I A computation of an integral on page 344 of Schutz's textbook

[MathematicalPhysicist]

On page 344 of "A First Course in GR" he writes the following:

Consider, next, $k=1$. Let us define a new coordinate $\chi(r)$ such that:
$$(12.16)\ \ \ d\chi^2 = \frac{dr^2}{1-r^2}$$
and $\chi =0$ where $r=0$. This integrates to
$$(12.17) \ \ \ r=\sin \chi,$$

When I do the integration I get the following: $\int_0^{\chi^2}d\chi^2= \int_0^{r^2}\frac{dr^2}{1-r^2}= \chi^2 = -\ln (1-r^2)$, after I invert the last relation I get: $r=\sqrt{1-\exp(-\chi^2)}$, where did I go wrong in my calculation?

Thanks!

 

[vanhees71]

You should of course integrate
$$\int \mathrm{d} \chi = \int \mathrm{d}r \frac{1}{\sqrt{1-r^2}}...$$

 

[MathematicalPhysicist]

You should of course integrate
$$\int \mathrm{d} \chi = \int \mathrm{d}r \frac{1}{\sqrt{1-r^2}}...$$

Ah, yes you are correct. It's $(dx)^2$ and not $d(x^2)$.
I am really getting old and sloppy... :-(

 

[vanhees71]

How do you define and Integral with $\mathrm{d} x^2$, and it's not $\mathrm{d} (x^2)=2x \mathrm{d} x$, because you want $\chi$ and not something else ;-).

 

[MathematicalPhysicist]

How do you define and Integral with $\mathrm{d} x^2$, and it's not $\mathrm{d} (x^2)=2x \mathrm{d} x$, because you want $\chi$ and not something else ;-).

I meant I thought it was $d(\chi^2)=\frac{d(r^2)}{1-r^2}$. But now I see my mistake, thanks for clearing this simple thing.