5.5 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 [95, 96]. We define the orbital radius r0 and the eccentricity in the same way as in the Schwarzschild case by
| ∂(R-∕r4)| ∂r |r=r0= 0, and R (r0(1 + e)) = 0. (193 )
We also assume e ≪ 1. In this case, Ω φ is given to 2 𝒪 (e ) by
[ ( ) ] 3 2 3- 9-2 9- 3 ( 2) 4 5 6 Ω φ = Ωc 1 − qv + e − 2 + 2v − 2 qv + 3 6 + q v − 60qv + 𝒪 (v ) . (194 )

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 + 4 π x3 − ------+ --q2 x4 − ---q − ----π 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 π x + − 189 + 16 q x ( ) ]} + + 3223-q − 773π x5 , (195 ) 168 3
UpdateJump To The Next Update Information
⟨ ⟩ ( ) { ( ) ( ) ( ) dJz- = dJz- 1 − 1247-x2 + − 11-q + 4 π x3 + − 44711-+ 33q2 x4 − 59q + 8191-π x5 dt dt N 336 4 9072 16 16 672 [23 3259 ( 371 209 ) +e2 ---− ----x2 + − ---q + ----π x3 8( 168 24 8 ) ]} 1041349 171 2 949 785 5 + − --------+ ---q + ----q − ---π x . (196 ) 18144 16 56 6
UpdateJump To The Next Update Information
  Go to previous page Go up Go to next page