0
$\begingroup$

How can you calculate how large the perihelion shift should be under Schwarzschild conditions in general relativity for minimally disturbed circular orbits in the strong field limit?

I think there is a functioning expression that works well in the weak fields of our solar system.

What would the perihelion shift be for a minimally disturbed circular orbit just above the minimum stable circular orbit at $r=6GM/c^2$ and, for instance, at $r=9GM/c^2$, $r=12GM/c^2$ and $r=15GM/c^2$ ?

$\endgroup$
1

1 Answer 1

1
$\begingroup$

The periapsis precession for a circular geodesic with radius $r$ in the Schwarzschild metric is

$$ 2 \pi \left( \sqrt{\frac{r}{r-6M}}-1\right)$$

The answer for a generic bound geodesic with (dimensionless) semilatus rectum $p$ and eccentricity $e$ is:

$$ 4\sqrt{\frac{p}{p-6+2e}}K\left(\sqrt{\frac{4e}{p-6+2e}}\right)-2\pi,$$ where $K$ is the complete elliptic integral of the first kind.

These formula's are, of course, assuming that the orbit is a geodesic and therefore that the orbiting body is a test particle whose own gravitational influence can be neglected.

Calculation of corrections due to the mass of the orbiting body can be found here and here.

$\endgroup$
3
  • $\begingroup$ Thank you. This is basically true for all r > 6M and not just a weak field approximation? $\endgroup$
    – Agerhell
    Nov 16, 2023 at 22:09
  • 1
    $\begingroup$ @Agerhell - Since a truly circular orbit doesn't have a periapsis it must be an approximation for small ellipticities. You can calculate it numerically as I did here, but I think there was an analytical solution as well, if I find or remember it I'll come back at it. I think it might work if you replace r with r_max or r_min or something like that, but don't quote me on that (yet). $\endgroup$
    – Yukterez
    Nov 16, 2023 at 22:46
  • $\begingroup$ @Yukterez This is the exact circular orbit limit of the periapsis shift for Schwarzschild geodesics $\endgroup$
    – TimRias
    Nov 17, 2023 at 7:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.