I am trying to introduce a tortoise coordinate for a modified Schwarzschild metric

$$\mathrm{d}s^2=\left(1-\frac{2M\mathop{}\!\mathrm{erf}(r)}{r}\right) \mathrm{d}t^2 + \left(1-\frac{2M\mathop{}\!\mathrm{erf}(r)}{r}\right)^{-1} \mathrm{d}r^2 +r^2\mathrm{d}\Omega^2,$$

where $\mathop{}\!\mathrm{erf}$ is the error function.

For standard Schwarzschild, one would make the transform

$$r*=r+2M \ln\left(\frac{r}{2M} -1\right),$$

to get the metric in the form

$$\mathrm{d}s^2=-\left(1-\frac{2M }{r}\right) \left(\mathrm{d}t^2 - \mathrm{d}r^{*2} \right) +r^2\mathrm{d}\Omega^2.$$

However, here there is a derivative of the error function which messes everything up if we try

$$r*=\frac{r}{\mathrm{erf}(r)} +2M \ln\left(\frac{r}{2M\mathop{}\!\mathrm{erf}(r)} -1\right).$$

To do the transform properly, we really need to find $r^*$ such that $dr^*=\left(1-\frac{2M\mathop{}\!\mathrm{erf}(r)}{r}\right)^{-1}dr$,

i.e. solve

$$\int \mathrm{d}r \left(1-\frac{2M\mathop{}\!\mathrm{erf}(r)}{r}\right)^{-1}$$

which I cannot calculate analytically.

Does anyone know how I can make a transformation that would get my modified Schwarzschild metric into the form

$$f(r) -(\mathrm{d}t^2 -\mathrm{d}r^{*2}) +r^2 \mathrm{d}\Omega^2?$$

  • 1
    $\begingroup$ What is the question? $\endgroup$ – user4552 Nov 6 '18 at 15:11
  • 4
    $\begingroup$ Yeah, it just sounds like you just need to define a function $g(r)$ equal to your integral. It's going to be some intractible hypergeometric function without a closed form or something. $\endgroup$ – Jerry Schirmer Nov 6 '18 at 15:18

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