metric has enough symmetry for a solution in quadrature. Here is the metric ds2 = dt2. (. 1 â. 2M r. ) â dr2. (. 1 â
1
General Relativity
What do paths in the Schwarzschild spacetime look like? It turns out that that metric has enough symmetry for a solution in quadrature. Here is the metric −1 2M 2M 2 − r2 cos2 θdφ2 − r2 dθ2 ds = dt 1 − − dr 1 − r r 2
2
(1)
Without loss of generality let’s look at paths in the plane θ = π/2. The metric does not depend on time and the angle φ so we have two integrals of motion: E = pt and L = pφ . Let’s calculate p2 using the metric to give 2
m =E
2
2M 1− r
−1
r 2
− (p )
2M 1− r
−1 −
L2 r2
and solve for pr in terms of the constants of motion L2 2M 2 (pr ) = E 2 − m2 + 2 1− . r r
(2)
(3)
Furthermore, we know that pφ
2
=
L2 r4
(4)
and combining these results yields,
dφ dr
2 =
−1 1 r2 2M m2 r2 − 1 + 1 + r2 b2 r L2
(5)
where b2 = L2 /(E 2 − m2 ). This equation is sufficient to calculate the path of any object around a static black hole. In particular for E 2 − m2 < 0 (therefore, b2 < 0 the expression in parentheses vanishes for two values of r and the object is bound and travels between those radii. Furthermore, the twice difference between the integral of dφ/dr and π gives the periastron advance over a single orbit. We are interested in the total angular deflection when the deflection is small and the object is not bound, so we are interested in the value of the integral between the minimum radius and infinity for small values of M . To make further progress, we can write 1 1 2M 1 m2 = − + (6) b2 r02 r0 r02 L2 which gives the value of the minimum radius implicitly terms of E, m and L and substitute this expression into the previous equation to yield
dφ dr
2 =
−1 1 r2 2M r2 1 m2 2M m2 r2 − 1 − + + 1 + r2 r02 r0 r02 L2 r L2
1
(7)
and define x = r/r0 to give
and
dφ dx
2
dφ dx
2
−1 1 2M 1 2M 2 m2 r02 m2 r02 2 x + − 1 − x 1 + + x x2 r0 L2 r0 x L2
(8)
−1 2 2 2M 1 1 2 2 m r0 . = 2 2 1+ − x + (x − x ) 2 x (x − 1) r0 x2 − 1 x L
(9)
=
1
Now we will expand dφ/dx in terms of M/r0 to first order dφ M 1 1 1 m2 r02 2 2 − ≈ √ − x + x − x . dx r0 x (x2 − 1)3/2 x L2 x x2 − 1 The angle through which the path is scattered is given by Z ∞ M 4GM GM m2 r02 c2 M m2 r02 dφ dx − π ≈ π + 4 − π = + 2 2 +2 dx r0 r0 L2 r0 c2 r0 c2 L2 1
(10)
(11)
where in the final step we have inserted the powers of G and c to make contact with the non-relativistic limit ∆θ ≈
2GM 2GM + 2 r0 c r0 v 2
(12)
where v is the velocity at closest approach relative to a static observer. The expression is valid for small angles and all velocities.
2
Periastron Advance
Let’s start with the equation 2 −1 dφ 1 r2 2M r2 1 m2 2M m2 r 2 = 2 2 −1− + + 1 + dr r r0 r0 r02 L2 r L2
(13)
and define L in terms of the maximum radius r1 as r1 r0 (r1 + r0 ) − 2M r12 + r02 + r1 r0 1 = 2 L 2m2 M r12 r02
(14)
yielding
dφ dr
2 =
−1 1 r1 r0 1 1 1 1 − 2M + + . r2 (r1 − r) (r − r0 ) r0 r1 r
Let us expand dφ/dr to first order in M to yield r dφ 1 r1 r0 1 1 1 ≈ 1+M + + . dr r (r1 − r) (r − r0 ) r0 r1 r 2
(15)
(16)
We require the following integrals Z r 1 r1 r0 (r1 + r0 ) r − 2r1 r0 dr = arcsin +C r (r1 − r) (r − r0 ) r (r1 − r0 )
(17)
and Z r r1 r0 1 (r1 − r)(r − r0 ) 1 r1 + r0 (r1 + r0 ) r − 2r1 r0 dr = + arcsin +C. r2 (r1 − r) (r − r0 ) r1 r0 2 r1 r0 r (r1 − r0 ) (18) These yield the following expression for the orbit, 3M 1 1 (r1 + r0 ) r − 2r1 r0 M (r1 − r)(r − r0 ) φ(r) ≈ 1 + + arcsin + 2 r1 r0 r (r1 − r0 ) r1 r0 (19) The argument of the arcsine is 1 for r = r1 and −1 for r = r0 , yielding the following result Z r1 3π 1 dφ 1 dr ≈ π + M + (20) 2 r1 r0 r0 dr so the advance of periastron through one orbit is 3πGM 1 1 6πGM ∆θ ≈ + = 2 c2 r1 r0 c a(1 − e2 )
(21)
where a is the semi-major axis of the elliptical orbit and e is its eccentricity. If we specialize to Mercury, we have GM /c2 = 1.48 km, a = 5.791 × 107 km and e = 0.2056 so ∆θ ≈ 5.03 × 10−7 (22) or 43 arcseconds per century. Looking back at the expression for the orbit, let us define a new angle that accounts for the precession of the orbit, 3M 1 1 φ(r) = 1 + + ψ(r) (23) 2 r1 r0 so −1 M (r1 − r)(r − r0 ) 3M 1 1 (r1 + r0 ) r − 2r1 r0 + 1+ + r (r1 − r0 ) r1 r0 2 r1 r0 (24) The exact solution is also tractable in terms of elliptic integrals. Looking at (dφ/dr)2 again yields 2 dφ c2 1 r0 r1 r2 = (25) dr 2GM r (r − r0 ) (r1 − r) (r − r2 )
ψ(r) ≈ arcsin
where
1 1 1 r2 = − − 2M r1 r0
−1
c2 1 1 = − − 2GM r1 r0
3
−1 .
(26)
The Newtonian limit obtains as c → ∞ so the exact relativistic expression is the minimal modification to the Newtonian result. For the value of (dφ/dr) to make sense we must have r1 ≥ r0 ≥ r2 , so the limiting orbit has r1 = r0 = r2 which allows us to conclude that r0 = r1 = r2 = 6M is the limiting stable circular orbit. There are circular orbits down to r0 = r1 = 3M . However, these orbits do not have neighbouring non-circular orbits; therefore, they are unstable.
3
Time Evolution
We have a very compact expression for the shape of the orbit. Can we develop an expression for how the position of the particle changes with time? We have pt = so
dt dr
2
=
pt pr
2
=
pφ pr
2
pt pφ
−1 2M E 1− r
2
E2
=
1−
2M 2 r
(27)
r0 r1 r2 r4 1 1 L2 2M r (r − r0 ) (r1 − r) (r − r2 ) (28)
We can clean this up a bit to yield
4
dt dr
2 =
E2 r5 r0 r1 r2 2M L2 (r − 2M )2 (r − r0 ) (r1 − r) (r − r2 )
(29)
Is space curved?
Let’s look at this in the context of an alternative metric with a different amount of spatial curvature ds2 =
−1 2M 2γM 1− dt2 − 1 − dr2 − r2 cos2 θdφ2 − r2 dθ2 . r r
(30)
Here γ is the PPN parameter that quantifies how much space curvature is produced by unit rest mass. Again without loss of generality let’s look at paths in the plane θ = π/2. The metric does not depend on time and the angle φ so we have two integrals of motion: E = pt and L = pφ . Let’s calculate p2 using the metric to give −1 −1 2γM L2 2M 2 − (pr ) 1 − − 2 m2 = E 2 1 − r r r
(31)
and solve for pr in terms of the constants of motion −1 L2 2M 2γM 2M 2 2 (p ) = E − m + 2 1− 1− 1− . r r r r r 2
4
(32)
This expression is a bit more complicated than the general relativistic result, Eq. 3. In general relativity the two terms outside the brackets cancel. However, the derivation of the equation for the shape of the orbit can proceed almost identically. Again, we know that 2 L2 (33) pφ = 4 r and combining these results yields −1 −1 1 r2 2M 2M m2 r2 2γM = 2 2 −1+ 1− 1+ 1− r b r L2 r r (34) where b2 = L2 /(E 2 − m2 ). Again we look at the minimum radius of r0 for an unbound trajectory. We can write E 2 in terms of the value of r0 as before, Eq. 6, we have 1 1 2M 1 m2 = − + (35) b2 r02 r0 r02 L2
dφ dr
2
which gives the value of the minimum radius implicitly terms of E, m and L and substitute this expression into the previous equation to yield −1 1 r2 2M r2 1 m2 m2 r 2 2M r − 2M − 1 − + 1 + + 2 2 2 2 2 r r0 r0 r0 L r L r − 2γM (36) as before but with the extra factor. Again we can define x = r/r0 and expand dφ/dx to lowest order in M/r0 to yield dφ 1 1 M m2 r02 1 M γ−1 2 2 √ ≈ √ − − x + . x − x + dx r0 x (x2 − 1)3/2 x L2 r0 x x2 − 1 x x2 − 1 (37) We can calculate the deflection as before but with the extra term to yield
dφ dr
2
=
∆θ ≈
2GM 2GM 2(γ − 1)GM 2γGM 2GM + 2 + = 2 + 2 2 2 c r0 v r0 c r0 c r0 v r0
(38)
so the fact that the bending of light is twice the Newtonian value due to the curvature of space as well as time. We can also look at the perihelion advance, starting with the modification of Eq. 15 to become 2 −1 dφ 1 r1 r0 1 1 1 r − 2M = 2 1 − 2M + + (39) dr r (r1 − r) (r − r0 ) r0 r1 r r − 2γM and to first order in M we have r dφ 1 r1 r0 1 1 γ ≈ 1+M + + . dr r (r1 − r) (r − r0 ) r0 r1 r 5
(40)
This yields the following expression for the orbit, γ 1 1 (r1 + r0 ) r − 2r1 r0 γM (r1 − r)(r − r0 ) φ(r) ≈ 1 + M 1 + + arcsin + 2 r1 r0 r (r1 − r0 ) r1 r0 (41) and the advance of periastron through one orbit of γ 1 γ 2πGM 1 4πGM 1 + 1+ (42) ∆θ ≈ + = 2 2 2 c 2 r1 r0 c a(1 − e ) 2 so one third of the periastron advance is due to the curvature of space and two-thirds from the curvature of time. The γ = 0 values of the periastron advance and the equivalent deflection of a distant passer by are called the classical values. The gravitational redshift is independent of γ because it only depends on the g00 part of the metric.
6
