How can Block Holes form in finite time?

[twhites]

I think this subject may have been discussed here already; please point to the thread if so.

I am unable to understand how Black Holes can form within the lifetime of the universe.

If nothing can cross an event horizon in finite time, then it seems clear that the horizon can't form (at least by a collapse mechanism; is there another way?) in finite time either. It seems to me that several arguments can be made that nothing (even light) can, in fact, cross an event horizon in a finite time (as we on Earth, i.e., far from the BH) would measure it.

I am aware that an object falling on a radial trajectory will cross an event horizon in finite proper time. But that is not the time that a far-field observer (ffo) uses, not only because he/she is in flat space-time, but because he/she is stationary w.r.t. the horizon. It appears to me that even the clock of a momentarily stationary observer near the horizon runs very much more slowly than that of the ffo (in the limit of r=2M, infinitely so). Conversely, the clock of the ffo runs faster than that of the poor soul near the horizon (again, infinitely so at r=2M). This statement is, I believe, equivalent to the observation of the gravitational red-shift/blue-shift of light exchanged between the two observers, which I understand to diverge as stated (in both directions).

Likewise while the proper distance between the ffo and the horizon is finite, this distance is not directly measurable (it is, after all, space-like). We measure distance by timing light transmission times, so the measured distance from the ffo to the horizon is determined by (say) measuring the time it takes for a light pulse emitted by the ffo to be reflected and returned to the ffo by a (momentarily stationary) observer near the horizon. This time diverges as r->2M; it is certainly the case that at r = 2M, no such light pulse ever returns to the ffo. This appears to me to mean that (even) light can't reach/cross the horizon in finite (far-field) time. The difference between this distance and that measured by the inertial observer (falling on a radial trajectory) I interpret as the result of a large Lorentz contraction; the inertial observer is moving at a velocity (relative to the ffo and the horizon) close to (approaching) c as he/she approaches r=2M.

What's wrong with these arguments? It seems to me to give a picture with certain benefits, since it avoids the necessity of worrying about what happens inside an event horizon--there is no such volume. There would appear to be no entropy-loss problem either.

 

[JesseM]

If nothing can cross an event horizon in finite time

Objects can certainly cross in finite "proper time", i.e. time according to their own clock. They can't cross in finite coordinate time in the Schwarzschild coordinate system, but that's just a feature of that particular coordinate system, in the Kruskal-Szekeres coordinate system (and others) objects do cross the horizon in finite coordinate time. And note that even in flat Minkowski spacetime with no gravity, you can come up with a non-inertial coordinate system where it takes an infinite amount of coordinate time to cross some arbitrary boundary, like in Rindler coordinates for accelerating observers where it takes an infinite coordinate time for objects to cross the Rindler horizon, even though they'd reach it in finite coordinate time in inertial frames.

 

[Naty1]

In general, black holes form BEFORE a horizon forms. The horizon is a reflection of the fact that a black hole has formed. But you can also have horizons without black holes.

The actual process with black holes depends on whether you consider absolute or apparent event horizons. The former shows continuous growth as the black hole forms; the latter is discontinuous. And of course they grow immediately as matter or energy passes the event horizon.

The universe has a horizon which was present almost as soon as space and time formed after inflation. An accelerating observer also creates a horizon...see Unruh effect explanations.

So there are multiple examples of "quick" horizons forming.

 

[skeptic2]

Objects can certainly cross in finite "proper time", i.e. time according to their own clock. They can't cross in finite coordinate time in the Schwarzschild coordinate system, but that's just a feature of that particular coordinate system, in the Kruskal-Szekeres coordinate system (and others) objects do cross the horizon in finite coordinate time.

Suppose we have a 10 solar mass BH (1.98892E+31 kg) with a free falling object 100 m above the event horizon (having free fallen from infinity), using Kruskal-Szekeres coordinates how long will it take the object to reach the event horizon in coordinate time?

 

[JesseM]

Suppose we have a 10 solar mass BH (1.98892E+31 kg) with a free falling object 100 m above the event horizon (having free fallen from infinity), using Kruskal-Szekeres coordinates how long will it take the object to reach the event horizon in coordinate time?

Not sure how to calculate this, perhaps someone else can help.

 

[twhites]

They can't cross in finite coordinate time in the Schwarzschild coordinate system, but that's just a feature of that particular coordinate system, in the

Thanks for replying.

Is it not the case that the ffo tells time by the Schwarzschild clock? In any case, it seems to me that the red-shift/blue shift argument still raises a problem. Is it not true that the red-shift of light from the momentarily stationary observer near the horizon, as viewed by us in flat space time, decreases and approaches zero frequency? Is the converse not also true (i.e., that light from us, received by the observer near the BH, diverges to high frequency)? These frequencies are clocks (by definition), and the divergences are (presumably) real experimental facts--I've never read any statement that they are not, at least the red-shift part. If so, then at the horizon, an infinite number of clicks of the far-field clock occur between clicks of the local clock (as counted by the horizon clock). In the far-field, the clock at the horizon stops altogether, while the local one keeps going, with the same result. Both observers agree on this, don't they? If the Schwartzchild metric predicts this experimental result correctly, then it seems to me that (from our viewpoint) the fellow trying to get into the BH can't do it except in an infinite amount of time (by the external clock). By his own clock, there's no problem, except, in my view, the universe has died in the meantime.

I've looked at the Kruskal-Szekeres coordinates, and it seems to me that these coordinates do not reflect actual measurements in any direct way, since they mix spatial and temporal components in a way that instruments (clocks and rulers) do not. Don't you have to reconvert from u and v (in my text) back to t and r in order to relate to an actual clock or ruler? And doesn't that get you back to Schartzchild coordinates? Or am I completely off-base here?

Clearly, the coordinates are the root of my problem here, but what about the (thought) experiment?

 

[JesseM]

Is it not the case that the ffo tells time by the Schwarzschild clock?

He can use any coordinate system he likes, but his actual subjective experience of time will be determined by his proper time.

In any case, it seems to me that the red-shift/blue shift argument still raises a problem. Is it not true that the red-shift of light from the momentarily stationary observer near the horizon, as viewed by us in flat space time, decreases and approaches zero frequency?

Yes, as viewed by us.

Is the converse not also true (i.e., that light from us, received by the observer near the BH, diverges to high frequency)?

No, that's not true for a freefalling observer (though I believe if you have a series of observers hovering at constant Schwarzschild radius, than the blueshift they'd experience from the outside universe would approach infinity in the limit as their radius approaches the radius of the event horizon). See this section of the Usenet Physics FAQ:

Will you see the universe end?

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.

Incidentally, something is similarly true about the Rindler horizon seen by accelerating observers in flat spacetime; if I am an accelerating observer and I see you approaching my own Rindler horizon, I'll see you move slower and slower as you approach it and your redshift will approach infinity from my perspective, but you experience crossing this boundary (which is basically just the boundary of the future light cone of an event at the vertex of the "Rindler wedge", see the diagram here depicting the Rindler observers and the Rindler horizon from the perspective of an inertial coordinate system) in finite time and don't see me as infinitely blueshifted as you cross it.

I've looked at the Kruskal-Szekeres coordinates, and it seems to me that these coordinates do not reflect actual measurements in any direct way, since they mix spatial and temporal components in a way that instruments (clocks and rulers) do not.

No, it's true that the space coordinate (as well as the time corodinate) of KS coordinates mix together the space and time coordinates of Schwarzschild coordinates, but there is nothing "special" about Schwarzschild coordinates, you could equally well define the Schwarzschild space coordinate as a mix of the space and time coordinates of KS coordinates. The most physical notion of time in GR is proper time along timelike worldlines (time as measured by a clock moving along that worldline), and likewise the most physical notion of distance is proper distance along spacelike worldlines (which can be thought of in terms of the sum of measurements along a series of short inertial rulers, made at the moment their timelike worldlines cross the spacelike worldline, and such that the two worldlines are orthogonal when they cross). Neither Schwarzschild time nor KS time would match up with the proper time along an arbitrary timelike worldline, and likewise for distance. But by integrating the metric, you can use either coordinate system to calculate proper time or proper distance along any worldline, both would be proportional to the integral of ds in the metric equation.

 

[skeptic2]

Suppose astronomers were to discover a black hole whose angular diameter they are able to measure. Knowing their distance from the black hole, they are able to calculate its apparent diameter. How would they use the black hole's apparent diameter to calculate the black hole's mass?

 

[JesseM]

Suppose astronomers were to discover a black hole whose angular diameter they are able to measure. Knowing their distance from the black hole, they are able to calculate its apparent diameter. How would they use the black hole's apparent diameter to calculate the black hole's mass?

See the formula for Schwarzschild radius--rearranging gives M = r*c²/2G, where r is the radius of the event horizon.

 

[skeptic2]

Yes, but that's not the problem. Just as the measurement of time depends of one's reference frame, so does the measurement of distance. Outside a black hole space is increasingly contracted in a radial direction resulting in objects appearing closer to the event horizon from a distance than they are in their own reference frame. But when you calculate the radius of a black hole from its mass you are calculating distance from singularity to event horizon INSIDE the black hole. To what extent is space contracted inside the event horizon and how is the distance from singularity to event horizon translated to the apparent diameter astronomers see?

 

[JesseM]

Yes, but that's not the problem. Just as the measurement of time depends of one's reference frame, so does the measurement of distance. Outside a black hole space is increasingly contracted in a radial direction resulting in objects appearing closer to the event horizon from a distance than they are in their own reference frame.

Don't understand what you mean by "in their own reference frame", unlike with inertial frames in SR there's no unique way to construct a "reference frame" for a given object in curved spacetime, there are an infinite number of coordinate systems where they are at rest. The radius of a black hole is normally defined in Schwarzschild coordinates, I presume it wouldn't be a problem to figure out theoretically the relation between apparent visual size to an observer at a given distance and the radius in Schwarzschild coordinates, this should just be an optics problem in GR (and you can find animations created by physicists, like the ones here, showing exactly what a black hole of some known Schwarzschild radius would look like to an observer at various distances).

But when you calculate the radius of a black hole from its mass you are calculating distance from singularity to event horizon INSIDE the black hole.

Not really, from the perspective of someone inside the black hole the singularity is at a timelike separation from them, not a spacelike separation--it doesn't lie in any particular spatial direction, rather it lies in their future, like the Big Crunch in a closed universe. The radius is just the difference between the radial coordinate of the singularity and the radial coordinate of the event horizon in Schwarzschild coordinates, but the radial coordinate is actually timelike, not spacelike, inside the horizon.

 

[skeptic2]

Don't understand what you mean by "in their own reference frame", unlike with inertial frames in SR there's no unique way to construct a "reference frame" for a given object in curved spacetime, there are an infinite number of coordinate systems where they are at rest.

I did not say inertial reference frame and used that term instead of proper spatial dimension to mean the spatial equivalent of proper time because I thought it would be easier to understand. I will use proper distance from now on.

The radius of a black hole is normally defined in Schwarzschild coordinates, I presume it wouldn't be a problem to figure out theoretically the relation between apparent visual size to an observer at a given distance and the radius in Schwarzschild coordinates, this should just be an optics problem in GR (and you can find animations created by physicists, like the ones here, showing exactly what a black hole of some known Schwarzschild radius would look like to an observer at various distances) .
Not really, from the perspective of someone inside the black hole the singularity is at a timelike separation from them, not a spacelike separation--it doesn't lie in any particular spatial direction, rather it lies in their future, like the Big Crunch in a closed universe. The radius is just the difference between the radial coordinate of the singularity and the radial coordinate of the event horizon in Schwarzschild coordinates, but the radial coordinate is actually timelike, not spacelike, inside the horizon.

This is exactly the problem. The radius from the event horizon to the singularity is measured in seconds, not meters, yet your formula for converting the radius to mass assumes the radius is measured in meters.
Briefly, since the radius of a black hole lies inside of the event horizon, doesn’t it have to be described in units of time instead of distance? How is that radius calculated in proper time? How are Schwarzschild proper time coordinates inside the event horizon converted to coordinate distance coordinates outside the event horizon?

 

[twhites]

Further discussion on the points raised by JesseM in post #7.

First, I think there may be some confusion of acronyms. In my first post, "ffo" stands for "far-field observer" (i.e., us), and not "free-falling observer". In addition, the observer near the horizon my remarks referred to was intended to be a momentarily stationary inertial observer, so he/she/it is free falling, but instantaneously has no velocity relative to the ffo or the putative horizon. This observer will of course rapidly acquire a very high velocity, but has none at the instant that the measurements I am postulating are made.

For such a stationary inertial observer we appear to agree that the red-shift/blue-shift argument is valid, with divergences to zero and infinite frequency at r= 2M as we have discussed. I think we will also agree that the frequency of the light that each observer (i.e., the ffo and the stationary one near the "horizon") receives from the other reflects each observer's perception of the other's clock. These clocks (the t coordinate in the Schwartzchild Metric) measure the proper time for each of these observers in their particular environment--and one, by this argument, runs infinitely faster than the other (in the limit of r->2M). I would then say QED to the proposition that BH's can't form in our universe, but then I would be disagreeing with a lot of people very much smarter than I am.

(On the matter of coordinates: Surely there is only one clock that measures proper time for each observer. Is that clock not the coordinate referred to as t in the Schwartzchild metric? So while you can, indeed, use any convenient valid coordinate system in order to do calculations, don't you still have to transform the results back into (t, r, theta, phi) in order to relate to experimental data--actual clocks, for example?)

Then there's the question of distances, and their measurement. Proper times are measured things, with a local clock. On the other hand, proper distances can only be inferred, not measured directly, since they are space-like. While proper distances are intervals that all observers can agree on, by making the appropriate inference, that is not the issue here, it seems to me. Like the time measurements we have been discussing, what is important (to us, as observers watching a stellar collapse, say) is the distance that we measure. . Distances are defined by light transit times, more or less as I describe--this is why physicists can measure time in meters (I personally think it would be better to measure distances in seconds, as astronomers do, but the two units are equivalent, through the definition of the value of light-speed, so it doesn't matter). And the measured distance, in this sense, from an ffo to the horizon diverges at 2M, if I've done the math correctly. Since the velocity of the freely falling observer with respect to the ffo (us) can't exceed the speed of light (i.e., his/her/its velocity remains finite), once again it appears to me that even such an observer, whose velocity approaches c at r->2M, can't reach the horizon in a finite (proper) time from the standpoint of the ffo (that is, us). Other observers might view things differently, especially the freely-falling one him/her/itself, who measures a very much smaller distance (a finite one, in fact) to the horizon because of the Lorentz contraction resulting from his/her/its high velocity relative to it. He/she/it gets there, as everyone agrees, in a finite time, where the time is the proper time by his/her/its own clock.

In other words, I still don't get it. And not having to worry about the interior of the black hole is, I hope you will agree, a very attractive feature of this viewpoint.

 

[Trenton]

I am baffled by the notion that an outside, safe distance observer would see an object fall towards a black hole and freeze at the event horizon. Is this an established interpretation of GR or is it a popular myth?

If it were true it would mean that the infalling object would have attained the speed of light at the point of the swartzchild radius and this could only happen if the gravitational potential at Rs was infinite. Unless I am mistaken and I fully accept I could be, grav pot can't be infinite at a finite distance from a point mass unless the mass was infinite - which nobody is suggesting.

Where am I going wrong?

 

[skeptic2]

Shouldn't the freefalling velocity of a mass infalling from infinity be equal in magnitude to the escape velocity at every point along its trajectory? This means that the infalling object does reach the speed of light at the event horizon without the gravitational potential being infinite.

 

[JesseM]

First, I think there may be some confusion of acronyms. In my first post, "ffo" stands for "far-field observer" (i.e., us), and not "free-falling observer". In addition, the observer near the horizon my remarks referred to was intended to be a momentarily stationary inertial observer, so he/she/it is free falling, but instantaneously has no velocity relative to the ffo or the putative horizon. This observer will of course rapidly acquire a very high velocity, but has none at the instant that the measurements I am postulating are made.

For such a stationary inertial observer we appear to agree that the red-shift/blue-shift argument is valid, with divergences to zero and infinite frequency at r= 2M as we have discussed. I think we will also agree that the frequency of the light that each observer (i.e., the ffo and the stationary one near the "horizon") receives from the other reflects each observer's perception of the other's clock. These clocks (the t coordinate in the Schwartzchild Metric) measure the proper time for each of these observers in their particular environment--and one, by this argument, runs infinitely faster than the other (in the limit of r->2M).

But this limit seems unphysical, since no massive observer can be even instantaneously at rest relative to the event horizon in a local sense, as the velocity of the event horizon is c as measured locally. And if the free-falling observer instantaneously comes to rest in Schwarzschild coordinates at any finite distance above the event horizon, no matter how small, the blueshift of the far-field observer will always be finite at that moment, and will remain finite as they fall down to the event horizon and pass through it.

Incidentally, did you look at the stuff on Rindler coordinates and the Rindler horizon in flat spacetime? From the perspective of an inertial frame, the Rindler horizon is a light-like worldline moving at c, and observers who are at rest in Rindler coordinates at positions closer and closer to the horizon will have greater and greater proper accelerations, so as seen in an inertial frame their velocities approach c more and more rapidly. See the diagram from this page, drawn from the perspective of an inertial frame, with the Rindler horizon as a dotted line and observers at rest in Rindler coordinates as black hyperbolas:

So, now consider an inertial observer who is momentarily at rest relative to one of these Rindler observers, i.e. an observer with a straight worldline whose slope is tangent to one of those hyperbolas at the moment it crosses it. You can see that if we consider the limit of hyperbolas closer and closer to the horizon, then in this limit the slope of the inertial observer whose worldline is tangent to a given hyperbola would approach the slope of a light ray, yes? And of course in the limit as your velocity in this inertial frame approaches c, the blueshift you will see from an observer approximately at rest in this frame (like a "far-field" hyperbola whose acceleration is negligible) approaches infinity too. But surely you wouldn't claim that this demonstrates that no object can actually cross the Rindler horizon, since the Rindler horizon is just an arbitrary light-like worldline in flat spacetime.

(On the matter of coordinates: Surely there is only one clock that measures proper time for each observer.

Yes, the clock that moves along with them on their own worldline--that's the physical definition of proper time.

Is that clock not the coordinate referred to as t in the Schwartzchild metric?

No, only for the "far-field observer" does their own proper time match the coordinate time, for observers closer to the black hole, whether they are falling or hovering at a constant radius or something else, the difference in t-coordinate between two events on their worldline will not usually be the same as the proper time along their worldline between those same two events.

Then there's the question of distances, and their measurement. Proper times are measured things, with a local clock. On the other hand, proper distances can only be inferred, not measured directly, since they are space-like.

As I said you can measure proper distance by using a series of (infinitesimally) short free-falling rulers, such that the ends of the rulers all line up at the same moment the (timelike) worldline of each end crosses the spacelike worldline, and such that each ruler's timelike worldline is orthogonal to the spacelike worldline at the moment they cross.

For example, if you just wanted to measure your proper distance to the event horizon along a spacelike worldline of constant t-coordinate in Schwarzschild coordinates, you could imagine a chain of short freefalling rulers which are all instantaneously at rest at the same t-coordinate, with their ends all lined up at that t-coordinate, and the total proper distance would just be the sum of all the short distances along each ruler. This proper distance would have a finite value: if your radius is r, then the proper distance to the horizon along a line of constant t-coordinate is calculated on p. 824 of MTW's Gravitation to be $$\int_{2M}^{r} | g_{rr} |^{1/2} \, dr = [r(r - 2M)]^{1/2} + 2M \, ln |(r/2M - 1)^{1/2} + (r/2M)^{1/2} |$$.

 

[JesseM]

I am baffled by the notion that an outside, safe distance observer would see an object fall towards a black hole and freeze at the event horizon. Is this an established interpretation of GR or is it a popular myth?

It's correct in principle, though it depends on the idealization that the falling object emits light in a completely continuous way (instead of in discrete photons, which would imply there is a last photon emitted before crossing the horizon) and that the outside observer can detect light of arbitrarily high redshift. See this section of the Usenet Physics FAQ, particularly the part with the title "Won't it take forever for you to fall in? Won't it take forever for the black hole to even form?"

If it were true it would mean that the infalling object would have attained the speed of light at the point of the swartzchild radius

Why would you think it has anything to do with attaining the speed of light? Gravitational time dilation and redshift is a separate phenomenon from velocity-based time dilation and redshift.

 

[Trenton]

Shouldn't the freefalling velocity of a mass infalling from infinity be equal in magnitude to the escape velocity at every point along its trajectory? This means that the infalling object does reach the speed of light at the event horizon without the gravitational potential being infinite.

Yes these are equal at every point along the trajectory which means there has to be a problem with the other assertion - that the escape velocity is c at Rs.

It would take an infinite amount of energy to accelerate matter to c and infinity exceeds the mass energy equivilence of a black hole no matter how large it is. We therefore have something of a conundrum.

As a possible resolution to this, it seems to me that the formula for Rs, which does not have a term for the lorentz contraction, could do with one. Either way the gravitational potential from infinity to Rs does equate to the energy required to accelerate an object to 0.86c.

Can anyone point out in precise terms, if I have this wrong, how and why. This has been driving me mad for quite a while!

 

[JesseM]

Yes these are equal at every point along the trajectory which means there has to be a problem with the other assertion - that the escape velocity is c at Rs.

It would take an infinite amount of energy to accelerate matter to c and infinity exceeds the mass energy equivilence of a black hole no matter how large it is. We therefore have something of a conundrum.

Why do you think the fact that "the escape velocity is c at Rs" implies that the matter itself has been accelerated to c once it crosses the horizon? That's not true at all, from the perspective of any local freefalling frame (where the laws of physics locally look the same as in SR, thanks to the equivalence principle) the matter still has a sub-c velocity as it crosses the horizon, while the horizon itself will be moving outwards at c in this frame. And if you're using some different type of coordinate system like Schwarzschild coordinates, you have to be careful in your statement about escape velocities, light itself does not have a constant coordinate speed in Schwarzschild coordinates and in fact the speed of light approaches zero as you approach the horizon in these coordinates, though it's still true that in the limit as the distance from the horizon approaches zero the speed needed for a freefalling object to escape approaches the speed of light at that distance (it might be more intuitive to use Kruskal-Szekeres coordinates where light does have a constant coordinate speed at all points, but here again we find that the horizon itself is moving outwards at the speed of light).

 

[twhites]

I can't figure out how to use the quote buttons, but all the quotes are from JesseM in the following.

"But this limit seems unphysical, since no massive observer can be even instantaneously at rest relative to the event horizon in a local sense, as the velocity of the event horizon is c as measured locally. And if the free-falling observer instantaneously comes to rest in Schwarzschild coordinates at any finite distance above the event horizon, no matter how small, the blueshift of the far-field observer will always be finite at that moment, and will remain finite as they fall down to the event horizon and pass through it."

Yes, of course, both of these statements are true; the unphysical limit you speak of is exactly the first point that lies infinitely far in the future (from the standpoint (always) of a far-field observer), so it is indeed unphysical. It is a limit point, however; the divergences we have discussed apply in that limit. This is basically the idea that convinces me that the horizon can't form--so you are absolutely correct, in my view.

"Yes, the clock that moves along with them on their own worldline--that's the physical definition of proper time."


Once again, of course this is true. I believe that it is also true that the proper time of other observers at smaller (r>2M) have proper times that are proportional to the (square root of) the temporal part of the line element, not so? This is what I actually meant, though it is not what I said. Sorry for the confusion.

"As I said you can measure proper distance by using a series of (infinitesimally) short free-falling rulers, such that the ends of the rulers all line up at the same moment the (timelike) worldline of each end crosses the spacelike worldline, and such that each ruler's timelike worldline is orthogonal to the spacelike worldline at the moment they cross."


Your procedure will give, it is true, the proper distance, since it corresponds to integrating the dt = 0 interval. But I maintain that it is in fact an inference, since at each stage, you have to know the length of the ruler, and you have to calculate (infer) that length; you don't just measure it. By definition, in flat space-time, you measure the length by measuring the time it takes for a photon to go from one end of the ruler to the other. But that does not give the proper length of the ruler in a curved space-time like the one we are talking about. In order to get the proper length you have to use the metric to correct for the space-time curvature between one end of the ruler and the other. Since your procedure is stated to yield the proper distance from the ffo to the horizon, implicitly this is what you have done, and I agree that the proper distance (calculated this way) is finite. However, if you don't make the inference (essentially an assumption about simultaneity), and measure what I have come to understand is called the "radar distance", i.e., by simply measuring the light transmission time and then dividing by c, the distance you get is infinite (this is really pretty obvious, since it's clear that light that gets as far as the horizon will never get back to the far field). However, it's less obvious that the transit time for the light pulse smoothly increases to infinity as the limit as r->2M, but I believe it does.
I take this to mean that even light takes forever to reach the horizon, as timed by a distant observer. And since everything else moves slower than that, no massive object gets there either.

One more comment, about the universe ending in a blue flash according to the observer falling into the BH. I think I disagree that this is not what will be seen. The reason is that everything outside of the horizon has shrunk as far as the observer near the horizon is concerned. The radial line element (1-2M/r)^-1 dr^2 is the length of the spatial basis vector (one-form?) at position r in terms of that in the far-field (true?); it gets very large as r->2M (space stretches near the mass). Just turn this around (invert the coordinate transformation). This inversion implies that, in terms of the line element near the horizon, the line element far away is relatively very small. The result is that light from the end of the universe doesn't have to travel very far. And besides, according to me, it has forever to do it, since time is very, very slow at r just slightly larger than 2M. Thus, blue flash. But of course, it never actually happens, according to me, since the horizon is never achieved.

(I have looked at the Rindler coordinates, and have some comments about them also, but this reply is too long already) Once again, however, thanks for your continued patience and input.

 

[skeptic2]

Why would you think it has anything to do with attaining the speed of light? Gravitational time dilation and redshift is a separate phenomenon from velocity-based time dilation and redshift.

Why do you think the fact that "the escape velocity is c at Rs" implies that the matter itself has been accelerated to c once it crosses the horizon? That's not true at all, from the perspective of any local freefalling frame (where the laws of physics locally look the same as in SR, thanks to the equivalence principle)...

This is an interesting pair of quotes. In the first notes that gravitational dilation is different form velocity based time dilation. The second one refers to the equivalence principle which says that there is essentially no difference between acceleration due to gravity and acceleration due to a rocket.

Yes these are equal at every point along the trajectory which means there has to be a problem with the other assertion - that the escape velocity is c at Rs.

It would take an infinite amount of energy to accelerate matter to c and infinity exceeds the mass energy equivilence of a black hole no matter how large it is. We therefore have something of a conundrum.

As a possible resolution to this, it seems to me that the formula for Rs, which does not have a term for the lorentz contraction, could do with one. Either way the gravitational potential from infinity to Rs does equate to the energy required to accelerate an object to 0.86c.

Can anyone point out in precise terms, if I have this wrong, how and why. This has been driving me mad for quite a while!

An object freely falling toward a black hole it is not being accelerated in the usual sense. The object is at rest during its fall. If energy is expended "accelerating" an object toward a black hole, where does that energy come from? Certainly not from the black hole. [If one assumes that the mass or momentum of an object increases as the object approaches c, it would make no difference because the gravity would accelerate that relativistic mass just as fast.] So yes, the object does reach c at the event horizon, if it ever in fact reaches the event horizon, and no it doesn't take an infinite amount of energy to do it.

When someone says that an object slows down and stops as it approaches the event horizon from the perspective of a distant observer, you must realize that this is because of both time dilation AND the fact that space is increasingly contracted close to the event horizon. This means that to a distant observer, a meter of radial distance appears to be much shorter the same way objects are contracted in length when traveling at relativistic speeds. A distant observer sees a contracted distance traveled in dilated time which results in the falling object coming to a halt at the event horizon. Don't forget that in order to reach or cross the event horizon the object must overcome the complete dilation of time and the radial distance being contracted to zero.

 

[JesseM]

This is an interesting pair of quotes. In the first notes that gravitational dilation is different form velocity based time dilation. The second one refers to the equivalence principle which says that there is essentially no difference between acceleration due to gravity and acceleration due to a rocket.

The equivalence principle only works in an infinitesimally small region of curved space time, while gravitational time dilation deals with clock time vs. coordinate time in some coordinate system covering a non-infinitesimal region of curved spacetime, like Schwarzschild coordinates.

When someone says that an object slows down and stops as it approaches the event horizon from the perspective of a distant observer, you must realize that this is because of both time dilation AND the fact that space is increasingly contracted close to the event horizon. This means that to a distant observer, a meter of radial distance appears to be much shorter the same way objects are contracted in length when traveling at relativistic speeds. A distant observer sees a contracted distance traveled in dilated time which results in the falling object coming to a halt at the event horizon. Don't forget that in order to reach or cross the event horizon the object must overcome the complete dilation of time and the radial distance being contracted to zero.

In Schwarzschild coordinates the time dilation (amount of coordinate time for a clock to tick forward by a given amount) does approach infinity at the horizon but this is just a feature of this particular coordinate system, other coordinate systems in the same spacetime don't have this feature. It is likewise true in flat spacetime that you can construct coordinate systems like this, I already mentioned the example of Rindler coordinates, in these coordinates the time dilation of a clock approaching the Rindler horizon approaches infinity too, but obviously nothing problematic happens at the Rindler horizon (which is just a lightlike worldline) in an inertial coordinate system. Would you say about an object approaching the Rindler horizon that "Don't forget that in order to reach or cross the event horizon the object must overcome the complete dilation of time and the radial distance being contracted to zero"?

Incidentally, I don't really understand that comment about the radial distance being contracted to zero, either--as I noted earlier, the proper distance to the horizon to the horizon is finite, and that proper distance represents measurements on a series of short rulers, each moving inertially in locally inertial frame, so I don't think it could be true that the ratio between the rulers' length in Schwarzschild coordinates and their length in a locally inertial rest frame is approaching zero.

 

[JesseM]

"But this limit seems unphysical, since no massive observer can be even instantaneously at rest relative to the event horizon in a local sense, as the velocity of the event horizon is c as measured locally. And if the free-falling observer instantaneously comes to rest in Schwarzschild coordinates at any finite distance above the event horizon, no matter how small, the blueshift of the far-field observer will always be finite at that moment, and will remain finite as they fall down to the event horizon and pass through it."

Yes, of course, both of these statements are true; the unphysical limit you speak of is exactly the first point that lies infinitely far in the future (from the standpoint (always) of a far-field observer), so it is indeed unphysical.

Why "infinitely far in the future"? This wasn't supposed to be a temporal series, but rather a spatial series of freefalling objects which instantaneously achieve a velocity of zero in Schwarzschild coordinates at varying distances above the horizon, the time that each object achieves that velocity can be anything we want. Why can't you arrange things so that object #0 comes to rest at r=2Rs at t=2, and object #1 comes to rest at r=1.5Rs at t=1.5, and object #2 comes to rest at r=1.25Rs at t=1.25, and in general object N comes to rest at r=(1 + (1/2^N))*Rs at t=(1 + (1/2^N))?

It is a limit point, however; the divergences we have discussed apply in that limit. This is basically the idea that convinces me that the horizon can't form--so you are absolutely correct, in my view.

Please don't be "convinced that the horizon can't form", or you will be falling victim to the Dunning-Kruger effect where people with little knowledge of a subject routinely overestimate their own ability to make sound judgments about it. Do you really think it is likely that you have hit on a brilliant argument that proves that thousands of physicists over the decades have been wildly mistaken in their views about black holes? Even if the argument seems convincing to you at first glance, a little basic intellectual humility should lead you to think it's very likely to be flawed even if you can't quite understand why, and your purpose on this forum should be to try to figure out where the flaw is rather than trying to make arguments in favor of this "view".

But if that alone isn't enough to convince you there is probably a flaw in your reasoning, once again I urge you to consider the issue of Rindler coordinates. Everything I said above about the possibility of a series of free-falling objects which temporarily come to rest above the horizon in Schwarzschild coordinates would also apply to a series of inertial objects which temporarily come to rest above the Rindler horizon in Rindler coordinates--the limit in this case would be equally unphysical since no massive object can even instantaneously be at rest relative to the Rindler horizon (since the Rindler horizon is a lightlike worldline). It would be possible for a photon to remain on the Rindler horizon, but the same is true in the case of a black hole where a photon could remain on the event horizon. This is more obvious if you switch from Schwarzschild coordinates to Kruskal Szekeres coordinates, indeed if you look carefully you may notice that when lines of constant R and t coordinates in Schwarzschild coordinates are drawn in a Kruskal diagram, they look exactly like lines of constant R and t coordinates in Rindler coordinates when they're drawn in a Minkowski diagram, and in both cases the horizon is drawn as a straight line at 45 degrees. So there are quite a lot of ways in which Kruskal coordinates are to Schwarzschild coordinates as Minkowski coordinates are to Rindler coordinates.

"Yes, the clock that moves along with them on their own worldline--that's the physical definition of proper time."


Once again, of course this is true. I believe that it is also true that the proper time of other observers at smaller (r>2M) have proper times that are proportional to the (square root of) the temporal part of the line element, not so?

If you're just talking about objects hovering at a constant Schwarzschild radius that's true, but obviously it's more complicated for an object with changing radius, such as an object in free-fall.

"As I said you can measure proper distance by using a series of (infinitesimally) short free-falling rulers, such that the ends of the rulers all line up at the same moment the (timelike) worldline of each end crosses the spacelike worldline, and such that each ruler's timelike worldline is orthogonal to the spacelike worldline at the moment they cross."


Your procedure will give, it is true, the proper distance, since it corresponds to integrating the dt = 0 interval. But I maintain that it is in fact an inference, since at each stage, you have to know the length of the ruler, and you have to calculate (infer) that length; you don't just measure it. By definition, in flat space-time, you measure the length by measuring the time it takes for a photon to go from one end of the ruler to the other. But that does not give the proper length of the ruler in a curved space-time like the one we are talking about.

But remember I said I was talking about "short" free-falling rulers--what I was really getting at was free-falling rulers of infinitesimal length (or considering the limit as the length of the rulers goes to zero) so that by the http://www.einstein-online.info/spotlights/equivalence_principle you can find a local free-falling frame covering the region occupied by each ruler when it takes its measurement, and the laws of physics in that local frame will be just like those of SR. Hopefully you agree that in SR there is a standard way to define the length of an inertial object in its own rest frame, the same definition can be used here.

One more comment, about the universe ending in a blue flash according to the observer falling into the BH. I think I disagree that this is not what will be seen.

Again, a little intellectual humility, do you really think that your qualitative argument is likely to be right when it differs from the opinion of all those physicists who have analyzed the problem in detail?

The reason is that everything outside of the horizon has shrunk as far as the observer near the horizon is concerned.

What do you mean as far as they are concerned? What measurement procedure are you suggesting they use to measure the length of distant objects relative to the length of their own ruler, if they can't compare them locally?

The radial line element (1-2M/r)^-1 dr^2 is the length of the spatial basis vector (one-form?) at position r in terms of that in the far-field (true?)

The radial line element tells you how, if you have a purely radial spacelike worldline, the proper distance along this worldline is varying relative to the radial coordinate. It gets large as r approaches the Schwarzschild radius, which means if you have two equal-sized increments of the r-coordinate, the proper distance along the increment closer to the horizon is larger than the proper distance along the increment farther away. I don't see how this would have anything to do with comparing lengths near the horizon with those of the far-field, I'm not a GR expert but I don't think you can really compare lengths in different regions of curved spacetime in any coordinate-independent way.

Just turn this around (invert the coordinate transformation).

Invert what coordinate transformation? To have a coordinate transformation you need two different coordinate systems, you've only been talking about a single coordinate system, Schwarzschild coordinates. I suppose one could do a transformation on the radial coordinate so that one had a new coordinate system similar to Schwarzschild coordinates but where the radial distance from the horizon matched the proper distance...

This inversion implies that, in terms of the line element near the horizon, the line element far away is relatively very small. The result is that light from the end of the universe doesn't have to travel very far.

Again a purely qualitative argument, do you imagine that every physicist in history who's done the detailed calculations got the math wrong? Also note that Schwarzschild coordinates are provably identical to Kruskal-Szekeres coordinates in all their physical predications about local events outside the horizon (and inside too, but Schwarzschild coordinates break down on the horizon) like whether light from a given event reaches a given observer before he crosses the horizon...if you familiarize yourself with the way light worldlines work in Kruskal-Szekeres coordinates it's easy to see that there are plenty of events whose light doesn't reach the observer before he reaches the horizon.

(I have looked at the Rindler coordinates, and have some comments about them also, but this reply is too long already)

OK, please do, and also at Kruskal-Szekeres coordinates (besides the wiki page you might also look at some of the textbook diagrams I posted [post=2336347]here[/post]).

 

[skeptic2]

And if you're using some different type of coordinate system like Schwarzschild coordinates, ...

I'm an RF engineer and RF engineers deal with impedances which sometimes go to infinity (an open circuit). Infinities are difficult to work with mathematically so Phillip Smith invented a coordinate conversion to make it easier http://en.wikipedia.org/wiki/Smith_chart. The conversion from impedance (Z) to a Smith Chart coordinate (ρ) is ρ = (Z-1)/(Z+1). This transforms the infinity of an open circuit to the value of 1. But no engineer really believes that when the conversion is done, an open circuit actually has a value of 1. After calculations are done using the Smith Chart coordinates, they are converted back to Z to obtain working values.

Likewise Kruskal-Szekeres coordinates are just a transformation of Schwarzschild coordinates that may not show time dilation or space contraction. U and V in Kruskal-Szekeres coordinates are not the same as r and t, afterall. That doesn't mean that time dilation and space contraction aren't real, they just don't show up in K-S coordinates. It would be a mistake to assume that because they don't exist in K-S coordinates, they don't exist.

 

[JesseM]

I'm an RF engineer and RF engineers deal with impedances which sometimes go to infinity (an open circuit). Infinities are difficult to work with mathematically so Phillip Smith invented a coordinate conversion to make it easier http://en.wikipedia.org/wiki/Smith_chart. The conversion from impedance (Z) to a Smith Chart coordinate (ρ) is ρ = (Z-1)/(Z+1). This transforms the infinity of an open circuit to the value of 1. But no engineer really believes that when the conversion is done, an open circuit actually has a value of 1. After calculations are done using the Smith Chart coordinates, they are converted back to Z to obtain working values.

Likewise Kruskal-Szekeres coordinates are just a transformation of Schwarzschild coordinates that may not show time dilation or space contraction. U and V in Kruskal-Szekeres coordinates are not the same as r and t, afterall. That doesn't mean that time dilation and space contraction aren't real, they just don't show up in K-S coordinates. It would be a mistake to assume that because they don't exist in K-S coordinates, they don't exist.

In relativity time dilation and length contraction are inherently coordinate-dependent quantities, there's no way to measure them with physical measuring-devices in a way that doesn't involve programming the device to keep track of its position/time coordinates in some (arbitrarily chosen) coordinate system. And all coordinate systems are equally valid in general relativity, see the discussion of "diffeomorphism invariance" on this page. I'm pretty sure it's not true of impedance that one can't measure it using physical devices in a way that doesn't require an arbitrary choice of coordinate system, and as long as that's not true of impedance your analogy doesn't work, the Phillip Smith coordinate chart would just be one that distorts the "true" impedance as measured by physical devices.

 

[Trenton]

Having been advised not to be suduced by this thory or that, I will set out exactly what I am convinced of so any flaws are more visible!

My starting points are a) that matter cannot under any circumstances, travel as fast as light in vacuo and b) the speed of light will always be measured the same.

Using the SR based example of a rocket that can forever accelerate at 1g or any other value, (a) implies it would take forever to reach a state of affairs in which it stopped being overtaken by photons following it's exact trajectory - even though according to it's own clocks it could hop from one galaxy to another in absurdly short periods of time. And (b) implies that the rocket would measure the speed of the overtaking photons as c.

In the GR black hole, is there anything that delivers the killer blow so that an infalling object stops being overtaken by photons on the same trajectory? Yes the gravity gets stronger and stronger but it does not have forever to do this before Rs is breached. Only when the object falls to the center of the hole do we get the eqivilent of the SR rocket having traveled forever.

It seems to me that in just the same way that an object can never reach c, that time cannot actually be stopped altogether (except perhaps at the center of the black hole). And nor can light be red-shifted to zero.

And I see no reason why black holes need this assertion that escape velocity reaches c at Rs, in order to stay black. The red-shift of escaping light would be extreme enough to turn extreme gamma rays into mains transformer hum. That would look pretty close to black to most observers.

Please can someone point out how GR specifically overrides (a) and/or (b) ?

 

[JesseM]

My starting points are a) that matter cannot under any circumstances, travel as fast as light in vacuo and b) the speed of light will always be measured the same.

Using the SR based example of a rocket that can forever accelerate at 1g or any other value, (a) implies it would take forever to reach a state of affairs in which it stopped being overtaken by photons following it's exact trajectory - even though according to it's own clocks it could hop from one galaxy to another in absurdly short periods of time. And (b) implies that the rocket would measure the speed of the overtaking photons as c.

Yes, both are correct.

In the GR black hole, is there anything that delivers the killer blow so that an infalling object stops being overtaken by photons on the same trajectory?

An infalling object with nonzero mass will always find that photons can overtake and pass it. Again I recommend familiarizing yourself with Kruskal-Szekeres coordinates where light rays always move at 45 degrees while massive objects always have worldlines whose slope is less than 45 degrees from the vertical, see the discussion on the wiki page here and the diagrams from the textbook Gravitation I posted [post=2336347]here[/post].

It seems to me that in just the same way that an object can never reach c, that time cannot actually be stopped altogether (except perhaps at the center of the black hole).

What does "time cannot actually be stopped" even mean? You don't seem to understand that all statements about time dilation are relative to your choice of coordinate system, there is no objective frame-independent sense in which any clock slows down or stops in any situation whatsoever in relativity. I could design a coordinate system where you, walking from one end of your room to another, move slower and slower relative to coordinate time as you cross some arbitrary boundary at the center of your room, such that it takes you an infinite coordinate time to actually cross this boundary, and the rate your clock ticks approaches zero as your distance from the boundary approaches zero. This coordinate system would be just as valid as any other in general relativity!

And nor can light be red-shifted to zero.

I presume you mean redshifted to infinity? Anyway, for some light to be redshifted to infinity basically just means the light can never catch up with you. If you consider an observer with constant proper acceleration, as seen in an inertial frame his worldline will look like a hyperbola, and the asymptote of this hyperbola (the line it approaches but never quite touches) is just the worldline of a light beam, a light beam that never manages to catch up with the accelerating observer even though the distance is constantly shrinking (because the accelerating observer's own velocity is getting closer and closer to c so the rate at which the distance is shrinking keeps getting slower), as seen in this diagram the Rindler horizon page:

The light ray's worldline forms the "Rindler horizon" for that accelerating observer. If you imagine other light rays from events just to the right of the Rindler horizon (more straight lines at 45 degrees originating from events along the x-axis, slightly to the right of the origin where the x and t axes cross), then these light rays would catch up with the accelerating observer eventually. However, the closer the events that originated the light rays were to the Rindler horizon, the longer it would take for the light rays to reach the worldline of the accelerating observer--and since the accelerating observer's velocity is getting closer and closer to c, the longer it takes for light to catch up with him the more redshifted the light will be from his perspective when it reaches him. In fact, if you look at the limit as the events creating the light rays approach zero distance from the horizon, the redshift experienced by the accelerating observer approaches infinity. So in this sense we can say that the light from an event on the Rindler horizon would be infinitely redshifted, but that doesn't actually mean any "infinitely redshifted light" ever catches up with the accelerating observer, you can see from the diagram that a light ray traveling along the Rindler horizon just never reaches the observer. On the other hand, a normal observer crossing the horizon wouldn't see a light ray traveling along the horizon as infinitely redshifted, since this observer would have some finite velocity smaller than c as he crossed the horizon. The increasing redshift from events closer to the horizon experienced by the accelerating observer is not an intrinsic property of the light from those events, it has to do with the fact that the closer the event is to the horizon, the higher along the t-axis you have to go before the light from that event intersects the worldline of the accelerating observer, and the higher you go along the t-axis the closer the accelerating observer's own velocity is to c.

And if you look at the Kruskal-Szekeres diagram it works exactly the same way--the event horizon is a straight line at 45 degrees, and the worldline of an observer hovering at constant Schwarzschild radius is a hyperbola which never touches the event horizon, and all light rays are also straight lines at 45 degrees. So it's for similar reasons that light from events closer and closer to the horizon will be increasingly redshifted as seen by the constant-Schwarzschild-radius observer, and "infinite redshift at the horizon" is just a statement about limits, a light ray from an event actually at the horizon will just never reach this observer, but an infalling observer who does cross the horizon will see this light as having a finite redshift.

The point is, you can't really reject statements about "infinite redshift" in the case of the black hole without also rejecting these statements in the case of the Rindler observer and the Rindler horizon. But as long as you understand that these are just statements about limits, and that there is never a situation where a light ray actually hitting a given observer can be said to have infinite redshift by that observer, then you should be able to see why there isn't really anything unphysical going on here.

And I see no reason why black holes need this assertion that escape velocity reaches c at Rs, in order to stay black. The red-shift of escaping light would be extreme enough to turn extreme gamma rays into mains transformer hum. That would look pretty close to black to most observers.

It's not that this assertion is made because physicists for some arbitrary reason want them to "stay black", it's because if you actually analyze the curvature of spacetime predicted by general relativity, you see that at the horizon something moving at exactly c will stay on the horizon forever, while something moving slower than c will inevitably fall inward and hit the singularity.

Please can someone point out how GR specifically overrides (a) and/or (b) ?

It doesn't override them! It's still true that any observer at or inside the horizon will measure photons to overtake and pass them, and to have a local velocity of c. If you think that conflicts with the notion that the "escape velocity is c" at the horizon, your reasoning has gone wrong somewhere. Did you read my earlier comment to you about this in [post=3243152]post 19[/post]?

Why do you think the fact that "the escape velocity is c at Rs" implies that the matter itself has been accelerated to c once it crosses the horizon? That's not true at all, from the perspective of any local freefalling frame (where the laws of physics locally look the same as in SR, thanks to the equivalence principle) the matter still has a sub-c velocity as it crosses the horizon, while the horizon itself will be moving outwards at c in this frame.

 

[Trenton]

JesseM,

I am very baffled as to what sense the horizon moves outward in this or any other frame but for now I shall limit my response to Rindler.

Can anything really be the other side of the rindler horizon in practice? Is there anywhere in the universe one can be where relative to you, other objects are moving away at such a speed that their light will never reach you?

This is one reason why I picked falling into a black hole as a scenario. An observer falling in and looking backwards still sees the stars because as you have confirmed, the photons following your trajectory still overtake you. And if you can't put yourself the other side of rindler by jumping into a black hole then how could it be done?

The idea that a light beam that never manages to catch up with the accelerating observer even though the distance is constantly shrinking (because the accelerating observer's own velocity is getting closer and closer to c so the rate at which the distance is shrinking keeps getting slower). This sounds like Zeno who defined all travel as a asymptotic process.

Yes I did mean red-shifted to infinity (so much for my lingustic pendantry). But to do it you have to attain c relative to the light source. I don't think you can achieve c relative to anything, ever.

Of course this all centers around what is meant by 'in the universe'. I will get into that later but for now I have an awful lot of reading to do ...

 

[pervect]

I have one remark that might be useful about an object crossing the event horizon at the speed of light.

The event horizon can be thought of as "trapped light" - or a "trapped null surface" to use the technical language.

A beam of light, a "photon" if you will, at the event horizon, traveling outward, will stay there. It's the only thing that CAN stay there - matter isn't able to stay at the event horizon - only light can.

Thus it's not really terribly surprising that the relative velocity of this trapped light relative to matter is equal to 'c'.

Thinking of the event horizon as a "place", a timelike world line, is the problem. It's not. It's a null surface, a trapped beam of light. The "point of view" of the event horizon doesn't exist anymore than the "point of view" of a photon does - the event horizon doesn't experience time, being a null surface.

So a better way to think of the event horizon relative to matter falling is to use the view of the infalling observer, who has a timelike worldline. And the light trapped at the event horizon has a relative velocity of 'c' relative to such an infalling observer, just as any other light does.

 

[JesseM]

JesseM,

I am very baffled as to what sense the horizon moves outward in this or any other frame but for now I shall limit my response to Rindler.

Do you understand the difference between inertial frames and non-inertial ones in SR, and that only in inertial frames it is a rule that light has a coordinate speed of c? If so, you need to also understand that in GR no coordinate covering a larger region of spacetime (like Schwarzschild coordinates) can be an "inertial" one, and light does not necessarily have a constant coordinate speed in non-inertial coordinate systems. However, an observer in free fall can construct a local inertial frame in a small patch of spacetime around them, and in this local inertial frame the laws of inertial frames in SR still apply (this is known as the equivalence principle, a very important idea in GR), such as the rule that light always moves at c. Because large coordinate systems in GR like Schwarzschild coordinates are non-inertial, it is quite possible that something which is at rest in such a non-inertial coordinate system could actually be moving at c as measured in the local inertial frames of observers passing it, and this is true of the black hole's event horizon.

Note that exactly the same is true of the Rindler horizon in the non-inertial Rindler coordinate system--it is at rest in these coordinates, but relative to any inertial frame in SR, the Rindler horizon is moving at c, as you can see from the diagram in my earlier post which was drawn from the perspective of an inertial frame.

Can anything really be the other side of the rindler horizon in practice?

Sure--any light ray in SR would be the Rindler horizon of some Rindler coordinate system, and it would always be possible to have an observer with constant proper acceleration who has this light ray as his own Rindler horizon (meaning the light ray gets arbitrarily close to him but never quite catches up as long as he maintains the same acceleration). So anytime a slower-than-light object crosses the path of a light ray anywhere in spacetime, it has crossed beyond a Rindler horizon.

Is there anywhere in the universe one can be where relative to you, other objects are moving away at such a speed that their light will never reach you?

In SR this depends on how you are moving! If you are undergoing constant proper acceleration forever, then you will have a Rindler horizon, and no light from events on or beyond that horizon will reach you. But at any time you could easily see light from events on or beyond the horizon just be ceasing your acceleration.

This is one reason why I picked falling into a black hole as a scenario. An observer falling in and looking backwards still sees the stars because as you have confirmed, the photons following your trajectory still overtake you. And if you can't put yourself the other side of rindler by jumping into a black hole then how could it be done?

Again, you can put yourself on the other side of your Rindler horizon just be stopping your acceleration. The Rindler horizon is something seen in special relativity, where gravity isn't taken into account, so it has nothing to do with black holes.

The idea that a light beam that never manages to catch up with the accelerating observer even though the distance is constantly shrinking (because the accelerating observer's own velocity is getting closer and closer to c so the rate at which the distance is shrinking keeps getting slower). This sounds like Zeno who defined all travel as a asymptotic process.

It's sort of like Zeno's paradox but a little different, Zeno's paradox is for constant-velocity motion (like an arrow) so the idea that the arrow will never cross a certain distance would actually be wrong, although the arrow's path can be divided up into an infinite series of shorter and shorter intervals the time for the arrow to cross each interval also grows shorter and shorter, and from calculus we know the sum of an ever-decreasing series can be a finite number, like 1 + 1/2 + 1/4 + 1/8 + 1/16 + ... = 2. In contrast, because of the acceleration of the Rindler observer, the light can never actually catch him in any finite amount of time, as long as he maintains the same acceleration.

Yes I did mean red-shifted to infinity (so much for my lingustic pendantry). But to do it you have to attain c relative to the light source. I don't think you can achieve c relative to anything, ever.

But again, in both the Rindler case and the black hole case it's a limit. There's no finite moment when you can say a particular light beam is redshifted by infinity relative to you, but the light from events closer and closer to the Rindler horizon/event horizon would be more and more redshifted by the time it reaches you (that time is further and further in the future the closer to the horizon it was emitted), and in the limit as the distance from the horizon approaches zero, both the redshift and the time needed for the light to catch up with you approach infinity.

 

[skeptic2]

An observer falling in and looking backwards still sees the stars because as you have confirmed, the photons following your trajectory still overtake you.

If an object falls towards a black hole from infinity, at large distances from the black hole its infalling velocity is equal in magnitude to the escape velocity at its location, is it not? If its infalling velocity is less than c at the event horizon, what factor has reduced the acceleration of the object?

If you were falling at c at the event horizon, you would still be able to see the stars behind you. Normally you would think that traveling at c, the light from the stars behind you would be red shifted to zero but the light from those stars is also falling towards the black hole and has been blue shifted by the same amount that your velocity would have red shifted it.

 

[Trenton]

skeptic2,

There might actually be a discrepancy. If we consider the case of particles accelerated in an accelerator, the velocity could never get close to c because the particle just gets heavier and so requires more energy per unit increase in velocity. If we then tried to extract energy from the particle we would find (except for losses in the equiptment) that it had as much energy as was put in.

When matter falls into a black hole though, increases in mass due to velocity are matched by extra gravitational attraction. This will not allow acceleration to c (in my view) because the falling process occurs over finite space and time.

But it makes escape impossible for matter because no matter how fast it's initial upward velocity, it will have to climb against a force that matches it. In this sense and only in this sense, can it be said that the escape velocity is c - and then this would be hopelessly misleading.

It is far better to say escape for matter is impossible because for all values of energy (an infinite scale), the gravity will always just overwhelm.

Saying the escape velocity is c is a nonsense in two ways. It suggest that if something could reach c (which would make it more massive by an infinite factor that the entire observable universe). And it leads to the notion that light cannot escape (this would only be true if time stands still at Rs which I think is wrong)

The idea that time stops at Rs where the gravitational field is finite seems to me to be wrong. The field equations do have two solutions for singularities, one at Rs and one at R=0. I think (much to the horror of others) that singularity means a region of stopped time and infinite density etc which CAN ONLY occur at R=0. I question if such a thing can actually form but I consider singularity at Rs to be invalid solutions. After all, if matter cannot reach c with finite energy how can time stop in a finite gravitational field?

 

[JesseM]

Trenton, the problem is that you try to import ideas from Newtonian gravity directly into GR and it doesn't work, GR is a very different theory. For example:

But it makes escape impossible for matter because no matter how fast it's initial upward velocity, it will have to climb against a force that matches it. In this sense and only in this sense, can it be said that the escape velocity is c - and then this would be hopelessly misleading.

There is no such thing as a gravitational "force" in general relativity! Gravitational effects are due to the fact that matter curves spacetime, and free-falling objects follow "geodesic" paths through this curved spacetime. Some intros to the basic concepts of GR, including curved spacetime and geodesics, can be found in this thread.

Saying the escape velocity is c is a nonsense in two ways. It suggest that if something could reach c (which would make it more massive by an infinite factor that the entire observable universe).

No, I've already told you it doesn't suggest that. In Newtonian physics it's true that an object falling in from infinity will have a velocity equal to the escape velocity when it hits the surface of a planet (or star or whatever), as seen in the inertial frame where the planet is at rest. But in GR escape velocity doesn't have this meaning! For one thing there is no possibility of an "inertial frame" in a large region of curved spacetime, you have an infinite variety of equally valid non-inertial frames and they all measure "velocity" differently (they also don't necessarily say that light itself has a constant velocity). As I explained to you, in a very small patch of spacetime you can have a "local inertial frame" where an object in free-fall is at rest, and where the laws of physics are like those of SR (including the idea that light moves at c), thanks to the equivalence principle. But in any local inertial frame at the horizon, it will be the horizon itself that's moving outward at c, not the free-falling object.

And it leads to the notion that light cannot escape (this would only be true if time stands still at Rs which I think is wrong)

There is no need for time to stand still at the horizon, you can pick coordinate systems where the time dilation experienced by clocks falling through the horizon is finite, and they cross the horizon at a finite coordinate time, like Eddington-Finkelstein coordinates (if you're weirded out by coordinate systems where the horizon is moving outward, you may like this one better since it does have the horizon at a fixed radial coordinate). In these coordinates, the fact that light can't escape from a black hole is explained by the fact that the future light cones of events closer to the horizon become increasingly tilted, until at the horizon the entire future light cone is inside or on the horizon, as illustrated in this diagram from the textbook Gravitation:

[PLAIN]http://www.valdostamuseum.org/hamsmith/DFblackIn.gif

The idea that time stops at Rs where the gravitational field is finite seems to me to be wrong.

Please note that the question of whether "time stops" has no coordinate-independent meaning, it simply depends on how you choose to define your coordinate system. As I mentioned earlier, one can define a coordinate system where "time stops" at any arbitrary boundary like the center of your room, in the sense that in this coordinate system the ratio of your clock time to coordinate time would approach zero as you approached the center, so the event of your crossing the boundary of the center would not be assigned any finite time-coordinate. Please read up on "diffeomorphism invariance" in this article if you have trouble with the idea that non-inertial coordinate systems can be defined in absolutely any way we want them to.

 

[Trenton]

As regard to co-ordinates, I accept a lack of fluency in transforming these at will. The co-ordinates I use by default are spherical and have the center of the black hole rather than any point on the event horizon as the origin. I can't add the time progressions to the coordinates at the moment because all the ones inside Rs seem not to progress according to the formula. This seems to have horrible implications for frame dragging.

I am fully aware that in making this statement I will get a barrage of complaints that I just don't get GR and to a certain extent I don't. I deny the charge of being stuck on Newton. I might be prepared to admit being stuck with SR. I my defence I would say I only did a module in SR. There wasn't one on GR on the course I did!

However, I think the GR concepts are simple enough but the language used to describe them are too easy to be taken out of context. In just the same way as one can choose any co-ordinate regime you want you can also choose how something is described. You can translate between co-ordinate systems and you can translate between terms. In fact doing so, highlights exactly where GR's advantage over SR and SR's advantage in Newtonian mechanics lie - and it serves to check concepts have been appropriately or inapproapriately mapped between the three different regimes.

Take the 'curvature of spacetime' and the geodesic. Light is said to travel in a straight line but it has to follow a geodesic so it describes a curved path. Such is the language of GR but this is surely not the only valid way of explaining it even if it might be the most convient method of applying the math. It is not invalid to say light falls under gravity even in so thinking, one might have trouble deriving the location of such as the photon sphere.

Another GR term that has been brought to my attention is that there is no such thing as the force of gravity - only the curvature of spacetime due to the presence of mass. This distinction is quite correct of course but why is it correct? Probably only because time dilation is a function of gravitational potential and not of field strength as might be tempted to think if the word 'force' were to leave one's lips instead of the word 'curvature'. Or is there a more fundamental reason for the distiction?

One of the things I have studied is NLP (neuro-linguistic programing). This makes very clear why it is important to explain the terms of a superior model in terms of an inferior one - it is to easy to think one has a complete theory in one's mind when in fact key tennets have been learned by rote or are the results of unsound conceptual leaps.

I want to make it absoutely clear I am NOT suggesting that anyone in this forum is making that mistake but I have in the past managed to tie up in knots, more than one public speaker at meetings on astronomy. This was not my intention as all I wanted was to understand certain assertions. In one example the speaker asserted that at the exact point where the radius of a nuetron star becomes equal to Rs (having been hit by one last cosmic ray) - the whole entity collapses to a point. This did not seem to me to be proven and since the equation of state and the degeneracy limits of neutonium is not fully known, I asked if the rusult could simply be a cloaked neutron star. He started out thinking he could answer the question but quickly descended into incoherence with several members of the audiance baying for blood (none of whom understood more than SR). It was a shame because it ruined what had otherwise been a well executed and absolutely exhillirating presentation.

All that said, is there anyting wrong with the idea that (except where curvature is such that time has stopped), it is impossible for matter to effect displacement at the same rate as a photon? (in any set of co-ordinates)

Likewise does matter having realized some gravitational potential, accuire mass by virtue of velocity (relative to the center of the black hole if I must)? And if so, could this be said to account for any descrepancy between velocity and escape velocity along the infalling trajectory. Or similarly, could degrading orbits be explained in these terms just as they can in GR speak? Or is there no deviation between velocity and escape velocity along the trajectory? And is matter falling into a black hole somehow bereft of mass aquisition?

In a similar vane, what happens in erergy terms to light escaping very dense objects? If a neutron star emits say 1 million tonnes of light per second (energy equivelence) but only a proportion of this escapes due to the red shift, I am guessing the net loss of mass to the star is only what escapes? What are the machanisums involved here?

And in reverse, of infalling CBR microwaves blue shifted to far gamma rays by black holes, what would the rate of mass aquisition be?

And what of the interior of the black hole, of R < Rs. Does matter and light move to the center at the same rate (of c)?

And finally (for today), how does the accepted model of a black hole fit with the rather tempting idea that the observable universe is all contained within the interior of a very large black hole?

 

[Trenton]

Having read up on this some more, it is clear that matter cannot follow lightlike paths into a black hole. They have to follow timelike paths which might be asymtopes of lightlike paths but are not actual lightlike paths. This directly equates to matter failing to quite reach the speed of light.

The only real question remaining is why the horizon is said to move outward (at c or any other speed). I can't see any reason why this is the case. If time stands still at the horizon so will light and so will the horizon - but does it? If so isn't time inside the horizon going backwards since the gravitational potential is much higher the further in you go? From what I have read this is not allowed.