A Black Hole's Event Horizon: No Particles Have Crossed...Yet

[alphachapmtl]

It takes an infinite amount of time for any particle to cross the event horizon of a black hole, from our point of view as an outside observer. Which means that since the big bang, not a single particle has ever crossed the event horizon of a black hole. They just come closer and closer to it, piling up onto the surface into a quasi-2D shell.
Am I wrong? What does this all means? Any comment?

 

[sri sharan]

I had the same kind of doubt
https://www.physicsforums.com/showthread.php?t=455677"

 

[alphachapmtl]

I had the same kind of doubt
https://www.physicsforums.com/showthread.php?t=455677"

Interesting. Those commenting on your post don't seem to get the point.
Since object falling onto a black hole are just piling up onto the surface into a quasi-2D shell, you would expect the entropy of a black hole to be proportional to it's surface area, which is exactly the case.

Let me also quote from: http://www.straightdope.com/columns/read/2072/can-black-holes-in-space-ever-become-normal-again"

And black holes also don't have infinite density. When we talk about a black hole, we generally mean the entire region inside the event horizon, the surface of no return--to put it another way, the region from which no light escapes. This horizon has a radius called the Schwarzschild radius, which is directly proportional to the mass of the black hole (specifically, Rs = 2GM/c^2, where G is the gravitational constant, M is the mass of the object, and c is the speed of light). You can calculate the Schwarzschild radius for any mass, whether it's a black hole or not. For instance, the sun has a Schwarzschild radius of about 3 kilometers.
If a black hole has a nonzero volume, then it also has a finite density. Since the radius of a hole is proportional to its mass, its density is inversely proportional to the mass squared. Bigger holes, in other words, have lower densities. For a hole of a few solar masses, this density is greater than that of the nucleus of an atom, but a supermassive black hole such as one finds in the core of a galaxy can have a density more like water, or even air.

 

[Nabeshin]

There's nothing new or shocking here, and it does not imply that, say, a black hole can never be created! The event horizon of a BH expands in ANTICIPATION of matter falling into it. Therefore, the EH has begun expanding even before the outside observer sees the matter fall into where he thought the EH was.

And what's your point about the density? It's all a matter of how you define density. When you define the volume of the hole to be the volume of the event horizon, density is definitely finite. If however you wish to regard the singularity as the volume...

 

[Ich]

Am I wrong? What does this all means?

The posts in the other thread are not bad.

Those commenting on your post don't seem to get the point.

Let me add some information:
The redshifting mentioned there is exponential in time. After a finite (and normally very short) time, there is nothing substantial left to see.
Then, yes, the objects seem to "pile up" in Schwarzschild coordinates. But that does not mean that there is any physical observer to watch this: Those far away see them fade to black, those near the horizon see them fall quickly down and then fade to black, and infalling observers will only encounter the most recently gone objects, if any.
This "piling" is more like a coordinate artifact, a bookkeeper's way of telling us that we lost contact with the events inside the horizon. You shift the last observable event to +infinity, so all the later events are outside our scope.

 

[Chronos]

A singularity is not mathematically necessary to create an event horizon.

 

[twofish-quant]

It takes an infinite amount of time for any particle to cross the event horizon of a black hole, from our point of view as an outside observer. Which means that since the big bang, not a single particle has ever crossed the event horizon of a black hole. They just come closer and closer to it, piling up onto the surface into a quasi-2D shell.
Am I wrong?

Yes you are wrong. It *appears* from a distant observers that an object going into a black hole never falls in, but from the point of view of the object, you fall in at a finite time.

 

[twofish-quant]

You can calculate the Schwarzschild radius for any mass, whether it's a black hole or not. For instance, the sun has a Schwarzschild radius of about 3 kilometers. If a black hole has a nonzero volume

Ummmmm no...

A black hole has a finite radius but an undefined volume. One thing that gravity does is to mess with the equations for volume. If you measure the radius of the earth, and then measure it's volume, you'll find it slightly larger than what you would get from the standard math formulas.

 

[sri sharan]

Let me add some information:
The redshifting mentioned there is exponential in time. After a finite (and normally very short) time, there is nothing substantial left to see.
Then, yes, the objects seem to "pile up" in Schwarzschild coordinates. But that does not mean that there is any physical observer to watch this: Those far away see them fade to black, those near the horizon see them fall quickly down and then fade to black, and infalling observers will only encounter the most recently gone objects, if any.
This "piling" is more like a coordinate artifact, a bookkeeper's way of telling us that we lost contact with the events inside the horizon. You shift the last observable event to +infinity, so all the later events are outside our scope.

As you have said,the redshift is exponential in time.after you get a wave you have to wait longer to get the next one,but it never stops right.I mean mathematically we can say that the wave is getting red-shifted so that as time tends to infinity the wavelength will also tend to infinity ,but in reality it doesn't happen.the wave dopes get stretched and with the using the difference in wavelengths between two corresponding waves we can even may be even calculate the distance the object moved
The gravity of the BH is what stretches the wave ,so if we get 2 corresponding waves(when the object is very close to the horizon) with some wavelength difference,then can't we say that the object has moved that tiny bit closer towards the event horizon?so as it approaches the horizon,the wavelength emitted by it exponentially increases and the and the distance moved by it closer to the horizon exponentially decreases(compared to the previous instant)

 

[Ich]

As you have said,the redshift is exponential in time.after you get a wave you have to wait longer to get the next one,but it never stops right.

The object will emit a finite number of waves before it enters the horizon. There is a last wave. The phase of said last wave will asymptotically reach a fixed value.
It's a mapping of a finite proper time to an infinite observer/coordinate time.

but in reality it doesn't happen.

Right, in reality, after a few milliseconds (for a stellar BH), there is nothing more to see. The Black Hole is black again.

Could you please rephrase the rest of your post, in case your questions are not answered yet?

 

[sri sharan]

ok,this makes the issue clear (sorry,it got no math in it,but i liked the explanation)
http://www.xs4all.nl/~johanw/PhysFAQ/Relativity/BlackHoles/fall_in.html"

 

[Naty1]

"...Which means that since the big bang, not a single particle has ever crossed the event horizon of a black hole. They just come closer and closer to it, piling up onto the surface into a quasi-2D shell..."

Not so...
different observers make different observations...most black holes ARE believed to be accreting and growing larger, like the massive black holes that may power most galactic centers...

things are NOT always as they appear...What you describe is as "real" (or not) as two observers passing each other at high speed...each sees the other's time dilated and length foreshortened...that is, different from their own...and that changes as either moves faster or slower...but at each observer's location, neither sees any anamoly locally...

In the case of a black hole horizon, a distant inertial observer apparently "sees" something like you describe; yet a free falling observer sees no horizon nor any particles building up there. Likewise, the free falling observer does not see the thermal radiation that fries nearby inertial observer.

Wikipedia says it this way:

The popular notion of a black hole "sucking in everything" in its surroundings is therefore only correct near the black hole horizon; far away, the external gravitational field is identical to that of any other body of the same mass.[

34]
and

To a distant observer, clocks near a black hole appear to tick more slowly than those further away from the black hole.[40] Due to this effect, known as gravitational time dilation, an object falling into a black hole appears to slow down as it approaches the event horizon, taking an infinite time to reach it.[41] At the same time, all processes on this object slow down causing emitted light to appear redder and dimmer, an effect known as gravitational redshift.[42] Eventually, at a point just before it reaches the event horizon, the falling object becomes so dim that it can no longer be seen.

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

 

[Russinnj]

I have a question that's been Puzzling me if we know that energy frequency to an outside observer exhibits doppler shift as space speeds up near the event horizon then isn't it also possible that at some point the energy expands to a volume of space greater then that in which the black hole is located in. I mean as the wave approaches the speed of light its time would be shifted so greatly that it may eventually end up with a frequency that's measured in cycles per billions of years to the outside observer which would appear as well next to nothing. I guess what I'm asking is unless a black hole can contain all space and all time and have an infinite life it must fizzle out just by loosing things so week at a massive scale. if it ran out of stuff to suck in it would still be loosing mass this way just ever so slowly but at an accelerated pace equal but opposite to its growth and we would not be able to observe it. It would be so slow you would almost be able to call it dark energy?

 

[Limitless777]

Russinnj question is very interesting and from my point of view is definitely messing with the quantum effects. Extremely low observed frequencies (red shift) of that order mean that the energy/matter the black hole consists of become actually unobservable due to the quantum uncertainty. It looks like it has something to do with the Hawking black hole evaporation.

 

[mrspeedybob]

This observation also raises questions about the formation of black holes.

As an object such as a star contracts gravity, and therefore time dilation, at its surface will increase. As the surface contracts toward the schwartzchild radius it will grow dimmer until it cannot be seen, but from a distant observers perspective (like ours) it will never shrink to the schwartzchild radius. This would mean that seeing the surface of the object would never be impossible, only limited by how long a wavelength of light you can detect. This would mean that from our frame of reference, or any distant observer, black holes do not exist, only very very dim holes.

 

[Cosmo Novice]

The object will emit a finite number of waves before it enters the horizon. There is a last wave. The phase of said last wave will asymptotically reach a fixed value.
It's a mapping of a finite proper time to an infinite observer/coordinate time.

Right, in reality, after a few milliseconds (for a stellar BH), there is nothing more to see. The Black Hole is black again.

Would it be true to say that after a few milliseconds there is something to see, as any perception must be from an objective viewpoint of the object outside the EH. At this point time is ticking differently for the observer and for the object crossing into the EH. So at any given space/time from the observer the object does exist - from the frame of reference of the observer, however the object exists within its own timeframe reference which is much slower. Both timeframe references are accurate and enforced by the gravitational time dilation of the BH.

This being the case would the oldest (If that term can be applied) particles in the universe be photons that have been traveling (subjectively in their own timeframe) for an almost infinite amount of time as they were emmitted the shortest possible length (a Plancke length?) from the edge of the EH?

**Any assumptions I have made are in no way founded and purely what I assume to be correct, apologies in advance and any corrections welcome. Thanks

 

[Limitless777]

As the surface contracts toward the schwartzchild radius it will grow dimmer until it cannot be seen, but from a distant observers perspective (like ours) it will never shrink to the schwartzchild radius. This would mean that seeing the surface of the object would never be impossible, only limited by how long a wavelength of light you can detect. This would mean that from our frame of reference, or any distant observer, black holes do not exist, only very very dim holes.

All of this is correct (for naked BH, meaning that no new matter is falling except those that forms the horizon) only up to the certain point, which I mentioned before. Quantum particles of falling matter are waves too, so time dilation will make their wavelength to increase (in our distant resting frame) to the point when they become comparable with the BH horizon radius - and then these waves will diffract on the BH, escaping from its gravity. It will be just an infinitely small part of all falling matter, but anyway. Also because of the backwards diffraction some matter will go back too. So I think that even naked BH will still be visible. Not sure but it looks like that it is just one more way to describe Hawking BH evaporation, which also makes BH visible.