- #36
fzero
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latentcorpse said:Where did you get this from?
You can compute the distance between points with the metric.
And also, I don't understand how the proof of Penrose's Theorem (given on page 51/52) actually proves anything to do with the theorem!
And on p52, he says that null geodesics may enter [itex]\cal{H}^+[/itex] but never leave it. Does this mean that for the time reversal, null geodesics may leave [itex]\cal{H}^-[/itex] but never enter it? I guess that makes sense since null geodesics go from [itex]\mathfrak{I}^-[/itex] to [itex]\mathfrak{I}^=[/itex] so in order for them to enter [itex]\cal{H}^-[/itex] they would have to follow a path at an angle greater than 45 degrees i.e. they'd be going faster than light which is forbidden. However, how could there ever be a situation where they could leave [itex]\cal{H}^-[/itex]? We know that all null geodesics start on [itex]\mathfrak{I}^-[/itex] and end on [itex]\mathfrak{I}^+[/itex]. So I guess if they started at the point that [itex]\mathfrak{I}^- , \cal{H}^-[/itex] share then they could travel up [itex]\cal{H}^-[/itex], leaving it at some point and then heading back to [itex]\mathfrak{I}^+[/itex] at 45 degrees. Did I get this right?
Townsend talks about time-reversibility right on that page.
Finally, on p53, he says that the singularity at r=0 which occurs in spherically symmetric collapse is hidden in the sense that no signal can reach it from [itex]\mathfrak{I}^+[/itex]. Surely, he means [itex]\mathfrak{I}^-[/itex], no? Although, I'm guessing not because the Kruskal diagram below quite clearly shows that a signal from [itex]\mathfrak{I}^-[/itex] can reach the r=0 singularity. Why are we so concerned with [itex]\mathfrak{I}^+[/itex] here?
No, he says that no signal can reach [itex]\mathfrak{I}^+[/itex] from the singularity at r=0.