Event Horizons & Falling Objects: Will Bob See It?

[Smoggyrob]

Hi everyone:

Do event horizons appear to be surrounded by all the stuff that has ever fallen in? As I understand it, if you watch something fall into a black hole's event horizon, it will appear to take forever.

To A/B it, if Alice starts watching the horizon at t=0, and something falls in at t=1, Alice will observe it taking forever to fall in.

But what if Bob starts watching the horizon at t=2 (after the object fell in)? Will he see the object taking forever to fall in, or will he not see the object at all (because it already fell in)? Thanks for your time!

Rob

 

[Simon Bridge]

Welcome to PF.
It's true that you get lots of stuff around a BH that has never fallen in and still has a long time yet before it does - these are called accretion disks, and are the main way we detect BHs.

However, this is not what you mean.

Recall that the radius of the event horizon is proportional to the mass enclosed.
When an object gets close enough to the even horizon to be part of a black-hole, what happens is the event horizon expands to cover the object. The BH sort-of "gulps down" things that get too close.

Be aware - when we try to describe things like black holes using simple analogies - we are going to get it wrong in important ways. The math for black holes is not as bad as you think so if you are really really interested, they make a decent study.)

10 things you don't know about black holes (Bad Astronomy blog.)

 

[tom.stoer]

... if you watch something fall into a black hole's event horizon, it will appear to take forever.

As seen by an outside, stationary observer it will take forever. But note that in reality he will see nearly nothing (absolutely nothing) i.e. nearly no (no) light from the objects closed to (at) the horizon b/c there is a large (infinite) red shift for light emitted near (at) the horizon.

For the objects themselves it takes finite proper time to cross the horizon and to fall into the singularity, so they will not stay at the horizon (they will cross the horizon at the speed of light!)

The horizon marks the boundary between a region where light rays can escape to infinity and a region where all light rays will converge towards the singularity. That means that light emitted radially outwards exactly at the horizon will neither fall into the singularity nor escape to infinity - it will stay at the horizon! But b/c light always moves at c the entire horizon will move at c. That means that a free falling observer will cross the horizon at c simply b/c the entire horizon is a so-called light-like surface moving with the speed of light.

No back to the accretion disks: b/c of the tidal forces becoming large at the horizon of solar black holes, infalling matter becomes extremely hot and radiates X- and γ-rays. Rays emitted radially outwards directly at the horizon will stay at the horizon. Rays emitted tangentially at a larger radius (1.5 * Schwarzschild radius for Schwarzschild black holes) can orbit the black hole at constant radius (like satellite) and form the so-called photon-orbit.

 

[PeterDonis]

For the objects themselves it takes finite proper time to cross the horizon and to fall into the singularity, so they will not stay at the horizon (they will cross the horizon at the speed of light!)

The part in parentheses is not correct. The infalling object (assuming it is not itself an ingoing light beam) moves on a timelike worldline, so it never moves "at the speed of light".

The horizon will move outward *past* the object at the speed of light, but that's because the *horizon* is a lightlike surface, as you note:

light emitted radially outwards exactly at the horizon will neither fall into the singularity nor escape to infinity - it will stay at the horizon! But b/c light always moves at c the entire horizon will move at c. That means that a free falling observer will cross the horizon at c simply b/c the entire horizon is a so-called light-like surface moving with the speed of light.

It's the *horizon* that's moving at c, not the object. We don't say we are "moving at c" when a light ray passes us going in the exact opposite direction; that's what the horizon is doing relative to the infalling object.

 

[tom.stoer]

The infalling object will cross the horizon at the speed of light i.e. it moves at the speed of light w.r.t. the horizon - and the horizon moves at the speed of light w.r.t. the object.

So the part in parentheses is correct; the next sentences provide the explanation.

 

[netheril96]

The infalling object will cross the horizon at the speed of light i.e. it moves at the speed of light w.r.t. the horizon - and the horizon moves at the speed of light w.r.t. the object.

So the part in parentheses is correct; the next sentences provide the explanation.

Nothing light-like can act as an observer, so "at the speed of light w.r.t. the horizon" is not a valid expression.

 

[tom.stoer]

Nothing light-like can act as an observer

It cannot act as a physical observer, that's correct; the definition of reference frames becomes 'singular'.

"at the speed of light w.r.t. the horizon" is not a valid expression.

If there is a flash of light then the speed of the flash w.r.t. to me is c and my speed w.r.t. the flash is c, too. This is a perfectly valid statement. In GR I always have to say "speed w.r.t. something" and if this "something" is light, it will be clear that my speed is c.

But I think this discussion is quibbling and totally irrelevant for the original question.

My idea was to explain why it is difficult to say that an object 'stays at the horizon' b/c a) the horizon is moving at c and therefore no massive body can stay there and b/c b) this statement would depend on the reference frame and a stationary observer outside the horizon will soemthing totally dfferent than the free-falling observer who crosses the horizon

Btw.: the observer will cross the surface of the horizon, so there is a relative speed of the observer w.r.t. to the horizon; what is this speed? c? something else? no speed at all? not defined?

 

[netheril96]

It cannot act as a physical observer, that's correct; the definition of reference frames becomes 'singular'.


If there is a flash of light then the speed of the flash w.r.t. to me is c and my speed w.r.t. the flash is c, too. This is a perfectly valid statement. In GR I always have to say "speed w.r.t. something" and if this "something" is light, it will be clear that my speed is c.

But I think this discussion is quibbling and totally irrelevant for the original question.

My idea was to explain why it is difficult to say that an object 'stays at the horizon' b/c a) the horizon is moving at c and therefore no massive body can stay there and b/c b) this statement would depend on the reference frame and a stationary observer outside the horizon will soemthing totally dfferent than the free-falling observer who crosses the horizon

Btw.: the observer will cross the surface of the horizon, so there is a relative speed of the observer w.r.t. to the horizon; what is this speed? c? something else? no speed at all? not defined?

How is horizon moving at the speed of light? It stays (in Schwarchild coordinates) at r=\frac{2GM}{c^2}. And it is not an object with a null worldline.

 

[tom.stoer]

... it is not an object with a null worldline

It is!

We start with the Schwarzschild metric

ds^2 = \left(1-\frac{r_s}{r}\right)\,dt^2 - \left(1-\frac{r_s}{r}\right)^{-1}\,dr^2 - r^2\,d\Omega^2

Now we look at radial, light-like motion, i.e. d\Omega = 0 and ds² = 0

0 = \left(1-\frac{r_s}{r}\right)\,dt^2 - \left(1-\frac{r_s}{r}\right)^{-1}\,dr^2

This can be solved for dr/dt

\frac{dr}{dt} = 1-\frac{r_s}{r}

From this differential equation all radial null-geodesics can be derived.

A special solution is

r(t) = r_s = \text{const.}

for which both l.h.s. and r.h.s. of the diff eq. are zero

 

[netheril96]

It is!

We start with the Schwarzschild metric

ds^2 = \left(1-\frac{r_s}{r}\right)\,dt^2 - \left(1-\frac{r_s}{r}\right)^{-1}\,dr^2 - r^2\,d\Omega^2

Now we look at radial, light-like motion, i.e. d\Omega = 0 and ds² = 0

0 = \left(1-\frac{r_s}{r}\right)\,dt^2 - \left(1-\frac{r_s}{r}\right)^{-1}\,dr^2

This can be solved for dr/dt

\frac{dr}{dt} = 1-\frac{r_s}{r}

From this differential equation all radial null-geodesics can be derived.

A special solution is

r(t) = r_s = \text{const.}

for which both l.h.s. and r.h.s. of the diff eq. are zero

It is a surface, not a point however. I don't know how you can define the speed of an extended non-rigid entity.

BTW: isn't there a difference between null geodesics and null curves?

 

[tom.stoer]

It is a surface, not a point however. I don't know how you can define the speed of an extended non-rigid entity.

I derived a family of null-geodesics for all points r(t) = r_S with arbitrary \Omega defining the horizon. Therefore the horizon is defined by

x^\mu = (t,r_s,\Omega)

So for each t you have a 2-surface of fixed r(t) = r_S with coordinates \Omega on the surface

 

[netheril96]

I derived a family of null-geodesics for all points r(t) = r_S with arbitrary \Omega defining the horizon. Therefore the horizon is defined by

x^\mu = (t,r_s,\Omega)

So for each t you have a 2-surface of fixed r(t) = r_S with coordinates \Omega on the surface

Whatever.
We were simply arguing about semantics, which is neither fun nor meaningful.

 

[tom.stoer]

We were simply arguing about semantics, which is neither fun nor meaningful.

Sorry, but here I totally disagree!

What we are doing here is to start investigating what 'horizons' are. There are several different definitions (apparent horizons in terms of trapped null-surfaces = closed surfaced on which outward-pointing light rays are converging), absolute horizons, isolated horizons, ...

This is highly non-trivial and still subject to current research.

You may want to have a look at

• Hawking & Ellis: The large scale structure of space-time
• Wald: General Relativity
• Thorne, Misner & Wheeler: Gravitation
• Ashtekar & Krishnan: Isolated and Dynamical Horizons and Their Applications; http://relativity.livingreviews.org/Articles/lrr-2004-10/

Of course all these guys neither had fun nor did they work on something meaningful (sorry for the irony ;-)

 

[netheril96]

Sorry, but here I totally disagree!

What we are doing here is to start investigating what 'horizons' are. There are several different definitions (apparent horizons in terms of trapped null-surfaces = closed surfaced on which outward-pointing light rays are converging), absolute horizons, isolated horizons, ...

This is highly non-trivial and still subject to current research.

You may want to have a look at

• Hawking & Ellis: The large scale structure of space-time
• Wald: General Relativity
• Thorne, Misner & Wheeler: Gravitation
• Ashtekar & Krishnan: Isolated and Dynamical Horizons and Their Applications; http://relativity.livingreviews.org/Articles/lrr-2004-10/

Of course all these guys neither had fun nor did they work on something meaningful (sorry for the irony ;-)

My point was that null surface as it is, the "speed" of event horizon is not a well-defined and physically useful concept. And you were trying to prove that the horizon is moving at the speed of light without defining the speed of a surface. It's not about the properties of event horizon; it's about the definition of speed.

 

[tom.stoer]

My point was that null surface as it is, the "speed" of event horizon is not a well-defined and physically useful concept. And you were trying to prove that the horizon is moving at the speed of light without defining the speed of a surface. It's not about the properties of event horizon; it's about the definition of speed.

Sorry, but I think the math in post #9 is rigorous. The horizon moves at the speed of light in just the manner as a spherical wave front in Minkowski space does. Where's your problem? Which equation or derivation is wrong?

 

[Simon Bridge]

To drag this back to OP's question ... there is a lot of talk about objects "falling in" to a black hole and of black-holes growing over time ... of "feeding".
[eg. Bad Astronomy - 10 things ... about Black Holes #9
... Even the Milky Way has a black hole at its core ... it can’t really harm us. Unless it starts actively feeding. Which it isn’t. But it might start sometime, if something falls into it.]

Since it takes an infinite amount of time, from the POV of an observer a long way from the BH, surely you'd never see one grow and the stuff "fallen in" never actually does this but gets stuck just "outside"... so how can it be said to "grow"?

Considering the level the question is being asked for it is probably safe to take "event horizon" as the locus in space about the BH where the escape velocity is equal to the speed of light in a vacuum.

The use of the word suggests a Schwartzchild Hole rather than a Kerr Hole.

The question appears to be trying for is, given it takes an infinite amount of time to fall into a black hole, how come black holes can be said to grow? Note: I'm being deliberately sloppy with language here - the question is hard to define in sensible terms for all the reasons you guys have been discussing and more. I'm setting up the following, more accessible, description, to be found here.

So if you, watching from a safe distance, attempt to witness my fall into the hole, you'll see me fall more and more slowly as the light delay increases. You'll never see me actually get to the event horizon. My watch, to you, will tick more and more slowly, but will never reach the time that I see as I fall into the black hole. Notice that this is really an optical effect caused by the paths of the light rays.

This is also true for the dying star itself. If you attempt to witness the black hole's formation, you'll see the star collapse more and more slowly, never precisely reaching the Schwarzschild radius.

Now, this led early on to an image of a black hole as a strange sort of suspended-animation object, a "frozen star" with immobilized falling debris and gedankenexperiment astronauts hanging above it in eternally slowing precipitation. This is, however, not what you'd see. The reason is that as things get closer to the event horizon, they also get dimmer. Light from them is redshifted and dimmed, and if one considers that light is actually made up of discrete photons, the time of escape of the last photon is actually finite, and not very large. So things would wink out as they got close, including the dying star, and the name "black hole" is justified.

As an example, take the eight-solar-mass black hole I mentioned before. If you start timing from the moment the you see the object half a Schwarzschild radius away from the event horizon, the light will dim exponentially from that point on with a characteristic time of about 0.2 milliseconds, and the time of the last photon is about a hundredth of a second later. The times scale proportionally to the mass of the black hole. If I jump into a black hole, I don't remain visible for long.

Also, if I jump in, I won't hit the surface of the "frozen star." It goes through the event horizon at another point in spacetime from where/when I do.

(Some have pointed out that I really go through the event horizon a little earlier than a naive calculation would imply. The reason is that my addition to the black hole increases its mass, and therefore moves the event horizon out around me at finite Schwarzschild t coordinate. This really doesn't change the situation with regard to whether an external observer sees me go through, since the event horizon is still lightlike; light emitted at the event horizon or within it will never escape to large distances, and light emitted just outside it will take a long time to get to an observer, timed, say, from when the observer saw me pass the point half a Schwarzschild radius outside the hole.)

... which should be a reasonably complete answer for OP.

Note: All observers are moving at c wrt light. Ergo: meaningless observation.
FWIW: I'm going to have to agree with that one.

 

[George Jones]

Sorry, but I think the math in post #9 is rigorous. The horizon moves at the speed of light in just the manner as a spherical wave front in Minkowski space does. Where's your problem? Which equation or derivation is wrong?

So, you're saying that in special relativity, when an observer and a photon are coincident, the observer moves with speed of light with respect to the frame of the photon? Also in special relativity, that when an observer crosses a Rindler Killing horizon, the observer moves with the speed of light with respect to the Rindler horizon?

Aren't these examples of the type of language that we try to discourage on this forum?

 

[timmdeeg]

It's the *horizon* that's moving at c, not the object. We don't say we are "moving at c" when a light ray passes us going in the exact opposite direction; that's what the horizon is doing relative to the infalling object.

PeterDonis, sorry if I come back to this, but perhaps I can improve my understanding.
The Newtonian gravitational force is proportional to (1 - Rs/R)^-1/2, implying that an infalling mass is accelerated to c in the moment Rs = R and consequently it's future light cone is tangent to the event horizon.

Why can we not say something moves at c, in the very special case it is accelerated to c?

 

[Simon Bridge]

Why can we not say something moves at c, in the very special case it is accelerated to c?

With respect to what exactly?
(Is Newtonian Gravity a useful model for behavior close to the event horizon of a black-hole?)

 

[tom.stoer]

My point was that null surface as it is, the "speed" of event horizon is not a well-defined and physically useful concept. And you were trying to prove that the horizon is moving at the speed of light without defining the speed of a surface. It's not about the properties of event horizon; it's about the definition of speed.

Sorry, but I think the math in post #9 is rigorous. The horizon moves at the speed of light in just the manner as a spherical wave front in Minkowski space does. Where's your problem? Which equation or derivation is wrong?

So, you're saying that in special relativity, when an observer and a photon are coincident, the observer moves with speed of light with respect to the frame of the photon? Also in special relativity, that when an observer crosses a Rindler Killing horizon, the observer moves with the speed of light with respect to the Rindler horizon?

Aren't these examples of the type of language that we try to discourage on this forum?

Again, I tried to rephrase and explain mathematically – see post #9 which is mathematically rigorous – please tell me if I am wrong – what it means that “the horizon moves at c – a radially emitted photon at the horizon does exactly this! I tried to explain that this can be compared to an ordinary wave front in Minkowski space – the main difference is that in Minkowski space we have r=ct whereas for the EH we have r = const. I tried to explain that a free-falling observer will cross the horizon at the speed of light – horizon and observer move at c w.r.t. each other. I explained that the horizon (a null-surface) cannot be realized as a physical reference frame i.e. cannot be realized as a massive body. I tried to explain what the “speed of a surface” means – every point on the surface follows a null-line, i.e. the whole surface is a null-surface. I explained that the perception of a free-falling observer crossing the horizon in finite proper time and an asymptotic observer seeing objects to never cross the horizon are different.

George, Peter, I am perfectly aware of the fact that words are dangerous in this context. Nevertheless we have to try to explain what the formulas mean. In a certain sense the observer crosses the horizon at c – you can do or find calculations showing exactly this in terms of free fall w.r.t. a ‘singular unphysical reference frame hovering at the horizon’. In a sense it’s the horizon that moves at c – this is the translation of null-surface – and it can be realized physically by photons emitted radially outwards exactly at the horizon.

 

[PAllen]

George, Peter, I am perfectly aware of the fact that words are dangerous in this context. Nevertheless we have to try to explain what the formulas mean. In a certain sense the observer crosses the horizon at c – you can do or find calculations showing exactly this in terms of free fall w.r.t. a ‘singular unphysical reference frame hovering at the horizon’. In a sense it’s the horizon that moves at c – this is the translation of null-surface – and it can be realized physically by photons emitted radially outwards exactly at the horizon.

Your last phrasing here is the only one I ever use, to avoid confusion. In the sense of relative velocities being only local in GR, then in any local frame (alternatively locally Minkowski coordinate patch) containing the horizon, the horizon moves at c and 'includes' the wave front of all outwardly emitted light from everything that has crossed the horizon. I never speak of the frame of the horizon, or speed of something relative to the horizon because this would be equivalent to talking about the frame of light.

 

[timmdeeg]

With respect to what exactly?
(Is Newtonian Gravity a useful model for behavior close to the event horizon of a black-hole?)

Yes, in their book "Exploring Black Holes" Edwin F. Taylor and John Archibald Wheeler write - commenting the equation dr_shell/dt_shell = -(2M/r)^1/2 - "This expression for radial velocity is the same as the result of the Newtonian analysis" (3-15).

And in the same context: "As the particle crosses the event horizon at r = 2M, nearby shell observers record it as moving at the speed of light." However add, as discussed in this Thread "Shells - and even shell observers- cannot exist inside the horizon or even at the horizon ...".

 

[Simon Bridge]

"of the shell" not the person crossing the shell.
they do not go so far as to say that Newtonian gravity is a good model for behavior close to the event horizon. On top of which, all observers see light traveling at the speed of light ... not a significant observation.

Why can we not say something moves at c, in the very special case it is accelerated to c?

"accelerated to c" with respect to what, exactly?

 

[dm4b]

I think what people have issue with is that massive objects cannot travel at c. The post above can be read ( perhaps mistakenly?), as if they can.


Also, in highly curved, dynamic spacetimes, relative velocities lose "meaning". So, in this regime, to say something is moving at a certain velocity with respect to something else is a loosely worded statement, to begin with.

 

[Naty1]

From the OP

Do event horizons appear to be surrounded by all the stuff that has ever fallen in?

Simple answer...yes, BUT!...it depends on just what you mean by "surrounded' and also your frame of [observer] reference. One viewpoint is that everything that has ever been enclosed within the black hole event horizon is represented as information residing just outside the even horizon...maybe at Planck length outside (per Leonard Susskind).

But there is a lot more to it:

Here are several descriptions which cover the basics:


Two 'simple' ones first:
from Roger Penrose:

There is no mass as we know it (inside); inside all particles have been destroyed and gravitational effects remain outside the event horizon along with a few characteristics (electric charge, spin, etc).

Mitchell Porter posts:

... the idea is that the interior of the black hole has a dual (holographic) description in terms of states on the horizon; a lot like AdS/CFT, with the horizon being the boundary to the interior. So when someone crosses the horizon from outside, there's a description which involves them continuing to fall inwards, until they are torn apart by tidal forces and their degrees of freedom redistributed among the black hole's degrees of freedom, all of which will later leak away via Hawking radiation; but there's another description in which, when you arrive at the horizon, your degrees of freedom get holographically smeared across it, once again mingling with all the black hole's prior degrees of freedom (also located on the horizon), which all eventually leak away as Hawking radiation

and moving on to the more complex:

[Two concepts first. If you are familiar with special relativity, no gravity, you'll note that two high speed observers do not in general measure identical distances (lengths) nor even the same passage of time. In general relativity, with gravity, gravitational potential also affects the passage of time. So different observers in general make different observations; there is no absolute 'reality'.

From Kip Thorne in BLACK HOLES AND TIME WARPS

when the star forms a black hole:

Finkelstein's reference frame was large enough to describe the star's implosion ...simultaneously from the viewpoint of far away static observers and from the viewpoint of observers who ride inward with the imploding star. The resulting description reconciled...the freezing of the implosion as observed from far away with (in contrast to) the continued implosion as observed from the stars surface...an imploding star really does shrink through the critical circumference without hesitation...That it appears to freeze as seen from far away is an illusion...General relativity insists that the star's matter will be crunched out of existence in the singularity at the center of the black...

[How can this be?? It seems crazy! (It is!) Think about a near light speed observer looking at a clock tick off time back on earth,,,it seems almost stopped...but not to the person on earth...nothing is unusual in the appropriate local frame.]


[This quote shows why nothing we observe can be inside the horizon: if it were nobody outside could observe it!]
Kip Thorne says (Lecture in 1993 Warping Spacetime, at Stephan Hawking's 60th birthday celebration, Cambridge, England,)

The flow of time slows to a crawl near the horizon, and beneath the horizon time becomes so highly warped that it flows in a direction you would have thought was spacial: it flows downward towards the singularity. That downward flow, in fact, is why nothing can escape from a black hole. Everything is always drawn inexorably towards the future, and since the future inside a black hole is downward, away from the horizon, nothing can escape back upward, through the horizon.

Black Hole Complementarity
Leonard Susskind, THE BLACK HOLE WAR (his arguments with Stephen Hawking)

Complementarity

(p238) Today a standard concept in black hole physics is a stretched horizon which is a layer of hot microscopic degrees of freedom about one Planck length thick and a Planck length above the event horizon. Every so often a bit gets carried out in an evaporation process. This is Hawking radiation. A free falling observer sees empty space.

(p258) From an outside observer’s point of view, an in falling particle gets blasted apart….ionized….at the stretched horizon…before the particle crosses the event horizon. At maybe 100,000 degrees it has a short wavelength and any detection attempt will ionize it or not detect it!

(p270)…. eventually the particle image is blurred as it is smeared over the stretched horizon and….and the image may (later?) be recovered in long wavelength Hawking radiation.




FAQ black Holes Virginia Tech
http://www.phys.vt.edu/~jhs/faq/blackholes.html#q11

Will an observer falling into a black hole be able to witness all future events in the universe outside the black hole?

The normal presentation of these gravitational time dilation effects can lead one to a mistaken conclusion. It is true that if an observer (A) is stationary near the event horizon of a black hole, and a second observer (B) is stationary at great distance from the event horizon, then B will see A's clock to be ticking slow, and A will see B's clock to be ticking fast. But if A falls down toward the event horizon (eventually crossing it) while B remains stationary, then what each sees is not as straight forward as the above situation suggests.

 

[tom.stoer]

Also, in highly curved, dynamic spacetimes, relative velocities lose "meaning". So, in this regime, to say something is moving at a certain velocity with respect to something else is a loosely worded statement, to begin with.

This is true for two bodies distant from each other; here relative velocity may become meaningless. But locally e.g. for two colliding objects velocity is still a valid concept.

 

[Naty1]

tom.stoer and PAllen seem to agree here:
"... In a sense it’s the horizon that moves at c – this is the translation of null-surface – and it can be realized physically by photons emitted radially outwards exactly at the horizon."

and that comports with many descriptions I have read.


tom.stoer posts:

That means that a free falling observer will cross the horizon at c simply b/c the entire horizon is a so-called light-like surface moving with the speed of light.

But there is no observable horizon for a free falling observer,only for an accelerating [stationary outside the horozon] observer, so saying one moves at a certain speed with respect to it seems rather abiguous to me.

 

[Simon Bridge]

Perhaps what is being imagined is that observer can measure the speed the singularity approaches - and by reference, the speed that space is passing? Said observer could note when his separation from the singularity is consistent with the distance to the even horizon observed by someone not free falling.

Thing is, I don't see why the speed the singularity approaches has to be c under these conditions.With reference to post #1
http://www.jimhaldenwang.com/black_hole.htm

Physical Interpretation of the Event Horizon

We found earlier that the Schwarzschild metric has a coordinate singularity at the event horizon, where the coordinate time becomes infinite. Recall that the coordinate time is approximately equal to the far away observer's proper time. However, a calculation using transformed coordinates shows that the infalling observer falls right through the event horizon in a finite amount of time (the infalling observer's proper time). How can we interpret solutions in which the proper time of one observer approaches infinity yet the proper time of another observer is finite?

The best physical interpretation is that, although we can never actually see someone fall through the event horizon (due to the infinite redshift), he really does. As the free-falling observer passes across the event horizon, any inward directed photons emitted by him continue inward toward the center of the black hole. Any outward directed photons emitted by him at the instant he passes across the event horizon are forever frozen there. So, the outside observer cannot detect any of these photons, whether directed inward or outward.

Consider two observers far from the black hole. Suppose they synchronize their watches, then one of them remains far from the black hole while the other descends slowly (at first) toward the event horizon. Then the time on the watch of the descending observer as he reaches and falls through the event horizon will be approximately equal to the time on the watch of the far away observer as she sees her companion disappear very near the event horizon.

... has a lot of description backed by math about an observer crossing the event horizon.

 

[tom.stoer]

There is no observation or measurement possible which shows that a (free falling) observer crosses the horizon. In that sense "crossing the horizon at c" is a mathematical statement only, not a physical (phenomenological) one. The problem is that even so the light cones are tilted globally, locally they appear as ordinary light cones, therefore the "photons emitted radially outwards exactly at the horizon" appear by no means special w.r.t. to (free falling) observers.

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

 

[Naty1]

My understanding of this explanation:

The best physical interpretation is that, although we can never actually see someone fall through the event horizon (due to the infinite redshift), he really does. As the free-falling observer passes across the event horizon, any inward directed photons emitted by him continue inward toward the center of the black hole. Any outward directed photons emitted by him at the instant he passes across the event horizon are forever frozen there. So, the outside observer cannot detect any of these photons, whether directed inward or outward.

is Schwarzschild metric based...a somewhat 'idealized' view from infinity. similar to my earlier posts...[As always what you "see" depends on your model and your reference frame. ]

Here is why I think that:

Essentially observers at finite distance from black holes DO see things disappear...and black holes form...If not, how could we observe (the effects of) ANY black holes today??

I have these ever so brief comments in my notes from other discussions:

According to the book Quantum Fields in Curved Space by Birrell and Davies, pages 268-269,

... the event horizon is a global construct, and has no local significance, so it is absurd too conclude that it acts as physical barrier to the falling observer.

(which to me means such an inertial observer is unaware of the horizon as she passes.)


Jessem:

There's no coordinate-independent way to define the amount of time dilation for a clock at various distances from the horizon--what we're talking about is the rate a clock is ticking relative to coordinate time, so even if that rate approaches zero in Schwarzschild coordinates which are the most common ones to use for a nonrotating black hole, in a different coordinate system like Kruskal-Szekeres coordinates it wouldn't approach zero at the horizon at all.

Dalespam:

In the Schwarzschild metric a stationary worldline is not a geodesic and therefore cannot be inertial….

[The Schwarzschild metric is flat only at infinite distance. It is not flat at any finite distance.]


Q: So, are you saying, that an observer, suspended in intergalactic space, or one in the center of Milky Way (if that space was not already occupied) is non-inertial in any significant/measurable/noticeable way?

Dalespam:

Is intergalactic space or the center of the Milky Way well described by the Schwarzschild metric and is the observer stationary in those coordinates? If so, then yes.

Clarifications and other interpretations welcome...

 

[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 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 ;-)