General Relativity Fall 2019 Lecture 20: Geodesics of Schwarzschild Yacine Ali-Ha¨ımoud November 7, 2019 In this lecture we study geodesics in the vacuum Schwarzschild metric, at r > 2M. Last lecture we derived the following equations for timelike geodesics in the equatorial plane (θ = π=2): d' L 1 dr 2 M L2 ML2 = ; + Veff (r) = ;Veff (r) + ; (1) dτ r2 2 dτ E ≡ − r 2r2 − r3 where (E2 1)=2 can be interpreted as a kinetic + potential energy per unit mass. The radial equation can also be rewrittenE ≡ as− d2r M 3 h i = V 0 (r) = r~2 L~2r~ + 3L~2 ; r~ r=M; L~ = L=M: (2) dτ 2 − eff − r4 − ≡ CIRCULAR ORBITS AND THE ISCO We show the effective potential in Fig. 1. In contrast to the Newtonian effective potential for orbits around a central 2 2 2 3 2 mass (i.e. Veff M=r + L =2r , without the last term ML =r ), which always has a minimum at rNewt = L =M, ≡ − − the relativistic effective potential has both a maximum and a minimun for L > p12 M, an inflection point for L = p12 M, and is strictly monotonic for L < p12 M. 0 Circular orbits (with r = constant) are such that Veff (r) = 0. Solving this equation, one finds that such orbits exist only for L > p12 M. When this condition is satisfied, the radii of circular orbits are L2 p r± = 1 1 12M 2=L2 : (3) c 2M ± − The Newtonian limit is obtained for L M, in which case r+ L2=M. c ! 2 2 From classical mechanics, we know that circular orbits where d Veff =dr > 0 are stable, whereas those with 2 2 d Veff =dr < 0 are unstable. This can simply be understood graphically: for a given energy, an orbit is confined to the interval in radii for which Veff (r) . At a circular orbit, this is an exact equality. Perturb the orbit away by ≤ E 2 2 2 2 giving it a bit of extra energy, and it will start oscillating between two radii if d Veff =dr > 0. If d Veff =dr 0, it will either plunge towards r 2M or diverge to infinity. ≤ ! The smallest radius of a stable circular orbit is obtained for L = p12 M, and has value rc = 6M { strictly speaking, only orbits with rc > 6M are actually stable. This orbit is called the innermost stable circular orbit (ISCO). The energy associated with a circular orbit at the ISCO is given by 1 p p isco = Veff (6M) = ; Eisco = 1 + 2 isco = 8=9 0:94: (4) E −18 ) E ≈ This number is often invoked to quantify the efficiency of accretion around Schwarzschild black holes: suppose material is orbiting around a central mass (typically a black hole) on a circular orbit, and slowly looses energy and angular momentum due to dissipative forces, while remaining on a quasi-circular orbit. If it starts at large distances with E 1, it will lose about 6% of its energy to dissipation before reaching the ISCO and plunging towards r 2M. Thus if≈ the rate of mass-energy accretion is M_ , the luminosity radiated by dissipation is at most 0:06 M_ .! This is usually quoted as a maximal efficiency of 6% for accretion onto Schwarzschild black holes. Spinning black holes (which we will explore in the last classes) can∼ be much more efficient. 2 ��� L/M = Newtonian effective potential 5 ��� circular orbits ) � 4.2 ( ��� ��� � 2p3 -��� 3 -��� � �� �� ��� �/� p FIG. 1. Effective potential for orbits in the Schwarzschild metric. The numbers label the value of L=M. For L=M < p12, the potential has no extremum, and there are no circular orbits { in fact, all orbits are plunging to r ! 2M. For L=M > 12, the effective potential has a local maximum, correspondingp to an unstable circular orbit, and a local minimum, corresponding to a stable circular orbit. The two merge together at L = 12M, for which the marginally stable circular orbit is at r = 6M. This is the innermost stable circular obit (ISCO). PRECESSION OF PERICENTER In this section we consider bound orbits, i.e. for which < 0, and the coordinate r is bound within a range E rmin r rmax, called the pericenter and apocenter, respectively. We adjust the polar angle ' such that, at some ≤ ≤ arbitrary pericenter passage, r(' = 0) = rmin, and dr/dτ(' = 0) = 0. Note that the next pericenter passage need not be at ' = 2π if the orbit does not close, as we will see now! From the angular equation, we have d/dτ = (L=r2)d=d', so the radial equation can be rewritten as 1 L2 dr 2 + Veff (r) = : (5) 2 r4 d' E Let us define the variable u L2=(Mr). It satisfies the following differential equation: ≡ 1 du 2 1 1 2L2 u + u2 (M=L)2u3 = (1 e2); 1 e2 (6) 2 d' − 2 − −2 − − ≡ − M 2 E Differentiating again, we get d2u + u 1 = 3(M=L)2u2: (7) d'2 − Recall that we have adjusted ' such that dr/dτ = 0 at ' = 0, hence du=d' = 0 at ' = 0. We will solve this equation perturbatively in (M=L)2 1, corresponding to the Newtonian limit: 2 u(') = u0(') + (M=L) u1(') + ::: (8) Note that since (L=M)2 (consider, for instance, circular orbits), the quantity 1 e2 is of order unity. The lowest-order termE satisfies ∼ − 2 d u0 + u0 = 1: (9) d'2 3 This has a simple solution: u0(') = 1 + A cos ', where we have adjusted the origin of ' such that du0=d' = 0 at ' = 0. To find the integration constant A, we plug into 2 1 du0 1 2 1 2 u0 + u = (1 e ); (10) 2 d' − 2 0 −2 − implying A2 = e2. For definiteness, we pick A = +e, corresponding to L > 0. The zero-the order solution is therefore the well-known ellipse, L2=M r0(') = ; (11) 1 + e cos ' where e is the eccentricity. This orbit happens to be periodic, which is a unique feature of the 1=r potential. Let us now find the next-order correction: 2 d u1 2 du1 + u1(') = 3u ('); (' = 0) = 0: (12) d'2 0 d' The solution is 2 2 2 2 u1(') = (3 + e ) + (C 3 e ) cos ' + e sin ' + 3e ' sin '; (13) − − where C u1(0) is an integration constant, which we won't need to calculate explicitly for the problem of pericenter precession,≡ but is easy to find from Eq. (6): C = (1 + e)3=e. Obviously this solution cannot be correct for arbitrary ', because of the growing term ' sin '. This is because deviations away from a Keplerian orbit accumulate over time. So we should only consider this perturbative solution to be valid over one orbital period 0 ' 2π { after that, one can reset the origin of ' and reiterate the perturbative expansion. ≤ ≤ We want to know at what angle the next pericenter passage will be. In the Kepler problem, it is at ' = 2π, but, because of the additional term in the effective potential, the next pericenter passage will be at ' = 2π + ∆', where we expect ∆' (M=L)2. The pericenter passage is defined by dr=d' = 0, hence u0(') = 0. Thus we must solve for ∼ 0 2 0 u0(2π + ∆') + (M=L) u1(2π + ∆') = 0: (14) We Taylor-expand this around ∆' = 0, and find, to lowest order, 0 2 u1(2π) 2 ∆' = (M=L) 00 = 6π(M=L) : (15) − u0 (2π) Thus we conclude that the pericenter (as well as apocenter) precesses by ∆' = 6π(M=L)2 per orbit. This is usually expressed in terms of the semi-major axis a and eccentricity. The semi-major axis is a = (rmax + 2 2 2 2 rmin)=2 = L =M=(1 e ), i.e. L = Ma(1 e ), thus − − 6πM ∆' = : (16) a(1 e2) − We now apply this to the orbit of Mercury around the Sun, which was one of the earliest successfull tests of 12 GR. Here we have M = M , aM 6 10 cm, e 0:2. This leads to a precession rate of ∆' 0:1 arcsec per orbital period, hence Ω = d∆'=dt ≈ 43× arcsec/century.≈ ≈ It is interesting to compare this to≈ the effect of Jupiter on the orbit. The tidal acceleration due to Jupiter is of order 3 MJaM=aJ. The amplitude of this perturbed acceleration relative to the relativistic correction is therefore of order 3 3 δfJupiter MJaM=aJ aM MJ aM −6 aM 2 3 10 10; (17) δfGR ∼ M =aM ∼ M M aJ ∼ M ∼ −3 13 where we used MJ 10 M and aJ 7 10 cm 10 aM. Therefore we see that the relativistic precession is a factor of 10 times∼ weaker than the∼ precession× due∼ to Jupiter's tidal force. To be aware of this tiny \anomalous precession"∼ in the early 20-th century, physicisists therefore had to model the effect of Jupiter (and Solar-sytem planets) to exquisite precision! 4 RADIAL ORBITS Consider a massive particle that is radially infalling, so that L = 0, and dr/dτ < 0. For L = 0 the radial geodesic equation reduces to dr p (dr/dτ)2 2M=r = E2 1 = E2 + 2M=r 1: (18) − − ) dτ − − The proper time elapsed between two radii ri > rf is Z ri dr Z ri=2M dx Z ri=2M px dx ∆τ = = 2M = 2M ; (19) p 2 p 2 2 r E + 2M=r 1 r =2M E + 1=x 1 r =2M p1 x + xE f − f − f − 2 2 For E > 0, this integral is clearly convergent, for any rf 2M and even for E , it converges and remains finite ≥ ! as rf 2M.