Change in coordinates for a metric in a given form

Question

With a change in coordinates, transform

\begin{align} ds^2 = -z^2dt^2 + dz^2 \end{align}

to

\begin{align} ds^2 = -dT^2 + dX^2. \end{align}

My attempt. It is clear that incoming null geodesics $\dot{z}<0$ have the form $z=ce^{-t}$, and outgoing null geodesics have $z=de^{t}$. In terms of $(T,X)$, incoming geodesics take the form $T = -X + a$, and outgoing geodesics take the form $T=X+b$. So I try to map

\begin{align} T = \log z,\quad X = t. \end{align}

But I obtain $$ ds^2 = \frac{1}{z^2}dz^2 + dt^2, $$ which is close but not the right answer.

Answer 1

Look up Rindler coordinates. They're the Lorentzian analogue of polar coordinates. Set $T = z \sinh t$ and $X = z\cosh t$. So $$\begin{align*} -{\rm d}T^2 + {\rm d}X^2 &= -(\sinh t\,{\rm d}z + z\cosh t\,{\rm d}t)^2 + (\cosh t\,{\rm d}z + z\sinh t\,{\rm d}t)^2 \\ &= -\sinh^2t\,{\rm d}z^2 - 2z\cosh t\sinh t\,{\rm d}z\,{\rm d}t - z^2\cosh^2t\,{\rm d}t^2 \\ &\qquad + \cosh^2t\,{\rm d}z^2 + 2z\cosh t\sinh t\,{\rm d}t\,{\rm d}z + z^2\sinh^2t\,{\rm d}t^2 \\ &= {\rm d}z^2 - z^2\,{\rm d}t^2,\end{align*}$$as wanted.