



Be the first pioneers to continue the Astronomy Discussions at our new Astronomy meeting place... The Space and Astronomy Agora 
Re: Black Hole Endings
Forum List  Follow Ups  Post Message  Back to Thread Topics  In Response To Posted by Zephram Cochrane/">Zephram Cochrane on October 5, 1999 07:36:10 UTC 
As far as I know, the only analytic work supporting my statement about objects being remotely observed to cross event horizons of changing size is my own, and I've based it in the semistrong equivalence principle as applied to accelerating rockets. The math gets a bit deep, but in word my premise isn't too complicated so I'll describe it with a little of both. Most of this will be establishing a proof that I can use a certain spacetime geometry, which afterward makes my assertion obvious. Lets say an observer, S', is in an inertial frame with no gravitational fields. Let's say a rocket has an on board observer, S. The rocket will burn its fuel at a variable rate so that when S stands on a scale it always reads the same weight, W. In the proper frame ( rocket S observer's point of view ) an acceleration is calculated from the mass of S( I Never use mass to mean relativistic mass unless specified. I always use it to mean invariant mass which is the same thing as Local proper frame mass) by the equation a = W/m. This acceleration a is called the proper acceleration. Now we can look at this situation from three points of view, but I'll only discuss the two that are relevant. 1. We can look at it from the S' observer's point of view and say that the rocket is accelerating in no gravitational field. 2. We can look at it from the S observer's point of view and say that the rocket is being held still by the rocket engines in a universe that has a gravitational field that accelerates everything down the other way. The semistrong equivalence principle as applied to accelerating rockets would be the statement that both points of view are just as correct. Now for a moment lets look at this from the S' observer's frame. Using special relativity one can show that the S' frame position x' of the rocket as a function of proper time t is given by: x' = (c2/a)cosh(at/c)  c2/a Also special relativity shows that the S' observer's time t' at which the rocket S frame observer's clock reads t is given by the equation: t' = (c/a)sinh(at/c) Now keep this in the back of the mind for a moment and look at what happens according to the rocket S frame observer's point of view. The rocket frame observer says that the rocket is being held still in a gravitational field. Dropping balls, or doing whatever experiments confined to the domain of the rocket ship that is desired, the rocket frame observer determines that the gravitational acceleration is as far as could be measured to be constant throughout the ship. The rocket frame observer decides to do a weak field approximation to determine the geometry of the spacetime. Doing this results in the spacetime geometry looking like: ds2 = (1+2ax/c2)dct2  dx2/(1+2ax/c2)  dy2  dz2 This turns out to be a vacuum field solution with a zero Reimann tensor and so the rocket frame observer is happy in knowing that if space is vacuum outside the rocket (which it is in this scenario) that this solution would then be just fine to describe physical processes for anything down to the event horizon that exists at x = c2/2a. Now this is a key point. This vacuum field solution has an event horizon. The term dx2/(1+2ax/c2) goes infinite there. Now lets return to the S' frame for a moment. This frame has no gravitational field and so the event horizon does not exist from the S frame observer's point of view. However the S' frame observer can understand why the rocket frame observer says that there is one for the following reason. As the ship continues to accelerate near the speed of light in the S' frame it never quite reaches it and so it approaches the speed of light asymptotically. This implies that there is a distance behind the ship from which information sent at the speed of light toward the ship will never catch it. At t' = 0 this distance is: x' = c2/a. Now the first question that arises is why the factor of 1/2 difference between the location in the S frame and the t' = 0 location in the S' frame. Simply put, that is just a choice of spaceal coordinate scaling made by the rocket frame observer, when the rocket frame observer decided to use the above expression for ds2 globally. This scaling transformation doesn't effect the tensor form of the equations being used by the rocket frame observer and so in that sense it is unimportant, but it does come into importance in that this creates the same kind of infinity in the equation for ds2 at the event horizon that shows up in the solution for ds2 for black holes. Recall the equations for position of the rocket and time dilation above. Lets say for now that the two frame's decided to relate their coordinates by the equations consistent with the above by: x' = [x + (c2/a)]cosh(at/c)  c2/a t' = [(x/c) + (c/a)]sinh(at/c) y' = y z' = z And we asserted that the S' frame is special relativistic so we have: ds2 = dct'2  dx'2  dy'2  dz'2 Doing some trivial calculus results in: ds2 = (1 + ax/c2)2dct2  dx2  dy2  dz2 So using a "good" choice of coordinates that the rocket frame observer didn't use the event horizon doesn't correspond to an infinity in the spacetime, but instead corresponds to a zero in the time dilation component of the spacetime. It is at x = c2/a. But now lets look at the transformation that the rocket frame observer actually chose when asserting the ds2 geometry that was actually found from the weak field approximation in the first place. What was done by this corresponds to the transformation: (x' + c2/a)2 = (c2/a)2[1 + 2ax/c2]cosh2(at/c) t'2 = (c/a)2[1 + 2ax/c2]sinh2(at/c) y' = y z' = z Asserting that the S frame is special relativistic again with some trivial calculus results in: ds2 = (1+2ax/c2)dct2  dx2/(1+2ax/c2)  dy2  dz2. What the choice of scaling the rocket frame observer made resulted in was in putting events on both sides of the event horizon ( x' less than c2/2a and x' greater than c2/2a)onto the same side (x greater than c2/2a). This gives the over all topology of the spacetime a wormhole geometry in the same sense as Flamm's paraboloid does for black holes. (definitely is a curved topology with zero Reimann tensor which is why I say a zero Reimann tensor is bad to use for the definition of a state of spacetime curvature)As long as we use the metric ds2 = (1+2ax/c2)dct2  dx2/(1+2ax/c2)  dy2  dz2 we must consider this transformation and realize that the events on both sides do exist but that rocket coordinates x less than c2/2a Do NOT describe any events in the actual universe. Keeping this in mind and correcting in a breakdown in the rocket frame's geodesic equations descriptive ability at the event horizon that is then implied this spacetime will still be fine for the rocket frame to use in its description of physical processes. Finally I think I've established that I can use the equation ds2 = (1+2ax/c2)dct2  dx2/(1+2ax/c2)  dy2  dz2 Now let's not consider y or z motion: ds2 = (1+2ax/c2)dct2  dx2/(1+2ax/c2) Now make a close comparison of this with the Schwartzschield black hole equation not considering angular motion: ds2 = (1  2GM/rc2)dct2  dr2/(1  2GM/rc2) If the rocket would suddenly stop, the event horizon would drop away to infinity and the objects on the other side would then emerge from the horizon as this occurs. This would correspond to the black holes event horizon decreasing to zero as objects from the other side of Flamm's paraboloid emerge. If the rocket waits a while and then fires back up the rockets the event horizon would then rush back to its position near the rocket and object in the S' frame that where at a location x' less than c2/2a would be emmersed by it. this would correspond to the black hole's event horizon increasing in size as it engulfs objects that where outside of it. So I say objects can be remotely (like the rocket frame) observed to cross the horizon when it is changing in size. Now the geometrodynamics prevents the grandfather paradox for this kind of black hole in the following way. The geodesic equation is where the time travel paradox is implied. Fortunately the wormhole geometry of the hole or the rocket frame spacetime implies that a falling object can only obey the geodesic equation down to the event horizon where it then crosses to the other side of Flamm's paraboloid instead of to the inside of the hole. This is like a probe dropped from the rocket falling to a coordinate in the S' frame x' less than c2/2a which is not at a coordinate x less than c2/2a. Because the geodesic equation only implies a time reversal for objects on the inside of the hole and not on the other side of the hole, the wormhole geometry of the black hole saves us from the grandfather paradox ( for this kind of hole). 

Additional Information 

About Astronomy Net  Advertise on Astronomy Net  Contact & Comments  Privacy Policy 
Unless otherwise specified, web site content Copyright 19942021 John Huggins All Rights Reserved Forum posts are Copyright their authors as specified in the heading above the post. "dbHTML," "AstroGuide," "ASTRONOMY.NET" & "VA.NET" are trademarks of John Huggins 