Kepler's Third Law in Schwarzschild

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.

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