University of California at San Diego – Department of Physics – Prof. John McGreevy
General Relativity (225A) Fall 2013
Assignment 4 – Solutions
Posted October 16, 2013
Due Monday, October 28, 2013
1. Non-relativistic limit of a perfect fluid
The stress-energy tensor for a perfect fluid in Minkowski space is
T µν = (( + p) uµ uν + pη µν ) .
Consider the continuity equation ∂µ T µν = 0 in the nonrelativistic limit, p (recall
that includes the rest mass!). Show that it implies the conservation of mass, and
Euler’s equation:
~
~
ρ ∂t~v + ~v · ∇ ~v = −∇P.
(See section 4.2 of Wald for more on this. Note that he uses ρ for and sets c = 1. )
The continuity equation says
0 = ∂µ T µν = (∂µ ( + p)) uµ uν + ( + p) ((∂µ uµ ) uν + uµ ∂µ uν ) + (∂µ p) η µν .
(1)
The vector field u satisfies
−1 = uν uν =⇒ 0 = ∂µ (uν uν ) = 2 (∂µ uν ) uν
(in flat space). Projecting the conservation equation onto uν gives
0 = −uµ ∂µ ( + p) − ( + p)∂ν uν + 0 + ∂µ puµ
= −uµ ∂µ − ∂µ uµ ( + p)
(2)
To find the content of the conservation law ⊥ to uν add uν times (2)
0 = uν (−( + p)∂ν uν + 0 + ∂µ puµ )
to both sides of (1) to get:
0 = +uµ uν ∂µ p + ( + p) uµ ∂µ uν + ∂µ pη µν
= ( + p) uν ∂µ uν + ∂µ p (η µν + uµ uν ) .
(3)
~ v.
Now take the nonrelativistic limit: p, uµ = (1, ~v )µ in which case ∂µ v µ → −0+ ∇·~
Eqn (2) becomes
~ +∇
~ · v 1 + p ' + 1 ∂ + ∇
~ · (~v )
0 = uµ ∂µ + ∂µ uµ ( + p) ' +∂x0 + ~v · ∇
c ∂t
1
which is energy conservation. Eqn (3) becomes
~ uν + (∂0 + ~v · ∇)p
~
0 = ( + p) uµ ∂µ uν +∂µ p (η µν + uµ uν ) ' ∂0 + ~v · ∇
uν +∂µ pη µν
Now look at the time and space components:
1
~ − 1 ∂t p = ~v · ∇p
~
ν = 0 : 0 = (0) + ∂t p + ~v · ∇p
c
c
1
1
~ ~v + ~v
~
~
ν=
6 0 :0=
∂t~v + ~v · ∇
∂t p + ~v · ∇p
+ ∇p
c
c
In the NR limit v c, so the important terms are:
1
~ ~v + ∇p
~
ν 6= 0 : 0 = ∂t~v + ~v · ∇
c
which is the Euler equation.
2. Stress tensors for fields in Minkowski space
(a) Given a (translation-invariant) lagrangian density L(φ, ∂µ φ) for a scalar field φ,
define the energy-momentum tensor as
Tνµ = −
∂L
∂ν φ + δνµ L.
∂ (∂µ φ)
Show that the equation of motion for φ implies the conservation law ∂µ Tνµ .
The EoM for φ is
δS
∂L
∂L
0=
=
(x) − ∂µ
(x)
δφ(x)
∂φ
∂∂µ φ
(see the lecture notes section 3.1), so
∂L
∂L
−
∂µ ∂ν φ +
∂ν L
∂µ Tνµ = − ∂µ
|{z}
∂ (∂µ φ) ∂ (∂µ φ)
| {z }
+∂ν (∂µ φ)
=∂ν φ ∂L
∂φ
=− ∂L
∂φ
= −
∂L
∂ (∂µ φ)
∂L
∂L
∂L
∂L
∂ν φ −
∂µ ∂ν φ + ∂ν φ
+ ∂µ ∂ν φ
= 0.
∂φ
∂ (∂µ φ)
∂φ
∂ (∂µ φ)
(b) Show that the energy-momentum tensor for the Maxwell field
1
1 µν 2
µν
µρ ν
F F ρ− η F
TEM =
4πc
4
is traceless, that is (TEM )µµ = 0.

Tµµ =

1  µρ
1
F Fµρ − 4F 2  = 0.
4πc | {z } 4
=F 2
2
(4)
An apology: the notation Fρν in the pset statement is ambiguous: when raising
and lowering indices on an object with multiply (non-symmetrized) indices, you
have to keep track of which one was raised or lowered. I should have written F ν ρ
(which is harder to TeX).
(c) Show that the energy-momentum tensor for the Maxwell field
1 µν 2
1
µν
µρ ν
F F ρ− η F
TEM =
4πc
4
in the presence of an electric current j µ obeys
µν
∂µ TEM
= −j ρ F ν ρ .
Explain this result in words.
I am setting c = 1. Recall that Maxwell’s equations with a source are
0 = µνρσ ∂ν Fρσ ,
∂ ρ Fµρ = 4πjµ .
We will need both.
µν
∂µ TEM
1
2 µν
µρ
ν
µρ
ν
αγ βδ
=
(∂µ F ) F ρ + F ∂µ Fρ − η
∂µ Fαβ Fγδ η η
4π 4
1 ν
1
µ
νρ
µρ
µρ
ν
− (∂ Fρµ ) F + F
∂µ Fρ − (∂ Fµρ ) F
=
4π
2
1 µρ νσ
1
= −jρ F νρ +
F η
∂µ Fσρ − ∂σ Fµρ
4π
2
(5)
But by antisymmetry of F : F µρ (∂µ Fσρ ) = 12 F µρ (∂µ Fσρ − ∂ρ Fσµ ), so
µν
∂µ TEM
= −jρ F νρ +
1 µρ νσ
F η (∂µ Fσρ + ∂ρ Fµσ + ∂σ Fρµ ) = −j ρ F ν ρ .
|
{z
}
8π
∝µσρν ναβγ ∂α Fβγ =0
The EM field can do work on the currents and vice versa; therefore the energy
and momentum in the EM field can turn into energy of the charges and vice versa,
and is therefore not conserved. The RHS of the ν = 0 component is minus the
work done on the particles by the fields; the RHS of the ν = i component is minus
the force (change in momentum per unit time) in the i direction exerted on the
particles by the Maxwell field.
(d) Optional: show that tracelessness of Tµµ implies conservation of the dilatation
µ
ν µ
Rcurrent 0D = x Tν . Convince yourself that the associated conserved charge
D is the generator of scale transformations.
space
Z
0
Z
D =
space
space
∂µ Dµ = Tµµ + xν ∂µ Tνµ = Tµµ .
0
i 0
i
i ∂
tT0 + x Ti = tH + x P = −i t∂t + x
∂xi
which actions on functions of t, x by : (t∂t + x∂x ) tα xβ = (α + β)tα xβ and exponentiates to eat∂t f (t) = f (ea t) (it shifts log t by a).
3
3. Polyakov form of the worldline action
(a) Consider the following action for a particle trajectory xµ (t):
Z
dxµ dxν
S? [x] = −m dt
gµν (x) .
dt dt
(Here gµν is some given metric. You may set gµν = ηµν if you like.) Convince
yourself that the parameter t is meaningful, that is: reparametrizing t changes
S? .
Reparemetrizing by t̃ = t̃(t) changes the measure by one factor of J ≡ ∂t t̃, but
multiplies ẋ2 by J −2 .
Now consider instead the following action
Z
1 dxµ dxν
2
gµν (x) − m e(s) .
S[x, e] = − ds
e(s) ds ds
The dynamical variables are xµ (s) (positions of a particle) and e(s); e is called an
einbein1 :
ds21d = e2 (s)ds2 .
(b) Show that S[x, e] is reparametrization invariant if we demand that ds21d is an
invariant line element.
R√
Notice that the second term is just
g our usual way of making a coordinate∂
s each transform like e, so
invariant integration measure. In the first term the ∂s
1 2
ẋ transforms like e.
e
(c) Derive the equations of motion for e and xµ . Compare with other reparametrizationinvariant actions for a particle.
0=
√
δS
= −e−2 ẋ2 + m2 =⇒ e = m ẋ2
δe(s)
Z
√
S[x, e] = −m ds ẋ2
is the reparam-invariant action we discussed before, so obviously the EoM agree.
Even the values of the action agree when we set e equal to the solution of its EoM.
(d) Take the limit m → 0 to find the equations of motion for a massless particle.
In this limit, e is a Lagrange multiplier which enforces that the tangent vector is
lightlike:
δSm=0
0=
∝ ẋ2 .
δe(s)
The EoM for x is still the geodesic equation. Now we can’t choose s to be the
proper time, but we have to demand that the lengths of the tangent vectors
are independent of s – they are all zero. So any parametrization satisfying the
lightlike-constraint is affine.
1
that’s German for ‘the square root of the metric in one dimension’
4
4. Show that S 2 using stereographic projections (aka Poincaré maps)for the coordinate
charts (see the figure)
xN : S 2 − {north pole} → IR2
xS : S 2 − {south pole} → IR2
is a differentiable manifold of dimension two. More precisely:
(a) Write
the Poincaré maps explicitly in terms of the embedding in IR3 ({(x1 , x2 , x3 ) ∈
3 P
2
IR | i xi = 1} → IR2 ).
If we place the center of the unit sphere at the origin, the plane in the picture is
{x3 = −1}. The line LN in the picture passes through i(N ) = (0, 0, 1) (the embedding
of the north pole) and i(p) = (x, y, z), the embedding of the point of interest p. The
N
intersection of LN with z = −1 determines xN (p) = (xN
1 , x2 , −1). We can parametrize
~ N (t) so L
~ N (0) = i(N ) = (0, 0, 1) and L
~ N (1) = (x, y, z), which gives
it as L
~ N (t) ≡ (xN (t), xN (t), xN (t)) = (tx, ty, 1 − t + tz).
L
1
2
3
2
The value of t = tN where −1 = xN
3 (tN ) = 1 − (1 − z)tN tN is then tN = 1−z . Notice
that this is a nice function as long as z 6= 1, and that’s why we must exclude the north
pole (where (x, y, z) = (0, 0, 1)) from this patch. The coordinates of the stereographic
projection from N are then
N
xN (p) = (xN
1 (tN ), x2 (tN )) = (tN x, tN y) =
2
(x, y).
1−z
For the south pole, the plane is at x3 = +1 and the line in question is
LS (t) = (tx, ty, (1 − t)(−1) + tz)
so the value of t where x3 (tS ) = +1 is tS =
−2
1+z
and
N
xs (p) = (xN
1 (tS ), x2 (tS )) =
−2
(x, y).
1+z
2
2
(b) Show that the transition function xS ◦ x−1
N : IRN → IRS is differentiable on the
overlap of the two coordinate patches (everything but the poles).)
5
The overlap UN ∩ US where both coord systems are defined is everywhere but the north
xN
xN
2
. The transition functions are nice and
and south poles. Notice that x1S = z+1
=
z−1
xS
1
2
diagonal
N
S
S
z+1 1 0
x1
x1
x
S
S
=
≡ M (x1 , x2 ) 1S
N
S
x2
x2
x2
z−1 0 1
– notice that we must think of z here as a function of the coordinates xS1,2 . Explicitly
it is findable by
2
2
x2 + y 2
1 − z2
r2 ≡ xS1 + xS2 = 4
=
4
(1 + z)2
(1 + z)2
p
−2r2 ± 4r4 − 4(r2 − 4)(r2 + 4)
r2 ∓ 4
=
−
.
=⇒ z =
2(r2 + 4)
r2 + 4
The bottom root is just z = −1 which is not what we want – −1 < z < 1 on the
overlap of patches. So the top root it is:
2
2
4 − xS1 − xS2
4 − r2
z=
=
2
2 .
4 + r2
4 + (xS1 ) + (xS2 )
And
z+1
z−1
=
−4
r2
=
−4
2
2
(xS1 ) +(xS2 )
and therefore the transition map is
N
S
−4
x1
x1
=
N
S
2
2
S
S
x2
(x1 ) + (x2 ) x2
2
2
The inverse map is obtainable noticing that xN
+ xN
=
1
2
16
2
2
(xS1 ) +(xS2 )
exactly the same form, but with N ↔ S. The transition function
, so it has
φ = xN ◦ x−1
S : xS (US ∩ UN ) → xN (US ∩ UN )
is then perfectly smooth away from the points where 0 = xN
1
S 2
S 2
x1 + x2 which are just S and N .
2
+ xN
2
2
or 0 =
5. Verify explicitly that if ωµ is a one-form (cotangent vector), then ∂µ ων −∂ν ωµ transforms
as a rank-2 covariant tensor.
Under the coordinate change xµ → x̃a (x), ωµ (x) → ω̃a (x̃) = Jaµ ωµ (x), and ∂µ → ∂˜a =
µ
= ∂˜a xµ .
Jaµ ∂µ with Jaµ ≡ ∂x
∂ x̃a
(dω)µν = ∂µ ων − ∂ν ωµ
∂˜a ω̃b − ∂˜b ω̃a
∂˜a ∂˜b xµ ωµ − (a ↔ b)
µ
µ
˜
˜
˜
˜
∂a ∂b x − ∂b ∂a x ωµ + ∂˜b xµ ∂˜a ωµ − ∂˜a xµ ∂˜b ωµ
|
{z
}
→
=
=
=0
chain rule
=
relabel dummy indices
=
=
factorize
6
∂˜b xµ ∂˜a xν ∂ν ωµ − ∂˜a xµ ∂˜b xν ∂ν ωµ
∂˜b xν ∂˜a xµ ∂µ ων − ∂˜a xµ ∂˜b xν ∂ν ωµ
∂˜b xµ ∂˜a xν (∂µ ων − ∂ν ωµ )
(6)
6. Lie brackets. The commutator or Lie bracket [u, v] of two vector fields u, v on M is
defined as follows, by its action on any function on M :
[u, v](f ) = u(v(f )) − v(u(f )) .
(a) Show that its components in a coordinate basis are given by
[u, v]µ = uν ∂ν v µ − v ν ∂ν uµ .
[u, v](f )
=
product rule
=
rename dummy indices
=
=
uµ ∂µ (v ν ∂ν (f )) − v µ ∂µ (uν ∂ν (f ))
uµ (∂µ v ν )∂ν f − v µ (∂µ uν )∂ν f + uµ v ν ∂µ ∂ν f − v µ uν ∂µ ∂ν f
(uµ (∂µ v ν )∂ν − v µ (∂µ uν )∂ν ) f + uµ v ν ∂µ ∂ν f − v ν uµ ∂µ ∂ν f
(7)
(uµ (∂µ v ν )∂ν − v µ (∂µ uν )∂ν ) f
So the components of the bracket are as given above.
(b) Using the fact that uµ , v µ transform as contravariant vectors, show explicitly that
[u, v]µ also transforms this way.
This is very similar to the problem about the transformation of dω above. I will
do this one in a more geometric way, by studying the vector itself, rather than
the components :
uµ (∂µ v ν )∂ν − v µ (∂µ uν )∂ν
∂ x̃a
(Jµa ≡
)
∂xµ
a
−1 σ
σ
(Jµ (J )a = δµ )
→
ũa (∂˜a ṽ b )∂˜b − (u ↔ v)
=
Jµa uµ ((J −1 )σa ∂σ Jνb v ν )(J −1 )ρb ∂ρ − (u ↔ v)
=
uσ (∂σ v ρ )∂ρ − (u ↔ v)
rename dummy indices
=
uµ (∂µ v ν )∂ν − v µ (∂µ uν )∂ν .
(c) (Optional extra bit) Convince yourself from the general definition of Lie derivative
given in lecture that Lu v = [u, v].
See the footnote in section 5.3.1 of the lecture notes.
7
(8)