Department of Physics, UCSD
Physics 225B, General Relativity
Winter 2015
Homework 1, solutions
1. (a) From the Killing equation, ∇ρ Kσ ` ∇σ Kρ “ 0 taking one derivative,
∇µ ∇ρ Kσ ` ∇µ ∇σ Kρ “ 0
´∇σ ∇µ Kρ ´ ∇σ ∇ρ Kµ “ 0
∇ρ ∇µ Kσ ` ∇ρ ∇σ Kµ “ 0.
Adding these we have
p∇µ ∇ρ ` ∇ρ ∇µ qKσ “ r∇σ , ∇µ sKρ ` r∇σ , ∇ρ sKµ .
Add r∇µ , ∇ρ sKσ to both sides to obtain (twice) what we are looking for in terms of
commutators of derivatives:
2∇µ ∇ρ Kσ “ r∇σ , ∇µ sKρ ` r∇σ , ∇ρ sKµ ` r∇µ , ∇ρ sKσ .
Fro the very definition of curvature we have, for any vector field Aλ ,
r∇µ , ∇σ sAρ “ Rρλµσ Aλ
so we have immediately that
∇µ ∇ρ Kσ “ 12 pRρλσµ ` Rµλσρ ` Rσλµρ qK λ .
Now we just need to shuffle indices around a bit. The second term is already of the form
we want, Rµλσρ “ Rσρµλ . Write the first term as Rρλσµ “ Rσµρλ and combine with the third
term using anti-symmetry in the last three indices, Rσµρλ ` Rσλµρ “ ´Rσρλµ “ Rσρµλ , which
doubles the second term and the result follows,
∇µ ∇ρ Kσ “ Rσρµλ K λ .
(0.1)
(b) We are going to use the Bianchi identity 12 ∇µ R “ ∇λ Rλ µ . Contracting indices in
the solution to part (a), Eq. (0.1) , we have ∇µ ∇ρ Kµ “ Rρλ K λ , and taking ∇ρ of this,
∇ρ ∇µ ∇ρ K µ “ ∇ρ pRρλ K λ q “ p∇ρ Rρλ qK λ ` Rρλ ∇ρ K λ .
1
(0.2)
Since the Ricci tensor is symmetric in its indices we can replace 12 ∇pρ K λq for ∇ρ K λ in the
last term on the right hand side, buy by Killing’s equation ∇pρ K λq “ 0.
So to prove that p∇ρ Rρλ qK λ “ 0 we have to show that the left hand side in (0.2)
vanishes:
∇ρ ∇µ ∇ρ Kµ “ r∇ρ , ∇µ s∇ρ Kµ
(again, since ∇pρ Kµq “ 0)
“ r∇ρ , ∇µ s∇ρ K µ
“ Rµ ρµλ ∇ρ K λ ` Rλ ρµλ ∇µ K ρ
“ Rρλ ∇ρ K λ ´ Rρµ ∇µ K ρ
(once again, since ∇pρ Kµq “ 0)
“ 0.
So p∇ρ Rρλ qK λ “ 0 , or, by the Bianchi identity, p∇λ RqK λ “ 0
2.(a) The pullback is
ĝij “
Bxµ Bxν
gµν ,
By i By j
where xµ are coordinates on R3 and y i are coordinates on P, and the map xµ py i q is
x “ ρ cos φ
y “ ρ sin φ
z “ ρ2
Compute the derivatives:
Bx
“ cos φ
Bρ
Bx
“ ´ρ sin φ
Bφ
By
“ sin φ
Bρ
By
“ ρ cos φ
Bφ
Bz
“ 2ρ
Bρ
Bz
“0
Bφ
Using gµν “ δµν we have:
Bxµ Bxν
δµν “ cos2 φ ` sin2 φ ` 4ρ2 “ 1 ` 4ρ2
Bρ Bρ
Bxµ Bxν
“ ĝφρ “
δµν “ cos φp´ρ sin φq ` sin φpρ cos φq ` 2ρp0q “ 0
Bρ Bφ
Bxµ Bxν
“
δµν “ p´ρ sin φq2 ` pρ cos φq2 ` p0q2 “ ρ2
Bφ Bφ
ĝρρ “
ĝρφ
ĝφφ
2
which is summarized as
dŝ2 “ p1 ` 4ρ2 qdρ2 ` ρ2 dφ2
(b)As a matrix, the inverse of ĝij is
˜
ĝ ij “
1
1`4ρ2
0
0
1
ρ2
¸
.
The push-forward, which i will denote by g̃ µν , generally, is given by
g̃ µν “
Bxµ Bxν ij
ĝ .
By i By j
Computing:
„
ˆ ˙2

ˆ ˙2
1
1
cos2 φ
1
x2
Bx
Bx
2
2
xx
`
“
` sin φ “
`y
g̃ “
Bρ
1 ` 4ρ2
Bφ ρ2
1 ` 4ρ2
z 1 ` 4z
g̃
xy
g̃ xz
g̃ yz
g̃ yy
g̃ zz
Bx By
1
Bx By 1
sin φ cos φ
4ρ2 sin φ cos φ
4xy
“
`
“
´ sin φ cos φ “ ´
“´
2
2
2
2
Bρ Bρ 1 ` 4ρ
Bρ Bρ ρ
1 ` 4ρ
1 ` 4ρ
1 ` 4z
Bx Bz 1
2ρ cos φ
2x
Bx Bz
1
`
“
“
“
2
2
2
Bρ Bρ 1 ` 4ρ
Bρ Bρ ρ
1 ` 4ρ
1 ` 4z
By Bz 1
2ρ sin φ
2y
By Bz
1
`
“
“
“
2
2
2
Bρ Bρ 1 ` 4ρ
Bρ Bρ ρ
1 ` 4ρ
1 ` 4z
ˆ ˙2
ˆ ˙2
„

By
1
By
1
sin2 φ
1
y2
2
2
“
`
“
` cos φ “
`x
Bρ
1 ` 4ρ2
Bφ ρ2
1 ` 4ρ2
z 1 ` 4z
ˆ ˙2
ˆ ˙2
Bz
1
Bz
1
4ρ2
4z
“
`
“
“
2
2
2
Bρ 1 ` 4ρ
Bφ ρ
1 ` 4ρ
1 ` 4z
Keep in mind that these are defined only on the sub-manifold P (so, in a sense, it is better
to keep the expressions for g̃ as given in terms of ρ and φ.
(c) Not much to do here. g µν “ δ µν is very different from g̃ µν , but there was no reason
to expect them to be the same.
3. (a) The integral curve of V µ pxq “ xµ is the solution to
dxµ ptq
“ V µ pxptqq “ xµ ptq.
dt
We want a solution that satisfies xµ p0q “ xµo . The integral is simple:
xµ ptq “ xµo et .
3
For an integral curve through the origin, xµo “ 0, the solution above is not a curve but just
a map from the real line to a single point. The reason is that at that point the tangent
field vanishes: the integral curve needs to know in what direction to move!
(b) The map φt : Rn Ñ Rn takes xµo to y µ “ xµ ptq “ xµo et .
(c) The push forward is given by
pφ´t˚ W qµ |po “
By µ
|p W ν |p
Bxν
Explicitly, this is
pφ´t˚ W qµ pxo q “
Bpxµ e´t q ν
W pxo et q “ e´t W µ pxo et q.
Bxν
Notice the minus sign in the exponential. This is because the map φ´t takes y µ “ xµ ptq ÞÑ xµo
which we are writing as xµ ÞÑ e´t xµ . The Lie derivative is, by definition,
1
B
rpφ´t˚ W qµ pxo q ´ W µ pxo qs “ e´t W µ pxo et q “ ´W µ pxo q ` xνo Bν W µ pxo q
tÑ0 t
Bt
LV W “ lim
(d) rV, W sµ “ xν Bν W µ ´ W ν Bν xµ “ xν Bν W µ ´ W µ .
This, of course, agrees with the result of part (c).
4. First calculate integral curves of
A“
y´x B
y`x B
´
r Bx
r By
That is, we look for solutions to
dx
y´x
“
dt
r
Note that the vector field A has magnitude
y`x
dy
“´
dt
r
?
2 everywhere, is not defined at the origin and
is tangential to a circle about the origin, pointing in the clockwise direction. So we expect
the integral curves to grow towards the origin as they circulate clockwise.
Now, the fact that Aµ depends on x{r and y{r cries out for a description in a polar
coordinate system, exactly what the whole formalism is suppose to do for us automatically.
That is, if ξ µ is a new coordinate system, with ξ µ “ ξ µ pxν q and if we denote the vector
field A components in the new coordinate system by õ , then
õ “
Bξ µ ν
A
Bxν
4
So we take for new coordinates ξ µ “ pr, φq defined so that x “ r cos φ and y “ r sin φ, that
is
φ “ arctanpy{xq
r“
a
x2 ` y 2
If I can still compute derivatives,
Bξ µ
“
Bxν
˜
cos φ
sin φ
´ 1r sin φ
1
r
¸
cos φ
.
I have written the result in terms of the coordinates ξ so we can write õ in terms of those
coordinates:
Ãr “ cos φpsin φ ´ cos φq ` sin φp´ sin φ ´ cos φq “ ´1
1
1
1
Ãφ “ ´ sin φpsin φ ´ cos φq ` cos φp´ sin φ ´ cos φq “ ´
r
r
r
The equations for the integral curve are now simple,
dr
“ ´1,
dt
dφ
1
“´ .
dt
r
If the initial point is pr0 , φ0 q at t “ 0, the solution is rptq “ r0 ´t and φptq “ φ0 `lnp1´t{r0 q.
In terms of the original coordinates we have then
xptq “ pr0 ´ tq cospφ0 ` lnp1 ´ t{r0 qq,
yptq “ pr0 ´ tq sinpφ0 ` lnp1 ´ t{r0 qq.
Here is a plot of the curves. I show three curves, going through p0, 1q, p0, 2q and p0, 3q:
5
4
3
2
1
-4
-3
-2
-1
1
-1
5
Things are much simpler for the B field, because it is easy to integrate directly. We
have
dx
dy
“ xy,
“ ´y 2 .
dt
dt
The equation for yptq can be integrated immediately, yptq “ y0 {p1 ` y0 tq. Inseting this in the
equation for xptq we have xptq “ x0 p1 ` y0 tq. Note that this has xptqyptq “ x0 y0 “ constant,
so the parametric plot is easily drawn:
2.0
1.5
1.0
0.5
0.5
1.0
1.5
2.0
Here I have taken curves that go through x “ 0.1 at y “ 1, 2, 3. Of course, there are also
analogous integral curves on the other three quadrants of the cartesian plane. We can also
see both sets of integral curves together:
3
2
1
-0.4
-0.2
0.2
0.4
-1
6
0.6
0.8
1.0
Finally, compute C “ LA B “ rA, Bs, or C µ “ Aν Bν B µ ´ B ν Bν Aµ :
´ y ` x¯
”
´y ´ x¯
´ y ´ x ¯ı
y´x
x
2
C “
Bx pxyq ` ´
By pxyq ´ xyBx
` p´y qBy
r
r
r
r
2
ypy ´ xq xpy ` xq 2xy py ` xq
´
`
“
r
r
r3
4
2 2
3
4
y ` 2x y ´ 2x y ´ x
“
r3
´ y ` x¯
”
´ y ` x¯
´ y ` x ¯ı
y´x
y
2
2
2
C “
Bx p´y q ` ´
By p´y q ´ xyBx ´
` p´y qBy ´
r
r
r
r
2
2ypy ` xq 2xy py ´ xq
`
“
r
r3
2ypx3 ` 2xy 2 ` y 3 q
“
r3
Below is a plot of the vector field C superimposed on the integral curves of A and B.
You can sketch the integral curves of C in an obvious way (by stringing vectors together):
3
2
1
-0.4
-0.2
0.2
0.4
-1
7
0.6
0.8
1.0