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View Full Version : (Alpha) The Existence of Black Holes
RJBeery 10-08-10, 10:34 AM I preface this post by saying that I am not a Physicist and am very likely to mangle terminology, so I ask for your patience. I've made this an Alpha thread because I want to avoid ad hominems and other derailments. If you ask me questions as a point of clarification of something I've written or asked I shall do my best to get my meaning across. In return, I ask that you make a sincere attempt to understand my layman's terminology. Any posts, even PART of a post, not directly related to the OP (i.e. the existence of black holes and the various attempts to explain that existence) shall be reported to the moderator for deletion.
In the past I've asked (what I feel are) some pretty basic, logical questions about black holes to many people that I would assume would be able to supply straight-forward, well established, community-accepted answers. Here's a version of my logical conundrum:
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#1 Do all frames outside of the BH calculate that no mass crosses the EH using Schwarzschild coordinates (or, more specifically, mass crosses the EH at t=infinity)?
#2 Does the EH expand only after mass has crossed it?
If you concur with #1 and #2, then run the clock backwards in your mind and describe to me how this theoretical black hole formed in the first place...
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To this point I've received many responses including the following:
1) Backreaction occurs before EH crossing
2) Unruh radiation
3) Vaidya metric
4) Kruskal coordinates
5) The finite proper time of the in-falling body proves the BH's existence
6) "A very famous man told me so" (this is my favorite)
7) Speculative "mystery mechanisms"
8) "Here, check out my class notes" (implying, to me, that they feel the answer was explained to them in class but they didn't understand it either)
...and in most cases, when I ask more questions, I generally get the feeling that the respondent is speaking with less authority than I would expect from someone that was 100% sure of what they were talking about. If nothing else, I find it odd that there isn't more consistency in the explanations.
Anyway, my most recent discussion involved #4 (Kruskal coordinates), although posters are welcome to discuss any explanation they wish. If you believe that Kruskal coordinates allow for the existence of black holes, and you feel up to the challenge of explaining this to a layman, please explain to me how they do so. I understand that Kruskal coordinates avoid the mathematical singularity associated with the Schwarzschild event horizon by selecting new space and time components, U and V, to replace R and T. This, I am OK with and believe I understand. Next, the claim is made that, [I]because these new components are finite at the crossing of the EH, infalling bodies cross that EH in finite time from the outsider's perspective. This, I am not getting...a change in coordinates should not change conclusions about whether an infalling body makes it beyond a certain threshold...please help. :shrug:
Guest254 10-08-10, 10:45 AM Next, the claim is made that, because these new components are finite at the crossing of the EH, infalling bodies cross that EH in finite time from the outsider's perspective.
Who said this?
Next, the claim is made that, because these new components are finite at the crossing of the EH, infalling bodies cross that EH in finite time from the outsider's perspective.
I addressed this in my last post in the other thread. Here's the relevant part reproduced:
Anyway, it appears to me that you're trying to make the case that, Schwarzschild calculations be damned, Kruskal "proves" that in-falling bodies do so in finite outside observer's time.
No, I'm saying that the question of when an in-falling body crosses the event horizon according to an outside observer isn't a meaningful one. You need to explain what convention you're using to attribute the horizon crossing event a time on the outside observer's clock. And I don't know any such convention that wouldn't be arbitrary. That's why, when you raise the question, I keep substituting it with different questions formulated in terms of concepts such as trajectories (eg. of light signals) and light cones, which are better defined (ie. independent of the choice of coordinate system).
For example, take the fact that an in-falling object eventually becomes irretrievable to an outside observer, which I illustrated on a Kruskal chart. This result isn't meaningful just because I happen to like the Kruskal chart. It's meaningful because I was careful to pick a question whose answer was independent of any particular choice of coordinates. I then used the Kruskal chart to illustrate the result only because light cones have a particularly simple appearance in that chart. Proper times and light cones are invariant concepts, so a complete calculation in some other chart is guaranteed to produce the same result.
Your response to this is that the phrase "time required" to do something is simply a label and apparently doesn't actually measure anything..?
Depends on the context. In general, it's only meaningful to consider the "clock time" of an event occurring at the same place as the clock (ie. the event has to occur on the clock's world line). Attributing a time on a clock to an event that takes place somewhere else isn't so straightforward.
RJBeery 10-08-10, 01:31 PM For example, take the fact that an in-falling object eventually becomes irretrievable to an outside observer, which I illustrated on a Kruskal chart. This result isn't meaningful just because I happen to like the Kruskal chart. It's meaningful because I was careful to pick a question whose answer was independent of any particular choice of coordinates.
Here you used the word eventually (emphasis mine), yet you also said
You don't need to calculate anything. Kruskal coordinates have the nice feature that light rays travel along straight diagonal lines, just like on a Minkowski diagram in SR. This makes the Kruskal chart particularly well suited to studying causal relationships: you can just draw the past and future light cones of an event like you would in SR. If you consider the situation on the Kruskal chart, you'll see that as B advances along his world line, the point where A crosses the event horizon soon drops out of B's future light cone. Beyond that, B has no hope of pleading with or rescuing A.
Maybe this is my problem. You are saying that there exists a time, call it T1, that a distant observer would read on his clock after which time he cannot rescue the infalling observer. We don't need to calculate it because you can see it on the graph (is that what you meant?). Whether or not we "need" to calculate it, are we able to calculate it? I'm asking because frankly this contradicts my understanding of black holes.
Another question: were U and V assigned the labels of time and space components arbitrarily, or could the labels be reversed?
RJBeery 10-08-10, 01:40 PM The question of when an in-falling body crosses the event horizon according to an outside observer isn't a meaningful one
This seems to me to contradict the notion that there exists a time T1 on the the distant observer's clock after which he can no longer rescue the infalling body. In other words, changing the question as you did to that of one involving light cones and causal relationships should be able to provide us with a definitive time, if it exists.
You are saying that there exists a time, call it T1, that a distant observer would read on his clock after which time he cannot rescue the infalling observer. We don't need to calculate it because you can see it on the graph (is that what you meant?).
Yes. At least, the Kruskal chart allows us to see that T1 is finite without calculating it.
Whether or not we "need" to calculate it, are we able to calculate it?
Yes.
Another question: were U and V assigned the labels of time and space components arbitrarily, or could the labels be reversed?
They're not arbitrary. I also gave an answer to this in my last post in the other thread:
Let me ask you this: in Kruskal coordinates, is V given the label of "time component" arbitrarily?
No. General relativity provides a clear distinction between space-like, time-like, and light-like vectors (and therefore coordinates) at every point. It inherits these notions from SR. From the perspective of any locally inertial frame at a point, a particle following the Kruskal V coordinate (ie. U = cst) would be travelling at less than the speed of light as it passed that point. A point following the U coordinate (V = cst) would conversely be moving faster than light. Basically, if you imagine drawing the "coordinate grid", a material point could travel along the time-like coordinate lines but not the space-like ones.
This seems to me to contradict the notion that there exists a time T1 on the the distant observer's clock after which he can no longer rescue the infalling body. In other words, changing the question as you did to that of one involving light cones and causal relationships should be able to provide us with a definitive time, if it exists.
There's no contradiction. To clarify, consider the case in flat space-time. Suppose someone tells you there's going to be a Big Event one light-minute from where you live, at exactly 12 o'clock (all in your rest frame). Then:
At 11:59, it's too late to decide you want to participate, because you can't travel faster than light.
At 12:00, the Event is supposedly occurring "right now".
At 12:01, light emitted or reflected by the Big Event reaches you and you see it on your telescope. You can confirm that the Event took place.
Getting back to the in-falling observer crossing the event horizon, I'm saying that for the outside observer there's an analogue of (a) which occurs after a finite time on his clock, no well defined analogue of (b), and he never experiences the analogue to (c) unless he himself crosses the event horizon.
prometheus 10-08-10, 05:00 PM I preface this post by saying that I am not a Physicist and am very likely to mangle terminology, so I ask for your patience. I've made this an Alpha thread because I want to avoid ad hominems and other derailments. If you ask me questions as a point of clarification of something I've written or asked I shall do my best to get my meaning across. In return, I ask that you make a sincere attempt to understand my layman's terminology.
If you went to Japan and spoke to them in Afrikaans the Japanese people (probably) wouldn't understand you. The same applies here. If you want to order a drink from a bar in "physicsland," you need to learn to speak the language, at least to a passable level.
#1 Do all frames outside of the BH calculate that no mass crosses the EH using Schwarzschild coordinates (or, more specifically, mass crosses the EH at t=infinity)?
The answer to this question is no. An observer in the "Painleve coordinate frame," that is one either falling towards or moving away from the black hole sees matter crossing the horizon at a finite time.
#2 Does the EH expand only after mass has crossed it?
That depends on what you mean by "after." In special relativity, simultaneity is tricky, but in general relativity as you might imagine it's even worse. I will admit now that I don't know too much about this, but I know of a good example where the black hole grows as matter crosses the horizon.
Since this is an alpha I shall give you a reference: here (http://arxiv.org/abs/gr-qc/0502040).
I'm almost certain this is a waste of the time I just spent to look that up again because I've provided that link to you before and I'm certain you haven't got it. If you're super lazy go straight to figure 2. That shows the collapse of a continuous set of light shells, and the line labelled R shows the horizon.
To this point I've received many responses including the following:
I think a lot of these have been lost in translation, for example what does
2) Unruh radiation
Have to do with black hole formation? Unruh radiation is the radiation observed by an accelerating observer. It is related to Hawking radiation but only because the mathematical technique you use to calculate it is very similar.
Also, the
3) Vaidya metric
is the metric of a radiating star.
...and in most cases, when I ask more questions, I generally get the feeling that the respondent is speaking with less authority than I would expect from someone that was 100% sure of what they were talking about. [It goes without saying that my authority is exactly ZERO, save for rational thought and the ability to ask questions; if black holes cannot be explained rationally then I have a problem with them] If nothing else, I find it odd that there isn't more consistency in the explanations.
Who's digging out the ad hominems now? Consider this post reported.
RJBeery 10-08-10, 05:20 PM (ie. U = cst)
I'm sorry but what does cst stand for?
1. At 11:59, it's too late to decide you want to participate, because you can't travel faster than light.
2. At 12:00, the Event is supposedly occurring "right now".
3. At 12:01, light emitted or reflected by the Big Event reaches you and you see it on your telescope. You can confirm that the Event took place.
Getting back to the in-falling observer crossing the event horizon, I'm saying that for the outside observer there's an analogue of (a) which occurs after a finite time on his clock, no well defined analogue of (b), and he never experiences the analogue to (c) unless he himself crosses the event horizon.
Excellent, good analogy. OK, let's discuss #3, because confirming the Big Event in your telescope here is analogous to no longer seeing the in-falling body due to the fact that it has crossed the event horizon. The problem is that I contend this "never" happens. What is actually predicted to happen is that the in-falling body appears dimmer and dimmer, and more and more red-shifted. If we had perfect-precision instruments that could detect "arbitrarily red-shifted" photons then I'm not sure there is ever a time that they would ever stop coming, even if we were required to wait longer and longer and longer for them to arrive. (I could be wrong on this point, but even so it could be explained by the quantum nature of photons rather than the fact that the light is no longer there)
I'm sorry but what does cst stand for?
Constant.
Excellent, good analogy. OK, let's discuss #3, because confirming the Big Event in your telescope here is analogous to no longer seeing the in-falling body due to the fact that it has crossed the event horizon. The problem is that I contend this "never" happens. What is actually predicted to happen is that the in-falling body appears dimmer and dimmer, and more and more red-shifted. If we had perfect-precision instruments that could detect "arbitrarily red-shifted" photons then I'm not sure there is ever a time that they would ever stop coming, even if we were required to wait longer and longer and longer for them to arrive. (I could be wrong on this point, but even so it could be explained by the quantum nature of photons rather than the fact that the light is no longer there)
This is correct. Why do you consider this a problem? It wouldn't be much of an event horizon otherwise.
rpenner 10-08-10, 05:42 PM #1 Do all frames outside of the BH calculate that no mass crosses the EH using Schwarzschild coordinates (or, more specifically, mass crosses the EH at t=infinity)?
A Schwarzschild chart is a set of spherically static coordinates -- an artifice which may or may not be useful in describing the physics of space time, depending on the question you are asking.
Assuming we are talking about static, uncharged and finite black holes, the coordinates you speak of allow hypersurfaces of constant time coordinate values which far from the black hole approximate the hypersurfaces of constant time of a Lorentzian flat-space frame at rest with respect to the black hole. But since even in flat-space the stationary observer has no privilege to refer to his chosen time coordinates as "time" itself. In Special Relativity, different states of motion have different time coordinates. In General Relativity, even observers with constant separation may experience different time coordinates. And in both, two observers may synchronize clocks, take different paths through space-time, and rendezvous at some future point in space and time with different values on their clocks.
In the usual Schwarzschild coordinates, the exterior of the black hole is labeled with (r_A, \theta, \phi, t) with r_A \in \left( + \frac{2 G M}{c^2}, +\infty \right), \theta \in \left[0, 2 \pi \right), \phi \in \left[0, \pi \right] and t \in \left(-\infty,+\infty \right). But, even though the angles seem to make sense, and the circumference for every radius is 2 \pi r, the speed of light is not the same in every direction with the line element of c^2 \left(\Delta \tau \right)^2 = 0 = \left(c^2 - \frac{2 G M}{r_A} \right) \left(\Delta t \right)^2 - \frac{1}{1 - \frac{2 G M}{c^2 r_A}} \left(\Delta r_A \right)^2 - r_A^2 \left(\Delta\phi\right)^2 - r_A^2 \sin^2\phi \left(\Delta\theta\right)^2 .
Even if you use Eddington's isotropic Schwarzschild coordinates, the coordinate speed of light is now the same in every direction for every point, but now the speed is different at different radii, when physically the speed of light is an assumed physical constant. The physics are distinct from the coordinates, and sticking to one global chart is going to be crippling to your understanding of General Relativity.
It's even crippling to use one chart on Earth, since Australians don't enforce their traffic regulations based on what a chartist in England calculates their coordinate speed to be.
So perhaps your first question needs clarification.
RJBeery 10-08-10, 11:05 PM Excellent, good analogy. OK, let's discuss #3, because confirming the Big Event in your telescope here is analogous to no longer seeing the in-falling body due to the fact that it has crossed the event horizon. The problem is that I contend this "never" happens. What is actually predicted to happen is that the in-falling body appears dimmer and dimmer, and more and more red-shifted. If we had perfect-precision instruments that could detect "arbitrarily red-shifted" photons then I'm not sure there is ever a time that they would ever stop coming, even if we were required to wait longer and longer and longer for them to arrive. (I could be wrong on this point, but even so it could be explained by the quantum nature of photons rather than the fact that the light is no longer there)
This is correct. Why do you consider this a problem? It wouldn't be much of an event horizon otherwise.
OK...now, if photons continue to reach the outside observer indefinitely, regardless of their degree of red-shifting or the amount of time it takes between photon arrival events, then we cannot say that a time T1 exists on the outside observer's clock after which he cannot rescue the infalling body. The reason is because if photons can travel from the infalling body to the outside observer then that body can escape with a sufficiently powerful jetpack. In a trivial example we could define "rescuing" the infalling observer as simply shining a light toward him that would ignite it. (I presume we are not going to disagree that the photons shining toward the infalling body shall never have a problem reaching him.) We've already agreed that photons emitted near the EH, eventually reaching the outside observer, will do so indefinitely from the observer's perspective (albeit requiring more and more patience and precise detection instrumentation), and this forces us to admit that the return path away from the EH back to the outside observer is also available indefinitely from the outsider's perspective.
What is actually predicted to happen is that the in-falling body appears dimmer and dimmer, and more and more red-shifted. If we had perfect-precision instruments that could detect "arbitrarily red-shifted" photons then I'm not sure there is ever a time that they would ever stop coming, even if we were required to wait longer and longer and longer for them to arrive. (I could be wrong on this point, but even so it could be explained by the quantum nature of photons rather than the fact that the light is no longer there)
No, this isn't quite correct. There is a point at which we no longer receive photons from the infalling body. It's not that the photons are no longer perceptible due to red-shifting. There are no more photons, and what we would detect is a frozen image which fades to nothing.
I'd give you a link but I haven't looked it up anywhere on the web. The best I can do is refer you to pages 244-249 in Kip Thorne's Black Holes & Time Warps
Given that, then
OK...now, if photons continue to reach the outside observer indefinitely, regardless of their degree of red-shifting or the amount of time it takes between photon arrival events, then we cannot say that a time T1 exists on the outside observer's clock after which he cannot rescue the infalling body
is not valid.
OK...now, if photons continue to reach the outside observer indefinitely, regardless of their degree of red-shifting or the amount of time it takes between photon arrival events, then we cannot say that a time T1 exists on the outside observer's clock after which he cannot rescue the infalling body. The reason is because if photons can travel from the infalling body to the outside observer then that body can escape with a sufficiently powerful jetpack.
This argument seems incomplete.
If the infalling body could escape with a sufficiently powerful jetpack at the time it emitted a photon, then how does it follow that the outside observer who receives that photon at a later time can reach the infalling body before it crosses the horizon?
I had a vague idea that photons emitted from arbitrarily close to the event horizon (in schwarzschild coordinates) take an arbitrarily long time to escape... so the outside observer will keep receiving photons from the infalling object for an arbitrarily long time after it has crossed the event horizon.
I presume we are not going to disagree that the photons shining toward the infalling body shall never have a problem reaching him.
Is there a negative too many in that sentence?
I do question whether photons shining toward the infalling body can reach it before it crosses the horizon.
M00se1989 10-09-10, 01:36 AM On black holes... what physical elements do we believe it is inherently made of?
is it a larger element than is observable to us? It takes in mass when mass is to delicate to pierce through its entire volume. It also emits light at a constant, relative to its mass and volume. Light is the only "constant" that we can travel both toward and away from. Any atom has electrons that create space inherently. Is it not possible to say the elements within its composition are greater than those of our known values?
If the elements were making themselves they would have done it in secret. That means there is a potential to do it in front of you, but also without you knowing.
On black holes... what physical elements do we believe it is inherently made of?
is it a larger element than is observable to us? It takes in mass when mass is to delicate to pierce through its entire volume. It also emits light at a constant, relative to its mass and volume. Light is the only "constant" that we can travel both toward and away from. Any atom has electrons that create space inherently. Is it not possible to say the elements within its composition are greater than those of our known values?
If the elements were making themselves they would have done it in secret. That means there is a potential to do it in front of you, but also without you knowing.
It's not made of any physical elements. What we commonly call the black hole is the event horizon, that point where gravitational forces are strong enough so that the escape velocity is the speed of light. Everything and anything can be swallowed by the BH. NO light is emitted.
Friends don't let friends drink and post.
OK...now, if photons continue to reach the outside observer indefinitely, regardless of their degree of red-shifting or the amount of time it takes between photon arrival events, then we cannot say that a time T1 exists on the outside observer's clock after which he cannot rescue the infalling body. The reason is because if photons can travel from the infalling body to the outside observer then that body can escape with a sufficiently powerful jetpack.
True but irrelevant. I specifically attempted to clarify this in an edit to this post:
B never receives any confirmation that A fell past the event horizon (unless B himself crosses the event horizon). B could indeed always find out later that A turned his jetpack on at the last minute, or that someone else rescued A. What I'm saying is that, assuming A stayed on course into the black hole, there would be a finite time on B's clock after which A's fate would be out of B's hands. If A accidentally left his jetpack behind, B would have a finite amount of time to notice and arrange a rescue mission before A became irretrievable.
In a trivial example we could define "rescuing" the infalling observer as simply shining a light toward him that would ignite it. (I presume we are not going to disagree that the photons shining toward the infalling body shall never have a problem reaching him.)
I specifically showed the opposite: the point where the in-falling observer crosses the event horizon drops out of the outside observer's future light cone. After that point, the outside observer can't send signals that will intercept the in-falling observer before he crosses the horizon.
If you program a probe to fire its engines and return to you only when it receives a signal from you, and allow that probe to fall toward a black hole, you only have a finite time to send that signal before the probe becomes irretrievable.
RJBeery 10-10-10, 10:19 AM If you program a probe to fire its engines and return to you only when it receives a signal from you, and allow that probe to fall toward a black hole, you only have a finite time to send that signal before the probe becomes irretrievable.
If the infalling body could escape with a sufficiently powerful jetpack at the time it emitted a photon, then how does it follow that the outside observer who receives that photon at a later time can reach the infalling body before it crosses the horizon?
I had a vague idea that photons emitted from arbitrarily close to the event horizon (in schwarzschild coordinates) take an arbitrarily long time to escape... so the outside observer will keep receiving photons from the infalling object for an arbitrarily long time after it has crossed the event horizon.
If you both agree that photons will continue to be received by the outside observer indefinitely then it does not follow that there is a finite time after which there is no source of those photons. There cannot be some sort of "infinite cache" of light waiting to make its journey to the outside observer (unless that "cache" is the infalling body itself either reflecting or emitting those photons, which is what the Schwarzschild coordinate calculates).
If you both agree that photons will continue to be received by the outside observer indefinitely then it does not follow that there is a finite time after which there is no source of those photons. There cannot be some sort of "infinite cache" of light waiting to make its journey to the outside observer (unless that "cache" is the infalling body itself either reflecting or emitting those photons, which is what the Schwarzschild coordinate calculates).
The apparent freezing of the infalling body is an illusion generated by the difference in relativistic frames.
http://casa.colorado.edu/~ajsh/schwp.html
http://casa.colorado.edu/~ajsh/schwp.html
Schwarzschild Geometry
Gravitational slowing of time
In general relativity, clocks at rest run slower inside a gravitational potential than outside.
In the case of the Schwarzschild metric, the proper time, the actual time measured by an observer at rest at radius r, during an interval dt of universal time is (1 - rs/r)1/2 dt, which is less than the universal time interval dt. Thus a distant observer at rest will observe the clock of an observer at rest at radius r to run more slowly than the distant observer's own clock, by a factor
( 1 - rs / r )1/2 .
This time dilation factor tends to zero as r approaches the Schwarzschild radius rs, which means that someone at the Schwarzschild radius will appear to freeze to a stop, as seen by anyone outside the Schwarzschild radius.
Gravitational redshift
The gravitational slowing of time produces a gravitational redshift of photons. That is, an outside observer will observe photons emitted from within a gravitational potential to be redshifted to lower frequencies, or equivalently to longer wavelengths.
Conversely, an observer at rest in a gravitational potential will observe photons from outside to be blueshifted to higher frequencies, shorter wavelengths.
In the case of the Schwarzschild metric, a distant observer at rest will observe photons emitted by a source at rest at radius r to be redshifted so that the observed wavelength is larger by a factor
( 1 - rs / r )-1/2
than the emitted wavelength. The redshift factor tends to infinity as r approaches the Schwarzschild radius rs, which means that someone at the Schwarzschild radius will appear infinitely redshifted, as seen by anyone outside the Schwarzschild radius.
RJBeery 10-10-10, 11:54 AM The apparent freezing of the infalling body is an illusion generated by the difference in relativistic frames.
This technically doesn't eliminate my objection. All photons must have an emission source, and I doubt the author referred to above is suggesting a "ghost source" that hangs around indefinitely providing any supposed illusions.
Also, in all other aspects of Relativity, light itself is the ultimate arbiter of measurement (e.g. in SR, "seeing" a passing clock move slowly and deducing that it is moving slowly are equivalent; in GR, "seeing" the light being bent by the sun in the eclipse of 1919 is what confirmed Einstein's theory over Newton). Given this, it is difficult for me to accept any such explanation as being a mere illusion.
This technically doesn't eliminate my objection. All photons must have an emission source, and I doubt the author referred to above is suggesting a "ghost source" that hangs around indefinitely providing any supposed illusions.
Yes, but the photons are not received indefinitely. The last photon emitted before the body passes the event horizon is frozen in time, as seen from the external frame, but it is the last photon received.
Given this, it is difficult for me to accept any such explanation as being a mere illusion
Research it more. The site I linked to is a place to start. If you have difficulty accepting an explanation which has universal acceptance among those who have studied the subject, you might consider that the problem is a deficiency in your understanding, rather than the explanation is wrong.
If you both agree that photons will continue to be received by the outside observer indefinitely then it does not follow that there is a finite time after which there is no source of those photons.
What's the relevance of this? I don't even know whether I should agree with "there is a finite time after which there is no source of those photons" or not. As presented, that statement is too vague and ill-defined to even discuss in the context of GR.
I've always been specific about what I was claiming: the point where the in-falling observer crosses the event horizon drops out of the outside observer's future light cone after a finite time. Even with that settled, your statement doesn't make sense: let
X = "light emitted by the in-falling observer will continue to be received by the outside observer indefinitely"
and
Y = "the in-falling observer becomes irretrievable to the outside observer in a finite time on the outside observer's clock".
If you say "X, and it doesn't follow that Y", then read literally you're saying "X doesn't imply Y". Now that's true, but irrelevant: nobody is trying to argue that Y is true *because* X is. I'm claiming that Y is true, independently of X, because I've depicted the situation on a Kruskal chart.
That's obvious, so I think it's more likely you meant "X implies Y is false". And that simply isn't true. X and Y together mean that the horizon crossing event falls out of the outside observer's future light cone after a finite time, but never falls into his past light cone. There's really no reason X should invalidate Y. Both are visible on the Kruskal diagram for instance.
There cannot be some sort of "infinite cache" of light waiting to make its journey to the outside observer
Why not? What's happening is simply that light emitted by the in-falling observer has a progressively harder time climbing out of the black hole's gravitational well and reaching the outside observer. Another point of view is that the outside observer is accelerating away from the light, which is how the situation appears on a Kruskal diagram. He'd get all the light in a finite time if he simply allowed himself to free-fall past the event horizon.
RJBeery 10-10-10, 01:37 PM Yes, but the photons are not received indefinitely. The last photon emitted before the body passes the event horizon is frozen in time, as seen from the external frame, but it is the last photon received.
With respect, this is not what others in this thread are saying, but I did leave this possibility open when I said
If we had perfect-precision instruments that could detect "arbitrarily red-shifted" photons then I'm not sure there is ever a time that they would ever stop coming, even if we were required to wait longer and longer and longer for them to arrive. (I could be wrong on this point, but even so it could be explained by the quantum nature of photons rather than the fact that the light is no longer there)
What I was saying is that there is a big difference between no longer seeing photons because the light source has passed the EH vs no longer seeing them because the light source's emitted energy does not suffice to produce a photon.
Research it more. The site I linked to is a place to start. If you have difficulty accepting an explanation which has universal acceptance among those who have studied the subject, you might consider that the problem is a deficiency in your understanding, rather than the explanation is wrong.
This is quite likely, but in fairness I don't feel I've been given "the" explanation, but rather many disparate ones, which is one of the confounding reasons I keep asking questions. You say the photons will stop coming; Pete and przyk say they won't; Guest appears to question whether przyk's argument is valid (by implication, here, when he asked who made that statement in post #2); prometheus says the answer lies in the "Painleve coordinate frame". I am only really able to concentrate on a single line of reasoning at a time which is why I want my discussion with przyk and Pete to run its course before moving on to something else (such as "illusions" and "Painleve coordinate frame", which I guess I shall need to add as #9 and #10 in my OP :))
This is quite likely, but in fairness I don't feel I've been given "the" explanation, but rather many disparate ones, which is one of the confounding reasons I keep asking questions. You say the photons will stop coming; Pete and przyk say they won't; Guest appears to question whether przyk's argument is valid (by implication, here, when he asked who made that statement in post #2); prometheus says the answer lies in the "Painleve coordinate frame". I am only really able to concentrate on a single line of reasoning at a time which is why I want my discussion with przyk and Pete to run its course before moving on to something else (such as "illusions" and "Painleve coordinate frame", which I guess I shall need to add as #9 and #10 in my OP :))
Which is why I suggest you research the issue yourself. Try websites ending in .edu.
in SR, "seeing" a passing clock move slowly and deducing that it is moving slowly are equivalent
No they're not. If a clock is moving with a speed \beta = \frac{v}{c} its apparent time dilation factor could be anything between \frac{\sqrt{1 - \beta}}{\sqrt{1 + \beta}} and \frac{\sqrt{1 + \beta}}{\sqrt{1 - \beta}} depending on whether it's approaching or receding from you. In the former case, an approaching clock will appear to be ticking faster than yours. It's only when you correct for the Doppler effect that you get the familiar factor of \frac{1}{\sqrt{1 - \beta^{2}}}.
M00se1989 10-10-10, 02:12 PM what's all this nonsense about a big blob of energy?
Don't blame the clock for getting stretched out a bit. It runs off light which slows down the time when it gets a little "bent".
and so does any theory.
Larger means more heat, and therefore more gravity, and more Light.
Just because you don't see a photon does not mean it isn't still there.
but it would swallow you up and spit you right back out in "space":D
its only when your in its "sphere of influence" that you are part of it. And that is determined entirely by your relative speed inside the sphere. So if you have stopped you probably went too far too fast and crashed.
This is quite likely, but in fairness I don't feel I've been given "the" explanation, but rather many disparate ones
They aren't all so disparate actually:
You say the photons will stop coming; Pete and przyk say they won't;
He's right in at least one sense: if the in-falling observer emits N photons before crossing the horizon, the outside observer will receive N photons. One of these will be the last photon and the outside observer will detect it after a finite time. How long that time is depends on how close the in-falling observer was to the horizon before emitting it. The time approaches infinity as the point of emission approaches the event horizon.
So far I've just been discussing in terms of continuous, classical light rays.
Guest appears to question whether przyk's argument is valid (by implication, here, when he asked who made that statement in post #2);
Guest asked who claimed "infalling bodies cross that EH in finite time from the outsider's perspective". I've never claimed this. I've repeatedly stated claims of this sort aren't well defined.
prometheus says the answer lies in the "Painleve coordinate frame".
I'm not familiar with Painleve coordinates, but (emphasis added):
An observer in the "Painleve coordinate frame," that is one either falling towards or moving away from the black hole sees matter crossing the horizon at a finite time.
This sounds like it probably is consistent with what I'm saying: a free-falling outside observer does see matter cross the horizon after a finite time (in the sense of receiving light emitted or reflected by the in-falling mass). He sees the mass cross the horizon when he himself crosses the horizon.
prometheus may also have meant the mass crosses the horizon in finite coordinate time in Painleve coordinates, in the same way the mass crosses the horizon in finite Kruskal time. I don't know.
RJBeery 10-10-10, 05:00 PM So far I've just been discussing in terms of continuous, classical light rays.
...and my argument is even clearer in that context. A discrete event horizon crossing would result in an abrupt end to the classical continuous light ray from the outsider's perspective. This is not what is predicted from Schwarzschild coordinates. This directly exposes my (unsophisticated) view that there is a problem here, because unless someone is claiming that Schwarzschild coordinates are invalid or wrong (meaning specifically that they would hold inaccurate predictive power), then I am left with a cognitive disconnect.
Guest asked who claimed "infalling bodies cross that EH in finite time from the outsider's perspective". I've never claimed this. I've repeatedly stated claims of this sort aren't well defined.
You're right, this is my mistake. Despite you trying to clarify this point, I still have trouble seeing the difference between saying that something happens in finite time on my watch vs saying that something is eventually outside of my future light cone. There should be a specific point on my watch after which that event is outside my light cone, correct?
...and my argument is even clearer in that context. A discrete event horizon crossing would result in an abrupt end to the classical continuous light ray from the outsider's perspective. This is not what is predicted from Schwarzschild coordinates.
It's not predicted period. It doesn't matter which coordinate system you use. You get the same result using the Kruskal chart for instance.
There really isn't anything remarkable about this result. It's even possible to describe a very similar situation in SR: suppose you start from rest at (t = 0, x = 1) and travel along the trajectory
x(t) \,=\, \sqrt{1 \,+\, c^{2} t^{2}}
(in the coordinates of some given inertial frame), and suppose I emit a signal from the spatial origin x = 0 starting at ct = -1 up until ct = 0. Then you'll receive the beginning of my signal just as you start to leave, but your curve accelerates away from me in such a way that you'll never, by your watch, receive the end of it.
Roughly, here's what the situation looks like on a Minkowski diagram:
http://img89.imageshack.us/img89/2324/accel.png
Here the red dotted lines represent the propagation of the beginning and the end of my signal. The blue curve gives an idea of your hyperbolic trajectory. It asymptotically approaches, but never touches, the world line depicting the end of the signal. The upper diagonal line is effectively an event horizon to you: no signal emitted from above and to the left of it will ever reach you.
In Kruskal coordinates, a diagram of the signal sent by the in-falling observer approaching the event horizon and received by the outside observer would look very similar to this.
Despite you trying to clarify this point, I still have trouble seeing the difference between saying that something happens in finite time on my watch vs saying that something is eventually outside of my future light cone. There should be a specific point on my watch after which that event is outside my light cone, correct?
Yes. The reason I'm picking on this is that, referring to the analogy I gave in post #6, saying that a distant event happens in finite time on your watch would normally mean the analogue of (b).
RJBeery 10-11-10, 05:38 PM It's not predicted period. It doesn't matter which coordinate system you use. You get the same result using the Kruskal chart for instance.
OMG przyk, I'm so grateful you said this. This talk of alternate coordinate choices to solve problems seriously had me feeling like higher mathematics demanded a forfeiture of reasoning. I've been interpreting some posters' comments as implying that a change in coordinates allows for a change in prediction of what "actually happens", and this was deeply troublesome for me to understand.
Your illustration is very, well, illustrative by the way. I see your point and after seeing your graph did I realize something that I should've thought of earlier - the proper time of the infalling object is indisputably finite before crossing the EH. This in itself necessitates (in my mind) a finite number of photons to be emitted, period.
It's a busy week for me but I'm going to try posting a picture of my own later. I'd really like to see an application of the Kruskal coordinates with, say, a 1 solar mass BH to calculate T1 after which the outside observer could not rescue the infalling body. I won't ask you to do it (unless it's trivial); I'll try to make it to campus and read what I can on the subject.
AlphaNumeric 10-11-10, 05:51 PM OMG przyk, I'm so grateful you said this. This talk of alternate coordinate choices to solve problems seriously had me feeling like higher mathematics demanded a forfeiture of reasoning. I've been interpreting some posters' comments as implying that a change in coordinates allows for a change in prediction of what "actually happens", and this was deeply troublesome for me to understand.This demonstrates the point made by myself, Guest and others, that because you haven't had any experience with this area of mathematics nor anything which is considered a requirement for said area of mathematics you lack the ability to make an informed opinion or view on the matter.
One of the founding principles of theoretical physics, in fact pretty much any area of physics striving for quantitative models, is that the outcomes should not depend on our method of description. Its from this you get the notion of covariant formalisms, both in terms of space-time coordinates and gauge potentials. High level general relativity is often done in a way which doesn't even need you to define a set of coordinates and the best derivations or proofs are those which don't require them. Most advanced courses on GR go down that route because it presents more powerful and rigorous approaches the problems.
If you'd done any reading on the role of coordinates in physics you'd have found this out. Instead it's several threads and dozens of posts for you to find out something which you should have known before forming an opinion.
I don't think its against the Alpha rules to say that perhaps it would be wise if you spent a little less time arguing about something you haven't done any reading on and a little more time actually doing the reading. That isn't an insult, its a piece of honest advice.
RJBeery 10-11-10, 06:01 PM One of the founding principles of theoretical physics, in fact pretty much any area of physics striving for quantitative models, is that the outcomes should not depend on our method of description.
So, what you mean to say is that I was right to trust my gut rather than my (possible) misinterpretation of what others were telling me? Thanks for the encouragement!
Also, what I'm doing with przyk is not arguing, it's questioning and answering. I'm very impressed with his style, and others on this forum would do well to take note.
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