Time to Fall Into Black Hole?

[69dodge]

Originally posted by new drkitten
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)

The source of that equation is a web page entitled "acceleration in special relativity." Special relativity doesn't deal with gravity.

a-prime is the object's acceleration relative to an inertial reference frame in which the object is momentarily stationary. But an object freely falling in a gravitational field is not accelerating at all relative to a local inertial reference frame (though it is accelerating relative to a distant one), because local inertial reference frames are "falling" too.

You're considering the (special) relativistic effects of the object's motion relative to a distant observer, but you're not considering the (general) relativistic effects of its proximity to the gravitating black hole. In empty space, two clocks that are stationary relative to each other run at the same rate; but if they're on earth, one at sea level and the other on a mountaintop, they run at different rates, even though they are still stationary relative to each other.

 

[Kiless]

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!

Argh! That made sense! Thank you! Wow! Where'd you want your Wordsworth then? Hope you're not allergic to daffodils....

 

[catbasket]

I'm no physicist, but it appears to me you are all making this far too complex. My immediate thoughts were along the lines of manny's comment above -

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)?

Stars have collapsed into black holes.
There's one nearby but I can't "see" it.
Therefore the statement "... and relativistic time-dilation implies that the entire collapse takes infinintely long when seen by an outside observer" appears wrong - otherwise we'd see all black holes in the universe as collapsing, not collapsed?

Does that make any sense at all to physicists?

 

[SpaceFluffer]

Does that make any sense at all to physicists?

Absolutely.

jmercer, I don't see where these weird effects would occur during a star's collapse, since there is no event horizon until the star has collapsed and become a black hole. So why would there be any strange effects during the collapse when it's still just a large massive object? It can't act like a black hole until it's a black hole!

 

[gnome]

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.

Bump? Sorry to be a pest, I've been dying to sort this out with someone that actually knows what they're talking about (instead of an armchair physicist like me )

 

[jmercer]

Absolutely.

jmercer, I don't see where these weird effects would occur during a star's collapse, since there is no event horizon until the star has collapsed and become a black hole. So why would there be any strange effects during the collapse when it's still just a large massive object? It can't act like a black hole until it's a black hole!

Are we talking stationary observer, or one falling toward the star at the same rate that it's collapsing? (Pre-black-hole formation.)

The "infalling" observer effects are due to two different factors.

The lensing effect occurs as the density of the collapsing star increases. The gravity well around the star should become smaller in area and more intense in effect, creating a visible gravitational lense effect... It should be noticeable long before the actual black hole forms.

The forward-looking blue-shift and rearward-looking red-shift are strictly due to the increasing velocity as the surface of the star falls inward and the "infalling" observer accelerates with it. Light from the stars behind the collapsing star are seen blue-shifted because the observer is falling "toward" them, while stars behind the observer are red-shifted because the observer is falling "away" from them.

Does that help explain my thinking any better?

 

[SpaceFluffer]

Where'd you want your Wordsworth then?

In the cupboard under the stairs, natch.

 

[Interesting Ian]

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 that's right. The whole of eternity would elapse for the rest of the Universe by the time you hit the singularity.

 

[gnome]

Yes that's right. The whole of eternity would elapse for the rest of the Universe by the time you hit the singularity.

If this is true, black holes can be thought of not as something that matter falls into, but that matter collects around... for all of time no matter will ever be observed to enter it, instead slowly freezing in time just outside the event horizon.

Which is where the "density" issue I'm bringing up comes from... does the pressure of the matter building up around the event horizon increase faster than the force necessary to escape the black hole just barely outside the point of no return?

 

[69dodge]

Originally posted by gnome
Bump? Sorry to be a pest, I've been dying to sort this out with someone that actually knows what they're talking about (instead of an armchair physicist like me )

Who, me? I don't actually know what I'm talking about; I just read web pages written by those who do.

But seriously, the page I mentioned earlier is really good. Here it is again. Everyone go read it.

About your density question: I would guess that a distant observer doesn't see falling objects as touching if they don't really touch; probably, he just sees them as flattening out as they approach the event horizon. I'm not at all sure of this; it's just a guess. But I don't think it's very important, anyway. What an observer sees depends as much on what happens to the light after it leaves the objects as it does on what was happening to the objects themselves as they emitted the light. For example, it you look through a telescope, things look big. That doesn't mean they really are big, of course; the telescope just messes with the light coming from them before it gets to your eyes.

 

[Ziggurat]

I'm no physicist, but it appears to me you are all making this far too complex. My immediate thoughts were along the lines of manny's comment above -

Stars have collapsed into black holes.
There's one nearby but I can't "see" it.
Therefore the statement "... and relativistic time-dilation implies that the entire collapse takes infinintely long when seen by an outside observer" appears wrong - otherwise we'd see all black holes in the universe as collapsing, not collapsed?

Does that make any sense at all to physicists?

Not quite. From the perspective of someone sitting on the surface of the star, collapse takes a finite amount of time. During the time of collapse but before getting smaller than the event horizon, then, the star will emit a finite amount of light. OK so far?

Now, from outside, if the collapse appears to take infinitely long, what must be happening to the intensity of light? It MUST be decreasing, because there's a finite amount of it in total. That decrease in brightness is the part you're missing. So the surface of the star will get dimmer and dimmer (and more and more red-shifted) as time goes on. From a practical standpoint, it will become "black" in a finite time period, since the intensity will just continue to decrease, and shift towards lower wavelength. But assuming you somehow COULD detect arbitrarily low intensities and arbitrarily long wavelengths, you could still see it. In reality, we can't. At some point, it will become darker than the surrounding interstellar gas, and we will never be able to detect it.