# Question about a black hole formation scenerio.

1. Mar 29, 2013

### mrspeedybob

This scenario is described by Lenard Susskind in a lecture on general relativity starting at time stamp 44:23 in the following video.

The scenario is that a spherical shell of radiation is directed inward to a point. The shell contains enough energy to form a black hole.

He states that inside the shell is just flat space-time. This part I'm comfortable with because it is simply the GR version of Newton's shell theorem.

He then goes on to say that from the outside the shell looks like a Schwartzchild geometry. I question this assertion because a person just outside the shell is still causally disconnected from the energy on the opposite side of the shell. Suppose an observer is 1 light-second from the center focal point of the shell. It seems to me that the transition from flat space-time (when the shell is > 1 light-second from the center) to a Schwartzchild geometry would be a gradual transition taking 2 seconds. 1 second after the shell passed you until it collapses to a singularity, and 1 second for the geometrical ramifications of the collapse to reach you 1 second away.

Put another way, if I am 1 light-second away from the center of the sphere when the shell passes me, I am 2 seconds away from causal contact with the energy on the other side of the shell.

I can't imagine that this he really missing this. It seems like such an obvious flaw I can't imagine that he doesn't see it. Is he deliberately overlooking it in in order to simplify the lesson he's trying to teach or is there something wrong with MY reasoning? Much of the logic from the rest of the lecture seems to rest on this transition from flat geometry to Schwartzchild geometry. It doesn't seem to me like an insignificant detail.

Susskind is such an intelligent physicist I'm going to assume that either my reasoning is flawed or the difference between an instantaneous and a gradual transition is an insignificant detail.

Could someone please explain either why I'm wrong or why it's insignificant?

Last edited by a moderator: Sep 25, 2014
2. Mar 29, 2013

### Staff: Mentor

Yes.

This is shown by a mathematical theorem called Birkhoff's theorem:

http://en.wikipedia.org/wiki/Birkhoff's_theorem_(relativity)

Most relativity textbooks give a proof of this theorem at some point. I have posted a short version of the proof in MTW on my PF blog:

https://www.physicsforums.com/blog.php?b=4211 [Broken]

Until the shell collapses enough to form a horizon, this isn't true. Even after the shell collapses enough to form a horizon, the past history of the shell before it fell through the horizon still provides enough information to the rest of the spacetime to produce the Schwarzschild geometry there. See further comments below.

No; the "transition" happens instantly as soon as the shell falls past you--at least, it does in the idealized case of an infinitely thin shell. Yes, this is not really physically reasonable, but neither is an infinitely thin shell.

In the more realistic case of a shell with finite thickness, the transition would happen more gradually as the shell fell past you; it would start when the shell's inner surface passed you and end when the shell's outer surface passed you, at which point you would be sensing the full Schwarzschild geometry.

The reason there isn't a time delay for the effects from the far side of the shell to get to you is that those effects aren't coming from the far side of the shell "now"; they are coming from the far side of the shell 1 light speed time delay in the past. The reason this doesn't make the effect from the far side of the shell look different from the effect of the near side of the shell (the one that is just falling past you), or "spread out" in time instead of being instantaneous, is that the effect of the energy in the shell is not just the Newtonian "gravity" force you are thinking it is; it is more complicated than that.

Last edited by a moderator: May 6, 2017