Kepler’s third law in classical mechanics states that the square of the orbital period, \(T\), of a planet or particle is proportional to the cube of the semi-major axis of its orbit. The classical equation describing this is

\[\left(\frac{d\phi}{dt_{\text{Newt}}}\right)^2=\frac{G(M+m)}{a^3}\]

where \(a\) is the semi-major axis, \(t_{\text{Newt}}\) the Newtonian time, and \(m\) the mass of the test particle. We are interested in test particles, so \(M+m\approx M\).

An interesting question is how this generalizes to test particles moving around a very massive body in general relativity, such as a black hole. The metric in Schwarzschild coordinates that describes this geometry is

\[ds^2=-(1-2GM/r)dt^2+\frac{dr^2}{1-2GM/r}+r^2d\Omega^2\]

where \(M\) is the mass of the black hole. The question is whether Kepler’s third law applies to the Schwarzschild time, \(t\), or the proper time, \(\tau\), of the test particle? To answer this question let us assume that \(\theta=\pi/2\) and that the motion is circular, so \(\frac{dr}{d\tau}=\frac{d^2r}{d\tau^2}=0\).

The radial geodesic equation is given by (see, for example, Sean Carroll, or Misner, Thorne, and Wheeler)

\[\frac{d^2r}{d\tau^2}+\frac{GM}{r^3}(r-G2M)\left(\frac{dt}{d\tau}\right)^2-\frac{GM}{r(r-2GM)}\left(\frac{dr}{d\tau}\right)^2-(r-2GM)\left[\left(\frac{d\theta}{d\tau}\right)^2+\sin^2(\theta)\left(\frac{d\phi}{d\tau}\right)^2\right]=0\]

Using our assumptions this simplifies to

\[\frac{GM}{r^3}\left(\frac{dt}{d\tau}\right)^2-\left(\frac{d\phi}{d\tau}\right)^2=0\]

From this we see that

\[\left(\frac{d\phi}{dt}\right)^2=\frac{GM}{r^3}\]

and that it is the Schwarzschild time coordinate that is analogous to the Newtonian time in Kepler’s third law.