JPL and others calculating ephemerides in the solar system are using a method based on taking the Schwarzschild solution, not as expressed in the most common Schwarzschild coordinates but in the isotropic coordinates and then taking a low-order expansion of that solution to calculate a few approximate "disturbance terms" to add to the classical Newtonian gravitational acceleration term to account for relativistic effects on the orbits in the weak field of our solar system. Their method is outlined in the official documentation, "Formulation forObserved and Computed Values ofDeep Space Network Data Typesfor Navigation".

On page 2-9 there is en expression 2-16 that except from a scale factor looks like:


which I find to be is what they use to calculate the orbits under Schwarzschild conditions.

There are several places on the Internet where orbits around a Schwarzschild black hole are shown graphically using Schwarzschild coordinates but I have not really found any using isotropic coordinates.

  1. What does the strong field orbits around a Scwarzschild black hole look like if the isotropic coordinates are used to calculate the orbits?

2.What does the strong field orbits around a Schwarzschild black hole look like if the metric/coordinates that is a low order expansion of the isotropic metric, shown above, is used?

(I understand that to get to the Schwarzschild solution in isotropic coordinates from the solution in Schwarzschild coordinates you make the transformation $r=r'(1+\frac{GM}{2r'c^2})^2$ so maybe it is obvious that, for the isotropic coordinates, that you can find the orbits in Schwarzschild coordinates first and then make the substitution?)

It would be nice to see what the orbits would look like because Nasa is using a low expansion of the isotropic metric to get the orbits and interprets the "r-coordinate" as a real euclidian radial distance.

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    $\begingroup$ You can certainly just apply the coordinate transformation to the orbits expressed in Schwarzschild coordinates. $\endgroup$ – Javier May 4 '19 at 14:50
  • $\begingroup$ Related post by OP: physics.stackexchange.com/q/473577/2451 $\endgroup$ – Qmechanic May 13 '19 at 4:40

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