Question

Show, by perturbing the geodesic in the equatorial plane, that the circular orbit of a photon at r= = 3m is unstable.

Answer 1

Schwarzschild Solution: Geodesics ř - Zvi? – re”ő? – re“ sin? 062 +22" )t? = 0 (1) = r = 0 = 2 2 Ö+r0 sin 0 cos 002 0 (2) cos e 21+2 00 (3) sin e Ï +v'ri 0 (4) and from gruxitxik = 1 (or = c^) (here we assume a massive particle) we get : -e"? + e^;? + r20 +r2 sin? 062 = 1. (5) If a geodesic is passing through a point P in the equatorial plane (0 = 7/2) and has a tangent at P situated also in this plane (0 = 0 at P) then from (2) we get ö = 0 at P and all higher derivatives are also vanishing at P. That is, the geodesic lies entirely in the plane defined by P, the tangent at P and the center of symmetry of the space. Since the symmetry planes are equivalent to each other, it will be sufficient to discuss the geodesics lying on one of these planes e.g. the equatorial plane 0 = 7/2.

The geodesics on the equatorial plane are: 1 1 ï v'j2 12 – reu + e2v j't? = : 0 (6) 2 2 2 + = 0 r = 0 Ï +v'rt -e į² tel; 2 + r2d2 (7) (8) (9) 1 Then from equations (7) and (8) we can easily prove (how?) that: d =0 1 120 = L = const : (Angular Momentum) (10) dr d (126) (e"i) = 0 1 e” i = Ec? = const : (Energy) (11) dr For a massive particle with unit rest mass, by assuming that t is the affine parameter for its motion we get pl = xH or Pue = fux". Thus po = gooi = e' i = Ec2 and Pø = gobo = –120=-L (12)

An observer with 4-velocity UP will find that the energy of a particle with 4-momentum pH is: E = Pu UM An observer at rest at infinity, will have UM = (1,0,0,0) leading to E = po = Ec2. Actually, for a particle with rest mass mo we get E = E/(moc?) i.e. it is the energy per unit rest-mass. For a particle at infinity we assume E = moc?. Finally, by substituting eqns (11) and (10) into (9) we get the "energy equation" for the r-coordinate: er ;2 + -L2 + e" = E2 (13) r2 which suggests that at r * oo (and r = 0) we get that E = 1.

By combining the 2 integrals of motion and eqn (9) we can eliminate the proper time to derive an equation for a 3D path of the particle (how?) (cos") * + c E2 – 1 L2 + 2 Mu + 2Mp3 L2 (14) where we have used u = 1/r and dr do L dr dr di = = do di r2 do The previous equation can be written in a form similar to Kepler's equation of Newtonian mechanics (how?) i.e. d2 u + = d02 M L2 +3 Mu? (15) The term 3 Mu2 is the relativistic correction to the Newtonian equation. This term for the trajectories of planets in the solar system, where M= Mo = 1.47664 km, and for orbital radius r ~ 4.6 x 107km (Mercury) and r~1.5 x 108 km (Earth) gets very small values ~ 3 x 10-8 and ~ 10-8 correspondingly.

Trajectories of photons Photons, as any zero rest mass particle, move on null geodesics. For photons we cannot use the proper time t as the parameter to characterize the motion and thus we will use some affine parameter o. We will study photon orbits on the equatorial plane and the equation of motion will be in this case = E e į e” i? – e-v;2 – 1262 r20 (37) (38) 0 L = (39) The equivalent to eqn (13) for photons is 2+12= (40) while the equivalent of eqn (15) is d2 d824 + u = 3Mv2 (41)

Radial motion of photons For radial motion 0 = 0 and we get e” į? - e-V;2= 0 Particle Observer ci from which we obtain dr dt =+1 +(1-2) (42) Photon integration leads to: Outgoing Photon t=r+ 2M In Jam - 11 1 + const Incoming Photon r=2 PER t=-r – 2M In - 11 + + const 2M Figure: Radially infalling particle emitting a radially outgoing photon.

Circular motion of photons For circular orbits we have r=constant and thus from eqn (41) we see that the only possible radius for a circular photon orbit is: r= 3M or ZGM 3 GM c2 (43) There are no such orbits around typical stars because their radius is much larger than 3M (in geometrical units). Only outside black holes photons can follow such orbits.

Stability of photon orbits We can rewrite the "energy" equation (40) as eu 1 12 L2 + (44) r2 D2 where D=L/E and Veff = eu/r Veff-1/27M2 9.01 Vell 0.01 10 /M -0.01 Figure: The effective potential for photon orbits. We can see that Veff has a single maximum at r = 3M where the value of the potential is 1/(27M2). Thus the circular orbit r = 3M is unstable.