Event Horizons & Falling Objects: Will Bob See It?

[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 happend.

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 definately 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 defintion that "no null ray can escape to infinity" seems to become useless; Hawking radiation consists of massles particles at infinity ;-)