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Crossing the Event Horizon of a Black Hole 
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#55
Dec1109, 12:21 PM

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To be clear, DiamondGeezer, are you claiming that any coordinateinvariant quantities determined by the metric, like the proper time between two events on a given worldline, become imaginary once you cross the horizon? (of course if you try to calculate the proper time along a spacelike path you get an imaginary number, but this is equally true outside the horizon)



#56
Dec1109, 02:18 PM

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(Specifically, we are talking here about near the horizon, above, at and below, and not what happens near the central singularity, which, of course, exists only in the case of the black hole.) In the case of a black hole, we have already seen, earlier in this thread, that different observers hovering at different constant heights above a horizon will measure the energy of an infalling particle with a number that gets ever greater the closer the observer is to the horizon. Remember, energy is an observerdependent quantity. Even in nonrelativistic theory, ½mv^{2} depends on what you measure v relative to. So there's no need to ask "where the extra energy comes from" when you switch from one hovering observer's measurement to another, as it's an irrelevant question: energy isn't conserved when you are comparing one observer's measurement against another's. (N.B. the observers at different heights really are in different frames because although they are at fixed distances from each other, each has a different notion of radial distance and time  gravitational timedilation. Therefore their definitions of energy disagree.) If you consider a sequence of hovering observers getting closer to the horizon, yes, in the limit as the distance tends to zero then the "physical speed" (as defined earlier in the thread) of the infalling particle tends to c and its energy tends to infinity. But these are mathematical limits and can never actually be observed by a hovering observer. No observer can actually hover on the horizon itself. And if you want, instead, to imagine an oberver who slowly descends to the horizon, that observer isn't hovering, so the equations derived earlier are no longer valid. The difference may be tiny at a significant height, but very close to the horizon the difference will be substantial (in fact, diverging to infinity). Any observer will only ever measure a finite energy (and a "physical speed" less than c). George Jones chose his words carefully in post #16 to avoid saying that the particle's physical speed was equal to c. Everything I've said above also applies to my example of an accelerating rocket. The only difference is that whenever I speak of a "hovering observer" above, that should be interpreted as meaning "stationary relative to the rocket, as measured by the rocket". In the diagram I referred to in my previous post, an apple dropped from the rocket when T=t=0 follows a vertical worldline on the spacetime diagram. As the apple gets closer to the horizon at x=ct, the local Rindler observers are travelling faster and faster relative to the apple, approaching c in the limit. These observers, each at rest in the rocket's frame, measure a kinetic energy of the apple that diverges to infinity in the limit. So ask yourself, where does this energy come from? Does the question make sense? In both cases, the rocket and the black hole, the reason the falling object "gains infinite energy" (as you put it) is because the observer is accelerating towards the speed of light, relative to the nonaccelerating object, not because of something happening to the falling object. But this is the point. The construction of the (T,R) coordinate system is such that an object cannot cross the Rindler horizon at finite coordinate values, but it certainly does cross the horizon at finite (t,x) coordinates. This is an artefact of the (T,R) coordinate system. Similarly, an object cannot cross a black hole's event horizon at finite Schwarzschild coordinates, but it certainly does cross the horizon at finite coordinates in other coordinate systems. Again, this is an artefact of the Schwarzschild coordinate system. Incidentally the "Rindler horizon" has something equivalent to a black hole's Hawking radiation. It is called the Unruh effect. A final comparison between my rocket example and a black hole. My inertial coords (t,x) are equivalent to KruskalSzekeres coords. My (T,R) coords are equivalent to Schwarzschild coords. All of this is discussed in Rindler's book which I referenced in my old post that I previously linked to. 


#57
Dec1109, 04:49 PM

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It's a fudge, George. MTW fudges it by rolling out two different equations which avoid the fact that the line integral becomes complex when r<2M. Taylor and Wheeler do the same. Kruskal tried it with a different coordinate transformation but got the same result. Are you seriously arguing that thousands of professional scientists cannot be wrong and therefore the laws of mathematics can be suspended by popular vote? If the line integral is continuous then "The coordinate transformation which you have given is valid only when [itex]r > 2m[/itex]" and there's another one for [itex]r < 2M[/itex] then the line integral is discontinuous and you are talking about two different universes. Or that I cannot argue that this is mathematically invalid lest I get a from George Jones? 


#58
Dec1109, 05:24 PM

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The relativistic energy as you know is: [itex]E = \gamma mc^2[/itex] so as [itex]r \rightarrow 2M[/itex] then [itex]\gamma[/itex] reaches [itex]\infty[/itex] I do know the difference. I'm not stupid. Neither will they or anything else reach c (which requires infinite energy that the Universe does not have). They will not reach r=1/2, which isn't on their worldline. On the other hand, any infalling particle (or apple) apparently gets infinite energy from falling into a black hole. Every free falling particle reaches r=2M. Or if you like, as the apple approaches the event horizon, the kinetic energy of the apple rises without limit (even past the total energy of the Universe itself). I argue that that points to a fundamental flaw in black hole theory, one that cannot be transformed away. I question those fudges that avoid those questions. I think there IS a transformation of the Schwarzschild Metric which leads to a mathematically consistent noncontradictory solution for the region about a mass which undergoes infinite collapse. But if so, then black holes do not exist. Something else does. 


#59
Dec1109, 06:40 PM

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While we're discussing energy conservation, can someone tell me please tell me if this is true:
1) Outside the horizon there is a timelike Killing vector so energy is conserved for freely falling particles. 2) Inside the horizon there is no timelike Killing vector, so energy is not conserved for freely falling particles. ? 


#60
Dec1209, 04:11 PM

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A very, very large energy can be observed only by someone travelling at very, very high velocity relative to the apple being measured. The source of the high velocity is the rocket motor of the observer who has accelerated to that high velocity. Noone measures the apple's energy as infinite because noone's rocket motors can accelerate the observer to high enough velocity. This applies to both my rocket example and a black hole. When you are close to an event horizon (or an apparent horizon) the natural tendency is to fall into it, and to resist that you need to expend a huge amount of energy. Or to put it another way, there's no single coordinate system in which energy conservation applies but in which the apple's energy changes from finite to infinite as it falls. The infinite value arising in this thread was a limit of energies in lots of different coordinate systems. 


#61
Dec1209, 04:25 PM

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So either the black hole possesses infinite energy or there's something fundamentally wrong with the theory of black holes because they possess such unphysical properties. 


#62
Dec1209, 05:30 PM

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#63
Dec1209, 05:35 PM

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#64
Dec1209, 05:52 PM

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#65
Dec1209, 06:01 PM

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#66
Dec1209, 06:07 PM

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Unless you think that a black hole is massless, it too possesses an energy which is invariant. By the way, have you spotted why the fudge of using two different metrics inside and outside the event horizon is incorrect? Because the square root of a negative number [itex]\neq[/itex] minus the square root of a positive number. They might meet at zero, but they are strangers after that. 


#67
Dec1209, 06:23 PM

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I have no more time to comment today. But I've just time to ask this:



#68
Dec1209, 06:29 PM

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#69
Dec1409, 02:46 PM

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Also your assertion that you can't measure tidal forces inside the event horizon is wrong; on the contrary, the tidal forces get ever stronger the further you fall in. I have to guess that what you really meant was "4momentum is conserved". Conservation means something quite different from invariance. Conserved quantities don't change over time (to put it simply; or more sophisticatedly they satisfy certain differential equations). Invariant quantities take the same value when measured by different coordinate systems. If a particle has a 4momentum P and an observer has a 4velocity U, then the particle's energy relative to the observer is [itex]g(\textbf{P},\textbf{U}) = P_\alpha U^\alpha[/itex]. The same argument used earlier in this thread to show the physical speed approaches c (as the height above the horizon of a hovering observer drops to zero) also shows that the energy, as given by this formula, diverges to infinity. That isn't because the 4momentum becomes infinite, it's because of the change in U. In 4Dgeometrical terms it's because of the "change in angle" between P and U. The 4momentum of a freefalling particle is constant (free falling means no external force, and force is rate of change of momentum). ("Constant" means the covariant derivative along the worldline is zero.) If you choose a family of Us approaching the speed of light (relative to some single inertial observer), then [itex]P_\alpha U^\alpha[/itex] will diverge to infinity even though P remains constant. The 4momentum of the falling particle never changes but the 4velocities of each of the local hovering observers are enormously different as you get very close to the horizon. This is best understood through the analogous Rindler example; the same method applies in both cases. The Rindler example (see my earlier posts) gives explicit formulas for converting between each of the several local (T,R) coordinates and the single global Minkowski (t,x) coordinates. The formulas are different inside and outside the horizon, but the same metric equation applies to all of the different (T,R) coordinate systems. I'm not sure why you think someone is implying that [itex]\sqrt{x} = \sqrt{x}[/itex], which obviously isn't true; you'll have to provide an explicit quote to where you think that's happening. 


#70
Dec1509, 03:22 PM

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a) there is no space to be measured b) all light cones point in the time direction only In other words, any two spacially separated objects would be beyond the light horizon of the other. How to measure the spacial separation of two objects like that? Impossible. According to you, the singularity occurs at s=2M (using the Schwarzschild coordinate system where zero is the EH), whereas the continuous worldline of the Schwarzschild Metric shows the singularity at s=2iM They are not the same place, and your "inner" curve may meet at zero, but is discontinuous with the Schwarzschild solution. 


#71
Dec1509, 04:43 PM

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DiamondGeezer,
By your logic, it's impossible for anyone to travel through the Earth's North Pole because the latitude/longitude coordinate system has a discontinuity there. All the issues you raise can be resolved if you make the effort the understand the analogous set of disjoint (T,R) coordinate systems for the Rindler rocket (follow the links in post #50), which also exhibit discontinuities at the horizon, where R is time and T is space behind the horizon, where the metric equation expressed in these coordinates makes no sense exactly at the horizon. And yet this is just a weird set of coordinate systems that, between them, describe the flat, continuous, gravityfree Minkowski spacetime of special relativity. I've explained this the best I can. If you're not getting it, I'm not sure there's much more I can say. If anyone else is still reading this thread and would like to contribute, maybe they can find some different way of explaining it that will get the point across. 


#72
Dec1709, 03:28 AM

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I'm tired. 


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