# Thread: Can a falling object actually enter a black hole?

1. Yes, that's a nice description of "parallel transport", one of those crucial concepts GR types understand and just seems weird to the rest of us.

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One more little cute thing about the Schwarzschild metric. For light itself, the v(r)/c expression is just:

(v(r)/c)^2 = (1 - R/r)^2. You can get that from the original form with the specific energy by noting the specific energy of light is infinite. Specific energy is equivalent to the ratio of relativistic mass to rest mass, and for light, the rest mass is zero. So this eliminates the additional R/r term on this.

Now, let's take the derivative of this and get the acceleration of light. Let x = R/r, and move the c^2 over to the right: d/dx = dr/dx*d/dr, and we'll take the derivative of both sides with respect to x, then mulitply the right side by dx/dr:

2v*dv/dr = 2 dv/dt = c^2[ -2 + 2x]*dx/dr.

dx/dr is just -R/r^2. Factor that in, and we have:

dv/dt = Rc^2/2r^2[ -2 + 2R/r ] Now R = 2GM/c^2, so this is just

dv/dt = -GM/r^2 * [ -2 + 2R/r ]

The acceleration of radial light is twice the Netwonian value (and in the opposite direction -- it is slowing down as it falls, not speeding up!)

So there is the famous factor of 2 on the acceleration of light. Here, for a radial "fall", it is in the opposite direction. But if we looked at the general case, with L nonzero and considered the theta coordinate equation, we'd find the factor of 2 exactly toward the center for light at infinity at a tangential initial path.

We will find that highy relativistic particles (high specific energy) follow similiar paths to light. As that Felber/Farber/Whatever guy found out and made such a big deal about (well, it was hard for me to find it out, but everything he found is right there in the Schwarzschild metric and no big deal, IMHO), a radial free faller moving at faster than Sqrt(1/3)c is going to slow down, not speed up (relative to our distant stationary frame).

Does this really mean a "repulsive" force? Not at all. Gravity is still going to pull relativisitic particles and light toward it, as can be easily seen in the tangential initial case. It's just that is you're alreading travelling toward the mass greater than a certain velocity, it's going to "brake" you a bit. It still "grabs you" and pulls your world lines toward it, but there is a velocity dependent braking effect as well.

-Richard

3. Originally Posted by publius
The acceleration of radial light is twice the Netwonian value (and in the opposite direction -- it is slowing down as it falls, not speeding up!)
But if I understand correctly, this is more a statement about the chosen reference frame than it is about light. Remind me again, what is the special value of this frame, that makes this statement carry additional importance? (Or is the absence of additional importance the point you are making, as below: )
Originally Posted by publius
Does this really mean a "repulsive" force? Not at all. Gravity is still going to pull relativisitic particles and light toward it, as can be easily seen in the tangential initial case.
Excellent point, you are saying that "repulsiveness" should have some kind of absolute quality (such as not depending on the direction of motion), and should not just be an arbitrary attribute of a particular reference frame. This appears to be what Felber/Farber/Whatever is missing when he oversells his conclusions. You could define repulsive as meaning that the inward speed is smaller than it would have been in the same coordinates but in the absence of any force, i.e., if the coordinates had somehow magically been globally inertial.

4. Originally Posted by Ken G
No doubt you are aware that in the frame of the pebble, it falls into the hole in short order....
I agree with this. So the answer to the OP question, "Can a falling object actually enter a black hole?" is essentially YES. The only oddity, as I understand it, as much of this thread has focused on, is viewing such an event from the outside. Such viewing is severely affected by the extreme gravity on the light that one is observing, as Publius has quantified.

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Originally Posted by Cougar
... Such viewing is severely affected by the extreme gravity on the light that one is observing, as Publius has quantified.
We are not viewing the pebble, we are calculating its trajectory according to GR which seems to indicate that it never enters the blackhole. Viewing involves seeing the photons that travel between the pebble and the viewer - after a photon has reflected off of the pebble it has no effect whatsoever upon the continued path/trajectory of the pebble.

No one will ever see the pebble enter the blackhole and according to the GR calculation it never enters the blackhole but according to observations it does enter the blackhole: we have different sized blackholes as measured by their calculated Schwarzschild radii.

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Originally Posted by Squashed
We are not viewing the pebble, we are calculating its trajectory
according to GR which seems to indicate that it never enters the
blackhole. Viewing involves seeing the photons that travel between
the pebble and the viewer - after a photon has reflected off of
the pebble it has no effect whatsoever upon the continued
path/trajectory of the pebble.

No one will ever see the pebble enter the blackhole and according
to the GR calculation it never enters the blackhole but according
to observations it does enter the blackhole: we have different sized
blackholes as measured by their calculated Schwarzschild radii.
This short passage has some of everything: Parts I agree with,
parts I think are wrong, parts that may be true but misleading,
parts that are ambiguous.

What do you mean by "calculating its trajectory according to GR"?
If the result of the calculation seems to indicate that the pebble
never enters the black hole, then you made a mistake somewhere.

What do you mean by "according to observations it does enter the
blackhole"? There are no observations of anything entering a BH.

-- Jeff, in Minneapolis

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Originally Posted by Squashed
We are not viewing the pebble, we are calculating its trajectory according to GR which seems to indicate that it never enters the blackhole. Viewing involves seeing the photons that travel between the pebble and the viewer - after a photon has reflected off of the pebble it has no effect whatsoever upon the continued path/trajectory of the pebble.

No one will ever see the pebble enter the blackhole and according to the GR calculation it never enters the blackhole but according to observations it does enter the blackhole: we have different sized blackholes as measured by their calculated Schwarzschild radii.
Squashed,

All this is for a classic, static Schwarzschild black hole/point mass, and even more importantly, we are assuming the mass of the stuff falling in is negligible.

If the falling mass is not neglible, it changes things as it falls, and makes a dynamic, time-varying space-time. It is no longer a simple Schwarzschild black hole, but is some complicated, time varying thing.

What happens roughly is the event horizon expands as the matter falls in. As we're watching from our far away frame, we would see the event horizon expanding to meet the matter falling in, and that matter would appear to stop at a greater radial distance. And as the horizon expands, it pushes all the other "frozen stuff" with it.

In these time-varying metrics, there is something they call the "apparent horizon" which is different from the "real" event horizon somehow. In a static situation the two are the same, but in a dynamic one, the two will be different and only merge together after everything settles down. And that settling down may actually take infinite time, but they will be close after a sufficient time.

-Richard

8. But I think Squashed has a more fundamental problem, which is that he is trying to make absolute statements like "the pebble crosses the event horizon in a finite time". Statements like that, although they sound absolute, are actually relative to the reference frame. The most obvious example of this concept that comes up even in special relativity is a statement like "a supernova occured a million light years away shortly before I was born". If a passing alien, a million years from now, wishes to verify this statement, it is actually going to depend on how fast they are moving. There are certainly reference frames where the supernova has not even occurred by "now", the time we call 2006, from the point of view of that alien frame. Talking about when things happen depends on the choice of reference frame.

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Originally Posted by Ken G
But if I understand correctly, this is more a statement about the chosen reference frame than it is about light. Remind me again, what is the special value of this frame, that makes this statement carry additional importance? (Or is the absence of additional importance the point you are making, as below: )

Excellent point, you are saying that "repulsiveness" should have some kind of absolute quality (such as not depending on the direction of motion), and should not just be an arbitrary attribute of a particular reference frame. This appears to be what Felber/Farber/Whatever is missing when he oversells his conclusions. You could define repulsive as meaning that the inward speed is smaller than it would have been in the same coordinates but in the absence of any force, i.e., if the coordinates had somehow magically been globally inertial.
Ken,

About the factor of two. In gravitational lensing discussions, it will often be noted that the trajectory of light (around a spherical mass distribution) has an acceleration of twice the Netwonian value, calculated by the trajectory curve. I was wondering if this would come out in any way for the simple radial trajectories I was considering. And it did, although "backwards", and I thought that was cute.

If pressed, I couldn't really say there was any special about this frame compared to others. However, I do consider it somewhat special as this is what we, distant observers watching stuff falling in, would see.

-Richard

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## The Earth's Reference Frame

Originally Posted by Ken G
But I think Squashed has a more fundamental problem, which is that he is trying to make absolute statements like "the pebble crosses the event horizon in a finite time". Statements like that, although they sound absolute, are actually relative to the reference frame. ....
There has been a reference frame attached to the earth since it came into existence - this reference frame may have varied due to conditions around the earth but basically the earth-based reference frame is. According to this earth-based reference frame, that I and everyone on the planet live in, a blackhole can never accrete matter because time stops at the event horizon and so it takes forever to enter the blackhole.

Jeff Root says: "There are no observations of anything entering a BH." which, if true, then all blackholes should have the same radius for the event horizon because it is impossible from the earth-based reference frame to see a blackhole deviate from its initial size.

I state that we have witnessed blackholes accrete because we know of blackhole candidates that are of various sizes.

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Originally Posted by Squashed
Jeff Root says: "There are no observations of anything entering a
BH." which, if true, then all blackholes should have the same
radius for the event horizon because it is impossible from the
earth-based reference frame to see a blackhole deviate from its
initial size.
Ackkkkk!!!

Don't do that!

I said that there are no observations of anything entering a
black hole. I most certainly did not say that nothing enters
a black hole!

I have never observed anything entering my stomache. That does
not mean that nothing enters my stomache!

Originally Posted by Squashed
I state that we have witnessed blackholes accrete because we
know of blackhole candidates that are of various sizes.
Candidates, you say. They are candidates because they have
never been directly observed. We observe stars and gas orbiting
at speeds which indicate large concentrations of mass. We also
see gas ejected at enormous speed from some of these locations.
But the ejection of gas is believed to occur at a considerable
distance from the event horizon. Nothing has ever been seen to
actually get close to an event horizon, simply because of the
limits of our observing apparatus at such great distances.

I think the idea that objects even appear to slow down as they
approach the event horizon is wrong. While the passage of time
in those objects does appear to slow to a stop (the hands of a
clock would appear to stop as it reaches the event horizon, in
our view), the speed of the object away from us would never be
reduced or appear to be reduced.

Absurdly, I think those may be two separate things: The speed
away from us keeps increasing, but we can't see it, while the
apparently-increasing speed away from us is an illusion caused
by the distorted geometry near the event horizon.

-- Jeff, in Minneapolis

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Jeff,

Let's look at it this way. I assume you agree that the stationary external observer (and this actually includes *all* such stationary observers, no matter how close they are to the horizon relative to a distant frame) does not see the falling object cross the horizon in finite time?

Now that means the position of the falling object must never become smaller than R in finite time. IOW, r -->R as t -> infinity. r(t) must asymptotically approach R. That curve, r(t), must start at the point r0 we drop it from. It can do a lot of things between R and r0, but it must eventually converge to R, never equal or smaller.

Now, the velocity is the derivative of r(t), dr/dt, and is the slope of the r(t) curve. As r --> R, that curve must flatten and it's slope must go to zero. The velocity must go to zero as r --> R, if the object is not to enter the horizon in finite time.

If the velocity does not go to zero, the r would have to drop below R, meaning the object did cross the horizon in finite time in our frame.

If that were the case, then the "freezing" would be a light-travel illusion. We would never see its image cross the horizon, but we could calculate that it did actually cross. That is not the case.

The Schwarzschild geodesics in the frame of the distant stationary observer are for the actual position, the "instantaneous position" of the object, not its light image. If we wanted to calculate when we would actually see it, when its photons would reach our eyes, we'd have to add the light travel time.

It is perfectly fine for us to say that objects never enter the hole. They don't in any stationary frame. We just see a bunch of mass frozen near the event horizon, gravitating like crazy, but it is not inside as far as we're concerned.

In its own frame, it does enter the hole and meet its fate at the singularity in finite time.

The problem here is our own minds see this as a contradiction. We want to *demand* that everyone agree the object falls into the hole because our common sense says that something either falls in or doesn't. But curved space-time does not require that. Simultaneity is warped beyond anything in SR, and "reciprocity" is completely out the window.

-Richard

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Originally Posted by Ken G

Excellent point, you are saying that "repulsiveness" should have some kind of absolute quality (such as not depending on the direction of motion), and should not just be an arbitrary attribute of a particular reference frame. This appears to be what Felber/Farber/Whatever is missing when he oversells his conclusions. You could define repulsive as meaning that the inward speed is smaller than it would have been in the same coordinates but in the absence of any force, i.e., if the coordinates had somehow magically been globally inertial.
Ken,

Negative mass would be repulsive according to this, and would repel everything including other negative mass. I can't help but starting thinking about plugging in a negative point mass M in the Schwarzschild metric. If that is the kosher, and the minus sign doesn't change the solution and we can just plug "-M' into the metric, things are going to get very weird.

The factor will be (1 + R/r), and things are going to get very strange indeed. Things "dropped" at low initial velocity will be repelled. Light on a tangential path will be curved away. However, light will be speeding up, not slowing down, and the path of radially inward directed light is going to be really weird. Just as the attractive positive mass slows it down as it falls, the negative mass is going to *speed it up toward it*. This is just crazy of course. And that light will reach infinite speed at r = 0. I think it will act as some sort of crazy mirror, reflecting all that radial light back (and perhaps in finite time). And time will speed up reaching infinity at r = 0 as well.

So I think it would be "ultimately repulsive" for radial light, but it's going to pull it in then spit it back out in the radially outward direction. And a highly relativistic particle might well follow the same path, being pulled all the way in before it was spit back out.

This negative mass is quite different from the "white hole", I caution -- that's the time reversed black hole and quite different.

-Richard

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Originally Posted by publius
Let's look at it this way. I assume you agree that the stationary
external observer (and this actually includes *all* such stationary
observers, no matter how close they are to the horizon relative to
a distant frame) does not see the falling object cross the horizon
in finite time?
Although the idea that the image freezes at the event horizon
is widespread, my understanding is that it will continue to appear
to accelerate away from the viewer until it disappears at the
event horizon.

Originally Posted by publius
Now that means the position of the falling object must never
become smaller than R in finite time.
Did you mean the apparent position as seen by a stationary
observer?

Originally Posted by publius
IOW, r -->R as t -> infinity. r(t) must asymptotically approach R.
That curve, r(t), must start at the point r0 we drop it from. It
can do a lot of things between R and r0, but it must eventually
converge to R, never equal or smaller.

Now, the velocity is the derivative of r(t), dr/dt, and is the
slope of the r(t) curve. As r --> R, that curve must flatten and
its slope must go to zero. The velocity must go to zero as r -->
R, if the object is not to enter the horizon in finite time.

If the velocity does not go to zero, the r would have to drop
below R, meaning the object did cross the horizon in finite time
in our frame.
All that is pretty much a tautology. You are saying that if
something approaches the event horizon but never reaches it,
then its speed must fall to zero before it reaches the event
horizon, and it never reaches the event horizon. Obviously.
I disagree with the premise.

Originally Posted by publius
If that were the case, then the "freezing" would be a light-travel
illusion. We would never see its image cross the horizon, but we
could calculate that it did actually cross. That is not the case.

The Schwarzschild geodesics in the frame of the distant stationary
observer are for the actual position, the "instantaneous position"
of the object, not its light image. If we wanted to calculate when
we would actually see it, when its photons would reach our eyes,
we'd have to add the light travel time.
I'm making bald assertions without backing them up, so you have
every right to complain when I ask you to support an assertion
of yours, but...

Where do you get that the Schwarschild calculations give the
"actual, instantaneous position" in the frame of the distant
stationary observer? That sounds suspicious.

Originally Posted by publius
It is perfectly fine for us to say that objects never enter the
hole. They don't in any stationary frame. We just see a bunch of
mass frozen near the event horizon, gravitating like crazy, but
it is not inside as far as we're concerned.
Can you spell out in words what you mean by "objects never enter
the hole [from the point of view of a stationary frame]? What
is the difference, from our point of view, between a thing being
inside and that thing being outside?

Originally Posted by publius
In its own frame, it does enter the hole and meet its fate at the
singularity in finite time.
I have said in this thread that a thing falling into a BH sees
the event horizon move away as it approaches the center. The
falling object will quickly reach the vicinity of the singularity
(I don't know what happens very close to the singularity),
but it will of course not cross its own event horizon.

Originally Posted by publius
The problem here is our own minds see this as a contradiction.
We want to *demand* that everyone agree the object falls into the
hole because our common sense says that something either falls in
or doesn't. But curved space-time does not require that.
My sense says that the object rapidly falls to the center, and
the object sees itself rapidly fall to the center; we see the
object apparently accelerate into the black hole at increasing
speed and with increasing redshift and at rapidly-increasing
distance until it disappears at our event horizon.

-- Jeff, in Minneapolis

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Jeff,

Okay, so you disagree that a free-falling object never reaches the horizon in a stationary frame, and the freezing is therefore a trick of light travel time. And r(t) does indeed get less than R in finite time. Fair enough.

However, that is incorrect.

I do not mean "apparent position". I mean the position of the object as measured by the rulers of our distant stationary frame against the time defined by our clocks in that frame. r(t) --> R as t --> infinity,
and v(t) -->0 as r --> R.

I don't have the skill to derive the Schwarzschild metric from the Einstein Equation, nor go through solving the geodesic equations from that metric. I can only assure you that they are correct. I went to a lot of trouble digging this up and making sure I had the correct expressions.

The geodesic equations are derived in the coordinate system one is using and their solutions faithfully describe the equations of motions in terms of the rulers and clocks defined by those coordinates. These are inertial paths -- paths followed by objects with no "real" forces acting on them. The curve, in terms of velocity vs position for a particle dropped from infinity with zero initial velocity is simply:

(v/c)^2 = (1 - R/r)^2 * R/r. When r = R, that goes to zero. v(R) is zero. The local velocity takes out the (1-R/r)^2 factor, and you're amazingly left with the Newtonian result, v/c = Sqrt(R/r). That goes to c at R=r. But there is no stationary frame at R=r. Any stationary frame at r > R will see the free-falling object whiz by it at the local speed, but it will see it slow down and stop as well, no matter how close. The local rulers are getting very short at r near R, and local stationary observers think the horizon is a lot farther away than we would in our distant frame..

The maximum speed of the object dropped with zero initial velocity occurs at the photon sphere, r = 3/2R. The local free-fall speed there would be Sqrt(2/3)c. A stationary observer there would see the object whizzing by him at over 80%c, but he would see that thing start to slow down and stop. The distance it travelled would be a lot greater than 1/2R in his frame, though.

-Richard

16. Well, this seemed like a good question for mathematical physicist and GR guru John Baez, and the UCR.edu site What happens to you if you fall into a black hole? seems to have our answers. One of the points made there is as follows:
So if you, watching from a safe distance, attempt to witness my fall into the hole, you'll see me fall more and more slowly as the light delay increases. You'll never see me actually get to the event horizon. My watch, to you, will tick more and more slowly, but will never reach the time that I see as I fall into the black hole. Notice that this is really an optical effect caused by the paths of the light rays.

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Cougar,

I'm going to have disagree with that view expressed by this quote: "Notice that this is really an optical effect caused by the paths of the light rays." That is like saying the time dilation of SR is just a trick of the doppler shift of light. When you correct for the doppler effect, you find the frequency is still slower than the emitting observer thinks it is. And that is just by the time dilation factor.

The reason the "optical illusion" is so appealing is because common sense seems to demand that if something crosses the horizon in one frame, it must cross in all frames. And GR does not require that.

He says this in that same page:
At large distances t does approach the proper time of someone who is at rest with respect to the black hole. But there isn't any non-arbitrary sense in which you can call t at smaller r values "the proper time of a distant observer," since in general relativity there is no coordinate-independent way to say that two distant events are happening "at the same time." The proper time of any observer is only defined locally.
Now, t is the time coordinate of an observer at infinity, but the proper time of closer stationary observers is about the same where R/r is very small. The author is arguing that even though this time goes to infinity, it's not real, and only the proper time of the free-faller means anything. He is actually getting close to violating a cardinal rule of relativity, which is that there is no preferred frame. He is claiming the frame of the free-faller is the one that is "real", and not that of distant stationary observers.

That is not something I agree with at all. It does take infinite time in the frame of any stationary observer, whether close or distant, and that view is a good and correct as that of the free-falling frame itself.

-Richard

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## You Nailed It

Originally Posted by publius
...The author is arguing that even though this time goes to infinity, it's not real, and only the proper time of the free-faller means anything. He is actually getting close to violating a cardinal rule of relativity, which is that there is no preferred frame. He is claiming the frame of the free-faller is the one that is "real", and not that of distant stationary observers.

That is not something I agree with at all. It does take infinite time in the frame of any stationary observer, whether close or distant, and that view is a good and correct as that of the free-falling frame itself.

-Richard
That is exactly what I have thought but was unable to say convincingly. To use the "illusion" description requires a real and not-real frame determination and that is just totally wrong. I have hated the "illusion" description but have never had the mathematical backing to prove that my dislike was justified, thank you publius.

But now back to blackhole candidates of different sizes; how can we explain them because either:

1.) The matter gets inside the event horizon or
2.) The matter stacks in shells on the outside of the event horizon or
3.) Blackholes never actually form or ...

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## Just a Thought

A thought crossed my mind while reading another thread: If the matter is converted into electromagnetic radiation at the event horizon then reference frames no longer apply to the radiation and irregardless of which way the momentum of the radiation points it will ultimately spiral into the blackhole - because the event horizon is inside the photon sphere and so the "matter" actually enters but only after it is converted into radiation.

20. Originally Posted by publius
He is actually getting close to violating a cardinal rule of relativity, which is that there is no preferred frame. He is claiming the frame of the free-faller is the one that is "real", and not that of distant stationary observers.
It might be more fair to say that he is arguing that the very concept of a reference frame is a local concept. This is also what I meant in a few other threads when I pointed out the difference between a reference frame and a global coordinate system. Global coordinate systems are absolutely arbitrary, as long as you know how to transform them into local reference frames (i.e., frames that have the correct spacetime structure locally). All the global coordinate system is, then, is a means for bookkeeping the transformations from local frame to local frame as you consider different events. These transformations must obey some rules to go from a reference frame to a reference frame (i.e., frames that actual local observations will obey the laws of physics), and those rules are specified in a coordinate-independent way by general relativity, but which ones you choose are arbitrary.

I would explain the above dispute as arising from the fact that what publius is doing is choosing a particular coordinate system, and its applied allowed transformations, to make statements about what is happening. That is one way to do relativity, but it is an usual way to do it, because you are never actually in a frame where any observations can be made to test your result-- you will always need to transform to a local reference frame where actual observations get done to check your answer. Put succinctly, all clocks measure their own proper time. Baez is not even considering other coordinate systems, presumably because they are not directly usable for measurements. If you restrict to reference frames (i.e., to somebody's local proper time), and you want to make inferences about somebody else's local proper time, then you do have to distinguish between what is real and what is an optical illusion (and note that in special relativity, proper time and reference frames can be extended globally, but not so in general relativity-- reference frames are local). So Baez is simply talking about something different from what publius has done.

By the way, the more standard approach (used by Baez) is to make transformations that always transform to the particle's local rest frame. Then not only is one always in a reference frame where measurements can be made, but one is also using "proper time" (the "proper" means owned by the particle, not "correct"). When you do that, the words you get to describe what is happening are totally different from the approach of publius. This is what I call pedagogy, and one should not expect the pedagogies arising from two different global coordinate systems to sound the same-- heck, they don't even sound the same between two different local reference frames.

21. Originally Posted by Squashed
If the matter is converted into electromagnetic radiation at the event horizon then reference frames no longer apply to the radiation and irregardless of which way the momentum of the radiation points it will ultimately spiral into the blackhole - because the event horizon is inside the photon sphere and so the "matter" actually enters but only after it is converted into radiation.
This is an interesting speculation, but note the matter is generally not turned into electromagnetic radiation as it falls into a black hole (just ask someone who has...!). However, you may be on to a similar idea-- there are ways of looking at this (and I can't put this on a rigorous foundation) in which the rest mass of the particle reaches zero at the event horizon. In such a system (which is not a reference frame, see my above comment), the particle is not radiation, yet does have zero rest mass so acts like radiation. So I'm not sure, but there may be some value in saying that particles can cross an event horizon as their rest mass reaches zero. (Note it will not be zero in their own frame, of course, that's how you can tell they are not converted to radiation-- they still have a rest frame).

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## X-rays

Originally Posted by Ken G
This is an interesting speculation, but note the matter is generally not turned into electromagnetic radiation as it falls into a black hole (just ask someone who has...!). However, you may be on to a similar idea-- there are ways of looking at this (and I can't put this on a rigorous foundation) in which the rest mass of the particle reaches zero at the event horizon. In such a system (which is not a reference frame, see my above comment), the particle is not radiation, yet does have zero rest mass so acts like radiation. So I'm not sure, but there may be some value in saying that particles can cross an event horizon as their rest mass reaches zero. (Note it will not be zero in their own frame, of course, that's how you can tell they are not converted to radiation-- they still have a rest frame).
One of the ways of identifying a blackhole is by the x-rays that matter "screams" when it spirals into a blackhole - this fact indicates to me that a change is taking place: from matter to radiation.

As far as rest frames goes it should be annihilated because when we destroy a small amount of matter in a nuclear explosion the rest frame of that matter is destroyed when it is turned into radiation and so the same would hold true for matter that is converted into radiation upon entering a blackhole.

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I think it should be noted that a Matt McIrvin wrote the FAQ Cougar posted not Baez himself. I just noticed this.

-Richard

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Originally Posted by Ken G
This is an interesting speculation, but note the matter is generally not turned into electromagnetic radiation as it falls into a black hole (just ask someone who has...!). However, you may be on to a similar idea-- there are ways of looking at this (and I can't put this on a rigorous foundation) in which the rest mass of the particle reaches zero at the event horizon. In such a system (which is not a reference frame, see my above comment), the particle is not radiation, yet does have zero rest mass so acts like radiation. So I'm not sure, but there may be some value in saying that particles can cross an event horizon as their rest mass reaches zero. (Note it will not be zero in their own frame, of course, that's how you can tell they are not converted to radiation-- they still have a rest frame).
Ken,

This may actually be close to what the ECO/MECO theory says happens, actually. Wiki has a pretty good article on the theory:

http://en.wikipedia.org/wiki/Magneto...lapsing_object

Abhas Mitra is the Indian physcist who first proposed this

http://en.wikipedia.org/wiki/Abhas_Mitra

An ECO would pretty much look like a black hole from afar, but things would be very different "inside". It becomes a radiation dominated "soup" as it collapses and indeed may "burn" much matter into radiation.

I'm beginning to like this ECO theory, and wouldn't be surprised if black holes "evaporate" as possible real objects.

-Richard

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Ken,

I certainly agree with what you say generally. Except in this case, we can never be in the local frame past the horizon unless we wish to share its fate. Past the horizon, there is no return, and the two "twins" cannot get back together to compare their clocks or whatever.

The events, the space-time coordinates of the local frame once it crosses the horizon do not transform to anything but "nonsense" (infinite time or even imaginary time or other nonsense past the horizon) in the external stationary frames. We can say the internal events do not exist in the accessible space-time of any observer who is not himself going into the hole.

So for us outside, the only meaningful thing is our own reference frames, and in those frames it takes infinite time for the object to cross the horizon.

I see this question of whether an object "really falls in" has been debated endlessly on other forums as well. Someone (on my side) made an argument I like.

He posed the question: Is there any time on our (external stationary) clock at which we can be sure the free-faller has crossed the horizon and can never come back? The answer is no. We'll say the faller crosses the horizon at 12:00. Until we see his clock tick 12:00, we can never be sure he didn't fire a powerful rocket and stop his fall at the last second and turn around. The closer his clock gets to 12, the force required would increase without limit, but remain finite until exactly 12. He could save himself 1000 years after he effectively "froze" in our frame, and only a twinkling of an eye passed in his.

So, there is no time on our clock in which can conclude the faller is gone forever.

-Richard

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Originally Posted by Squashed
That is exactly what I have thought but was unable to say convincingly. To use the "illusion" description requires a real and not-real frame determination and that is just totally wrong. I have hated the "illusion" description but have never had the mathematical backing to prove that my dislike was justified, thank you publius.

But now back to blackhole candidates of different sizes; how can we explain them because either:

1.) The matter gets inside the event horizon or
2.) The matter stacks in shells on the outside of the event horizon or
3.) Blackholes never actually form or ...
Squashed,

The "growing" is not really a problem, as I mentioned in another post. Remember, here we're talking about free-fallers of negligible mass that do not change space-time as they fall in. When the mass in not negligible, the Schwarzschild metric no longer applies and we have a dynamic space-time that is changing with time.

Roughly what we see in our stationary frame is the event horizon (and the all the "frozen stuff" near it) expanding to meet the mass falling it. R, the Schwarzschild radius, expands and both meet in the middle so to speak.

Externally, the gravity of a spherical shell is the same as of a point mass so it doesn't matter.

Now, according the ECO theory, classical black holes can never form in finite *proper time* in their own frame. I'm almost a convert to the ECO theory now.

-Richard

27. Originally Posted by publius
So for us outside, the only meaningful thing is our own reference frames, and in those frames it takes infinite time for the object to cross the horizon.
Well, I think one could argue that in fact the only thing meaningful for us is even less than that-- it is our own local reference frame, since that's the only frame we can use to make any measurements. All other coordinate system that extend to include distant objects (like those falling into black holes) are then completely arbitrary-- they only need to give the right results when the light reaches our frame. So there is really no practical difference, at the end of the day, between what is real and what is an illusion.

Originally Posted by publius
Is there any time on our (external stationary) clock at which we can be sure the free-faller has crossed the horizon and can never come back? The answer is no. We'll say the faller crosses the horizon at 12:00. Until we see his clock tick 12:00, we can never be sure he didn't fire a powerful rocket and stop his fall at the last second and turn around. The closer his clock gets to 12, the force required would increase without limit, but remain finite until exactly 12.
But this leads to a paradox, because the black hole is not in a steady state over an infinite amount of time. What if, when the universe gets to age 20 billion years, a giant hand swoops down and causes everything to vanish everywhere. Where will the rocket be when that happens, for the rocket crew? Where will they be for us? Can that be different? It comes down to the issue of using our time for them is just a bad idea. We can calculate exactly when we would need to release a light signal that says "good luck" and reaches the rocket just when it crosses the event horizon (you can probably do this now!). If you were the one sending out this signal, would you really think they never actually get to the event horizon, even though you know your friends will receive your wish?

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Originally Posted by Ken G
We can calculate exactly when we would need to release a light
signal that says "good luck" and reaches the rocket just when it
crosses the event horizon.
I believe we had this discussion before, just a few weeks ago,
but the people aboard the rocket could still receive a signal
from us after they have crossed our horizon. However, it will
stretch out more and more as it falls in after them, so the
length of the message has to be limited if they are to receive
it all before they are spaghettified.

The opposite of what you suggested is more clear-cut. We can
calculate exactly when those aboard the rocket need to send a
signal that can reach us. The end of the message would be just
before they cross our event horizon.

"My God, it's full of starrrrrrssssssssssssssssssssssssssssss!"

-- Jeff, in Minneapolis

29. Without all this posturing and scientific dribble, The answer is yes.
You can fly into a black hole.
BUT... This is a very bad idea. You would be torn apart, radiated, crushed, spaghetti fide. stretched. and very dead well before you actually got to it.
Maybe you could use its enormous gravity to sling shot you away.
I am not coming, It sounds very dangerous

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Originally Posted by Ken G
Well, I think one could argue that in fact the only thing meaningful for us is even less than that-- it is our own local reference frame, since that's the only frame we can use to make any measurements. All other coordinate system that extend to include distant objects (like those falling into black holes) are then completely arbitrary-- they only need to give the right results when the light reaches our frame. So there is really no practical difference, at the end of the day, between what is real and what is an illusion.

But this leads to a paradox, because the black hole is not in a steady state over an infinite amount of time. What if, when the universe gets to age 20 billion years, a giant hand swoops down and causes everything to vanish everywhere. Where will the rocket be when that happens, for the rocket crew? Where will they be for us? Can that be different? It comes down to the issue of using our time for them is just a bad idea. We can calculate exactly when we would need to release a light signal that says "good luck" and reaches the rocket just when it crosses the event horizon (you can probably do this now!). If you were the one sending out this signal, would you really think they never actually get to the event horizon, even though you know your friends will receive your wish?
Ken,

Ah, let's take the second one first. Remember the free-faller is going to see us accelerating away from him, and our clocks slow and our light redshifted (a stationary observer far down would see fast clocks and high blue shift).

We can calculate the time by our clock to send a signal that will reach him just as he crosses the horizon. But guess what? It will take infinite time, by our clock for that signal to reach him! If he fires a rocket and saves himself, he will receive that signal in our frame "before infinity", since he stops his fall. If we plot the course of that last bon voyage message chasing him down the hole, we'll see that signal itself slow to a crawl near the horizon, and essentially see two little radially compressed ants crawling at incredibly slow speed. The rear ant is travelling such a tad faster, and they will meet at the horizon after infinite time has passed.

Now, a signal sent just a tad after could reach the free faller in his frame *after* he crossed but before he hit the singularity. But that would take "more than infinite" time in our frame.

Now, about the hand swooping down and wiping everything out. If the hand appears near us, roughly in our frame, then any causal influences of that hand cannot travel faster down in the hole that our light signal, and will therefore take infinite time for that hand to have any effect on stuff down near the horizon.

If we say this is "magic wand" hand of some sort, then we're talking something outside of space-time as we know it.......................

-Richard

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