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Physics , General Relativity
Homework
Due Wednesday, th^ December
Jacob Lewis Bourjaily
Problem 1 Consider a gyroscope moving in circular orbit of radius R about a static, spherically-symmetric planet of mass m. a. We are to derive the equations of motion for the gyroscopic spin vector as a function of azimuthal angle and show that the spin precesses about the direction normal to the orbital plane.
This calculation will be far from elegant, and will probably not give rise to much insight. Nevertheless, we start by recalling the Lagrangian describing a particle’s worldline(in the θ = π 2 plane) in a static, isotropic spacetime,
L = −gabuaub^ = f (r)(ut)^2 −
f (r)
(ur^ )^2 − r^2 (uϕ)^2 , (1.1)
where ua^ ≡ dx
a dτ for some affine parameter^ τ^.^ Because our analysis will be limited to circular geodesics, we will not have much use for the ur^ coordinate; however, its equation of motion will be necessary to relate the various integrals of motion. First observe that uϕ^ is non-dynamical in the Lagrangian and so it gives us our first integral of motion,
J ≡ r^2 uϕ. (1.2)
For circular geodesics, ua^ will of course only have 0 and ϕ components; ut^ is also non- dynamical, and so we are free to set ut^ by the normalization of the affine parameter τ :
u^2 = −gabuaub^ = f (R)(ut)^2 −
J^2
R^2
≡ 1 , =⇒ ut^ =
f (R)
J^2
R^2
Now, it is easy to see that the equation of motion for the r-component is
¨r f (r)
r˙^2 f 2 (r)
f ′(r) = −
r˙^2 f 2 (r)
J^2
r^3
Because we are looking for solutions where both ˙r and ¨r vanish—and r = R—we see at once that this implies the relation
J^2 =
f ′(R)(ut)^2 R^3 =
m R^2
R^4
(R − 2 m)
J^2
R^2
mR^2 R − 2 m
1 − (^) R−m 2 m
mR^2 R − 3 m
Above, we made use of the definition of the Schwarzschild metric’s f (r) = 1 − 2 mr. We have now completely specified the circular geodesic of radius R in which we are interested. The direction of a gyroscope’s spin is therefore simply a vector Sa^ which satisfies the orthogonality condition uaSbgab = 0 along the geodesic. Recall that two parallelly- transported vectors have the property that the gradient of their scalar product van- ishes. This immediately allows us to write down the equation for the evolution of the components of Sa^ along τ ,
dSa dτ
= ΓbacSbuc, (1.6) 1
2 JACOB LEWIS BOURJAILY
which, upon using the Christoffel symbols for the Schwarzschild metric^1 , becomes
dSt dτ
= ΓrttSr ut^ =
f (R)f ′(R)Sr ut^ =
f (R)
J^2
R^2
Sr ; (1.7)
dSr dτ
= ΓtrtStut^ + ΓϕrϕSϕuϕ^ =
f ′(R) 2 f (R)
f (R)
J^2
R^2
St +
J
R^3
Sϕ; (1.8)
dSθ dτ
= ΓϕθϕSϕuθ^ + Γθθr Sθur^ + Γrθθ Sr uθ^ = 0; (1.9) dSϕ dτ
= ΓθϕϕSθ uϕ^ + ΓrϕϕSr uϕ^ = −
J
R
f (R)Sr. (1.10)
This almost completes our analysis. Indeed, notice that the above system of equations implies that the θ-component of the gyroscope’s spin is fixed. All the motion of Sa as it is transported along τ is confined to the plane normal to θˆ. Therefore, we may conclude that the gyroscope will precess about the axis normal to its orbital plane. The finicky reader may object that the system of equations (1.6-9) are over-specified. To be thorough we should eliminate redundancy. The first of the relations among these expressions comes from the orthogonality condition on the spin vector Saua^ = 0. In components this reads
Stut^ + Sϕuϕ^ = 0 =⇒ St
f (R)
J^2
R^2
J
R^2
Sϕ. (1.11)
Also, it is more physically interesting to compute evolution relative to the angle ϕ as observed by a stationary observer on the planet. Replacing St in favour of Sϕ and making us of the fact dτdϕ = R
2 J ,
dSt dϕ
R^2
2 J
f (R)
J^2
R^2
Sr ;
dSr dϕ
R
f ′(R) 2 f (R)
Sϕ;
dSϕ dϕ
= −Rf (R)Sr ;
dSθ dϕ
The last redundancy to take care of comes from the geodesic equation for SaSbgab— namely, that this scalar is preserved. Let us choose to normalize SaSbgab = +1 so that
1 = −
f (R) S^2 t + f (R)S^2 r +
R^2
S θ^2 +
R^2
S ϕ^2 ,
R^2
S^2 ϕ
J^2
R^2
1 + J
2 R^2
R^2
S θ^2 ,
S ϕ^2 (R^2 + J^2 )
R^2
S θ^2.
Bearing in mind that Sθ is a constant of motion, me may therefore write
S^2 ϕ = f (R)
R^2 + J^2
f (R)
f (R)R^2
S θ^2 − S r^2
or S r^2 =
f (R) (R^2 + J^2 )
(R^2 + J^2 ) −
(R^2 + J^2 )
R^2
S^2 θ − S^2 ϕ
The two substantive equations of motion are clearly dS dϕr and dS dϕϕ. Squaring the equations derived above, and using the normalization condition to reexpress unlike components,
(^1) And specializing to the obvious coordinate choice θ 7 → π 2 everywhere it is encountered.
4 JACOB LEWIS BOURJAILY
Problem 2 Consider a +-dimensional AdS spacetime described by the metric
ds^2 = −f (r)dt^2 +
dr^2 f (r)
- r^2 dΩ^23 , where f (r) = 1 + r^2 −
μ r^2
We are to determine the radial coordinate of the black hole horizon, calculate the proper time of a massive object to free-fall from the surface of the black hole to the singularity at r = 0, and determine the radius and period of the null-circular orbit.
The horizon radius is that for which f (r) vanishes. A child’s experience with the qua- dratic formula is sufficient to see that there is exactly one real root of f (r) and this corresponds to a horizon radius of
∴ rh =
4 μ^2 + 1 − 1
‘ ´oπ≤ρ ´≤δ≤ι πoι’ ησαι
To calculate the proper time for free-fall from the horizon, we need to quickly derive the equations for motion in only the r-direction. Because we’ll need the angular dependence later, we’ll start a bit more generally. First, look at the Lagrangian for the particle’s motion (its worldline),
L = −gabuaub^ = f (r) ˙t^2 −
f (r)
r˙^2 − r^2 ϕ˙^2. (2.3)
Now, as always, a ‘˙’ indicates differentiation with respect to an affine parameter, say τ , along the worldline of the particle. We will eventually impose the normalization condition (think Lagrange multipliers)
κ ≡ −gabuaub, (2.4)
where κ = 1 for time-like worldlines and κ = 0 for null. The first thing that should be apparent form the Lagrangian is that there are two non-interacting degrees of freedom, ˙t and ϕ˙, giving rise to two integrals of motion^3 E ≡ f (r) ˙t, and J ≡ r^2 ϕ.˙ (2.5)
Now, in the case of purely radial motion of a massive object, J = 0 and κ = 1; so we are left with only
1 =
f (r)
E^2 −
f (r)
r˙^2 , =⇒ r˙^2 = E^2 − f (r). (2.6)
Notice that this means that E must be chosen so that ˙r^2 = 0 = E^2 − f (R) for some R. In the problem under consideration, we want to find the motion of an object dropped from rest at R = rh—and rh is defined to be such that f (rh) = 0. Therefore, E^2 = 0 for our present problem, and dr dτ
−f (r); (2.7)
which is easy enough to formally invert:
τ =
∫ (^) rh
0
dr √ −f (r)
Our computer algebra software had no difficulty evaluating this, showing that
∴ τ = π 4
arccot (
μ). (2.9)
‘ ´oπ≤ρ ´≤δ≤ι πoι’ ησαι
(^3) We could have framed this discussion in terms of Killing fields, but we’ll stick to Euler while we can.
PHYSICS : GENERAL RELATIVITY HOMEWORK 5
Lastly, we are asked to find the radius at which light can orbit circularly, and determine the coordinate time of this orbit’s period. To do this, we need only to re-instate J into our expression for ˙r^2 and set κ → 0 for null geodesics. Written suggestively, this gives 1 2
r˙^2 +
f (r)
J^2
r^2
E^2. (2.10)
Reminiscent of effective potentials, we are inspired to consider an analogue problem in +-dimensions governed by the effective potential
Veff =
J^2
2 r^2
1 + r^2 −
μ r^2
This effective potential has only one turning point, at
−
J^2
r^3
μJ^2 r^5
= 0 =⇒ r =
2 μ. (2.12) ‘ ´oπ≤ρ ´≤δ≤ι πoι’ ησαι
This is the radius at which there are circular, null geodesics—as evidenced by the fact that ˙r = 0 at this radius. Inserting r =
2 μ into (2.10),
E^2 = f (
2 μ)
J^2
2 μ
= J^2
4 μ
J^2
E^2
4 μ 4 μ + 1
This is needed for us to compute the coordinate-time orbit period. Recall from our definitions of E and J that dϕ dt
dϕ dτ
dτ dt
J
r^2
f (r) E
—which when combined with the above implies dϕ dt
4 μ + 1 μ
This is trivially integrated. We find that the coordinate time of one orbit is
∴ tp = 4π
μ 4 μ + 1
‘ ´oπ≤ρ ´≤δ≤ι πoι’ ησαι
Problem 3 Consider a clock in circular orbit at radius R = 10m about a spherically symmetric star. a. We are to determine the proper time of the R = 10m orbit. We can draw heavily on our work above. Using the notation and conventions of problem one, we see that
τp =
dτ =
dτ dϕ dϕ =
R^2
J
dϕ = 2π
R^2
J
Using our equation (1.5) for J at a given R, we find
∴ τp = 2π
R
R − 3 m √ m
. (3.a.1) ‘ ´oπ≤ρ ´≤δ≤ι πoι’ ησαι
For the particular question at hand, r = 10m, we find the period to be τp = 20
7 πm. (3.a.2)
