5.5 Circular orbit with a small inclination from the equatorial plane around a Kerr black hole

5.5 Circular orbit with a small inclination from the equatorial plane around a Kerr black hole

Next, we consider a particle in a circular orbit with small inclination from the equatorial plane around a Kerr black hole [52 ]. In this case, apart from the energy $E$ and $z$ -component of the angular momentum $lz$ , the particle motion has another constant of motion, the Carter constant $C$ . The orbital plane of the particle precesses around the symmetry axis of the black hole, and the degree of precession is determined by the value of the Carter constant. We introduce a dimensionless parameter $y$ defined by

C y = --2, Q2 = l2z + a2(1 - E2). (187) Q

Given the Carter constant and the orbital radius $r0$ , the energy and angular momentum is uniquely determined by $R(r) = 0$ , and $@R(r)/@r = 0$ . By solving the geodesic equation with the assumption $y « 1$ , we find that $1/2 y$ is equal to the inclination angle from the equatorial plane. The angular frequency $_O_f$ is determined to $O(y)$ and $4 O(v )$ as