I want to compute the general relativity prediction for the difference in period between clockwise and countercloskwise orbits of a planet around a star which has small mass $M$ and small angular momentum $J$.

I have not found this calculation online, although I think it should be around.

The metric is a first order expansion of the Kerr metric, $$ ds^2=-\left(1-\frac{R}{r}\right)dt^2-\frac{2J\sin^2\theta}{r}dtd\phi+\left(1+\frac{R}{r}\right)dr^2+r^2d\Omega,$$ where $d\Omega=d\theta^2+\sin^2\theta d\phi^2$ is the solid angle and $R=2M$ is the Schwarzschild radius.

I think the period can be computed as $$\frac{dt}{d\phi}=\frac{dt/d\tau}{d\phi/d\tau}=\frac{p^0}{p^\phi},$$ for an equatorial circular orbit with $p^r=p^\theta=0$ and $\theta=\pi/2$.

The contravariant momenta are obtained from the covariant ones $p_0=E$ and $p_\phi=L$, which are constant, as $$ p^0=g^{00}p_0+g^{0\phi}p_\phi,\quad p^\phi=g^{\phi 0}p_0+g^{\phi\phi}p_\phi.$$

I get $$ p^0 =-\left(1+\frac{R}{r}\right)E+\frac{2J}{r^3}L$$ and $$ p^\phi =\frac{L}{r^2}+\frac{2J}{r^3}E.$$

[The quantity $g^{0\phi}$ was originally wrong, it was corrected following Wikipedia as per the answer below by Paul T.]

I read that the period difference should not depend on $r$. I can't find that result. I think one should use that $L=\sqrt{Rr/2}$, but I am not sure.

Would appreciate some help.


2 Answers 2


Remember that $g_{\mu\alpha}g^{\nu\alpha} = {\delta_\mu}^\nu$. For a diagonal metric, like the Schwarzschild case, you can easily invert the metric easily to arrive at the contravariant components. For example: $$ g^{tt} = \frac{1}{g_{tt}} = -\frac{1}{1-R/r}.$$

But with the off diagonal terms in the Kerr metric you need to be careful calculating the contravariant components. A good place to look them up is via the expression for the Kerr wave operator, where I can see that: $$ g^{\phi t} = -\frac{r_s r a}{\Sigma \Delta} \rightarrow -\frac{2J}{r^3},$$ taking the limit in your notation. It appears all three of the metric components you used may need to be fixed.

  • $\begingroup$ Actually, I think only $g^{\phi t}$ is in disagreement. But thank you for pointing this out. $\endgroup$
    – thedude
    Commented Jan 17, 2023 at 21:52

The period of a circular (equatorial) orbit in (full) Kerr is:

$$P = 2\pi\left(\frac{r^{3/2}}{\sqrt{M}}\pm \frac{J}{M}\right)$$

with the $\pm$ for the prograde and retrograde cases, respectively.

Their difference is therefore $4\pi J/M$ for any angular momentum $J$ and any radius $r$.

To get this result, you need to eliminate $E$ and $L$ from your equations. You do this by calculating $dr/d\tau$ from the norm of the 4-velocity and setting both $dr/d\tau$ and $d^2r/d\tau^2$ to zero. (The latter is necessary to ensure that your orbit is actually circular.) Working through this is elementary but tedious, which I will leave to you (or your favorite computer algebra package).


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