B5: Uno
cial Past Paper Solutions (2011-2016)
Toby Adkins
May 23, 2018
Disclaimer: These solutions are produced for student use, and have no connection to the
department what-so-ever. As such, they cannot be guaranteed to be correct, nor have any
bearing on the actual mark schemes that were used in the exams. Furthermore, they are
almost literally word for word my practise solutions, so they are far from 'model' solu-
tions. The formatting may suer somewhat as a result. As to the question of why I typed
them, as supposed to just scanning my solutions; anyone that knows me personally will
attest to the dire state of my handwriting. For queries and corrections, please email me at
2011
Question 1
Consider the zeroth component of the geodesic equation:
d
(2c2 αṫ) = 0
dλ
As ṙ = 0,
2c2 αṫ = constant −→ ṫ = constant
Clearly, with appropriate choice of integration constants, we are at liberty to choose t = λ.
Now, consider the metric along dr = dφ = 0 (ie. at
xed r and φ) for circular orbits:
2 2
2 2 2 2 2 2 2 2
rs dt 2 dθ
−c dτ = −c αdt + r dθ −→ −c = −c 1− +r
r dτ dτ
Dierentiating with respect to r:
2 2
c2 rs
dt dθ
0=− 2 + 2r
r dτ dτ
such that
2
dθ GM
=
dτ r3
as required.
Again, consider the metric:
rs 2 2
−c2 dτ 2 = − 1 − c dt + r2 dθ2
r
1
,Toby Adkins B5 Past Papers
For the station on the North Pole, dθ = 0, such that
rs 2 2 rs 1/2
−c2 dτN
2
P = − 1 − c dt −→ dτN P = 1 − dt
R R
where dt is the coordinate time associated with some observer in asymptotically
at space-
time, and we have evaluated at r = R. Similarly, for Satellite A, we have that
1/2
rs GM
dτA = 1− − 2 dt
r c r
Taking the ratio of these two times, we have that
rs
1/2 3GM 1/2
− GM
∆τA 1− r c2 r
1− 2
c r
= =
∆τN P rs 1/2 2GM 1/2
1− R 1− c2 R
However, GM/(c2 r)
1 as we are in asymptotically
at spacetime, and we are on large
spatial scales in comparison to the Earth's Schwarzchild radius. This means that we can
expand our expression as
∆τA 3GM GM GM 3GM
≈ 1− 1+ 2 ≈1+ 2 −
∆τN P 2c2 r c R c R 2c2 r
where we have only kept terms
rst order in rs /r. We have thus obtained the desired result.
In the second case, the station on the equator has
rs 1/2
dτEQ = 1 − dt
R
as we are again stationary with respect to θ at r = R. Now for the satellite, θ = π/2 and
θ̇ = 0, meaning that we are simply left with
rs 1/2
dτB = 1 − dt
r
Taking the ratio and expanding,
rs 1/2
∆τB 1− r
rs rs GM GM
= ≈ 1− 1+ =1+ 2 − 2
∆τEQ rs 1/2 2r 2R c R c r
1− R
as required. The dierence between these results is that the
st includes both the gravita-
tional redshift and Doppler shift due to the relative motion of the satellite and the observer,
while the second includes only the gravitational redshift (as the observer and satellite are
stationary with respect to one another in this case).
Question 2
Consider the Euler-Lagrange equations
d ∂L ∂L
=
dλ ∂ ẋρ ∂xρ
Substituting our Lagrangian in
∂L
= 2gµν ẋµ δ νρ = 2gµρ ẋµ
∂ ẋρ
∂L
= ∂ρ gµν ẋµ ẋν
∂xρ
2
,Toby Adkins B5 Past Papers
Thus, geodesics in this spacetime satisfy
d
(2gµρ ẋµ ) = ∂ρ gµν ẋµ ẋν
dλ
We are considering a metric for which g00 = −1, and gij = −a2 .
• Consider the µ = 0 component. As the metric is diagonal, we clearly have that
d
2g00 ẋ0 = ∂0 gij ẋi ẋj
dλ
d
(−2c2 ṫ) = 2aa0 ṙ2 + r2 θ̇2 + r2 sin2 θφ̇2
dλ
where a0 = da/dt.
• µ = r:
d
(2grr ṙ) = ∂r gij ẋi ẋj
dλ
d
2a2 ṙ = 2a2 r θ̇2 + sin2 θφ̇2
dλ
• µ = θ:
d
2gθθ θ̇ = ∂θ gij ẋi ẋj
dλ
d 2 2
2a r θ̇ = 2a2 r2 sin θ cos θφ̇2
dλ
• µ = φ:
d
(2gφφ ) = ∂φ gij ẋi ẋj
dλ
d 2 2 2
2a r sin θφ̇ = 0
dλ
as the metric is independent of φ. We have thus derived the required geodesic equa-
tions.
Now, consider the form of the geodesic equations
ẍρ + Γρµν ẋµ ẋν = 0
If we dierentiate the LHS of the previous expressions that we have derived, we can compare
the resultant expressions with this form to read o the connection coe
cients.
• µ = 0:
aa0 2 2 2 2 2 2
ẗ + ṙ + r θ̇ + r sin θ φ̇ =0
c2
We thus have
aa0
Γtrr =
c2
aa0
Γtθθ = 2 r2
c
aa0 2 2
Γtφφ = 2 r sin θ
c
3
, Toby Adkins B5 Past Papers
• µ = r:
2a2 r̈ + 2aȧṙ = 2a2 r θ̇2 + sin2 θφ̇2
By the chain rule,
da da dt
ȧ = = = a0 ṫ
dλ dt dλ
We thus have that
a0
r̈ + ṫṙ − r θ̇2 + sin2 θφ̇2 = 0
a
such that the connection coe
cients are
a0
Γrtr =
a
Γrφφ = −r sin2 θ
Γrθθ = −r
• µ = θ:
2a2 r2 θ̈ + 2aa0 ṫθ̇r2 + 2a2 (2rṙ)θ̇ = 2a2 r2 sin θ cos θφ̇2
a0 2
θ̈ + ṫθ̇ + ṙθ̇ − sin θ cos θφ̇2 = 0
a r
where we have again made use of the chain rule. The associated connection coe
-
cients are
a0
Γθtθ =
a
1
Γθθr =
r
Γθθφ = − sin θ cos θ
• µ = φ:
2aa0 ṫr2 sin2 θφ̇ + 2a2 r2 sin2 θφ̈ + 2a2 (2rṙ) sin2 θφ̇ + 2a2 r2 cos θ sin θ θ̇φ̇ = 0
a0 2
φ̈ + ṫφ̇ + ṙφ̇ + cot θ θ̇φ̇ = 0
a r
Thus,
a0
Γφtφ =
a
1
Γφrφ =
r
Γφφθ = cot θ
We have thus derived all non-zero connection coe
cients for this spacetime.
Let us now consider radial geodesics in the orbital plane, such that θ̇ = φ̇ = 0. Then our
remaining geodesic equations are
d
−2cṫ = 2aa0 ṙ2
dλ
d
2a2 ṙ = 0
dλ
4
cial Past Paper Solutions (2011-2016)
Toby Adkins
May 23, 2018
Disclaimer: These solutions are produced for student use, and have no connection to the
department what-so-ever. As such, they cannot be guaranteed to be correct, nor have any
bearing on the actual mark schemes that were used in the exams. Furthermore, they are
almost literally word for word my practise solutions, so they are far from 'model' solu-
tions. The formatting may suer somewhat as a result. As to the question of why I typed
them, as supposed to just scanning my solutions; anyone that knows me personally will
attest to the dire state of my handwriting. For queries and corrections, please email me at
2011
Question 1
Consider the zeroth component of the geodesic equation:
d
(2c2 αṫ) = 0
dλ
As ṙ = 0,
2c2 αṫ = constant −→ ṫ = constant
Clearly, with appropriate choice of integration constants, we are at liberty to choose t = λ.
Now, consider the metric along dr = dφ = 0 (ie. at
xed r and φ) for circular orbits:
2 2
2 2 2 2 2 2 2 2
rs dt 2 dθ
−c dτ = −c αdt + r dθ −→ −c = −c 1− +r
r dτ dτ
Dierentiating with respect to r:
2 2
c2 rs
dt dθ
0=− 2 + 2r
r dτ dτ
such that
2
dθ GM
=
dτ r3
as required.
Again, consider the metric:
rs 2 2
−c2 dτ 2 = − 1 − c dt + r2 dθ2
r
1
,Toby Adkins B5 Past Papers
For the station on the North Pole, dθ = 0, such that
rs 2 2 rs 1/2
−c2 dτN
2
P = − 1 − c dt −→ dτN P = 1 − dt
R R
where dt is the coordinate time associated with some observer in asymptotically
at space-
time, and we have evaluated at r = R. Similarly, for Satellite A, we have that
1/2
rs GM
dτA = 1− − 2 dt
r c r
Taking the ratio of these two times, we have that
rs
1/2 3GM 1/2
− GM
∆τA 1− r c2 r
1− 2
c r
= =
∆τN P rs 1/2 2GM 1/2
1− R 1− c2 R
However, GM/(c2 r)
1 as we are in asymptotically
at spacetime, and we are on large
spatial scales in comparison to the Earth's Schwarzchild radius. This means that we can
expand our expression as
∆τA 3GM GM GM 3GM
≈ 1− 1+ 2 ≈1+ 2 −
∆τN P 2c2 r c R c R 2c2 r
where we have only kept terms
rst order in rs /r. We have thus obtained the desired result.
In the second case, the station on the equator has
rs 1/2
dτEQ = 1 − dt
R
as we are again stationary with respect to θ at r = R. Now for the satellite, θ = π/2 and
θ̇ = 0, meaning that we are simply left with
rs 1/2
dτB = 1 − dt
r
Taking the ratio and expanding,
rs 1/2
∆τB 1− r
rs rs GM GM
= ≈ 1− 1+ =1+ 2 − 2
∆τEQ rs 1/2 2r 2R c R c r
1− R
as required. The dierence between these results is that the
st includes both the gravita-
tional redshift and Doppler shift due to the relative motion of the satellite and the observer,
while the second includes only the gravitational redshift (as the observer and satellite are
stationary with respect to one another in this case).
Question 2
Consider the Euler-Lagrange equations
d ∂L ∂L
=
dλ ∂ ẋρ ∂xρ
Substituting our Lagrangian in
∂L
= 2gµν ẋµ δ νρ = 2gµρ ẋµ
∂ ẋρ
∂L
= ∂ρ gµν ẋµ ẋν
∂xρ
2
,Toby Adkins B5 Past Papers
Thus, geodesics in this spacetime satisfy
d
(2gµρ ẋµ ) = ∂ρ gµν ẋµ ẋν
dλ
We are considering a metric for which g00 = −1, and gij = −a2 .
• Consider the µ = 0 component. As the metric is diagonal, we clearly have that
d
2g00 ẋ0 = ∂0 gij ẋi ẋj
dλ
d
(−2c2 ṫ) = 2aa0 ṙ2 + r2 θ̇2 + r2 sin2 θφ̇2
dλ
where a0 = da/dt.
• µ = r:
d
(2grr ṙ) = ∂r gij ẋi ẋj
dλ
d
2a2 ṙ = 2a2 r θ̇2 + sin2 θφ̇2
dλ
• µ = θ:
d
2gθθ θ̇ = ∂θ gij ẋi ẋj
dλ
d 2 2
2a r θ̇ = 2a2 r2 sin θ cos θφ̇2
dλ
• µ = φ:
d
(2gφφ ) = ∂φ gij ẋi ẋj
dλ
d 2 2 2
2a r sin θφ̇ = 0
dλ
as the metric is independent of φ. We have thus derived the required geodesic equa-
tions.
Now, consider the form of the geodesic equations
ẍρ + Γρµν ẋµ ẋν = 0
If we dierentiate the LHS of the previous expressions that we have derived, we can compare
the resultant expressions with this form to read o the connection coe
cients.
• µ = 0:
aa0 2 2 2 2 2 2
ẗ + ṙ + r θ̇ + r sin θ φ̇ =0
c2
We thus have
aa0
Γtrr =
c2
aa0
Γtθθ = 2 r2
c
aa0 2 2
Γtφφ = 2 r sin θ
c
3
, Toby Adkins B5 Past Papers
• µ = r:
2a2 r̈ + 2aȧṙ = 2a2 r θ̇2 + sin2 θφ̇2
By the chain rule,
da da dt
ȧ = = = a0 ṫ
dλ dt dλ
We thus have that
a0
r̈ + ṫṙ − r θ̇2 + sin2 θφ̇2 = 0
a
such that the connection coe
cients are
a0
Γrtr =
a
Γrφφ = −r sin2 θ
Γrθθ = −r
• µ = θ:
2a2 r2 θ̈ + 2aa0 ṫθ̇r2 + 2a2 (2rṙ)θ̇ = 2a2 r2 sin θ cos θφ̇2
a0 2
θ̈ + ṫθ̇ + ṙθ̇ − sin θ cos θφ̇2 = 0
a r
where we have again made use of the chain rule. The associated connection coe
-
cients are
a0
Γθtθ =
a
1
Γθθr =
r
Γθθφ = − sin θ cos θ
• µ = φ:
2aa0 ṫr2 sin2 θφ̇ + 2a2 r2 sin2 θφ̈ + 2a2 (2rṙ) sin2 θφ̇ + 2a2 r2 cos θ sin θ θ̇φ̇ = 0
a0 2
φ̈ + ṫφ̇ + ṙφ̇ + cot θ θ̇φ̇ = 0
a r
Thus,
a0
Γφtφ =
a
1
Γφrφ =
r
Γφφθ = cot θ
We have thus derived all non-zero connection coe
cients for this spacetime.
Let us now consider radial geodesics in the orbital plane, such that θ̇ = φ̇ = 0. Then our
remaining geodesic equations are
d
−2cṫ = 2aa0 ṙ2
dλ
d
2a2 ṙ = 0
dλ
4
