As one wants to jump to Isotropic coordinates in order to write the Schwarzschild metric in terms of them, one does this coordinate transformation:


So we start with the very well-known form:

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

And arrive at $$ ds^2 = -\left(\frac{1-M/2r'}{1+M/2r'}\right)^2dt^2 +(1+M/2r')^4[dr'^2 +r'^2( d\theta^2 +\sin^2\theta d\phi^2)]$$

My question is: Where did this coordinate transformation come from?

  • 1
    $\begingroup$ What happens when $r=2m$ or $r=0$? $\endgroup$
    – Kyle Kanos
    Nov 7, 2014 at 14:35
  • $\begingroup$ Singularity. @KyleKanos $\endgroup$ Nov 7, 2014 at 14:36
  • 1
    $\begingroup$ What you you mean with your question? Coordinate transformations don't "come" from anywhere - they are simply (suitably nice) functions of coordinates. $\endgroup$
    – ACuriousMind
    Nov 7, 2014 at 14:41
  • $\begingroup$ @PhilosophicalPhysics: Does that happen with the transformation? $\endgroup$
    – Kyle Kanos
    Nov 7, 2014 at 14:42
  • 1
    $\begingroup$ No it doesn't. So there's your answer: the transformation comes from the desire to not have a singularity at $r'=2m$ (there would still be one at $r'=0$ though). $\endgroup$
    – Kyle Kanos
    Nov 7, 2014 at 15:10

1 Answer 1


The aim of the isotropic coordinates is to write the metric in the form where the spacelike slices are as close as possible to Euclidean. That is, we try to write the metric in the form:

$$ ds^2 = -A^2(r)dt^2 + B^2(r)d\Sigma^2 $$

where $d\Sigma^2$ is the Euclidean metric:

$$ d\Sigma^2 = dr^2 + r^2(d\theta^2 + \sin^2\theta d\phi^2) $$

So let's use the substitution $r\rightarrow r'$ and write down our metric:

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

If we compare this with the Schwarzschild metric:

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

Then for the angular parts to be equal we must have:

$$ B^2(r')r'^2 = r^2 $$

And for the radial parts to be equal we must have:

$$ B^2(r')dr'^2 = \frac{dr^2}{1 - 2M/r} $$

Divide the second equation by the first to eliminate $B$ and we end up with:

$$ \frac{dr'^2}{r'^2} = \frac{dr^2}{r^2 - 2Mr} $$

Some Intermediate Steps are Integrate this $$ \frac{dr'}{r'} = \frac{dr}{\sqrt{r^2 - 2Mr}} $$ Integration Will give $$ \log{r'}=\log{(\frac{r-M+\sqrt{r(r-2M)}}{M})}+\log{K} $$ Where K is The Integration Constant and set K$=\frac{M}{2}$ $$ r'=\frac{r-M+\sqrt{r(r-2M)}}{2}$$ And then just take the square root and integrate and we get the substitution you describe:

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

  • 5
    $\begingroup$ the third line in the derivation is not correct it should read r(r') only then the correct result also for the dt^2-term in isotropic coordiantes is obtained $\endgroup$ Sep 10, 2017 at 9:39
  • $\begingroup$ Upon performing this integration, I am getting $r=\frac{r'}{2K} {(1+\frac{MK}{r'})^{2}}$, where K is the integration constant. Any idea on how to show that $K=\frac{1}{2}$? $\endgroup$
    – V Govind
    Mar 6 at 6:40

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