Given the two-dimensional metric $$ds^2=-r^2dt^2+dr^2$$ How can I find a coordinate transformation such that this metric reduces to the two-dimensional Minkowski metric?
I know that $g_{\mu\nu}=\begin{pmatrix}-r^2&0\\0&1\end{pmatrix}$ (this metric) and $\eta_{\mu\nu}=\begin{pmatrix}-1&0\\0&1\end{pmatrix}$ (Minkowski). Obviously, the matrix transformation is $\begin{pmatrix}1/r^2&0\\0&1\end{pmatrix}g_{\mu\nu}=\eta_{\mu\nu}$, but how is that related to the coordinate transformation itself?
EDIT: would the following transformation be acceptable? $$r'=r\cosh t$$ $$t'=r\sinh t$$ Such that: $dr'=\cosh t\ dr+r\sinh t\ dt,\quad dt'=\sinh t\ dr+r\cosh t\ dt$
And: $ds'^2=-dt'^2+dr'^2=-r^2dt^2+dr^2=ds^2$
Where we have: $ds'^2=\eta_{\mu\nu}dx^{\mu}dx^{\nu}$ as requested.
Is that correct? Also, is there a formal way of "deriving" the proper change of coordinates (since mine is more of an educated guess)?
Answer
If you were to Wick rotate $t \rightarrow i \theta$, the metric would be $ds^2 = dr^2 + r^2 d\theta^2$, which is just flat space in polar coordinates. The standard cartesian coordinates can be obtained by $x=r\cos\theta$, $y=r\sin\theta$. The same procedure works in the original Lorentzian signature metric, but with hyperbolic trig functions instead of sines and cosines.
By the way, this is two-dimensional Rindler space, which is just a patch of two-dimensional Minkowski space: http://en.wikipedia.org/wiki/Rindler_coordinates.
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