I am trying to derive the angular velocity of a circular orbit in Kerr geometry, eqn.(2.16) in [Bardeen et al (1972)](https://ui.adsabs.harvard.edu/abs/1972ApJ...178..347B/abstract) which reads $$\Omega=\dfrac{1}{r^{3/2}+a}$$ (Note that I am using the units in which $M=1$ and only direct orbits are considered i.e. only the upper sign is used)

**MY APPROACH TO THE SOLUTION:**

Now using the angular momentum per unit mass $(\mathscr{L})$ and energy per unit mass $(\mathscr{E})$, the conserved specific angular momentum can be defined as $$l=\dfrac{\mathscr{L}}{\mathscr{E}}$$

Using this definition, I had obtained that $$\Omega=\dfrac{d\phi}{dt}=\dfrac{\dot{\phi}}{\dot{t}}=-\dfrac{g\_{t\phi}+lg\_{tt}}{g\_{\phi\phi}+lg\_{t\phi}}$$

which for $l=0$ is simply $\Omega=-\dfrac{g\_{t\phi}}{g\_{\phi\phi}}=\omega$, i.e. the angular velocity that corresponds to frame dragging.

But, I am getting trouble how to find the result as given in the above paper. Any suggestions regarding the approach to derive the equation would be helpful.