the ane parameter

= 0
...


•

For a geodesic,

2

|v| = const
...
If this this geodesic relation holds for the whole curve, then it is timelike (or
spacelike) everywhere
...
In Minkowski coords, gµν =
which implies that there are no derivatives of the metric
...
e
...


Newton's laws
•

and

•

2
dxµ
µ
α
µ
α v (4-acceleration)
dτ and |v| = −1, a = v
µ
ma (4-force)
...


For a timelike particle at

fµ =

xµ (τ )

where

vµ =

Geometrically, 4-acceleration is the deviation of a 4-vector from being a geodesic
...
For a geodesic
and 4-acceleration is the change of this tangent wrt time
...
The matter
in question are massive
...
Neutron stars, which can cause space-time to bend
...
I
...
Regardless of what
coordinate system they are used in, they must all show that matter is conserved
...


•

In the usual case, objects with 4-indices are thought of as tensors and they must be Lorentz
invariant i
...




∂µ →

µ
...
This can be written in two ways:

∗ ∂[ν Fαβ] = 0
αβν

∂µ F µ = j ν

or

∂β Fµν = 0
...


•

As shown in a, cut a continuous piece of material in half on surface S, or remove a small spherical
piece from the centre, as shown on b
...


•

To main the separated shapes, internal forces must be applied on

•

As deduced by Cauchy, in
area
...


Ed in cartesian coords, there is a local relationship between force and
i
ij
i
is: df = σ dSj where f is the force acting on surface element dSi and

is the stress tensor, where this tensor can be diagonalised where the eigenvalues represent

the force per unit area i
...
pressure
...
In words, the stress tensor is dened as the momenA⊥j
time×A⊥j
tum ux in the i-direction, through a surface normal to j
...


´

´

 ∂i σ ij = 0 is true in cartesian coords since 0 = S df i = S σ ij dSj
...


Relativistic Stress Tensor
ij
• σREL is dened as a symmetric tensor, which can be written as Tµν
...


The Stress tensor is dened to be in the frame of rest for the matter, at some point
frame is such that the total 3-momentum vanishes at

P,

P
...


This is the denition of

the usual stress tensor with the exception that this denition is for a surface that does not
have to be in equilibrium
...


26

 T 00 =Energy density (A combination of K
...
and Rest energy)
 T 0i = T i0 = The momentum density in the i-direction
...


•

The observer at

P,

density of the form:

ν=0

and

µ=0

with 4-velocity, observes matter at the same point

pµ = −v ν T µν
...


•

In Minkowski space, in Minkowski coordinates, consider a spatial volume

•

The energy and momentum observed in the frame

´

µ

V

to have a 4-momentum

To obtain the general tensor expressions as above, substitute

d3 xT 0µ =⇒ dp =
dt
´ 3
 = − V d x∂i T iµ

´

V

xµ

V , which is deemed x
...


 =s−
•

Since

´

S

dSi T iµ

- This is a direct consequence of the divergence theorem

T i0 is dened as the momentum density in the i-direction (or
⊥ j, the rate of change of p0 is given by the the integral of

surface

the energy ux through the
energy ux over S, through

the surface
...
e
...


Tµν = Fµα Fνα −
1
αβ

...

Example: In Minkowski space, in Minkowski coordinates, the Tensor is:

Example(2): To show that

µT

µν

the Faraday tesnor is required
...


∂µ T µν = 0, the use of Maxwell's equations for
00
results in T
≈ E 2 + B 2 and T 0i ≈ E × B , which is

=0

i
...


27

Perfect Fluid Matter
•

Fluid matter is dened as Perfect if it is governed by a local energy density
4-velocity

•

uµ (x)

where

ρ(x)

and a local

uµ uµ = −1
...

•

These two points give the Stress Tensor of the form:

T µν = (ρ + P )uµ uν + P g µν

- This is valid

in general
...
g
...
This implies that
uµ = (1, v i ) where v i
1
...

0
...


•

Compressing matter increases the density, thereby increasing radiation - Also blueshift contribution involved
...
, u
•

µT

The equation of motion for a perfect uid:

µ

= 0
...


At a point

0i

µν

ρ, T = 0
T ij ≈ 0
...
For the latter equation, one can assume that

v

i

T 00 =

is very small and so,

Curved Spacetime
•

The metric is a

(0, 2)

symmetric (can diagonalise metric in order to give real values) tensor

gµν ,

as in Riemannian space
...
Consequently, det(gµν )

•

<0

implying that

gµν

is invertible
...
Note: Any curved
A geodesic is an ane parameterisation

geometry, on a small scale, can be modelled as Monkowskian
...

µ
µ αβ

so, at

x = 0:

= 0
...


Note: At

x = 0,

µ

= ∂µ
...

=

•

The laws in Minkowski coordinates, in minkowski geometry, generalised to arbitrary spacetime
is achieved through the principle of covarainace where:

 An object with 4 indices → tensor
 ∂µ → µ
 Example: Newton's force law aµ = v α
µ
curve, has a tangent

µ

v =

µ
α v
...
e
...


•

The Newtonian spacetime is dened as a small deformation of Minkowski spacetime
...


can be expressed in terms of the

with a slight perturbation added to it
...




Note: The equation above is not a tensor equation and it is only true in Newtonian frame
of reference
...


The coordinates are dened in the Newtonian frame of reference i
...


The object

moves slows but with reference to another object
...
This potential
˜ = 4πGN ρ which obey the Poisson equation, and is proportional to density
...

The Newtonian potential is dened as

where

ij

•

One requires that the matter density,

GN ρR

2

1

˜
Φ ≤ 1
...


•

Note: For the following situations, assume that there is a static potential i
...




A particle follows a curve with tangent
µ

dx
dτ =
and the right space
...
,

1
2

∂t Φ = 0
...
)

where the left side of the vector represents time

is used in, it must be able to contract with something else e
...


and

v

ν

contracted with

gµν

gives

f=

1
i j
2 δij v v

−Φ

v i → |v|

2


...
implies that gravity, along with
basic redshift, eects time dilation
...

µ

1 dv i
df
d2 xµ
i
i
2
dτ 2 = ( dτ +
...
) so a =
dτ where
is removed, the gravitational eect on f vanishes
...


If

1
2

29

Geodesics
•

d 2 xi
i
dτ 2 = a = −∂i Φ
...


•

For non-accelerated particles, we require

•

The inverse metric of

µ
g µα gαν = δν + O(

3
2

)
...


and this is given by:

g µν = η µν + 2 Φδ µν + O(

3
2

)

where

- The order of the indices of the Kronecker delta or Minkowski metric do

not matter, as long as they both equal

•

xµ (τ )

By working through ES6:

diag(1, 1, 1, 1)
...
and Γkij = (∂k Φδij −∂i Φδjk −∂j Φδik )+
...


• ηµν = diag(−1, 1, 1, 1)
•

δµν

+1 µ = ν
0
µ=ν

For the non-accelerated particle dened at the end of Lecture 21:

(
•

and

df
dτ
...


Note:

√

→

dv i
because
dτ

=

d2 xµ
dτ 2

Consequently, the geodesic equation:

√

d 2 xµ
dτ 2

=

df √ dv i
dτ ,
dτ

+
...




The time component of the geodesic equation becomes:

d 1
i j
dt ( 2 δij v v

+ Φ) = 0 Conservation

of energy in Newtonian spacetime
...


The curvature deviation eect is so small that only very massive matter, or the deviation eect
extended over a large time frame, will signify that the curvature eects are real
...


In this situation, the 4-

velocity of the two particles is:

µ
 v(1,2) =

dxµ (τ(1,2) )
dτ(1,2)



= 1 − Φ(x(1,2) ) +
...


dt
dτ(1,2)

= (1 − Φ(x(1,2) ), 0, 0, 0)
hence

represented on the following diagram:

dτ1
dτ2

where

= 1+

v(1,2)

2

= −1

Φ(x1 )−Φ(x2 )
c2

and so
...


This can be

30

•

Since spacetime is static, not including cosmological redshift, regardless of trajectory of light ray,
if shown on spacetime reference frame, the time dierences are
by

•

If particle


...


(1) emits at frequency ω1

and

(2) receives a frequency ω2 ;

ω(2)
ω(1)

≈ 1+

Φ(x1 )−Φ(x2 )
c2

+

where the frequencies are relative to the particles own proper time and the result shows

gravitational redshift
...


Note: Since

and this is dened as redshift
...
g
...


Light bending
•

Null geodesics are also deected by the curvature of spacetime
...


•

A large gravitational force can cause light to bend around it however, after moving away from
the source, the light continues in a straight line again
...


The laser/light source has wavelength

hµ → 0

whereas, at the end,

The potential is dened as

˜
Φ= √

GN M
and
x2 +(y+R)2 +z 2

continue onto

∆θ =

λ → +∞
...
g
...


•

Gravitational redshift and light bending causes deviation from the Newtonian model of spacetime
...


In 1919, Eddington (Founder to the Eddington luminosity) could see the star eld behind the
sun during the solar eclipse - evidence of light bending
...


Correction laws in curved spacetime
•

In general, there are no correction laws in curved spacetime whilst, for very small spacetimes,
that are Minkowskian, the genereal laws are reproduced
...


•

In this situation, the isometry, with the killing vector

vµ ,

is then given by:

 0 = Lie(v, g)αβ = v µ ∂µ gαβ + gαµ ∂β v µ + gµβ ∂α v µ = 2 (α vβ)
 This implies that (α vβ) is antisymmetric
...


and a timelike killing vector

d
µ
α
µ
dτ (u vµ ) = u ∂α (u vµ )
µ α
)vµ + u u (α vµ)
...
The second term

=0

by the

rule of symmetry/antisymmetry pairs
...
The timelike killing vector eld conserves
particle energy whilst the spacelike eld conserves the momentum
...
Isometries of
rotation in curved spacetime would result in conservation of angular momentum
...


killing vector eld

v µ = (1, 0, 0, 0)
...
To solve these equations, Newtonian spacetime must be used only
...
= 0
...


•

However,

[

α,

β ] vµ

ν
= Rαβµ vν = 0

where

R

is the Riemann tensor
...
The Riemann tensor is used to deduce whether the spacetime is

Minkowski or curved
...


Ideally, solve the expression from outside to in
...


The commutator acting on any tensor can be given in terms of Riemann
...


v

and

ω
...


The rst term can be chosen such that it cancels with the second term, whilst the second
term



δ
= Rαβµ vδ
...
Γ

δ
β ] Aµν = Rαβµ Aδν
k
and Γ µν = O(x) then
α,

+ Rαβνδ Aµδ
...


Symmetries of Riemann tensor
•

ν
ν
Rαβµ = −Rβαµ

The rst symmetry:

(Anti-symmetric in the rst index pair) by the denition

of the Riemann tensor
...


 [

α,

β ] gµν

δ
= Rαβµ gδν + Rαβνδ gµδ = 0 = Rαβµν + Rαβνµ

- The commutation of a

(0, 2)

tensor
...


R[αβµ] ν = 0

The third symmetry:

- The square brackets imply anti-symmetrisation of the lower

indices
...


This is due to the symmetry of the

lower indices
...


1 + 2 + 3 = 1 + 2 + Rµναβ = Rαβµν - The

last two terms indicate an interchange symmetry
...


•

Using the interchange property, the Ricci tensor is found to be symmetric i
...


•

The Ricci scalar is the contraction to the Ricci tensor

α
R ≡ Rα
...


with

33

Einstein tensor:

Gµν ≡ Rµν − 1 gµν R
2

•

The Einstein tensor is a rearrangement of the trace of the Ricci tensor
...


where

G

is the contraction of the Einstein tensor:

G = Gµν g µν
...


The latter are even permutations of

the 1st and 2nd indices
...
Can prove in general, but require
the Christofell symbols however, geodesically, can prove by using the denition:

•

[

]ν = Rν
...
R
...
R
...
∂Γ

whilst the last two terms vanish in the LIF
...


The contractions of the Bianchi identity are important






Firstly, contract
Then, contract

ν

µ

and

and

µ
µ Rναβ

ρ:

µ
µ Rα

β:

Compiling together gives:
This gives:

µ

Gµν = 0
...
e
...


•

Physically, the Riemann tensor looks like the below
...


In general,

v µ around the loop gives v ν

where

v µ −v µ =

vµ = v µ
...


•

Recalling Newtonian spacetime:

•

The Einstein equations explain why Newtonian spacetime has to be used to reproduce the given
equation above
...


This gives:

˜
δ ij ∂i ∂j Φ = 4πGN ρ
...


•

The LHS of that equation suggests that the curvature (2 derivatives of the metric - like Riemann
- Christofell requires 1) can be related to the stress tensor
...
and
K = 8πGN
...

˜
spatial laplacian of Φ
...


For a uid,

Therefore,

Gµν = 8πGN Tµν

Tµν = (ρ + P )uµ uν + P gµν where,
µ
= (1, 0, 0, 0)):

where

for non-relativistic matter (P

ρ)

moving slowly

in the Newtonian frame (u

 Ttt = ρ, Tti = Tij = 0
•

To write down an Einstein equation, it must be of the form: Curvature

Tµν

is a symmetric and conserved tensor i
...


µ

=

Const
...


1
Gµν = Rµν − 2 gµν R is symmetric and conserved µ Gµν = 0,
8πGN
it can be derived from the second derivative of the Bianchi identity
...
=8π
...
It contradicts the idea of an expanding universe but, though it is true,

Gµν =

it is negligible in size
...





From small scale laboratory measurements,

Tµν = 0

is true for all matter
...
e
...
However, this is
µ

not true
...


The Bianchi identity and the conservation of stress tensor law prove that Einsteins initial
assumption was wrong
...




For a perfect uid, the solving is easier since you only need to solve the Einstein's equations
- since they follow the Bianchi identity
...


•

The stress tensor does not fully govern the spacetime metric because it only gives information
about the trace - just like Ricci
...


35

•

Tµν = 0 therefore, the Minkowski metric is a solution since the curvature
Gµν = 0 =⇒ Rµν = 0
these do not imply that Rµναβ = 0
...
However, it is not the only solution
...


gµν = ηµν + hµν where the latter gives a small correction, as a result of
1
...


Gravity waves gravity and
the wave

Cosmology (the geometry behind it)
•

The most common cosmological principle states that there is no preferred spatial point or spatial direction in the universe
...


•

Geometrically, on the scale of large matter i
...
galaxy size (Averaging out planets and stars),
this principle imples that spacetime is homogenous (spatial translations are an isometry) and
isotropic (spatial rotations are an isometry)
...


•

t coordinates, the above initial conditions implies that: ds2 = −dt2 +a(t)2 δij dxi dxj
...
(The FLRW - Friedmann Lemoutre Robertson Walker - metric describes the universe
on a large scale)
...


It has easier

geometry
...




Since

a(t)

∗

∗



da
dt

d2 a
dt2

is a scale factor:

>0
<0

> 0

= <0


=0

=

Expanding Universe
Contracting Universe
Accelerating Expansion of Universe
Decelerating Expansion of Universe
Constant Expansion of Universe

The Hubble constant

H=

da
dt

a

is not a true constant, since it changes with

a is more important than a itself
...


this signies that the rate of change of
the approximate age of the universe

•

For a conformal time

τ

perspective:

ds2 = a(τ )2 −dτ 2 + δij dxi dxj


...


Furthermore,

Finding

a(τ )

H −1

gives

is the conformal

deformation of Minkowski space therefore the null geodesics are the same as Minkowski
...

˙

For the relationship of a perfect uid

equation is used to



•

P = wρ where w
determine a by using ρ =

Assume an equation of state

At

w = 0:

is some constant where

a0
a

3(1+w)

where

a0

ρ∝

The conservation

1
a3

Non-relativistic matter is present i
...
dust:

1
3 : Relativistic matter is present i
...
radiation also:
to redshift
...


w=

is some other constant
...
τ − τ0

a(τ ) =

is required since there is no special

time i
...
The 2 equations are invariant thus the translation is symmetric
...


Geodesics in FLRW
•

The easiest way to solve the geodesic is to not solve the equations but to vary the Lagrangian:

L=



´

dλgµν xµ xν
˙ ˙

where

L = gµν xµ xν
˙ ˙

and

To do this, parameterise the curve:

xµ =
˙

dxµ
dλ
...

˙ ˙



Vary the Lagrangian by using the E-L equation for
nothing depends on

d
dλ



∂L
∂ xµ
˙

=

x,

x

with respect to

τ
...


d
∂L
∂L
dλ ∂ xi = ∂xi
˙
This result implies momentum conservation
...


37

•

Considering the timelike geodesic

impose




xµ xν gµν
˙ ˙


+1

= −1


0

τ

component by using the E-L is possible but you need to

with additional factors
...

˙
µ
Consequently, the position is spacetime x = (τ, 0) where x x gµν = −1 (Timelike geodesic
˙µ ˙ν
dτ
1
equation) and
dt = a
...


coordinates
...
This is

cosmological horizon is dened as the furthest point of our observable universe for someone on
Earth to identify
...
In this situation:

xi
(1)

is dened as the co-moving emitter whilst

x2
(2)

is dened as Us - i
...
the observer
...







dt0 = a(t0 )dτ
is: dte = a(te )dτ

The proper time for the observer is:
The proper time for the co-emitter
The ratio of the two is dened as
If the universe is expanding,

dt0
dte

=

i

v
δij v i v j , a2

a(t0 )
a(te )

a(te ) < a(t0 ) and so, dt0 > dte which means, from the principles

of redshift, the frequency of the light at the observer is lower than when it was emitted
...


38

Schwarzschild metric
•

Rs
dt2
r
metric denes a static, uncharged black hole
...


This

Tµν = 0 →A

vacuum solution and

=⇒ Rµν = 0
...


spherically symmetric

•

ds2 = − 1 −

Rµναβ Rµναβ =

the solution is that of a BH singularity which has

∞

2
12Rs
r6

curvature
...
e
...
At this

region, curvature is nite but, when

r < Rs

the spacelike coordinates become time like and

vice versa
...
e
...




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