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Old 18-January-2007, 08:12 PM
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Default On Rindler horizon, black holes, and singularities

Well, publius has been helping me with this in another thread, and we're not quite done yet, so I may be a little too quick in posting this, but it's looking to me so far that the Rindler metric is very different from what we would otherwise think it is. The ratio of the expected velocity, a*t, and the observed relative velocity is the same as that for the time dilation, which would mean that there is actually no distance contraction at all, as far as I can tell anyway, but I guess I would still have to find the field of view to be sure.

So this is what appears to be happening. As we accelerate from t=0 and v=0, a horizon immediately forms behind us from which we will never receive any more light within a finite time of travel. This is because as we accelerate, faster and faster, the light behind us dilates more and more, redshifting equally in the process, and even though the light that was emitted at our time of departure past the Rindler horizon keeps catching up, it just can't quite make it in a finite time. But the catch is, this only applies for light that was emitted from beyond that original distance behind us when or before we began to accelerate. But all of the pulses of the light that are already in transit on our side of the horizon will indeed catch up, but will be more and more dilated with time. The field of view will continue to expand as we travel, so by the time we actually receive the light and any particular point in time, we will still observe it as coming from the actual position it came from, so there is no distance contraction, but it will just be redshifted more and more, redshifting into oblivion with the instantaneous velocities over an eternity, but never completely diminish. The Rindler horizon, then, is just that original distance from which we cannot receive new light.

So how does this compare to the event horizon of a black hole? It would mean that matter falls into a black hole in due time, but we simply won't observe it doing so within a finite time. It would mean black holes aren't really black, but only redshift into oblivion with time. This also means that the space below the horizon isn't "elsewhere" at all, but real space after all, just like the space beyond the Rindler horizon is real. The complex result of the equations just means we can't observe it. It has always bothered me how a black hole could be formed if the matter that forms it never actually falls below the event horizon, since the gravity that is produced at the event horizon only depends on what actually lies below that radius, not above. Now it all makes sense (almost). Also, if a ray of light were to fall below that horizon, it would not be lost forever. That would only apply to light originating there. It would just become tremendously gravitationally lensed. This is due to the law of symmetry. Whatever conditions occur while the pulses are falling inward will also occur identically after they pass through the plane containing the center of mass, and so the path in and then out again will also be identical, unless something continually acts in the line of motion, such as friction, or it strikes something within and is absorbed.

So what about singularities? Well, where would that singularity lie? At the center of mass, right? But what is the gravity at the center of mass? Zero. That means that the first particles that reach it within the center of the body will just keep going, pass right through. The rest would not even reach that point or pass right through as well. Worst case scenario, the whole thing turns inside out, over and over. Now, this is if the internal pressure of the mass can collapse this far to begin with. A whole body of matter cannot collapse to a point anyway, since that means the surface reaches the singularity, but the gravity on the surface is inversely proportional to the square of the distance and the internal pressure is inversely proportional to its cube. That means the internal pressure that keeps it from collapsing will always become greater than the gravity at some point. Even if someone wants to argue that this is not enough to prevent the collapse, then we are only right back to our worst case scenario, where the matter passes right through and keeps going. Matter would not even reach that exact point anyway. Peculiar kinetic energies will always cause a slight miss, however slight. Individual particles might come to orbit that point or something (the center of mass), or the kinetic energies might combine to create an overall rotation. That is all.
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Let's put together the pieces of The Grand Puzzle . (website - now revised)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

Last edited by grav; 18-January-2007 at 08:55 PM..
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Old 19-January-2007, 04:40 AM
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...The Rindler horizon, then, is just that original distance from which we cannot receive new light.
You sound excited and I’m sure you will get learned responses, but in the meantime can you explain this concept a little further.

You say, “The Rindler horizon, then, is just that original distance from which we cannot receive new light.” That is true as long as you keep accelerating, which you can do in a thought experiment. But when you relate the Rindler Horizon to the event horizon of a black hole, how do you achieve continued acceleration?

Or what is the point of relating the Rindler Horizon to the event horizon?
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Old 19-January-2007, 06:01 AM
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You sound excited and I’m sure you will get learned responses, but in the meantime can you explain this concept a little further.

You say, “The Rindler horizon, then, is just that original distance from which we cannot receive new light.” That is true as long as you keep accelerating, which you can do in a thought experiment. But when you relate the Rindler Horizon to the event horizon of a black hole, how do you achieve continued acceleration?

Or what is the point of relating the Rindler Horizon to the event horizon?
An observer accelerating from a point would be the same as the point accelerating away from the observer, as toward a black hole for a stationary distant observer, according to the equivalence principle. Some of the metrics for the Rindler horizon and an event horizon are obviously different, such as an increasing acceleration upon approaching a black hole and such, but the principle is the same. The horizon itself is just the point beyond which we can receive no light.

The point of relating the two is that the Rindler horizon can tell us much about the event horizon of a black hole as well. The space beyond it is still real, but we simply cannot observe it. The complex results of equations really just show that something is impossible within the given parameters, not that it is experiencing another level of space-time or something. It just means that we cannot ever observe anything that originates from past either horizon.
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Let's put together the pieces of The Grand Puzzle . (website - now revised)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."
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Old 05-February-2007, 11:16 PM
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This should have gone here, I think.

Quote:
Originally Posted by grav
Ironically, although the Rindler horizon is meant for relativistic effects, it is really found only by Euclidean means. In flat space, it is half that of Rindler. By using relativistic effects for the instantaneous relevant velocities, but still using the speed of light as c, instead of that relative to the observer, then we get the same distance as that of Rindler. If we do it correctly, however, and consider the light to be travelling at c+v in respect to the observer travelling at each instantaneous velocity of v, for (c+v)-v=c, then we actually get no horizon at all, but only that same time that the light would have taken to travel to the observer from the original distance, as if he had never begun accelerating. The reason is that in Euclidean, we are considering that the observer can obtain any speed over time, but the speed of light remains at c, so the observer can actually catch up to the light in front of him and pass it. With relativity, however, he can never catch the light that moves away in front of him, and therefore, by the same token, he can never escape the light coming from behind him, and so he can never outrun it, and will always receive light from any finite distance in some finite time, regardless of how fast he travels.
And this.

Quote:
Originally Posted by grav
Before relativity, the apparent motion of bodies depended on how they each moved in respect to a medium. The formulas related them as such, and what an observer at K watching K' might see would be different than what K' saw at K. By removing that medium, as relativity does, then all motions are only seen in respect to each other, and so there is always a reciprocity between them. That is what Einstein did with SR. He did have trouble with GR, however, and could never completely let loose of the idea of an ether. But either there is an ether or there isn't. We can't have it both ways. If there is not, which is what the whole idea of relativity is based upon, then there is necessarily reciprocity for accelerating observers as well, since there is nothing else to relate to except each other. In that case, what an observer "stationary" at K sees of an accelerating observer at K' will be exactly the same as what the accelerating observer sees at K. If we invoke the Mach principle here as Einstein attempts to do with this, then we go right back to relating to a stationary frame of reference, as with an ether, and we must consider velocities in relation to this frame, just as with the ether. But by considering it from the frame of reference of the stationary observer, we eliminate the potential mishap of misinterpreting for the speed of light within the accelerating observer's frame, by unintentionally relating it to some absolute frame in space (stationary). And so what the stationary observer sees of the Rindler horizon in relation to the accelerating observer (or if everything else accelerated away together), which is none (but an increasing redshift and time dilation still occur), is exactly as it would be for the accelerating observer for everything else that remains stationary.
__________________
Let's put together the pieces of The Grand Puzzle . (website - now revised)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."
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