Coordinate transform

Question

Can anyone see what transformation $$r\to f(r)$$ transforms $$\exp(2\phi(r))(dr^2+r^2d\theta^2)$$ to $$df^2+\sinh^2(f)d\theta^2$$? I there a systematic way to attack such a problem -- rather than just hoping that I somehow spot it?

Thank you!

Answer 1

Well, $\phi$ can't be very creative. Since the curvature of this surface must be invariant under the change of coordinates, with a bit of computation we'll get (from the latter) $K=-1$ and (from the former) $K=-e^{-2\phi(r)}\big(\phi''(r)+\frac1r\phi'(r)\big)$. Thus, $\phi$ must satisfy the ODE $$\phi''(r)+\frac1r\phi'(r)=e^{2\phi(r)}\,.\tag{$\star$}$$ (This should look somewhat familiar if you've computed curvature of a conformally flat metric before.) It also makes us think we're looking at some hyperbolic metric, of course.

Starting at the other end of the problem, you need \begin{align*} f'(r) &= e^{\phi(r)} \\ \sinh f(r) &= re^{\phi(r)}\,, \end{align*} and we can actually integrate patiently to obtain $$\cosh f(r) = \frac{Cr^2+1}{Cr^2-1}\,,$$ so $$f(r) = \log\left(\frac{(Cr+1)^2}{Cr^2-1}\right)\,.$$ Amazingly, if we now set $\phi(r) = \log f'(r)$, we do in fact (!) get a solution of the ODE ($\star$) above.