9
$\begingroup$

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

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

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?

$\endgroup$
  • 1
    $\begingroup$ What happens when $r=2m$ or $r=0$? $\endgroup$ – Kyle Kanos Nov 7 '14 at 14:35
  • $\begingroup$ Singularity. @KyleKanos $\endgroup$ – PhilosophicalPhysics Nov 7 '14 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 '14 at 14:41
  • $\begingroup$ @PhilosophicalPhysics: Does that happen with the transformation? $\endgroup$ – Kyle Kanos Nov 7 '14 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 '14 at 15:10
10
$\begingroup$

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:

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

And then just take the square root and integrate and we get the substitution you describe:

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

$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – Norbert Schwarzer Sep 10 '17 at 9:39

protected by Qmechanic Sep 10 '17 at 12:14

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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