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Space dynamics in GR anaylysis of black holes

by yanniru
Tags: anaylysis, black, dynamics, holes, space
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yanniru
#1
Dec6-03, 07:19 PM
P: 108
I have a question concerning the treatment of black holes in General Relativity. As I understand it, the force of gravity is replaced in GR by distortion of space, and the distortion may be dynamic, a function of time. In particular, space may be sucked into black holes.

Now without the force of gravity, my expectation is that the event horizon of a black hole is given by the surface where space is being sucked into the black hole at the speed of light. As such light cannot escape the event horizon.

This thinking has been criticied on the Astronomy forum. But I do not understand how it could be otherwise. Can someone explain it for me.

Richard
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Ambitwistor
#2
Dec6-03, 07:27 PM
P: 837
Originally posted by yanniru
Now without the force of gravity, my expectation is that the event horizon of a black hole is given by the surface where space is being sucked into the black hole at the speed of light.
An event horizon is the boundary of a region of space from which light cannot escape to infinity. It is a null surface, meaning that it's (theoretically) possible for a light ray emitted at the horizon to remain at the horizon. Thus, for any body that falls through the horizon, one could say that the horizon passes them by at the speed of light (although they cannot locally tell that this has happened).

I can't think of a meaningful way in general relativity to say that space is "sucked into" a black hole, or indeed, what it would mean to say that "space moves". (Moves with respect to what?? When we say that something moves, we say that it changes its position in space. What does it mean for space itself to move?)
yanniru
#3
Dec7-03, 10:37 AM
P: 108
You offered a reply in semantics rather than physics, to wit:

"Moves with respect to what?? When we say that something moves, we say that it changes its position in space. What does it mean for space itself to move?)"

Obviously we are talking about movement with respect to the location of the black hole.

Back to physics. GR, given a distribution of mass, determines the distortion of the geometry of space.

Now if that geometric distortion is not a function of time, I want to know what stops the light from escaping the event horizon of a black hole. And do not say the force of gravity cause there ain't any in GR.

My expectation is that the geometric distortion is time varying and at the event horizon, and that the movement of space is at the speed of light for a distant observer; so that again for the distant observer, light coming out of the black hole is stopped at the event horizon, so to speak. The actual space distortion is probably greater within the event horizon.

Richard

Ambitwistor
#4
Dec7-03, 11:01 AM
P: 837
Space dynamics in GR anaylysis of black holes

Obviously we are talking about movement with respect to the location of the black hole.
You missed the point. This is not mere semantics. What does it mean for space to "move with respect to" anything, a black hole or otherwise? That is a physical question: what experiment can I perform that will determine "the speed of space" with respect to anything, a black hole or otherwise? How do I measure it?

Now if that geometric distortion is not a function of time, I want to know what stops the light from escaping the event horizon of a black hole.
Spacetime is curved in a way that all trajectories that are not faster-than-light are curved into the hole.

What does whether the curvature is changing in time have to do with anything?

My expectation is that the geometric distortion is time varying
It can be, but a dynamic black hole will rapidly settle down into a stationary state (generally, a Kerr black hole) in which the geometry does not change with time. It is definitely not true that a black hole must have a time-varying geometry.

and at the event horizon, and that the movement of space is at the speed of light for a distant observer;
Nobody who works in the field of gravitation speaks of "space moving at the horizon". I don't even know what that could mean in GR. If you can define "the movement of space" (in operational terms, in the form of an experiment one could perform to measure it), I'll tell you whether general relativity predicts it.
yanniru
#5
Dec7-03, 02:47 PM
P: 108
Measure the speed of light in the local frame at the event horizon of a black hole. My assumption is that the speed of light is c in every local frame regardless of geometric distortion.

What you seem to be telling me is that the speed of light is zero in the local frame at the event horizon. You said trajectory. So for a photon traveling normal to the event horizon, it then travels outward to the event horizon, stops and turns around there and goes back into the black hole. Is this what you mean?

And if that is correct, then my presumption that the speed of light is c in every local frame regardless ofgeometric distortion is incorrect.

Richard
Ambitwistor
#6
Dec7-03, 02:54 PM
P: 837
Originally posted by yanniru
Measure the speed of light in the local frame at the event horizon of a black hole. My assumption is that the speed of light is c in every local frame regardless of geometric distortion.
In general relativity, the speed of light in a local inertial frame is always c, regardless of the curvature, whether you are at a horizon, etc. (However, there is no local inertial frame at rest with respect to the horizon, which is a null surface.)

So for a photon traveling normal to the event horizon, it then travels outward to the event horizon, stops and turns around there and goes back into the black hole. Is this what you mean?
No.
yanniru
#7
Dec7-03, 08:48 PM
P: 108
You just will not admit it. You said no to a trajectory of light turning at the event horizon, and you also said "there is no local inertial frame at rest with respect to the horizon, which is a null surface".

In my book space is therefore moving, and as you agree that the speed of light is always c in the local frame. The local frame must be moving at the speed of light towards the center of the black hole at every event horizon.

Thank you very much.

Richard
Ambitwistor
#8
Dec7-03, 09:58 PM
P: 837
Originally posted by yanniru
You just will not admit it.
Admit what?

You said no to a trajectory of light turning at the event horizon, and you also said "there is no local inertial frame at rest with respect to the horizon, which is a null surface".
So?


In my book space is therefore moving,
Well, fine, but if you won't tell me what you think it means for space to move, I can neither agree nor disagree with you.

The local frame must be moving at the speed of light towards the center of the black hole at every event horizon.
What do you mean, "the" local frame? There are infinitely many local frames at any point. For that matter, if you want to talk about the speed of a local frame, that is only defined with respect to some other local frame at the same point. Which other frame are you talking about? (It can't be a local frame at rest with respect to the horizon, at the location of the horizon, because as I already said, no such frame exists.)
yanniru
#9
Dec8-03, 07:32 AM
P: 108
"What do you mean, "the" local frame? There are infinitely many local frames at any point. For that matter, if you want to talk about the speed of a local frame, that is only defined with respect to some other local frame at the same point. Which other frame are you talking about? (It can't be a local frame at rest with respect to the horizon, at the location of the horizon, because as I already said, no such frame exists.)"


This is so self-contradictory. First you say that there are an infinite number of local frames at ANY point. Then you say thyere are none at the event horizon.

I wish someone who made sense would address these my question.
Ambitwistor
#10
Dec8-03, 09:48 AM
P: 837
Originally posted by yanniru
This is so self-contradictory. First you say that there are an infinite number of local frames at ANY point. Then you say thyere are none at the event horizon.
I said that there are no local frames at the horizon which are at rest with respect to the horizon. This is obvious: no timelike observer can remain at the horizon.

I wish someone who made sense would address these my question.
My answers would make more sense to you if you paid attention to what they say.


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