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Thread: How are black holes able to move?

  1. #1 How are black holes able to move? 
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    GR predicts that time would be observed to stop by at the event horizon of a black hole from the perspective of any outside observer, so doesn't that imply that a black hole must stay put (by which I mean be observed to stay put by observers outside the event horizon) forever in the same spot after it is formed?

    I suppose galaxies with black holes at their center can still orbit each other in a sense. Maybe the black holes perceive themselves to still be orbiting each other (since they perceive time to be progressing normally at the event horizon), but they're actually not moving at all from our perspective (because we're not at the event horizon)?


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    Since motion is relative, how do you distinguish between the BH moving relative to you and you moving relative to the black hole? For something to stay put forever in the same spot, implies that there is an absolute frame of rest for it to stay put with respect to. Relativity disavows such a reference frame.


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    How does one interpret gravitational time dilation, then? I only have to concern myself with whether objects in the event horizon are moving relative to one another then? If they're all moving relative to me as a group (but remaining in the same position relative to one another) that's not an issue?

    However one thing does remain, if there is more than one black hole, then those black holes all have event horizons with the same strength of gravitational field. So they should all share the same reference frame and therefore cannot move relative to one another (by which I mean, cannot be observed by a person standing in the Earth's reference frame, to be moving relative to one another.) - Or can they?
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    Quote Originally Posted by kojax View Post
    GR predicts that time would be observed to stop by at the event horizon of a black hole from the perspective of any outside observer, so doesn't that imply that a black hole must stay put (by which I mean be observed to stay put by observers outside the event horizon) forever in the same spot after it is formed?

    I suppose galaxies with black holes at their center can still orbit each other in a sense. Maybe the black holes perceive themselves to still be orbiting each other (since they perceive time to be progressing normally at the event horizon), but they're actually not moving at all from our perspective (because we're not at the event horizon)?
    This effect is perceived by on outside observer observing an object falling towards the event horizon and has to do with the way light is bend and red-shifted in the immediate vicinity of the black hole; it isn't applicable to the black hole itself, nor is it physically real in the sense that time actually stops for the infalling object. It doesn't. An observer "riding" on the object would notice nothing special as he crosses the event horizon.
    If you look at the BH in its entirety, it really is just another body with a finite mass, and therefore moves around and behaves just like any other body would.
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    Here's some of what I'm working off of.

    Time dilation - Wikipedia, the free encyclopedia

    here is another case of time dilation, where both observers are differently situated in their distance from a significant gravitational mass, such as (for terrestrial observers) the Earth or the Sun. One may suppose for simplicity that the observers are at relative rest (which is not the case of two observers both rotating with the Earth—an extra factor described below). In the simplified case, the general theory of relativity describes how, for both observers, the clock that is closer to the gravitational mass, i.e. deeper in its "gravity well", appears to go slower than the clock that is more distant from the mass (or higher in altitude away from the center of the gravitational mass). That does not mean that the two observers fully agree: each still makes the local clock to be correct; the observer more distant from the mass (higher in altitude) measures the other clock (closer to the mass, lower in altitude) to be slower than the local correct rate, and the observer situated closer to the mass (lower in altitude) measures the other clock (farther from the mass, higher in altitude) to be faster than the local correct rate. They agree at least that the clock nearer the mass is slower in rate and on the ratio of the difference.


    If this is correct, then a person standing at the event horizon of a black hole should agree with me that he's moving much slower than I am moving. Perhaps that means he perceives my world to be moving at infinite speed and his at normal speed, but that still means my perception that he isn't moving at all remains valid. If there are two black holes in a binary orbit, then certainly he thinks they're orbiting each other. But they should appear to be stationary (relative to one another) from my perspective, oughtn't they?

    I try to read the plain English when I'm studying these things, because I think that's the best way to go about it. If the words of a source clearly say something, I try to just take them at face value. It's the only approach I can think to use. But when I do that it seems that everyone starts to disagree with me, so I don't know what's going on.

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    Quote Originally Posted by kojax View Post
    If this is correct, then a person standing at the event horizon of a black hole should agree with me that he's moving much slower than I am moving. Perhaps that means he perceives my world to be moving at infinite speed and his at normal speed, but that still means my perception that he isn't moving at all remains valid. If there are two black holes in a binary orbit, then certainly he thinks they're orbiting each other. But they should appear to be stationary (relative to one another) from my perspective, oughtn't they?
    If the two observers were to compare their clocks, then they would agree that the clock located closer to the black hole does run slower due to gravitational time dilation. This does not mean that the clock closer to the BH stops and that observer sees the rest of the universe at infinite speed. It just runs slower, and the universe outside the BH would appear to be speeding up and/or be blue-shifted.
    Now, the thing is, this has no connection to the movement of the BH itself. The BH as a macroscopic object behaves like any other object since it has finite mass. So, two black holes orbiting each other would not appear stationary.
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    The formula for gravitational time dilation is:

    Gravitational time dilation - Wikipedia, the free encyclopedia



    • t0 is the proper time between events A and B for a slow-ticking observer within the gravitational field,
    • tf is the coordinate time between events A and B for a fast-ticking observer at an arbitrarily large distance from the massive object (this assumes the fast-ticking observer is using Schwarzschild coordinates, a coordinate system where a clock at infinite distance from the massive sphere would tick at one second per second of coordinate time, while closer clocks would tick at less than that rate),
    • G is the gravitational constant,
    • M is the mass of the object creating the gravitational field,
    • r is the radial coordinate of the observer (which is analogous to the classical distance from the center of the object, but is actually a Schwarzschild coordinate),
    • c is the speed of light, and
    • r0 = 2GM / c2 is the Schwarzschild radius of M. If a mass collapses so that its surface lies at less than this radial coordinate (or in other words covers an area of less than G2M2 / c4), then the object exists within a black hole.
    The Schwarzchild radius, which is the formula that defines the radius of a black hole's event horizon is:



    where:
    is the Schwarzschild radius;
    is the gravitational constant;
    is the mass of the gravitating object;
    is the speed of light in vacuum.
    If you insert the Schartzchild radius into the first equation, you get the square root of 1 - (2GM/C^2)/(2GM/C^2) or in other words, the square root of 1-1, which is zero. In other words, yes, time actually is completely halted at the event horizon from the perspective of anyone else looking in. The ratio between the rate of time as perceived by an outside observer to the rate of time as seen to flow in the event horizon is 0/(some finite number). For an observer located at the event horizon looking out, the ratio is (some finite number)/0.
    Last edited by kojax; January 23rd, 2012 at 01:28 PM.
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    Hmmm..... if something approaching the speed of light is time-dilated, such that at 99.99% of light speed, their time is dilated by a factor of 70, how can they move? For every minute that passes here, less than one second passes for that object, and yet it is travelling at nearly the speed of light.

    If we take it even further and try to imagine the (impossible and invalid) frame of reference of a photon, then time-dilation is infinite - time has stopped for that photon. And yet it moves at the speed of light.

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    I was going to point out that, the Lorentz contraction formula is:

    Length contraction - Wikipedia, the free encyclopedia



    where
    L0 is the proper length (the length of the object in its rest frame),
    L is the length observed by an observer in relative motion with respect to the object,
    is the relative velocity between the observer and the moving object,
    is the speed of light,
    So, if you are an object moving at C, then L0 is the square root of 1 - C^2/C^2 = 1-1 = zero. So from their own perspective photons shouldn't even have to move. Everything in front of them is already right next to them.

    Clearly it's not a problem for them to move relative to us, because we've already established that we can move relative to a black hole also. However, we also observe photons to move with respect to each other. So I guess if beams of light can move relative to one another at C, then black holes must be able to do the same.

    Now..... if only I could understand it.
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    Or wait.... I guess that's the trick to it then. We observe time to have frozen for the photon, but also have to allow that the distance is zero from the photon's perspective, allowing the photon to traverse that distance in no time at all. Consider this example from wiki, about Muon decay:

    http://en.wikipedia.org/wiki/Length_..._verifications

    Another example is the observed lifetime of muons in motion and thus their range of action, which is much higher than that of muons at low velocities. In the proper frame of the atmosphere, this is explained by the time dilation of the moving muons. However, in the proper frame of the muons their lifetime is unchanged, but the atmosphere is contracted so that even their small range is sufficient to reach the surface of earth.[
    We don't observe contraction of our own atmosphere to occur as a result of the Muon's relativistic speed, but it still happens from the Muon's perspective, and the effects lead to the same combined result. For light, the problem is more complicated because we have to think about which zero value matters more.


    So, now there's another question: is there a length contraction associated with gravitational time dilation also, like there is with time dilation from moving at relativistic speeds? Or, is the effect only temporal?Another question: is there a length contraction associated with gravitational time dilation also, like there is with time dilation from moving at relativistic speeds? Or, is the effect only temporal?
    Last edited by kojax; January 23rd, 2012 at 03:17 PM.
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    Quote Originally Posted by kojax View Post
    Or wait.... I guess that's the trick to it then. We observe time to have frozen for the photon, but also have to allow that the distance is zero from the photon's perspective, allowing the photon to traverse that distance in no time at all. Consider this example from wiki, about Muon decay:

    Length contraction - Wikipedia, the free encyclopedia

    Another example is the observed lifetime of muons in motion and thus their range of action, which is much higher than that of muons at low velocities. In the proper frame of the atmosphere, this is explained by the time dilation of the moving muons. However, in the proper frame of the muons their lifetime is unchanged, but the atmosphere is contracted so that even their small range is sufficient to reach the surface of earth.[
    We don't observe contraction of our own atmosphere to occur as a result of the Muon's relativistic speed, but it still happens from the Muon's perspective, and the effects lead to the same combined result. For light, the problem is more complicated because we have to think about which zero value matters more.
    Strictly speaking we don't observe time to have frozen for a photon, we incorrectly infer it if we assume the photon to have a valid reference frame, which it doesn't! I'm beginning to regret introducing that particular concept as an example, but what the heck, in for a penny, in for a pound!

    Theoretically, we would observe both length contraction in the direction of travel and time-dilation in the muon, and an observer in the frame of the muon would observe both length contraction in the direction of travel and time-dilation for the universe around it. In Special Relativity, length contraction and time dilation come hand in hand, they are two sides of the same coin, as it were.

    The point I was trying to make was that, even if something is theoretically time-dilated such that time "stops" for it, it can still be moving.

    But Special Relativity doesn't really cover what we are talking about here - for black holes we need to look to General Relativity.

    As an object approaches the event horizon of a black hole, a distant observer will calculate that object to be time dilated, such that time stops for that object at the event horizon. Simply put, to the outside universe (at a distance, and not falling towards the black hole), time stops at the event horizon. But to the in-faller, time passes normally as they cross the event horizon and they reach the gravitational singularity in a finite time.

    So, we call the event horizon a "coordinate" singularity - it exhibits singular behaviour when using some coordinate systems (a distant observer at rest in relation to the black hole, for instance), but is not a singularity when using a different coordinate system (an observer close to the Schwarzschild radius, falling into it).

    Now then, inside the event horizon things get a little tricky. For the in-faller, the link between space and time becomes more apparent than it ever has before, as they swap some of their properties with each other! They find they have only one direction in which they can move, and that is towards the singularity, whilst accelerations in different directions only change the time it will take to reach it! (Or something like that, anyway). And as for the outside observer, there is no "inside the event horizon"...

    Theoretically, time ends at a gravitational singularity. This is distinctly different to time simply "stopping" relative to someone else whose time carries on as normal. But luckily, the event horizon protects the rest of the universe from this phenomenon! And that is the point I am leading towards here.

    The space-time inside the black hole is separated from the space-time outside the black hole by the event horizon. To the outside universe, there is no "inside the black hole", inside the universe! Whatever goes on inside the event horizon is not causally connected to the outside universe - it is not part of the universe, any more.

    Also, it is worth considering whether a gravitational singularity, where time "ends", can actually form in the centre of black hole. There are schools of thought that point out that the uncertainty principle should come into play, along with other quantum effects, if we are indeed talking about something infinitely small!

    The mass of an object has only to gravitationally collapse to within its own Schwarzschild radius for a black hole to form, the gravitational singularity is only the end result of the gravitational collapse predicted by GR and we may well find GR to be only an approximation in that domain.

    The upshot of all this waffling is that, to the outside universe at a distance, the mass simply acts gravitationally like it acted before it collapsed. It is only when things approach the event horizon that the difference in the curvature of space-time becomes apparent, and any effects that might actually cause time to end are hidden from the rest of the universe, so the black hole can act as normal in that respect. Well, normal for an object with its mass, that is!

    Quote Originally Posted by kojax View Post
    So, now there's another question: is there a length contraction associated with gravitational time dilation also, like there is with time dilation from moving at relativistic speeds? Or, is the effect only temporal?
    I tried to answer this seemingly simple question three times and restarted, because it all depends on what level you look at it, and what relationships you consider to be equivalent.

    If length contraction in the direction of travel goes along with time-dilation due to relative motion, then presumably you are looking for an associated length contraction in the direction of gravity?
    Last edited by SpeedFreek; January 23rd, 2012 at 06:35 PM. Reason: added upshot
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    It seems that the issue here is that, in Special Relativity, every observer believes their own clock is the fast one. But, in General Relativity, the observer with the slower clock knows their clock is slower (or rather, knows the other observer's clock is the fast one). The two situations are different animals.

    The picture I'm starting to envision of black holes is that none of them have ever quite formed yet. The matter is all still sitting there right at the event horizon waiting to fall in. They think it will happen in 0.0000000......01 seconds. We observe that it may be billions of years away (or never happen). If there's matter already inside the outer event horizon, the matter outside the horizon doesn't count toward the Schwarzchild radius for it, so it could still be outside the event horizon that applies to it. You know, like in the same way as how the gravity is less and less as you get closer to the center of the Earth, until you reach the core and it is zero because of all the mass canceling out.

    At this point, I just don't get what part of the picture I'm missing. Taking all the information available, it seems likely that this would be the way of things, but some very prominent scientists have said there is such a thing as matter collapsing inside the Scharzchild Radius. I'm just curious to understand how it is that they think this is possible.
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    Quote Originally Posted by kojax View Post
    It seems that the issue here is that, in Special Relativity, every observer believes their own clock is the fast one. But, in General Relativity, the observer with the slower clock knows their clock is slower (or rather, knows the other observer's clock is the fast one). The two situations are different animals.

    The picture I'm starting to envision of black holes is that none of them have ever quite formed yet. The matter is all still sitting there right at the event horizon waiting to fall in. They think it will happen in 0.0000000......01 seconds. We observe that it may be billions of years away (or never happen). If there's matter already inside the outer event horizon, the matter outside the horizon doesn't count toward the Schwarzchild radius for it, so it could still be outside the event horizon that applies to it. You know, like in the same way as how the gravity is less and less as you get closer to the center of the Earth, until you reach the core and it is zero because of all the mass canceling out.

    At this point, I just don't get what part of the picture I'm missing. Taking all the information available, it seems likely that this would be the way of things, but some very prominent scientists have said there is such a thing as matter collapsing inside the Scharzchild Radius. I'm just curious to understand how it is that they think this is possible.
    kojax, thank you for bringing this up, it is an interesting conundrum, isn't it ??
    The main point here is that we are dealing with a full GR scenario; as such it needs to be clear that there is no absolute frame of reference on which our two observers ( the stationary one and the in-falling one ) can agree. This means, as we shall see, the two observers seeing different outcomes of the experiment.
    The most important concept in this case is that time cannot be defined globally, but only locally - time as the observers see it ( clocks ) will not agree. The stationary observer ( let us assume he is far from the black hole ) sees coordinate time, whereas the in-falling observer measures proper time. Coordinate time, as the name implies, depends only the specific choice of coordinate system, whereas proper time depends on the physical geometry of space-time ( mathematically it is a function of the metric tensor ); it is immediately obvious now that these only coincide if space-time is sufficiently flat, which unfortunately isn't the case for a black hole. Let's do the maths, starting with the stationary observer who measures coordinate time t(coord) :



    wherein F and C are constants following from the starting conditions of the observer's state of motion. The derivation of the above integral is beyond the scope of this post, but I can give the required references if anyone needs them; basically it follows from a solution of the geodesic equations from of the Schwarzschild Metric. What the above means is that, from the frame of reference of a far-away observer, anything falling into the black hole will appear to need infinite time to reach the event horizon; this is just like kojax has pointed out.
    For the in-falling observer, on the other hand, we get the following proper time, as measured by a clock falling together with the observer :



    which is finite. One can verify this by plugging the integral into, e.g. Wolfram Alpha. This means that for an in-falling observer, he will reach and cross the event horizon in some finite time, and will notice nothing special when doing so. This means that the event horizon is not a physical singularity, but rather one introduced through choice of coordinates. So, a clock falling towards an event horizon will not physically stop; what the outside observer sees is only an apparent effect.

    A much more elaborate explanation of this can be found below - I urge all visitors to this thread to have a read through this, it is a very good explanation :

    What happens to you if you fall into a black holes

    F
    or the original point of this thread, why black holes are able to move - this is simply because the proper time at the event horizon does not actually physically stop; in-falling matter and energy crosses the event horizon ( and reaches the singularity ) on a Schwarzschild geodesic, and within finite proper time. The outside observer will disagree, but that is only because of the way light traveling upward from the event horizon towards the stationary observer behaves.
    So who is right ? The in-falling or the stationary observer ? Well, they are both right within their own frames of reference. That's the fun and joy of General Relativity !
    Last edited by Markus Hanke; January 24th, 2012 at 03:42 AM.
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    Quote Originally Posted by kojax View Post
    At this point, I just don't get what part of the picture I'm missing. Taking all the information available, it seems likely that this would be the way of things, but some very prominent scientists have said there is such a thing as matter collapsing inside the Scharzchild Radius. I'm just curious to understand how it is that they think this is possible.
    You aren't really missing anything, it is just that in GR what the distant observer measures and what an in-falling particle sees at the event horizon are not the same. They disagree because there no longer is a global, common frame of reference. So far as the original question re: movement of the BH is concerned however, it is the proper time of the in-falling observer that is relevant, because here is where the relevant physics happen for gravitational collapses etc etc. And that one is finite and doesn't stop at the event horizon, even though a distant observer will disagree.

    The picture I'm starting to envision of black holes is that none of them have ever quite formed yet. The matter is all still sitting there right at the event horizon waiting to fall in.
    As explained, this is only true in the reference frame of a distant observer, but not for the matter approaching the event horizon itself. You need to accept that these two observers will disagree on the outcome.
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    Quote Originally Posted by Markus Hanke View Post
    The most important concept in this case is that time cannot be defined globally, but only locally - time as the observers see it ( clocks ) will not agree. The stationary observer ( let us assume he is far from the black hole ) sees coordinate time, whereas the in-falling observer measures proper time. Coordinate time, as the name implies, depends only the specific choice of coordinate system, whereas proper time depends on the physical geometry of space-time ( mathematically it is a function of the metric tensor ); it is immediately obvious now that these only coincide if space-time is sufficiently flat, which unfortunately isn't the case for a black hole. Let's do the maths, starting with the stationary observer who measures coordinate time t(coord) :


    Alright. So proper time is the wrist watch an observer is wearing, right? I've been reading this article trying to make sense of the concept of "coordinate time".

    Coordinate time - Wikipedia, the free encyclopedia


    I'm assuming, you mean something to do with relativity of simultaneity, which is something I hadn't considered. If light can't escape the black hole, then nothing inside of it is truly simultaneous with anything that happens outside, at least not from the perspective of Special Relativity, and without simultaneity it's really not possible to say "when" anything occurs, according to the outside observer's clock.

    However, if we can examine the black hole the moment before it falls within its own event horizon, when it is infinitesimally near collapse but hasn't quite collapsed yet, then light is still escaping, but only barely, so it arrives red shifted to the point of an extremely low frequency making it all-but-impossible to even detect. At that point, the light simply takes a long time to escape, so the question of what is happening right "now" appears to drag on and on. I can see how that could lead to the conclusion that what the outside observer is seeing is merely an artifact of simultaneity.



    wherein F and C are constants following from the starting conditions of the observer's state of motion. The derivation of the above integral is beyond the scope of this post, but I can give the required references if anyone needs them; basically it follows from a solution of the geodesic equations from of the Schwarzschild Metric. What the above means is that, from the frame of reference of a far-away observer, anything falling into the black hole will appear to need infinite time to reach the event horizon; this is just like kojax has pointed out.
    For the in-falling observer, on the other hand, we get the following proper time, as measured by a clock falling together with the observer :



    which is finite. One can verify this by plugging the integral into, e.g. Wolfram Alpha. This means that for an in-falling observer, he will reach and cross the event horizon in some finite time, and will notice nothing special when doing so. This means that the event horizon is not a physical singularity, but rather one introduced through choice of coordinates. So, a clock falling towards an event horizon will not physically stop; what the outside observer sees is only an apparent effect.
    Now I'm a bit confused about "proper time" in this context. Do you mean time as measured on the falling person's watch?

    I'm not contesting that an observer falling into the black hole perceives time to continue flowing in their own vicinity, nor that they perceive them self to arrive at the event horizon in a finite amount of time.



    A much more elaborate explanation of this can be found below - I urge all visitors to this thread to have a read through this, it is a very good explanation :

    What happens to you if you fall into a black holes

    F
    or the original point of this thread, why black holes are able to move - this is simply because the proper time at the event horizon does not actually physically stop; in-falling matter and energy crosses the event horizon ( and reaches the singularity ) on a Schwarzschild geodesic, and within finite proper time. The outside observer will disagree, but that is only because of the way light traveling upward from the event horizon towards the stationary observer behaves.
    So who is right ? The in-falling or the stationary observer ? Well, they are both right within their own frames of reference. That's the fun and joy of General Relativity !
    I still don't understand this part. You're saying the observer at the event horizon still perceives events occurring outside the black hole to transpire in longer than zero lengths of time? Also in falling debris appears to take longer than zero amounts of time to reach him/her?
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    Now I'm a bit confused about "proper time" in this context. Do you mean time as measured on the falling person's watch?
    Yes, that's right. "Coordinate time" is what the stationary observer far away from the black hole measures ( stationary clock ); "proper time" is what the in-falling observer measures ( moving clock ). In the case of a BH these watches will not agree because relativistic effects are significant.

    I can see how that could lead to the conclusion that what the outside observer is seeing is merely an artifact of simultaneity.
    Yes, except that the concept of "simultaneity" is now no longer defined because there is no global frame of reference.

    You're saying the observer at the event horizon still perceives events occurring outside the black hole to transpire in longer than zero lengths of time? Also in falling debris appears to take longer than zero amounts of time to reach him/her?
    Yes, because proper time is finite for him, therefore his watch hasn't stopped. It does, however, run slower, so the outside world should appear to him as if it was speeding up the closer he gets to the singularity. Other debris falling in together with him should appear and behave normally.
    Have a look at this :

    What would it look like to fall into a black hole? - YouTube

    I am not a fan of YouTube for scientific purposes, but I think this video is pretty accurate so far as I can tell.
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    Hi Markus!

    Thank you for providing the formalism for the difference between the proper and coordinate time of the in-faller and showing in more detail how the event horizon is only a coordinate singularity. I only mentioned these things in my earlier post, but I could not explain them in anything like such a rigorous manner as you did!

    So...
    Quote Originally Posted by Markus Hanke View Post

    F
    or the original point of this thread, why black holes are able to move - this is simply because the proper time at the event horizon does not actually physically stop; in-falling matter and energy crosses the event horizon ( and reaches the singularity ) on a Schwarzschild geodesic, and within finite proper time. The outside observer will disagree, but that is only because of the way light traveling upward from the event horizon towards the stationary observer behaves.
    So who is right ? The in-falling or the stationary observer ? Well, they are both right within their own frames of reference. That's the fun and joy of General Relativity !
    I'm not sure that really explains why black holes are able to move. As the issue seems to be time, what happens at the gravitational singularity?
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    I'm not sure that really explains why black holes are able to move. As the issue seems to be time, what happens at the gravitational singularity?
    I assume that from the outset we were talking about non-rotating, uncharged black holes. Assuming this, it is irrelevant what happens at the singularity, because it is hidden behind an event horizon and therefore not causally connected to the rest of the universe. The BH's gravitational field does not depend on the exact nature of the region of space behind the event horizon, it only depends on the total mass of the BH.
    On the outside, a black hole - despite all its strange features - is just a very heavy body with a finite mass. As such it moves just like any other mass, because there is nothing special about it. An example is this :

    Cygnus X-1 - Wikipedia, the free encyclopedia

    in which it would appear a small black hole is locked into a binary system with a star, and they both orbit their common center of gravity every six days or so.

    I think the question is ill posed - it should rather be : why do you think a black hole would not be able to move ?
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    Thank you for providing the formalism for the difference between the proper and coordinate time of the in-faller and showing in more detail how the event horizon is only a coordinate singularity. I only mentioned these things in my earlier post, but I could not explain them in anything like such a rigorous manner as you did!
    No problem
    It should be noted though that these integrals tend to get really ugly once you move to anything more complicated than the trivial Schwarzschild Metric; even in the trivial case they often become elliptic and are only solvable with numerical methods. If it was only a matter of wanting to test for physical singularities, I actually would have gone another route :



    This curvature invariant is independent of the choice of coordinates; it is immediately obvious that the only physical singularity is at r=0. The event horizon is not physically special in terms of curvature.
    Last edited by Markus Hanke; January 24th, 2012 at 03:50 PM.
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    As the issue seems to be time, what happens at the gravitational singularity?
    No one knows
    This would depend on the physical nature of the region of spacetime behind the event horizon, which in turn would require a theory of quantum gravity to describe, which we don't really have yet.

    Here's one interesting model though, for those of you who are curious :

    http://en.wikipedia.org/wiki/Fuzzball_(string_theory)

    Pretty cool, ey ?
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    Quote Originally Posted by Markus Hanke View Post
    I'm not sure that really explains why black holes are able to move. As the issue seems to be time, what happens at the gravitational singularity?
    I assume that from the outset we were talking about non-rotating, uncharged black holes. Assuming this, it is irrelevant what happens at the singularity, because it is hidden behind an event horizon and therefore not causally connected to the rest of the universe. The BH's gravitational field does not depend on the exact nature of the region of space behind the event horizon, it only depends on the total mass of the BH.
    On the outside, a black hole - despite all its strange features - is just a very heavy body with a finite mass. As such it moves just like any other mass, because there is nothing special about it.
    Exactly. Thank you.
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    Quote Originally Posted by Markus Hanke View Post
    Quote Originally Posted by kojax View Post
    It seems that the issue here is that, in Special Relativity, every observer believes their own clock is the fast one. But, in General Relativity, the observer with the slower clock knows their clock is slower (or rather, knows the other observer's clock is the fast one). The two situations are different animals.

    The picture I'm starting to envision of black holes is that none of them have ever quite formed yet. The matter is all still sitting there right at the event horizon waiting to fall in. They think it will happen in 0.0000000......01 seconds. We observe that it may be billions of years away (or never happen). If there's matter already inside the outer event horizon, the matter outside the horizon doesn't count toward the Schwarzchild radius for it, so it could still be outside the event horizon that applies to it. You know, like in the same way as how the gravity is less and less as you get closer to the center of the Earth, until you reach the core and it is zero because of all the mass canceling out.

    At this point, I just don't get what part of the picture I'm missing. Taking all the information available, it seems likely that this would be the way of things, but some very prominent scientists have said there is such a thing as matter collapsing inside the Scharzchild Radius. I'm just curious to understand how it is that they think this is possible.
    kojax, thank you for bringing this up, it is an interesting conundrum, isn't it ??
    The main point here is that we are dealing with a full GR scenario; as such it needs to be clear that there is no absolute frame of reference on which our two observers ( the stationary one and the in-falling one ) can agree. This means, as we shall see, the two observers seeing different outcomes of the experiment.
    The most important concept in this case is that time cannot be defined globally, but only locally - time as the observers see it ( clocks ) will not agree. The stationary observer ( let us assume he is far from the black hole ) sees coordinate time, whereas the in-falling observer measures proper time. Coordinate time, as the name implies, depends only the specific choice of coordinate system, whereas proper time depends on the physical geometry of space-time ( mathematically it is a function of the metric tensor ); it is immediately obvious now that these only coincide if space-time is sufficiently flat, which unfortunately isn't the case for a black hole. Let's do the maths, starting with the stationary observer who measures coordinate time t(coord) :



    wherein F and C are constants following from the starting conditions of the observer's state of motion. The derivation of the above integral is beyond the scope of this post, but I can give the required references if anyone needs them; basically it follows from a solution of the geodesic equations from of the Schwarzschild Metric. What the above means is that, from the frame of reference of a far-away observer, anything falling into the black hole will appear to need infinite time to reach the event horizon; this is just like kojax has pointed out.
    It took me a while to figure out why this formula uses an integral. My guess at this point is because we're attempting to calculate the time a beam of light would need to traverse the distance? It's just like if an automobile is continuously accelerating up to 100 km/h and arrives in say 80 seconds, and we want to calculate how far it went. We'd need to take a summation of all the speeds it traveled over the time it spent accelerating. I'm hoping that's right, because then the consequences make sense to me more easily.

    I'm not sure if all of this is correct, but this is what I think I'm seeing in the equations:

    ----

    For a beam of light departing from a point very near (but not quite inside) the event horizon headed toward an observer outside the gravitational field, the outside observer would perceive that the summation of the times it requires to traverse the distances would add up to nearly (but not quite) infinity. Or at least a very very long time that almost seems to approximate to infinity because it's so large.

    For a beam of light departing from the singularity itself, the time of travel would appear to be infinite.

    For an observer already located very near (but not quite inside) the event horizon, a beam of light approaching from outside the event horizon would appear to take a very short, but non-zero time to arrive. Partly that's because some of the intervening space has a level of time dilation almost equal to the observer's own level, which makes the beam of light appear to be moving only slightly faster than it should be. Granted that this is a very short area, and the beam of light is moving through it at C, so the time to travel it would be very nearly zero.

    For an observer located at the event horizon, there are two possibilities.

    1) - The time dilation effect is continuously defined at all points. In that case, there should be regions of space near the event horizon in which time dilation is not quite equal to the observer's own time, but also kind of not quite infinite by comparison to the observer's own time, and so the time required for light to travel through that region would not quite be observed to be zero by this observer's local clock. Just as before, it's a small region, and the light is moving at C through that region, so it's pretty close to zero.

    2) - The time dilation effect is quantized in some way. In that case, there must exist a point very near the event horizon where the time dilation effect jumps suddenly from zero, to a non-zero quantity (whatever the value of a quanta of the effect is). In that event, (one quanta)/0 is still infinity. An integral can't be used to accurately describe regions of a function where it is not continuously defined, so the integral is giving us an inaccurate reading.

    ---
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    So, .... if I'm right about the above post, then if gravity is quantized, singularities shouldn't be able to form. ... or maybe they could form, but they couldn't collapse any further below the event horizon.
    Some clocks are only right twice a day, but they are still right when they are right.
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    The in-falling particle is described by a curve, which is parametrized by time. The metric of space-time is described as infinitely small "slices" of space and/or time; to get the total amount of time it takes for the particle to travel along the curve we need to integrate all these infinitesimal differentials along the curve - that's why the integral. In general, proper time is defined as a path integral of the form



    with



    For the falling observer spacetime always appears flat locally; however, for the stationary observer who sees the free falling particle move along, a coordinate transformation for the path coordinate r is needed, as described by the Schwarzschild Metric, like so :



    This is where the second factor in the fraction under the integral for coordinate time comes from. Remember that the stationary observer measures the falling particle's time with his own stationary watch. That's why this adjustment is required.
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    For an observer located at the event horizon, there are two possibilities.

    1) - The time dilation effect is continuously defined at all points. In that case, there should be regions of space near the event horizon in which time dilation is not quite equal to the observer's own time, but also kind of not quite infinite by comparison to the observer's own time, and so the time required for light to travel through that region would not quite be observed to be zero by this observer's local clock. Just as before, it's a small region, and the light is moving at C through that region, so it's pretty close to zero.

    2) - The time dilation effect is quantized in some way. In that case, there must exist a point very near the event horizon where the time dilation effect jumps suddenly from zero, to a non-zero quantity (whatever the value of a quanta of the effect is). In that event, (one quanta)/0 is still infinity. An integral can't be used to accurately describe regions of a function where it is not continuously defined, so the integral is giving us an inaccurate reading.
    To be honest I do not really understand what you are saying; you are confusing me now ( sorry ).
    For an observer at the event horizon, the proper time formula given in post 13 still applies ( it's the second, shorter integral that needs to be used ! ). Nothing special would happen.
    There is also no quantization involved here; one can mathematically show that there is no physical singularity at the event horizon by forming a full contraction of the Riemann curvature tensor. This invariant shows that the only singularity happens at r=0 ( see post 19 ). The time dilation singularity at the event horizon is only there because the formula is written in Schwarzschild coordinates; if you choose another arbitrary coordinate system, this artifact would disappear.
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    Quote Originally Posted by kojax View Post
    So, .... if I'm right about the above post, then if gravity is quantized, singularities shouldn't be able to form. ... or maybe they could form, but they couldn't collapse any further below the event horizon.
    Yes they could. An observer standing on the surface of a collapsing star would measure a finite proper time until he reaches the newly formed singularity. So does all the rest of the star's matter. Therefore the collapse is complete from any point on the surface or within the volume of the star.
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    For a beam of light departing from a point very near (but not quite inside) the event horizon headed toward an observer outside the gravitational field, the outside observer would perceive that the summation of the times it requires to traverse the distances would add up to nearly (but not quite) infinity. Or at least a very very long time that almost seems to approximate to infinity because it's so large.

    For a beam of light departing from the singularity itself, the time of travel would appear to be infinite.
    Yes, this is correct, but only for a stationary observer far enough away from the black hole.

    For an observer already located very near (but not quite inside) the event horizon, a beam of light approaching from outside the event horizon would appear to take a very short, but non-zero time to arrive. Partly that's because some of the intervening space has a level of time dilation almost equal to the observer's own level, which makes the beam of light appear to be moving only slightly faster than it should be. Granted that this is a very short area, and the beam of light is moving through it at C, so the time to travel it would be very nearly zero.
    Also correct, but only for an observer close to the Black Hole.

    For an observer located at the event horizon
    Anything with non-zero mass located at the event horizon would inevitably fall into the black hole. Photons ( having zero rest mass ) could conceivable get "trapped" on the surface of the event horizon, meaning they neither fall into the singularity nor do they escape. Note however, that their proper time along a curve on the horizon surface would still be finite !
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    If the observer falling through the event horizon perceives the whole universe outside to progress all the way to heat death in the (very short) time it takes them to cross the event horizon, that means they are in agreement with the observer outside the black hole's gravitational field who perceives that the event of the in falling observer's crossing takes an infinite amount of time.

    If they agree, then either perspective is valid, and both perspectives are also (more or less) synchronized. Both observers should be able to agree that the in falling observer hasn't fallen through yet. They only disagree as to how long it took for him to get there, and as to how much longer it will be before he finally does fall in.
    Some clocks are only right twice a day, but they are still right when they are right.
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    Quote Originally Posted by kojax View Post
    If the observer falling through the event horizon perceives the whole universe outside to progress all the way to heat death in the (very short) time it takes them to cross the event horizon, that means they are in agreement with the observer outside the black hole's gravitational field who perceives that the event of the in falling observer's crossing takes an infinite amount of time.

    If they agree, then either perspective is valid, and both perspectives are also (more or less) synchronized. Both observers should be able to agree that the in falling observer hasn't fallen through yet. They only disagree as to how long it took for him to get there, and as to how much longer it will be before he finally does fall in.
    If only it were that simple! Have a read of that page Markus linked earlier.

    If an external observer sees me slow down asymptotically as I fall, it might seem reasonable that I'd see the universe speed up asymptotically—that I'd see the universe end in a spectacular flash as I went through the horizon. This isn't the case, though. What an external observer sees depends on what light does after I emit it. What I see, however, depends on what light does before it gets to me. And there's no way that light from future events far away can get to me. Faraway events in the arbitrarily distant future never end up on my "past light-cone," the surface made of light rays that get to me at a given time.

    That, at least, is the story for an uncharged, nonrotating black hole. For charged or rotating holes, the story is different. Such holes can contain, in the idealized solutions, "timelike wormholes" which serve as gateways to otherwise disconnected regions—effectively, different universes. Instead of hitting the singularity, I can go through the wormhole. But at the entrance to the wormhole, which acts as a kind of inner event horizon, an infinite speed-up effect actually does occur. If I fall into the wormhole I see the entire history of the universe outside play itself out to the end. Even worse, as the picture speeds up the light gets blueshifted and more energetic, so that as I pass into the wormhole an "infinite blueshift" happens which fries me with hard radiation. There is apparently good reason to believe that the infinite blueshift would imperil the wormhole itself, replacing it with a singularity no less pernicious than the one I've managed to miss. In any case it would render wormhole travel an undertaking of questionable practicality.
    But basically there is no requirement for the two observers to agree on simultaneity.

    I hope I didn't mislead you when I earlier wrote that time "ends" at the singularity - I didn't mean you see the end of time, as it were. I meant that all paths or world-lines inside the event horizon end at the singularity.
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    The link is good, but the writer doesn't explain why things are happening the way he says.

    I can't tell if it's because of things like this:

    Won't it take forever for you to fall in? Won't it take forever for the black hole to even form?

    Not in any useful sense. The time I experience before I hit the event horizon, and even until I hit the singularity—the "proper time" calculated by using Schwarzschild's metric on my worldline—is finite. The same goes for the collapsing star; if I somehow stood on the surface of the star as it became a black hole, I would experience the star's demise in a finite time.


    Where clearly the reason you don't experience time coming to a stop as you fall in is because you're coming to a stop with it.


    If an external observer sees me slow down asymptotically as I fall, it might seem reasonable that I'd see the universe speed up asymptotically—that I'd see the universe end in a spectacular flash as I went through the horizon. This isn't the case, though. What an external observer sees depends on what light does after I emit it. What I see, however, depends on what light does before it gets to me. And there's no way that light from future events far away can get to me. Faraway events in the arbitrarily distant future never end up on my "past light-cone," the surface made of light rays that get to me at a given time.


    I'm not seeing how this is so. It doesn't show up in the maths that have been used so far. Maybe there are some other maths that describe it?

    I guess it would make sense if the gravitational effect is continuously defined, because then every in falling photon increases the black hole's mass by one quantum and therefore moves the event horizon out an infinitesimally small amount from where it was, causing itself to come to a stop just barely before reaching you. Since you're on the inside of where it stopped, you're still in a gravitational field that's no stronger than it was before. On the other hand, if the gravitational field is quantized, then probably photons keep reaching you (but not interacting with you since you're frozen in time) until enough of them arrive to cause a move in the event horizon.

    Or .... even better, your own arrival at the event horizon causes the event horizon to grow, so any new in falling photons will stop at the new boundary, and never go on to reach you at the old boundary.

    I don't want to just believe stuff because "smart people said". Folks do too much of that these days. I'd rather understand exactly why the old model is wrong, the one where:

    Now, this led early on to an image of a black hole as a strange sort of suspended-animation object, a "frozen star" with immobilized falling debris and gedankenexperiment astronauts hanging above it in eternally slowing precipitation. This is, however, not what you'd see. The reason is that as things get closer to the event horizon, they also get dimmer. Light from them is redshifted and dimmed, and if one considers that light is actually made up of discrete photons, the time of escape of the last photon is actually finite, and not very large. So things would wink out as they got close, including the dying star, and the name "black hole" is justified.


    He says these gedankenexperiment astronaut theories didn't end up being correct, but why aren't they correct?

    Everything I can see points to them being quite accurate, but surely someone discovered that they weren't, and had a good argument for it.




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    He says these gedankenexperiment astronaut theories didn't end up being correct, but why aren't they correct?
    Everything I can see points to them being quite accurate, but surely someone discovered that they weren't, and had a good argument for it.
    kojax, this has already been explained to you; the far-away observer and the in-falling particle measure time in a different way. This is why they don't agree on what happens when you fall into the black hole. The far-away observer notices that in-falling objects appear to asymptotically approach the horizon surface. The in-falling objects themselves however notice nothing special and just fall through and into the singularity. The mathematical formalism is a line integral along a geodesic curve ( post 13 & 24 ), which in itself is a solution of GR equations and reflects the curvature of space-time ( which is why it is not a straight line ). It is mathematically obvious that one of these integrals is finite, whereas the other one diverges.
    Purely physically, objects appear to stop at the horizon, but the object actually falls through into the singularity. Which one you see depends only on your frame of reference.
    You need to let go of the notion of simultaneity - this is no longer defined here in the absence of a global reference frame. Time can only be meaningfully defined locally.

    I guess it would make sense if the gravitational effect is continuously defined,
    It is continuously defined from infinity all the way down to the point of singularity. The apparent special status of the event horizon is only an artifact of the chosen coordinate system, but not a real physical singularity.
    Did you watch the video I linked to ? In many ways that visualization is much more intuitive than the maths used.
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    Quote Originally Posted by Markus Hanke View Post
    For the in-falling observer, on the other hand, we get the following proper time, as measured by a clock falling together with the observer :


    So this integral is really where I should be looking to understand this? I haven't been able to figure out where this equation comes from. I've been looking around for it. Are you sure it's the right formula?

    Looking at the equation, it doesn't look like it's capable of giving values larger than which is a pretty small number.




    which is finite. One can verify this by plugging the integral into, e.g. Wolfram Alpha. This means that for an in-falling observer, he will reach and cross the event horizon in some finite time, and will notice nothing special when doing so. This means that the event horizon is not a physical singularity, but rather one introduced through choice of coordinates. So, a clock falling towards an event horizon will not physically stop; what the outside observer sees is only an apparent effect.
    I think there's some confusion arising from our collective choices of observers. Let's list the ones that seem to matter.

    1) - An observer who is in the process of falling toward the event horizon.

    2) - An observer who is presently located at the event horizon, but hasn't fallen in yet, but will soon.

    3) - An observer who is presently located at (or very near) the event horizon and has a super powered rocket ship that can hover there without falling in.

    4) - An observer very far away from the black hole, presumably also far from any other gravitational fields, and stationary with respect to the black hole.


    Clearly observer #1 doesn't notice time slowing around him/her, if only because his/her brain, and any clocks they may have on their person are also slowing down to keep pace with the effect. The thing I wonder is if, when they look out into space outside the event horizon, they think time is speeding up out there?

    Observer #2 doesn't notice time speeding up outside the black hole because he/she is probably moving at or near the speed of light (toward the black hole) by the time he/she has fallen to that distance, and so nothing from outside the event horizon can catch up.

    Observer #3 ....... ? What would observer #3 see? We know the light arriving would be extremely blue shifted. If they had a TV set on the ship, and were able to adjust its receiver to accommodate the effect (interpreting super high energy gamma rays instead of radio waves...) they would notice that many years of TV programming was reaching their TV set in mere seconds (or shorter). If the TV programming carried a news channel, then they'd receive many years worth of news broadcasts in a few seconds (or less).
    Some clocks are only right twice a day, but they are still right when they are right.
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    Observer #3 ....... ? What would observer #3 see?
    Nothing really special. He would still continue to see the outside universe in a fairly normal manner, apart from distortions caused by the strong gravity of the black hole, and of course the blue-shifting ( this would actually irradiate the observer with gamma rays !!! ). I have found another excellent resource that goes into a lot of detail about what happens in different scenarios in and around a black hole :

    Falling into a Black Hole

    And specifically the page

    Falling to the Singularity of the Black Hole

    I strongly urge you to have a good read of all these pages, it is a very good resource ! Even I found things on there which I wasn't aware of, specifically in the Reissner-Nordstroem BH section

    If the TV programming carried a news channel, then they'd receive many years worth of news broadcasts in a few seconds (or less)
    That may be possible, but the amount of news received would be finite. That is the main point. The clock doesn't stop, and he does not see the entire future of the universe on TV.

    I haven't been able to figure out where this equation comes from.
    I will give you a quick explanation in a separate post.

    [EDIT: Irrelevant info.]
    Last edited by Markus Hanke; January 26th, 2012 at 06:47 AM.
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    So this integral is really where I should be looking to understand this? I haven't been able to figure out where this equation comes from. I've been looking around for it. Are you sure it's the right formula?
    Firstly, yes, I am pretty sure the maths are correct, unless I am getting this whole thing completely wrong conceptually, which I don't think is the case. Let me explain to you how to get this :

    1. Firstly one obtains the Schwarzschild Metric by solving the Einstein Field Equations. Obviously I haven't done this myself; in any case the result is



    2. Given this metric one can now solve the geodesic equations of motion :



    The result of which gives the geodesic that a particle will follow in a gravitational field as described by the SM. I haven't done this myself either, and solving these equations, even for something as simple as the Schwarzschild Metric, is actually a lot of work and anything but trivial. I would refer you to any GR textbook for details and the end result ( also online at http://www2.warwick.ac.uk/fac/sci/ph.../lecture15.pdf, page 63 ). After doing all the maths, and assuming that the particle is non-rotating, we get the following :



    3. From here on in I did my own maths; firstly we isolate the differential d(tau) :



    then plug it into the general formula for proper time ( Proper time - Wikipedia, the free encyclopedia ) :



    which is the integral that I gave earlier. Actually evaluating this integral is quite a pain, but a good computer algebra system such as Maple or WolframAlpha immediately confirm that this is finite.

    4. For the other integral ( the one for coordinate time ), one looks at the relation between dt ( coordinate time ) and d(tau) ( proper time ), which can be seen from the first coefficient in the Schwarzschild Metric :



    Coordinate time is then defined as



    wherein both F and C are just constants depending on the initial conditions of the observers; we can set them to F=C=1 without loss of generality. This time it is obvious that the integral diverges towards infinity for r -> event horizon, even without actually evaluating it.

    Does this answer your questions ?
    Last edited by Markus Hanke; January 26th, 2012 at 01:08 PM.
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    Actually I am very stupid - there is a much easier way to illustrate this in a general way. The following relationship exists between proper time d(tau) and coordinate time dt :



    Thus proper time becomes



    wheres coordinate time is



    Because the Schwarzschild Metric contains terms of the form



    The proper time integral will be finite, whereas the coordinate time integral will diverge if r -> event horizon.

    That's all there is to it !
    Last edited by Markus Hanke; January 26th, 2012 at 01:09 PM.
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