3
$\begingroup$

The Schwarzschild metric is

$$ ds^2 = -\left(1-\frac{2M}{r}\right)dt^2 + \frac{dr^2}{1 - 2M/r} + r^2(d\theta^2 + sin^2\theta d\phi^2) $$

and to make it isotropic we'd like to get it into the form:

$$ ds^2 = -A^2(r') dt^2 + B^2(r')\left(dr'^2 + r'^2(d\theta^2 + sin^2\theta d\phi^2)\right) $$

This can be done with the coordinate transformation:

$$ r = r'\left(1 + \frac{M}{2r'}\right)^2 $$

Is there a good way to physically interpret this new $r'$ coordinate?

It is no longer related to the circumference in a simple way like $r$ was. Nor does it appear to be simply related to the distance from the spherical mass this space-time is outside of.

$\endgroup$
1
$\begingroup$

I can tell you that much that as when JPL and others are calculating the orbits of celestial objects in the solar system they are using the first order of the "post-Newtonian expansion". They take a first order expansion of the Schwarzschild metric in isotropic coordinates from which they get relativistic "correction terms" that they add to the classical Newtonian gravitational acceleration. This is explained in their official documentation, "Formulation for Observed and Computed Values of Deep Space Network Data Types for Navigation"

So basically NASA/JPL interprets the new $r'$ coordinate as an ordinary euclidian distance in order to find relativisic correction terms to add to the classical Newtonian gravitation to account for relativistic effects in the weak fields of our solar system.

This interpretation might introduce errors, but that is what they do.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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