Does an event horizon ever exist?

[Passionflower]

The only claims I've seen about a portion of the Kerr spacetime being "unphysical" refer to the region inside the inner horizon where CTCs are predicted.)

I just love to have a pair of batteries with negative energy, then I can finally get my anti-gravity rockets to work.

By the way I really do not see the big deal about CTC's, so the state of an object keeps rotating: "rock, paper, scissors", rendezvous-ing allover again, so what?

 

[photonkid]

I just love to have a pair of batteries with negative energy, then I can finally get my anti-gravity rockets to work.

http://www.stuff.co.nz/the-press/christchurch-life/8372542/Bright-sparks-and-black-holes

<quote> He says there is this little idea he has about negative particles that would have repulsive gravity. If he can refine the mathematical detail, he might just alarm his hosts with a presentation when he collects his Einstein Medal in May. </>

 

[pervect]

And now that I look at the "talk page" for the Wikipedia article,

The Wiki isn't a terribly reliable source. The talk page is probably slightly less reliable than the wiki.

I think it's strange that most articles on black holes never mention that the idea of a singularity is just a theory and that what actually happens when mass collapses into itself is completely unknown.

This is a bit of a non-sequitur. The central singularity is not the event horizon.

I don't really agree with the remarks made about the singularity either but that seems like a topic for a different thread.

Sticking to the topic of the event horizon and not getting sidetracked:

The event horizon is both predicted by GR, and confirmed well by experiment. One recent paper:

http://arxiv.org/abs/0903.1105

Black hole event horizons, causally separating the external universe from compact regions of spacetime, are one of the most exotic predictions of General Relativity (GR). Until recently, their compact size has prevented efforts to study them directly. Here we show that recent millimeter and infrared observations of Sagittarius A* (Sgr A*), the supermassive black hole at the center of the Milky Way, all but requires the existence of a horizon. Specifically, we show that these observations limit the luminosity of any putative visible compact emitting region to below 0.4% of Sgr A*'s accretion luminosity. Equivalently, this requires the efficiency of converting the gravitational binding energy liberated during accretion into radiation and kinetic outflows to be greater than 99.6%, considerably larger than those implicated in Sgr A*, and therefore inconsistent with the existence of such a visible region. Finally, since we are able to frame this argument entirely in terms of observable quantities, our results apply to all geometric theories of gravity that admit stationary solutions, including the commonly discussed f(R) class of theories.

The short version of this is that our black hole candidate is -- black. If it were any sort of object with an observable surface, we'd expect to see radiative emissions from said surface due to infalling matter. For instance, we can easily detect the surface of a neutron star if matter is falling on it.

The primary astrophysical importance of a horizon is
that the gravitational binding energy liberated by ma-
terial as it accretes can be advected into the black hole
without any further observational consequence. This is
very different from accretion onto other compact objects,
e.g., neutron stars, in which this liberated energy ulti-
mately must be emitted by the stellar surface.

There may still be a few small experimental loopholes, but the bulk of the evidence very storngly suggests that event horizons are very real, and that Sag. A has an event horizon.

Event horizons are also firm theoretical predictions of SR.

The fact that one can reach the event horizon in a finite proper time is another firm theoretical prediction of GR.

 

[PeterDonis]

But if physicists are still debating whether event horizons actually exist, how can you say we know what actually happens?

I didn't say we know what actually happens, period. I said we know a lot more than nothing. We know that classically a horizon is predicted to form. We know that quantum effects, if they are going to prevent the horizon from forming, would have to be large: just small changes in the classical behavior won't do it. And we know that, for black holes of astronomical size, the spacetime curvature at the horizon is small compared to the expected scale of quantum gravity (which would be a radius of curvature comparable to the Planck length); it's hard to see how quantum corrections could be large in such a regime.

As I understand it, arguments like these are why the current mainstream view is that horizons form. (AFAIK arguments like Susskind's for why this does not violate quantum unitarity are also mainstream, but that's really a separate question since it has to do with how quantum corrections affect the singularity as well as the horizon.)

Btw, it's important to distinguish two things: whether or not a horizon forms, and whether or not a singularity forms (meaning a singularity at r = 0, the "center" of the black hole). All the stuff I said above (and in earlier posts) was about whether or not a horizon forms. But even if a horizon forms (because quantum corrections are too small to prevent it for a black hole of astronomical size), we still expect quantum corrections to the classical behavior to be large as the singularity is approached, because the classical prediction is that spacetime curvature increases without bound in that regime, so at some point it will certainly reach the Planck regime.

I'm not a physicist and it's hard for me to grasp these arguments about "proper time" continuing for an object falling into a black hole. How could mass that falls past the event horizon ever make it to the singularity in "our time"?

The classical prediction is made using the same sort of math that predicts a finite proper time to fall to the horizon. However, the use of the term "our time" is not correct. The coordinates that are "natural" to an observer far away from the black hole, which are where the concept of "our time" comes from, simply do not cover the region of spacetime inside the horizon. Many people get hung up over this because they simply can't conceive how that can be; but once again, the math is unambiguous, and it is not controversial at all.

If it was possible to measure it, would we see the photon slowing down inside the event horizon?

This question doesn't really have a meaningful answer, because there's no way to define what "slowing down" means inside the horizon. Outside the horizon, there is a set of "hovering" observers that stay at the same altitude above the horizon, and these observers can be used as a reference to define what "time slowing down" means (observers closer to the horizon have clocks that "run slow" compared to observers higher up). Inside the horizon, there are no such observers, so there's no way to construct a reference for "time" that works the way the reference system outside the horizon does.

(This lack of "hovering" observers inside the horizon is related to the fact I noted above, that the natural time coordinate in the exterior region does not cover the interior region.)

Does light slow down in "our time" when it "goes past" a large mass?

If you mean light that stays outside the horizon (including the case where the mass doesn't have a horizon, like an ordinary planet or star), then yes (where "slow down" means relative to the time reference I described above, that only works outside the horizon). This is called the Shapiro time delay, and it has been measured:

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

 

[photonkid]

The Wiki isn't a terribly reliable source. The talk page is probably slightly less reliable than the wiki.

So is there any detail on this page that you dispute and if so, why don't you correct it? It seems to me that the people contributing to the article are mainstream physicists and considering the unpleasantness and number of kooks on the usenet relativity forum, it's just as well.

The short version of this is that our black hole candidate is -- black. If it were any sort of object with an observable surface, we'd expect to see radiative emissions from said surface due to infalling matter. For instance, we can easily detect the surface of a neutron star if matter is falling on it.

But does time slow down close to the neutron star as it does close to the "almost" event horizon of an almost black hole.

The fact that one can reach the event horizon in a finite proper time is another firm theoretical prediction of GR.

You mean in the falling object's frame of reference? Can the falling object reach the singularity? The closer you get to the singularity, the slower time goes?

 

[photonkid]

As I understand it, arguments like these are why the current mainstream view is that horizons form. (AFAIK arguments like Susskind's for why this does not violate quantum unitarity are also mainstream, but that's really a separate question since it has to do with how quantum corrections affect the singularity as well as the horizon.)

Btw, it's important to distinguish two things: whether or not a horizon forms, and whether or not a singularity forms (meaning a singularity at r = 0, the "center" of the black hole). All the stuff I said above (and in earlier posts) was about whether or not a horizon forms. But even if a horizon forms (because quantum corrections are too small to prevent it for a black hole of astronomical size), we still expect quantum corrections to the classical behavior to be large as the singularity is approached, because the classical prediction is that spacetime curvature increases without bound in that regime, so at some point it will certainly reach the Planck regime.

ok, so this page
http://en.wikipedia.org/wiki/Black_hole#Singularity
says <quote> there exist attempts to formulate such a theory of quantum gravity. It is generally expected that such a theory will not feature any singularities. </>
Do you know if this is a correct statement?

Many people get hung up over this because they simply can't conceive how that can be; but once again, the math is unambiguous, and it is not controversial at all.

(This lack of "hovering" observers inside the horizon is related to the fact I noted above, that the natural time coordinate in the exterior region does not cover the interior region.)

ok, I haven't managed to understand special relativity yet so I'm not likely to understand this in the near future.

If you mean light that stays outside the horizon (including the case where the mass doesn't have a horizon, like an ordinary planet or star), then yes (where "slow down" means relative to the time reference I described above, that only works outside the horizon). This is called the Shapiro time delay, and it has been measured:

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

Yes, that's what I meant. Thanks.

 

[photonkid]

If you DO like my suggested definition, hopefully I have already answered your question, and you just neeed to read it and think it over a bit.

Well for your information, your answer comes across as arrogant and convoluted. Nobody else in this thread saw any need to debate what "exists" means.

Also, note that although the subject was "does an event horizon ever exist", in the content I said
<quote> Is it true that the event horizon never comes into existence - or at least, if time slows down like general relativity suggests, would an event horizon and a singularity never come into existence? </>

If you wanted me to take your answer seriously you should have said
"mainstream physicists believe that event horizons actually do exist because..."

"mainstream physicists do/ do-not believe that singularities exist because..."

since it turns out that the answer is quite complicated evidenced by the fact that people in this thread are debating what happens to time near an event horizon.

 

[PeterDonis]

<quote> there exist attempts to formulate such a theory of quantum gravity. It is generally expected that such a theory will not feature any singularities. </>
Do you know if this is a correct statement?

AFAIK it is, yes. I note that there is a statement later on on that Wiki page, in the "Alternatives" section, to the effect that a quantum gravity will not feature any event horizons either. I wasn't aware that that was a mainstream view (as I've said in this thread), but the footnote there references a review article in Annalen der Physik that I haven't read. I'll take a look at it.

 

[Nugatory]

But does time slow down close to the neutron star as it does close to the "almost" event horizon of an almost black hole.

Yes. Also close to the surface of the earth, although there the effect is much smaller because the gravitational field of the Earth is much weaker than that of a neutron star. It's been measured on Earth.

 

[Nugatory]

So is there any detail on this page that you dispute and if so, why don't you correct it?

Editing some wikipedia pages is a thankless and Sisyphean task.
Answering questions here is merely thankless.

 

[Nugatory]

The fact that one can reach the event horizon in a finite proper time is another firm theoretical prediction of GR.

You mean in the falling object's frame of reference?

(Be aware that "frame of reference" is a treacherous concept in GR; usually you're better off thinking in terms of local inertial frames)

Pervect specifically said "proper time"; proper time is frame-independent. Intuitively, proper time is the amount of time that passes for a single clock that travels on some path between two points in space-time. It is the same for all observers and is in no way affected by changing frames of reference.

For example, if my airplane takes off at noon according to my wristwatch and lands at 1:00 according to that same wristwatch, all observers everywhere will agree about three facts: the watch read noon at takeoff; the watch read 1:00 at landing; I experienced a one-hour journey and aged one hour between takeoff and landing. Thanks to time dilation, relativity of simultaneity, and other relativistic effects, the observers may have measured very different times for my journey, but they all agree that for me it was a one-hour journey. That's proper time.

It's important to understand that proper time only works for a single clock that's only at a single place at any moment. My wristwatch measures the proper time that I experience, but it tells me nothing about the experience of other observers outside the airplane.

Can the falling object reach the singularity?

Yes, in a finite amount of proper time according to classical GR - see #12 in this thread. It is likely that classical GR stops working very close to the singularity, in which case the answer might be different... but if so, that happens long after the infalling object has passed through the event horizon.

 

[PeterDonis]

I note that there is a statement later on on that Wiki page, in the "Alternatives" section, to the effect that a quantum gravity will not feature any event horizons either. I wasn't aware that that was a mainstream view (as I've said in this thread), but the footnote there references a review article in Annalen der Physik that I haven't read. I'll take a look at it.

Having looked at it, I don't see anything in that article that says that quantum corrections are expected to prevent a horizon from forming. I do see references to the fact that, in order to show unitarity, you have to include amplitudes for spacetime histories where a black hole does not form, as well as for histories where one does form. But it also says that macroscopic black holes have classical behavior that emerges from the underlying quantum amplitudes in the same way as for any other macroscopic object, and that classical behavior includes a horizon.

So I'm sticking with what I said earlier in this thread: it looks like the mainstream physics view is that quantum corrections will remove the singularity, but not the event horizon.

 

[photonkid]

So I'm sticking with what I said earlier in this thread: it looks like the mainstream physics view is that quantum corrections will remove the singularity, but not the event horizon.

ok, well having re-read most of this entire thread including what you said here it's hard for me to understand how an outside observer will see a falling object frozen indefinitely at the edge of the event horizon yet the object does actually reach the event horizon. It seems that this is not analogous to trying to accelerate an object of non zero mass to the speed of light. It also seems that the evidence that black holes do actually exist is strong and mainstream physics view is that event horizons are most likely real.

I suggest you don't try to explain any further but do you know if there are any authoritative articles or books that explain what happens when a star collapses e.g. that the event horizon starts out very small and gets gradually bigger or whatever happens, and why an object is frozen indefinitely at the edge of the event horizon from an outside observer's view?

Anyway, thanks for the considerable time you've spent posting in this thread, and everyone elses.

 

[PAllen]

An authoritative description of collapse including the evolution of the event horizon is the following:

http://www.aei.mpg.de/~rezzolla/lnotes/mondragone/collapse.pdf

As they discuss, the event horizon grows from the center of collapsing body, and stop growing as all the mass is encompassed.

As for what a distant observer sees for, e.g., a rod falling through a horizon, see if this helps:

- light emitted at one millimeter (say) above the horizon is received eventually as long microwaves.
- light emitted at 1/2 millimeter is received as radio waves.
- light emitted at .1 millimeter above the horizon is so long wave that no plausible detector can detect it (even ignoring being swamped by CMP radiation).

So, what you actually see is the rod getting redder and redder (including radio waves in this), finally the front disappears at radius of last detectability. Then further up the rod disappears as it reached this radius; etc. until the whole rod has disappeared. This process will actually happen relatively fast for the distant observer, despite the extreme time dilation of the near horizon region relative to the distant region. Visually, with our ideal detector, it looks for all the world like the rod has progressively vanished into a black hole in space.

 

[photonkid]

As for what a distant observer sees for, e.g., a rod falling through a horizon, see if this helps:

Thanks. It helped from the point of view that now I know what "pancaked" means. Also it helped that I no longer think the in-falling object remains visible to an outside observer indefinitely, but... at first glance my brain is still seeing a paradox because
1. it sounds like the wavelength coming from the in-falling observer approaches infinity and that to an outside observer, the wave never stops coming
and
2. the closer the in-falling object gets to the event horizon, the slower time goes, so he never quite makes it - just like trying to reach the speed of light

But for 1, it probably does stop coming because after a certain time, you have to wait an infinite amount of time to see any more "oscillations in the wave". For 2, it's impossible for me to come up with any kind of numbers that show no matter how close you get you never quite make it because the external observer can't see what's happening, whereas for accelerating to the speed of light, the object being accelerated can do the measuring and no external observer is needed.

So if it's a paradox, it's a very complicated paradox and I can't actually tell if there's a paradox or not...

Anyway, I've been wondering why this isn't in an FAQ somewhere and then today I found that it's here
http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/fall_in.html


Is there any possibility that when a star collapses, that time slows down so much that the density of the mass never goes past a certain limit and that even though the star keeps on contracting, the contraction rate gets smaller and smaller? Does the maths show this can't happen - or is it because this would look like a neutron star and black hole candidates don't look like neutron stars? (A short answer will do).

 

[PeterDonis]

Is there any possibility that when a star collapses, that time slows down so much that the density of the mass never goes past a certain limit and that even though the star keeps on contracting, the contraction rate gets smaller and smaller?

No.

Does the maths show this can't happen

Yes. This case is treated in all of the major relativity textbooks.

or is it because this would look like a neutron star and black hole candidates don't look like neutron stars?

Sort of. There is a theorem (originally due to Einstein) that says that an object in a static equilibrium, like a neutron star (or a white dwarf or anything else that holds itself up statically against its own gravity) can't have a radius smaller than 9/8 of the Schwarzschild radius for its mass. The time dilation factor at that radius is only 3 (i.e., time at the surface of such an object "flows" 1/3 as fast as it does at infinity); it certainly isn't approaching infinity.

One of the key things about black hole candidates is that at least some of them appear to be confined within a radius smaller than the above limit. (At least, that's my understanding.) So whatever is in there, it can't be something in static equilibrium like a neutron star.

 

[A.T.]

There is a theorem (originally due to Einstein) that says that an object in a static equilibrium, like a neutron star (or a white dwarf or anything else that holds itself up statically against its own gravity) can't have a radius smaller than 9/8 of the Schwarzschild radius for its mass. The time dilation factor at that radius is only 3 (i.e., time at the surface of such an object "flows" 1/3 as fast as it does at infinity); it certainly isn't approaching infinity.

What about the time dilation at the center of such an object?

 

[PeterDonis]

What about the time dilation at the center of such an object?

It depends on the object's internal structure, but it will in general be somewhat larger than at the surface. It can't go to infinity because the gradient of the time dilation factor gets smaller as you go inward from the surface, until it becomes zero at the center.

 

[photonkid]

Sort of. There is a theorem (originally due to Einstein) that says that an object in a static equilibrium, like a neutron star (or a white dwarf or anything else that holds itself up statically against its own gravity) can't have a radius smaller than 9/8 of the Schwarzschild radius for its mass. The time dilation factor at that radius is only 3 (i.e., time at the surface of such an object "flows" 1/3 as fast as it does at infinity); it certainly isn't approaching infinity.

So when we hear about the age of the universe being 13 billion years, is that 13 billion years of "earth proper time"? Since planet Earth hasn't been around all that time, where does "13 billion years" come from - is it 13 billion years of "zero gravity/ zero velocity" time?

13 billion years is presumably long enough for a black hole to form. Is it possible to calculate how long it would take (in Earth years) for an infalling object to fall the last one meter before it reaches the event horizon and if so, is this likely to be longer than 13 billion "earth years". Hopefully you can see what I'm trying to get at.

 

[PeterDonis]

So when we hear about the age of the universe being 13 billion years, is that 13 billion years of "earth proper time"?

Not really. It's the proper time that would be elapsed since the Big Bang for an observer whose current spatial location is Earth, but who has always seen the universe as homogeneous and isotropic. Such observers are called "comoving" observers. We don't see the universe as isotropic on Earth: we see a dipole anisotropy in the CMBR, for example, indicating that we are not "comoving" observers, even when the effects of the Earth's rotation and orbit about the Sun are corrected for.

13 billion years is presumably long enough for a black hole to form.

Way more than enough, yes.

Is it possible to calculate how long it would take (in Earth years) for an infalling object to fall the last one meter before it reaches the event horizon

Yes.

is this likely to be longer than 13 billion "earth years".

It's a lot shorter, even for the largest black holes that we think are likely to exist in the universe (the current estimate, I believe, is black holes of billions of solar masses at the centers of quasars). For a black hole of that size, the time for an infalling object to fall the last one meter to the horizon is much less than one Earth year. Even the time to fall from an astronomically significant distance, such as a million light-years (i.e., from well outside the quasar the black hole is at the center of), is less than 13 billion years.

Note that the "time" I'm referring to here is the proper time experienced by the infalling object, i.e., the time elapsed on that object's clock.

 

[A.T.]

It depends on the object's internal structure, but it will in general be somewhat larger than at the surface. It can't go to infinity because the gradient of the time dilation factor gets smaller as you go inward from the surface, until it becomes zero at the center.

In this previous thread:
https://www.physicsforums.com/showthread.php?p=1840160#post1840160
there were claims that proper-time at the center would become negative for R < 9/8 Rs. If that's correct, wouldn't it approach zero when R approaches 9/8 Rs from above?ETA: See also:
https://www.physicsforums.com/showthread.php?p=2430580#post2430580

Solutions of the Oppenheimer--Volkoff Equations Inside 9/8ths of the Schwarzschild Radius
http://deepblue.lib.umich.edu/bitstream/handle/2027.42/41992/220-184-3-597_71840597.pdf

We refine the Buchdahl 9/8ths stability theorem for stars by describing quantitatively the behavior of solutions to the Oppenheimer–Volkoff equations when the star surface lies inside 9/8ths of the Schwarzschild radius. For such solutions we prove that the density and pressure always have smooth profiles that decrease to zero as the radius r→ 0, and this implies that the gravitational field becomes repulsive near r= 0 whenever the star surface lies within 9/8ths of its Schwarzschild radius.

 

[PeterDonis]

In this previous thread:
https://www.physicsforums.com/showthread.php?p=1840160#post1840160
there were claims that proper-time at the center would become negative for R < 9/8 Rs. If that's correct, wouldn't it approach zero when R approaches 9/8 Rs from above?

ETA: See also:
https://www.physicsforums.com/showthread.php?p=2430580#post2430580

A brief comment: the time dilation factor given in those threads is derived from a highly idealized solution for an object with constant density, not a general solution for a static object. It has some heuristic value, but I don't think any physicist considers it physically realistic.

That said, yes, one way of stating the theorem I stated is that, if an object could be static with a radius of 9/8 of its Schwarzschild radius, the time dilation factor at its center would go to infinity (or zero, depending on how you define it). However, if you compute the proper acceleration at the center, you will see that it diverges, indicating that it's impossible for a timelike object to actually follow such a worldline. In other words, the curve R = 0 in such a solution would actually be null, not timelike. And if you do the math for a solution with R < 9/8 of the Schwarzschild radius, the curve R = 0 is spacelike (and there is a null surface at some R > 0)--to say that "time flows backwards" is a serious misstatement, since that would imply that the R = 0 curve is timelike when it's actually spacelike. So the R = 9/8 Schwarzschild radius solution is a limiting case, not an actual physically possible solution, and solutions with R < 9/8 Schwarzschild radius are not physically possible either.

What I just said does depend on some assumptions, one or more of which are violated in the solutions considered in this paper:

Solutions of the Oppenheimer--Volkoff Equations Inside 9/8ths of the Schwarzschild Radius
http://deepblue.lib.umich.edu/bitstream/handle/2027.42/41992/220-184-3-597_71840597.pdf

The paper describes the assumptions which lead to the conclusions I gave above, and what happens if they are violated. (One particular thing that makes the violations seem physically unrealistic to me: the solutions are all singular at r = 0, or, to put it another way, they require a point source at r = 0 with negative mass.) One particular thing to note: *none* of the violations lead to an infinite (or zero) time dilation factor anywhere inside the object. So even if you consider some of the solutions in the paper to be physically possible, what I said in post #68 still holds.

 

[Opti_mus]

Hi guys!

I've read all that it's said in this post, as I'm interested in clarifying a question, but I haven't been able to do it reading all the answers:

Suppose an observer A falling into a Schwarzschild's BH. It's clear he'll pass through the Horizon in a proper FINITE time. Suppose an static observer B, situated in the same vertical as A and over A, with a FINITE distance over the BH. It's clear for me too that B cannot see A passing through the Horizon, as the redshift goes to infinity. Despite that fact, I've a question I wouldn't been able to find an answer for: does A pass through the Horizon in a FINITE B's proper time or not? I'm not talking about A reaching the Horizon, I'm talking about A passing the Horizon.

Thank you, and sorry for my poor english!

 

[Nugatory]

I've a question I wouldn't been able to find an answer for: does A pass through the Horizon in a FINITE B's proper time or not? I'm not talking about A reaching the Horizon, I'm talking about A passing the Horizon.

You haven't been able to find an answer because there isn't any answer to find, both for reaching and for passing through. I'm going to try restating the situation a bit more precisely, so the problem with the question will be a bit clearer:

A and B start at the same place above the event horizon; B is somehow hovering there while A falls into the black hole. So their world lines are touching at the starting point but diverge as A and B separate, with A's world line passing through the event horizon and ending at the central singularity.

Now, when we say that some amount of an observer's proper time has passed between two events, we're talking about the time along that observer's world line between the two events. It's only meaningful if the observer's world line actually passes through the events. Thus, it's easy for us to know the proper time for A to reach the event horizon: He looks at his wristwatch at the starting point, he looks at his wristwatch again as he reaches the event horizon, the difference between the two readings is the proper time that passed between leaving B and reaching the horizon.

But what two events does B use to calculate a proper time for him? Clearly he looks at his wristwatch at the starting point, just like A; but when should he look at it again to get a second reading so that he can find the elapsed proper time between the two readings?

The problem here is that there is no point on B's world line for which we can say "this and only this is when A reached the horizon". Therefore the thing that you're asking about, the amount of B's proper time elapsed for A to reach the horizon, is undefined and your question has no good answer.

 

[A.T.]

So the R = 9/8 Schwarzschild radius solution is a limiting case, not an actual physically possible solution, and solutions with R < 9/8 Schwarzschild radius are not physically possible either.

Thanks for your reply Peter. Since a static sphere is not possible, I wonder what the collapse would look like from the center (assuming you can look through the infalling matter). How would an observer at the center see the universe when the horizon forms?

 

[photonkid]

The problem here is that there is no point on B's world line for which we can say "this and only this is when A reached the horizon". Therefore the thing that you're asking about, the amount of B's proper time elapsed for A to reach the horizon, is undefined and your question has no good answer.

ok, but in this post PeterDonis said that it's possible to calculate how long (in Earth years) it will take to fall the last one meter - so the answer to
<quote> does A pass through the Horizon in a FINITE B's proper time or not? </>

would be yes wouldn't it?

Anyhow, I don't think it makes sense to distinguish between "reaching" and "passing through" because what happens after something reaches the event horizon is just speculation.

 

[PeterDonis]

How would an observer at the center see the universe when the horizon forms?

In a spherically symmetric collapse, such as that originally modeled in the 1939 Oppenheimer-Snyder paper, the event horizon forms at the center, r = 0, and moves outward at the speed of light. (As it moves outward, the light cones get tilted inward more and more by the increasing density of the collapsing matter, so the horizon moves outward more slowly, until the horizon becomes fixed at r = 2M just as the surface of the collapsing matter reaches r = 2M and intersects the horizon.)

An observer at the center would see the density around him increasing, and would see (I believe--I haven't run a computation to check) an increasing blueshift of light coming to him from the rest of the universe; but both the density and the blueshift would still be finite (and not necessarily very large) at the moment when the horizon forms at r = 0. The density and blueshift at r = 0 would only diverge to infinity later, when the surface of the collapsing matter reaches r = 0 and the singularity forms there.

 

[PeterDonis]

so the answer to
<quote> does A pass through the Horizon in a FINITE B's proper time or not? </>

would be yes wouldn't it?

It depends on how you match up events on B's worldline with events on A's worldline, since A and B are spatially separated. In a curved spacetime there is no unique way to do that in general; the only case in which there is a unique way to do it is if both worldlines are static. B's worldline is static (he stays at the same height above the horizon forever), but A's is not; so there is no unique answer to the question of when, by B's clock, A reaches the horizon.

 

[Nugatory]

ok, but in this post PeterDonis said that it's possible to calculate how long (in Earth years) it will take to fall the last one meter - so the answer to
<quote> does A pass through the Horizon in a FINITE B's proper time or not? </>

would be yes wouldn't it?

That's still A's proper time that we're talking about, even if we're choosing to measure it in years.

 

[Opti_mus]

You haven't been able to find an answer because there isn't any answer to find, both for reaching and for passing through. I'm going to try restating the situation a bit more precisely, so the problem with the question will be a bit clearer:

A and B start at the same place above the event horizon; B is somehow hovering there while A falls into the black hole. So their world lines are touching at the starting point but diverge as A and B separate, with A's world line passing through the event horizon and ending at the central singularity.

Now, when we say that some amount of an observer's proper time has passed between two events, we're talking about the time along that observer's world line between the two events. It's only meaningful if the observer's world line actually passes through the events. Thus, it's easy for us to know the proper time for A to reach the event horizon: He looks at his wristwatch at the starting point, he looks at his wristwatch again as he reaches the event horizon, the difference between the two readings is the proper time that passed between leaving B and reaching the horizon.

But what two events does B use to calculate a proper time for him? Clearly he looks at his wristwatch at the starting point, just like A; but when should he look at it again to get a second reading so that he can find the elapsed proper time between the two readings?

The problem here is that there is no point on B's world line for which we can say "this and only this is when A reached the horizon". Therefore the thing that you're asking about, the amount of B's proper time elapsed for A to reach the horizon, is undefined and your question has no good answer.

Thanks for your answer and sorry, 'cause it's totally impossible for me to understand that facts. If I (B in this case) can't define a proper time between 2 facts, even though I can't see one of them, it seems physics are over, in the sense that determinism is over: I can't know by sure if A falls or not into the BH, so I can't make any prediction about the dinamics of A and the BH!

 

[Nugatory]

Thanks for your answer and sorry, 'cause it's totally impossible for me to understand that facts. If I (B in this case) can't define a proper time between 2 facts, even though I can't see one of them, it seems physics are over, in the sense that determinism is over: I can't know by sure if A falls or not into the BH, so I can't make any prediction about the dynamics of A and the BH!

You are confusing yourself by jumping straight into a fairly subtle general relativity problem before understanding how the concepts of proper time and coordinate time work in the much simpler (no gravitational effects, no black holes, no curvature, inertial frames cover the entire universe) case of special relativity.

I'm not sure how far back to start explaining, but I'll try...

There are two ways in which we can say that two things happened at the same time:
1) They happened at the same place. If two cars collide at a street corner we know that they both were at the street corner at the same time - that's why there was a collision.
2) They happen at different places, but they both have the same time coordinate. For example this event happened at 12:37 PM; so did that one; so they happened at the same time.

Now, have you seen Einstein's thought experiment with the lightning flashes at each end of the train, the one that demonstrates the relativity of simultaneity? That shows that there is a fundamental difference between #1 and #2 above. All observers, regardless of their state of motion, will agree that in #1 both cars were at the street corner at the same time - either there's a crash or there isn't, but it can't be that some observers see the cars destroyed in a collision while others see them slipping safely past. However, as Einstein's train experiment shows, not all observers will agree about #2 - the two events at two different locations are simultaneous using one observer's coordinates but not simultaneous using the coordinates of another observer moving with reference to the first.

Study this thought experiment until you understand it thoroughly and the paragraph above makes sense. Note that physics is by no means over: determinism is intact, there's no uncertainty about whether the cars did collide, whether Einstein's lightning flashes did strike, and what time the various observers saw the various events happen. That last is different for the different observers, but in a perfectly understandable, predictable, and deterministic way.

OK, still with me? Then we can talk about "coordinate time", which is what people mean when they say "time" without further specifying what they mean. It's what we're using any time that we say something happens at a particular time and place. You've been using coordinate time all your life; it's the only kind we ever deal with day-to-day. And because of relativity of simultaneity, different observers use different time coordinates so have a different notion of time; but at least they can talk about events that are separated in space (the #2 case above).

We also have "proper time", which has a very specific and more restricted meaning, which I gave in my post earlier:

When we say that some amount of an observer's proper time has passed between two events, we're talking about the time along that observer's world line between the two events. It's only meaningful if the observer's world line actually passes through the events. Thus, it's easy for us to know the proper time for A to reach the event horizon: He looks at his wristwatch at the starting point, he looks at his wristwatch again as he reaches the event horizon, the difference between the two readings is the proper time that passed between leaving B and reaching the horizon.

The problem with asking how much of B's proper time passes while A falls to the event horizon (your original question) is that A reaching the horizon and B looking at his watch to see what it reads are happening at different places. So if we're going to say that B looks at his watch "when/at the same time" that A reaches the event horizon, we have to use the #2 definition of "at the same time"; the event of B looking at his watch must have the same time coordinate as the event of A reaching the horizon.

And what time coordinates are we going to use to make that determination? Because of the curvature of spacetime between A and B and the impossibility of exchanging light signals between them, there's no good answer to that question.

 

[Opti_mus]

Bufff

Well, first of all, thanks for your very extense reply.

I understand you but, at the same time, it's very difficult for me to handle something like the impossibility of B to say anything about A: he can't even say if A finally falls or not into the BH! This is very difficult to assimilate for me...

 

[Passionflower]

Bufff

Well, first of all, thanks for your very extense reply.

I understand you but, at the same time, it's very difficult for me to handle something like the impossibility of B to say anything about A: he can't even say if A finally falls or not into the BH! This is very difficult to assimilate for me...

It is really not that hard:

As I hope you would see there is no way for light or anything else from the event horizon to go to observer B, then how would observer B empirically verify the time something happens on the event horizon on his clock?

Even for something extremely close to the event horizon light takes 'ages' to return to observer B, of course for B it seems it takes forever but that is simply because the information takes a long time on his watch to return while locally it already zoomed by the event horizon a very long time ago.

 

[Nugatory]

he can't even say if A finally falls or not into the BH

But he can say that A falls into the black hole... It's easy.

He calculates A's proper time to reach the black hole - this describes what A experiences - he gets a finite value, and knows from that finite value that A will experience falling into the black hole. True, B doesn't actually see it happen, but we calculate things that we don't see all the time.

 

[Opti_mus]

But he can say that A falls into the black hole... It's easy.

He calculates A's proper time to reach the black hole - this describes what A experiences - he gets a finite value, and knows from that finite value that A will experience falling into the black hole. True, B doesn't actually see it happen, but we calculate things that we don't see all the time.

That's true, but insatisfactory for B, isn't it? What is interesting for B is to know what time takes an object to enter with the time of B, and he can't calculate it!

 

[Opti_mus]

And one last question guys (and thanks for all your answers): How can you council all that facts you told to me with what Kip Thorne tells in "BH and Time Warps", referenced to an observer (Arnold) who is falling into a BH as seen by another observer? (the translation is mine, I'm sure the words are other in the original version, but I'll try to do my best):

Does it means Arnold hasn't cross yet the horizon and that he'll never do it? Not at all. Those last signals which are duplicating constantly need an infinite amount of time to escape of the Hole's gravitatory power. Arnold crossed the horizon, moving with the speed of light, many minutes ago. It keeps arriving weak signals simply due to the fact that they've been traveling a lot of time. They're relics of the past

 

[Nugatory]

And one last question guys (and thanks for all your answers): How can you council all that facts you told to me with what Kip Thorne tells in "BH and Time Warps"

That sounds pretty much like what a bunch of people have been saying here: A crosses the horizon; B never sees (receives light from) the crossing event but that doesn't mean it doesn't happen.

 

[photonkid]

ok, but in this post PeterDonis said that it's possible to calculate how long (in Earth years) it will take to fall the last one meter - so the answer to
<quote> does A pass through the Horizon in a FINITE B's proper time or not? </>

would be yes wouldn't it?

That's still A's proper time that we're talking about, even if we're choosing to measure it in years.

yikes, I'm not sure I caught on to that even though PeterDonis said it. I can't get my head round this.

Well what about this then. If I drop a stone from my hand to the ground, I can calculate within some small error margin, what the time on my clock will be when the stone hits the ground. The fact that time is going slightly slower at ground level doesn't stop me from calculating what the time on *my* clock will be when the stone hits the ground. So if I drop a stone from one meter above an event horizon, why can we not calculate what the time on my clock will be when the stone reaches the event horizon? What's so special about the event horizon?

 

[PeterDonis]

If I drop a stone from my hand to the ground, I can calculate within some small error margin, what the time on my clock will be when the stone hits the ground.

But your clock is spatially separated from the ground. So the very concept of "the time on my clock when the stone hits the ground" assumes that you have specified a way of assigning times on your clock to events that are spatially separated from you.

In general, as I've said before, there is no unique way to do that; there are many different possible ways, and none of them are picked out by any feature of the spacetime, so none of them have any physical meaning. However, in the case of your hand and the ground, assuming that your hand is at rest with respect to the ground, there does happen to be a way to do it that is picked out by a feature of the spacetime and does have physical meaning. The feature is that the spacetime has a time translation symmetry--i.e., for observers following one of a particular family of curves, the metric looks the same at every moment of time. Your hand and the ground happen to be following such curves. And observers following such curves can set up a notion of simultaneity (i.e., of events happening "at the same time") that they all can share, and which you can use to assign a time on your clock to the event of the stone hitting the ground.

However, this method breaks down at the horizon, because at the horizon, that symmetry of the spacetime is no longer a time translation symmetry. The special family of observers are the ones following worldlines of constant r (and constant theta, phi as well, but we are leaving those coordinates out of the analysis here by assuming purely radial motion). But in order for observers to follow such worldlines, they have to be timelike; and the horizon itself is a curve of constant r which is null, not timelike. (And inside the horizon, curves of constant r are spacelike.) In other words, the metric still looks the same along curves of constant r <= 2M (i.e., at or inside the horizon), but those curves aren't timelike, so observers can't follow them, and can't use them to set up a notion of simultaneity. That means that, even if you're hovering just one meter above the horizon, and you drop a stone, there is no way to assign a time on your clock to the event of the stone hitting the horizon.

 

[Dale]

That's true, but insatisfactory for B, isn't it? What is interesting for B is to know what time takes an object to enter with the time of B, and he can't calculate it!

what B considers satisfactory is a matter of personal philosophy and preference, not physics. B can calculate the physics just fine.

 

[Jonathan Scott]

However, this method breaks down at the horizon, because at the horizon, that symmetry of the spacetime is no longer a time translation symmetry.

I don't understand why what happens "at the horizon" is relevant to the problem.

At any finite distance from the horizon, in a static metric relative to the source, the normal relativity method of assigning simultaneous times by the half-way point of a light-speed signal gives an unambiguous result which matches the coordinate time. It is symmetrical and there seems to be no reason to assume that it would be biased in any special way. This time coordinate however goes to infinity as the object gets closer to the horizon.

This seems to imply that although we don't have a way to describe the event of crossing the horizon in the static coordinate system, we can at least say that it definitely occurs AFTER our coordinate time becomes infinite.

It is also theoretically possible (with an appropriate hypothetical method of propulsion such as a relativistic bungee cord) for the falling object to reverse course and return along a path with similar velocity to its original path, in which case it can return to its starting point, and that too can follow a symmetric path and hence show that the static time coordinate is a physically valid way to describe its path from the static point of view.

I am therefore quite puzzled that people seems prepared to assert that something which clearly does not happen from a normal point of view until after an infinite time has elapsed - not even the end of the universe, but an INFINITE time - "actually happens".

This seems to violate standard physical assumptions about sequence and causality so strongly that I can't see how the maths could be considered to "prove" anything.

As far as I'm concerned, it seems perfectly physical reasonable for the maths to show a continuous solution from the point of view of the falling observer, but for it to be impossible for that solution to be reached because the falling observer's time rate relative to the normal universe approaches zero, as in my "stasis box" example. I consider that any statement that the crossing "actually happens" is not necessarily disproved by this model, but rather that such an extraordinary idea requires extraordinary evidence, and the continuity of the maths doesn't constitute such evidence.

 

[Nugatory]

If I drop a stone from my hand to the ground, I can calculate within some small error margin, what the time on my clock will be when the stone hits the ground.

No, actually, you cannot. What you know is the time on your watch when the light from the stone hitting the ground reaches your eyes. You can then subtract the light travel time between the ground and your eyes and say that that is the time on your watch when the stone hit the ground: "My wristwatch read exactly 12:00 noon when the stone hit the ground; I know this because the light hit my eyes at 6 nanoseconds past noon and my eyes are 180 centimeters off the ground". But notice that you've slipped a claim about simultaneity in there: you're saying that the stone hitting the ground and the watch reading 12:00 noon are simultaneous. An observer moving relative to you will not see these two events as simultaneous; so you have not actually calculated "when" the stone hits the ground in any absolute sense.

I agree that you what have calculated is when the stone hits the ground using the only definition of simultaneity that makes any sense to you: If at time T you observe something happening at distance X away from you, you say that it happened at time T-(X/c). It would be totally perverse to use any other definition of simultaneity. So...

What's so special about the event horizon?

The method I described above, the one for which there is no non-perverse alternative, doesn't work with an event horizon.

 

[Passionflower]

At any finite distance from the horizon, in a static metric relative to the source, the normal relativity method of assigning simultaneous times by the half-way point of a light-speed signal gives an unambiguous result which matches the coordinate time. It is symmetrical and there seems to be no reason to assume that it would be biased in any special way. This time coordinate however goes to infinity as the object gets closer to the horizon.

This seems to imply that although we don't have a way to describe the event of crossing the horizon in the static coordinate system, we can at least say that it definitely occurs AFTER our coordinate time becomes infinite.

I disagree with that, the term 'after' here does not make any sense.

The fact that it takes an (near) infinite time for light to reach the outside observer informing him that a probe crossed the event horizon does not contradict the fact that the probe simply crosses the event horizon more or less as a non-event. Why would you want to call it 'after' for light from inside the event horizon to reach the outside observer as this light would never be able to reach him?

By the way the internal Schwarzschild solution is not static, unlike the external solution.

I am therefore quite puzzled that people seems prepared to assert that something which clearly does not happen from a normal point of view until after an infinite time has elapsed - not even the end of the universe, but an INFINITE time - "actually happens".

It is simply because it takes light an (near) infinite time to reach the observer, he simply does not see it or at least very late. Think of sound when the wind goes in opposite direction, it would take longer to reach you, taken that to the extreme would you claim someone did not shout 'hello' because the wind blew the sound waves the wrong way and they never reached your ears?

 

[Jonathan Scott]

...

It is simply because it takes light an (near) infinite time to reach the observer, he simply does not see it or at least very late. Think of sound when the wind goes in opposite direction, it would take longer to reach you, taken that to the extreme would you claim someone did not shout 'hello' because the wind blew the sound waves the wrong way and they never reached your ears?

The problem with that analogy is that, as I mentioned earlier, you can bounce light off a mirror at a lower potential and the situation is completely symmetrical with respect to time.

 

[Passionflower]

The problem with that analogy is that, as I mentioned earlier, you can bounce light off a mirror at a lower potential and the situation is completely symmetrical with respect to time.

It is not.

If you take two static observers A and B at different r-coordinate values then light going from ABA will show a different elapsed time then going from BAB.

 

[Dale]

This seems to imply that although we don't have a way to describe the event of crossing the horizon in the static coordinate system, we can at least say that it definitely occurs AFTER our coordinate time becomes infinite.

Yes, there are coordinate systems which can assign a definite coordinate time to the crossing event. In all such coordinate systems that I am aware of the event crossing does occur after infinite time according to the Schwarzschild coordinates.

In the same way, in flat spacetime the inertial coordinates claim that a horizon crossing occurs after an infinite amount of Rindler time.

Note, as Passionflower noted you cannot use either Schwarzshild coordinates nor Rindler coordinates to make such a statement, you have to use a different coordinate chart, one which actually covers the event horizon.

I am therefore quite puzzled that people seems prepared to assert that something which clearly does not happen from a normal point of view until after an infinite time has elapsed - not even the end of the universe, but an INFINITE time - "actually happens".

Again, by that same logic anything in flat spacetime occurs after some Rindler horizon and therefore occurs after an INFINITE time by some observer's radar coordinates. Your logic would therefore say that no event can ever be considered to have "actually happened", even if we ourselves observed it, simply because some other observer might not be able to assign a finite time to it using radar coordinates.

 

[A.T.]

It is not.

If you take two static observers A and B at different r-coordinate values then light going from ABA will show a different elapsed time then going from BAB.

You mean the time elapsed for sender and final receiver?

 

[Jonathan Scott]

If you take two static observers A and B at different r-coordinate values then light going from ABA will show a different elapsed time then going from BAB.

What do you mean here?

If you mean that observers at different potentials see different proper times for the round trip from their point of view, that's perfectly normal, as that depends on the time dilation at the observer potential. However, the coordinate times for the light times for the trip AB and BA are equal as plotted by any observer, regardless of which end they are on.

Anyway, that seems irrelevant here. What I'm pointing out is that from the point of view of ANY static observer, the assumption that the turnaround point occurs at a time which is "simultaneous" with the half way time at the start and end location (as in SR) gives a consistent definition of "simultaneous" for all static observers at all potentials. This means that any static observer can naturally extend their time coordinate in the usual way down to any point arbitrarily close to the horizon.

The argument about a Rindler horizon seems unhelpful, as that is a theoretical concept which also involves infinities.

 

[Passionflower]

What I'm pointing out is that from the point of view of ANY static observer, the assumption that the turnaround point occurs at a time which is "simultaneous" with the half way time at the start and end location (as in SR) gives a consistent definition of "simultaneous" for all static observers at all potentials.

You can make that assumption if you pick the appropriate coordinates but that does not say anything about the physics.

Do you really think it is a natural view to assume it takes light just as much time to go down into a gravitational well as it takes to come back out? Remember the speed of light is always c locally, once you look at the speed of light over an extended curved region you can show just about anything with the appropriate coordinate chart.

 

[PAllen]

Would you consider the simultaneity of a thrown baseball compared to your own unphysical, or implausible? Then consider a static observer choosing to consider, moment to moment, the simultaneity of fastball thrown toward the BH. Whether you use Fermi-Normal simultaneity for the baseball at that moment, or computed radar (assuming extension of inertial trajectory of baseball the moment it leaved your hand, extended to the distant past and future) simultaneity, you would find by this slightly different point of view:

- at every point of the static world line, there are simultaneous events at and inside the horizon
- in the case of SC geometry, even the pure vacuum case, there is an earliest moment on your world line where thrown baseball simultaneity reaches the singularity (before this, it (Normal simultaneity) goes inside the horizon to some minimum r>0 then r grows (on the way to alternate KS sheet we are not considering here). Thus you have a sense of 'when' the singularity formed.
- all later history of static world line, your baseball 'now', includes events all the way to the singularity.

Whether one believes such predictions of classical GR apply to our universe, the above is unambiguously true of classical GR.