General Relativity: An Introduction for Physicists · Problem 17.24

Linearised general relativity — Hobson, Efstathiou & Lasenby

Physics · frame-dragging-circular-orbit-period

Problem

Consider a particle moving under gravity at speed

v

in a circular orbit of radius

r

in the equatorial plane of the body in Exercise 17.23. Show that the vector acceleration of the particle is given by

\vec{a} = -\frac{GM}{r^2}\hat{r} \pm \frac{2GJv}{c^2 r^3}\hat{r},

where

\vec{r}

is the position vector of the orbiting particle and the plus and minus signs corresponding to prograde and retrograde orbits respectively. Hence show that the angular velocity

\omega

of the particle is given to first order in

J

\omega^2 = \frac{GM}{r^3} \mp \frac{2GJ}{c^2 r^4}\sqrt{\frac{GM}{r}},

where the minus and plus signs now correspond to prograde and retrograde orbits respectively. Thus show that the retrograde orbit has a shorter period than the prograde orbit.

Intuitive explanation

In this problem, you're seeing how a rotating massive body (characterized by its angular momentum parameter $J$ ) creates a subtle correction to orbital motion through frame-dragging—a relativistic effect. The key insight is that circular orbits require the net radial acceleration (gravitational pull plus the relativistic correction) to equal the centripetal acceleration $v^2/r$ . By setting up this force balance and using $v = \omega r$ for circular motion, you can solve for how $\omega$ depends on $J$ , revealing that retrograde and prograde orbits experience opposite corrections that flip the sign relationship between them.

Intuitive explanation

Related theory

Ergosphere And Frame Dragging

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