# How to get the Bertotti-Robinson metric into its conformally flat version?

The Bertotti-Robinson metric is a very nice exact solution to the Einstein-Maxwell field equations in vacuum. It is usually stated into these equivalent versions (where $$r = \sqrt{x^2 + y^2}$$ and $$\bar{r} = \sqrt{\bar{x}^2 + \bar{y}^2 + \bar{z}^2}$$ and $$R$$ is just a constant): $$\begin{gather} ds^2 = dt^2 - \frac{1}{(1 + r^2 / 4 R^2)^2} (dx^2 + dy^2) - \sin^2 {\!(t/R)} \, dz^2, \tag{1} \\[12pt] ds^2 = dt^2 - R^2 (d\vartheta^2 + \sin^2 \vartheta \: d\varphi^2) - \sin^2 {\!(t/R)} \, dz^2, \tag{2} \\[12pt] ds^2 = (1 + \tilde{z}^2 / R^2) \, d\tilde{t}^2 - R^2 (d\vartheta^2 + \sin^2 \vartheta \: d\varphi^2) - \frac{1}{(1 + \tilde{z}^2 / R^2)} \, d\tilde{z}^2, \tag{3} \\[12pt] ds^2 = \frac{R^2}{\bar{r}^2} (d\eta^2 - d\bar{x}^2 - d\bar{y}^2 - d\bar{z}^2). \tag{4} \end{gather}$$ While I know how to find the coordinates transformations that change (1)-(2)-(3) into each other, I'm unable to find the coordinates transformations that give version (4) (the conformally flat version).

So how to get metric (4) from (1), or (2), or (3) ?

If one applies the standard coordinate transformation from 3D Cartesian coordinates $$(\bar{x},\, \bar{y},\, \bar{z})$$ in (4) to spherical coordinates $$(r,\theta,\phi)$$, then the metric (4) becomes: $$ds^2= \frac{R^2}{\xi^2}(d\eta^2-d \xi^2)-R^2 (d\theta^2 + \sin^2 \theta d\phi^2),$$ where we renamed the radial coordinate to $$\xi$$ to underscore that it no longer enters into the spherical part. This forms explicitly demonstrates the $$AdS_2 \times S_2$$ structure of the solution and has the same “angular” part as metrics (2) and (3).
So, what is left is the coordinate tranformation of the $$AdS_2$$ factor, which should not pose a problem, since the metric $$\xi^{-2}(d\eta^2-d \xi^2)$$ is simply a Wick-rotated standard metric of Poincaré half-plane model of the hyperbolic plane.
• I don't see how your radial part could be transformed into $(\dots) \, d\tilde{t}^2 - \frac{1}{(\dots)} \, d\tilde{z}^2$ from (3). Could you give the coordinates transformation in this case? I don't think it's possible. – Cham Aug 21 at 18:43
(4) can be written this way, using spherical coordinates, with $$\bar{r} = \sqrt{\bar{x}^2 + \bar{y}^2 + \bar{z}^2}$$: \begin{align} ds^2 &= \frac{R^2}{\bar{r}^2} \, d\eta^2 - \frac{R^2}{\bar{r}^2} \, \big( \, d\bar{r}^2 + \bar{r}^2 \, (d\vartheta^2 + \sin^2 \! \vartheta \: d\varphi^2) \big) \\[12pt] &= \frac{R^2}{\bar{r}^2} \, (d\eta^2 - d\bar{r}^2) - R^2 \, (d\vartheta^2 + \sin^2 \! \vartheta \: d\varphi^2). \tag{5} \end{align} The angular part is identical as the one appearing in (2). Now, I need to get this "radial" part: $$$$\tag{6} ds_{\text{rad}}^2 = dt^2 - \sin^2 \! (t / R) \, dz^2.$$$$ I'll substitute the following coordinates transformation into (5), where $$F(z / R)$$ is some function to find: \begin{align} \eta &= R \, F(z / R) \sinh{(\sigma / R)}, \\[12pt] \bar{r} &= R \, F(z / R) \cosh{(\sigma / R)}. \end{align} Then the "radial" part of (5) becomes: $$$$\tag{7} ds_{\text{rad}}^2 = \frac{1}{\cosh^2 {\! (\sigma / R)}} \, d\sigma^2 - \frac{F^{\prime \, 2}}{F^2 \cosh^2 {\! (\sigma / R)}} \, dz^2.$$$$ To get rid of the pesky factor $$F^{\prime \, 2} / F^2$$, I'll use $$F(z / R) = e^{z / R}$$ : $$$$\tag{8} ds_{\text{rad}}^2 = \frac{1}{\cosh^2 {\! (\sigma / R)}} \, d\sigma^2 - \frac{1}{\cosh^2 {\! (\sigma / R)}} \, dz^2.$$$$ Now, I introduce a new time variable: $$$$\tag{9} d\theta = \frac{1}{\cosh{\! (\sigma / R)}} \, d\sigma \qquad \Rightarrow \qquad \theta = 2 R \arctan{\big(\tanh{(\sigma / 2 R)} \big)}.$$$$ Some algebra gives $$\cosh{(\sigma / R)} = 1/\cos{(\theta / R)}$$. Then (8) becomes: \begin{align} ds_{\text{rad}}^2 &= d\theta^2 - \cos^2{\! (\theta / R)} \, dz^2 \\[12pt] &\equiv d\theta^2 - \sin^2 {\! (\theta / R + \pi / 2)} \, dz^2. \tag{10} \end{align} Finally, I define $$t = \theta + \pi R / 2$$, so I recover (6). So the complete transformation that take (4) and gives (2) is this: \begin{align} \eta &= R \, \frac{\cos{(t / R)}}{\sin{(t / R)}} \, e^{z / R}, \\[12pt] \bar{r} &= R \, \frac{1}{\sin{(t / R)}} \, e^{z / R}. \end{align} This transformation covers only a part of spacetime since \begin{align} \frac{\eta}{\bar{r}} &= \cos{(t / R)}, & \eta^2 - \bar{r}^2 &= -\, R^2 \, e^{2 z / R} < 0. \end{align}