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Kepler's second law, that planets in orbits sweep equal area in equal time, is a consequence of orbital angular momentum conservation. In the case of Schwarzschild spacetime, the angular orbital momentum is still conserved. Is Kepler's second also valid in Schwarzschild spacetime? Is there mathematical proof confirming the law or otherwise? Can any general remark be made regarding other astrophysical metrics.

Edit:

To make the question more precise and to clarify the points in the comments: if the following coordinate system is used, $$ds^2=-\left(1-\frac{2M}{r}\right)dt^2+\frac{1}{\left(1-\frac{2M}{r}\right)}dr^2+d\Omega^2$$

Then the time is the coordinate time and the area is the area on the $\theta=\frac{\pi}{2}$ plane and calculated with the induced metric on the plane.

Using Schwarzschild metric would give us correction to the Kepler's laws even for the Second law. I a wish to know how to go about calculating that correction. Hints will suffice.

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  • $\begingroup$ You have to ask "which time" and "which area" $r^{2}{\dot \phi}$ is a constant of the motion, but the dot is relative to arc length, which is proper time. If you convert to something like schwarzschild time, or try to compute out the geometric areas, you'll have to worry about factors of the metric tensor to do the conversions. $\endgroup$ Dec 25, 2020 at 4:17
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    $\begingroup$ Also, defining the spacelike surface interior to the orbit is complicated because the orbit is not closed, and the horizon lives inside that surface. $\endgroup$ Dec 25, 2020 at 4:31

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You can find it in Chandrasekhar's textbook ("The mathematical theory of black holes" Chap 3, sec. 19).

Only you need is to show the conservation laws $r^2 d\phi/d\tau$ of particle in Schwarzschild metric.

Edit 1: How to derive approximate Kepler's second law from GR

The static metric on $\theta=\pi/2$ $$ ds^2=\frac{1}{1-\frac{1}{r}}dr^2+r^2 d\phi $$ the mesure $\sqrt{\gamma}\sim r$ as $r\to\infty$, thus the area element is $$ d A\sim \frac{1}{2}r^2 d\phi $$ which implies $2d A/d\tau\sim r^2 d\phi/d\tau ={\rm const}$.

Edit 2: There is no exact Kepler's second law in GR.

Now, you should consider the full integral $$ \int_{1}^{r_0} \frac{r dr}{\sqrt{1-\frac{1}{r}}} $$ If you want see the correction, I suggest to take a limit again, such that the sun can be treat as point (don't understand me wrong, I have used the metric of sun, not the black hole, even there is only a little tiny difference.) The full expansion of integrand is $$ \sum _{k=0}^{\infty } \frac{(2 k-1)!!}{\left(2^k k!\right) r^{k-1}} $$ after integration, you will see that the leading term is Kepler's second law, the remaining term is "correction".

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  • $\begingroup$ Showing $r^2\frac{d\phi}{d\tau}$ is not sufficient. See @JerrySchirmer 's comment. $\endgroup$ Dec 28, 2020 at 6:21
  • $\begingroup$ @Faber Bosch, I think that is sufficient. $\endgroup$
    – user142288
    Dec 28, 2020 at 6:49
  • $\begingroup$ State the reason why you think @JerrySchirmer 's comment is irrelevant? How do you calculate infinitesimal area on the $r$-$\phi$ plane? and how do you adjust for the $\sqrt{-g_{00}}$ factor for considering coordinate time instead of proper time? $\endgroup$ Dec 28, 2020 at 6:55
  • $\begingroup$ @Faber Bosch see the renewed answer. $\endgroup$
    – user142288
    Dec 28, 2020 at 6:56
  • $\begingroup$ Okay, I see. I didn't refresh the page earlier. Please write down the precise formula for $dA$ before taking the limit. Because I want to know the order of the error. $\endgroup$ Dec 28, 2020 at 7:01

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