= Solution
For a regular <causal curve>, orient its tangent $Z$ toward the future; reversing its parametrization does not change $d\phi/dt$. Since $dt(Z)>0$, use $t$ as parameter and set $\Omega=d\phi/dt$. The causal inequality gives
$$
g_{tt}+2g_{t\phi}\Omega+g_{\phi\phi}\Omega^2+\frac\Sigma\Delta\left(\frac{dr}{dt}\right)^2+\Sigma\left(\frac{d\theta}{dt}\right)^2\le0.
$$
Both transverse terms are nonnegative in the exterior, so necessarily
$$
g_{\phi\phi}\Omega^2+2g_{t\phi}\Omega+g_{tt}\le0.
$$
Because $g_{\phi\phi}>0$ and the discriminant is $4D>0$, this is equivalent to
$$
\boxed{\frac{-g_{t\phi}-\sqrt D}{g_{\phi\phi}}\le\frac{d\phi}{dt}\le\frac{-g_{t\phi}+\sqrt D}{g_{\phi\phi}}.}
$$
Equality requires $dr/dt=d\theta/dt=0$ and a <null tangent>. The limiting vectors are the two local azimuthal <null directions>; a curve following one at constant radius need not be a <null geodesic>. For a curve with radial or polar motion these angular bounds are necessary, while the full causal inequality can impose stricter bounds.
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