# How to find a coordinates transformation on $A\,\mathrm dS_2$?

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

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): $$$$ds^2 = (R^2 + x^2) \, d\vartheta^2 - \frac{R^2}{(R^2 + x^2)} \, dx^2,$$$$ which is the same as (2) after a simple scale change of $$\theta$$ and $$x$$.