<i>Couldn’t I plunge towards a “healthy” black hole (ETA to evaporation: trillions of years), and, a few seconds later by my watch, before reaching the event horizon, receive a message from a distant observer that the hole is gone?</i> The distant observer could not, in this case, know that the hole was gone before you did. You'd both agree the hole had gone before you reached the event horizon.
James R, Agreed. Yet since a longer time than that needed for the hole to evaporate always occurs in the outside universe before I reach the event horizon, I can always receive a message from a distant observer that the hole is gone before I cross. It follows that I will always see the hole evaporate before I cross. This is in conflict with your previous statement: A distant observer sees the hole evaporate before you cross the horizon. You see yourself cross the horizon before the hole evaporates. Can you be more specific as to how the 2nd sentence could be true? It seems that, by the plunger’s clock, the hole evaporates in an arbitrarily short amount of time. The faster I plunge the faster the hole evaporates. If I plunge towards the horizon with an ETA of one hour by my watch, the hole will disappear in less than one hour. Any attempt to cross the horizon is foiled by the evaporation. Likewise a collapsing star evaporates before the surface reaches the event horizon, as observed by anyone. The star collapses not into a singularity, but into oblivion.
James R, Did I stump you or am I on ignore? Please Register or Log in to view the hidden image! I went back to my books for assistance. Alas, they contradict on this issue.
zanket, Sorry, I didn't see your post. <i>Agreed. Yet since a longer time than that needed for the hole to evaporate always occurs in the outside universe before I reach the event horizon, I can always receive a message from a distant observer that the hole is gone before I cross.</i> No. How could the distant observer see the hole evaporate before you see that? There's no way. <i>A distant observer sees the hole evaporate before you cross the horizon. You see yourself cross the horizon before the hole evaporates.</i> Hmm... did I say that? It seems I made a mistake, then. If the distant observer sees the hole evaporate, then either you've already crossed the horizon and you're gone with the hole, or you've hovered outside the horizon and you see the hole collapse before the distant observer sees it collapse. <i>It seems that, by the plunger’s clock, the hole evaporates in an arbitrarily short amount of time.</i> How do you figure that? <i>If I plunge towards the horizon with an ETA of one hour by my watch, the hole will disappear in less than one hour.</i> No. In one hour, by your watch (and that is important, since a distant observer's watch is different), you will cross the horizon and soon hit the singularity. You won't see the hole evaporate.
James R, Sorry, I didn't see your post. No problem. Your input is worth the wait. No. How could the distant observer see the hole evaporate before you see that? There's no way. True, but I didn’t say that. To my comment “It follows that I will always see the hole evaporate before I cross” I might have added “and before the distant observer sees the hole evaporate” to be more specific. I meant to show the former but not deny the latter. If the distant observer sees the hole evaporate, then either you've already crossed the horizon and you're gone with the hole, or ... I disagree. This is along the lines of the contradiction in my books I mentioned. In one chapter they'll say that a distant observer would view an infalling object (even the surface of the collapsing star itself) coast to rest at the event horizon, never crossing, if not for the increasing redshift that makes the object unviewable. In the next chapter they'll say that a distant observer sees an infalling object "wink out" as it “really does” cross the event horizon, although any signals that the object emitted before it crossed may continue to arrive forever with an ever-increasing redshift. It cannot be both ways. The object cannot be observed in two places simultaneously by anyone. The object cannot both "coast to rest at the event horizon" and "really cross the event horizon" as viewed by the same observer, if only they could see it. The first explanation makes the most sense to me, as the second explanation leads to other contradictions, which I can put forth if needed. Moreover, contradictions arise unless, from the perspective of any observer above the event horizon, an infalling object really doesn’t cross the event horizon and never will. So I believe your initial statement that “A distant observer sees the hole evaporate before you cross the horizon.” And if true, it seems that your subsequent statement “You see yourself cross the horizon before the hole evaporates” must be false. How do you figure that? Like this: Suppose you are the distant observer and your clock runs “normally.” From your perspective, an object falling towards the hole approaches a dead stop at the event horizon. It will never reach the event horizon; rather, it will forever approach at an ever-slower pace. From your perspective the object’s clock also slows to approach--but never attain--a dead stop. There is some point above the event horizon where, from your perspective, time runs at “one-gazillionth” the rate of your clock. Suppose an object exists at that point. From the object’s perspective, its clock runs “normally” and your clock runs a gazillion times faster than its clock. Now assume that you the distant observer will observe the hole to evaporate in a gazillion hours. From the object’s perspective, in two proper hours it will observe your clock to have ticked off a gazillion hours (I added one proper hour to allow time for the information about your clock to travel to the object). Since the object is closer to the hole than you are, from the object’s perspective as well the hole must have evaporated. In this way, the closer to the horizon an observer is, the shorter the proper time to evaporation. The plunger can in principle observe the hole to evaporate in an arbitrarily short amount of proper time because the plunger can get to any point above the event horizon in any proper time > 0. Another way to put this: If any observer will see the hole evaporate in x hours, then an observer having a relatively different clock rate will see the hole evaporate in something other than x hours, where the limits are “never” and “approaching instantaneously.” (An example of an observer who never witnesses the evaporation is one who accelerates away from the hole such that the information about the evaporation never reaches him, as described here.) No. In one hour, by your watch (and that is important, since a distant observer's watch is different), you will cross the horizon and soon hit the singularity. Given evaporation, this leads to a contradiction. I will elaborate on my comment that “I can always receive a message from a distant observer that the hole is gone before I cross.” Suppose the distant observer will see the hole evaporate when her clock marks 10<sup>70</sup> years. Before I cross, I will see her clock mark 10<sup>70</sup> years. Before I cross, I will see her clock mark 10<sup>7000</sup> years. The distant observer has plenty of proper time, an infinite amount of time really (because from her perspective I will never cross), to see the evaporation and send me a message before I cross. I have plenty of “distant observer time,” an infinite amount of time really (because from my perspective an eternity of “distant observer time” will transpire by the moment I cross), to receive the message before I cross, no matter how short my proper ETA to the horizon is. Since I always have the opportunity to receive such a message before I cross, and since I would see the hole evaporate before the distant observer does, it follows that there will be no hole when I reach the “horizon,” nor a singularity beyond. Can you see a hole in my logic? Please Register or Log in to view the hidden image!
zanket, You have made me think about this. Thanks for that. I'll jump straight to where I think the problem is, which is here: <i>The distant observer has plenty of proper time, an infinite amount of time really (because from her perspective I will never cross), to see the evaporation and send me a message before I cross. I have plenty of “distant observer time,” an infinite amount of time really (because from my perspective an eternity of “distant observer time” will transpire by the moment I cross), to receive the message before I cross, no matter how short my proper ETA to the horizon is.</i> To keep this to one example, I will consider the situation where you free-fall towards the hole and a distant observer watches (rather than the case where you hover above the event horizon). It is true that in this case, the distant observer will never see you cross the horizon. However, the observer will observe light from you to be red-shifted until you are invisible to him or her. The part I'm not sure about is what happens when the hole evaporates (according to the distant observer). The problem is that to evaporate, the hole gradually loses mass, and therefore the spacetime curvature gradually decreases until the spacetime is flat where the hole used to be. My problem is that I don't know whether the singularity persists all the time, and if it does not then how does it disappear? The devil is in the detail. I <i>suspect</i> that something must happen so that when the distant observer sees the hole as having evaporated, you are no longer there - you have fallen into the hole and been evaporated along with the rest of the mass which fell in. The other half of the story, I think, is easier. From your point of view, you cross the horizon and, a short time later, hit the singularity. The horizon is nothing special as far as you are concerned - it is an invisible line in spacetime. I don't think it is true that you will see the entire future history of the outside universe before you cross the horizon. Certainly it would be a logical contradiction for you to see the evaporation of the black hole when it is clear that you will fall into the hole and hit the singularity in a few fractions of a second of proper time. Therefore, I suspect that you could not receive a message from the outside observer telling you that he has seen the black hole evaporate. These are my thoughts on this at present. I will do a little checking and see if I can solidify my thoughts a bit more.
James R The part I'm not sure about is what happens when the hole evaporates (according to the distant observer The distant observer should see zanket cross the event horizon at the same moment the distant observer sees the black hole vanish. Although the light emitted by zanket could take an arbitrarily, if not infinite amount of time to climb out and reach the distant observer, black hole evaporation takes a finite amount of time, and once the black hole evaporates, nothing is left to stop the light from climbing out and reaching the distant observer. My problem is that I don't know whether the singularity persists all the time, and if it does not then how does it disappear? I've heard theories that the black hole does not completely evaporate and that a small remnant is left over after the last burst of Hawking radiation. I find that hard to believe considering that as the black hole evaporates, its temperature increases.
James R, The part I'm not sure about is what happens when the hole evaporates (according to the distant observer). The problem is that to evaporate, the hole gradually loses mass, and therefore the spacetime curvature gradually decreases until the spacetime is flat where the hole used to be. My problem is that I don't know whether the singularity persists all the time, and if it does not then how does it disappear? The devil is in the detail. Let’s examine the detail. Singularities never exist from the perspective of someone above the event horizon (hereafter a “distant observer” although a nanometer above the event horizon will do). You said, “It is true that in this case, the distant observer will never see you cross the horizon.” It is also true that the distant observer will never see the surface of the collapsing star itself cross the event horizon; the surface of the star remains always above the horizon. The black hole is really a wannabe-hole, ever approaching but never attaining full blackness. The distant observer could prove this to herself by bouncing signals off of a reflector on the star’s surface. She might need billions of years to collect the return signals using equipment beyond our ken, but it shows that the black hole never fully forms from a perspective above the horizon. That the distant observer can see a black hole vanish is an exaggeration for the sake of convenience. She really sees only a wannabe-hole vanish. No singularity ever existed or will exist from her perspective. The moment during the evaporation at which the hole “vanishes” can be defined as the moment the hole loses its grip, which is to say the moment that I may cross the event horizon and be able to cross back. At that moment the event horizon ceases to have significance and becomes simply a place where the event horizon used to be. The spacetime curvature continues to decrease, but that is immaterial to significant predictions along the lines we’re discussing. From the distant observer’s perspective, just before the vanishing moment I am above the event horizon. In the next moment the hole has lost its grip, and she will clock some finite time for me to traverse the remaining distance to where the event horizon used to be. Another way to visualize this: From my perspective plunging towards the event horizon, her clock runs faster than mine, running ever faster the closer I get to the event horizon. At time x on my clock, when I see that her clock is well beyond the moment at which she sees the hole vanish, I could choose to reverse course and then join her. She would tell me that the sequence of events she saw was: the hole vanished, and then I reversed course. Now had I chosen not to reverse course at my time x, there is only one sequence of events she could have seen: the hole vanished, and then I crossed where the event horizon used to be. It follows that I cannot reach the event horizon before the hole vanishes, because whatever the distant observer sees about the hole I must see first, being closer to it. Have the distant observer hold up a giant clock and a giant mirror. Before I cross the event horizon, at some moment of mine I will see a time on her clock that is the moment at which she sees the hole vanish. In the mirror I will see the hole vanish; I will see what she saw at that moment on her clock. And when I turn around to look towards the hole, it can no longer be there. Gravitational time dilation ensures that events a distant observer sees happening to the hole are seen to happen faster from vantages closer to the hole. In the limit, at the event horizon, these events happen instantaneously, which is to say infinitely fast. As I plunge towards the event horizon, the rate of evaporation increases without limit. Because the rate of evaporation needs increase only to some finite level for the hole to lose its grip in any given proper time > 0, my proper ETA to the event horizon is always sufficient time for the hole to lose its grip. In this way the hole’s evaporation foils any attempt to reach the event horizon. From your point of view, you cross the horizon and, a short time later, hit the singularity. The horizon is nothing special as far as you are concerned - it is an invisible line in spacetime. I don't think it is true that you will see the entire future history of the outside universe before you cross the horizon. (boldface mine) Many of my books repeat the boldfaced statements, but it seems they must be false. For now I will ignore the logic above that refutes the ability to cross the horizon at all. I will talk only about what the theory says or strongly implies even if it contradicts the theory itself. The gravitational time dilation (redshift) formula shows that I must indeed observe the entire hereafter of the outside universe by the moment I reach the event horizon. One book puts it this way: “The observer on the surface of the collapsing star would see time go on quite normally, but in the brief moments prior to the severing of his universe from the outside world, he would see all of eternity in that world pass by.” All of my books explicitly say that, from the perspective of someone above the event horizon, time is stopped at the event horizon. More specifically: the surface of the collapsing star that ever-more-slowly falls towards but never reaches the event horizon is falling towards a place where--if only the surface ever reached it--time would be stopped. Only some of my books mention that the opposite perspective must be true: When I reach the event horizon I must see time in the outside universe passing at an infinite rate, an infinite blueshift effectively “wringing out” all the remaining time in the outside universe. None of my books mention what else must be true and which is strongly implied by the formula: By the moment I reach the event horizon I also wring out all the space--the distance--in the outside universe. All the spacetime gets wrung out, not just the time. Until I reach the event horizon, the hole is really a wannabe-hole and all the spacetime that the hole occupies is part of the outside universe. As I approach the event horizon, spacetime gets increasingly wrung out. When I reach the event horizon, all spacetime collapses to a point, including that spacetime which the hole formerly occupied. From my perspective only, and only when I reach the event horizon, the event horizon is where the singularity is, not at the center of the hole, because all spacetime collapses to a point at the point on the event horizon that I reach. Having no remaining spacetime, I cannot continue on to plunge into a singularity at the center of the hole. Because the singularity at the center of the hole never exists from anyone’s perspective, it becomes an artifact and may be dismissed as a misinterpretation. If you doubt this, ask yourself: How could I otherwise witness all of a distant star’s remaining time by the moment I reach the horizon? Embedded in the star’s light is information about the star’s clock (the redshift/blueshift). All the light or information the star outputs in the remainder of eternity must reach me by the moment I reach the event horizon, or else some time will remain in the outside universe. If there remains any space between the star and myself when I reach the event horizon, there will be a transmission delay because information takes time to traverse space. Some time would then remain in the outside universe when I reach the event horizon. The answer is that the distance between the star and myself must reduce to zero by the moment I reach the event horizon. Let’s come back now to include evaporation into the picture. Here are a few of my thoughts as to how, if evaporation prevents me from reaching the event horizon, not only is the theory more logical but it makes more intuitive sense as well. The theory shows that I will see all eternity in the outside universe transpire by the moment I reach the event horizon. I further showed that all spacetime must collapse by then. This would seem to require infinite mass-energy on the part of the hole. Not only is a finite mass sufficient to collapse the entire universe to a single point from my perspective, it is also sufficient to sever me from the outside universe, drop me into a new universe, and allow me to exist after the hereafter, at least until I witness all spacetime collapse again when I reach the center of the hole. With evaporation preventing me from reaching the event horizon, my attempt to reach the event horizon only throws me into the future, so to speak, analogous to riding an elevator down to the ground floor to partake of the relatively slower time there. Spacetime does not collapse to a single point. No new universes are created. Nothing becomes infinite in a finite time. According to the theory, holes having larger masses affect me significantly differently only in that they prolong my proper time to reach the singularity after I cross the event horizon. With evaporation preventing me from reaching the horizon, the larger the mass of the hole the further I am thrown into the future, which I find more intuitively satisfying. Certainly it would be a logical contradiction for you to see the evaporation of the black hole when it is clear that you will fall into the hole and hit the singularity in a few fractions of a second of proper time. That is clear to me only from the standpoint that it is predicted by the theory or stated in books. I have offered here counter-reasoning so simple that it would seem to need refutation if the theory is to be satisfactory. I have resolved the logical contradiction by fixing the widely accepted interpretation of the gravitational time dilation formula. Can you find a flaw in my reasoning per se? BTW, thanks for keeping up with me on this. I’ve been thinking about it for months, and it’s nice to get some feedback.
zanket, I've discussed this question with a friend and I think I have it worked out now. First, the explanation. Again, I will restrict myself to the situation where somebody free-falls into a black hole. Call the distant observer (a long way from the hole) A and the falling person B, for convenience. Let's suppose that B sends out one photon per second to A as he falls (by his own clock). Then A sees these photons arrive at a slower and slower rate. There is a "last photon" which B emits, just before he reaches the event horizon. A will eventually see that photon. All photons emitted by B after he crosses the horizon are lost to the hole. The photon right at the horizon is "frozen" at the horizon. So, as A watches B, A sees B approaching the horizon at a slower and slower rate (according to A's clocks). Eventually, the photons from B are so few and far between that B is approximately invisible to A. After a long time, A watches the hole evaporate. As it does that, the event horizon shrinks bit by bit and the hole emits Hawking photons. However, none of these photons is identifiable as having originated with B, because the hole itself destroys that information. When the hole stops evaporating, B is gone, along with his last photon. Note: as the hole evaporates, photons from B gradually start arriving at A at a faster rate than before, but only for a while, until B's last photon is received by A. Now consider B's point of view. This time, let's say A is broadcasting photons at B at a rate of 1 per second according to A's clocks. Then as B approaches the horizon he sees photons arriving more and more frequently from A. However, importantly, the rate of arrival of photons from A never goes to infinity, even as B crosses the horizon. Hence, it is not true to say that B sees the entire future of the universe before he crosses the horizon. In fact, B only sees some of that "future", and he certainly can't be informed by A about the evaporation of the hole, because by that time he is long gone. The event horizon for A is a kind of singularity, but only because of the reference frame of A. Change the frame of reference and all the "singular" characteristics of the horizon vanish. That is why the event horizon is often called a mathematical singularity rather than a physical singularity. For B the event horizon is nothing particular special. Time for B still passes at a normal rate. There is no infinite blue shift or infinite time dilation for B, even though A sees infinite red-shift and infinite time dilation. So, these parts of your post are incorrect: <i>From my perspective plunging towards the event horizon, her clock runs faster than mine, running ever faster the closer I get to the event horizon. At time x on my clock, when I see that her clock is well beyond the moment at which she sees the hole vanish, I could choose to reverse course and then join her.</i> As explained, you will not see the time on her clock when the hole vanishes. <i>It follows that I cannot reach the event horizon before the hole vanishes, because whatever the distant observer sees about the hole I must see first, being closer to it.</i> Hopefully you can see why this is false. You cross the horizon. <i>Gravitational time dilation ensures that events a distant observer sees happening to the hole are seen to happen faster from vantages closer to the hole. In the limit, at the event horizon, these events happen instantaneously, which is to say infinitely fast.</i> Again, not true. The dilation is not infinite for an observer at the horizon. <i>The gravitational time dilation (redshift) formula shows that I must indeed observe the entire hereafter of the outside universe by the moment I reach the event horizon. One book puts it this way: “The observer on the surface of the collapsing star would see time go on quite normally, but in the brief moments prior to the severing of his universe from the outside world, he would see all of eternity in that world pass by.”</i> It seems the book is incorrect on this point. <i>All of my books explicitly say that, from the perspective of someone above the event horizon, time is stopped at the event horizon.</i> That is true, in a manner of speaking. Think of the broadcasting of photons; it makes it clearer. <i>Only some of my books mention that the opposite perspective must be true: When I reach the event horizon I must see time in the outside universe passing at an infinite rate, an infinite blueshift effectively “wringing out” all the remaining time in the outside universe.</i> Those books are wrong, it seems. (Which books, by the way?)
James R, First of all, I appreciate that you answered so completely and in layman’s terms. It has been surprisingly difficult to find good explanations like that. I think it is just what my more detailed books are getting at. I understand your explanation in full. If I can agree with the first few paragraphs then I’d agree with it all. A piece of it still doesn’t smell right, but I’ll have to think about it some more (that’s the fun part). I’ll let you know. Those books are wrong, it seems. (Which books, by the way) These two at least: Understanding Einstein's Theories of Relativity contains the quote about “all of eternity” transpiring in the outside universe by the moment the event horizon is reached. Time: A Traveler’s Guide has this formula: t<sub>2</sub> = t<sub>1</sub> / sqrt(1 – C<sub>h</sub>/C<sup>2</sup>) Accompanied by this explanation: “Here t<sub>1</sub> is the elapsed time you experience when you hover close to the hole as compared with the elapsed time t<sub>2</sub> far from the hole. C<sub>h</sub> is the circumference of the black hole’s event horizon. C is the circumference around the hole at which you’d hover, slightly above the event horizon.” (Actually the quote has t<sub>1</sub> and t<sub>2</sub> erroneously reversed, but the mistake is clear from other context.) You can see that as I hover closer to the hole, the distant observer’s elapsed time approaches infinity. Maybe the difference between hovering and freefalling radically alters the picture? I don’t see how. Do you think the formula is compatible with “B’s point of view” in your post?
The gotcha, is that for B to stay at the event horizon then from B's point of view, B would have to be travelling at C, hence blue shift to infinity and see all time. In practice B would be falling in and traveling at less than C from B's point of view. Aint English and maths fun ? - C is the speed of light Please Register or Log in to view the hidden image!
James R, I gave some thought to your explanation. It seems to be missing a crucial piece of info. You said “As A watches B, A sees B approaching the horizon at a slower and slower rate (according to A's clocks).” Given the context, this is due to the increasing delay of the arrival of B’s photons at A (hereafter just “increasing delay”). Due to this effect, A sees the rate of B’s clock get slower and slower. However (this is the missing piece), A also sees that the rate of B’s clock gets slower and slower due to relativistic time dilation. This effect is additive to the increasing delay effect and has the same magnitude, so the total time dilation that A sees about B is double what either effect alone would show. (Perhaps the doubling is reflected by the 2 in the gravitational redshift formula.) Another way to put this: As B approaches the hole, B’s velocity relative to A ever increases, so from A’s perspective B’s clock gets ever slower from that fact alone. That gravity increasingly delays the receipt of B’s photons by A is an additional effect. If B held up a giant clock, A would see images of an ever slowing clock arriving ever slower. In your explanation, from A’s perspective B “really does” cross the horizon, although A never sees that due to the increasing delay. When A sees the hole evaporate, B “really is” gone along with the hole. (quotes are mine) When relativistic time dilation is included, however, the situation is altered. From A’s perspective: B “really does” approach the horizon slower and slower and “really never does” reach the horizon. When the hole evaporates, B “really does” remain above the horizon. Even if B could reach the horizon, B’s clock “really would” be stopped there and B “really would” move no further. It seems that my previous explanation accounted only for relativistic time dilation, whereas yours accounts only for the increasing delay. The former effect, though, is a crucial one. It means that B doesn’t evaporate along with the hole. It means that the complementary perspective of A is seen by B, which is that an eternity transpires at A by the moment the horizon is reached. And that puts us back at square one. [Edited to add:] Thinking about it some more, it seems either effect alone “really does” slow B’s progress towards the horizon from A’s perspective, and that the combined effect is necessary to prevent B from “really” reaching the horizon from A’s perspective. The gravitational redshift formula accounts for both effects I think. And all of the time dilation indicated by the formula is a real effect, meaning that if A applies the formula regarding B and finds that the redshift is 50%, then from A’s perspective B “really is” moving at half of the velocity that would otherwise be expected (also the rate of B’s biological processes, the speed of B’s light, etc. is halved). This has been observed in experiments of course, where clocks elapsed less time than counterparts as given by the formula. If I were wrong in this post it seems that B would still find the horizon to be something special. Forget what I said in this post for now. You said “All photons emitted by B after he crosses the horizon are lost to the hole. The photon right at the horizon is ‘frozen’ at the horizon.” The collapsing star also leaves photons frozen at the horizon. These photons compose an image of the surface of the star when it fell through the horizon. Now imagine B plunging toward the horizon. When B reaches the horizon, he will see himself “hit” the “surface” of the star and punch through it. (We wade through images every waking moment, but, if no mirrors or holograms are involved, when we encounter an image of something located at our current location it means we are touching it.) At the risk of rambling (too late I guess), after re-reading your post I want to elaborate upon the difference between what you and I are saying, because the difference is subtle. Me: All of my books explicitly say that, from the perspective of someone above the event horizon, time is stopped at the event horizon. You: That is true, in a manner of speaking. Think of the broadcasting of photons; it makes it clearer. I’m saying that is true, period. From the perspective of someone above the event horizon, anything at the event horizon is immobilized. It isn’t just a net effect of, for example, photons moving outward at c against gravity flowing inward at c. Given your explanation, from A’s perspective B crosses the horizon before the hole evaporates. So A could bounce signals off B’s ship for only a limited proper time. When the signal doesn’t return when expected, A would know that B crossed even though A can still see B’s ship above the horizon. Given my explanation, A can bounce signals off B’s ship for an unlimited proper time; that is, for ever. The signal would always return eventually, indicating to A that B remains above the horizon. Given your explanation, the horizon exists for both A and B; it just can’t be seen by A. Given my explanation, the horizon exists for nobody; the surface of the collapsing star “really” never reaches the horizon. The “proof” that my explanation is the correct one is that a clock at the top of a building elapses more time than its ground counterpart. That is a “real” effect, as opposed to “in a manner of speaking.” The reality of the effect means that what the top clock “observes” about the ground clock’s rate, the reciprocal is observed by the ground clock about the top clock. When B is just a bit above the horizon according to A, she observes B’s rate of time as one-gazillionth the proper rate. When B finds himself the same bit above the horizon, he observe’s A’s rate of time as a gazillion times the proper rate. In the limit, by the moment B reaches the horizon, B observes an eternity transpire at A.
zanket, <i>However (this is the missing piece), A also sees that the rate of B’s clock gets slower and slower due to relativistic time dilation. This effect is additive to the increasing delay effect and has the same magnitude, so the total time dilation that A sees about B is double what either effect alone would show.</i> General relativity already takes both effects into account - gravitational time dilation and time dilation due to relative motion. Remember that special relativity is a subset of general rel. <i>Another way to put this: As B approaches the hole, B’s velocity relative to A ever increases, so from A’s perspective B’s clock gets ever slower from that fact alone. That gravity increasingly delays the receipt of B’s photons by A is an additional effect.</i> Not really, though you might choose to look at it that way. GR lumps both effects together, effectively. <i>When relativistic time dilation is included, however, the situation is altered. From A’s perspective: B “really does” approach the horizon slower and slower and “really never does” reach the horizon.</i> It depends what you mean by "really does". Reality is different for each observer, and I stick by what I said before. From B's point of view, B really does cross the horizon. From A's point of view, A can never see B cross the horizon. Whether that means that B "really does" cross from A's point of view is arguable. You can really only judge the end result - when the hole evaporates, B is no longer there. <i>When the hole evaporates, B “really does” remain above the horizon. Even if B could reach the horizon, B’s clock “really would” be stopped there and B “really would” move no further.</i> I disagree with the first sentence. As for the second, you must specify a reference frame. The second sentence is true from A's point of view, but not B's. <i>The gravitational redshift formula accounts for both effects I think. And all of the time dilation indicated by the formula is a real effect, meaning that if A applies the formula regarding B and finds that the redshift is 50%, then from A’s perspective B “really is” moving at half of the velocity that would otherwise be expected (also the rate of B’s biological processes, the speed of B’s light, etc. is halved). This has been observed in experiments of course, where clocks elapsed less time than counterparts as given by the formula.</i> Bringing velocity into this somewhat clouds the issue. In the curved-spacetime picture, B's 4-velocity is constant since he is in free fall. Whether A judges the velocity to be constant or increasing depends on how A is measuring it (i.e. taking into account spacetime curvature or not). <i>If I were wrong in this post it seems that B would still find the horizon to be something special. Forget what I said in this post for now. You said “All photons emitted by B after he crosses the horizon are lost to the hole. The photon right at the horizon is ‘frozen’ at the horizon.” The collapsing star also leaves photons frozen at the horizon. These photons compose an image of the surface of the star when it fell through the horizon. Now imagine B plunging toward the horizon. When B reaches the horizon, he will see himself “hit” the “surface” of the star and punch through it.</i> Photons at the horizon are invisible to somebody outside the horizon, because they cannot propagate outwards. At the horizon, B would see nothing. No photons could propagate outwards to him. Also, for B, the surrounding space at the horizon would seem quite normal - until he tried to turn around and go the other way. <i>All of my books explicitly say that, from the perspective of someone above the event horizon, time is stopped at the event horizon. ... I’m saying that is true, period. From the perspective of someone above the event horizon, anything at the event horizon is immobilized.</i> I'm not sure if that's the case. I think that the frozen time thing only applies technically to an observer at an infinite distance from the hole. In a real situation, the closer you get to the horizon, the more "normal" time will seem at the horizon. <i>Given your explanation, from A’s perspective B crosses the horizon before the hole evaporates. So A could bounce signals off B’s ship for only a limited proper time. When the signal doesn’t return when expected, A would know that B crossed even though A can still see B’s ship above the horizon.</i> No. A can send signals down to B, but they will bounce back with ever-increasing delays. The same applies to signals emitted by B towards A; the mirror thing makes no fundamental difference. A will never be able to see B's ship and at the same time receive no bounced signals from B. <i>Given my explanation, A can bounce signals off B’s ship for an unlimited proper time; that is, for ever. The signal would always return eventually, indicating to A that B remains above the horizon.</i> I don't think so. If A fires one signal per second at B (according to A's clock), then those signals will arrive back with ever-increasing delays (and greatly red-shifted). There will be a "last signal" bounced back from B, just as there is a last signal emitted by B.
First I must admit I'm not an expect on relativity. But to get back to the original topic... If sound is propagated by random motion and collision of particles, then wouldn't these particles have to go faster than the speed of light in order to propogate the sound? If the speed of sound is approximately 1000 ft/sec in air that means that the air particles are actually moving much faster than that. So wouldn't it make sense, that in order for another observer to hear a sound created by observer 1, the air particles had to move faster than the speed of light? Would this create problems with the sound arriving before it left? edit: I just realized that it might make a difference if one one traveling through stationary air, or if the air was traveling with the traveler at the speed of light. How does this affect things?
James R, General relativity already takes both effects into account - gravitational time dilation and time dilation due to relative motion. Are these the same two effects that took Einstein some years to realize must both be considered to account for, for example, the deflection of a ray of light as it grazes the sun? I can’t be sure from the context in my books. From A's point of view, A can never see B cross the horizon. Whether that means that B "really does" cross from A's point of view is arguable. You can really only judge the end result - when the hole evaporates, B is no longer there. Confusing! I don’t see how it is arguable. It seems factual, a yes or no question like “is your arm broken?” From A’s point of view, if B is no longer above the horizon when the hole evaporates, then how can it be argued that B really didn’t cross? More on my confusion below. As for the second [sentence], you must specify a reference frame. The second sentence is true from A's point of view, but not B's. I tried to take a shortcut on specifying the reference frame, applying “From A’s perspective:” to the remaining sentences in the paragraph. Didn’t work. Bringing velocity into this somewhat clouds the issue. In the curved-spacetime picture, B's 4-velocity is constant since he is in free fall. Whether A judges the velocity to be constant or increasing depends on how A is measuring it (i.e. taking into account spacetime curvature or not). I’m talking about the layman’s idea of velocity, that which does not take spacetime curvature into account. It seems to me that the question as to whether the event horizon is a mathematical or physical singularity can be answered by answering “Does a clock on the ground elapse less time than a clock on a tower?” If yes, then the singularity is physical; otherwise, it’s mathematical. Experiments have shown the answer is yes. The clock on the ground ages slower, and that’s a physical difference. From A’s point of view, when B is close to the horizon, in every physical way B approaches the horizon ever slower, because B ages ever slower. At the limit, the horizon, B would no longer be aging and could move no further. But, from A’s point of view, because B’s aging slows at an increasing rate with the horizon as a limit, B never physically attains the horizon. A doesn’t have to see that to know it; it can be deduced from the gravitational redshift formula (more on this below). At the horizon, B would see nothing. No photons could propagate outwards to him. I don’t understand how B would see nothing. The photons I’m referring to don’t need to propagate outwards to him; they are those left at the horizon by the collapsing star. The star collapses through the horizon whilst emitting photons from its surface. The photons emitted from the surface at the moment the surface reaches the horizon stay perpetually at the horizon. (From the photons’ point of view, the surface falls away at c.) Why wouldn’t B see these photons when B reaches the horizon and encounters these photons there? I'm not sure if that's the case. I think that the frozen time thing only applies technically to an observer at an infinite distance from the hole. Wish me luck on the terminology to follow. The gravitational redshift formula tells you the percentage by which light emitted at the horizon is redshifted when received at the specified radius. Alternatively the formula tells you the rate of a clock at the specified radius, as a percentage of the rate of a clock at an infinite distance. At the horizon, the formula returns 0%. At an infinite distance, the formula returns 100%. (In support of the alternate usage of the formula, I offer this quote from Exploring Black Holes, regarding the difference in rates between a clock on a tower and a clock on the ground: “Do the intrinsic rates of the emitter and receiver or of the clock change, or is it the light signal that changes frequency during its flight? The answer is that it doesn’t matter. Both descriptions are physically equivalent. Put differently, there is no operational way to distinguish between the two descriptions.”) To find the factor by which light is redshifted as it climbs from a lower to a higher radius, you divide the result of the formula for the higher radius by the same for the lower radius. So to find the factor by which light is redshifted as it rises from the horizon to any higher radius, you divide by zero, making the result indeterminate (didn’t we just have this discussion with ProCop?). The reason the result is indeterminate is because time is stopped at the horizon, as given by the alternate usage of the formula. The light from there goes nowhere from the perspective of anyone at a higher radius. The light never reaches a higher radius. The redshift factor is not infinite, it is “does not apply.” And it means that the frozen time thing applies to observers at any distance above the horizon. (Notice that I contradicted myself. I said “... light emitted at the horizon is redshifted ...” and then I said that the light from the horizon goes nowhere. I may have done the best I can terminology-wise considering that the horizon is a physical limit in the calculus sense. Nobody sees the collapsing star’s surface reach the horizon. The formula remains applicable for radii above the horizon.) In a real situation, the closer you get to the horizon, the more "normal" time will seem at the horizon. As shown above, the closer you get to the horizon, the more “normal” time will seem at lower radii above the horizon, but time remains stopped at the horizon. A will never be able to see B's ship and at the same time receive no bounced signals from B. If A fires one signal per second at B (according to A's clock), then those signals will arrive back with ever-increasing delays (and greatly red-shifted). There will be a "last signal" bounced back from B, just as there is a last signal emitted by B. I translate the first statement to: A will always receive bounced signals from B when A is able to see B’s ship. But why wouldn’t that “last signal” occur at A’s forever? Consider: B can emit photons arbitrarily close to the horizon. The closer to the horizon, the longer they’ll take to arrive at A from her point of view. They could take 10<sup>70</sup> years or 10<sup>7000</sup> years to arrive, there’s no limit. And if A will always receive bounced signals from B when A is able to see B’s ship, then from A’s point of view, B never crosses, right? After re-reading your posts, I realize I may not understand your explanation like I thought. You said: “Now consider B's point of view. This time, let's say A is broadcasting photons at B at a rate of 1 per second according to A's clocks. Then as B approaches the horizon he sees photons arriving more and more frequently from A.” What is the reason for this blueshift in your explanation? The reason in my explanation is that the relative slowdown of time closer to the hole makes the photons relatively decelerate and bunch closer together (like the distance between the cars of an decelerating motorcade decreases), so to speak, from either A or B’s perspective. Let me emphasize that I want to understand what happens, not prove my point. I do realize that what you say matches my most detailed books (like the one linked to above), but it doesn't yet make sense to me and I can’t accept it until it does.
empennage, As I understand it, the particles that transmit sound need not go anywhere. They pass the “information” from one particle to another at the speed of sound. Picture a ripple on a pond. The water particles stay approximately in place, but the ripple moves. If you are in a spaceship, the speed of the ship is irrelevant as to whether or not you can hear the other passengers, because the ship is stationary relative to you. If a spaceship passes you at close to the speed of light, its passengers could still hear each other, because the ship is stationary relative to them. If you added the speed of the ship plus the speed of the sound within the ship, it would always be less than the speed of light, as given by the addition of velocities formula, which applies to situations involving velocities close to the speed of light.
zanket, Me: General relativity already takes both effects into account - gravitational time dilation and time dilation due to relative motion. You: <i>Are these the same two effects that took Einstein some years to realize must both be considered to account for, for example, the deflection of a ray of light as it grazes the sun? I can’t be sure from the context in my books.</i> Special relativity was built into general relativity from the beginning. Use the GR equations and you automatically account for both effects. <i>Confusing! I don’t see how it is arguable. It seems factual, a yes or no question like “is your arm broken?” From A’s point of view, if B is no longer above the horizon when the hole evaporates, then how can it be argued that B really didn’t cross?</i> It's just a matter of reference frame. From B's point of view, B certainly crossed. From A's point of view, A did not see B cross. Does that mean that from A's point of view B did not cross? It's a question of interpretation. <i>It seems to me that the question as to whether the event horizon is a mathematical or physical singularity can be answered by answering “Does a clock on the ground elapse less time than a clock on a tower?” If yes, then the singularity is physical; otherwise, it’s mathematical. Experiments have shown the answer is yes.</i> I am aware of the experiments. An event horizon is somewhat different. There is a simple test which can be used to determine if a singularity in spacetime is physical or mathematical (i.e. real or "removable"). We simply ask: is it possible that a change to a different frame of reference will show no singularity? In the case of the event horizon, the answer is "yes"; the event horizon singularity is removed by transforming to the freefall frame. Compare the central singularity of a black hole - that is a physical singularity. <i>From A’s point of view, when B is close to the horizon, in every physical way B approaches the horizon ever slower, because B ages ever slower.</i> I might well ask - how does A know B is "really" going slower? All A has to go on are signals received from B. So again, what is "real" for A is not necessarily real for B. <i>I don’t understand how B would see nothing [at the horizon]. The photons I’m referring to don’t need to propagate outwards to him; they are those left at the horizon by the collapsing star. The star collapses through the horizon whilst emitting photons from its surface. The photons emitted from the surface at the moment the surface reaches the horizon stay perpetually at the horizon.</i> This is difficult. A photon travelling radially outwards at the horizon would be "frozen", as you say. A photon travelling in any other direction at the horizon would fall into the hole. So, as B fell through the horizon, he might see some photons - but only the ones travelling outwards exactly tangentially. So, I guess I agree with you on this point. Quote: <i>“Do the intrinsic rates of the emitter and receiver or of the clock change, or is it the light signal that changes frequency during its flight? The answer is that it doesn’t matter. Both descriptions are physically equivalent. Put differently, there is no operational way to distinguish between the two descriptions.”</i> I agree with this. <i>To find the factor by which light is redshifted as it climbs from a lower to a higher radius, you divide the result of the formula for the higher radius by the same for the lower radius. So to find the factor by which light is redshifted as it rises from the horizon to any higher radius, you divide by zero, making the result indeterminate (didn’t we just have this discussion with ProCop?).</i> Agreed. Another way to say this is that no photons can move outwards from the horizon, as you say. Saying that they are "infinitely redshifted" is a similar way to say essentially the same thing. <i>As shown above, the closer you get to the horizon, the more “normal” time will seem at lower radii above the horizon, but time remains stopped at the horizon.</i> Yeah, ok. I prefer to think in terms of the photons; it makes it easier. <i>I translate the first statement to: A will always receive bounced signals from B when A is able to see B’s ship.</i> Yes. <i>But why wouldn’t that “last signal” occur at A’s forever? Consider: B can emit photons arbitrarily close to the horizon. The closer to the horizon, the longer they’ll take to arrive at A from her point of view. They could take 10<sup>70</sup> years or 10<sup>7000</sup> years to arrive, there’s no limit.</i> Yes, but going back to your original question, the important thing is that they <b>must</b> arrive at A before A sees the hole evaporate. <i>And if A will always receive bounced signals from B when A is able to see B’s ship, then from A’s point of view, B never crosses, right?</i> The sequence of events is: 1. A receives B's second last photon. 2. The event horizon shrinks even more. 3. A receives B's last photon. 4. A sees the hole evaporate. 5. A receives no more photons from B (who is gone). <i>What is the reason for this blueshift in your explanation? The reason in my explanation is that the relative slowdown of time closer to the hole makes the photons relatively decelerate and bunch closer together...</i> As I said, I don't like thinking of this in terms of velocities. As far as B is concerned, all the photons arriving from A arrive at the usual speed of light (as measured by B). However, their frequencies are shifted due to the gravitational blueshift created by the curved spacetime near the hole. <i>Let me emphasize that I want to understand what happens, not prove my point.</i> That's ok. You're asking good questions. I'm not claiming to have all the answers either. I'm not setting myself up as an expert here; other people know more about this stuff than me. I might be wrong. I'm just telling you what I think, given the extent of my knowledge on these matters. If you find out more from a better authority, please let me know. I'm here to learn just like everybody else.
James R, There is a simple test which can be used to determine if a singularity in spacetime is physical or mathematical (i.e. real or "removable"). We simply ask: is it possible that a change to a different frame of reference will show no singularity? In the case of the event horizon, the answer is "yes"; the event horizon singularity is removed by transforming to the freefall frame. That’s what I wish to understand. I can easily see how B plunges “normally” toward the horizon, unlike what A sees about B. What I don’t grasp is how B avoids a physical singularity at the horizon; that is, an infinite blueshift and the concomitant collapse of spacetime. More on this below. I might well ask - how does A know B is "really" going slower? All A has to go on are signals received from B. So again, what is "real" for A is not necessarily real for B. A can deduce it from the gravitational redshift formula. A knows that a clock at a lower radius ages slower, as predicted by the formula, and as can be physically verified. A accepts B’s slow time as reality just like A accepts the reality of the Andromeda galaxy, about which A has only received photons to go on. I agree that B has his own reality. Another way to say this is that no photons can move outwards from the horizon, as you say. Saying that they are "infinitely redshifted" is a similar way to say essentially the same thing. Isn’t that like saying 4/0 = infinity? Yes, but going back to your original question, the important thing is that [the last photon from B] <b>must</b> arrive at A before A sees the hole evaporate. Then A cannot see the hole evaporate, in defiance of Hawking. As shown, in the limit A receives B’s last photon at A’s forever. The sequence of events becomes: 1. A receives a photon from B. 2. The event horizon shrinks even more. 3. Go to step 1. As I said, I don't like thinking of this in terms of velocities. As far as B is concerned, all the photons arriving from A arrive at the usual speed of light (as measured by B). However, their frequencies are shifted due to the gravitational blueshift created by the curved spacetime near the hole. OK, let’s examine the blueshift per se. Unfortunately my books don’t have the formulas, but I can deduce them given simple logic I’ll provide if requested. The blueshift factor formula must be: To find the factor by which light is blueshifted as it falls from a higher to a lower radius, you divide the result of the gravitational redshift formula for the higher radius by the same for the lower radius. So to find the factor by which light is blueshifted as it falls from any higher radius to the horizon, you divide by zero, making the result indeterminate. As B plunges toward the horizon, the blueshift from every higher radius approaches infinity. At the horizon the blueshift doesn’t apply because an infinite blueshift has occurred by that point. Although B can plunge “normally” toward the horizon, the infinite blueshift requires the collapse of all spacetime onto B by the moment B reaches the horizon, as I previously described. When B reaches the horizon, no more space exists whence light may blueshift. An infinite blueshift by the moment the horizon is reached means the horizon is a physical singularity. Can you show that B would not experience an infinite blueshift by the moment the horizon is reached? Or give me a reference to look up? My books artfully dodge the issue. [Edited to add:] Hey I found a site that addresses evaporation and the blueshift: What happens to you if you fall into a black hole? And other questions.
James R, After reading the site linked to above, I may better understand your question “how does A know B is ‘really’ going slower?” I said it can be physically verified, as when two clocks once separated by altitude are compared side-by-side. The site’s author points out the distinction: “If I then return home [from hovering close to the horizon], I'll have aged less than you. In this case [as opposed to the case of A and B], general relativity can say something about the difference in proper time experienced by the two of us, because our ages can be compared locally at the start and end of the journey.” The author implies that the aging difference is real only if A and B are compared locally. If compared from afar, the aging difference is an optical illusion (that seems to be appropriate terminology). Because the comparison is from afar, from A’s point of view B really does cross the horizon. A sees the illusion that B doesn’t cross. But if that’s true, then Mercury’s perihelion advance (that anomalous part of it addressed by Einstein) must also be an illusion. The advance is predicted by the same effects that are accounted for by the gravitational redshift formula. If Mercury isn’t really slowed when it approaches the sun, from our perspective, then the slowdown is an illusion. And if the perihelion advance is illusory then the perihelion never really advanced. So if we visit Mercury, as we approach it we should see all the perihelion advances that ever happened to Mercury from our perspective be rolled back. This is analogous to A visiting B. You and the author would, I think, say that A will find that B is gone, assuming that A waited long enough before visiting. I say that A will always find that B remains above the horizon, no matter how close A gets to the horizon, for the reason that I showed where no matter how close you get to the horizon, time remains stopped at the horizon. What do you think?
zanket, I agree with the explanation you linked to above. I said: <i>Yes, but going back to your original question, the important thing is that [the last photon from B] must arrive at A before A sees the hole evaporate.</i> You replied: <i>Then A cannot see the hole evaporate, in defiance of Hawking. As shown, in the limit A receives B’s last photon at A’s forever. The sequence of events becomes: 1. A receives a photon from B. 2. The event horizon shrinks even more. 3. Go to step 1.</i> That sequence would only apply if B emitted an infinite number of photons before crossing the horizon. He doesn't. B can count how many photons he emits, and A must agree with that number. Therefore, there is a "last photon" which it is possible for A to receive. A actually receives that photon prior to seeing the hole evaporate. <i>OK, let’s examine the blueshift per se. Unfortunately my books don’t have the formulas, but I can deduce them given simple logic I’ll provide if requested. The blueshift factor formula must be: To find the factor by which light is blueshifted as it falls from a higher to a lower radius, you divide the result of the gravitational redshift formula for the higher radius by the same for the lower radius.</i> I'm not sure that this is correct. <i>Can you show that B would not experience an infinite blueshift by the moment the horizon is reached?</i> Maybe, but I'll have to think about it some more. <i>The author implies that the aging difference is real only if A and B are compared locally.</i> Yes, that's true in GR whenever you compare two clocks. <i>If compared from afar, the aging difference is an optical illusion (that seems to be appropriate terminology). Because the comparison is from afar, from A’s point of view B really does cross the horizon. A sees the illusion that B doesn’t cross.</i> That's more or less correct, but I don't like calling it an illusion. What A sees is real <b>for A</b>. He has no way of determining any other reality except by applying known laws of physics (namely, relativity) to calculate what B would observe. And B's reality is no more privileged than A's. <i>But if that’s true, then Mercury’s perihelion advance (that anomalous part of it addressed by Einstein) must also be an illusion.</i> No. The perihelion advance can be thought of as resulting from the curvature of space near the sun. Every observer agrees on what that curvature is, and therefore all agree on the perihelion advance observed. <i>This is analogous to A visiting B. You and the author would, I think, say that A will find that B is gone, assuming that A waited long enough before visiting.</i> Yes. <i>I say that A will always find that B remains above the horizon, no matter how close A gets to the horizon, for the reason that I showed where no matter how close you get to the horizon, time remains stopped at the horizon.</i> I disagree. The linked article again clarifies this a little. B would not see the surface of the collapsed star as he crossed the horizon, because that surface crosses the horizon at a different point in spacetime, and that point in spacetime is never a part of B's past light cone. Similarly, in our example, when A gets to a point where B was at one point in time, B will no longer be there, since B has moved in space since then.
