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物理代写-PX4124/PTX221

时间：2021-01-12

PX4124/PTX221 Introduction to General

Relativity

Autumn 2020

Continuous Assessment Assignment

16 November 2020

You must finish and return the solutions for this assessment by 14:00 on

Friday 4 December. The maximum mark obtainable on this assignment

is 60. This will be rescaled to contribute 30% of your final mark for this

module. The mark obtainable for a question or part of a question is shown

in brackets alongside the question.

Instructions for Completing the Assessment:

• Answer 6 of the 8 questions. If more than 6 are attempted, please

indicate which ones should be marked; otherwise the first 6 attempted

will be counted.

• Use black/blue ink and dark pencils, on white paper. You may use

graph paper where necessary.

• Before the end of the exam duration, allow time to use Microsoft-Lens,

Adobe PDF scan (or equivalent software) to photograph your answers

and convert into a PDF to upload.

• Make sure the pages of your answers are in the correct order and

orientation.

• Include your Student Number clearly on the first page, and in the

filename of your PDF.

• Submit your PDF answer file before 2 PM on 4 December

by:

uploading as assignment on PX4124 Learning Central module,

or emailing to SchutzBF(at)cardiff.ac.uk if you think the upload did

not work.

1

Notice to Students:

• This is an “open book” assessment, you may make use of any resource.

• The use of calculators / computers is permitted.

• You are expected to work without communication with any other per-

son.

• By submitting your answers, you declare that your submission (and

all the material contained in it), is your own work, and you have not

shared it with anyone else.

• The examiners reserve the right to further question you, if necessary,

by pre-arranged telephone calls, so you can further explain the an-

swers you have provided in your submission. The Cardiff University

plagiarism detection software may also be used to check originality of,

and any collusion in, the submitted work.

• By submitting your answers, you declare you understand that deceiv-

ing, or attempting to deceive the examiners by passing off the work of

another as your own is plagiarism. You also declare that you under-

stand that plagiarising another’s work, or knowingly allowing another

student to plagiarise from your work, is against University Regula-

tions and that doing so will result in loss of marks and disciplinary

proceedings.

2

1. (a) Transform the line element of special relativity from the usual

(t, x, y, z) rectangular coordinates to new coordinates (t′, x′, y′, z′)

related by

t =

(

c

g

+

x′

c

)

sinh

(

gt′

c

)

x = c

(

c

g

+

x′

c

)

cosh

(

gt′

c

)

− c

2

g

y = y′

z = z′

for a constant g with the dimensions of acceleration. [3]

SOLUTION: The differentials are

dt =

dx′

c

sinh(

gt′

c

) + (

c

g

+

x′

c

)

g

c

cosh(

gt′

c

)dt′

dx = dx′ cosh(

gt′

c

) + g(

c

g

+

x′

c

) sinh(

gt′

c

)dt′

dy = dy′

dz = dz′ .

Therefore, the line element in the new coordinates is

ds2 = −c2dt2 + dx2 + dy2 + dz2

= −(dx′)2 sinh2(gt

′

c

)− ( c

g

+

x′

c

)2g2 cosh2(

gt′

c

)(dt′)2

− 2g( c

g

+

x′

c

) sinh(

gt′

c

) cosh(

gt′

c

)dx′dt′

+ (dx′)2 cosh2(

gt′

c

) + g2(

c

g

+

x′

g

)2 sinh2(

gt′

c

)(dt′)2

+ 2g(

c

g

+

x′

c

) sinh(

gt′

c

) cosh(

gt′

c

)dx′dt′

+ (dy′)2 + (dz′)2

= −g2( c

g

+

x′

c

)2

[

cosh2(

gt′

c

)− sinh2(gt

′

c

)

]

(dt′)2

+

[

cosh2(

gt′

c

)− sinh2(gt

′

c

)

]

(dx′)2 + (dy′)2 + (dz′)2

= −g2( c

g

+

x′

c

)2(dt′)2 + (dx′)2 + (dy′)2 + (dz′)2

MARKING SCHEME:

3

1 Computing differentials.

1 Substituting into flat-space line element.

1 Obtaining correct result.

(b) For gt′/c 1, show that this corresponds to a transformation to

a uniformly accelerated frame in Newtonian mechanics. [3]

SOLUTION: Expand the sinh and cosh functions for gt′/c 1

and keep only terms that do not vanish as c→∞:

sinhx = x+O(x3)

coshx = 1 +

1

2

x2 +O(x4) .

We get

t =

(

c

g

+

x′

c

)(

gt′

c

)

+ . . .

= t′ + . . .

x = c

(

c

g

+

x′

c

)(

1 +

1

2

g2(t′)2

c2

)

− c

2

g

+ . . .

=

c2

g

+ x′ +

1

2

g(t′)2 − c

2

g

+ . . .

= x′ +

1

2

g(t′)2 + . . .

Combined, these give

x(t) = x(0) +

1

2

gt2 ,

which is the Newtonian equation of motion for a particle under-

going a constant acceleration g.

MARKING SCHEME:

2 Expansions of t and x.

1/2 Combining to get x(t).

1/2 Identifying result as Newtonian EOM.

(c) Show that a clock at rest in this frame at x′ = h runs fast com-

pared to a clock at rest at x′ = 0 by a factor (1 + gh/c2). How is

this related to the equivalence principle idea? [4]

4

SOLUTION: The proper time is given by

dτ2 = −ds

2

c2

=

g2

c2

(

c

g

+

x′

c

)2(dt′)2 − 1

c2

[(dx′)2 + (dy′)2 + (dz′)2] .

Therefore a clock at fixed x′, y′, z′ measures

dτ =

g

c

(

c

g

+

x′

c

)dt′ = (1 +

gx′

c2

)dt′ .

The rate of a clock at x′ = h compared to one at x′ = 0 is thus

dτh

dτ0

=

(1 + gh/c2)dt′

dt′

= 1 +

gh

c2

.

Therefore the clock at x′ = h runs faster by a factor 1+gh/c2. By

the equivalence principle, a clock at rest in a uniform gravitational

field g at a height h will run faster than one at height 0 by the

same factor,

1 +

Φ(h)− Φ(0)

c2

= 1 +

gh

c2

,

where Φ(x′) = gx′.

MARKING SCHEME:

1 Proper time for a stationary observer.

1 Computing rates of higher clock relative to lower clock.

2 Interpretation in terms of equivalence principle.

5

2. Spacetime geometry in the vicinity of the Earth is given to good ac-

curacy by the static weak-field line element

ds2 = −(1 + 2Φ)dt2 + (1− 2Φ)[dx2 + dy2 + dz2] ,

where Φ is the Newtonian potential. Consider a satellite in a circular

orbit around the Earth in the equatorial plane with a ∆t = 12 hour

period, and an observer at rest on the surface of the Earth at the

equator.

(a) Show that the components of the four-velocity of the observer in

(t, r, θ, φ) coordinates are

uα =

(

1− 2M⊕

R⊕

− 4pi

2R2⊕

T 2⊕

)−1/2(

1, 0, 0,

2pi

T⊕

)

,

where M⊕, R⊕, and T⊕ are the mass, radius, and rotation period

of the Earth, respectively. [3]

SOLUTION: In the slow-motion approximation the Φ[dx2 +

dy2 + dz2] term in the line element is negligible. In (t, r, θ, φ)

coordinates the line element becomes

ds2 = −(1 + 2Φ)dt2 + dr2 + r2(dθ2 + sin2 θdφ2) ,

where Φ = −M⊕/r.

The observer is not moving in the r or θ directions. His four-

velocity components are

uα = (ut, 0, 0, uφ) =

dt

dτ

(

1, 0, 0,

dφ

dt

)

=

dt

dτ

(

1, 0, 0,

2pi

T⊕

)

,

where T⊕ = 24 hr is the rotation period of the Earth. The nor-

malization condition u · u = −1 then gives

−1 = −

(

1− 2M⊕

R⊕

− 4pi

2R2⊕

T 2⊕

)(

dt

dτ

)2

dt

dτ

=

(

1− 2M⊕

R⊕

− 4pi

2R2⊕

T 2⊕

)−1/2

.

The four-velocity components are thus

uα =

(

1− 2M⊕

R⊕

− 4pi

2R2⊕

T 2⊕

)−1/2(

1, 0, 0,

2pi

T⊕

)

.

MARKING SCHEME:

6

1 Ignoring Φd~x2 terms.

1 ur = 0, uθ = 0 and relating uφ to T⊕.

1 Using u · u = −1 to solve for dt/dτ .

(b) Show that the components of the four-velocity of the satellite are

uα =

[

1− 3

(

2piM⊕

Tsat

)2/3]−1/2(

1, 0, 0,

2pi

Tsat

)

,

where Tsat = 12 hr is the orbital period of the satellite. (You may

use Newtonian mechanics to model the orbit.) [4]

SOLUTION: Kepler’s law relates the radius of the satellite’s

orbit to its period:

M = Ω2r3 ,

where Ω is the angular velocity. We thus have

Rsat =

(

M⊕

Ω2sat

)1/3

=

(

M⊕T 2sat

4pi2

)1/3

.

The satellite is not moving in the r or θ directions. Its four-

velocity components are

uα = (ut, 0, 0, uφ) =

dt

dτ

(

1, 0, 0,

dφ

dt

)

=

dt

dτ

(

1, 0, 0,

2pi

Tsat

)

.

The normalization condition u · u = −1 then gives

−1 = −

(

1− 2M⊕

Rsat

− 4pi

2R2sat

T 2sat

)(

dt

dτ

)2

dt

dτ

=

(

1− 2M⊕

Rsat

− 4pi

2R2sat

T 2sat

)−1/2

=

(

1− 2M⊕

(

4pi2

M⊕T 2sat

)1/3

− 4pi

2

T 2sat

(

M⊕T 2sat

4pi2

)2/3)−1/2

=

[

1− 3

(

2piM⊕

Tsat

)2/3]−1/2

.

The four-velocity components are thus

uα =

[

1− 3

(

2piM⊕

Tsat

)2/3]−1/2(

1, 0, 0,

2pi

Tsat

)

,

MARKING SCHEME:

7

1 Ignoring Φd~x2 terms.

1 ur = 0, uθ = 0.

1 Relating uφ to Tsat with Kepler’s law.

1 Using u · u = −1 to solve for dt/dτ .

(c) Compute the difference in proper time measured by a clock on the

satellite and one carried by the observer on the ground over one

day. (The mass of the Earth is 6.0 × 1024 kg, and its equatorial

radius is 6,400 km.) [3]

SOLUTION: The proper time τ is related to the coordinate

time t by

ut =

dt

dτ

.

Since ut is a constant for both the satellite and the observer on

the ground, we have over one day (∆t = 24 hr)

∆τ =

∆t

ut

=

24 hr

ut

for each. For the satellite we have (inserting a factor of G/c3 to

get the units right)

∆τsat = 24 hr

[

1− 3

(

2piGM⊕

c3Tsat

)2/3]1/2

= 24× 3600 s

[

1− 3

(

2pi 6.67× 10−11m3kg−1s−26× 1024kg

(3× 108ms−1)3(12× 3600s)

)2/3]1/2

= 86400 s

[

1− 5.0× 10−10]1/2

= 86400 s

[

1− 2.5× 10−10] .

For the observer on the ground we have (inserting factors of G/c2

and 1/c2 to get the units right)

∆τground = 24 hr

[

1− 2GM⊕

c2R⊕

−

(

2piR⊕

cT⊕

)2]1/2

= 86400 s

[

1− 2× 6.67× 10

−11m3kg−1s−26× 1024kg

(3× 108ms−1)2(6.4× 106m)

−

(

2pi 6.4× 106m

(3× 108ms−1)86400s

)2]1/2

= 86400 s

[

1− 1.4× 10−9 − 2.4× 10−12]1/2

= 86400 s

[

1− 6.9× 10−10] .

8

The clock on the satellite therefore runs faster each day by an

amount

∆τsat−∆τground = 86400 s [6.9− 2.5]×10−10 = 3.8×10−5s = 38µs .

MARKING SCHEME:

1 Formula for τ from ut.

1 G/c3 factor for units.

1 Computing ∆τsat and ∆τground and difference.

9

3. (a) Mark each of the following expressions as good (G) or bad (B)

tensor expressions, according to whether or not they follow the

rules for constructing index expressions. [5]

i. (xα)2 + (xβ)2 = (xc)2

ii. αβγA

α + gβγ = BβCγ

iii. AαβB

β + Cα

βDβ = 0

iv. gαβ ∂f

∂xβ

= V α.

v. MµµC

α +NνKαν = 0.

SOLUTION:

i. B.

The repeated indices in terms like (xα)@ = xαxα are both up,

so they violate the Einstein summation convention. Even in

special relativity, you need the minus sign in the sum when

α = 0, but not for the other values of α, and that is provided

by the metric tensor, which is not mentioned explicitly in

this expression.

ii. G

iii. B

There is a free index α in both terms but one is up and one

is down, which does not make sense. You can add a vector

to a vector but you can’t add a vector to a co-vector in a

coordinate-independent way. A coordinate transformation

applied to this expression would not be consistent for both

terms, since a co-vector transforms the way the coordinate

basis vectors transform but a vector transforms in the inverse

way.

iv. G

This is good because the partial derivative with respect to

xβ is the β component of the gradient co-vector, which has

a downstairs (covariant) index. So the Einstein summation

convention on β is satisfied here. In addition the position of

the free index α on both sides is the same.

v. G

Don’t be worried by the 0 on the right-hand-side. Every

kind of coordinate transformation will just turn 0 into 0,

leaving it invariant. That is because such transformations

are always multiplications. We could have put the second

term on the left-hand side of this expression over on the right

by subtracting it. Then we would have a tensor equation

10

without a zero. But it would still satisfy the rules for tensor

expressions.

MARKING SCHEME: One mark for each correct answer

(b) Classify each of the following hypersurfaces within Minkowski

spacetime as spacelike (S), timelike (T), or null (N). [5]

i. t = 1

ii. x = 1

iii. t+ x = 1

iv. The Lorentz hyperboloid t2 − x2 − y2 − z2 = a2 for some

constant a.

v. The future light cone of the event (0, 1, 0, 0).

SOLUTION: Remember that the classification of a hypersur-

face depends on the direction of its normal vector. If the vector

is timelike, the hypersurface is spacelike; if the vector is space-

like, the hypersurface is timelike. If the vector is null, so is the

hypersurface.

i. S

The normal vector is parallel to the t-axis, so is timelike.

That makes the hypersurface spacelike.

ii. T

By analogy with the last item, the normal is parallel to the

x-axis, so is spacelike. That makes the hypersurface timelike.

iii. N

The surfaces of constant t+x are 45-degree lines in the t−x

plane, then extended in the y- and z-directions. The nor-

mal vector is a null vector parallel to the surface, since it is

orthogonal to itself. So the hypersurface is null.

iv. S

Do not be fooled by the fact that this represents a negative

squared-interval, hence a timelike interval. That does not

make the hypersurface timelike.

v. N

MARKING SCHEME: One mark for each correct answer

11

4. (a) Consider the coordinate transformation u = t − x, v = t + x

in Minkowski spacetime with c = 1. Define eu to be the vec-

tor pointing from the event {u = 0, v = 0, y = 0, z = 0} to

{u = 1, v = 0, y = 0, z = 0}, and define ev to point from

{u = 0, v = 0, y = 0, z = 0} to {u = 0, v = 1, y = 0, z = 0}.

These are the basis vectors associated with the coordinates u and

v, respectively.

i. Show that eu = (et − ex)/2, ev = (et + ex)/2, and draw eu

and ev in a spacetime diagram of the t− x plane. [2]

ii. Show that {eu, ev, ey, ez} are a basis for vectors in Minkowski

spacetime. [1]

iii. Find the components of the metric tensor on this basis. (Hint:

recall that eα ·eβ = gαβ.) This shows that eu and ev are null

and not orthogonal. (They are called a null basis for the t−x

plane.) [2]

SOLUTION:

i. For eu, solving t − x = 1 and t + x = 0 gives t = 1/2, x =

−1/2. This is what is required for the components of eu.

The calculation for ev goes similarly. Drawing them shows

they are along the null lines.

ii. They are clearly linearly independent so they are a basis.

iii. Writing out the dot products explicitly:

guu = eu · eu = 1

4

(ηtt + ηxx) = 0,

gvv = ev · ev = 1

4

(ηtt + ηxx) = 0,

guv = gvu = eu · ev = 1

4

(ηtt − ηxx) = −1

2

.

The components gyy and gzz are unaffected.

iv. The dot products in the previous section show that they are

null and not orthogonal.

MARKING SCHEME:

1 Understanding how to draw the new basis vectors

1 algebra to get basis vectors

1 Understanding what a basis for Minkowski space is

2 Application of ηαβ to the components to get the results.

12

(b) Consider the two-dimensional spacetime spanned by coordinates

(v, x) with the line element

ds2 = −xdv2 + 2dv dx ,

with c = 1. (This is NOT the same spacetime as in the previous

part of this question, despite the similarity of the names of the

coordinates!!)

i. Calculate the slopes of the two lines making up the light cone

at a general point (v, x). [1]

ii. Draw a (v, x) spacetime diagram showing how the light cones

change with x. [2]

iii. Explain from this diagram why it is that a particle can cross

from positive x to negative x but not from negative x to

positive x. (The world-line x = 0 is called a horizon.) [2]

SOLUTION: World lines of particles must lie inside or on the

light cone. There are wordlines that cross from positive x to

negative x, but not the other way.

The light cone is determined by ds2 = 0:

0 = dv(−xdv + 2dx) ,

which has solutions

dv

dx

= 0 and

dv

dx

=

2

x

.

The dv/dx = 2/x portion of the light curve is positive for x > 0

and negative for x < 0, and vertical at x = 0. For x > 0

the x =constant lines (vertical) are timelike, since they give

ds2 = −xdv2 < 0, and so lie inside the light cone. For x < 0 the

x =constant lines are spacelike, since they give ds2 = −xdv2 > 0,

and therefore lie outside the light cone. The light cones therefore

look as follows:

Thursday, 24 October 2013

13

MARKING SCHEME:

1 Explanation that worldlines lie inside light cones

1 Setting ds2 = 0

1 Computing the two solutions for dv/dx.

1 Plot showing the behaviour for x < 0 and x > 0.

1 Justification for which portion of spacetime lies inside the light

cone for x > 0 and x < 0.

14

5. In a certain spacetime geometry the metric is

ds2 = −(1−Ar2)2dt2 + (1−Ar2)2dr2 + r2(dθ2 + sin2 θdφ2) .

(a) Calculate the proper distance along a radial line from r = 0 to a

coordinate radius r = R. Assume AR2 < 1. [2]

SOLUTION: Along a radial line dt = 0, dθ = 0, dφ = 0, so the

distance is

distance =

∫ R

0

dr

√

grr =

∫ R

0

dr

√

(1−Ar2)2 =

∫ R

0

dr(1−Ar2) = R−1

3

AR3 .

MARKING SCHEME:

1 Setting up integral.

1 Evaluating integral.

(b) Compute the area of a sphere of coordinate radius r = R. [2]

SOLUTION: The surface of a sphere has dt = 0, dr = 0, so

the area is

area =

∫ pi

0

dθ

∫ 2pi

0

dφ

√

gθθgφφ .

Since gθθ and gφφ have the same values as in flat space, we get

the same result for the area of the sphere:

area =

∫ pi

0

dθ

∫ 2pi

0

dφR2 sin θ =

[∫ pi

0

dθ sin θ

] [∫ 2pi

0

dφ

]

R2 = 4piR2 .

MARKING SCHEME:

1 Setting up integral.

1 Evaluating integral.

(c) Calculate the 3-volume of a sphere of coordinate radius r = R.

[3]

15

SOLUTION:

volume =

∫ R

0

dr

∫ pi

0

dθ

∫ 2pi

0

dφ

√

grrgθθgφφ

=

[∫ R

0

dr r2(1−Ar2)

] [∫ pi

0

dθ sin θ

] [∫ 2pi

0

dφ

]

= 4pi

[

1

3

R3 − 1

5

AR5

]

=

4pi

3

R3

[

1− 3

5

AR2

]

.

MARKING SCHEME:

1 Setting up integral.

2 Evaluating integral.

(d) Calculate the 4-volume of a 4-dimensional tube bounded by a

sphere of coordinate radius R and two t = constant planes sepa-

rated by a coordinate time T . [3]

SOLUTION:

4− volume =

∫ t1+T

t1

dt

∫ R

0

dr

∫ pi

0

dθ

∫ 2pi

0

dφ

√−gttgrrgθθgφφ

=

[∫ t1+T

t1

dt

] [∫ R

0

dr r2(1−Ar2)2

] [∫ pi

0

dθ sin θ

] [∫ 2pi

0

dφ

]

= 4piT

[

1

3

R3 − 2

5

AR5 +

1

7

A2R7

]

=

4pi

3

R3T

[

1− 6

5

AR2 +

3

7

A2R4

]

.

MARKING SCHEME:

1 Setting up integral.

2 Evaluating integral.

16

6. Consider the following four different metrics, as given by their line

elements:

(i) ds2 = −dt2 + dx2 + dy2 + dz2;

(ii) ds2 = −(1−2M/r) dt2+(1−2M/r)−1 dr2+r2(dθ2+sin2 θ dφ2),

where M is a constant;

(iii)

ds2 =− ∆− a

2 sin2 θ

ρ2

dt2 − 2a2Mr sin

2 θ

ρ2

dt dφ

+

(r2 + a2)2 − a2∆ sin2 θ

ρ2

sin2 θ dφ2 +

ρ2

∆

dr2 + ρ2 dθ2,

where M and a are constants and we have introduced the short-

hand notation ∆ = r2 − 2Mr + a2, ρ2 = r2 + a2 cos2 θ;

(iv) ds2 = −dt2+R2(t) [(1− kr2)−1dr2 + r2(dθ2 + sin2 θ dφ2)], where

k is a constant and R(t) is an arbitrary function of t alone.

The first one is familiar to us as Minkowski spacetime. The names of

the others are, respectively, the Schwarzschild, Kerr, and Robertson–

Walker metrics.

(a) For metric (i), use symmetry arguments to identify by name all

the conserved quantities associated with the four-momentum pα

of a freely falling particle. [1]

SOLUTION: The metric coefficients are all constants, so there

is a conserved quantity associated with each of them. So p0 is

the negative of the energy; px is the x-component of the linear

momentum; and similarly for py and pz.

MARKING SCHEME:

1/2 For saying there are 4 conserved quantities

1/2 For getting the names of energy and linear momentum

(b) Metric (i) can also be put into the following form, using spherical

polar coordinates instead of Cartesian for the spatial coordinates:

(i′) ds2 = −dt2 + dr2 + r2(dθ2 + sin2 θ dφ2).

Does this increase the number of conserved components pα? If

so, name them all. [2]

SOLUTION: This shows that there is a conserved quantity pφ,

17

which is called angular momentum. But because the equator can

be taken in any plane, there are actually three independent angu-

lar momentum components, associated with rotational symmetry

about the three Cartesian coordinate axes.

MARKING SCHEME:

1 For identifying the angular momentum

1 For describing the other two angular momentum components

(c) How many of the metrics (i)-(iv) are spherically symmetric? [1]

SOLUTION: (i), (ii), and (iv). Three in total.

MARKING SCHEME:

1 Acceptable answer is either ’three’ or the list of the three. If

three are listed but one is wrong, then 1/2 mark.

(d) For each of the metrics (ii)-(iv), use symmetry arguments to iden-

tify by name all the conserved quantities associated with the four-

momentum pα of a freely falling particle. [6]

SOLUTION:

(ii): The symmetry coordinates of Schwarzschild are t and φ. By

spherical symmetry, there are also rotations about the other

two axes. So the names are −p0 = energy, pφ = angular

momentum about z-axis, and then the two other angular

momenta.

(iii): Again t and φ are symmetry coordinates, but there is no

spherical symmetry, so there are only two conserved quanti-

ties here, −p0 = energy and pφ = angular momentum about

the Kerr symmetry axis.

(iv): Similar to (ii) BUT now t is not a symmetry because R(t)

makes the metric time-dependent. So the only conserved

quantities that are obvious from this form of the metric are

the three angular momentum components. (Really, the spa-

tial hypersurfaces of constant t are homogeneous, so there is

also a translation symmetry, but this is not clear from the

given form of the metric and has not been studied in class,

so this is not expected in the answers. It is mentioned in the

next problem, in fact.)

MARKING SCHEME:

18

1 For (ii), naming the energy and z-angular momentum (1/2

mark for identifying t and φ without names).

1 For (ii), including the other two angular momentum compo-

nents.

1 For (iii), identifying t and φ as symmetry coordinates

1 For (iii), naming the associated quantities as energy and angu-

lar momentum

1 For (iv), understanding that t is not a symmetry

1/2 For (iv), stating that φ has a symmetry and leads to a con-

served angular momentum

1/2 For (iv), including the other two angular momentum com-

ponents

19

7. The three-dimensional line element

ds2 = dr2 + (r2 + a2)(dθ2 + sin2 θ dφ2),

where a is a constant and r has the range −∞ < r <∞, defines what

is called a wormhole.

(a) Show that this is a purely spatial line-element, i.e. that there are

no negative intervals. [1]

SOLUTION: All the terms in this line element are positive. So

the intervals are positive-definite, which means spacelike.

MARKING SCHEME:

1 Mention positive-definite or something equivalent.

(b) Show that this is a spherically symmetric space. [1]

SOLUTION: (i) It contains the term (dθ2+sin2 θ dφ2) involving

coordinates θ and φ. This is the line element of a sphere in flat

space.

(ii) The coordinates θ and φ do not appear anywhere else in the

line element, so that the symmetries of the sphere apply to this

spacetime.

MARKING SCHEME:

1 At least mention the presence of the spherical part of the line

element. If the absence of other terms involving these coor-

dinates is mentioned, give a 1 point bonus, but do not exceed

10 for total marks for the question.

(c) Show that r = 0 is not a point, but is rather a sphere of radius

a, and that this sphere has the minimum circumference of any of

the spheres in this space. [2]

SOLUTION: The circumference of each sphere of constant r is

given in terms of the coefficient of the spherical part, i.e. r2 + a2.

It is 2pi

√

(r2 + a2). So when r = 0 the circumference is 2pia,

which is non-zero and is the minimum of the circumferences.

MARKING SCHEME:

1 For knowing how to compute the circumference

1 For showing that r = 0 gives the minimum circumference.

20

(d) Compute all the non-vanishing Christoffel symbols for this met-

ric. You may compute them directly by taking derivatives of the

metric, or you may use the method of the Lagrangian principle

that we introduced in the notes, where in this case the Lagrangian

is the proper distance along a curve. [6]

SOLUTION: In the lectures we presented two ways to compute

the Christoffel symbols, either using the action principle (which

was recommended as the easier way for a general metric) or the

direct way using derivatives of the metric tensor. I first give the

solution using the action principle method, followed by its mark-

ing scheme, and then below I give the solution using derivatives

of the metric, and its marking scheme.

Because this is a three-dimensional spatial metric I will use Latin

letters (a, b, c, . . . , j, k, . . .) for the indices, just to emphasise the

difference from 4D.

Action Principle method:

i. The Lagrangian is L = [gjkx˙

j x˙k]1/2, where a dot indicates a

derivative with respect to proper distance s. (We shall use

this convention here for clarity, even though a dot normally

indicates a time-derivative, and there is no time coordinate

here. Sometimes space derivatives are denoted by a prime

(′), but elsewhere in this assignment we have used primes to

denote alternative coordinate systems, not derivatives. So

we shall stick with dots here for derivatives with respect to

proper distance s. Thus, A˙ := dA/ds for any A. ) As in

the case of the 4-dimensional action principle, we can scale

the parameter s so that the numerical value of L is 1, which

can be used in expressions when it is not being differentiated

with respect to the variables (r, θ, φ) and (r˙, θ˙, φ˙). So we

write out the Lagrangian as

L :=

[

r˙2 + (r2 + a2)(θ˙2 + sin2 θφ˙2)

]1/2

.

ii. The derivatives with respect to the velocities are (and using

L = 1 after the differentiation):

∂L

∂r˙

= r˙,

∂L

∂θ˙

= (r2 + a2)θ˙,

∂L

∂φ˙

= (r2 + a2) sin2 θ φ˙.

21

iii. Their derivatives with respect to s are

d

ds

(

∂L

∂r˙

) = r¨,

d

ds

(

∂L

∂θ˙

) = (r2 + a2)θ¨ + 2rθ˙r˙,

d

ds

(

∂L

∂φ˙

) = (r2+a2) sin2 θ φ¨+2r sin2 θ φ˙r˙+2(r2+a2) sin θ cos θ φ˙θ˙.

iv. The derivatives of the Lagrangian with respect to the coor-

dinates are

∂L

∂r

= rθ˙2 + r sin2 θ φ˙2,

∂L

∂θ

= (r2 + a2) sin θ cos θ φ˙2,

∂L

∂φ

= 0.

v. The Euler-Lagrange (E-L) equations from these,

d

ds

∂L

∂x˙a

− ∂L

∂xa

= 0,

is written in terms of the Christoffel symbols in this way:

x¨a + Γabcx˙

bx˙c = 0.

Putting the previous calculations together to get this gives

three equations, from each of which we can deduce both the

non-zero Christoffel symbols and the zero Christoffel sym-

bols. The three equations are:

A. For a = r, we have:

r¨ − rθ˙2 − r sin2 θ φ˙2 = 0.

This gives all the non-zero Γrbc.

Γrθθ = −r, Γrφφ = −r sin2 θ.

The ones with lower index combinations that don’t ap-

pear here all vanish.

B. For a = θ we have to clear a factor in front of θ¨ to get the

right form of the equation, and this is done by dividing

by r2 + a2. The E-L equation is then:

θ¨ +

2r

r2 + a2

θ˙r˙ − sin θ cos θ φ˙2 = 0.

22

For the first time here we encounter a case where the

indices b 6= c, so we have to recognize that such a term

enters the summation as both Γθbc and Γ

θ

cb. Then the

symmetry of the Christoffel symbol on exchange of b and

c means that these two terms are equal, and therefore

their sum (which is what appears in the E-L equation)

will contain double the amount from a single Christoffel

symbol of this type. Therefore we divide the appropriate

coefficients by 2 to get the right answer for a single sym-

bol. This consideration leads to the following non-zero

Γθbc terms:

Γθθr = Γ

θ

rθ =

r

r2 + a2

, Γθφφ = − sin θ cos θ.

C. Finally, for a = φ, the calculation proceeds in close anal-

ogy to the previous one, so that the E-L equation, after

dividing by (r2 + a2) sin2 θ, is

φ¨+

2r

r2 + a2

φ˙r˙ + 2

cos θ

sin θ

φ˙θ˙ = 0.

And again proceeding similarly to the previous one for

the cases where the bottom indices are different, we infer

that the non-zero Γφbc terms are:

Γφφr = Γ

φ

rφ =

r

r2 + a2

, Γφθφ = Γ

φ

φθ = cot θ.

D. We remark that the Christoffel symbols involving only

the angular coordinates are the same as we get in flat

space for spherical coordinates, which is not surprising.

This is a good check on whether mistakes have been made

along the way.

MARKING SCHEME: Marking scheme for the action prin-

ciple method of solution:

1 Correct Lagrangian. Can involve σ at first but student must

set σ = s when evaluating L.

1 Correct expressions for the derivatives of the Lagrangian with

respect to the “velocities”. Half-mark for at least two correct.

1 Correct derivatives with respect to s of the velocity derivatives.

Again half-marks possible.

1 Correct derivatives of the Lagrangian with respect to the co-

ordinates. Half-marks possible.

1 Correct form of the three E-L equations. Half-marks possible

23

1 Correct expressions for at least 4 of the 6 non-zero Christoffel

symbols. Earn 1/2 mark if 2-3 of them are right. Lose 1/2

for not remarking on or correctly handling the symmetry on

the lower two indices.

SOLUTION: The other way to approach this is to use deriva-

tives of the metric tensor in Eq (98) of the lecture notes (changed

to Latin indices):

Γabc =

1

2

gaj

(

∂gjb

∂xc

+

∂gjc

∂xb

− ∂gbc

∂xj

)

.

This is considerably simplified if one uses some key facts. First,

the metric is diagonal, so we will always have a = j in this. And

second, the metric is reflection-symmetric in φ, so that a Christof-

fel symbol that contains one or three φ-indices will automatically

vanish.

The calculations are not difficult, so I will just illustrate two.

First, the calculation of Γrθθ:

Γrθθ =

1

2

grj(

∂gjθ

∂θ

+

∂gθj

∂θ

− ∂gθθ

∂xj

).

But because the metric is diagonal, so is the inverse, and indeed

grr = 1/grr = 1. Only j = r survives in this summation, leaving

Γrθrθ =

1

2

(2

∂grθ

∂θ

− ∂gθθ

∂r

) = −r.

This is the same as we had in the other method, of course.

Next, the calculation of Γθrθ. We note from the start that g

θθ =

1/(r2 + a2) and that gθr = 0, so that we only have

Γθrθ = Γ

θ

θr =

1

2

1

r2 + a2

(

∂gθθ

∂r

) =

r

r2 + a2

.

The rest of the Christoffel symbols can be obtained in this way,

and their values need to match up with the ones we had above.

So do the ones that vanish.

MARKING SCHEME: Marking scheme for the second method

of deriving the Christoffel symbols:

1 Correctly observing that the metric is diagonal

1 Correctly calculating the thee terms in the inverse metric. 1/2

for getting two right.

24

up to 1 A bonus of up to 1 for noting that reflection symmetry

on φ implies that all terms with one or three φ-indices van-

ish, so don’t need to be computed. The total marks for the

question must not exceed 10.

4 For correctly computing at least 4 of the 6 non-vanishing Christof-

fel symbols. Earn 2 marks for getting 2-3 of them right. Lose

1/2 mark for each algebraic error in the calculation.

25

8. The static, weak-field line element that describes the geometry of a

Newtonian star with gravitational potential Φ is given by

ds2 = −(1 + 2Φ)dt2 + (1− 2Φ)[dx2 + dy2 + dz2] (1)

Consider a hypothetical massless particle that moves at the speed of

light but does not interact with normal matter. The particle crosses

the Earth’s orbit around the Sun and goes straight through the centre

of the Sun. If the Sun had not been there, then the crossing would

have taken a time equal to T0 = 2Rorbit, where Rorbit is equal to 1

Astronomical Unit (AU), or 1.5× 1011 m.

(a) Show by using the line element that the difference between the

true time through the Sun and the time T0 is given by

∆T = −4

∫ R

0

Φ(r)dr, (2)

where R is the radius of the Sun, 7× 108 m. [4]

SOLUTION: The particle moves on a radial line, which without

loss of generality we can take to be the x-axis. Then dy = dz = 0.

It moves at the speed of light, so ds2 = 0. This gives us a relation

between dt and dx:

dt

dx

=

[

1− 2Φ

1 + 2Φ

]1/2

.

Since in the Newtonian case we can take |Φ| 1, we can re-write

this keeping terms of first order in Φ:

dt

dx

= [(1−2Φ)(1−2Φ+ . . .)]1/2 = (1−4Φ+ . . .)1/2 = 1−2Φ+ . . . .

In this part of the problem, we are only asked about the time

crossing the Sun itself. We can get this by integrating dt/dx. If

we integrate this along x, for a spherical Sun, that is equivalent

to integrating radially along a diameter of the Sun. The inward

and outward segments will contribute the same, so the result is

T = 2

∫ R

0

[1− 2Φ(r)]dr = 2R − 4

∫ R

0

Φ(r)dr.

From this it is clear that the excess time crossing the sun is as

given in Eq. 2 in the problem. (Unfortunately there was a mis-

print in the problem, which asked for the difference to the time

26

T0, which had been defined as the time to cross the Earth’s orbit.

The problem should have asked for the crossing time of the whole

orbit.)

MARKING SCHEME:

1 using ds2 = 0

1 reducing the dimensionality to just a motion along x (or equiv-

alent)

1 getting the right equation for dt/dr in the approximation of

small Φ.

1 doing the integral correctly. Marking is flexible to account for

possible confusion caused by the error in the question.

(b) We know that outside the Sun, the potential is Φ = −GM/r,

where GM = 1.5 × 103 m is the geometrised mass of the Sun

in units where c = 1. Rather than develop a good model of the

potential inside the Sun, we shall make the crude approxima-

tion that it is constant at the value it has at the Sun’s surface,

−GM/R. The error induced by this approximation will be

small since the majority of our particle’s journey is outside the

Sun. Using this approximation, show that the time difference is

∆T = 4M(1 + ln

Rorbit

R

).

[4]

SOLUTION: Here the time-difference across the whole orbit

is asked for. The result is to integrate Eq 2 for both the interior

of the Sun and the exterior. The interior is simple because the

integrand is constant:

∆Tinterior = −4−GM

R

R = 4GM.

The exterior is an integral fromR toRorbit, and the only variable

inside the integral is 1/r. So the integration is

∆Texterior = −4(−GM)

∫ Rorbit

R

1

r

dr = 4GM ln

Rorbit

R

.

The full answer is the sum of these two, and gives us the answer

required in the question. That answer is framed with G = 1, but

we move between different unit systems in this course, so that

should not give a problem.

MARKING SCHEME:

27

1 Setting up the problem correctly as an integral from the Sun’s

centre to the Earth’s orbit

1 Correct interior integration

1 Correct exterior integration

1 Combining the two for the correct answer.

(c) Since this is positive, the particle suffers a delay on passing through

the Sun. Compute, in seconds, the size of the delay. This is called

the Shapiro delay. [2]

SOLUTION: Since M = 1.5 km, its conversion to time is

dividing by c = 3× 105 km/s, which gives M = 5µs. So 4M =

20µs. We are given that Rorbit = 1.5 × 1011 m, and R = 7 ×

108 m, so their ratio is 214, and the log is 5.4. This makes ∆T =

127µs. This is a time difference which can be measured very

accurately with atomic clocks, which are accurate to picoseconds

or better.

MARKING SCHEME:

1 Correct conversion of M to seconds.

1 Correct arithmetic to get final answer.

28

Relativity

Autumn 2020

Continuous Assessment Assignment

16 November 2020

You must finish and return the solutions for this assessment by 14:00 on

Friday 4 December. The maximum mark obtainable on this assignment

is 60. This will be rescaled to contribute 30% of your final mark for this

module. The mark obtainable for a question or part of a question is shown

in brackets alongside the question.

Instructions for Completing the Assessment:

• Answer 6 of the 8 questions. If more than 6 are attempted, please

indicate which ones should be marked; otherwise the first 6 attempted

will be counted.

• Use black/blue ink and dark pencils, on white paper. You may use

graph paper where necessary.

• Before the end of the exam duration, allow time to use Microsoft-Lens,

Adobe PDF scan (or equivalent software) to photograph your answers

and convert into a PDF to upload.

• Make sure the pages of your answers are in the correct order and

orientation.

• Include your Student Number clearly on the first page, and in the

filename of your PDF.

• Submit your PDF answer file before 2 PM on 4 December

by:

uploading as assignment on PX4124 Learning Central module,

or emailing to SchutzBF(at)cardiff.ac.uk if you think the upload did

not work.

1

Notice to Students:

• This is an “open book” assessment, you may make use of any resource.

• The use of calculators / computers is permitted.

• You are expected to work without communication with any other per-

son.

• By submitting your answers, you declare that your submission (and

all the material contained in it), is your own work, and you have not

shared it with anyone else.

• The examiners reserve the right to further question you, if necessary,

by pre-arranged telephone calls, so you can further explain the an-

swers you have provided in your submission. The Cardiff University

plagiarism detection software may also be used to check originality of,

and any collusion in, the submitted work.

• By submitting your answers, you declare you understand that deceiv-

ing, or attempting to deceive the examiners by passing off the work of

another as your own is plagiarism. You also declare that you under-

stand that plagiarising another’s work, or knowingly allowing another

student to plagiarise from your work, is against University Regula-

tions and that doing so will result in loss of marks and disciplinary

proceedings.

2

1. (a) Transform the line element of special relativity from the usual

(t, x, y, z) rectangular coordinates to new coordinates (t′, x′, y′, z′)

related by

t =

(

c

g

+

x′

c

)

sinh

(

gt′

c

)

x = c

(

c

g

+

x′

c

)

cosh

(

gt′

c

)

− c

2

g

y = y′

z = z′

for a constant g with the dimensions of acceleration. [3]

SOLUTION: The differentials are

dt =

dx′

c

sinh(

gt′

c

) + (

c

g

+

x′

c

)

g

c

cosh(

gt′

c

)dt′

dx = dx′ cosh(

gt′

c

) + g(

c

g

+

x′

c

) sinh(

gt′

c

)dt′

dy = dy′

dz = dz′ .

Therefore, the line element in the new coordinates is

ds2 = −c2dt2 + dx2 + dy2 + dz2

= −(dx′)2 sinh2(gt

′

c

)− ( c

g

+

x′

c

)2g2 cosh2(

gt′

c

)(dt′)2

− 2g( c

g

+

x′

c

) sinh(

gt′

c

) cosh(

gt′

c

)dx′dt′

+ (dx′)2 cosh2(

gt′

c

) + g2(

c

g

+

x′

g

)2 sinh2(

gt′

c

)(dt′)2

+ 2g(

c

g

+

x′

c

) sinh(

gt′

c

) cosh(

gt′

c

)dx′dt′

+ (dy′)2 + (dz′)2

= −g2( c

g

+

x′

c

)2

[

cosh2(

gt′

c

)− sinh2(gt

′

c

)

]

(dt′)2

+

[

cosh2(

gt′

c

)− sinh2(gt

′

c

)

]

(dx′)2 + (dy′)2 + (dz′)2

= −g2( c

g

+

x′

c

)2(dt′)2 + (dx′)2 + (dy′)2 + (dz′)2

MARKING SCHEME:

3

1 Computing differentials.

1 Substituting into flat-space line element.

1 Obtaining correct result.

(b) For gt′/c 1, show that this corresponds to a transformation to

a uniformly accelerated frame in Newtonian mechanics. [3]

SOLUTION: Expand the sinh and cosh functions for gt′/c 1

and keep only terms that do not vanish as c→∞:

sinhx = x+O(x3)

coshx = 1 +

1

2

x2 +O(x4) .

We get

t =

(

c

g

+

x′

c

)(

gt′

c

)

+ . . .

= t′ + . . .

x = c

(

c

g

+

x′

c

)(

1 +

1

2

g2(t′)2

c2

)

− c

2

g

+ . . .

=

c2

g

+ x′ +

1

2

g(t′)2 − c

2

g

+ . . .

= x′ +

1

2

g(t′)2 + . . .

Combined, these give

x(t) = x(0) +

1

2

gt2 ,

which is the Newtonian equation of motion for a particle under-

going a constant acceleration g.

MARKING SCHEME:

2 Expansions of t and x.

1/2 Combining to get x(t).

1/2 Identifying result as Newtonian EOM.

(c) Show that a clock at rest in this frame at x′ = h runs fast com-

pared to a clock at rest at x′ = 0 by a factor (1 + gh/c2). How is

this related to the equivalence principle idea? [4]

4

SOLUTION: The proper time is given by

dτ2 = −ds

2

c2

=

g2

c2

(

c

g

+

x′

c

)2(dt′)2 − 1

c2

[(dx′)2 + (dy′)2 + (dz′)2] .

Therefore a clock at fixed x′, y′, z′ measures

dτ =

g

c

(

c

g

+

x′

c

)dt′ = (1 +

gx′

c2

)dt′ .

The rate of a clock at x′ = h compared to one at x′ = 0 is thus

dτh

dτ0

=

(1 + gh/c2)dt′

dt′

= 1 +

gh

c2

.

Therefore the clock at x′ = h runs faster by a factor 1+gh/c2. By

the equivalence principle, a clock at rest in a uniform gravitational

field g at a height h will run faster than one at height 0 by the

same factor,

1 +

Φ(h)− Φ(0)

c2

= 1 +

gh

c2

,

where Φ(x′) = gx′.

MARKING SCHEME:

1 Proper time for a stationary observer.

1 Computing rates of higher clock relative to lower clock.

2 Interpretation in terms of equivalence principle.

5

2. Spacetime geometry in the vicinity of the Earth is given to good ac-

curacy by the static weak-field line element

ds2 = −(1 + 2Φ)dt2 + (1− 2Φ)[dx2 + dy2 + dz2] ,

where Φ is the Newtonian potential. Consider a satellite in a circular

orbit around the Earth in the equatorial plane with a ∆t = 12 hour

period, and an observer at rest on the surface of the Earth at the

equator.

(a) Show that the components of the four-velocity of the observer in

(t, r, θ, φ) coordinates are

uα =

(

1− 2M⊕

R⊕

− 4pi

2R2⊕

T 2⊕

)−1/2(

1, 0, 0,

2pi

T⊕

)

,

where M⊕, R⊕, and T⊕ are the mass, radius, and rotation period

of the Earth, respectively. [3]

SOLUTION: In the slow-motion approximation the Φ[dx2 +

dy2 + dz2] term in the line element is negligible. In (t, r, θ, φ)

coordinates the line element becomes

ds2 = −(1 + 2Φ)dt2 + dr2 + r2(dθ2 + sin2 θdφ2) ,

where Φ = −M⊕/r.

The observer is not moving in the r or θ directions. His four-

velocity components are

uα = (ut, 0, 0, uφ) =

dt

dτ

(

1, 0, 0,

dφ

dt

)

=

dt

dτ

(

1, 0, 0,

2pi

T⊕

)

,

where T⊕ = 24 hr is the rotation period of the Earth. The nor-

malization condition u · u = −1 then gives

−1 = −

(

1− 2M⊕

R⊕

− 4pi

2R2⊕

T 2⊕

)(

dt

dτ

)2

dt

dτ

=

(

1− 2M⊕

R⊕

− 4pi

2R2⊕

T 2⊕

)−1/2

.

The four-velocity components are thus

uα =

(

1− 2M⊕

R⊕

− 4pi

2R2⊕

T 2⊕

)−1/2(

1, 0, 0,

2pi

T⊕

)

.

MARKING SCHEME:

6

1 Ignoring Φd~x2 terms.

1 ur = 0, uθ = 0 and relating uφ to T⊕.

1 Using u · u = −1 to solve for dt/dτ .

(b) Show that the components of the four-velocity of the satellite are

uα =

[

1− 3

(

2piM⊕

Tsat

)2/3]−1/2(

1, 0, 0,

2pi

Tsat

)

,

where Tsat = 12 hr is the orbital period of the satellite. (You may

use Newtonian mechanics to model the orbit.) [4]

SOLUTION: Kepler’s law relates the radius of the satellite’s

orbit to its period:

M = Ω2r3 ,

where Ω is the angular velocity. We thus have

Rsat =

(

M⊕

Ω2sat

)1/3

=

(

M⊕T 2sat

4pi2

)1/3

.

The satellite is not moving in the r or θ directions. Its four-

velocity components are

uα = (ut, 0, 0, uφ) =

dt

dτ

(

1, 0, 0,

dφ

dt

)

=

dt

dτ

(

1, 0, 0,

2pi

Tsat

)

.

The normalization condition u · u = −1 then gives

−1 = −

(

1− 2M⊕

Rsat

− 4pi

2R2sat

T 2sat

)(

dt

dτ

)2

dt

dτ

=

(

1− 2M⊕

Rsat

− 4pi

2R2sat

T 2sat

)−1/2

=

(

1− 2M⊕

(

4pi2

M⊕T 2sat

)1/3

− 4pi

2

T 2sat

(

M⊕T 2sat

4pi2

)2/3)−1/2

=

[

1− 3

(

2piM⊕

Tsat

)2/3]−1/2

.

The four-velocity components are thus

uα =

[

1− 3

(

2piM⊕

Tsat

)2/3]−1/2(

1, 0, 0,

2pi

Tsat

)

,

MARKING SCHEME:

7

1 Ignoring Φd~x2 terms.

1 ur = 0, uθ = 0.

1 Relating uφ to Tsat with Kepler’s law.

1 Using u · u = −1 to solve for dt/dτ .

(c) Compute the difference in proper time measured by a clock on the

satellite and one carried by the observer on the ground over one

day. (The mass of the Earth is 6.0 × 1024 kg, and its equatorial

radius is 6,400 km.) [3]

SOLUTION: The proper time τ is related to the coordinate

time t by

ut =

dt

dτ

.

Since ut is a constant for both the satellite and the observer on

the ground, we have over one day (∆t = 24 hr)

∆τ =

∆t

ut

=

24 hr

ut

for each. For the satellite we have (inserting a factor of G/c3 to

get the units right)

∆τsat = 24 hr

[

1− 3

(

2piGM⊕

c3Tsat

)2/3]1/2

= 24× 3600 s

[

1− 3

(

2pi 6.67× 10−11m3kg−1s−26× 1024kg

(3× 108ms−1)3(12× 3600s)

)2/3]1/2

= 86400 s

[

1− 5.0× 10−10]1/2

= 86400 s

[

1− 2.5× 10−10] .

For the observer on the ground we have (inserting factors of G/c2

and 1/c2 to get the units right)

∆τground = 24 hr

[

1− 2GM⊕

c2R⊕

−

(

2piR⊕

cT⊕

)2]1/2

= 86400 s

[

1− 2× 6.67× 10

−11m3kg−1s−26× 1024kg

(3× 108ms−1)2(6.4× 106m)

−

(

2pi 6.4× 106m

(3× 108ms−1)86400s

)2]1/2

= 86400 s

[

1− 1.4× 10−9 − 2.4× 10−12]1/2

= 86400 s

[

1− 6.9× 10−10] .

8

The clock on the satellite therefore runs faster each day by an

amount

∆τsat−∆τground = 86400 s [6.9− 2.5]×10−10 = 3.8×10−5s = 38µs .

MARKING SCHEME:

1 Formula for τ from ut.

1 G/c3 factor for units.

1 Computing ∆τsat and ∆τground and difference.

9

3. (a) Mark each of the following expressions as good (G) or bad (B)

tensor expressions, according to whether or not they follow the

rules for constructing index expressions. [5]

i. (xα)2 + (xβ)2 = (xc)2

ii. αβγA

α + gβγ = BβCγ

iii. AαβB

β + Cα

βDβ = 0

iv. gαβ ∂f

∂xβ

= V α.

v. MµµC

α +NνKαν = 0.

SOLUTION:

i. B.

The repeated indices in terms like (xα)@ = xαxα are both up,

so they violate the Einstein summation convention. Even in

special relativity, you need the minus sign in the sum when

α = 0, but not for the other values of α, and that is provided

by the metric tensor, which is not mentioned explicitly in

this expression.

ii. G

iii. B

There is a free index α in both terms but one is up and one

is down, which does not make sense. You can add a vector

to a vector but you can’t add a vector to a co-vector in a

coordinate-independent way. A coordinate transformation

applied to this expression would not be consistent for both

terms, since a co-vector transforms the way the coordinate

basis vectors transform but a vector transforms in the inverse

way.

iv. G

This is good because the partial derivative with respect to

xβ is the β component of the gradient co-vector, which has

a downstairs (covariant) index. So the Einstein summation

convention on β is satisfied here. In addition the position of

the free index α on both sides is the same.

v. G

Don’t be worried by the 0 on the right-hand-side. Every

kind of coordinate transformation will just turn 0 into 0,

leaving it invariant. That is because such transformations

are always multiplications. We could have put the second

term on the left-hand side of this expression over on the right

by subtracting it. Then we would have a tensor equation

10

without a zero. But it would still satisfy the rules for tensor

expressions.

MARKING SCHEME: One mark for each correct answer

(b) Classify each of the following hypersurfaces within Minkowski

spacetime as spacelike (S), timelike (T), or null (N). [5]

i. t = 1

ii. x = 1

iii. t+ x = 1

iv. The Lorentz hyperboloid t2 − x2 − y2 − z2 = a2 for some

constant a.

v. The future light cone of the event (0, 1, 0, 0).

SOLUTION: Remember that the classification of a hypersur-

face depends on the direction of its normal vector. If the vector

is timelike, the hypersurface is spacelike; if the vector is space-

like, the hypersurface is timelike. If the vector is null, so is the

hypersurface.

i. S

The normal vector is parallel to the t-axis, so is timelike.

That makes the hypersurface spacelike.

ii. T

By analogy with the last item, the normal is parallel to the

x-axis, so is spacelike. That makes the hypersurface timelike.

iii. N

The surfaces of constant t+x are 45-degree lines in the t−x

plane, then extended in the y- and z-directions. The nor-

mal vector is a null vector parallel to the surface, since it is

orthogonal to itself. So the hypersurface is null.

iv. S

Do not be fooled by the fact that this represents a negative

squared-interval, hence a timelike interval. That does not

make the hypersurface timelike.

v. N

MARKING SCHEME: One mark for each correct answer

11

4. (a) Consider the coordinate transformation u = t − x, v = t + x

in Minkowski spacetime with c = 1. Define eu to be the vec-

tor pointing from the event {u = 0, v = 0, y = 0, z = 0} to

{u = 1, v = 0, y = 0, z = 0}, and define ev to point from

{u = 0, v = 0, y = 0, z = 0} to {u = 0, v = 1, y = 0, z = 0}.

These are the basis vectors associated with the coordinates u and

v, respectively.

i. Show that eu = (et − ex)/2, ev = (et + ex)/2, and draw eu

and ev in a spacetime diagram of the t− x plane. [2]

ii. Show that {eu, ev, ey, ez} are a basis for vectors in Minkowski

spacetime. [1]

iii. Find the components of the metric tensor on this basis. (Hint:

recall that eα ·eβ = gαβ.) This shows that eu and ev are null

and not orthogonal. (They are called a null basis for the t−x

plane.) [2]

SOLUTION:

i. For eu, solving t − x = 1 and t + x = 0 gives t = 1/2, x =

−1/2. This is what is required for the components of eu.

The calculation for ev goes similarly. Drawing them shows

they are along the null lines.

ii. They are clearly linearly independent so they are a basis.

iii. Writing out the dot products explicitly:

guu = eu · eu = 1

4

(ηtt + ηxx) = 0,

gvv = ev · ev = 1

4

(ηtt + ηxx) = 0,

guv = gvu = eu · ev = 1

4

(ηtt − ηxx) = −1

2

.

The components gyy and gzz are unaffected.

iv. The dot products in the previous section show that they are

null and not orthogonal.

MARKING SCHEME:

1 Understanding how to draw the new basis vectors

1 algebra to get basis vectors

1 Understanding what a basis for Minkowski space is

2 Application of ηαβ to the components to get the results.

12

(b) Consider the two-dimensional spacetime spanned by coordinates

(v, x) with the line element

ds2 = −xdv2 + 2dv dx ,

with c = 1. (This is NOT the same spacetime as in the previous

part of this question, despite the similarity of the names of the

coordinates!!)

i. Calculate the slopes of the two lines making up the light cone

at a general point (v, x). [1]

ii. Draw a (v, x) spacetime diagram showing how the light cones

change with x. [2]

iii. Explain from this diagram why it is that a particle can cross

from positive x to negative x but not from negative x to

positive x. (The world-line x = 0 is called a horizon.) [2]

SOLUTION: World lines of particles must lie inside or on the

light cone. There are wordlines that cross from positive x to

negative x, but not the other way.

The light cone is determined by ds2 = 0:

0 = dv(−xdv + 2dx) ,

which has solutions

dv

dx

= 0 and

dv

dx

=

2

x

.

The dv/dx = 2/x portion of the light curve is positive for x > 0

and negative for x < 0, and vertical at x = 0. For x > 0

the x =constant lines (vertical) are timelike, since they give

ds2 = −xdv2 < 0, and so lie inside the light cone. For x < 0 the

x =constant lines are spacelike, since they give ds2 = −xdv2 > 0,

and therefore lie outside the light cone. The light cones therefore

look as follows:

Thursday, 24 October 2013

13

MARKING SCHEME:

1 Explanation that worldlines lie inside light cones

1 Setting ds2 = 0

1 Computing the two solutions for dv/dx.

1 Plot showing the behaviour for x < 0 and x > 0.

1 Justification for which portion of spacetime lies inside the light

cone for x > 0 and x < 0.

14

5. In a certain spacetime geometry the metric is

ds2 = −(1−Ar2)2dt2 + (1−Ar2)2dr2 + r2(dθ2 + sin2 θdφ2) .

(a) Calculate the proper distance along a radial line from r = 0 to a

coordinate radius r = R. Assume AR2 < 1. [2]

SOLUTION: Along a radial line dt = 0, dθ = 0, dφ = 0, so the

distance is

distance =

∫ R

0

dr

√

grr =

∫ R

0

dr

√

(1−Ar2)2 =

∫ R

0

dr(1−Ar2) = R−1

3

AR3 .

MARKING SCHEME:

1 Setting up integral.

1 Evaluating integral.

(b) Compute the area of a sphere of coordinate radius r = R. [2]

SOLUTION: The surface of a sphere has dt = 0, dr = 0, so

the area is

area =

∫ pi

0

dθ

∫ 2pi

0

dφ

√

gθθgφφ .

Since gθθ and gφφ have the same values as in flat space, we get

the same result for the area of the sphere:

area =

∫ pi

0

dθ

∫ 2pi

0

dφR2 sin θ =

[∫ pi

0

dθ sin θ

] [∫ 2pi

0

dφ

]

R2 = 4piR2 .

MARKING SCHEME:

1 Setting up integral.

1 Evaluating integral.

(c) Calculate the 3-volume of a sphere of coordinate radius r = R.

[3]

15

SOLUTION:

volume =

∫ R

0

dr

∫ pi

0

dθ

∫ 2pi

0

dφ

√

grrgθθgφφ

=

[∫ R

0

dr r2(1−Ar2)

] [∫ pi

0

dθ sin θ

] [∫ 2pi

0

dφ

]

= 4pi

[

1

3

R3 − 1

5

AR5

]

=

4pi

3

R3

[

1− 3

5

AR2

]

.

MARKING SCHEME:

1 Setting up integral.

2 Evaluating integral.

(d) Calculate the 4-volume of a 4-dimensional tube bounded by a

sphere of coordinate radius R and two t = constant planes sepa-

rated by a coordinate time T . [3]

SOLUTION:

4− volume =

∫ t1+T

t1

dt

∫ R

0

dr

∫ pi

0

dθ

∫ 2pi

0

dφ

√−gttgrrgθθgφφ

=

[∫ t1+T

t1

dt

] [∫ R

0

dr r2(1−Ar2)2

] [∫ pi

0

dθ sin θ

] [∫ 2pi

0

dφ

]

= 4piT

[

1

3

R3 − 2

5

AR5 +

1

7

A2R7

]

=

4pi

3

R3T

[

1− 6

5

AR2 +

3

7

A2R4

]

.

MARKING SCHEME:

1 Setting up integral.

2 Evaluating integral.

16

6. Consider the following four different metrics, as given by their line

elements:

(i) ds2 = −dt2 + dx2 + dy2 + dz2;

(ii) ds2 = −(1−2M/r) dt2+(1−2M/r)−1 dr2+r2(dθ2+sin2 θ dφ2),

where M is a constant;

(iii)

ds2 =− ∆− a

2 sin2 θ

ρ2

dt2 − 2a2Mr sin

2 θ

ρ2

dt dφ

+

(r2 + a2)2 − a2∆ sin2 θ

ρ2

sin2 θ dφ2 +

ρ2

∆

dr2 + ρ2 dθ2,

where M and a are constants and we have introduced the short-

hand notation ∆ = r2 − 2Mr + a2, ρ2 = r2 + a2 cos2 θ;

(iv) ds2 = −dt2+R2(t) [(1− kr2)−1dr2 + r2(dθ2 + sin2 θ dφ2)], where

k is a constant and R(t) is an arbitrary function of t alone.

The first one is familiar to us as Minkowski spacetime. The names of

the others are, respectively, the Schwarzschild, Kerr, and Robertson–

Walker metrics.

(a) For metric (i), use symmetry arguments to identify by name all

the conserved quantities associated with the four-momentum pα

of a freely falling particle. [1]

SOLUTION: The metric coefficients are all constants, so there

is a conserved quantity associated with each of them. So p0 is

the negative of the energy; px is the x-component of the linear

momentum; and similarly for py and pz.

MARKING SCHEME:

1/2 For saying there are 4 conserved quantities

1/2 For getting the names of energy and linear momentum

(b) Metric (i) can also be put into the following form, using spherical

polar coordinates instead of Cartesian for the spatial coordinates:

(i′) ds2 = −dt2 + dr2 + r2(dθ2 + sin2 θ dφ2).

Does this increase the number of conserved components pα? If

so, name them all. [2]

SOLUTION: This shows that there is a conserved quantity pφ,

17

which is called angular momentum. But because the equator can

be taken in any plane, there are actually three independent angu-

lar momentum components, associated with rotational symmetry

about the three Cartesian coordinate axes.

MARKING SCHEME:

1 For identifying the angular momentum

1 For describing the other two angular momentum components

(c) How many of the metrics (i)-(iv) are spherically symmetric? [1]

SOLUTION: (i), (ii), and (iv). Three in total.

MARKING SCHEME:

1 Acceptable answer is either ’three’ or the list of the three. If

three are listed but one is wrong, then 1/2 mark.

(d) For each of the metrics (ii)-(iv), use symmetry arguments to iden-

tify by name all the conserved quantities associated with the four-

momentum pα of a freely falling particle. [6]

SOLUTION:

(ii): The symmetry coordinates of Schwarzschild are t and φ. By

spherical symmetry, there are also rotations about the other

two axes. So the names are −p0 = energy, pφ = angular

momentum about z-axis, and then the two other angular

momenta.

(iii): Again t and φ are symmetry coordinates, but there is no

spherical symmetry, so there are only two conserved quanti-

ties here, −p0 = energy and pφ = angular momentum about

the Kerr symmetry axis.

(iv): Similar to (ii) BUT now t is not a symmetry because R(t)

makes the metric time-dependent. So the only conserved

quantities that are obvious from this form of the metric are

the three angular momentum components. (Really, the spa-

tial hypersurfaces of constant t are homogeneous, so there is

also a translation symmetry, but this is not clear from the

given form of the metric and has not been studied in class,

so this is not expected in the answers. It is mentioned in the

next problem, in fact.)

MARKING SCHEME:

18

1 For (ii), naming the energy and z-angular momentum (1/2

mark for identifying t and φ without names).

1 For (ii), including the other two angular momentum compo-

nents.

1 For (iii), identifying t and φ as symmetry coordinates

1 For (iii), naming the associated quantities as energy and angu-

lar momentum

1 For (iv), understanding that t is not a symmetry

1/2 For (iv), stating that φ has a symmetry and leads to a con-

served angular momentum

1/2 For (iv), including the other two angular momentum com-

ponents

19

7. The three-dimensional line element

ds2 = dr2 + (r2 + a2)(dθ2 + sin2 θ dφ2),

where a is a constant and r has the range −∞ < r <∞, defines what

is called a wormhole.

(a) Show that this is a purely spatial line-element, i.e. that there are

no negative intervals. [1]

SOLUTION: All the terms in this line element are positive. So

the intervals are positive-definite, which means spacelike.

MARKING SCHEME:

1 Mention positive-definite or something equivalent.

(b) Show that this is a spherically symmetric space. [1]

SOLUTION: (i) It contains the term (dθ2+sin2 θ dφ2) involving

coordinates θ and φ. This is the line element of a sphere in flat

space.

(ii) The coordinates θ and φ do not appear anywhere else in the

line element, so that the symmetries of the sphere apply to this

spacetime.

MARKING SCHEME:

1 At least mention the presence of the spherical part of the line

element. If the absence of other terms involving these coor-

dinates is mentioned, give a 1 point bonus, but do not exceed

10 for total marks for the question.

(c) Show that r = 0 is not a point, but is rather a sphere of radius

a, and that this sphere has the minimum circumference of any of

the spheres in this space. [2]

SOLUTION: The circumference of each sphere of constant r is

given in terms of the coefficient of the spherical part, i.e. r2 + a2.

It is 2pi

√

(r2 + a2). So when r = 0 the circumference is 2pia,

which is non-zero and is the minimum of the circumferences.

MARKING SCHEME:

1 For knowing how to compute the circumference

1 For showing that r = 0 gives the minimum circumference.

20

(d) Compute all the non-vanishing Christoffel symbols for this met-

ric. You may compute them directly by taking derivatives of the

metric, or you may use the method of the Lagrangian principle

that we introduced in the notes, where in this case the Lagrangian

is the proper distance along a curve. [6]

SOLUTION: In the lectures we presented two ways to compute

the Christoffel symbols, either using the action principle (which

was recommended as the easier way for a general metric) or the

direct way using derivatives of the metric tensor. I first give the

solution using the action principle method, followed by its mark-

ing scheme, and then below I give the solution using derivatives

of the metric, and its marking scheme.

Because this is a three-dimensional spatial metric I will use Latin

letters (a, b, c, . . . , j, k, . . .) for the indices, just to emphasise the

difference from 4D.

Action Principle method:

i. The Lagrangian is L = [gjkx˙

j x˙k]1/2, where a dot indicates a

derivative with respect to proper distance s. (We shall use

this convention here for clarity, even though a dot normally

indicates a time-derivative, and there is no time coordinate

here. Sometimes space derivatives are denoted by a prime

(′), but elsewhere in this assignment we have used primes to

denote alternative coordinate systems, not derivatives. So

we shall stick with dots here for derivatives with respect to

proper distance s. Thus, A˙ := dA/ds for any A. ) As in

the case of the 4-dimensional action principle, we can scale

the parameter s so that the numerical value of L is 1, which

can be used in expressions when it is not being differentiated

with respect to the variables (r, θ, φ) and (r˙, θ˙, φ˙). So we

write out the Lagrangian as

L :=

[

r˙2 + (r2 + a2)(θ˙2 + sin2 θφ˙2)

]1/2

.

ii. The derivatives with respect to the velocities are (and using

L = 1 after the differentiation):

∂L

∂r˙

= r˙,

∂L

∂θ˙

= (r2 + a2)θ˙,

∂L

∂φ˙

= (r2 + a2) sin2 θ φ˙.

21

iii. Their derivatives with respect to s are

d

ds

(

∂L

∂r˙

) = r¨,

d

ds

(

∂L

∂θ˙

) = (r2 + a2)θ¨ + 2rθ˙r˙,

d

ds

(

∂L

∂φ˙

) = (r2+a2) sin2 θ φ¨+2r sin2 θ φ˙r˙+2(r2+a2) sin θ cos θ φ˙θ˙.

iv. The derivatives of the Lagrangian with respect to the coor-

dinates are

∂L

∂r

= rθ˙2 + r sin2 θ φ˙2,

∂L

∂θ

= (r2 + a2) sin θ cos θ φ˙2,

∂L

∂φ

= 0.

v. The Euler-Lagrange (E-L) equations from these,

d

ds

∂L

∂x˙a

− ∂L

∂xa

= 0,

is written in terms of the Christoffel symbols in this way:

x¨a + Γabcx˙

bx˙c = 0.

Putting the previous calculations together to get this gives

three equations, from each of which we can deduce both the

non-zero Christoffel symbols and the zero Christoffel sym-

bols. The three equations are:

A. For a = r, we have:

r¨ − rθ˙2 − r sin2 θ φ˙2 = 0.

This gives all the non-zero Γrbc.

Γrθθ = −r, Γrφφ = −r sin2 θ.

The ones with lower index combinations that don’t ap-

pear here all vanish.

B. For a = θ we have to clear a factor in front of θ¨ to get the

right form of the equation, and this is done by dividing

by r2 + a2. The E-L equation is then:

θ¨ +

2r

r2 + a2

θ˙r˙ − sin θ cos θ φ˙2 = 0.

22

For the first time here we encounter a case where the

indices b 6= c, so we have to recognize that such a term

enters the summation as both Γθbc and Γ

θ

cb. Then the

symmetry of the Christoffel symbol on exchange of b and

c means that these two terms are equal, and therefore

their sum (which is what appears in the E-L equation)

will contain double the amount from a single Christoffel

symbol of this type. Therefore we divide the appropriate

coefficients by 2 to get the right answer for a single sym-

bol. This consideration leads to the following non-zero

Γθbc terms:

Γθθr = Γ

θ

rθ =

r

r2 + a2

, Γθφφ = − sin θ cos θ.

C. Finally, for a = φ, the calculation proceeds in close anal-

ogy to the previous one, so that the E-L equation, after

dividing by (r2 + a2) sin2 θ, is

φ¨+

2r

r2 + a2

φ˙r˙ + 2

cos θ

sin θ

φ˙θ˙ = 0.

And again proceeding similarly to the previous one for

the cases where the bottom indices are different, we infer

that the non-zero Γφbc terms are:

Γφφr = Γ

φ

rφ =

r

r2 + a2

, Γφθφ = Γ

φ

φθ = cot θ.

D. We remark that the Christoffel symbols involving only

the angular coordinates are the same as we get in flat

space for spherical coordinates, which is not surprising.

This is a good check on whether mistakes have been made

along the way.

MARKING SCHEME: Marking scheme for the action prin-

ciple method of solution:

1 Correct Lagrangian. Can involve σ at first but student must

set σ = s when evaluating L.

1 Correct expressions for the derivatives of the Lagrangian with

respect to the “velocities”. Half-mark for at least two correct.

1 Correct derivatives with respect to s of the velocity derivatives.

Again half-marks possible.

1 Correct derivatives of the Lagrangian with respect to the co-

ordinates. Half-marks possible.

1 Correct form of the three E-L equations. Half-marks possible

23

1 Correct expressions for at least 4 of the 6 non-zero Christoffel

symbols. Earn 1/2 mark if 2-3 of them are right. Lose 1/2

for not remarking on or correctly handling the symmetry on

the lower two indices.

SOLUTION: The other way to approach this is to use deriva-

tives of the metric tensor in Eq (98) of the lecture notes (changed

to Latin indices):

Γabc =

1

2

gaj

(

∂gjb

∂xc

+

∂gjc

∂xb

− ∂gbc

∂xj

)

.

This is considerably simplified if one uses some key facts. First,

the metric is diagonal, so we will always have a = j in this. And

second, the metric is reflection-symmetric in φ, so that a Christof-

fel symbol that contains one or three φ-indices will automatically

vanish.

The calculations are not difficult, so I will just illustrate two.

First, the calculation of Γrθθ:

Γrθθ =

1

2

grj(

∂gjθ

∂θ

+

∂gθj

∂θ

− ∂gθθ

∂xj

).

But because the metric is diagonal, so is the inverse, and indeed

grr = 1/grr = 1. Only j = r survives in this summation, leaving

Γrθrθ =

1

2

(2

∂grθ

∂θ

− ∂gθθ

∂r

) = −r.

This is the same as we had in the other method, of course.

Next, the calculation of Γθrθ. We note from the start that g

θθ =

1/(r2 + a2) and that gθr = 0, so that we only have

Γθrθ = Γ

θ

θr =

1

2

1

r2 + a2

(

∂gθθ

∂r

) =

r

r2 + a2

.

The rest of the Christoffel symbols can be obtained in this way,

and their values need to match up with the ones we had above.

So do the ones that vanish.

MARKING SCHEME: Marking scheme for the second method

of deriving the Christoffel symbols:

1 Correctly observing that the metric is diagonal

1 Correctly calculating the thee terms in the inverse metric. 1/2

for getting two right.

24

up to 1 A bonus of up to 1 for noting that reflection symmetry

on φ implies that all terms with one or three φ-indices van-

ish, so don’t need to be computed. The total marks for the

question must not exceed 10.

4 For correctly computing at least 4 of the 6 non-vanishing Christof-

fel symbols. Earn 2 marks for getting 2-3 of them right. Lose

1/2 mark for each algebraic error in the calculation.

25

8. The static, weak-field line element that describes the geometry of a

Newtonian star with gravitational potential Φ is given by

ds2 = −(1 + 2Φ)dt2 + (1− 2Φ)[dx2 + dy2 + dz2] (1)

Consider a hypothetical massless particle that moves at the speed of

light but does not interact with normal matter. The particle crosses

the Earth’s orbit around the Sun and goes straight through the centre

of the Sun. If the Sun had not been there, then the crossing would

have taken a time equal to T0 = 2Rorbit, where Rorbit is equal to 1

Astronomical Unit (AU), or 1.5× 1011 m.

(a) Show by using the line element that the difference between the

true time through the Sun and the time T0 is given by

∆T = −4

∫ R

0

Φ(r)dr, (2)

where R is the radius of the Sun, 7× 108 m. [4]

SOLUTION: The particle moves on a radial line, which without

loss of generality we can take to be the x-axis. Then dy = dz = 0.

It moves at the speed of light, so ds2 = 0. This gives us a relation

between dt and dx:

dt

dx

=

[

1− 2Φ

1 + 2Φ

]1/2

.

Since in the Newtonian case we can take |Φ| 1, we can re-write

this keeping terms of first order in Φ:

dt

dx

= [(1−2Φ)(1−2Φ+ . . .)]1/2 = (1−4Φ+ . . .)1/2 = 1−2Φ+ . . . .

In this part of the problem, we are only asked about the time

crossing the Sun itself. We can get this by integrating dt/dx. If

we integrate this along x, for a spherical Sun, that is equivalent

to integrating radially along a diameter of the Sun. The inward

and outward segments will contribute the same, so the result is

T = 2

∫ R

0

[1− 2Φ(r)]dr = 2R − 4

∫ R

0

Φ(r)dr.

From this it is clear that the excess time crossing the sun is as

given in Eq. 2 in the problem. (Unfortunately there was a mis-

print in the problem, which asked for the difference to the time

26

T0, which had been defined as the time to cross the Earth’s orbit.

The problem should have asked for the crossing time of the whole

orbit.)

MARKING SCHEME:

1 using ds2 = 0

1 reducing the dimensionality to just a motion along x (or equiv-

alent)

1 getting the right equation for dt/dr in the approximation of

small Φ.

1 doing the integral correctly. Marking is flexible to account for

possible confusion caused by the error in the question.

(b) We know that outside the Sun, the potential is Φ = −GM/r,

where GM = 1.5 × 103 m is the geometrised mass of the Sun

in units where c = 1. Rather than develop a good model of the

potential inside the Sun, we shall make the crude approxima-

tion that it is constant at the value it has at the Sun’s surface,

−GM/R. The error induced by this approximation will be

small since the majority of our particle’s journey is outside the

Sun. Using this approximation, show that the time difference is

∆T = 4M(1 + ln

Rorbit

R

).

[4]

SOLUTION: Here the time-difference across the whole orbit

is asked for. The result is to integrate Eq 2 for both the interior

of the Sun and the exterior. The interior is simple because the

integrand is constant:

∆Tinterior = −4−GM

R

R = 4GM.

The exterior is an integral fromR toRorbit, and the only variable

inside the integral is 1/r. So the integration is

∆Texterior = −4(−GM)

∫ Rorbit

R

1

r

dr = 4GM ln

Rorbit

R

.

The full answer is the sum of these two, and gives us the answer

required in the question. That answer is framed with G = 1, but

we move between different unit systems in this course, so that

should not give a problem.

MARKING SCHEME:

27

1 Setting up the problem correctly as an integral from the Sun’s

centre to the Earth’s orbit

1 Correct interior integration

1 Correct exterior integration

1 Combining the two for the correct answer.

(c) Since this is positive, the particle suffers a delay on passing through

the Sun. Compute, in seconds, the size of the delay. This is called

the Shapiro delay. [2]

SOLUTION: Since M = 1.5 km, its conversion to time is

dividing by c = 3× 105 km/s, which gives M = 5µs. So 4M =

20µs. We are given that Rorbit = 1.5 × 1011 m, and R = 7 ×

108 m, so their ratio is 214, and the log is 5.4. This makes ∆T =

127µs. This is a time difference which can be measured very

accurately with atomic clocks, which are accurate to picoseconds

or better.

MARKING SCHEME:

1 Correct conversion of M to seconds.

1 Correct arithmetic to get final answer.

28

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