Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 75 / 3 / e

Set $z=e^y$. The metric becomesso these are Poincare coordinates on anti-de Sitter spacetime with unit curvature radius. They cover one patch, not the whole global extension.

On the two-dimensional section with the three transverse $x_i$ suppressed, choose embedding coordinates $X_{-1}=\sec \rho \cos T$, $X_0=\sec \rho \sin T$, $X_d=\tan \rho$. The Poincaré transformation isChoose the connected component containing $T=\rho =0$. Its denominator is positive, givingIts two sloping edges are Poincaré horizons, null coordinate boundaries at $z\to \infty$. The right vertical edge is the $z\to 0$ timelike conformal boundary. Substituting the transformation into $z^{-2}(d{z}^2-d{t}^2)$ gives ${\sec }^2\rho (d{\rho }^2-d{T}^2)$, verifying the chart and its null edges.

The full coordinates in this part have $y\in \mathbb{R}$, hence the whole shaded $z>0$ patch. If one additionally retains only the Randall–Sundrum bulk $y\ge 0$ as in Question4, keep $z\ge 1$. The brane $y=0$, or $z=1$, has $\cos T=\cos \rho -\sin \rho$ and is drawn in orange; the retained side contains $T=0$, $\rho <0$. A patch horizon is not a curvature singularity or the timelike conformal boundary. A geodesic can cross it in finite proper time while the patch coordinate $t$ diverges.