Black Holes computation homework

[fzero]

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 $\cal{H}^+$ but never leave it. Does this mean that for the time reversal, null geodesics may leave $\cal{H}^-$ but never enter it? I guess that makes sense since null geodesics go from $\mathfrak{I}^-$ to $\mathfrak{I}^=$ so in order for them to enter $\cal{H}^-$ 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 $\cal{H}^-$? We know that all null geodesics start on $\mathfrak{I}^-$ and end on $\mathfrak{I}^+$. So I guess if they started at the point that $\mathfrak{I}^- , \cal{H}^-$ share then they could travel up $\cal{H}^-$, leaving it at some point and then heading back to $\mathfrak{I}^+$ 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 $\mathfrak{I}^+$. Surely, he means $\mathfrak{I}^-$, no? Although, I'm guessing not because the Kruskal diagram below quite clearly shows that a signal from $\mathfrak{I}^-$ can reach the r=0 singularity. Why are we so concerned with $\mathfrak{I}^+$ here?

No, he says that no signal can reach $\mathfrak{I}^+$ from the singularity at r=0.

 

[latentcorpse]

You can compute the distance between points with the metric.

So you used the conformally compactified Kruskal metric (2.169)? And we would have to use that here rather than the normal Kruskal metric because it's the compactified metric that relates to the Penrose diagram.

Townsend talks about time-reversibility right on that page.

I don't understand what you mean here. Was my interpretation of how the time reversibility of the event horizon works accurate?

No, he says that no signal can reach $\mathfrak{I}^+$ from the singularity at r=0.

So the definition of a naked singularity is if a null or timelike signal can reach $\mathfrak{I}^+$ from the singularity?
In that case, it's clear that the white hole r=0 singularity of Kruskal is naked but surely since the black hole r=0 singularity of Kruskal actually touches the hypersurface $\mathfrak{I}^+$, it will also be naked?

Thank you.