 Solution

With $r=R\sin \chi$, the static metric isThe radial coefficient is regular at $\chi =\pi /2$, but the time coefficient vanishes and these coordinates cease to be a chart. Part c supplies smooth horizon-crossing coordinates; the local Rindler coordinates also prove the degeneracy is a coordinate one. Extending $\chi$ while keeping the same static $t$ alone is not a regular coordinate extension.

The complete analytic extension is the hyperboloid $-Y_0^2+{\sum }_{i=1}^4{Y}_i^2=R^2$. Static coordinates are $Y_0=\sqrt{R^2-r^2}\sinh (t/R)$, $Y_4=\sqrt{R^2-r^2}\cosh (t/R)$, and $(Y_1,Y_2,Y_3)=r\mathbf{n}$. Global coordinates giveWith $\tan \eta =\sinh (T/R)$ this isThe Penrose diagram is a rectangle, with spacelike past/future infinity, regular pole lines $\psi =0,\pi$, and radial null rays at $45$ degrees. Observer horizons divide static patches from inaccessible regions; they are not curvature boundaries.