General Relativity: An Introduction for Physicists · Problem 13.18

The Kerr geometry — Hobson, Efstathiou & Lasenby

Physics · kerr-photon-effective-potential

Problem

Show that the effective potential for photon orbits in the equatorial plane of the Kerr geometry is given by Veff(r;b)=1r2[1−(ab)2−2μr(1−ab)2].V_{\rm eff}(r; b) = \frac{1}{r^2}\left[1 - \left(\frac{a}{b}\right)^2 - \frac{2\mu}{r}\left(1 - \frac{a}{b}\right)^2\right].
Intuitive explanation

Photon orbits in Kerr spacetime require you to use the null geodesic equations and the conserved quantities (energy and angular momentum) that arise from the spacetime's symmetries. The key is to combine the normalization condition gμνx˙μx˙ν=0g_{\mu\nu}\dot{x}^\mu\dot{x}^\nu = 0 for light with the definitions of impact parameter bb (angular momentum per unit energy) and the metric components to eliminate time and angular derivatives, leaving an effective radial potential. This reduces the problem to a one-dimensional mechanics picture where you can identify the turning points and orbital structure of light rays.

Intuitive explanation

Photon orbits in Kerr spacetime require you to use the null geodesic equations and the conserved quantities (energy and angular momentum) that arise from the spacetime's symmetries. The key is to combine the normalization condition gμνx˙μx˙ν=0g_{\mu\nu}\dot{x}^\mu\dot{x}^\nu = 0 for light with the definitions of impact parameter bb (angular momentum per unit energy) and the metric components to eliminate time and angular derivatives, leaving an effective radial potential. This reduces the problem to a one-dimensional mechanics picture where you can identify the turning points and orbital structure of light rays.

Related theory

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