Is time present in a black hole?

[Sorthal]

Is there time inside of a black hole?

 

[adjacent]

Actually, if you could survive a trip into a black hole (which you cannot) you would not be aware of any slowing down of any clock you carried as you fell in. However, if you could compare the speed of your clock with that of a reference clock kept far away, then then the clock falling into the black hole would appear to slow down relative to the clock far from the hole. The place where the falling in clock would appear to 'stop' is the spherical surface called the event horizon. This is also regarded as the boundary of the black hole, since nothing crossing this surface can escape.

 

[bapowell]

Notions of time and space reverse inside the black hole. Observers inside the black hole move through space towards the singularity as inevitably as observers outside the black hole move through time towards the future.

 

[DrGreg]

Notions of time and space reverse inside the black hole. Observers inside the black hole move through space towards the singularity as inevitably as observers outside the black hole move through time towards the future.

That's not really true. If you use Schwarzschild coordinates then inside the event horizon the time coordinate is called "r" and the space coordinate is called "t", but there are other coordinate systems (e.g. Kruskal coordinates) where there is no such "reversal".

Observers inside the black hole move through time towards the singularity (even though in Schwarzschild coordinates the time is labelled "r".)

 

[dauto]

That's not really true. If you use Schwarzschild coordinates then inside the event horizon the time coordinate is called "r" and the space coordinate is called "t", but there are other coordinate systems (e.g. Kruskal coordinates) where there is no such "reversal".

Observers inside the black hole move through time towards the singularity (even though in Schwarzschild coordinates the time is labelled "r".)

One important point is that the singularity is space-like

 

[ABunyip]

Observers inside the black hole move through time towards the singularity

...but never reach it (from the point of view of observers outside the BH). Once an object reaches the EH it continues to accelerate into the (our) future, but observed from outside, time starts slowing down until it pretty well stops altogether for objects at the singularity. Eventually everything that entered the BH will be evaporated out as Hawking Radiation zillions of years later. When the BH is so depleted that it finally allows light to escape it will most likely, IMHO, rapidly release the last remnants of its energy in a huge protracted pulse of radiation.

So the answer is yes, and no.

 

[bapowell]

Almost: to an observer outside, time dilation tends to infinity at the event horizon. So to outside observers, objects falling towards the black hole appear to come to rest asymptotically at the event horizon. The inside of the black hole is of course invisible to outside observers.

It's not clear that things that fall into the black hole are reconstituted as Hawking radiation -- this is the founding problem of the information paradox.

 

[ABunyip]

I think of the EH as an optical effect - like a rainbow. Its just the point at which everything beyond it becomes invisible. As objects approach the EH they are already being thrust into the future and will fade away as they go through it. As they get closer to the singularity over time, they are accelerated more and more into the future and less and less toward the singularity. So I beg to differ on this small point, bapowell (pun intended).

 

[bapowell]

This is physics, so it's really not a matter of opinion. It can be demonstrated objectively that objects appear to freeze at the event horizon to observers sufficiently far outside the black hole. If you beg to differ, please substantiate your claim formally.

 

[PeterDonis]

As they get closer to the singularity over time, they are accelerated more and more into the future and less and less toward the singularity.

The singularity is in the future, so "accelerating more into the future", to the extent that makes sense, is also accelerating more toward the singularity.

 

[PeterDonis]

Once an object reaches the EH it continues to accelerate into the (our) future, but observed from outside, time starts slowing down until it pretty well stops altogether for objects at the singularity.

You're confusing the singularity with the horizon. Observers far away from the hole see light from infalling objects more and more redshifted (which can also be interpreted as "time slowing down" for those objects) as the objects approach the EH. No light from the EH or anywhere inside it can ever reach any observer outside the EH; so to observers outside the EH, an object disappears when it reaches the EH, and its entire future trajectory inside the EH is invisible to observers outside the EH.

Eventually everything that entered the BH will be evaporated out as Hawking Radiation zillions of years later.

We don't actually know this; as bapowell pointed out, whether or not things that enter the hole get evaporated out is one of the key questions involved in the BH information paradox, which has not been fully resolved.

 

[pervect]

This is physics, so it's really not a matter of opinion. It can be demonstrated objectively that objects appear to freeze at the event horizon to observers sufficiently far outside the black hole. If you beg to differ, please substantiate your claim formally.

I beg to differ in that at minimum the statement needs further qualification. Specifically, it is true that objects appear to freeze at the event horizon for static observers using the conventional notion of simultaneity outside the black hole. In essence, no radar signal could return to a static observer from the event horizon.

It is also true that objects falling into a black hole take a finite proper time to reach the event horizion (and even beyond, to the singularity). This proper time could be measured, for instance, by an atomic clock they carry with them. If you need a reference for this, I can provide one, but most any GR textbook will mention this. The reference I'd provide would be looking up the specific page in MTW's Gravitation that tells you this, which may or may not be useful, in that you can find it easily yourself if you already have or can borrow the textbook, and if you don't you have to take my word for it anyway.)

Non-static observers who are themselves falling into a black hole will not see the freeze effect. Basically, when they reach the event horizon, they'll get their radar return signal.

 

[ABunyip]

This is physics, so it's really not a matter of opinion.
<SNIP>
If you beg to differ, please substantiate your claim formally.

What?!?

 

[skeptic2]

It can be demonstrated objectively that objects appear to freeze at the event horizon to observers sufficiently far outside the black hole.

I beg to differ in that at minimum the statement needs further qualification. Specifically, it is true that objects appear to freeze at the event horizon for static observers using the conventional notion of simultaneity outside the black hole.

Both statements need further qualification. To observers sufficiently far outside the black hole, objects do freeze, not just appear to freeze, at the EH and every experiment that could be done from that distance would support that observation.

The notion that objects only appear to freeze comes from the assumption that the viewpoint of the infalling observer is somehow more legitimate than that of a distant observer. Both observations are equally valid. One cannot pick one observer, e.g. the infalling one, and claim that his perspective is reality and what the distant observer sees and measures is not.

 

[ABunyip]

Is there time inside of a black hole?

Yes.

 

[PAllen]

This is physics, so it's really not a matter of opinion. It can be demonstrated objectively that objects appear to freeze at the event horizon to observers sufficiently far outside the black hole. If you beg to differ, please substantiate your claim formally.

A more accurate description is as follows. This is readily derived by following light rays in a Kruskal chart with Schwarzschild r and t lines represented.

Imagine a long ship heading toward an isolated (no accretion disc) supermassive BH. Imagine an observer hovering a good distance from the horizon, off somewhat to the side of the rocket's radial infall trajectory. For any detection technology available, there is a surface (just) outside the horizon such that any light reaching the hovering observer from this surface is red shifted to the point where it is unobservable - effectively as black as possible (e.g. much blacker than CMB).

Then, the hovering observe sees the following, in a time span very short compared to the life time of the universe:

The front of the ship vanishes at this surface, then as each part of the ship advances to this surface, it vanishes. There is no perception of the ship being pancaked or frozen. Instead, it would appear as if the ship were progressively vanishing as it went through a surface of invisibility. In vernacular, it would appear as if the ship smoothly and slowly passed into a hole in space.

 

[.Scott]

The front of the ship vanishes at this surface, then as each part of the ship advances to this surface, it vanishes. There is no perception of the ship being pancaked or frozen. Instead, it would appear as if the ship were progressively vanishing as it went through a surface of invisibility. In vernacular, it would appear as if the ship smoothly and slowly passed into a hole in space.

[But that's not what happens at the event horizon, that's what happens when light become trapped within the photon sphere.][STRIKE]No. That is a complete misunderstanding of an event horizon.[/STRIKE]
There is an event horizon shrouding the black hole from the observer hovering far above that BH. If a clock on the ship passed through noon just as it passed through that black hole, the observer would see the ship approach the event horizon but never cross it. He would also see the clock approach closer and closer to noon, but never reach it.

Event horizons are not specific to black holes. If you place yourself in a continuously accelerating spacecraft , you would observe an event horizon develop behind you [and you have objects "collecting" on its surface].

In that sense, the event horizon is like a rainbow (as described above). If you try to visit the end of the rainbow - it moves back from you or disappears altogether.

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

 

[PAllen]

No. That is a complete misunderstanding of an event horizon.
There is an event horizon shrouding the black hole from the observer hovering far above that BH. If a clock on the ship passed through noon just as it passed through that black hole, the observer would see the ship approach the event horizon but never cross it. He would also see the clock approach closer and closer to noon, but never reach it.

Event horizons are not specific to black holes. If you place yourself in a continuously accelerating spacecraft , you would observe an event horizon develop behind you.

In that sense, the event horizon is like a rainbow (as described above). If you try to visit the end of the rainbow - it moves back from you or disappears altogether.

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

No, what I said is correct. The event horizon itself is undetectable. The surface of last visibility is outside the event horizon. The same description would apply to a Rindler horizon. An extended object would be seen to finally disappear as each part reached the surface of last visibility. The process of vanishing would be slow, but finite for any given sensitivity of imaging device.

 

[Nugatory]

No. That is a complete misunderstanding of an event horizon.

Careful... PAllen isn't describing the event horizon, he is describing the behavior of light emitted from a free-falling object very near to but still outside the horizon. That doesn't mean his description is correct, but if it's wrong it's wrong for some reason deeper than "complete misunderstanding".
(I suspect that you and he may be working with a different threshold for "effectively as black as possible")

 

[.Scott]

Careful... PAllen isn't describing the event horizon, he is describing the behavior of light emitted from a free-falling object very near to but still outside the horizon. That doesn't mean his description is correct, but if it's wrong it's wrong for some reason deeper than "complete misunderstanding".
(I suspect that you and he may be working with a different threshold for "effectively as black as possible")

Yes. That is what's happening. He introduced his remark as a "more accurate description", but then what he described was not what happens at the event horizon.

He put the observer "off somewhat to the side" so that the observer would not be able to track the traveler all the way to the horizon. Instead, at a certain point in the descent light from the traveler headed towards the observer becomes trapped in the photon sphere and the traveler appears to simply vanish.

 

[Nugatory]

We are talking about how this would look to the outside observer - aren't we? As the traveler descends toward the observer's event horizon, that traveler will appear to slow in time. Since he would be seen as completely frozen in time at the horizon, how could he be seen as passing through it?

He can't be seen passing through the event horizon. But here PAllen is suggesting that there is a surface of last visibility outside the event horizon. Because it is outside, it can be reached in a finite amount of the observer's proper time; the infaller passes through this surface before time is "completely frozen".

 

[.Scott]

No, what I said is correct. The event horizon itself is undetectable. The surface of last visibility is outside the event horizon. The same description would apply to a Rindler horizon. An extended object would be seen to finally disappear as each part reached the surface of last visibility. The process of vanishing would be slow, but finite for any given sensitivity of imaging device.

Another way to look at that is that the event horizon is undetectable, but you can set things up to make an observation arbitrarily close to it.
I'm going to drop a matter-antimatter bomb into the black hole, set to detonate a meter above the event horizon. Assuming I can still detect the explosion, the extreme time dilation should be very apparent.

 

[.Scott]

He can't be seen passing through the event horizon. But here PAllen is suggesting that there is a surface of last visibility outside the event horizon. Because it is outside, it can be reached in a finite amount of the observer's proper time; the infaller passes through this surface before time is "completely frozen".

Yes, the photon sphere - or rather a sphere within the photon sphere at a distance based on how far "aside" the observer is.
I already edited the post you quoted - as well as the earlier one.

 

[PAllen]

Another way to look at that is that the event horizon is undetectable, but you can set things up to make an observation arbitrarily close to it.
I'm going to drop a matter-antimatter bomb into the black hole, set to detonate a meter above the event horizon. Assuming I can still detect the explosion, the extreme time dilation should be very apparent.

The extreme time dilation is certainly apparent. There is an easy way to ball park some quantities I only gave qualitatively. Imagine a hypothetical 1km rocket falling radially into supermassive BH, with the long dimension oriented radially. You are looking at this from far enough away to hover safely, and on a different radial line from the rocket so you can see its whole length. Let's say the rocket is luminescent, emitting bright blue light. Now, make some assumption about what is the longest wavelength you can possibly image. For example, pretend you can still image blue light red shifted by a factor of a million. There is a specific surface outside the horizon that corresponds to this redshift factor. A rocket in free fall starting from reasonably far away will experience (but not detect - there is no plausible way to do so) this surface cross them at near c. Thus, rocket clocks will experience crossing this surface of last visibility in about 10^-5 seconds. The red shift factor of a million means a distant observer will see this crossing time as appx. 10 seconds.

Thus, the distant observer sees the 1km ship vanish as each portion crosses this surface, the whole process (from front crossing to back crossing the same surface) lasting about 10 seconds.

Make different assumptions, and you get different numbers, but the general picture does not change. You see the long ship crossing into invisibility, with each part becoming invisible per some declared device sensitivity as it reaches this surface (there will be progressive dimming as each part of the rocket approaches this surface) (the location of the surface is function of assumed imaging technology). It will look exactly like the rocket crossed into a 'void' in 10 seconds.

Note, that what I was arguing against is the frequent statement that a distant observer sees objects frozen at the horizon. This is simply a false statement, in any plausible sense of the term 'see'. What I describe is what would actually be seen (for an isolated supermassive BH).

 

[skeptic2]

Let's plug some numbers into PAllen's example above.

For a 10 solar mass BH the radius at which there would be a dilation factor (gamma) of 1 million would be about 2.95x10^-8 meters above the event horizon. Both the time and length of the spaceship would be diminished by gamma as seen from a distance. So at about 30nm above the horizon the spaceship would be 1mm long traveling at a velocity of approximately 300m/s (as seen from a distance). It would disappear rather quickly. In fact, the spaceship will disappear in about 3.333uS.

If the spaceship were luminescent in gamma rays we might actually be able to see it freeze just above the horizon.

 

[PeterDonis]

the photon sphere

If by this you mean the radius (1.5 times the horizon radius) at which photons can orbit the black hole in free fall, this has nothing to do with what PAllen was talking about; the "surface of last visibility" he is describing is much closer to the horizon. It's perfectly possible to see objects falling through the photon sphere at 1.5 times the horizon radius; the fact that free-fall photon orbits exist there does not mean all photons get trapped there.

 

[PAllen]

Let's plug some numbers into PAllen's example above.

For a 10 solar mass BH the radius at which there would be a dilation factor (gamma) of 1 million would be about 2.95x10^-8 meters above the event horizon. Both the time and length of the spaceship would be diminished by gamma as seen from a distance. So at about 30nm above the horizon the spaceship would be 1mm long traveling at a velocity of approximately 300m/s (as seen from a distance). It would disappear rather quickly. In fact, the spaceship will disappear in about 3.333uS.

If the spaceship were luminescent in gamma rays we might actually be able to see it freeze just above the horizon.

I was specifically referring to a supermassive BH. For one thing, tidal forces would be severe well outside the horizon of 10 solar mass BH. However, we can choose to ignore that.

But I'm not sure about your 3 microseconds. If the surface crossing takes tc for the ship, it will be seen to take tc*gamma for the distant observer. Thus, the ship would appear to be going slow enough to require this duration.

As for gamma rays, it might seem frozen above the horizon for a while, but will eventually be seen to go through (the surface of last visibility just above the horizon). Patience (and time lapse imaging)...

 

[pervect]

Both statements need further qualification. To observers sufficiently far outside the black hole, objects do freeze, not just appear to freeze, at the EH and every experiment that could be done from that distance would support that observation.

The notion that objects only appear to freeze comes from the assumption that the viewpoint of the infalling observer is somehow more legitimate than that of a distant observer. Both observations are equally valid. One cannot pick one observer, e.g. the infalling one, and claim that his perspective is reality and what the distant observer sees and measures is not.

Unless I'm missing something obvious, the object crossing the horizon does NOT "actually freeze" in GP coordinates (http://en.wikipedia.org/wiki/Gullstrand-Painlevé_coordinates#Rain_coordinates), if one uses the GP coordinates notion of a surface of simultaneity to define the time at which the object crosses the event horizon remotely, which amounts to ignoring propagation delays.

I'd find the fermi-normal coordinates of an infalling observer to be more interesting than GP coordinates, unfortunately they'd be annoyingly difficult to compute - so I'm using the GP coordinates as an example.

The infalling object still "appears to freeze" of course, because the delay from any signal moving away from the horizon becomes infinite. This can be attributed to propagation delay effects however, not the common (and IMO misleading) viewpoint that time somehow stops.

If both viewpoints are considered to be valid, one cannot claim that "actual freezing" happens universally , becase there are some (many!) coordinates that don't exhibit "actual freezing".

The biggest hint that the "time stops" viewpoit is wrong is the finite proper time it takes to reach the horizon, IMO.

Note that the notion of "actual freezing" I mentioned doesn't really have any "actual" physical content, though, its a purely coordinate based notion, in particular the notion of simultaneity that is implied by the coordinate system.

 

[PAllen]

Both statements need further qualification. To observers sufficiently far outside the black hole, objects do freeze, not just appear to freeze, at the EH and every experiment that could be done from that distance would support that observation.

Can you describe one such experiment? I believe there are no experiments that could be performed by a distant observer that show objects frozen at the horizon. In particular, suppose you shoot a stream of contact triggered nuclear bombs down at the horizon of a supermassive BH. If you have a mental model of objects frozen at the horizon, you would think that one would touch another eventually, and detonate, and you could see this (assuming it happened just outside the horizon).

Instead you would see each bomb vanish in sequence, one behind another (hundreds of them, if desired). Further, the fact that they never touch each other (even through the horizon) is invariant physics, not coordinate dependent.

Or try deflect gamma rays off an object as it falls in. After a time you will not be able to detect the in-falling object, no matter what energy gamma rays you use (because they will only be able to catch it inside the horizon; and later not at all because they will reach the singularity after the object has).

So, one experiment please?

 

[skeptic2]

The biggest hint that the "time stops" viewpoit is wrong is the finite proper time it takes to reach the horizon, IMO.

The old thought experiment about two lights flashing on a railroad car traveling at nearly c. An observer on the car sees both lights flash simultaneously but an observer standing beside the tracks sees one light flash before the other. The above argument is similar to an argument that the viewpoint of the trackside viewer must be wrong because the observer on the train sees both lights flashing simultaneously.

Edit: Please refer to this video and note that although the lecturer refers to infalling objects slowing down asymptotically at the event horizon, he never uses the words "appears to" or "apparently". The example begins shortly after 9:00 minutes.

 

[PAllen]

The old thought experiment about two lights flashing on a railroad car traveling at nearly c. An observer on the car sees both lights flash simultaneously but an observer standing beside the tracks sees one light flash before the other. The above argument is similar to an argument that the viewpoint of the trackside viewer must be wrong because the observer on the train sees both lights flashing simultaneously.

Not unless you reject both SR and GR. In both, simultaneity is frame or coordinate dependent, while proper time is invariant. The event of an infaller crossing the horizon is point in the spacetime manifold. It exists, period. You can choose, for example, a Mercator projection of the Earth that does not include the poles. That has no bearing whatsoever on the existence of the poles. There is no such thing in GR as events that exist or not depending on coordinates, or depending on the observer. The most you might say is that some coordinates don't include an event, or that some observer can't detect a particular event.

 

[PAllen]

Edit: Please refer to this video and note that although the lecturer refers to infalling objects slowing down asymptotically at the event horizon, he never uses the words "appears to" or "apparently". The example begins shortly after 9:00 minutes.

A couple of comments on ths:

1) While Lenny Susskind is an expert, serious physicist, this is not a serious scientific presentation or paper. It is, as its introduction notes, a completely unprepared talk for a mixed audience.

2) He does state that the 'outside view' he is describing is a consequence of coordinates chosen to describe the outside.

3) Unfortunately, at around 10:37 he makes a pure and simple mistake. I hope this is a function of the nature of the talk, not a reflection of the author in more serious work. [The mistake is the claim that, classically, all the matter forming the BH is contained in shells asymptotically approaching the horizon. This is simply false for a collapse leading to a BH. During the collapse, just before any horizon forms, most the mass is already inside where the horizon will be and cannot possibly end up outside the horizon. In fact, for a classical collapse, the horizon grows from the center out, encompassing more and more of the collapsing matter, reaching the final event horizon at precisely the same moment as all the matter is inside (as viewed from inside). From outside, a better description of the asymptotic description is that the whole collapse (with density and pressure greatest in the center) never quite reaches the point where the apparent horizon forms. The whole collapsing mass appears to go dark - once it quiesces and all nearby matter captured - just before forming a horizon. You can, from the outside, describe new matter falling in as being in thin shells around the BH.]

 

[skeptic2]

This discussion seems to have ended. If what Leonard Susskind says isn't accepted then there's zero chance anything I say will be taken seriously.

 

[PeterDonis]

This discussion seems to have ended. If what Leonard Susskind says isn't accepted then there's zero chance anything I say will be taken seriously.

We don't accept arguments from authority here. Even eminent physicists can make mistakes, or misstate things, or not translate correctly from the math into ordinary layman's language. Even Einstein made errors like these sometimes. That's why we want to talk about the actual substance of what was said, which is what PAllen's post addressed. If you have a substantive point to make, it will be considered on its merits. But if your argument is "Susskind says so", then yes, there's no point in discussion, because we don't accept anything just on someone's say-so, even if it's Susskind or Einstein.