Gravitational Collapse of a Cylinder

  • Thread starter edgepflow
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  • #1
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Main Question or Discussion Point

The Schwarzschild solution to the field equations is a vacuum solution for a spherically symmetric mass and provides the Schwarzschild radius:

rs = 2Gm / c^2.

Is there a similar treatment for a cylinder that would give its radius and height as a function of mass?
 

Answers and Replies

  • #2
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Is there a similar treatment for a cylinder that would give its radius and height as a function of mass?
I think that any cylinder that collapsed would end up as a sphere (if it was not rotating). This is basically a consequence of the "no hair theorem". The only exception *might" be a cylinder with infinite length, but for non zero density it would have infinite mass, complicating things a tad. For example, one complication is that any point on the cylinder would only be "gravitationally aware" of the other particles within its visible horizon.
 
  • #3
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I think that any cylinder that collapsed would end up as a sphere (if it was not rotating). This is basically a consequence of the "no hair theorem". The only exception *might" be a cylinder with infinite length, but for non zero density it would have infinite mass, complicating things a tad. For example, one complication is that any point on the cylinder would only be "gravitationally aware" of the other particles within its visible horizon.
That makes sense.

But before the collapse, is there a way to figure out the maximum height to diameter (H/D) ratio to form an event horizon? Imagine a square cylinder (H=D) that is on the edge of forming an event horizon. Now add the same material to increase H for a fixed D. What ratio H/D does the event horizon finally form?
 
  • #4
PAllen
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I think that any cylinder that collapsed would end up as a sphere (if it was not rotating). This is basically a consequence of the "no hair theorem". The only exception *might" be a cylinder with infinite length, but for non zero density it would have infinite mass, complicating things a tad. For example, one complication is that any point on the cylinder would only be "gravitationally aware" of the other particles within its visible horizon.
I think this is an unsolved question. The black hole theorem applied iff the result is a conventional black hole. If it is some other type of singularity, the no hair theorem does not apply.
 
  • #5
JesseM
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I think that any cylinder that collapsed would end up as a sphere (if it was not rotating). This is basically a consequence of the "no hair theorem". The only exception *might" be a cylinder with infinite length, but for non zero density it would have infinite mass, complicating things a tad. For example, one complication is that any point on the cylinder would only be "gravitationally aware" of the other particles within its visible horizon.
But if an infinite cylinder could collapse to a line singularity, I wonder what happens if you just have a finite cylinder which is long enough that the event of the ends beginning to collapse lie outside the past light cone of the event of the center reaching infinite density? (of course the past light cone of this point would still look different than the past light cone of the singularity in the infinite-cylinder case, so that might be enough to explain why it behaves differently) The "no hair" theorem doesn't necessarily rule out the possibility of naked singularities, I wonder if it can be ruled out that a sufficiently long cylinder might produce one...
 
  • #7
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No matter what the shape of the mass, the requirements for an event horizon (no spin, and no charge) is R>=M. All of mass M must be within a sphere of radius R.
 
  • #8
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The cylindrical equivalent of the spherically symmetric vacuum is the Weyl vacuum. If you do a web search for 'Weyl vacuum' there are lots of hits.

I know some cylindrically symmetric solutions but not for a collapsing cylinder.
 

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