5.4 Slightly eccentric orbit around a Kerr black hole

5.4 Slightly eccentric orbit around a Kerr black hole

Next, we consider a particle in a slightly eccentric orbit on the equatorial plane around a Kerr black hole [53 ]. We define the orbital radius $r0$ and the eccentricity in the same way as in the Schwarzschild case by

@(R/r4) | --------|| = 0, and R(r0(1 + e)) = 0. (183) @r r=r0

We also assume $e « 1$ . In this case, $_O_f$ is given to $O(e2)$ by

[ ( ) ] 3 2 3 9 2 9 3 ( 2) 4 5 6 _O_f = _O_c 1- qv + e - 2 + 2v - 2-qv + 3 6 + q v - 60qv + O(v ) . (184)

We now give the energy and angular momentum luminosity that are accurate to $O(e2)$ and to $O(x5)$ beyond Newtonian order:

< dE > (dE ) { 1247 (11 ) (44711 33 ) (59 8191 ) --- = --- 1 - ----x2 - ---q + 4p x3 - ------+ --q2 x4 - ---q- ----p x5 dt dt N [336 4 ( 9072 ) 16 ( 16 672) 2 157- 6781- 2 2009- 2335- 3 14929- 281- 2 4 +e 24 - 168 x + - 72 q + 48 p x + - 189 + 16 q x ( ) ]} + - 2399-q- 773p x5 , (185) 56 3

< > ( ) { ( ) ( ) ( ) dJz- = dJz- 1 - 1247-x2 + - 11-q + 4p x3 + - 44711-+ 33q2 x4 - 59q + 8191-p x5 dt dt N 336 4 9072 16 16 672 [23 3259 ( 371 209 ) +e2 ---- ----x2 + - ---q + ----p x3 8( 168 24 8 ) ]} 1041349 171 2 243 785 5 + - --------+ ---q - ----q- ---p x . (186) 18144 16 8 6