# A particular coordinate transformation of a metric tensor

So, this was a problem set question for my GR class due yesterday, and I can't for the life of me solve it, it seems I am missing something very trivial. Either the given answer is wrong, or I am.

The given line element is,

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

where $\alpha > 0$.

Then I am supposed to do a coordinate transformation using,

$$r = \rho\exp(2\tau/\alpha) \\ t = \tau - \frac{\alpha}{2}\ln(-\alpha^2+\rho^2\exp(2\tau/\alpha))$$

And I am supposed to get the following metric,

$$ds^2 = d\tau^2 - \exp(2\tau/\alpha)\left(d\rho^2+\rho^2d\theta^2+\rho^2\sin^2(\theta)d\phi^2\right)$$

But I can't seem to arrive at that answer. To start off with baby steps, I tried the following:

$$g_{m'n'} = \frac{\partial x^m}{\partial x^{m'}} \frac{\partial x^n}{\partial x^{n'}} g_{mn}$$

$$g_{\tau\tau} = \left(\frac{\partial x^m}{\partial \tau}\right)^2 g_{mm} = 1$$

$$g_{\tau\tau} = \left(\frac{\partial t}{\partial \tau}\right)^2 g_{tt} + \left(\frac{\partial r}{\partial \tau}\right)^2 g_{rr} = 1$$

This seems to be the correct coordinate transformation law for a tensor, but doesn't seem to give the right result - that is $1$.

I am sure I'll understand what I did wrong if I can't simply solve for the first component of the transformed metric. But I can't see where I went wrong.

There appears to be a typo in the coordinate transformation for $r$: it should be $$r = \rho\,\exp(\tau/\alpha).$$ With this in mind, your reasoning is correct; if you work out the partial derivatives, you'll find that $g_{\tau\tau} = 1$, and you can work out $g_{\rho\rho}$ in a similar way.
P.S. also note that $t = \tau - \frac{\alpha}{2}\ln(r^2-\alpha^2)$.