Event Horizons & Falling Objects: Will Bob See It?

[HotBuffet]

Is it correct that the event horizon is smaller for an observer at 2x the Scwarzschild Radius then it is for an observer at infinity?
Light from the event horizon should not be able to reach infinity, but it could still reach the half-way observer, right?

 

[timmdeeg]

In that sense I agree that my statement regarding "speed at the horizon" is slightly confusing and must not be overinterpreted.

The equation in post #22 shows that the speed of an infalling mass approaches c relative to stationary observers asymptotically as r approaches 2M. So, at best one can talk about the "speed at the horizon" as the result of an extrapolation.

 

[tom.stoer]

The equation in post #22 shows that the speed of an infalling mass approaches c relative to stationary observers asymptotically as r approaches 2M. So, at best one can talk about the "speed at the horizon" as the result of an extrapolation.

Yes. The problem is - as I said in the beginning - that a stationary observer cannot exist at the horizon, therefore strictly speaking the speed at the horizon can only be defined w.r.t. an unphysical (mathematically idealized) observer.

 

[tom.stoer]

Is it correct that the event horizon is smaller for an observer at 2x the Schwarzschild Radius then it is for an observer at infinity?

How would you define 'smaller'? What is the size of the horizon? It cannot simply be r=2M b/c r is only a coordinate and has no invariant meaning (the funny thing is that 2M has an invariant meaning!) So how would you define "the size" in an invariant way?

Light from the event horizon should not be able to reach infinity, but it could still reach the half-way observer, right?

No; it can't. Light rays emitted at the horizon will either stay exactly at the horizon (which is the limiting case) or converge towards the singularity.

 

[HotBuffet]

How would you define 'smaller'? What is the size of the horizon? It cannot simply be r=2M b/c r is only a coordinate and has no invariant meaning (the funny thing is that 2M has an invariant meaning!) So how would you define "the size" in an invariant way?

No; it can't. Light rays emitted at the horizon will either stay exactly at the horizon (which is the limiting case) or converge towards the singularity.

I'm referring to something I've once read, and of course can't find right now. It said that when you are the traveler and plunge yourself into the black hole, you will never see yourself cross the Event Horizon, but the Horizon will keep always stay in front of you, until you hit the center.

Now there are more reasons I can think of, depending on some details which I don't know if they are true. You already said it's invariant, so some things can't be true according to that:

1) Event Horizon is calculated for light / matter to escape to infinity. So at infinity you will never see the light. But if you are closer by, the light would still be able to reach you, making the 'r' for the Event Horizon a little bit smaller. This keeps happening until you hit the center.
But this can't be true according to the above, whit a 'fixed' Event Horizon, where the Event Horizon makes everything completely stand still.

2) some other theory, which I'm quite fond of: because of Time Dialation you will never enter the Black Hole before infinity. Because of the Hawking Radiation the Black Hole will evaporate before infinity. So you will see the Black Hole evaporate in front of you, see the Event Horizon shrink until it's gone, never passing the Event Horizon. (if you survive tidal forces and live long enough).

 

[tom.stoer]

I'm referring to something I've once read, and of course can't find right now. It said that when you are the traveler and plunge yourself into the black hole, you will never see yourself cross the Event Horizon, but the Horizon will keep always stay in front of you, until you hit the center.

That's wrong.

You don't see the horizon itself. If there is something at the horizon (light rays emitted at the horizon) you will see them when you are crossing the horizon (crossing the shell of light emitted and staying at the horizon)

1) Event Horizon is calculated for light / matter to escape to infinity.

No, the horizon is not calculated for light to escape to infinity. The horizon can be defined w/o referring to "infinity" but using local expressions only. It's bit harder to do that but it's sound.

But if you are closer by, the light would still be able to reach you, ...

No, the horizon is a surface from which no light can escape outwards - regardless where you are sitting and trying to observe the light (in that sense the horizon can be defined geometrically w/o ever referring to an observer).

... where the Event Horizon makes everything completely stand still.

It's not that everything is standing still; the pure astronaut will not stand still but cross the horizon and hit the singularity in finite proper time as measured with his wristwatch. An observer outside the horizon will never see the astronaut crossing the horizon, but this does not mean that the astronaut does not cross it in reality (his own reality). The geometry is only curved such that no light will esacpe and tell you what happened.

because of Time Dialation you will never enter the Black Hole before infinity. Because of the Hawking Radiation the Black Hole will evaporate before infinity. ...

No. As we said the free falling observer / astronaut / you will cross the horizon in finite proper time and will hit the singularity in finite proper time. When I have time I will post the calculation - it's not so complicated.

 

[tom.stoer]

For a Schwarzschild black hole we define the Schwarzschildradius R_S; for a free falling obserber starting at R > R_S his proper time for the journey from r=R to r=0 (i.e. for hitting the singularity) is

$$\tau_R = \frac{\pi}{2}\frac{R}{c}\sqrt{\frac{R}{R_S}}$$

 

[Naty1]

good stuff above from tom.

Originally Posted by HotBuffet
Light from the event horizon should not be able to reach infinity, but it could still reach the half-way observer, right? No; it can't. Light rays emitted at the horizon will either stay exactly at the horizon (which is the limiting case) or converge towards the singularity

also good. What does "reach out" is stuff just infinitesimally outside the horizon...say one Planck Length outside and greater...where Leonard Susskind has developed the "stretched horizon"...Thats where gravity, spin and charge (the 'hair') information resides.

 

[tom.stoer]

What does "reach out" is stuff just infinitesimally outside the horizon...say one Planck Length outside and greater...where Leonard Susskind has developed the "stretched horizon"...Thats where gravity, spin and charge (the 'hair') information resides.

Regardless what it means, what you are saying is definitely not related to classical GR but in some sense to quantum theory. That does not mean that it may not be correct, but it requires a different context.

Regarding "horizons" in quantum theory: in a sense the standard definition of horizons (absolute horizons from which no null ray can escape to infinity) breaks down and has to be replaced by some local definition (in LQG they talk about isolated horizons, in string theory they may have something else, the details do not matter). The problem is that due to Hawking radiation the definition that "no null ray can escape to infinity" seems to become useless; Hawking radiation consists of massles particles at infinity ;-)