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I recently started reading about conformal (Penrose) diagrams and have since been faced with a couple of conceptual doubts. Based on the coordinate transformations, null curves in Penrose diagrams are represented by $45^o$ or $135^o$ lines. Consider the exterior Schwarzschild Penrose diagram below, Exterior Penrose diagram

where the symbols have the usual meanings. It can be shown that the constant 'r' surfaces would be hyperbolas as shown. Now, suppose I want to represent some null geodesics which are bounded by the surface $r=a>r_h$.

My Doubt:

Based on what I understood, they should be depicted like the red,green and blue curves (each curve denoting the geodesic path starting at the past horizon and ending at the future horizon). Note that we are considering $\textit{inextendible}$ curves. Is this correct? If yes, is the sharp turn at $r=a$ physically smooth (meaningful)?

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  • $\begingroup$ But the diamond side are $(U,V)$ coordinates (Kruskal-Szekeres) where the $dU =0$ or $dV=0$ denote null curves right ? There is no special horizon at $a$, but I can find geodesics which do not escape to infinity (past or future) and curve into the BH horizon. Is this not a correct representation of them ? $\endgroup$
    – Lelouch
    Commented Aug 26, 2020 at 6:29

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There are two more dimensions in 4D spacetime which are not depicted in your diagram.

This diagram has null curves at 45 and 135 degrees only when understood as geometry induced on submanifold given by constancy of angular coordinates.

The light that would reverse its movement in radial direction, would necessarily had some angular components, which means the "radial velocity" will not be null, i.e. light on the diagram (when understood as projection) will not be at 45 or 135 degrees, but it would be some angle between 45 and 135 degrees. In fact, the radial velocity gets less and less as light approaches $r=a$ and the curve on the diagram should smoothly touch the $r=a$ curve.

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  • $\begingroup$ Allow me to interpret just one more aspect. Since the $r_h$ boundary is a null hypersurface, the bound null geodesics, will be some smooth curve $C$ in the PD, which intersects the horizon (past and future) at right angles, but need not be $45^o$ lines in the interior region. Is that correct @Umaxo ? $\endgroup$
    – Lelouch
    Commented Aug 26, 2020 at 6:52
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    $\begingroup$ @Lelouch No. The right angles represent null geodesics with zero components of tangent vector in angular directions. Bounded null geodesics needs to have nonzero angular components and thus they would look like "timelike" and not "null" curves in the diagram. $\endgroup$
    – Umaxo
    Commented Aug 26, 2020 at 6:57

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