Appendix Galilean transformation of work done (section 2.2.6)In Newtonian physics, work done is a transfer of energy under the action of a force, governed

by the equation:

F ∆ x=12

mu22−1

2m u1

2

With F being the force acting, ∆ x=x2−x1 the distance the force acts over and u1 and u2 the

initial and final velocities respectively.

If this equation is to retain its form after transformation, then we must have:

F ' ∆ x '=12

m (u2' )2−1

2m (u1

' )2

Where:

F '=F

∆ x '=x2' −x1

' =( x2−v t 2 )−( x1−v t1 )=∆ x−v (t 2−t 1 )

(u2' )2

=(u2−v )2=u22+v2−2u2 v

(u1' )2

=(u1−v )2=u12+v2−2u1 v

So that the left-hand side is:

F ' ∆ x '=F ∆ x−Fv (t 2−t1 )

and the right-hand side:

12

m (u2' )2−1

2m (u1

' )2=( 12

m u22+1

2m v2−mu2 v )−( 1

2mu1

2+ 12

m v2−m u1 v)Simplifying and gathering terms together:

F ∆ x−Fv (t 2−t1 )=12

mu22−1

2mu1

2−mv (u2−u1 )

Now we notice that, given Newtons second law of motion, F=ma:

F ∆ x−Fv (t 2−t1 )=F ∆ x−mav (t2−t1 )=F ∆ x−mv (u2−u1 )

Hence:

F ∆ x−mv (u2−u1 )=12

mu22−1

2mu1

2−mv (u2−u1 )

or

F ∆ x=12

mu22−1

2m u1

2

which is the same relationship, without dashed variables. This confirms that the work done

equation is covariant under the Galilean transformation (covariant in this context meaning

that it retains its mathematical form).

Spherical light waves in two systems (section 2.4)A spherical light wave is spreading out from the origin of system S, so that by time t the

wave is cutting the x axis at x1=ct and x2=−ct .

Viewed from the S' system moving at speed v relative to S, these locations are:

x1' =γ ( x1−vt )=γ (ct−vt )=γt (c−v )

x2' =γ ( x2−vt )=γ (−ct−vt )=−γt ( c+v )

so they are not equally spaced from the origin in S'. This is because the two events are not

simultaneous in S'. In fact, they take place at:

t 1' =γ ( t−x1 v /c2 )=γ (t−ctv /c2)=γt (1−v /c )

and

t 2' =γ ( t−x2 v /c2)=γ (t+ctv /c2 )=γt (1+v /c )

Clearly t 1' is an earlier time than t 2

' , by an amount 2 γtv /c:

t 2' −t 1

' =γt (1+v /c )−γt (1−v /c )=2 γtv /c

In this time, the light wave will have moved an extra distance c × (2 γtv /c )=2 γtv , meaning

that it will now have progressed to a position:

x1' +2 γtv=γt (c−v )+2 γtv=γt (c+v )

which is the same distance away from the S'origin as x2' . Hence, the wave appears to be a

sphere centred on the S' origin, when each part of the wave is viewed at the same time in S'.

Rotating axes (section 3.1)

Figure A.1 – Rotating co-ordinate axes and the polar co-ordinates of a point

The most elegant way of confirming the rotation transformations is to consider the polar co-

ordinates for a point, as labelled in Figure A.1. The marked point has co-ordinates

( X , Y )=(7 ,7 ) in Cartesian co-ordinates and ( R , α ) in polar co-ordinates, related by:

X=R cos (α )Y =R sin (α )

In the rotated ( X ' , Y ' ) system, the polar co-ordinates are ( R ,α−θ ) so that:

X '=R cos (α−θ ) Y '=R sin (α−θ )

Using the trigonometric relationships:

sin ( A−B )=sin ( A ) cos ( B )−cos ( A ) sin (B )

cos ( A−B )=cos ( A ) cos (B )+sin ( A ) sin B

gives:

X '=R {cos (α )cos (θ )+sin (α )sin (θ ) }=X cos (θ )+Y sin (θ )

Y '=R {sin (α ) cos (θ )−cos (α )sin (θ ) }=Y cos (θ )−X sin (θ )

Properties of the Minkowski Metric (section 3.5)In chapter 3 we used the relationship:

LT ηL=ηor LηL=η

to find the matrix, η=(1 00 −1).

Now we need to check that the matrix does obey the relationship that we used to define its

properties:

LηL=( γ −βγ−βγ γ )(1 0

0 −1)( γ −βγ−βγ γ )=( γ −βγ

−βγ γ )( γ −βγβγ −γ )=( γ 2−β2 γ2 −β γ2+β γ2

−β γ2+β γ2 β2 γ 2−γ2 )=(γ 2 (1−β2) 00 −γ 2 ( 1−β2 ))=(1 0

0 −1)=η

as hoped.

Boosting 4-velocities (section 3.7.1)We are attempting to prove the Lorentz transformation of the 4-velocity:

γu' u'=γ v ( γu u−βv γu c )

Starting with the right-hand side:

γv (γ uu−βv γ uc )=γ v γu (u−βv c )=γ v γ u (u−v )

From the transformation of 3-velocities we know that:

u'= u−v

1−uvc2

=c2 (u−v )c2−uv

so (u−v )=( c2−uv )u '

c2

hence:

γv (γ uu−βv γ uc )=γ v γu (u−v )=γv γ u ( c2−uv )u '

c2

Now:

γv γ u=1

√1− v2

c2

× 1

√1−u2

c2

= 1

√ c2−v2

c2

× 1

√ c2−u2

c2

= c2

√( c2−v2 ) (c2−u2 )

Inserting this into the result so far, we obtain:

γv (γ uu−βv γ uc )=γ v γu (c2−uv ) u'

c2 =( c2−uv) u'

√ (c2−v2 ) ( c2−u2 )

Next, we explore (c2−v2 ) (c2−u2 ) by expanding it out:

(c2−v2 ) (c2−u2 )=c4−c2u2−c2 v2+u2 v2=( c4+u2 v2 )−c2u2−c2 v2

Taking the hint from the numerator of our previous result, we note that:

(c2−uv )2=c4+u2 v2−2c2 uv

making:

c4+u2 v2=(c2−uv )2+2 c2uv

Consequently, we can substitute to give:

(c2−v2 ) (c2−u2 )=(c2−uv )2+2 c2uv−c2 u2−c2 v2=( c2−uv )2−c2 (u2+v2−2 uv )= (c2−uv )2−c2 (u−v )2

Factorising by (c2−uv )2 we get:

(c2−v2 ) (c2−u2 )=(c2−uv )2 {1− (u−v )2

( c2−uv)2 }=( c2−uv )2{1− (u−v )2

c2(1−uvc2 )

2 }so that:

(c2−v2 ) (c2−u2 )=(c2−uv )2 {1− (u ')2

c2 }Returning to our previous result from a few lines ago:

γv (γ uu−βv γ uc )=( c2−uv )u '

√( c2−v2 ) (c2−u2 )=

(c2−uv ) u'

√ (c2−uv )2 {1− (u ' )2

c2 }=

(c2−uv ) u'

(c2−uv )√1−(u' )2

c2

=γu' u'

completing the proof.

We can also show that γu' c=γ v ( γu c−β v γ uu ) as:

γv (γ u c−βv γu u )=γ v γu(c−uvc )= c γ v γu (c2−uv )

c2

Remembering that:

γv γ u=c2

√( c2−v2 ) (c2−u2)

produces:

γv (γ u c−βv γu u )= c2× c (c2−uv )c2 √ (c2−v2 ) ( c2−u2 )

or:

γv (γ u c−βv γu u )= c ( c2−uv )√( c2−v2 ) (c2−u2 )

Bringing back the result:

(c2−v2 ) (c2−u2 )=(c2−uv )2 {1− (u ')2

c2 }gives:

γv (γ u c−βv γu u )= c ( c2−uv )

√( c2−uv )2{1− (u' )2

c2 }= c

√1−(u ' )2

c2

=γ u' c

Hence, we have shown that:

( γ u' cγu' u ')=( γ −βγ

−βγ γ )(γu cγu u)

confirming the 4-vector nature of (γu cγu u).

Confirming the covariant transformation of the metric (section 7.5.2)Using our summation notation, the transformation of the Minkowski metric, η, takes the

form:

ηkl' =∑

i , j( L−1)ki ( L−1 )ljηij

Expanding this out gives (restricting to a 2x2 version of the metric):

ηkl' =∑

i( L−1)ki {( L−1 )l 0 ηi 0+( L−1 )l 1 ηi 1 }=( L−1 )k 0 ( L−1 )l 0η00+( L−1)k 0 ( L−1) l1 η01+( L−1 )k 1 ( L−1 )l 0 η10+( L−1 )k 1 ( L−1) l1 η11

As η01=η10=0, this reduces to:

ηkl' =( L−1 )k 0 ( L−1 )l 0 η00+( L−1 )k1 ( L−1 )l 1 η11

The transformation L−1 has the following components:

( L−1 )00=γ ( L−1 )01=βγ ( L−1 )10=βγ ( L−1 )11=γ

so, we can pick our way through the metric component transformations:

η00' = ( L−1 )00 ( L−1)00η00+( L−1)01 ( L−1 )01 η11=( γ ) (γ ) (+1 )+ ( βγ ) ( βγ ) (−1 )=1

η01' =( L−1 )00 ( L−1 )10 η00+( L−1 )01 ( L−1 )11 η11=(γ ) ( βγ ) (+1 )+( βγ ) ( γ ) (−1 )=0

η10' =( L−1 )10 ( L−1 )00η00+( L−1 )11 ( L−1 )01η11=( βγ ) (γ ) (+1 )+( γ ) ( βγ ) (−1 )=0

η11' =( L−1 )10 ( L−1 )10η00+( L−1 )11 ( L−1 )11 η11=( βγ ) ( βγ ) (+1 )+(γ ) ( γ ) (−1 )=−1

confirming that the metric transforms covariantly.

Just to be double sure of this, we can try a covariant transformation instead. Now the

transformation components are:

L00=γ L01=−βγ L10=−βγ L11=γ

and the expansion is:

ηkl' =Lk 0 Ll 0η00+Lk 1 Ll 1 η11

term by term, this gives us:

η00' =L00 L00η00+L01 L00η11=(γ ) (γ ) (+1 )+(−βγ ) ( γ ) (−1 )=γ2+β γ 2

η01' =L00 L10η00+L01 L11 η11=(γ ) (−βγ ) (+1 )+(−βγ ) ( γ ) (−1 )=0

η10' =L10 L00η00+L11 L01η11=(−βγ ) ( γ ) (+1 )+(γ ) (−βγ ) (−1 )=0

η11' =L10 L10 η00+L11 L11 η11=(−βγ ) (−βγ ) (+1 )+( γ ) (γ ) (−1 )=−( γ2+β γ 2)

which is clearly incorrect.

Rates of change for a covariant vector (section 7.6)We start from the tensor product of a covariant and contravariant vector:

S=∑b ,c

Ab Bc

Then we find the rate of change of this quantity in general co-ordinates:

Da(∑b , cAb Bc )=∑

b ,c{Da ( Ab ) Bc+Ab Da (B c) }

The first rate of change in the bracket is what we are trying to find. The second we already

know how to do, so we carry this out:

Da(∑b , cAb Bc )=∑

b ,c {D a ( Ab ) Bc+ Ab [∆ Bc

∆ xa+∑

dΓ da

c Bd ]}Now we contract both sides by setting b=c:

Da(∑bAb Bb)=∑

b {Da ( Ab ) Bb+ Ab[ ∆ Bb

∆ xa+∑

dΓ da

b Bd ]}and split the right-hand side, to allow us to play the shell game again, by changing indices:

d →b, b → d in the final term:

Da(∑bAb Bb)=∑

bDa ( Ab ) Bb+∑

bAb

∆ Bb

∆ xa+∑

d∑

bAd Γ ba

d Bb

The quantity on the left-hand side, ∑b

Ab Bb, is a scalar, which has an ordinary rate of

change, so:

Da(∑bAb Bb)=∑

b {(∆ Ab

∆ xa )Bb+ Ab(∆ Bb

∆ xa )}Equating, we can see that one term cancels:

∑b {( ∆ Ab

∆ xa )Bb+Ab( ∆ Bb

∆ xa )}=∑b

Da ( Ab ) Bb+∑b

Ab∆ Bb

∆ xa+∑

d∑

bAd Γba

d Bb

so that:

∑b (∆ Ab

∆ xa) Bb=∑

bDa ( Ab ) Bb+∑

d∑

bAd Γ ba

d Bb

Pulling terms to the same side and factorizing:

∑b {( ∆ Ab

∆ xa )−Da ( Ab )−∑d

Ad Γbad }Bb=0

As this must be true no matter what we choose for B, the term in the bracket has to be zero:

( ∆ Ab

∆ xa)−Da ( Ab )−∑

dAd Γ ba

d =0

Hence:

Da ( Ab )=( ∆ Ab

∆ xa)−∑

dAd Γba

d

which is the result we were looking for.

Deriving the Riemann Tensor (section 7.6)We are intending to calculate:

DΔ xb ( D (V )

Δ xc )− DΔ xc (D (V )

Δ xb )which, at a component level is:

Db (Dc (V a ))−Dc ( Db (V a ) )=Db (tensor∈c ,a )−D c ( tensor∈b , a )

so, we need to know how to find the rates of change of a tensor: so far, we have only done

this for vectors. Firstly, we need the rule to find the rate of change of a covariant vector,

which is:

Dk ( Ai )=D ( Ai )∆ xk

=∆ A i

∆ xk−∑

mAm Γ ik

m covariant

Dk ( Ai )=D ( Ai )∆ xk

=∆ A i

∆ xk+∑

mAm Γ mk

i contravariant (Eq 7.1)

and I have placed Eq 7.1 underneath for comparison. This particular rule has caused me some

pain in the past, so I advise you to note the minus sign and the order of indices in the

Christoffel symbol.

Armed with this, we can now figure out how to find rates of change of tensors. After all, as

we found with the energy density tensor, you can build tensors from the tensor product of

vectors, and the rule must be the same no matter how the tensor gets assembled. Starting

from:

Dk (C i j )=D k ( A i B j )=Dk ( A i ) B j+ Ai Dk (B j )

the separate rates of change are:

Dk ( Ai )=∆ A i

∆ xk−∑

mAm Γ ik

m∧Dk ( B j )=∆ B j

∆ xk+∑

mBm Γ mk

j

so, we plug these in to get:

Dk (C i j )={∆ Ai

∆ xk−∑

mAm Γ ik

m}B j+ A i{∆ B j

∆ xk+∑

mBm Γ mk

j }Gathering terms together:

Dk (C i j )={∆ Ai

∆ xkB j+ A i

∆ B j

∆ xk}−∑

mAm B j Γ ik

m+∑m

Ai Bm Γ mkj

and re-assembling the tensor, shows the final rule:

Dk (C i j )=∆ C i j

∆ xk−∑

mCm j Γ ik

m+∑m

Ci m Γmkj

In other words, to find the rate of change of a tensor, you calculate the component rate of

change and then subtract a Christoffel symbol for every covariant index and add one for

every contravariant index.

Using this, we can construct the first part of the Riemann tensor:

Db (Dc (V i ))=∆ ( Dc ( V i ))

∆ xb−∑

mDm (V i ) Γcb

m +∑m

Dc (V m ) Γ mbi

Now things are about to get a little messy, as we have to plug in:

Dc (V i )=∆ V i

∆ xc+∑

nV n Γ nc

i

to give us:

Db (Dc (V i ))=∆( ∆ V i

∆ xc+∑

nV n Γ nc

i )∆ xb

−∑m {∆ V i

∆ xm+∑

nV n Γnm

i }Γ cbm +∑

m {∆ V m

∆ xc+∑

nV n Γ nc

m }Γ mbi

Splitting up the first term, and switching the sums over m and n in the last couple of terms

results in:

Dc ( Db (V i ))=∆

∆ xb (∆ V i

∆ xc )+∑n

∆ (V n )∆ xb

Γnci +∑

nV n

∆ ( Γnci )

∆ xb−∑

n {∆ V i

∆ xn+∑

mV m Γmn

i }Γ cbn +∑

n {∆ V n

∆ xc+∑

mV m Γ mc

n }Γnbi

Next, we calculate Dc ( Db (V i )), which is easy, as all we need to do is switch c and b in the

previous result:

Dc ( Db (V i ))= ∆∆ xc (∆ V i

∆ xb )+∑n

∆ (V n )∆ xc

Γ nbi +∑

nV n

∆ ( Γnbi )

∆ xc−∑

n {∆ V i

∆ xn+∑

mV m Γ mn

i }Γbcn +∑

n {∆V n

∆ xb+∑

mV m Γmb

n }Γnci

If we put the two expressions underneath each other, we can see which terms will cancel

when we subtract one from the other (remembering that Γcbn =Γ bc

n ):

Db (Dc (V i ))= ∆∆ xb (

∆ V i

∆ xc )+∑n

∆ (V n )∆ xb

Γnci +∑

nV n

∆ ( Γnci )

∆ xb−∑

n {∆ V i

∆ xn+∑

mV m Γmn

i }Γ cbn +∑

n {∆ V n

∆ xc+∑

mV m Γ mc

n }Γnbi

Dc ( Db (V i ))=∆

∆ xc (∆ V i

∆ xb )+∑n

∆ (V n )∆ xc

Γ nbi +∑

nV n

∆ ( Γnbi )

∆ xc−∑

n {∆ V i

∆ xn+∑

mV m Γ mn

i }Γbcn +∑

n {∆V n

∆ xb+∑

mV m Γmb

n }Γnci

Hence, when we do the subtraction we are left with:

Db (Dc (V i ))−Dc (Db (V i) )=∑n

V n

∆ (Γ nci )

∆ xb+∑

n∑

mV m Γ mc

n Γnbi −∑

nV n

∆ (Γ nbi )

∆ xc−∑

n∑

mV m Γ mb

n Γ nci

This can be tidied up by grouping terms:

Db (Dc (V i ))−Dc (Db (V i) )=∑n

V n{∆ (Γnci )

∆ xb−

∆ (Γ nbi )

∆ xc}+∑

n , mV m {Γmc

n Γ nbi −Γmb

n Γnci }

Switching the summation in the first term on the right-hand side so that it is over m rather

than n allows us to group things further:

Db (Dc (V i ))−Dc (Db (V i) )=∑m

{∆ (Γ mci )

∆ xb−

∆ (Γmbi )

∆ xc+∑

n( Γmc

n Γ nbi −Γmb

n Γnci )}V m

Pulling out our definition of the Riemann tensor:

DΔ xc

( D (V i )Δ xb

)− DΔ xb

( D ( V i )Δ xc

)=∑a

R i abcV a

and remembering that a sum over a is equivalent to a sum over m, we have, finally:

Ri abc=∆ (Γ ac

i )∆ xb

−∆ ( Γab

i )∆ xc

+∑n

(Γ acn Γnb

i −Γ abn Γ nc

i )

and breath….

Confirming the approximate inverse metric in the weak field limit (section 8.1)In section 8.1 I introduced the weak field metric gab=ηab+hab and suggested that the inverse

of this metric was given by ( g−1)ab=(η−1 )ab−hab. The best way to demonstrate this is to try

multiplying the two metrics together and show that the unit matrix is produced. Hence, we

start with:

δ ij=∑k

( g−1 )ik gkj=∑k

{(η−1 )ik−hi k }{ηkj+hkj }

where I have used the standard shorthand for the components of the unit matrix,δ ij, which has

δ ij=0 if i≠ j and δ ij=1 if i= j.

Expanding out the brackets gives:

δ ij=∑k

{( η−1 )ik−hi k } {ηkj+hkj }=∑k

{( η−1 )ik ηkj+(η−1 )ik hkj−hi k ηkj−hi k hkj }

Normally, we turn a contravariant index into a covariant one via the metric:

hi j=∑k

gkj h ik

however, in the weak field approximation we are entitled to reduce gkjto just ηkj as using the

full gkj=ηkj+hkj would generate terms h2 which we could in any case neglect.

A covariant index is turned into a contravariant via the inverse metric:

hij=∑k

( g−1 )kj hi k

If we take it that the hk j in ( g−1)kj=( η−1 )kj−hk j is h, then the same argument tells us that we

can get away with just using (η−1 )kj in this case.

Now we have for each term in the summation:

∑k

( η−1 )ik ηkj=δij definition of inverses∑

k( η−1 )ik hkj=hi jcreating a covariant index in the weak field

approximation∑

k−hi k ηkj=−hi j creating a contravariant index

And ∑k

hi k hkj h2 which by the rules of our approximation, can be ignored.

Hence, we have:

δ ij=δ ij+h i j−hi j+O (h2 )=δij

Rate of change of gravitational potential (section 8.1.4)

Calculating ∆∆ r (−2G M

c2 r ) involves the following steps:

∆∆ r (−2GM

c2r )=−2c2

( GMr+∆ r

−GMr )

∆ r=−2

c2

GMr−GM (r+∆ r )r (r+∆ r )

∆ r= 2

c2GM ∆ r

r (r+∆ r )∆ r

so that:

∆∆ r (−2GM

c2r )= 2c2

GM ∆ rr (r+∆ r ) ∆ r

= 2c2

GMr (r+∆ r )

= 2c2

GMr2

as r2+r ∆ r ≈ r2.

Linear Gravity (section 11.2)Picking up the argument from chapter 11, having defined the linear Riemann tensor:

Ri abc ≈∆ (Γ ac

i )∆ xb

−∆ ( Γab

i )∆ xc

and the linear Ricci tensor

Rab ≈∑i

Ri abi=∑i

{∆ (Γ aii )

∆ xb−

∆ ( Γabi )

∆ xi}

the next step is to insert the appropriate Christoffel symbols:

Γ aii =1

2∑mηℑ{∆ hmi

∆ xa+

∆ ham

∆ x i−

∆ hai

∆ xm}

Γ abi =1

2∑mηℑ {∆ hmb

∆ xa+

∆ ham

∆ xb−

∆ hab

∆ xm}

into the Ricci tensor:

Rab=12∑i

∆∆ xb (∑m

ηℑ{∆ hmi

∆ xa+

∆ ham

∆ xi−

∆ hai

∆ xm})−1

2∑i

∆∆ x i (∑m

ηℑ{∆ hmb

∆ xa+

∆ ham

∆ xb−

∆ hab

∆ xm})

Now we explore the following expression:

∆∆ x j (∑n

η¿ An)=∑n {∆ η¿

∆ x jAn+η¿

∆ An

∆ x j}=∑

nη¿

∆ An

∆ x j

Where the cancelled term is zero as the elements of η are constant. This allows us to move

the metric in and out of a rate of change as is convenient. Using this to make to some re-

arrangements produces:

Rab=12∑i , m

ηℑ∆

∆ xb( ∆ hmi

∆ xa+

∆ ham

∆ x i−

∆ hai

∆ xm)−1

2∑i , mηℑ

∆∆ x i

(∆ hmb

∆ xa+

∆ ham

∆ xb−

∆ hab

∆ xm)

where the indicated terms cancel.

If we examine ∑i ,m

ηℑ∆

∆ x i( ∆ hab

∆ xm), and explicitly display the sum over m:

∑i∑

mηℑ

∆∆ x i (

∆ hab

∆ xm )=∑i {ηi 0∆

∆ x i (∆ hab

∆ x0 )+ηi 1∆

∆ x i (∆ hab

∆ x1 )+ηi 2∆

∆ xi (∆ hab

∆ x2 )+ηi 3∆

∆ x i (∆ hab

∆ x3 )}we can see that when we carry out the sum over i, the terms will vanish unless i takes the

same value as the m in that term. In other words:

∑i∑

mηℑ

∆∆ x i (

∆ hab

∆ xm )={η00∆

∆ x0 (∆ hab

∆ x0 )+η11∆

∆ x1 (∆ hab

∆ x1 )+η22∆

∆ x2 (∆ hab

∆ x2 )+η33∆

∆ x3 (∆ hab

∆ x3 )}Substituting in the values of the various metric components converts the sum to:

∑i∑

mηℑ

∆∆ x i

( ∆ hab

∆ xm)= ∆

∆ x0( ∆ hab

∆ x0)− ∆

∆ x1(∆ hab

∆ x1)− ∆

∆ x2(∆ hab

∆ x2)− ∆

∆ x3( ∆ hab

∆ x3)

As lovers of compact notation, and because this particular combination of rates of change

comes up quite often, mathematicians write:

2 ( ϕ)=∑u , v

guv∆

∆ xu ( ∆ ϕ∆ xv )= ∆

∆ xo ( ∆ ϕ∆ xo )−∇2 ϕ

Using this notation, we have:

2 (hab )= ∆∆ x0

( ∆ hab

∆ x0)−∇2 (hab)= ∆

∆ x0( ∆ hab

∆ x0)−∑

i

∆∆ x i

( ∆ hab

∆ xi)

and a Ricci tensor that looks like:

Rab=12 [∑i , m

ηℑ∆

∆ xb (∆ hmi

∆ xa−

∆ hai

∆ xm )−∑i ,m

ηℑ∆

∆ x i (∆ hmb

∆ xa )+ 2 (hab )]We can simplify things even more by defining h=∑

i, mηℑhmi, which turns the first term into:

∆∆ xb ( ∆ h

∆ xa ) With the second term, we can take the metric inside the rate of change, like this:

∑i∑

mηℑ

∆∆ xb

(∆ hai

∆ xm)=∑i , m

∆∆ xb

(ηℑ

∆ hai

∆ xm)=¿∑

i ,m

∆∆ xb

( ∆ ηmi hai

∆ xm)=∑

m

∆∆ xb

( ∆ ha m

∆ xm)¿

and we have also exploited the symmetry of the metric. Now the Ricci tensor looks like:

Rab=12 [ ∆

∆ xb ( ∆ h∆ xa )+ 2 ( hab )−∑

m

∆∆ xb (

∆ ha m

∆ xm )−∑i ,m

ηℑ∆

∆ x i (∆ hmb

∆ xa )]We can pull the same trick with the last term:

∑i ,m

ηℑ∆

∆ x i( ∆ hmb

∆ xa)=∑

i , m

∆∆ x i

ηℑ(∆ hmb

∆ xa)=∑i , m

∆∆ xi

( ∆ ηmi hmb

∆ xa)=∑

i

∆∆ x i

(∆ h ib

∆ xa)

simplifying the tensor still further:

Rab=12 [ ∆

∆ xb ( ∆ h∆ xa )+❑2 (hab )−∑

i

∆∆ x i (

∆ h ib

∆ xa )−∑m

∆∆ xb (

∆ ha m

∆ xm )]Now we define a new metric by H ub=hub−

12

ηubh and then multiply by η and sum to convert

one of the indices:

∑u

ηau Hub=∑u

ηau hub−12∑u

ηau ηub h

resulting in:

H a b=hab−12∑u

ηauηubh

Next, we find a rate of change:

∆ Hab

∆ x j=

∆ hab

∆ x j−1

2∑uηauηub

∆ h∆ x j

and further exploit our co-ordinate freedom pick H such that:

∑a

∆ H ab

∆ xa=0

this helps us nicely as then:

∑a

∆ ha b

∆ xa=1

2∑u , aηau ηub

∆ h∆ xa

=12

∆ h∆ xb

We can deploy this in the last two terms of the Ricci tensor:

Rab=12 [ ∆

∆ xb( ∆ h∆ xa

)+❑2 (hab )−∑i

∆∆ x i

(∆ h ib

∆ xa)−∑

m

∆∆ xb

( ∆ ha m

∆ xm)]

as:

∑i

∆∆ x i

( ∆ hi b

∆ xa)= ∆

∆ xa(∑i

∆ hi b

∆ x i)=1

2∆

∆ xa( ∆ h

∆ xb)

and:

∑m

∆∆ xb

( ∆ ha m

∆ xm)= ∆

∆ xb(∑m

∆ ham

∆ xm)=1

2∆

∆ xb ( ∆ h∆ xa )=1

2∆

∆ xa ( ∆ h∆ xb )

giving:

Rab=12 [ ∆

∆ xb ( ∆ h∆ xa )+❑2 (hab )−1

2∆

∆ xb ( ∆ h∆ xa )−1

2∆

∆ xb ( ∆ h∆ xa )]=1

2❑2 (hab )

as required.

Field equation solution (section 11.4.1)In chapter 11, I asserted that ∇2 ( H ab )=0 has a solution:

H ab ≈Cab

r

The quickest way to justify this, albeit in a slightly hand-wavy manner, is to start from the

spherical-polar version of ∇2 H ab,which makes the equation:

∇2 ( H ab )= 1r2

∆∆ r (r2 ∆ H ab

∆ r )+ 1r 2sin θ

∆∆ θ (sinθ

∆ H ab

∆ θ )+ 1r2 sin2θ

∆∆ φ ( ∆ Hab

∆ φ )=0

Assuming that H ab=H ab (r ) (i.e. that H is only a function of r), the equation reduces to:

∇2 ( H ab )= 1r2

∆∆ r (r2 ∆ H ab

∆ r )=0

If we suggest that H ab ≈Cab

r , with the Cab being a collection of constants, then our job is to

show that:

∇2 ( H ab )=∇2( Cab

r )= 1r2

∆∆ r (r2 ∆ (Cab /r )

∆ r )=0

Firstly, if H ab=Cab

r, then:

∆ Hab

∆ r= ∆

∆ r (Cab

r )=−Cab

r2

which is a standard result. Using this we get:

−1r2

∆∆ r (r2 Cab

r2 )= 1r2

∆∆ r ( Cab )=0

as the Cab are constant. Hence, we have ∇2 ( H ab )=0 as required

Isotropic implies homogeneous (section 12.2.2)Let’s take two observers, A and B and construct a sphere about A . By isotropy, the density

must be the same at all points on the surface of that sphere. A similar sphere about B is

constructed to intersect with that about A (Figure A.2). Points C and D have the same density

of material due to the isotropy about A, so point E must have the same density as the others

due to the isotropy about B. Indeed, all points along the arc between C and D through E must

have the same density. If we then play with the radii of the spheres and the location of the

observers, we can sweep through the whole universe showing every point to be the same,

establishing homogeneity.

Figure A.2 - – isotropy implies homogeneity. Points C and D have the same density due to isotropy about observer A.

Isotropy about observer B shows that E must have the same density as C and D. Moving the overlapping spheres around

allow us to cover the universe.

A closed universe (section 12.5)Taking k>0, ρΛ=0 and assuming a matter dominated universe filled with non-interacting

dust, so that P=0, the first Friedmann equation is:

( 1A ( ∆ A

∆ t ))2

=8 πG3

ρ0

A3 −k c2

A2

Re-arranging gives:

∆ A∆ t =A√ 8 πG

3ρ0

A3 −k c2

A2 =c√ 8 πG3 c2

ρ0

A −k

Now write ∆ ζ= c ∆ tA (which is a trick known as changing to conformal time) and process the

changes:

∫ζ 0

ζ

d ζ =∫0

A da

A √ 8πG3c2

ρ0

A −k

=∫0

A dA

√ 8πG3 c2 Aρ0−k A2

=[ 1√k

sin−1( 2 k A−8πG3 c2 ρ0

8 πG3c2 ρ0 )]

0

A

Hence:

ζ −ζ 0=1√k [sin−1( 0−8 πG

3 c2 ρ0

8 πG3 c2 ρ0 )−sin−1(2k A−8 πG

3 c2 ρ0

8 πG3 c2 ρ0 )]=−1

√k [ π2+sin−1( 2k A−8 πG

3 c2 ρ0

8 πG3 c2 ρ0 )]

We impose ζ =0 at a=0, so that ζ 0=0 so that:

√k ζ + π2=−sin−1( 2 k A−

8 πG3 c2 ρ0

8 πG3c2 ρ0 )

Inverting this:

sin(√k ζ + π2 )=¿

8 πG3 c2 ρ0−2k A

8πG3c2 ρ0

¿

Massaging further:

8 πG3 c2 ρ0sin (√k ζ + π

2 )=¿ 8 πG3 c2 ρ0−2 k A ¿

A=4 πG3 k c2 ρ0−

4 πG3 k c2 ρ0sin (√k ζ + π

2 )

A=4 πG3 k c2 ρ0 (1−cos (√k ζ ) )

Then, going back to ∆ ζ= c ∆ tA in the form c∫ dt=∫ Adζ tells us that:

ct=∫ 4 πG3 k c2 ρ0 (1−cos (√k ζ ) )dζ = 4πG

3 k c2 ρ0∫ (1−cos (√k ζ ) ) dζ

so that:

ct= 4 πG3k c2 ρ0[ζ − 1

√ksin (√k ζ )]

ζ 0

ζ

= 4 πG3k c2 ρ0(ζ − 1

√ksin (√k ζ ))

Plotting A against ct shows that this universe is both spatially and temporally closed: it

expands to a maximum scale factor and then re-contracts back into a Big Crunch (Figure

A.3).

Figure A.3 – a closed universe, both axes are shown in units of 4 πG3k c2 ρ0

Incidentally, the switch to conformal time turns the metric:

( Δs )2=c2 ( Δt )2−A2 (t )[ (∆ r )2

1−k r2 +r 2 (∆ θ )2+r2sin2 (θ ) (∆ ϕ )2]into:

( Δs )2=A2[ ( Δζ )2− (∆ r )2

1−k r2 +r2 (∆ θ )2+r 2sin2 (θ ) (∆ ϕ )2]

Killing vector condition (section 13.3.1)Starting from the equation in the book:

gab ( x )=gab ( x ')+ε∑v

∆ K v

∆ xbgav ( x' )+ε∑

u

∆ K u

∆ xagub ( x ' )

From standard rate of change theory (up to first order) we have:

gab ( x ' )=gab ( x+εK ) ≈ gab ( x )+ε∑j

K j

∆ gab (x )∆ x j

Substituting this in gets a bit messy

gab ( x )=gab ( x )+ε∑j

K j

∆ gab (x )∆ x j

+ε∑v

∆ K v

∆ xb(gav (x )+ε∑

jK j

∆ gav ( x )∆ x j

)+ε∑u

∆ Ku

∆ xa(gub (x )+ε∑

jK j

∆ gub ( x )∆ x j

)but we can quickly clear up some of the mess by dropping terms in ε 2 to give:

gab ( x )=gab ( x )+ε∑j

K j

∆ gab (x )∆ x j

+ε∑v

∆ K v

∆ xbgav ( x )+ε∑

u

∆ Ku

∆ xagub (x )

Or, dropping the superfluous ε :

∑j

K j

∆ gab ( x )∆ x j

+∑v

∆ K v

∆ xbgav ( x )+∑

u

∆ K u

∆ xagub ( x )=0

Now

∑v

∆ K v

∆ xbgav ( x )=∑

v

∆ ( K v gav ( x ) )∆ xb

−¿∑v

K v

∆ gav ( x )∆ xb

¿

and similarly

∑u

∆ Ku

∆ xagub ( x )=∑

u

∆ (K u gub ( x ) )∆ xa

−∑u

K u

∆ gub ( x )∆ xa

so that:

∑j

K j

∆ gab ( x )∆ x j

+∑v

∆ ( K v gav ( x ) )∆ xb

−¿∑v

K v

∆ gav ( x )∆ xb

+∑u

∆ (Ku gub ( x ) )∆ xa

−∑u

Ku

∆ gub ( x )∆ xa

=0¿

Re-arranging and absorbing some terms into the same summation:

∑v

∆ ( Kv gav ( x ) )∆ xb

+∑u

∆ ( Ku gub ( x ) )∆ xa

+(∑jK j{∆ gab ( x )

∆ x j−

∆ gaj ( x )∆ xb

−∆ g jb (x )

∆ xa })=0

In the first two terms, the metric serves to make the Killing vector covariant:

∆ ( Ka )∆ xb

+∆ ( Kb )∆ xa

+(∑jK j {∆ gab ( x )

∆ x j−

∆ gaj (x )∆ xb

−∆ g jb ( x )

∆ xa })=0

Now we recall the definition (with indices changed to be appropriate):

Db ( K a )=∆ K a

∆ xb−∑

mKm Γab

m =¿∆ K a

∆ xb−∑

mKm(1

2∑j( g−1 )mj{∆ g jb

∆ xa+

∆ gaj

∆ xb−

∆ gab

∆ x j})¿

The inverse metric which is part of the Christoffel symbol converts the index of the Killing

vector from covariant to contravariant:

Db ( K a )=∆ K a

∆ xb−1

2∑jK j {∆ g jb

∆ xa+

∆ gaj

∆ xb−

∆ gab

∆ x j}

We also need:

Da ( K a )=∆ K b

∆ xa−1

2∑jK j {∆ g ja

∆ xb+

∆ gbj

∆ xa−

∆ gba

∆ x j}

Substituting these in:

Db ( K a )+ 12∑j

K j{∆ g jb

∆ xa+

∆ gaj

∆ xb−

∆ gab

∆ x j}+Da ( Ka )+ 1

2∑jK j{∆ g ja

∆ xb+

∆ gbj

∆ xa−

∆ gba

∆ x j}+∑j

K j {∆ gab ( x )∆ x j

−∆ gaj ( x )

∆ xb−

∆ g jb ( x )∆ xa

}=0

gives us

Db ( K a )+D a ( K a )=0

as required.