General Relativity
Homework 2 Solutions
1. Let’s work a little more with the metric in flat space. In class, we’ve only discussed
the Minkowski metric in Cartesian coordinates, which is just ηµν = diag(−1, 1, 1, 1).
But the metric for flat space can also look more complicated. For this problem, you
can work in Euclidean 3-space, IR3 . At any point in IR3 , the tangent space can be
described by coordinates (x, y, z) and basis vectors (~ex , ~ey , ~ez ). The set of all tangent
spaces of IR3 is called the tangent bundle, and consists of a copy of the vector space
spanned by the basis (~ex , ~ey , ~ez ) for each point in IR3 . For the rest of the problem, I’ll
drop the vector arrows on the basis vectors, and just denote them ei .
a) First, let’s define three vector fields er , eθ and eϕ . Suppose we have curves with
r = ζ, θ = const, and ϕ = const throughout the entire space. For each point in
space (except the origin), there is only one and only one such curve going through
it. Define er at each point as the tangent vector of that curve, and write down er
in terms of ex , ey , and ez , depending on (r, θ, ϕ). Similarly, define eθ as the tangent
vector of r = const, θ = ζ, and ϕ = const and each point. Finally, define eϕ as the
tangent vector of r = const, θ = const, and ϕ = ζ at each point. Express eθ and eϕ
in terms of ex,y,z .
b) Consider r, θ, and ϕ as functions of x, y, and z, and show that in fact the gradients
dr, dθ, and dϕ are dual basis vectors of er , eθ and eϕ .
c) For a point (r, θ, ϕ), write down the metric components, and the full expression for
gij as a combination of tensor products of the dual basis vectors dr, dθ, and dϕ.
(This is still a simple Euclidean space, but the metric becomes “complicated” when
we choose a complicated basis.)
d) Alternatively, show that one can rescale er , eθ , and eϕ at each point into êr , êθ , and
êϕ , such that the metric becomes the Kronecker delta. (Now the metric is simple
again.)
e) Restricting to a unit sphere centered at the origin (r = 1), eθ and eϕ , as well as êθ
and êϕ , are tangent to the unit sphere: there exist curves that live completely on the
sphere whose tangent vectors correspond to these vectors. Show that the metric,
when restricted to (eθ , eϕ ), is location-dependent on the sphere, yet when restricted
to (êθ , êϕ ) is location-independent and is the Kronecker delta. (The two-sphere is
obviously a curved space, but we have just defined basis vectors such that the metric
components are the Kronecker delta! These concepts will be formalized shortly.)
Solution:
a) We have the transformations
x = r sin θ cos ϕ
y = r sin θ sin ϕ
z = r cos θ
and
p
x2 + y 2 + z 2
z
θ = cos−1 p
x2 + y 2 + z 2
ϕ = tan−1 (y/x)
r=
The tangent vector to the curve parametrized by r with θ and ϕ constant is
∂x ∂
∂y ∂
∂z ∂
er =
+
+
∂r ∂x ∂r ∂y ∂r ∂z
= sin θ cos ϕ ex + sin θ sin ϕ ey + cos θ ez .
Similarly,
eθ = r cos θ cos ϕ ex + r cos θ sin ϕ ey − r sin θ ez
eϕ = −r sin θ sin ϕ ex + r sin θ cos ϕ ey .
b) First, the easy one:
∂r
∂r
∂r
dx +
dy + dz
∂x
∂y
∂z
x
y
z
=
dx + dy + dz.
r
r
r
For dθ, notice cos θ = z/r. So the derivative is − sin θdθ = dz
− rz2 dr.
r
x
dz
z
y
z dθ = −
+
dx + dy + dz .
r sin θ r2 sin θ r
r
r
dy
y
1
Similarly, tan ϕ = y/x, so cos2 ϕ dϕ = x − x2 dx.
dy
y
2
dϕ = cos ϕ
− 2 dx .
x
x
dr =
Ultimately, you need to show that dxi ej = δji , or in matrix



dr
1 0 0






 dθ  er eθ eϕ =  0 1 0



dϕ
0 0 1
form,





c) For i,j = r,θ,ϕ,

1 0
0


(gi,j ) = (ei · ej ) =  0 r2
0

0 0 r2 sin2 θ





So g = dr ⊗ dr + r2 dθ ⊗ dθ + r2 sin2 θ dϕ ⊗ dϕ. Note that here I’m using g for the
2 × 2 metric tensor.
d) Since the metric is diagonal, you can rescale er , eθ and eϕ :
1
er̂ = √ er = er
grr
1
1
eθ̂ = √ eθ = eθ
gθθ
r
1
1
eϕ̂ = √
eϕ =
eϕ .
gϕϕ
r sin θ
Then (gîĵ ) is the identity matrix and ĝ = dr̂ ⊗ dr̂ + dθ̂ ⊗ dθ̂ + dϕ̂ ⊗ dϕ̂.
e) This part you can just write down: When restricted to (eθ , eϕ ), the metric is
g = r2 dθ ⊗ dθ + r2 sin2 θdϕ ⊗ dϕ,
which is clearly location-dependent. When restricted to (eθ̂ , eϕ̂ ), the metric is
g = dθ̂ ⊗ dθ̂ + dϕ̂ ⊗ dϕ̂,
which is location-independent.
2. Carroll problem 1.10. Using the tensor transformation law applied to Fµν , show how
~ and B
~ transform under:
the electric and magnetic field 3-vectors E
a) a rotation about the y-axis
b) a boost along the z-axis
Solution: The field strength tensor is


0 −E1 −E2 −E3




E 1
0
B3 −B2 
.

Fµν = 

E2 −B3
0
B1 


E3 B2 −B1
0
There are many ways to do this calculation, so this is just an example. If you do it in
matrix form, always make sure that the contractions are working as they should!
a) For a rotation about the y-axis the relevant Lorentz transformation is:

Λµ µ0
1
0
0
0





0 cos θ 0 sin θ 


=

0
0
1
0 


0 − sin θ 0 cos θ
Fµν is a rank 2 covariant tensor, therefore the components transform via Λ:
Fµ0 ν 0 = Λµ µ0 Λν ν 0 Fµν
The components of the electric field are related to the electromagnetic field tensor
by
Ei = Fi0 ; i, j, k = 1, 2, 3.
Therefore we have
Ei0 = Fi0 00 = Λµ i0 Λν 00 Fµν .
First note that since for this transformation Λi 00 = 0, we can ignore the terms
Λi i0 Λj 00 Fij (remember i, j = 1, 2, 3). Now, by taking advantage of the antisymmetry of Fµν we can write
E10 = (Λ110 Λ000 − Λ010 Λ100 )F10 + (Λ210 Λ000 − Λ010 Λ200 )F20 + (Λ310 Λ000 − Λ010 Λ300 )F30
= ((1)(cos θ) − (0)(0)) E1 + ((1)(0) − (0)(0)) E2 + (1)(− sin θ)E3
= E1 cos θ − E3 sin θ
Continuing in the same manner for E20 and E30 we see how the components of the
electric field transform under a rotation about the y-axis:



E1 cos θ − E3 sin θ
 


 


−
7
→
E 2 


E2
 


E3
E1 sin θ + E3 cos θ
E1

To see how the components of the magnetic field transform, we want to look specifically at B1 = F23 , B2 = F31 , and B3 = F12 .
Again using the fact that Λ0i0 = 0, we have that the first component of the magnetic
field (B10 = F20 30 ) transforms as
B10 =
=
=
=
Λ120 Λ230 F12 + Λ220 Λ130 F21 + Λ220 Λ330 F23 + Λ320 Λ230 F32 + Λ120 Λ330 F13 + Λ320 Λ130 F31
(Λ120 Λ230 − Λ220 Λ130 )F12 + (Λ220 Λ330 − Λ320 Λ230 )F23 + (Λ120 Λ330 − Λ320 Λ130 )F13
((0)(0) − (1)(sin θ))F12 + ((1)(cos θ) − (0)(0))F23 + ((0)(cos θ) − (0)(sin θ))F13
B1 cos θ − B3 sin θ
Repeating this calculation for the remaining components, we find


 
B1 cos θ − B3 sin θ
B1


 


 
−
7
→

.
B2 
B2


 
B3
B1 sin θ + B3 cos θ
0
b) To transform Fµν we need to use Λµ µ0 , which is the inverse of Λµ µ . Taking equation
(1.32) from Carroll and adjusting it to boost along the z-axis, the transformation is


γ
0 0 −γv




 0
1 0
0 
µ0
,
Λ µ=


 0
0 1
0 


−γv 0 0 γ
where γ = sinh φ. The inverse of this transformation will just be a boost in the
opposite direction, i.e. v → −v, so

Λµ µ 0
γ
0 0 γv


0 1 0
=

0 0 1

γv 0 0



0
.

0

γ
Note that Λµ ν is nonzero for µ = ν or µ = 0, ν = 3 and µ = 3, ν = 0. So we can
calculate, for example, E10 ,
E10 =
=
=
=
Λµ 10 Λν 00 Fµν
Λ1 10 Λ0 00 F10 + Λ1 10 Λ3 00 F13
(γ)(1)(E1 ) + (1)(γv)(−B2 )
γ(E1 − vB2 )
B10 =
=
=
=
Λµ 20 Λν 30 Fµν
Λ2 20 Λ3 30 F23 + Λ2 20 Λ0 30 F20
(1)(γ)(B1 ) + (1)(γv)(E2 )
γ(B1 + vE2 )
or B10 ,
You should find
 


 


E
γ(E1 − vB2 )
B
γ(B1 + vE2 )
 1


 1


 


 


E2  7−→ γ(E2 + vB1 ) & B2  7−→ γ(B2 − vE1 )
 


 


E3
E3
B3
B3
3. A few more questions about the field strength tensor:
a) Calculate the inner product of the field strength tensor with itself, Fµν F µν .
b) Is the quantity you calculated in part a relativistically invariant? Use your answer
to problem 2b to verify.
Solution:
a) First, remember to raise the indices on Fµν to obtain F µν = η µα η νβ Fαβ . Since the
field strength tensor is antisymmetric, Fµν F µν = −Fµν F νµ .
−Fµν F νµ = − F0ν F ν0 + F1ν F ν1 + F2ν F ν2 + F3ν F ν3
= −(E12 + E22 + E32 ) − (E12 − B32 − B22 ) − (E22 + B32 + B12 ) − (E32 + B22 + B12 )
= 2(B 2 − E 2 )
b) The answer to part (a) is a scalar quantity. To check the transformation properties,
we want to calculate B 02 − E 02 in a boosted system to compare to B 2 − E 2 .
B 02 − E 02 = B102 + B202 + B302 − E102 − E202 − E202 .
From the previous problem, we know how the electric and magnetic fields transform
under boosts along the z-axis, so we will assume that transformation.
B 02 − E 02 = γ 2 (B1 + vE2 )2 + γ 2 (B2 − vE1 )2 + B32
−γ 2 (E1 − vB2 )2 − γ 2 (E2 + vB1 )2 − E32
= (B12 + B22 )γ 2 (1 − v 2 ) + B32 − (E12 + E22 )γ 2 (1 − v 2 ) − E32
= B2 − E2
So Fµν F µν is a relativistic invariant!
4. Carroll problem 2.7. Prolate spheroidal coordinates can be used to simplify the Kepler
problem in celestial mechanics. They are related to the usual Cartesian coordinates
(x, y, z) of Euclidean three-space by:
x = sinh χ sin θ cos φ
y = sinh χ sin θ sin φ
z = cosh χ cos θ
Restrict your attention to the plane y = 0 and answer the following questions.
0
(a) What is the coordinate transformation matrix ∂xµ /∂xν relating (x, z) to (χ, θ)?
(b) What does the line element ds2 look like in prolate spheroidal coordinates?
Solution:
(a) In the prelate spheroidal coordinates the y-plane corresponds to φ = 0, so we
have:
x = sinh χ sin θ
z = cosh χ cos θ
The transformation matrix is then

 

∂x
∂x
µ
cosh
χ
sin
θ
sinh
χ
cos
θ
∂x
 ∂χ ∂θ  = 
.
0 =
µ
∂z
∂z
∂x
sinh χ cos θ − cosh χ sin θ
∂χ
∂θ
(b)
dx =
∂x
∂x
dχ +
dθ = cosh χ sin θdχ + sinh χ cos θdθ
∂χ
∂θ
dz =
∂y
∂y
dχ +
dθ = sinh χ cos θdχ − cosh χ sin θdθ
∂χ
∂θ
dx2 = (cosh χ sin θdχ + sinh χ cos θdθ)(cosh χ sin θdχ + sinh χ cos θdθ)
⇒ dx2 = cosh2 χ sin2 θdχ2 +cosh χ sinh χ sin θ cos θ(dχdθ+dθdχ)+sinh2 χ cos2 θdθ2
dz 2 = (sinh χ cos θdχ − cosh χ sin θdθ)(sinh χ cos θdχ − cosh χ sin θdθ)
⇒ dz 2 = sinh2 χ cos2 θdχ2 −cosh χ sinh χ sin θ cos θ(dχdθ+dθdχ)+cosh2 χ sin2 θdθ2
dx2 + dz 2 = (cosh2 χ sin2 θ + sinh2 χ cos2 θ)dχ2 + (sinh2 χ cos2 θ + cos2 χ sin2 θ)dθ2
⇒ dx2 + dz 2 = (sinh2 χ + sin2 θ)(dχ2 + dθ2 )