# Orbital period around slowly rotating star in general relativity

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.

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.

• Actually, I think only $g^{\phi t}$ is in disagreement. But thank you for pointing this out. 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).