By Townsend P.K.

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**Example text**

V˜ = π/2, π χ −π χ−τ = π ⇔ ˜ = −π/2, U − Flatten the cylinder to get the Carter-Penrose diagram of Minkowski spacetime. 43 ... . .. . ... + .. . . .. .... ... ...... . ... .... . .. .. .... . .. .. ....... ... .... ...... ... . . .... ... ... .... ...... ... ......... . . .

U − ...................... V r=r III’ V r = constant 0 < r < r− VI II curvature singularity at r = 0 r = constant r− < r < r+ Region II is the same as the region II covered by the (U + , V + ) coordinates. The other regions are new. g. in r < r− . We know that region II of the diagram is connected to an exterior spacetime in the past (regions I, III, and IV), by time-reversal invariance, region III’ must be connected to another exterior region (isometric regions I’, II’, and IV’).

N timelike r q null p null Choosing S = +, then gives Penrose’s Theorem. Properties (iv) and (v) show that null geodesics may enter H+ but cannot leave it. This result may appear inconsistent with time-reversibility, but is not. The time-reverse statement is that null geodesics may leave but cannot enter the past event horizon, H− .