Kerr metric and dragging.

The Kerr-metric is given by

$\begin{array}{rcl} ds^2 &=& \displaystyle \left( 1 - \frac{ r_s r }{ r^2 + \jmath^2 \cos^2(\theta) } \right) c^2 dt^2 - \frac{ r^2 + \jmath^2 \cos^2(\theta) }{ r^2 - r_s r + \jmath^2 } dr^2 - \left( r^2 + \jmath^2 \cos^2(\theta) \right) d\theta^2\\ && \displaystyle - \left( r^2 + \jmath^2 + \frac{ r_s r \jmath^2 \sin^2(\theta) }{ r^2 + \jmath^2 \cos^2(\theta) } \right) \sin^2(\theta) d\phi^2 + \frac{ 2 r_s r \jmath \sin^2(\theta) }{ r^2 + \jmath^2 \cos^2(\theta) } c dt d\phi \end{array}$,

where $\displaystyle r_s = \frac{2GM}{c^2}$ and $\displaystyle \jmath=\frac{J}{Mc}$. We solve the geodesic equation using

$\displaystyle \frac{d}{ds} \left( g_{\alpha\mu} \frac{dx^\mu}{ds} \right) = \frac{1}{2} \left( \frac{\partial g_{\kappa\lambda}}{\partial x^\mu} \right) \frac{dx^\kappa}{ds} \frac{dx^\lambda}{ds}$

It is clear that $g_{\kappa\lambda} = g_{\kappa\lambda}(\sin^2(\theta))$, then for $\theta$ we obtain the equation

$\displaystyle \frac{d}{ds} \left( g_{\theta\theta} \frac{d\theta}{ds} \right) = \sin(\theta) \cos(\theta) \left( \frac{ \partial g_{\kappa\lambda} }{ \partial \sin^2(\theta) } \right) \frac{dx^\kappa}{ds} \frac{dx^\lambda}{ds}$

So there is a solution $\theta = \pi/2$. Next we look to the equation for $r$ and we have

$\displaystyle \frac{d}{ds} \left( g_{rr} \frac{dr}{ds} \right) = \frac{1}{2} \left( \partial_r g_{tt} \left( \frac{dt}{ds} \right)^2 + \partial_r g_{rr} \left( \frac{dr}{ds} \right)^2 + \partial_r g_{\phi\phi} \left( \frac{d\phi}{ds} \right)^2 + 2 \partial_r g_{t\phi} \frac{dt}{ds} \frac{d\phi}{ds} \right)$

An orbit with a constant $r$ gives the equation

$\displaystyle \left( \partial_r g_{tt} + \partial_r g_{\phi\phi} \dot\phi^2 + 2 \partial_r g_{t\phi} \dot\phi \right) \left( \frac{dt}{ds} \right)^2 = 0$

Thus

$\displaystyle \partial_r g_{\phi\phi} \dot\phi^2 + 2 \partial_r g_{t\phi} \dot\phi = - \partial_r g_{tt}$

So

$\displaystyle \dot\phi_\pm = - \frac{ \partial_r g_{t\phi} }{ \partial_r g_{\phi\phi} } \pm \sqrt{ \left( \frac{ \partial_r g_{t\phi} }{ \partial_r g_{\phi\phi} } \right)^2 - \frac{ \partial_r g_{tt} }{ \partial_r g_{\phi\phi} } }$

Now we can write

$\left[\begin{array}{rcl} g_{tt} &=& \left( 1 - \psi(r) \right) c^2\\ g_{\phi\phi} &=& -\left( r^2 + \jmath^2 + \jmath^2 \psi(r) \right)\\ g_{t\phi} &=& \jmath \psi(r) c \end{array}\right.$

where $\displaystyle \psi(r) = \frac{r_s r}{r^2+\jmath^2}$. Thus

$\left[\begin{array}{rcl} \partial_r g_{tt} &=& - \psi'(r) c^2\\ \partial_r g_{\phi\phi} &=& -2r - \jmath^2 \psi'(r)\\ \partial_r g_{t\phi} &=& \jmath \psi'(r) c \end{array}\right.$

Then we can write

$\displaystyle \dot\phi_\pm = \frac{\jmath \psi'(r) c}{2r + \jmath^2 \psi'(r)} \pm \sqrt{ \left( -\frac{\jmath \psi'(r) c}{2r + \jmath^2 \psi'(r)} \right)^2 - \frac{ \psi'(r) c^2 }{2r + \jmath^2 \psi'(r)} }$

which can be written as

$\displaystyle \dot\phi_\pm = \frac{ (\jmath/c) \Psi(r) \pm \sqrt{-\Psi(r)} }{1 + (\jmath/c)^2 \Psi(r)}$,

where $\displaystyle \Psi(r) = \frac{\psi'(r) c^2}{2r}$ and $\displaystyle \frac{\psi'(r) c^2}{2r} = -\frac{GM}{r^3} \frac{1 - (\jmath/r)^2}{\left( 1 + (\jmath/r)^2\right)^2}$. The case $\jmath=0$ gives

$\displaystyle \dot\phi_\pm = \pm \sqrt{\frac{GM}{r^3}}$,

which reduces to the Third law of Kepler, known as $\displaystyle \dot\phi^2 = \frac{GM}{r^3}$.

The first order of $\jmath$ gives

$\displaystyle \dot\phi_\pm \simeq -\frac{\jmath}{c} \frac{GM}{r^3} \pm \sqrt{\frac{GM}{r^3}}$

This can we written as

$\displaystyle T_\pm \simeq \pm \sqrt{ \frac{4 \pi^2 r^3}{GM} } - \frac{\jmath}{c}$,

this effect is very small and therefore difficult to test by experiment. For the Sun we have $\displaystyle \frac{\jmath}{c} \simeq 6.3 \times 10^{-6} \texttt{s}$, for the Earth we have $\displaystyle \frac{\jmath}{c} \simeq 1.3 \times 10^{-6} \texttt{s}$