Schwarzschild Geodesics

Schwarzschild geodesics

April 12, 2015

1 Schwarzschild spacetime

Spherically symmetric solutions to the Einstein equation take the form 2M dr2 ds2 = − 1 − dt2 + + r2dθ2 + r2 sin2 θdϕ2 r 2M 1 − r GM where G = c = 1 and the mass M = c2 is determined from the asymptotic acceleration. We also have the connection components, which now take the form: M Γ0 = Γ0 = 01 10 2 2M r 1 − r M 2M Γ1 = 1 − 00 r2 r M Γ1 = − 11 2 2M r 1 − r 1 Γ2 = Γ2 = 12 21 r 2M Γ1 = −r 1 − 22 r 1 Γ3 = Γ3 = 13 31 r 2M Γ1 = − 1 − r sin2 θ 33 r cos θ Γ3 = Γ3 = 23 32 sin θ 2 Γ33 = − sin θ cos θ Despite the coordinate singularity at r = 2M, the only true geometric singularity is at the origin, r = 0.

2 The geodesic equations

Now consider the geodesic equation, duα 0 = + Γα uµuν dτ µν This equation determines the paths of freely falling particles in the neighborhoods of a planet, star or other µ dxα spherically symmetric source. We also have the norm of the 4-velocity, u = dτ , 2M dt 2 1 dr 2 dθ 2 dϕ2 −1 = − 1 − + + r2 + r2 sin2 θ (1) r dτ 2M dτ dτ dτ 1 − r

1 α dxα We may also find spacelike geodesics by setting the norm of the unit tangent vector, t = dx to +1, α dxα respectively. For null geodesics we cannot use τ for the parameter since dτ = 0, but setting t = dλ for α any other suitable parameter, we have t tα = 0. Substituting for the connection coefficients, the geodesic equation for each component of the 4-velocity (or tangent vector tα) is then, du0 2M = − u0u1 dτ 2 2M r 1 − r du1 M 2M M 2M 2M = − 1 − u0u0 + u1u1 + r 1 − u2u2 + 1 − r sin2 θu3u3 dτ r2 r 2 2M r r r 1 − r du2 2 = − u2u1 + sin θ cos θu3u3 dτ r du3 2 2 cos θ = − u1u3 − u2u3 dτ r sin θ A judicious choice of coordinates can simplify these equations. Because the spacetime is spherically symmetric, we may rotate in the θ and ϕ coordinates arbitrarily. Let the initial point of the geodesic be π arbitrary. Then we may rotate the θ and ϕ coordinates so tht the initial point has θ = 2 and ϕ = 0. We may also choose the origin of the time coordinate arbitrarily, so the makes initial position is π xα = 0, r , , 0 0 0 2 Once this point is fixed there remains a single rotation about the initial point. If the initial 4-velocity has 2 3 2 components u0, u0 tangent to the spheres of symmetry, we may rotate the coordinates until u0 = 0, so the arbitrary initial velocity may be taken to be α 0 1 3 u0 = u0, u0, 0, u0 With these conventions, the initial u2 acceleration is 2 du 2 2 1 π π 32 = − u0u0 + sin cos u0 dτ 0 r 2 2 = 0 π 2 Since there is no initial acceleration or velocity, θ = 2 at all times. The u equation is solved and the remaining geodesic equations reduce to du0 2M = − u0u1 dτ 2 2M r 1 − r du1 M 2M M 2M = − 1 − u0u0 + u1u1 + 1 − ru3u3 dτ r2 r 2 2M r r 1 − r du3 2 = − u1u3 dτ r This simplification occurs because of conservation of angular momentum. The direction of the angular momentum vector is constant, so if the coordinates are chosen so that the motion starts in the equatorial plane, it must stay there. This is not true for a rotating black hole, which generically puts a torque on the particle. There is another simplification we can make by considering the norm of the 4-velocity, eq.1. Differentiating with respect to τ, eq.1 becomes 0 1 2M du 2M 2 2 du 0 = −2 1 − u0 − u1 u0 + u1 r dτ r2 2M dτ 1 − r 3 1 2M 3 du 2 − u1 + 2r2u3 + 2ru1 u3 2M 2 r2 dτ 1 − r

2 This holds for spacelike or null geodesics as well, with the appropriate parameterizations. Substituting the du0 du3 expressions for dτ and dτ from the geodesic equation and cancelling common factors,

1 4M 2 2M 2 2 du 0 = u0 u1 − u1 u0 + u1 r2 r2 2M dτ 1 − r 1 2M 3 2 2 − u1 − 4r u3 u1 + 2ru1 u3 2M 2 r2 1 − r 1 M 2 1 du 1 M 2 2 0 = u0 + − u1 − r u3 r2 1 − 2M dτ 2M 2 r2 r 1 − r so that

1 du M 2M 2 M 2 2M 2 = − 1 − u0 + u1 + r 1 − u3 dτ r2 r 2 2M r r 1 − r reproducing the radial geodesic equation. We may therefore replace the radial equation by the norm of the 4-velocity. The ﬁnal set of equations to solve is:

du0 2M = − u0u1 (2) dτ 2 2M r 1 − r du3 2 = − u1u3 (3) dτ r (timelike) −1   2M 2 1 2 2 (null) 0 = − 1 − u0 + u1 + r2 u3 (4) r 1 − 2M (spacelike) +1  r

3 Solution

0 1 dr Eqs.(2) and (3) are straightforward to integrate. For eq.(2), divide by u and write u = dτ , 1 du0 2M dr = − u0 dτ 2 2M dτ r 1 − r u0 r 1 2M du0 = − dr ˆ u0 ˆ r (r − 2M) 0 r0 u0 r u0 1 1 ln 0 = − dr u0 ˆ r r − 2M r0 r r − 2M = ln − ln r0 r0 − 2M r (r − 2M) = ln 0 r0 (r − 2M) so exponentiating, 1 − 2M ! 0 0 r0 u = u0 2M 1 − r

3 3 3 1 dr The solution for u is found the same way, dividing by u and writing u = dτ , 1 du3 2 dr = − u3 dτ r dτ r u3 2 ln 3 = − dr u0 ˆ r r0 r = −2 ln r0 Exponentiation yields 2 3 2 3 r u = r0u0 ≡ L 3 3 L This is the angular momentum per unit mass. We may replace u using u = r2 . Finally, choosing the timelike case and using the results for u0 and u3, eq.4 gives u1:

2M 2 1 2 2 −1 = − 1 − u0 + u1 + r2 u3 r 2M 1 − r 1 2 2M 2 2 u1 = 1 − u0 − 1 − r2 u3 2M r 1 − r 2 2 2M 2 2M 2M 2 u1 = 1 − u0 − 1 − − r2 1 − u3 r r r

1 dr 0 3 so that with u = dτ and substituting for u and u , s 2 2 2 dr 2M 0 2 2M L 2ML = 1 − (u0) − 1 + − 2 + 3 dτ r0 r r r Deﬁne the initial constant to be 2E " 2 # 1 2M 02 E ≡ − 1 − 1 − u0 2 r0

M 1 2 1 2M 2 = − + u1 + 1 − r2 u3 r 2 2 r so that E is the (proper time) energy per unit mass. The form now simplifies to r dr 2M L2 2ML2 = 2E + − + dτ r r2 r3 and we have reduced the problem to quadratures. Compare this to the Newtonian result (see Appendix): r L2 2GM r˙ = 2E − + r2 r √ dt 2E+1 In addition to the expression in terms of τ rather than t, where dτ = 2M , the results differ only by the 1− r 2ML2 final term r3 . This term is generally small, and we can use what we have learned about Newtonian orbits to obtain solutions. Restoring G and c, we compare the magnitudes of the terms in the line element. We have already seen that 2M 2GM v2 = = escape r rc2 c2

4 where vescape is the classical escape velocity (there is no escape velocity from within r = 2M!). This term is less than 1 until the particle reaches the horizon at r = 2M. We also estimate

L2 L2 dt 2 r2ϕ˙ 2 dt 2 v2 = = = ϕ r2 r2c2 dτ c2 dτ c2 which is also less than 1 for nonrotating black holes. For the ﬁnal term, 2GML2 2GM L2 = × r3c2 rc2 c2r2 v2 v2 ∼ escape × orbital c2 c2 For ordinary velocities and solar system gravity, this is a small correction. Any orbits for which this is a small correction may be treated perturbatively.

Summary of geodesics: We have, for timelike curves: ! dt 1 − 2M = u0 r0 dτ 0 2M 1 − r dϕ L = dτ r2 r dr 2M L2 2ML2 = 2E + − + dτ r r2 r3 where 2 dϕ L = r0 dτ 0 " 2 # 1 2M 02 E = − 1 − 1 − u0 2 r0

dr Exercise: Find the form of dτ for spacelike and null geodesics (see problems). 4 The radial integral

As with the Newtonian case (see Appendix), it is much more revealing to solve for r (ϕ)than for r (τ). Therefore, with dr dr/dτ = dϕ dϕ/dτ r2 = r˙ L we integrate r dr r2 L2 2GM = 2E − + dϕ L r2 r The complete radial integral is

ϕ r Ldr dϕ = q ˆ ˆ 2 2M L2 2ML2 r 2E + − 2 + 3 0 r0 r r r

5 This is exactly integrable in terms of elliptic integrals, but the general form is very lengthy (try it in Wolfram!). However, except in cases of extremely strong gravity or high velocities, measurable results may be predicted perturbatively. In cases where the motion is close to the classical solution, we factor the square root in the denominator,

r Ldr ϕ = q ˆ 2 2M L2 2ML2 r 2E + − 2 + 3 r0 r r r r Ldr = q r ˆ 2 2M L2 2ML2 r 2E + − 2 1 + 2 r0 r r r32E+ 2M − L r r2

2ML2 Then, for 2 1, we expand the second root in a Taylor series, r32E+ 2M − L r r2 r ! dr ML2 ϕ ≈ 1 − 2 q 2 3 2M L ˆ 2 2M L r 2E + − 2 r 2E + − 2 r r r0 r r r r dr ML2dr = − q 3/2 ˆ 2 2M L2 ˆ 5 2M L2 r 2E + − 2 r 2E + − 2 r0 r r r0 r r

The ﬁrst integral is the classical one,

r dr y = sin−1 ˆ q 2M L2 A 2E + − 2 r0 r r where M L y = − L r M 2 A2 = 2E + L2

1 For the second integral we make the same substitutions. Let r = u ,

r r ML2dr ML2u3du = − 3/2 3/2 ˆ 5 2M L2 ˆ 2 2 r 2E + − 2 (2E + 2Mu − L u ) r0 r r r0 Check:

r r Ldr Ldu q = − p ˆ 2 2M L2 2ML2 ˆ (2E + 2Mu − L2u2) + 2ML2u3 r 2E + − 2 + 3 r0 r r r r0 r Ldu = − q ˆ p 2 2 2ML2u3 (2E + 2Mu − L u ) 1 + 2 2 r0 2E+2Mu−L u r Ldu ML3u3du = − + p 3/2 ˆ (2E + 2Mu − L2u2) ˆ (2E + 2Mu − L2u2) r0

6 Now write M 2 M 2 2E + 2Mu − L2u2 = 2E + − Lu − L2 L = A2 − y2 M y = − Lu L 1 M u = − y L L so that r 3 dy M M − y dy ϕ = − L p 3/2 ˆ A2 − y2 L ˆ (A2 − y2) r0 Now let y = A sin ξ

r M 3 M 2 M 2 2 3 3 M L − 3 L A sin ξ + 3 L A sin ξ − A sin ξ A cos ξdξ ϕ = dξ − ˆ L ˆ A3 cos3 ξ r0 M 3 M 2 M 2 2 3 3 y MA L − 3 L A sin ξ + 3 L A sin ξ − A sin ξ dξ = sin−1 − A L ˆ A3 cos2 ξ M 3 M 2 M 2 2 y MA LA + 3 LA 1 − cos ξ dξ MA 3 LA + 1 − cos ξ sin ξdξ = sin−1 − + A L ˆ cos2 ξ L ˆ cos2 ξ ! y MA M 3 3M dξ 3M MA cos2 ξdξ = sin−1 − + + A L LA LA ˆ cos2 ξ LA L ˆ cos2 ξ ! MA M 2 sin ξdξ MA cos2 ξ sin ξdξ + 3 + 1 − L LA ˆ cos2 ξ L ˆ cos2 ξ ! y MA2 M 2 3M 2 = sin−1 − 3 + tan ξ + ξ A L LA L2 ! MA M 2 1 MA + 1 + 3 + cos ξ L LA cos ξ L Substituting back, y ϕ = sin−1 A " ! ! # 1 MA2 M 2 3M 2 MA M 2 MA + − 3 + sin ξ + ξ cos ξ + 1 + 3 + cos2 ξ cos ξ L LA L2 L LA L 3M 2 y = 1 + sin−1 L2 A " ! # 1 MA M 2 MA M 2 y y2 + 2 + 3 − 3 + − q y2 L LA L LA A A2 1 − A2 Check: r r Ldr ML3dr ϕ = − q 3/2 ˆ 2 2M L2 ˆ 5 2M L2 r 2E + − 2 r 2E + − 2 r0 r r r0 r r

7 y y 1 3M 2 3M M 3 3M = sin−1 0 − sin−1 + −M 1 + + M + sin ξ + M 1 + 1 − sin2 ξ p 2 2 3 3 A A 1 − sin2 ξ L A LA L A LA y y 1 3M 2 3M M 3 y 3M y2 = sin−1 0 − sin−1 + −M 1 + + M + + M 1 + 1 − A A q y2 L2A2 LA L3A3 A LA A2 1 − A2 y y 3M 2 M 3M 2 y LA 3M y2 = sin−1 0 − sin−1 + 1 − + 1 + − 1 + A A q y2 LA L2A2 A 3M LA A2 LA 1 − A2 Before putting in the initial conditions, we discard some small terms. The relevant constants are M GM GM = = L 2 r2ϕ˙ cr2ϕ˙ c c v v = escape escape vϕ c ≡ αv 1 r M 2 A = 2E + L2 s 2GM v v 2 ≈ − + escape escape r c vϕ s v v 2 ∼ escape escape − 1 c vϕ p = v α2 − 1 M L y = − L r v v v = escape escape − ϕ c vϕ c v v v = escape escape − ϕ c vϕ vescape 1 = v α − α

vescape M vescape 1 where we have set v = c 1. Let L = α c = αv, A = βv and y = α − α v. Then putting all constants into the ﬁnal term, y y 3M 2 M 3M 2 y LA 3M y2 ϕ = sin−1 0 − sin−1 + 1 − + 1 + − 1 + A A q y2 LA L2A2 A 3M LA A2 LA 1 − A2 " ! # y y 1 α2 α2 y2 = sin−1 0 − sin−1 + αβv2 2 + 3 − αv 3 + y − A A q y2 β β β2v2 1 − A2 " ! # y y 1 α2 α2 αy2 = sin−1 0 − sin−1 + 2αβv2 + 3αβv2 − αβv2αv 3 + y − A A q y2 β β β 1 − β2v2

so if we drop terms v2 or smaller, this reduces to " # 1 α 1 α 1 2 − y2 = − α − v2 q 2 r 2 y β α− 1 β α 1 − β2v2 ( α ) 1 − α2−1

8 3/2 = − α2 − 1 v2

2 vescape so this is also of order c2 and may be dropped. Therefore, 3M 2 y ϕ = 1 + sin−1 L2 A L2ϕ 1 1 L2 sin = √ M − √ L2 + 3M 2 2EL2 + M 2 2EL2 + M 2 r √ L2 2EL2 + M 2 p L2ϕ = √ M − 2EL2 + M 2 sin r 2EL2 + M 2 L2 + 3M 2 L2 1 r = M q 2EL2 L2ϕ 1 − 1 + M 2 sin L2+3M 2

Appendix: Keplerian orbits Reduction to quadratures For comparison, we ﬁrst compute the orbits in Newtonian gravity. We start from the conservation laws. Since the velocity is ~v =r ˙rˆ + rθ˙θˆ + r sin θϕ˙ϕˆ the angular momentum is −→ L = −→r × −→p = m−→r × r˙rˆ + rθ˙θˆ + r sin θϕ˙ϕˆ

= mrθ˙−→r × θˆ + mr sin θϕ˙−→r × ϕˆ −→ Since this is conserved in both magnitude and direction, the orbit remains in the plane perpendicular to L . π Without loss of generality, we may take the orbit to lie in the θ = 2 plane, so that −→ L = mr2ϕ˙kˆ ~v =r ˙rˆ + rϕ˙ϕˆ

The energy is also conserved, 1 GMm E = m r˙2 + r2ϕ˙ 2 − 2 r L 2 E Deﬁne the angular momentum per unit mass, L = m = r ϕ˙ and the energy per unit mass, E = m . Then L we have ϕ˙ = r2 so that 1 L2 GM E = r˙2 + − 2 r2 r r L2 2GM r˙ = − 2E − + r2 r The negative sign is because as time increases, r decreases. To ﬁnd r (t), we must integrate

t r dr dt = − ˆ ˆ q L2 2GM 2E − 2 + 0 r0 r r

9 This is not diﬃcult when there is no angular momentum, L = 0. The time for straight infall from rest at r0 is, setting L = 0,

r dr t = − q ˆ 2E + 2GM r0 r r √ 1 rdr = −√ q 2GM ˆ 1 + Er r0 GM

With these initial conditions, E = − GM , so this becomes r0 r √ 1 rdr t = −√ q 2GM ˆ 1 − r r0 r0

If we set r = sin2 ξ < 1 then r0

r p 2 1 r0 sin ξ2r0 sin ξ cos ξdξ t = −√ p 2GM ˆ 1 − sin2 ξ r0 √ r 2r r = −√ 0 0 sin2 ξdξ 2GM ˆ r0 3/2 r r0 1 = −√ ξ − sin 2ξ 2 2GM r0  s  r 3/2 r r 2 r0 −1 r r r = −√ sin − 1 −  2GM r0 r0 r0 r0  s  3/2 r 2 r r0 π r r −1 r = √  + 1 − − sin  2GM 2 r0 r0 r0

For the Newtonian case, we may take the ﬁnal position to be r = 0, and the time for the total fall is

π r3/2 t = √ 0 2 2GM in agreement with Kepler’s third law. The general case is also integrable, but the result (given below) is not very enlightening. The answer is dr dr much simpler if we change from dt to dϕ , and this gives us the shape of the orbit.

Orbit shape L To ﬁnd an equation for the orbit, r (ϕ), divide by ϕ˙ = r2 and integrate: ldr ϕ = q ˆ 2 L2 2GM r 2E − r2 + r

10 1 Now set u = r so that −Ldu ϕ = √ ˆ 2E − L2u2 + 2GMu −Ldu = q ˆ GM 2 G2M 2 − L − Lu + 2E + L2 Let GM y = − Lu L G2M 2 A2 = 2E + L2 so that dy ϕ = ˆ p−y2 + A2 Then with y = A sin θ we have A cos θdθ ϕ = ˆ A cos θ = θ y = arcsin A Solving for r we have, A sin ϕ = y GM = − Lu L GM L = − L r so 1 GM A = − sin ϕ r L2 L L2 r = GM q 2L2E 1 − 1 + G2M 2 sin ϕ To see that this describes an ellipse, deﬁne L2 a ≡ GM r 2L2E e ≡ 1 + G2M 2 a so that r = 1−e sin ϕ . Then, changing to Cartesian coordinates, r − er sin ϕ = a r − ey = a r2 = a2 + 2eay + e2y2 x2 + y2 = a2 + 2eay + e2y2 x2 + y2 − 2eay − ey2 = a2 ea 2 e2 x2 + (1 − e) y − = a2 1 + 1 − e 1 − e

11 ea 2 e2 2 b2 Finally, setting y0 = 1−e , b = a 1 + 1−e and c = 1−e we have the standard form for an ellipse centered at (x, y) = (0, y0): x2 (y − y )2 + 0 = 1 b2 c2 An examination of the magnitudes of the constants shows that this solution is valid for bound states, with E < 0. For positive energy, the ﬁnal integral gives a hyperbolic function and the equation describes a hyperbola.

Time dependence dϕ L To ﬁnd the time dependence, we have dt = r2 so that dϕ L = (1 − e sin ϕ)2 dt a2 ϕ Lt dϕ = a2 ˆ (1 − e sin ϕ)2 0

2 Wolfram integrator√ integrates√ this, but does not know that e < 1, so we must pull out factors of i everywhere, replacing e2 − 1 = i 1 − e2. We are left with ϕ ! ! dϕ 2i i e − tan ϕ ie e cos ϕ = tanh−1 √ 2 − tanh−1 √ + 1 − ˆ (1 − e sin ϕ)2 (1 − e2)3/2 1 − e2 1 − e2 1 − e2 1 − e sin ϕ 0 The presence of factors of i means we need to convert from tanh−1 to tan−1 to make the solution real. Now we need ix = − tanh iy eiy − e−iy −ix = eiy + e−iy = i tan y x = − tan y that is, i tanh−1 ix = − tan−1 x Therefore,

 (e−tan( ϕ ))  √ e √ 2 Lt 1 2 2 2 2 e cos ϕ  −1 1−e −1 1−e  = tan − tan 2 + 1 − a2 2 3/2  2 (e−tan( ϕ ))  1 − e2 1 − e sin ϕ (1 − e ) 1 − √ e 1 − 2 1−e2 1−e2 √ √ ! 1 2e 1 − e2 2 1 − e2 e − tan ϕ e cos ϕ = tan−1 − tan−1 2 + 1 − 2 3/2 1 − 2e2 2 ϕ 2 1 − e2 1 − e sin ϕ (1 − e ) 1 − e − e − tan 2 Even if we could invert to ﬁnd ϕ (t) in closed form, the expression would likely be unwieldy. A much simpler guide to the rate of progression is given by the conservation of angular momentum, L = r2ϕ˙ We can interpret this by noting that in time dt the orbiting particle passes through an inﬁnitesimal angle dϕ given by L dt = dϕ r2

12 It therefore moves through an arc length rdϕ, and since this is the inﬁnitesimal base of an isoceles triangle with sides r, r, rdϕ the inﬁnitesimal area swept out is 1 dA = r · rdϕ 2 Therefore, the area swept out per unit time is constant dA 1 L = r2dϕ = dt 2 2 This is Kepler’s second law.