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I have the following 2D metrics (describing the $AdS_2$ spacetime), which are supposed to be the same in different coordinates: \begin{align} \mathrm ds^2 &= \mathrm dt^2 - \sin^2{\!\omega t} \, \mathrm dz^2, \tag{1} \\[12pt] \mathrm ds^2 &= (1 + x^2) \, \mathrm d\theta^2 - \frac{1}{(1 + x^2)} \, \mathrm dx^2. \tag{2} \end{align} Metric (1) covers only some part of the $A\mathrm dS_2$ manifold, while metric (2) is supposed to cover the full manifold. How could I find the coordinates' transformation $(t, z) \Rightarrow (\theta, x)$, or the reverse $(\theta, x) \Rightarrow (t, z)$ ?


Maybe I should go the following route. I define this surface: \begin{equation}\tag{3} u^2 + v^2 - w^2 = R^2, \end{equation} with a flat 3D metric: \begin{equation}\tag{4} \mathrm ds^2 = \mathrm du^2 + \mathrm dv^2 - \mathrm dw^2. \end{equation} Here's a partial parametrization of (3), with $\omega = 1/R$: \begin{align} u &= R \cos{\omega t}, & v &= R \sin{\omega t} \, \cosh{\omega z}, & w &= R \sin{\omega t} \, \sinh{\omega z}. \end{align} Substituting this parametrization into (4) gives metric (1): \begin{equation} \mathrm ds^2 = \mathrm dt^2 - \sin^2 {\! \omega t} \; \mathrm dz^2. \end{equation} Now, I need to find another parametrization of (3) which would give metric (2).

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Doh! I just found the answer. It was easy with the route I described in my question.

Here's another parametrization of (3): \begin{align} u &= \sqrt{R^2 + x^2} \, \sin{\vartheta}, & v &= \sqrt{R^2 + x^2} \, \cos{\vartheta}, & w &= x. \end{align} Substituting this parametrization into (4) gives the metric (2): \begin{equation} ds^2 = (R^2 + x^2) \, d\vartheta^2 - \frac{R^2}{(R^2 + x^2)} \, dx^2, \end{equation} which is the same as (2) after a simple scale change of $\theta$ and $x$.

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