# Isotropic Schwarzschild coordinates

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.

• – Martino Mar 16 '17 at 11:24

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.