PHYSICS 750-001: General Relativity
Homework # 3 Solution Key
1. [20 pts] You’re on your honor to do this one by hand. I realize you can use a computer or simply look
it up. Please don’t.
In a flat space, the metric in spherical coordinates, r,θ,φ is:


1 0
0

0
gµν =  0 r2
0 0 r2 sin2 θ
(a) Please compute all non-zero Christoffel symbols for this system.
Sol.
First, you need to realize that the only non-zero derivatives are:
gθθ,r
=
2r
gφφ,r
=
2r sin2 θ
gφφ,θ
=
2r2 sin θ cos θ
Since the metric is diagonal, we’re only to get terms with those three combinations of parameters. If the non-zero derivative is with respect to the r coordinate and r appears upstairs in the
Christoffel symbol, we’re going to get a minus sign.
Γrθθ
=
Γθrθ
=
Γrφφ
=
Γφrφ
=
Γθφφ
=
−r
1
r
−r sin2 θ
1
r
− sin θ cos θ
Γφθφ
=
cot θ
(b) Compute the divergence
α
V;α
Sol.
The general expansion is:
α
α
β
V;α
= V,α
+ Γα
αβ V
So we only get to include those Christoffel symbols that have the same upstairs and downstairs
components. This means:
2
α
V;α
= V,rr + V,θθ + V,φφ + V r + cot θV θ
r
If you like, this can be simplified to:
2 r
θ
φ
~ = 1 ∂r V + ∂V + cot θV θ + + ∂V
∇·V
r2 ∂r
∂θ
∂φ
Quick note: if you’re trying to make this look the same as the spherical divergance that you might
find in the inside cover of Jackson, for example, you’re going to be disappointed. The components
of a vector described in Jackson are normalized, but ours are not.
2. [15 pts] Consider a vector in 2-d flat space:
~v = î
at the position
~ri = î
(a) Express the vector and position in polar coordinates.
Sol.
This is pretty straightforward. Both are:
~v = ~r = r̂
(b) Suppose you moved the vector to a position (in Cartesian coordinates):
~rf = î + ĵ
where 1. You move without changing the vector itself. What would the components of the
vector, ~v be in polar coordinates at the final position?
Method 1: Use a straight geometric approach. Turn the Cartesian coordinates into polar coordinates.
Sol.
The radial component is simply computed via:
~ · r̂ = √ 1
Vr =V
1 + 2
and the θ component via:
V µ Vµ = 1 =
1
+ (1 + 2 )(V θ )2
1 + 2
or
θ
V =−
2
1 + 2
1/2
' −
so
~f ' ~er − ~eθ
V
(c) Method 2: Use the relation:
V;βα = 0
along the curve.
Sol.
I’m going to be more precise than is necessary here, and will save my approximations until the
end. First, we need to write our differential equations:
V,rr
=
V,θr
= −Γrθθ V θ = rV θ
V,rθ
= −Γθrθ V θ = −
V,θθ
0
Vθ
r
Vr
θ
r
= −Γθr V = −
r
So, consider moving along a curve in the r-direction, a generic vector transforms as:
V2r = V1r
and
V2θ = V1θ ×
r1
r2
(It’s a pretty straightforward differential equation).
Now, suppose we go along a curve of constant r but vary θ? Generically, we get a coupled
differential equation with the solution:
V3r = V2r cos(δθ) + B sin(δθ)
And:
Vr
B
1 dV r
= − 2 sin(δθ) + cos(δθ)
r dθ
r
r
Since for δθ = 0, this must yield the original, we get:
Vθ =
B = rV2θ
and thus:
V3r
= V2r cos(δθ) + rV2θ sin(δθ)
= V1r cos(δθ) + r1 V1θ sin(δθ)
Vr
V3θ = V2θ cos(δθ) − 2 sin(δθ)
r2
r1 θ
Vr
=
V1 cos(δθ) − 1 sin(δθ)
r2
r1
√
This is exact. In our case, we’re moving from r1 = 1 to r2 = 1 + 2 , and from θ = 0 to
sin θ
=
cos θ
=
2
1 + 2
1
√
1 + 2
√
Plugging and chugging, we get:
Vr = √
1
1 + 2
and
V θ = −√
2
1 + 2
both of which are exactly what I got in the previous section.
3. [15 pts] 5.12
Sol.
We have the one-form with components:
pµ =
x2 + 3y
y 2 + 3x
(a) pα,β
pα,β =
2x
3
3
2y
(b) Derivatives and then transform.
The transformation matrix is:
Λα
α
∂xα
∂xα
cos θ
=
sin θ
=
−r sin θ
r cos θ
But in order to solve for this, we first need to express the derivatives in terms of polar coordinates:
2r cos θ
3
pα,β =
3
2r sin θ
So, multiplying it out, we get:
pr;r = 6 sin θ cos θ + 2r sin3 θ + 2r cos3 θ
pr;θ = pθ;r = 2r2 cos θ sin θ(sin θ − cos θ) + 3r(cos2 θ − sin2 θ)
and
pθ;θ = 2r2 sin θ cos θ(r sin θ + r cos θ − 3)
Hardly an elegant expression!
(c) pα;β
This is a straightforward application, which first requires us to compute
pα = r(6 sin θ cos θ + r cos3 θ + r sin3 θ) −r2 r sin θ(cos θ − sin θ) + 3 − 6 cos2 θ
From there, confirming the result from the previous part is straightforward. I could re-write it if
you like, but I’m not going to.
4. [15 pts] 5.14
Sol.
Firstly, the non-zero derivatives of the tensor Aµν are:
Arr
,r
Arθ
,r
Arθ
,θ
Aθr
,r
Aθr
,θ
Aθθ
,θ
= 2r
= sin θ
= r cos θ
= cos θ
= −r sin θ
= sec2 θ
Combined with the non-zero Christoffel symbols:
Γrθθ = −r
Γθθr = Γθrθ =
1
r
gives us the following terms in Eq. 5.65:
∇r Arr = 2r
∇r Arθ = 2 sin θ
∇r Aθr = 2 cos θ
∇θ Arr = r2 sin θ − r2 cos θ
∇θ Arθ = r + r cos θ − r tan θ
∇θ Aθr = r − r sin θ − r tan θ
∇θ Aθθ = sec2 θ + cos θ + sin θ
2
∇r Aθθ = tan θ
r
5. [15 pts] 7.3
Sol.
The weak field metric:

gµν
−(1 + 2φ)

0
=

0
0
0
1 − 2φ
0
0
0
0
1 − 2φ
0

0

0


0
1 − 2φ
Since any derivatives are going to be 1st order in φ, the 1st order expansion of the Christoffel symbols
can be approximated as:
g µν = η µν
Likewise, the only non-zero terms are:
gµν,λ = −2δµν φ,λ
In other words, we’re going to get non-zero Christoffel symbols if two of the indices are the same.
These will be of the following form:
Γ0ii
= −φ,0
Γi0i
= −φ,0
Γ000
= φ,0
Γi00
Γ00i
Γjii
Γiij
= φ,i
= φ,i
= φ,j
= −φ,j
6. [20 pts] A cosmic string is a theoretical construct (never observed in nature) which is infinitely long,
and has a mass density per unit length of λ. The coordinates describing the spacetime surrounding a
cosmic string are:


t
 R 

xµ = 
 φ 
z
and which has a metric:

gµν
−1
 0
=
 0
0

0
0
0
1
0
0 

2
0 R (1 − 4λ) 0 
0
0
1
(a) Compute the volume element, dV , near the cosmic string.
Sol.
First, take the negative square root of the determinant:
g = −R2 (1 − 4λ)
so
dV =
√
−gdx0 dx1 dx2 dx3 =
√
1 − 4λRdRdφdzdt
(b) Compute all non-zero Christoffel symbols. Please be smart, and start by noting that a) the metric
is diagonal (for easy inversion), and b) most derivatives cancel.
Sol.
The only non-zero derivatives are:
g22,1 = 2R(1 − 4λ)
and thus, the only non-zero Christoffel symbols are:
Γ212 = Γ221 =
1
R
and
Γ122 = −R(1 − 4λ)
(c) Compute the distance between two points separated by dxµ = dR, and all other coordinates equal
to zero. From that, compute the distance from the string itself out to a distance of R = 1.
Sol.
Very straightforward:
ds2 = dR2
so
ds = dR
and thus:
Z
1
s=
dR = 1
0
(d) Compute the distance between two points, each R = 1 from the string separated by an angle dφ
(with all other dxµ = 0).
Using that, what is the total distance traversed by a particle covering a circular orbit of R = 1
around the cosmic string?
Sol.
Here, the distance is:
√
ds = 1 − 4λdφ
and thus, the distance around a circular orbit is:
√
s = 2π 1 − 4λ
(e) Compare parts c) and d) in the context of the normal relationship between radius and circumference. That is, does C = 2πr? If not, what should it be replaced with?
Sol.
This is odd. It’s not 2π times the radius (unless λ = 0).