Gravitation & Cosmology. Exercises #
4.1 - Geodesics
a) Show that the Euler-Lagrange equations
d ∂L
∂L
− µ =0
µ
dτ ∂ ẋ
∂x
(1)
1
L = gµν ẋµ ẋν
2
(2)
d 2 xµ
+ Γµνλ ẋν ẋλ = 0
dτ 2
(3)
Γµνλ = g µρ Γρνλ
1
Γµνλ = (gµν,λ + gµλ,ν − gνλ,µ )
2
(4)
for the Lagrangian
are the geodesic equations
where, as usual,
b) Show that L is constant along any geodesic, i.e. that
d
(gµν ẋµ ẋν ) = 0
dτ
(5)
for xµ (τ ) a solution of the geodesic equation.
c) Show that the Euler-Lagrange equations of the Lagrangian
1
L = (θ̇2 + sin2 (θ)φ̇2 )
2
(6)
coincide with the geodesic equations for S 2 derived in Lecture 4.
Solution
a)
∂L
1
= ẋρ ẋν ∂µ gρν
µ
∂x
2
∂L
ρ ∂
=
g
ẋ
ẋν = gρν ẋρ δµν = gρµ ẋρ
ρν
∂ ẋµ
∂ ẋµ
1
(7)
(8)
d
dτ
∂L
∂ ẋµ
!
= gρµ ẍρ + ẋρ ẋν ∂ν gρµ
(9)
1
= gρµ ẍρ + (ẋρ ẋν ∂ν gρµ + ẋν ẋρ ∂ρ gνµ )
2
The Euler-Lagrange equations becomes :
1
[E. − L.] = gµρ ẍρ + (∂ν gρµ + ∂ρ gνµ − ∂µ gρν )ẋν ẋρ
2
= gµρ ẍρ + Γµνρ ẋρ ẋν = 0
(10)
and they can be written in the usual (geodesic equations) form by multiplying by g λµ to
move the index µ up :
g λµ (gµρ ẍρ + Γµνρ ẋρ ẋν ) = ẍλ + Γλνρ ẋρ ẋν = 0
(11)
b) First we compute :
d
1 d
1
L=
gµν ẋµ ẋν =
2gµν ẍµ ẋν + (ẋρ ∂ρ gµν ẋµ ẋν )
dτ
2 dτ
2
!
(12)
Now using the identity :
∂ρ gµν = Γµνρ + Γνµρ
(13)
together with the fact that xµ (τ ) is a solution to the geodesic equation, which means that
we also have :
ẍµ = −Γµνρ ẋν ẋρ
(14)
leaves us with :
1
d
L = (−2gµν Γµλρ ẋλ ẋρ ẋν + ẋρ (Γµνρ + Γνµρ )ẋµ ẋν )
dτ
2
1
= (−2Γνλρ ẋλ ẋρ ẋν + (Γµνρ + Γνµρ )ẋµ ẋν ẋρ ) = 0
2
(15)
which is obviously zero if we relabel the indices.
c) Using the Euler-Lagrange equations with L = 12 (θ̇2 + sin2 (θ)φ̇2 ) we get:
!
d ∂L ∂L
1 d
2
−
=
2θ̇ − 2 sin(θ) cos(θ)φ̇ = 0
dτ ∂ θ̇
∂θ
2 dτ
d ∂L ∂L
d
−
=
(sin2 (θ)φ̇) = sin2 (θ)φ̈ + 2 cos(θ) sin(θ)θ̇φ̇ = 0
dτ ∂ φ̇
∂φ
dτ
which are precisely the equations derived in Lecture 4.
2
(16)
4.2 - Killing vector fields
a) In flat Minkowski spacetime find ten Killing vectors that are linearly independent.
b) Give a physical interpretation of the what you found in a).
c) Show that Lorentz transformations of the Killing vector fields listed in b)
give rise to linear recombinations of the same fields with constant coefficients.
Solution
a) In the Minkowski spacetime all the Levi-Civita connection coefficients vanish. The
Killing equation becomes
∂ µ Xν + ∂ ν Xµ = 0
(17)
From this equation it follows that Xµ is, at most, of the first order in x1 . Then, we can
have the constant solutions
µ
X(i)
= δiµ (0 ≤ i ≤ 3)
(18)
Next, let Xµ = aµν xν , aµν being constant. Equation (17) implies that aµν is anti-symmetric
with respect to µ ↔ ν. There are six independent solutions of this forms, three given by
X(j)0 = 0
X(j)m = εjmn xn
(1 ≤ j, m, n ≤ 3)
(19)
(1 ≤ k, m ≤ 3)
(20)
while the others
X(k)0 = xk
X(k)m = −δkm x0
Let’s check that these are really solutions of the Killing equation. For (19) we only have
to check the case in which µ, ν 6= 0 since all the other cases are trivial
∂p Xq + ∂q Xp = ∂p εjqn xn + ∂q εjpn xn
= εjqn δpn + εjpn δqn = εj(pq) = 0
(21)
For (20) we have to check the case in which µ = 0, ν 6= 0 and viceversa
∂0 Xp + ∂p X0 = −δkp + δkp = 0
(22)
1
In 2D with coordinates (t, x) the Killing equation can be expressed by ∂t ξt = 0 , ∂x ξx = 0 and
∂t ξx + ∂x ξt = 0. This means that the Killing vector fields satisfy the following relations ξt = f (x) ,
ξx = g(t) and f 0 (x) + ġ(t) = 0 which prove the claim.
3
b) The ten independent Killing fields correspond to the ten generators of the Poincaré
algebra: four translations, three rotations, and three boosts. Given in terms of the basis
vectors ~ea , we the Killing vector fields are therefore
• Translations : ~et , ~ex , ~ey , ~ez ;
• Rotations : y~ex − x~ey , z~ey − y~ez , x~ez − z~ex ;
• Boosts : x~et + t~ex , y~et + t~ey , z~et + t~ez ;
Each of these ten vector fields manifestly satisfies Killing’s equation.
Remark
In m-dimensional Minkowski spacetime (m ≥ 2), there are m(m + 1)/2 Killing vector
fields, m of which generate translations, (m − 1), boosts and (m − 1)(m − 2)/2, space
rotations. Those spaces (or spacetimes) which admit m(m + 1)/2 Killing vector fields are
called maximally symmetric spaces.
c) Since every Lorentz transformation can be built from infinitesimal ones, it is sufficient
to demonstrate the claim for infinitesimal Lorentz transformations. And this makes our
work exceptionally easy. Infinitesimal Lorentz transformations are simply the identity plus
a constant multiple of the generators of the Lorentz algebra; but (the last six of) the Killing
fields listed in b) are nothing but these Lorentz generators. Therefore, any infinitesimal
Lorentz transformation of the Killing fields listed in b) is a linear combination of those
same Killing fields with constant coefficients. And by extension, the same is true for any
finite Lorentz transformation.
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4.3 - Killing vector fields II
a) Show that if ζa (x) is a Killing vector field and pa (λ) is the tangent vector to
a geodesic curve γ(λ), then pa ζa (x) is constant along γ.
b) Show that if ζa and ηa are Killing fields and α, β constants, then αζa + βηa
is Killing.
Solution
a) The derivative of pa ζa along γ is
pb ∇b (pa ζa ) = pa pb ∇b ζa + ζa pb ∇b pa
(23)
The first term vanishes because pa pb is symmetric while ∇b ζa is antisymmetric (because
it is Killing). The second term vanishes because pa is the tangent of a geodesic, which
practically by definition implies that it obeys the geodesic equation, pb ∇b pa = 0.
b) By direct check:
∇b (αζa + βηa ) = α∇b ζa + β∇b ηa = −α∇a ζb − β∇a ηb = −∇a (αζb + βηb )
(24)
because, being constants, α, β commute with the gradient and ζa , ηa are Killing. Therefore
equation (24) implies that (αζa + βηa ) is Killing.
5
Extra
Consider a pair of twins are born somewhere in spacetime. One of the twins
decides to explore the universe; she leaves her twin brother behind and begins
to travel in the x-direction with constant acceleration a = 10m/s2 as measured
in her rocket frame. After ten years according to her watch, she reverses the
thrusters and begins to accelerate with a constant −a for a while.
a) At what time on her watch should she again reverse her thrusters so she
ends up home at rest ?
b) According to her twin brother left behind, what was the most distant point
on her trip ?
c) When the sister returns, who is older, and by how much?
Solution
a) There is an obvious symmetry in this problem: if it took her 10 years by her watch
to go from rest to her present state, then 10 years of reverse acceleration will bring her
to rest, at her farthest point from home. Because of the constant negative acceleration,
after reaching her destination at 20 years, she will begin to accelerate towards home again.
In 10 more years, when her watch reads 30 years, she will be in the same state as when
her watch read 10 years, only going in the opposite direction. Therefore, at 30 years, she
should reverse her thrusters again so she arrives home in her homes rest frame.
b) To do this, we need only to solve the equations for the travelling twins position and
time as seen in the stationary twins frame. This was done in a previous exercise class but,
in brief, we know that her four-acceleration is normal to her velocity: aα uα = 0 everywhere
along her trip, and aα aα = a2 is constant. This leads us to conclude that
at =
∂ut
= aux
∂τ
ax =
∂ux
= aut
∂τ
(25)
where τ is the proper time as observed by the travelling twin. This system is quickly solved
for an appropriate choice of origin2 :
t=
1
sinh(aτ )
a
x=
1
cosh(aτ )
a
(26)
This is valid for the first quarter of the twins trip. All four legs can be given explicitly
by gluing together segments built out of the above. For the purposes of calculating, it is
necessary to make aτ dimensionless. This is done simply by
a=
2
10 m
= 1.053 year−1
2
s
We consider the twin to begin at (t = 0, x = 1).
6
(27)
An approximate result3 could have been obtained by thinking of c = 3 × 108 m/s and
3 × 107 s = 1 year.
So the distance at 10 years is simply
x(10 yr) =
1
cosh(10.53) = 17710 light years.
1.053
(28)
The maximum distance travelled by the twin as observed by her (long-deceased) brother
is therefore twice this distance,
max(x) = 35420 light years.
(29)
c) Well, in the brothers rest frame, his sisters trip took four legs, each requiring
t(10 yr) =
1
sinh(10.53) = 17710 years.
1.053
(30)
which means that
ttotal = 70838 years
(31)
In contrast, his sisters time was simply her proper time, or 40 years. Therefore the brother
who stayed behind is now 70798 years older than his twin sister.
3
Since know c and the number of seconds per year to rather high-precision, there is no reason not to
use the correct a. Indeed, the approximate value of a ∼ 1 year−1 gives an answer almost 40% below our
answer.
7