General Relativity
Homework 3 Solutions
1. Carroll Problem 3.5. Consider a 2-sphere with coordinates (θ, φ) and metric
ds2 = dθ2 + sin2 θdφ2 .
(a) Show that lines of constant longitude (φ = constant) are geodesics, and that the only line of
constant latitude (θ = constant) that is a geodesic is the equator (θ = π/2).
(b) Take a vector with components V µ = (1, 0) and parallel-transport it once around a circle of
constant latitude. What are the components of the resulting vector, as a function of θ?
Solution:
(a) To verify a curve is a geodesic we need to write the geodesic equation for this metric. First we
need the connection coefficients, which can be found in Carroll’s equation (3.154). The nonzero
coefficients are
Γθφφ = − sin θ cos θ and Γφθφ = Γφφθ = cot θ.
The geodesic equation is
σ
ρ
d2 xµ
µ dx dx
= 0,
+
Γ
ρσ
dλ2
dλ dλ
so the θ and φ components are
dφ 2
d2 φ
dθ dφ
d2 θ
− sin θ cos θ
= 0 and
+ 2 cot θ
= 0.
2
2
dλ
dλ
dλ
dλ dλ
If φ is constant, the first equation becomes
the curve xµ = (λ, φ) is a geodesic.
For θ constant, the two equations become
sin θ cos θ
d2 θ
dλ2
= 0, while the second equation is trivial. Thus
dφ 2
d2 φ
= 0.
= 0 and
dλ
dλ2
The second equation is satisfied for any φ = aλ + b for a, b ∈ R. The first equation is satisfied for
θ = nπ/2 for n = 0, 1, 2,... Since θ ∈ (0, π), the only solution is θ = π/2, i.e. the equator.
(b) We want to parallel transport V µ = (1, 0) around a path with θ = constant. We can choose any
parametrization for φ, so let’s use the general one above, with φ = 0 fixed to λ = 0: φ = aλ. That
is, the path is xµ (λ) = (constant, aλ + b). The parallel transport equation is
d µ
dxσ ρ
V + Γµσρ
V = 0.
dλ
dλ
The µ = θ, φ components are
dV θ
− sin θ cos θ a V φ = 0
dλ
(1)
dV φ
+ cot θ a V θ = 0,
(2)
dλ
To solve this system, differentiate (2) with respect to λ (keeping in mind that θ is a constant),
θ
solve for dV
dλ , plug it into (1) and do a little algebra to obtain
d2 V φ
+ cos2 θ a2 V φ = 0.
dλ2
1
Similarly, differentiating (1) and plugging into (2),
d2 V θ
+ cos2 θ a2 V θ = 0.
dλ2
These equations are just those for a simple harmonic oscillator with frequency a cos θ, so the
solutions are
V θ (λ) = A cos(aλ cos θ) + B sin(aλ cos θ)
V φ (λ) = C cos(aλ cos θ) + D sin(aλ cos θ).
The boundary condition is V (φ = 0) = V (aλ = 0) = (1, 0), so A = 1 and C = 0. To pin down
the last two constants, appeal to one of the original equations, say (2):
dV φ
= −a cot θ cos (aλ cos θ)
dλ
Z
V φ = a cot θ dλ cos (aλ cos θ)
sin (aλ cos θ)
a cos θ
sin (aλ cos θ)
Vφ =−
sin θ
So C = 0 and D = −1/ sin θ. The solution to the parallel transport equation is then
sin(aλ cos θ)
µ
V (λ) = cos(aλ cos θ), −
.
sin θ
V φ = −a cot θ
After parallel transporting V µ around a circle of constant latitude (φi = 0 to φf = 2π is (aλ)i = 0
to (aλ)f = 2π), the resulting vector is
sin(2π cos θ)
µ
V (2π) = cos(2π cos θ), −
.
sin θ
As a check we can see that the vector is unchanged (i.e. V µ (2π) = (1, 0)) if we parallel transport
around the equator (θ = π/2), as expected.
2. Carroll Problem 3.6. A good approximation to the metric outside the surface of the Earth is provided
by
ds2 = −(1 + 2Φ)dt2 + (1 − 2Φ)dr2 + r2 (dθ2 + sin2 θdφ2 ),
where
GM
r
may be thought of as the familiar Newtonian gravitational potential. Here G is Newton’s constant and
M is the mass of the Earth. For this problem Φ may be assumed to be small.
Φ=−
(a) Imagine a clock on the surface of the Earth at a distance R1 from the Earth’s center, and another
clock on a tall building at distance R2 from the Earth’s center. Calculate the time elapsed on
each clock as a function of the coordinate time t. Which clock moves faster?
(b) Solve for a geodesic corresponding to a circular orbit around the equator of the Earth (θ = π/2).
What is dφ/dt?
(c) How much proper time elapses while a satellite at radius R1 (skimming along the surface of the
Earth, neglecting air resistance) completes one orbit? You can work to first order in Φ if you like.
Plug in the actual numbers for the radius of the Earth and so on to get an answer in seconds.
How does this number compare to the proper time elapsed on the clock stationary on the surface?
2
Solution:
(a) For a clock on the surface of the Earth, the world-line (parameterized by t) is xµ = (t, R1 , θ, ωt),
π rad
where ω is the angular velocity of the Earth (ω = 12
hrs ). Now calculate the proper-time:
Z tr
Z tq
dxµ dxν 0
∆τ =
−gµν
dt =
(1 + 2Φ) − R12 sin2 θω 2 dt0
dt
dt
0
0
r
2GM
∆τ = 1 −
− R12 sin2 θω 2 ∆t.
R1
Now we need to reinstate the factors of c. 2GM
R1 has units of velocity squared, so we should divide
2
that by c . The same goes for the second term under the square root. Also note that R1 ω << c,
2GM
so we can drop that term (we also have R12 ω 2 << 2GM
R1 , which justifies leaving in R1 ). We are
left with
r
2GM
∆τ ≈ 1 − 2
∆t.
c R1
For the clock on a tall building we replace R1 with R2 and see that the proper time is decreased.
Note that this effect is purely due to the difference in gravitational potential (the effect due to
time dilation associated with the difference in speeds is negligible).
(b) The geodesic equation for xµ = r is
d2 r
dxρ dxσ
+ Γrρσ
= 0.
2
dλ
dλ dλ
For paths with constant r and θ, this becomes
Γrtt
dφ 2
dt 2
dφ dt
= 0.
+ Γrφφ
+ 2Γrφt
dλ
dλ
dλ dλ
The necessary connection coefficients are
1 rρ
1
g (∂t gtρ + ∂t gtρ − ∂ρ gtt ) = − g rr ∂r gtt
2
2
1
1
2GM
GM
Γrtt =
∂r (1 −
)= 2
2 (1 − 2Φ)
r
r (1 − 2Φ)
Γrtt =
1 rρ
1
g (∂φ gφρ + ∂φ gφρ − ∂ρ gφφ ) = − g rr ∂r gφφ
2
2
1
1
r sin2 θ
=−
∂r (r2 sin2 θ) = −
2 (1 − 2Φ)
(1 − 2Φ)
Γrφφ =
Γrφφ
1 rρ
g (∂φ gtρ + ∂t gφρ − ∂ρ gφt ) = 0.
2
So the relevant geodesic equation becomes
Γrφt =
2
r sin2 θ
dφ
−
= 0.
(1 − 2Φ) dλ
r
dφ
GM dt
⇒
=
(3)
dλ
r3 dλ
This does not give us enough information for the geodesic, so next consider the equation for
xµ = t,
ρ
σ
d2 t
t dx dx
+
Γ
= 0.
ρσ
dλ2
dλ dλ
GM
r2 (1 − 2Φ)
dt
dλ
2
3
Again, considering only paths such that r is constant and θ = π/2, this simplifies to
dt 2
dφ 2
dt dφ
d2 t
+ Γttt
+ Γtφφ
+ Γttφ
= 0.
2
dλ
dλ
dλ
dλ dλ
The necessary connection coefficients are
Γtµν =
1 tt
g (∂µ gνt + ∂ν gµt − ∂t gµν ),
2
so for µ, ν ∈ {t, φ} the connection coefficients will all be zero. So this geodesic equation is just
d2 t
= 0.
dλ2
The simplest non-trivial solution is t = λ. Taking this as our parameterization, (3) implies
r
GM
dφ
.
=
dt
r3
Putting it all together, the (up to reparameterizations) the geodesic corresponding to a circular
orbit (of radius r) around the equator (θ = π/2) is
r
GM
µ
x (t) = (t, r, π/2,
t).
r3
(c) For the satellite skimming the surface of the Earth (at θ = π/2), the elapsed proper time for a
complete orbit is
s
2GM
R2 ω 2 2π
∆τ = 1 − 2
,
− 12
c R1
c
ω
q
and we have included the term related to the motion of the satellite because
where ω = GM
R3
1
it’s of the same order as the gravitational potential.
s
r
r
2GM
R13
R13
3R2
R12 GM
∆τ = 2π 1 − 2
= 2π
− 21
− 2
3
c R1
c R1
GM
GM
c
s
∆τ = 2π
(6.371 × 106 m)3
(6.674 ×
2
10−11 Nm
)(5.972
kg2
×
1024 kg)
−
3(6.371 × 106 m)2
= 5061s
(2.998 × 108 ms )2
Now to calculate the the elapsed proper-time for the clock stationary on the surface of the Earth,
borrow the result from part (a), setting θ = π/2.
r
r
r
r
2R2
2GM 2π
2GM
R13
R13
= 2π 1 − 2
= 2π
− 21
∆τ = 1 − 2
c R1 ω
c R1 GM
GM
c
s
(6.371 × 106 m)3
2(6.371 × 106 m)2
∆τ = 2π
−
= 5061s
2
(2.998 × 108 ms )2
(6.674 × 10−11 Nm2 )(5.972 × 1024 kg)
kg
For a clock stationary on the surface the elapsed proper-time (during the orbit of the satellite) is
the same as that of the satellite! This is because both are the same distance from the center of
the Earth, and therefore they both have the same gravitational potential (and the effect due to
time dilation clearly has a negligible effect).
To fully convince ourselves that this effect is due purely to gravity, consider a stationary clock
at the top of Mount Everest (≈ 9000 m high). Repeating the previous calculation we obtain
∆τ = 5072 s. Gravity slows the passage of proper-time! A true means for slowing the effects of
aging.
4
3. In five-dimensional spacetime, xM = (xµ , y), the Randall-Sundrum metric is given by
ds2 = gM N dxM dxN = e−2ky ηµν dxµ dxν + dy 2 ,
where k is a constant and ηµν =diag(−1, 1, 1, 1).
(a) Calculate the Christoffel connection coefficients. (Rather than direct calculation via Eq. (3.27),
you will probably find it easier to use the variational technique!)
(b) Calculate the Riemann tensor, Ricci tensor, and Ricci Scalar.
Solution:
(a) Using the variational technique, look for stationary points in the integral
Z
Z dxµ dxν
1
1
dx1 2
dx2 2
dx3 2
dy 2
dx0 2
gµν
I=
dτ =
e−2ky −
+
+
+
+
dτ.
2
dτ dτ
2
dτ
dτ
dτ
dτ
dτ
First let’s consider variations x1 → x1 + δx1 . Under such variations the only change in the
integrand is
dx1 2
dx1
dx1 2
d(δx1 ) 2
dx1 d(δx1 )
7→
+
=
+2
+ O (δx1 )2 .
dτ
dτ
dτ
dτ
dτ dτ
Therefore, to first-order in δx1 , the variation of the integral is
Z
1 −2ky dx1 d(δx1 ) δI =
2e
dτ.
2
dτ dτ
1
1
d(δx )
yields
Integration by parts with u = e−2ky dx
dτ and dv =
dτ
Z 2 1
1
dx1 1
d x −2ky
−2ky dx dy
δI = e−2ky
δx |boundary −
e
−
2ke
δx1 dτ.
dτ
dτ 2
dτ dτ
Demanding that the variation of x1 vanish at the boundary, and setting the variation of the
integral equal to zero, we find
dx1 dy
d2 x1
−
2k
= 0,
dτ 2
dτ dτ
which implies the non-zero Γ1ij are
Γ114 = Γ141 = −k.
I’m using the convention y = x4 . Generalizing this calculation for other i’s we get
Γii4 = Γi4i = −k, i = 0, 1, 2, 3.
0 2
Note that the minus sign in front of dx
does not affect the result of the calculation because
dτ
there are no off-diagonal terms in the metric and we are setting the variation equal to zero. Now
consider variations y → y + δy. Every term in the integrand will change as a result of this
variation.
dy 2
dy
d(δy) 2
dy 2
dy d(δy)
7→
+
=
+2
+ O (δy)2
dτ
dτ
dτ
dτ
dτ dτ
The other terms involve the exponential factor, whose variation is
e−2ky 7→ e−2k(y+δy) = e−2ky (1 − 2k δy + O(δy 2 )).
The variation of the integral is therefore (dropping terms O(δy 2 ))
Z 1
dx0 2
dx1 2
dx2 2
dx3 2
dy d(δy)
δI =
− 2k e−2ky δy −
+
+
+
+2
dτ.
2
dτ
dτ
dτ
dτ
dτ dτ
5
Integrating the last term by parts and dropping the boundary term,
Z dx1 2
dx2 2
dx3 2
d2 y
dx0 2
−2ky
δI =
+
+
+
− 2 δy dτ
−k e
−
dτ
dτ
dτ
dτ
dτ
dx0 2
dx1 2
dx2 2
dx3 2
d2 y
δI = 0 ⇒ k e−2ky −
+
+
+
+ 2 =0
dτ
dτ
dτ
dτ
dτ
The connection coefficients are therefore
Γ400 = −ke−2ky
Γ4ii = ke−2ky , i = 1, 2, 3.
All other connection coefficients vanish.
(b) Since the Riemann tensor has so many terms, even if you include symmetries, it is often helpful
to turn this calculation into a matrix calculation (matrices are the natural way of manipulating
several equations at once). For this particular metric it’s only mildly painful to just calculate it
term by term, since there are not too many non-zero connection coefficients, however for more
complicated metrics the matrix technique can come in handy.
The definition of the Riemann tensor is
Rρ σµν = ∂µ Γρνσ − ∂ν Γρµσ + Γρµλ Γλνσ − Γρνλ Γλµσ .
Define the matrices
(Γµ )ρσ = Γρµσ
(Rµν )ρσ = Rρ σµν ,
where ρ is the row index and σ is the column index. Rµν should not be confused with the Ricci
tensor Rµν . What we have defined is merely a matrix that’s useful for doing computations. With
these definitions, the Riemann tensor can be recast as the matrix equation
Rµν = ∂µ Γν − ∂ν Γµ + Γµ Γν − Γν Γµ .
The components of the Γ matrices can be written as
(Γ0 )ρσ = −k δ0ρ δ4σ − ke−2ky δ4ρ δ0σ
(Γi )ρσ = −k δiρ δ4σ + ke−2ky δ4ρ δiσ ; i = 1, 2, 3
(Γ4 )ρσ = −k
3
X
δρi δσi .
i=0
Also note that because of the antisymmetry of the Riemann tensor in the last two indices, it
suffices to consider just the case where µ < ν.
Let’s do some calculations:
R01 = ∂0 Γ1 − ∂1 Γ0 + Γ0 Γ1 − Γ1 Γ0
Because none of the connection coefficients
we have

0
0 0

0
0 0

0
0 0
R01 = 


0
0 0
−ke−2ky 0 0
depend on x0 or x1 the

0
0
0 −k
0
0 0 
0

0
0 0 
0

0 0  0
0
0 0
0 ke−2ky
6
derivative terms drop out and

0 0 0
0 0 −k 

0 0 0 

0 0 0 
0 0 0

0
0

−
0
0
0
0
0
0
0
ke−2ky

0
0
0
0
0

0
0

0
−k 


0
0 

0
0 
−ke−2ky
0
0
0
0
0
0
0
−k e

0
=


0
0
2 −2ky
−k 2 e−2ky
0
0
0
0
0
0
0
0
0
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0

0 0
0 0

0 0
.
0 0
0 0

−k
0 

0 

0 
0
Doing the same calculation for R02 and R03 , you should find the following non-zero components
to the Riemann tensor:
0
i
Ri0i
= R00i
= −k 2 e−2ky ; i = 1, 2, 3.
The calculation is also very similar (only differing by a minus sign) for Rij ; i < j = 1, 2, 3 resulting
in the components
j
i
Rjij
= −k 2 e−2ky ; Riij
= k 2 e−2ky ; i < j = 1, 2, 3
Now let’s calculate R14 :

0
0

= −∂y 
0
0
0
0
0
0
0
ke−2ky
0
0
0
0
0

−k
 0

−
 0
 0
0
R14 = ∂1 Γ4 − ∂4 Γ1 + Γ1 Γ4 − Γ4 Γ1

 
0
0
0 0 0
0 0
−k 0
  0 −k
0
0
0
0
−k
0 −k 

 


0
0 0 0 
0 0 
0
 0
 + 0
0
0 0 0  0
0 0  0
0
0
0
0 ke−2ky 0 0 0
0 0


0
0
0 0 0
0
0
0 0
0
0
0 0 −k 
−k 0
0 0




0
0 0 0 
0 −k 0 0 0

0
0 0 0 
0
0 −k 0 0
0
0
0 0
0 ke−2ky 0 0 0


0
0
0 0
0
0
0
0 0 −k 2 


0
0 0
0 
=
.
0
0
0
0 0
0 
0 k 2 e−2ky 0 0
0
0
0
−k
0
0
0
0
0
−k
0

0
0

0

0
0
Repeating this method for the other components, you should find
0
4
R404
= −k 2 ; R004
= −k 2 e−2ky
i
4
R4i4
= −k 2 ; Rii4
= k 2 e−2ky ; i = 1, 2, 3.
All other terms not related to one of these by a symmetry vanish.
To calculate the Ricci tensor, first note that Rλµλν is zero unless µ = ν. So all of the off-diagonal
terms are zero and the work is almost done.
λ
1
2
3
4
R00 = R0λ0
= R010
+ R020
+ R030
+ R040
.
Rewrite this in terms of the components that have already been calculated,
1
2
3
4
R00 = −R001
− R002
− R003
− R004
= k 2 e−2ky + k 2 e−2ky + k 2 e−2ky + k 2 e−2ky = 4k 2 e−2ky .
Proceeding in a similar manner, you should find
λ
Rii = Riλi
= −4k 2 e−2ky ; i = 1, 2, 3
7
λ
R44 = R4λ4
= −4k 2 .
Finally, the Ricci scalar:
R = Rµ µ = g µν Rνµ ,
where
g
µν
 2ky
−e
 0

=
 0
 0
0
0
e2ky
0
0
0
0
0
e2ky
0
0
0
0
0
e2ky
0

0
0

0

0
1
So we find
R = −e2ky (4k 2 e−2ky ) + e2ky (−4k 2 e−2ky − 4k 2 e−2ky − 4k 2 e−2ky ) + 1(−4k 2 ) = −20k 2 .
8