Need Help with Black Hole Computation Homework?

[latentcorpse]

The geodesic equation is V\cdot D V^\mu =0. It looks easy enough to verify that.

V \\cdot D V^\mu = V^\nu D_\nu V^\mu = V^\nu ( V^\mu{}_{, \nu} + \Gamma^\mu{}_{\rho \nu} V^\rho )
But V^\mu{}_{, \nu}=0 obviously
and since we are considering the Rindler metric (essentially Minkowski spacetime but in funny coordinates), the spacetime is flat adnd so the Christoffel symbols will vanish.
This means it will satisfy the geodesic equation.
Is that ok?

Evidently the proper distances traveled are all the same. It's probably a nice exercise to work out starting from the action (2.1).

How would I go about doing this? What are the steps involved?


And the top diagram on page 50 shows the future event horizon, \mathscr{H}^+. Why is that line the boundary of the closure of the causal past of future null infinity? I don't really get why it is that line, and that line alone, that is the boundary? Is it because any other such lines that could be boundaries e.g. \mathscr{I}^+ or \mathscr{I}^- are contained in J^- ( \mathscr{I}^+ ) and therefore by the definition \mathscr{H}^+ = \bar{J}^- ( \mathscr{I}^+ ) - J^- ( \mathscr{I}^- ) they get taken away again and can therefore not be part of the horizon? I think that's right...but why can't the dotted vertical line be part of the horizon still?

Cheers.

 

[fzero]

V \\cdot D V^\mu = V^\nu D_\nu V^\mu = V^\nu ( V^\mu{}_{, \nu} + \Gamma^\mu{}_{\rho \nu} V^\rho )
But V^\mu{}_{, \nu}=0 obviously
and since we are considering the Rindler metric (essentially Minkowski spacetime but in funny coordinates), the spacetime is flat adnd so the Christoffel symbols will vanish.
This means it will satisfy the geodesic equation.
Is that ok?

Yes.

How would I go about doing this? What are the steps involved?

I spoke a little bit quickly. The analysis in the notes never treats points in the interior of the star, because the metric there depends strongly on the distribution of dust, which he doesn't try to specify. So if his statement is true, it relies on something that he hasn't really mentioned.

In any case, the type of analysis (were the full metric known) would be something similar to what he does in section 2.3.

And the top diagram on page 50 shows the future event horizon, \mathscr{H}^+. Why is that line the boundary of the closure of the causal past of future null infinity? I don't really get why it is that line, and that line alone, that is the boundary? Is it because any other such lines that could be boundaries e.g. \mathscr{I}^+ or \mathscr{I}^- are contained in J^- ( \mathscr{I}^+ ) and therefore by the definition \mathscr{H}^+ = \bar{J}^- ( \mathscr{I}^+ ) - J^- ( \mathscr{I}^- ) they get taken away again and can therefore not be part of the horizon? I think that's right...but why can't the dotted vertical line be part of the horizon still?

Cheers.

Yes, everything to the right of {\cal H}^+ is in J^- ( \mathfrak{I}^+ ). The point is that all of the points on {\cal H}^+ are spacelike separated from \mathfrak{I}^+. You could verify this by calculation and you can also compare to the diagram on page 18 that shows the light-cones explicitly.

 

[latentcorpse]

I spoke a little bit quickly. The analysis in the notes never treats points in the interior of the star, because the metric there depends strongly on the distribution of dust, which he doesn't try to specify. So if his statement is true, it relies on something that he hasn't really mentioned.

In any case, the type of analysis (were the full metric known) would be something similar to what he does in section 2.3.

So even though we can't do it here I'd like to make sure I understand the method. We would use the metric and what make assumptions about traveling on radial timelike (since they are massive free particles) geodesics. Then we would get an expression for \frac{dt}{d \tau} and work out t and try to show that t doesn't depend on r?

Yes, everything to the right of {\cal H}^+ is in J^- ( \mathfrak{I}^+ ). The point is that all of the points on {\cal H}^+ are spacelike separated from \mathfrak{I}^+. You could verify this by calculation and you can also compare to the diagram on page 18 that shows the light-cones explicitly.

This spacelike separation is easy to see from the diagram at least. How would one prove it by calculation though?

 

[fzero]

So even though we can't do it here I'd like to make sure I understand the method. We would use the metric and what make assumptions about traveling on radial timelike (since they are massive free particles) geodesics. Then we would get an expression for \frac{dt}{d \tau} and work out t and try to show that t doesn't depend on r?

I think that you'd want to compute \int d\tau for a path starting at r=R and ending at r=0. If Townsend's statement is true, this proper distance would be independent of R.

This spacelike separation is easy to see from the diagram at least. How would one prove it by calculation though?

The proper distance between two points in the plane is

s^2 = - 4 \left( 1-\frac{2M}{r}\right) \Delta\tilde{U} \Delta\tilde{V} .

To the right of {\cal H}^+, points are timelike separated from \mathfrak{I}^+, {\cal H}^+ they are null, and to the left of {\cal H}^+, points are spacelike.

 

[latentcorpse]

The proper distance between two points in the plane is

s^2 = - 4 \left( 1-\frac{2M}{r}\right) \Delta\tilde{U} \Delta\tilde{V} .

Where did you get this from?

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?

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?

 

[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.