I am trying to deep dive and study the isotropic Schwarzschild coordinates, whose line element is written for particles lying onto the equatorial plane $\theta=\pi/2$ as:

$$ds^2 = -\left(\dfrac{1-\dfrac{R_s}{4r'}}{1+\dfrac{R_s}{4r'}}\right)^2c^2 dt^2 +\left(1+\dfrac{R_s}{4r'}\right)^4 \left[dr'^2 + r'^2 d\phi^2\right].$$

So far what I have learned is that this new line element keeps invariant the angular part (or well, the flat, Euclidean line element) for the metric outside a spherically symmetric distribution. Also, it rises from the transformation:

$$r = r'\left(1+\dfrac{R_s}{4r'}\right)^2,$$

where $r$ is the former, original radial coordinate in the common Schwarzschild coordinates. Likewise in common Schwarzschild, this new $r'$ coordinate does not have the same physical meaning as a longitudinal distance of say, a segment from 0 to $r'$ as in Euclidean, flat space. And also, there is a useful inverse transformation between $r'$ and $r$ which is:

$$r' = \dfrac{1}{2}\left(r-\dfrac{R_s}{2}\pm\sqrt{r\left(r-R_s\right)}\right).$$

This last equation is valid always that $r\geq R_s$ but the physical meaning of the two possible positive roots from it is what makes for me this new treatment tricky.

The case for the Schwarzschild radius is simple and plain: $r=R_s$ yields $r'=R_s/4$. But for example, if someone takes the radius of the photon orbit $R_c=1.5\,R_s$, one gets two roots (AND BOTH OF THEM POSITIVE): $r'_1\simeq 0.933\,R_s$ and $r'_2\simeq 0.067\,R_s$. Actually, you can see after plotting that for $r>R_s$, all the two roots will be always positive. Which one is the correct one then?

I include a plot of the two possible roots below, the green line indicates the plus sign solution and the blue one the minus sign solution. The horizontal axis is simply $r/R_s$ and the vertical axis is $r'/R_s$.

enter image description here


2 Answers 2


The isotropic coordinates satisfy a simple isometry. If your map $r'\mapsto \frac{R_s^2}{16r'}$ you obtain the exact same expression for the metric. Consequently, both the $r'>R_s/4$ and $R_s/4 > r' >0$ patches describe the exterior ($r>R_s$) patch of the Schwarzschild solution. Both patches provide equally valid coordinates for this patch, although the $r'>R_s/4$ has the distinct advantage of manifestly approaching Minkowski space in the the $r'\to\infty$ limit.

Geometrically, you can interpret the two isotropic coordinate patches as gluing together regions I and III at the bifurcation surface in the Kruskal diagram. However, keep in mind that $r'=R_s/4$ is still a coordinate singularity in isotropic coordinates (at the bifurcation surface).

  • $\begingroup$ I liked your approach when talking about the $r'\to\infty$ limit retrieving back the Minkowski spacetime. Thank you very much. $\endgroup$
    – omivela17
    Commented Feb 22 at 1:24

Working with your equation:

$$r' = \dfrac{1}{2}\left(r-\dfrac{R_s}{2}\pm\sqrt{r\left(r-R_s\right)}\right).$$

And substituting the photon orbit radius $(3/2) R_s$ for r we can write:

$$r' = \dfrac{1}{2}\left(\frac{ 3 R_s}{2}-\frac{R_s}{2}\pm\sqrt{\frac{3 R_s} 2 \left(\frac{3 R_s} 2 -R_s\right)}\right).$$

$$r' = \dfrac{R_s}{4}\left(2 \pm \sqrt{3}\right).$$

Since $\sqrt{3} >1$ , the root with the negative square root is $< R_s/4$ which have stated is less than the Schwarzschild radius in the new coordinates and since the isotropic coordinates are only valid for $r > R_s$ the second root is not in the valid range. In other words, while both roots are positive, the roots that are less than $R_s/4$ are not valid.

  • $\begingroup$ Why the down vote? Is my calculation wrong? $\endgroup$
    – KDP
    Commented Feb 22 at 6:35
  • $\begingroup$ (Not my downvote.) The question comes down to why should one choose the roots with $r'>R_s/4$, rather than the other branch. Your answer comes down to "Because you need to choose the roots with $r'>R_s/4$ ". $\endgroup$
    – TimRias
    Commented Feb 22 at 8:56

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