 Solution

By spherical symmetry, rotate a geodesic into the equatorial plane $\theta =\pi /2$. The geodesic conserved quantities from Killing vectors arewhere the dot denotes a suitable affine parameter. Normalize the tangent by $g(\dot{x},\dot{x})=-ϵ$, with $ϵ=1$ for timelike proper time, $ϵ=-1$ for spacelike proper length, and $ϵ=0$ for null geodesics. ThenTherefore the geodesic potential of the magnetic dilaton black hole isTogether with the two conserved first integrals, this reduces the radial motion to ${\dot{r}}^2/2+\tilde{V}=0$. To cover turning points as well, the radial geodesic equation itself givesThus a constant-radius geodesic must satisfy both $\tilde{V}=0$ and ${\tilde{V}}^{\prime }=0$. The intermediate static-coordinate formulas apply where $V\ne 0$; the radial equation extends through a regular horizon in regular coordinates.