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