Time to Fall Into Black Hole?

[drkitten]

Bless those Canuck bookstores; they carry everything. But in this case, they carry something that I think I disagree with, and I hope some of the physics brains can help me out here.

From Pratchett, Steward, and Cohen's The Science of Discworld III : Darwin's Watch (and a great series it is, if y'all care): The authors are basically discussion the standard cosmology of black holes:

If you were to watch a black hole collapse from outside, you would see the star shrinking towards the Schwarzschild radius [aka the `event horizon], but you'd never se it get there. As it shrinks, its speed of collapse as seen from outside approaches that of light, and relativistic time-dilation implies that the entire collapse takes infinintely long when seen by an outside observer.

I've seen this statement in print before, but I don't buy it. And, in fact, I think it can be disproven fairly easily.

Consider a black hole exactly the same mass as the Sun (for convenience), and consider the case of an object 'falling from infinity.' At all points outside the radius of the actual Sun, the effect of the black hole's gravity would be identical to the Sun's gravity; the Earth would, for example orbit our hypothetical black hole at 1AU in an orbit of exactly 1 year, etc. Similarly, an object falling 'from infinity' would hit the Sun's radius at exactly the escape velocity of the real Sun (about 650 km/second).

The Sun's radius is about 650,000km, so it would take an unaccelerated object only about 1,000 seconds to cover the distance between the Sun's surface and its center (at uniform speed). Our hypothetical object, however, will be undergoing positive gravitational acceleration all the way between the Sun's radius and the event horizon. Even under Einstein, velocity under positive acceleration is still monotonic; an accelerating object will be seen from all observers to be moving with higher velocity. Therefore, it will take less than 1000 seconds for an object to fall from the Sun's radius into our hypothetical black hole.

More generally, if it takes more than 1000 seconds (and an infinite amount of time is longer than 1000 seconds) to hit the event horizon, then at some point our falling object will be observed to be moving at less than 650 km/sec, which means that at some point within the Sun's radius, the object will achieve an observed maximum speed. But since acceleration is monotonic, this isn't possible.....

Can someone please explain where I've gone wrong, and at what point the maximum speed of our hypothetical falling object is actually achieved?

 

[Diamond]

I think the answer is: Stop changing frame of reference.

From the reference frame of the infalling object, it would take less than 1000 seconds.

From an observer far away, the infalling object would appear to redden and dim, and take longer and longer to move (looking sideways on). From an observer outside the black hole it would take an infinite amount of time to reach the event horizon.

It all depends on frame of reference.

 

[gnome]

I think the answer is: Stop changing frame of reference.

From the reference frame of the infalling object, it would take less than 1000 seconds.

From an observer far away, the infalling object would appear to redden and dim, and take longer and longer to move (looking sideways on). From an observer outside the black hole it would take an infinite amount of time to reach the event horizon.

It all depends on frame of reference.

This might clear it up: from an observer approaching the event horizon (for which it would take only seconds) ... would it appear as though time "outside" were passing very rapidly?

For me this would cause it all to make sense... plus offer the enticing idea that someone falling into a black hole would get to witness the end of the universe.

 

[69dodge]

Originally posted by new drkitten
Even under Einstein, velocity under positive acceleration is still monotonic; an accelerating object will be seen from all observers to be moving with higher velocity.

"Accelerating" can mean "moving with a nonconstant velocity" or it can mean "moving noninertially". An object that's falling freely in a gravitational field might be accelerating in the first sense (depending on what its velocity is measured relative to), but it's not accelerating in the second sense. Probably, your statement about monotonicity is not true for such objects.

What makes general relativity trickier than special relativity is that, in the presence of a gravitational field, two objects that are both moving inertially (i.e., not accelerating in the second sense) may yet be accelerating (in the first sense) relative to each other.

Here's a relevant web page.

 

[Diamond]

This might clear it up: from an observer approaching the event horizon (for which it would take only seconds) ... would it appear as though time "outside" were passing very rapidly?

For me this would cause it all to make sense... plus offer the enticing idea that someone falling into a black hole would get to witness the end of the universe.

Yes. In a sense, looking into a black hole is to look into the infinite future.

From an infalling observer's point of view, the Universe ages faster and faster.

 

[jmercer]

As a non-physics-brain, please educate me on this...

I'm looking at a star collapsing into a black hole. The star is emitting light.

As the star collapses faster and faster, its collapse eventually comes very close to the speed of light. As this happens, the star becomes red-shifted. At some point, the star reaches the Schwarzschild radius, and no photon can escape the star even if it's still being emitted.

My eyes require photons to enter them to see anything at all. So...

1) What would I be seeing without photons entering my eyes? Wouldn't the star seem to simply disappear? (black hole)

2) What happens to the photons at the edge of the radius? Do they freeze? Or do they continue their outward journey, but veeeerrrryyyy slooooowwwwllly?

Thanks.

 

[gnome]

Yes. In a sense, looking into a black hole is to look into the infinite future.

From an infalling observer's point of view, the Universe ages faster and faster.

Ok... let me extend this further--from the infalling observer's point of view... they'll see more and more "stuff" being sucked into the black hole along with them, also approaching the event horizon, though slightly behind them. More and more matter packs in as they approach the event horizon.

From an outsider's point of view... all of the matter reaches the event horizon as time approaches infinity.... so it is getting packed closer and closer together right before the event horizon.

From the insider's point of view, that packing will still occur, just on a much faster time frame. Density will increase as all the matter converges on the event horizon. I imagine, with sufficient mass available, that the pressure may eventually exceed the gravitational pull (since the matter has technically not yet reached the event horizon) and the black hole will explode.

At one time I had thought of this as a universal explosion based on the idea that all matter in the universe might eventually suck into black holes, and then the black holes converge. It would allow a mechanism for the "cyclical" theory of big bang/big crunch/big bang. However, I believe the "big crunch" theory is discredited lately, so that you're looking more at just a local explosion.

With sufficient mass available, is it possible that none of the matter will actually pass the event horizon before the density builds up enough to explode? Or am I missing a point?

 

[69dodge]

Originally posted by gnome
From the insider's point of view, that packing will still occur, just on a much faster time frame. Density will increase as all the matter converges on the event horizon.

No. He just falls through the event horizon, and sees other stuff near him falling through. As far as he can tell, nothing really special happens at the event horizon, at least not regarding stuff nearby, though the image he sees of faraway stars does get greatly distorted.

 

[gnome]

No. He just falls through the event horizon, and sees other stuff near him falling through. As far as he can tell, nothing really special happens at the event horizon, at least not regarding stuff nearby, though the image he sees of faraway stars does get greatly distorted.

I think that creates an inconsistency of observed position. Think about it mathematically. The outside observer sees both objects approaching event horizon increasingly slowly. If it takes an infinite amount of time to see the object reach the horizon, that means that when the second object is 1 cm from the event horizon, the first object is still not at the event horizon, and so it appears less than 1cm away. When the second object is 1 mm from the event horizon, the first object is observed to still be squashed between it and the event horizon. So clearly the outside observer will eventually see the objects touching. Doesn't that mean that an inside observer will see the objects touching as well? That's the density increase I'm referring to.

 

[jmercer]

Guys, can someone please tell me what I'm missing in this - conceptually, please?

Stationary observer outside of a Schwarzschild radius watching a star collapse:

1) Sees the star surface red-shift as the collapse accelerates away from observer.

2) After a finite time, the star collapses to a point where the Schwarzschild radius is formed, and the star is within the radius.

3) At the instant the radius is formed, photons emitted by the collapsing star cannot escape the boundary, and the star disappears from view. Depending upon the distance of the observer from the star before it collapsed into the radius, the observer will eventually see the star disappear utterly.

What am I missing?

Now - observer falling into the collapse at the same rate as the star surface, near enough to be within the Schwarzschild radius when it forms, but prior to the formation of the radius:

1) An observer looking toward the star as they "fall" into the collapse will see the collapsing star as 'normal' - but all the other distant stars behind the collapsing star will be increasingly blue-shifted. Eventually the observer will see a gravitational lensing effect in front of them.

2) Everything behind the observer will be red-shifted as the observer "falls" away from the universe.

3) Everything lateral to the observer will appear "normal".

Again, am I missing anything?

Finally - observer falling into the collapse at and after the Schwarzschild radius forms behind them (and this one is the one I really am lost on!):

1) Everything directly behind the observer will appear red-shifted even further as the observer accelerates his/her fall. This should be a small "porthole" of strongly red-shifted stars, surrounded by blackness.

2) In front of the observer will be nothing but a "halo - a ring of strongly blue-shifted stars with utter blackness at it's center. The ring will contain many more stars than are directly in line with the observer's viewing direction due to strong gravitational lensing by the black hole.

3) Everything lateral to the observer will appear to "fade" at one end, with the disappearing part being the area closest to the center of the singularity. Objects may also appear to be elongated - dunno. (Reflected photons won't travel laterally, but will be pulled off-course into the singularity before they reach the retina.)

Help?

 

[drkitten]

I think the answer is: Stop changing frame of reference.

Perhaps I wasn't clear. I'm trying to do this from the frame of reference of a distant observer at rest relative to the black hole.

From the reference frame of the infalling object, it would take less than 1000 seconds.

From an observer far away, the infalling object would appear to redden and dim, and take longer and longer to move (looking sideways on).

This is my problem. The object will "take longer and longer to move" -- this means that the object's apparent velocity, as seen from the distant observer, is becoming lower and lower.

Unfortunately, I can't get the math to work out properly.

Perhaps an ASCII diagram may help. From the distant observer's perspective, we have the following


| O-> o
Solar radius object moving to the right event horizon


The distance between the event horizon and one solar radius away is, by definition, fixed at one solar radius. The object is currently moving rightwards at a speed of v, which, at the point at which it crossed the solar radius, was approximately 650 km/sec.
It is also being accelerated to the right by the gravity of the black hole.

The formula for relativistic acceleration says that the apparent acceleration a of an object moving at relative velocity u, and being accelerated in a frame in which the object is nearly stationary at acceleration a-prime is:

a = a-prime*(1-u^2/c^2)^(3/2)

But this means that the apparent acceleration a to an outside obsever is still positive. (Source: http://www.physics.nmt.edu/~raymond/classes/ph13xbook/node59.html)

So the outside observer, at rest w.r.t. the black hole, will see the falling object moving at finite and ever-increasing velocity, until such time as it hits the event horizon moving at the speed of light. But because the velocity is ever-increasing, it will require less time (as measured by our external observer) to hit the event horizon than it would travelling the same distance unaccelerated in free space.

Therefore, the infalling object will not "take longer and longer to move," but in fact hit the black hole at light speed and explode in a burst of infinitely red-shifted radiation less than fifteen minutes after crossing the solar radius.

 

[Manny]

Well, now wait. Maybe there's another way to skin the cat. Are there observable, collapsed black holes or not (OK, obviously they wouldn't be physically observable, but you know what I mean)?

 

[SpaceFluffer]

I really don't understand a lot of what you're saying, jmercer, but I can paint a picture of how a star-black hole system behaves.

For one thing, stars do not get 'swallowed whole' by a black hole. If you want to think of the size of black hole as the event horizon radius, then typical black holes formed from dying stars will be a few miles across. Now a star doesn't simply 'fall' into the black hole; the hole and the star will orbit around each other (typical for a binary star system when one star dies and forms a black hole).

The black hole gradually sucks stellar material off the star, which spirals down in the black hole. As it does so it loses potential energy, the gas becomes heated, and radiates X-rays.

 

[SpaceFluffer]

The falling body falls smoothly into the black hole as viewed from it's reference frame.

An outside observer will see the infalling body have a trajectory described by:

t(r) = (S/c) * [ (-2/3 * (r/S)^(3/2) ) - ( 2(r/S)^(1/2) ) + ln((1+(r/S)^(1/2))/(1-(r/S)^(1/2))) ]

S is the Schwarzschild radius (=2GM/c^2), c is the speed of light, r is the radius at time t.

I can derive this from the Schwarzschild metric, but it was painful enough to type out the answer. Perhaps I can TeX it and post it later.

 

[Kiless]

The falling body falls smoothly into the black hole as viewed from it's reference frame.

An outside observer will see the infalling body have a trajectory described by:

t(r) = (S/c) * [ (-2/3 * (r/S)^(3/2) ) - ( 2(r/S)^(1/2) ) + ln((1+(r/S)^(1/2))/(1-(r/S)^(1/2))) ]

S is the Schwarzschild radius (=2GM/c^2), c is the speed of light, r is the radius at time t.

I can derive this from the Schwarzschild metric, but it was painful enough to type out the answer. Perhaps I can TeX it and post it later.

I hate that you sound so damned impressive and I haven't got a hope in hell of understanding it. Trade? Two hour lecture on Wordsworth for one definition of a shwardtys... shwrar... that thing.

 

[jmercer]

I really don't understand a lot of what you're saying, jmercer, but I can paint a picture of how a star-black hole system behaves.

For one thing, stars do not get 'swallowed whole' by a black hole. If you want to think of the size of black hole as the event horizon radius, then typical black holes formed from dying stars will be a few miles across. Now a star doesn't simply 'fall' into the black hole; the hole and the star will orbit around each other (typical for a binary star system when one star dies and forms a black hole).

The black hole gradually sucks stellar material off the star, which spirals down in the black hole. As it does so it loses potential energy, the gas becomes heated, and radiates X-rays.

Poor phrasing on my part SF - sorry. I wasn't speaking of a star that was falling into a black hole; I was speaking of a star in the process of collapsing - of a star becoming a black hole.

All my observer comments are from that kind of event... although the black-hole death-spiral (which I've seen illustrations of) has got to be one of the coolest sights in the universe.

So - in the case of a stationary external observer, it appears as if the star's surface is "falling" away from them as it collapses. Where the observer is falling toward the star at the same rate as the surface collapses would be the second set of observations... and the third, of course, would be after the Schwarzschild radius forms and the observer is inside of it, still falling.

Granted, given the rapid nature of these events my hypothetical observer would have to be incredibly fast to perceive these things... but since this is strictly for discussion, I'm not too worried about that.

 

[drkitten]

An outside observer will see the infalling body have a trajectory described by:

t(r) = (S/c) * [ (-2/3 * (r/S)^(3/2) ) - ( 2(r/S)^(1/2) ) + ln((1+(r/S)^(1/2))/(1-(r/S)^(1/2))) ]

That is indeed ugly and I thank you for typing it.

At the risk of my eternal soul for even asking this : is there a graph of this trajectory available anywhere? Barring that, at what point(s) does t''(r) = 0?

 

[SpaceFluffer]

The Schwarzschild radius is the maximum radius that an object would have to be for it turn into a black hole, and it only depends on the mass of an object. For example, for the Earth it is ~1cm, for the Sun ~3km.

i.e. crush the Earth into a ball about 2cm across and hey presto! A black hole!

 

[jmercer]

The Schwarzschild radius is the maximum radius that an object would have to be for it turn into a black hole, and it only depends on the mass of an object. For example, for the Earth it is ~1cm, for the Sun ~3km.

i.e. crush the Earth into a ball about 2cm across and hey presto! A black hole!

Well, yeah... but I'm thinking in terms of a something more along the lines of a super-massive blue-white star that was too small to nova, so it went through the red giant phase, exhausted it's helium, and then collapsed. Still a very small Schwarzschild radius, but big enough for discussion purposes.

The duration of the collapse would still be very, very short, though.

 

[SpaceFluffer]

That is indeed ugly and I thank you for typing it.

At the risk of my eternal soul for even asking this : is there a graph of this trajectory available anywhere? Barring that, at what point(s) does t''(r) = 0?

Consider your eternal soul at risk...

I've attached a diagram taken from "Principles of Cosmology and Gravitation" by Michael Berry, showing the trajectory as seen by a distant observer. As you can see, the acceleration of the infalling body only reaches zero at t=+-infinity.