Why wouldn't black hole singularity evaporate before it can form?

[lukesfn]

I assume people much more knowledgeable then me must have already thought this through, but the following line of thought has me very curious. I'm making a lot of intuitive leaps here though, so I am sure there is a lot of places I could have gone drastically wrong.

Hawking radiation is predicted to be observed from out side the event horizon of a black hole, however, I would expect somebody just inside the horizon to also observe something similar to hawking radiation. There will still be another event horizon between the observer and the center of the BH from from which out side of, information will be able to be sent back to the observer (although, obviously it can't move away from the center of the BH, it can just move slower towards the center then the observer does so that the observer catches up to it)

I would expect this would work similarly to what happens optically if you cross a BH horizon, you would still optically see the BH in front of you, although it would appear to wrap around you as you fall towards the center. So, I am expecting you would also continue to be able to observe some hawking like radiation as you fall all the way to the center.

If this is so, then it may get more interesting towards the center. I think the amount of hawking radiation is generally calculated via the mass of a black hole, but for an observed hawking like radiation inside at a radius inside the horizon, I would expect that it would have to be increase exponentially as the radius decreases since the curvature of space would increase.

Maybe I am confused, but it seems to me that the rate of increase of this radiation would tend towards infinity faster then an observer would be accelerated towards the center.

By this logic, I start to think that an observer should see everything in the BH evaporate before they could reach the singularity.

Following this logic further no particle should ever be able to reach a singularity, so is it possible for one to form?

It seems a little counter intuitive that a singularity couldn't form given that things would have to be moving out from the center at faster then the speed of light trough space tiem to stop a singularity from forming, but the mechanism of hawking radiation allows for FTL transfer of mass from the BH at the horizon, so could it not allow a trickle up effect from all the way from the center, especially considering the much higher stress on space time towards the center effectively causing space time too pull itself apart that much stronger.

Like I said, I assume many much more knowledgeable people have spent much more time then me pondering such things, but it does seem like such an elegant solution to many problems to my naive mind that I probably need somebody else to tell me why it isn't so.

 

[Simon Bridge]

Hawking radiation is predicted to be observed from out side the event horizon of a black hole, however, I would expect somebody just inside the horizon to also observe something similar to hawking radiation.

What leads you to expect this?

There will still be another event horizon between the observer and the center of the BH

The math does not predict one.

I would expect this would work similarly to what happens optically if you cross a BH horizon,

What leads you to expect this?
One of the ways to gain understanding is to investigate your expectations - to do that, you need to make the basis for them explicit.

you would still optically see the BH in front of you, although it would appear to wrap around you as you fall towards the center.

What does this mean? "Wrap around you"?
Do you know what the event horizon is?

So, I am expecting you would also continue to be able to observe some hawking like radiation as you fall all the way to the center.

What leads you to expect this?
Do you know how Hawking radiation is expected to happen?

If this is so, then it may get more interesting towards the center.

Things should get interesting towards the center anyway.

I think the amount of hawking radiation is generally calculated via the mass of a black hole, but for an observed hawking like radiation inside at a radius inside the horizon, I would expect that it would have to be increase exponentially as the radius decreases since the curvature of space would increase.

http://www.jimhaldenwang.com/black_hole.htm

By this logic, I start to think that an observer should see everything in the BH evaporate before they could reach the singularity.

It gets a little tricky - you are talking about time as experienced by an observer falling towards a black-hole? The link above discusses this with math. It will help you sort out these ideas.

Following this logic further no particle should ever be able to reach a singularity, so is it possible for one to form?

Of course it is.
The black hole forms as the mass concentration passes a critical value.
Then is is expected to radiate via the Hawking mechanism.

You are thinking, perhaps, that the matter forming the Schwarzschild black hole "evaporates" (by the Hawking mechanism) before it can totally collapse - even by that logic, the black hole can still form without all the matter being concentrated in the center.

 

[lukesfn]

What leads you to expect this?

I tried to explain below but obviously could have done better.

The math does not predict one.

Do you then think the radiation observed coming from a black hole just disappears when you cross the horizon? Isn't a free falling observer meant to see nothing special as they cross the horizon?

If that is so, then the rest of my post is irrelevant but perhaps I should should try to explain my logic better.

If you are free falling into a Black Hole, and there is an object in front of you, if it is out side a certain radius of the black hole, you will be able to see light from it, however, any object with in that radius you will not. I am calling that an effective event horizon. (The light from that object of course would have been emitted from a position further out from the Black Hole center then where you are currently located.

Should you not also not also experience some apparent hawking like radiation in a similar way?

What does this mean? "Wrap around you"?

As you move through the horizon and towards the center, the everything outside the black hole will optically appear to be compressed into a single point, which could look like the black hole was wrapping around you.

 

[Dale]

If you are free falling into a Black Hole, and there is an object in front of you, if it is out side a certain radius of the black hole, you will be able to see light from it, however, any object with in that radius you will not. I am calling that an effective event horizon. (The light from that object of course would have been emitted from a position further out from the Black Hole center then where you are currently located.

Don't forget, the event horizon is a null surface, meaning that it is moving at the speed of light wrt any free-falling local frame. So the fact that you cannot see light from an object which has crossed the horizon until you cross the horizon is no different from the fact that you cannot see light from any event until a time d/c has passed for the light to reach you.

If Jack and Jill were crossing a supermassive black hole's event horizon and they were 1 light-second apart (ladies first, of course). Then when they were both outside the event horizon Jack would have to wait 1 s to receive a signal from Jill, when Jill was inside and Jack was out Jack would have to wait 1 s to receive a signal from Jill (by which time Jack would be in since the horizon moves at c), and when they are both inside Jack would have to wait 1 s to receive a signal from Jill. And vice versa.

 

[Naty1]

Do you then think the radiation observed coming from a black hole just disappears when you cross the horizon? Isn't a free falling observer meant to see nothing special as they cross the horizon?

What radiation do you think is observed?
Yes, to the second part.

Even Einstein did not believe the mathematical results thinking black holes did not REALLY exist. Math can give us insights 'logic' cannot.A distant stationary observer will never receive a signal from a source at or within the event horizon even after an infinite amount of proper time; An inertially falling observer will receive a signal from at or within the event horizon after a finite amount of proper time.

I'd call that a 'logical contradiction'... at least from classical reasoning.

[This should be consistent with Dalespam's post just above.]edit: for a related view, try reading about the Unruh effect.

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

 

[Naty1]

Why wouldn't black hole singularity evaporate before it can form?

The simple answer:
because once mass collapses to within the Schwarzschild radius, r = 2M,
nothing can get out...gravity is just too strong.

 

[lukesfn]

Don't forget, the event horizon is a null surface, meaning that it is moving at the speed of light wrt any free-falling local frame.

Understood.

So the fact that you cannot see light from an object which has crossed the horizon until you cross the horizon is no different from the fact that you cannot see light from any event until a time d/c has passed for the light to reach you.

Time d/c? Sorry, I can't figure out that term, although, I assume I still understand your meaning.

If Jack and Jill were crossing a supermassive black hole's event horizon and they were 1 light-second apart (ladies first, of course). Then when they were both outside the event horizon Jack would have to wait 1 s to receive a signal from Jill, when Jill was inside and Jack was out Jack would have to wait 1 s to receive a signal from Jill (by which time Jack would be in since the horizon moves at c), and when they are both inside Jack would have to wait 1 s to receive a signal from Jill. And vice versa.

I'm not quite sure if you are trying to explain something I am missing here, however, I am curious, what is keeping them 1 light second apart? Would 2 objects in free fall following the exact same motion, starting 1 light second stay 1 light second apart? Talking about a 1 second delay for 1 way communication is very easy to get confused with. Would it be better to talk about a 2 second delay for Jack to have to wait for an instantly returned message from Jill?

What radiation do you think is observed?

Something that would appear exactly identical to hawking radiation with respect to the type of particles being detected, except, I expect it would need a different name.

I'd call that a 'logical contradiction'... at least from classical reasoning.

You are getting into a semantic discussion. When I use the word logic, the logic is trying to follow the rules of GR (and QM with regard to QM), so if any of my logic is contradicted by the mathematics, please feel free to let me know.

edit: for a related view, try reading about the Unruh effect.

I am aware of the unruh affect, although, I am not sure weather that would be considered the same effect I am talking about or not, because from what I have previously read, it occures due to an object constantly accelerating in SR, where as I am talking about objects free falling in GR. Still, I am talking about something similar. It is interesting to consider that if you sit just outside the EH accelerating to stay near the Speed of light to hover, that you would see a lot of Unruh radiation, but by another logic, because of time dialation, you would expect to see the BH evaporate a lot faster via hawking radiation. Does that mean that in this case the Unrah and Hawking radiation are the same thing?

The simple answer:
because once mass collapses to within the Schwarzschild radius, r = 2M,
nothing can get out...gravity is just too strong.

Are you saying that mass can not be transferred out via hawking radiation?

 

[Dale]

Time d/c? Sorry, I can't figure out that term,

Sorry, I should have defined my variables. d is distance, distance divided by speed is time.

I'm not quite sure if you are trying to explain something I am missing here, however, I am curious, what is keeping them 1 light second apart?

Inertia.

Would 2 objects in free fall following the exact same motion, starting 1 light second stay 1 light second apart?

Yes. Remember that the black hole is supermassive so tidal effects are negligible.

Talking about a 1 second delay for 1 way communication is very easy to get confused with. Would it be better to talk about a 2 second delay for Jack to have to wait for an instantly returned message from Jill?

It doesn't matter which you are talking about. The point is that it all behaves exactly the same as normal in a regular local inertial frame without an event horizon.

 

[lukesfn]

Yes. Remember that the black hole is supermassive so tidal effects are negligible.

Sorry, I didn't pay proper attention to the word super massive, but thanks for clearing up that fact that you where ignoring small effects due to curvature of space in this case. This is something I already understand well, and I don't think contradicts my like of reasoning.

Anyway, I've been informed I've crossed the line of being too speculative for the forum rules, so I will try to constrain my discussion to weather any hawking-like radiation may exist inside the EH of a BH.

If an observer outside the EH of a BH detects Hawking radiation, and they cross the EH,
1) does the observer simply see the hawking abruptly disappear after crossing the horizon and replaced by nothing, or
2) would the observer continue to detect some "hawking-like radiation" during and after crossing, that would, from the observers point of view, be very difficult to distinguished from hawking radiation, so much so that no device would be able to measure a sudden change in radiation levels to detect the crossing of the EH.

 

[bcrowell]

This strikes me as a situation where Penrose diagrams come in very handy: http://en.wikipedia.org/wiki/Penrose_diagram . Personally I can't reason at all about the cause and effect relationships discussed in #1 without a Penrose diagram in front of me. For example:

There will still be another event horizon between the observer and the center of the BH from from which out side of, information will be able to be sent back to the observer (although, obviously it can't move away from the center of the BH, it can just move slower towards the center then the observer does so that the observer catches up to it)

AFAICT from a Penrose diagram, the second observer can receive the information if and only if s/he dives in after the first observer within a short enough time. The WP article is not that great. Anyone who wants to go to Penrose-diagram bootcamp should read the treatment in Penrose's popularization Cycles of Time.

An event horizon is usually defined as a surface from which light can't escape to an infinitely distant observer. The definition seems to make it clear that horizons have no special properties except in relation to some observer. Therefore I doubt that it's correct to expect Hawking radiation to be a special phenomenon that only exists at an event horizon. This is probably a case where those of us (including me) whose understanding of quantum gravity is at the pop-sci level should not expect popularizations to get us beyond a certain point. Baez has a writeup about his doubts re this particular way of explaining Hawking radiation in popularizations: http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/hawking.html

 

[lukesfn]

An event horizon is usually defined as a surface from which light can't escape to an infinitely distant observer.

Excuse me, I've used that term incorrectly before creating all types of confusion. Perhaps apparent horizon is better term but I am not sure that is right either, I should have just said horizon.

 

[Simon Bridge]

Isn't a free falling observer meant to see nothing special as they cross the horizon?

That would be correct.

If that is so, then the rest of my post is irrelevant

... that was the gist, but it was possible I misunderstood.

Perhaps apparent horizon is better term but I am not sure that is right either, I should have just said horizon.

... no, that would just have lead to people asking you to be specific.
The "event horizon" is the distance from the center of mass where the escape velocity is equal to the speed of light[*].

I think the main thing to take away from bcrowells comment there is the bit about how far pop-science can take your understanding. Did you have a look at the links I gave you at all?

--------------------------
[*] A more accurate description could be that within this horizon, all lightlike paths are warped so as to fall farther into the hole.

 

[lukesfn]

This strikes me as a situation where Penrose diagrams come in very handy: http://en.wikipedia.org/wiki/Penrose_diagram . Personally I can't reason at all about the cause and effect relationships discussed in #1 without a Penrose diagram in front of me.

I like to visualize curved space as a flowing river with special relativity occurring over the top. With this analogy, inside a BH EH, space flows faster then the speed of light. I can stay pretty close to intuitive classical thinking this way, which makes things much easier to visualize for me, except when time dilation confuses everything, and I think I understand SR reasonably well.

no, that would just have lead to people asking you to be specific.

Which, from my point of view, would be a much better out come. I find it is usually much better to be asked for clarification then completely miss understood.

An event horizon is usually defined as a surface from which light can't escape to an infinitely distant observer. The definition seems to make it clear that horizons have no special properties except in relation to some observer.

I was actually aware of this kind of definition for an EH of a BH. It makes sense to me that there would always be some kind of observer dependent apparent horizon between an observer and the singularity, however, I have previously been told that an "apparent horizon" has some different definition. I think I need to describe better exactly how the horizon in my mind is defined, however, I might leave that for another occasion.

Therefore I doubt that it's correct to expect Hawking radiation to be a special phenomenon that only exists at an event horizon.

I've also taken a similar line of reasoning which brings us close back to the point.

Perhaps we could avoid this horizon talk and get back to the question of hawking-like radiation inside the event horizon?

To Quote my self:

If an observer outside the EH of a BH detects Hawking radiation, and they cross the EH,
1) does the observer simply see the hawking abruptly disappear after crossing the horizon and replaced by nothing, or
2) would the observer continue to detect some "hawking-like radiation" during and after crossing, that would, from the observers point of view, be very difficult to distinguished from hawking radiation, so much so that no device would be able to measure a sudden change in radiation levels to detect the crossing of the EH.

 

[Dale]

If an observer outside the EH of a BH detects Hawking radiation, and they cross the EH,

One of the nice things about discussing supermassive black holes is that the tidal gravity is negligible at the EH, so you can analyze the situation using accelerating reference frames (per the equivalence principle).

If Hawking radiation emanates from the black hole's horizon then there must be some equivalent from the Rindler horizon of the accelerating reference frame. This is known as Unruh radiation. Unruh radiation is not seen by an inertial observer. So, by the equivalence principle, I would expect that an observer free-falling across the EH of a supermassive BH would also not detect any Hawking radiation.

What I am less certain of is if an observer were not free falling across the EH, but passed the EH while accelerating (but not enough to "hover", or even accelerating towards the singularity).

 

[lukesfn]

One of the nice things about discussing supermassive black holes is that the tidal gravity is negligible at the EH, so you can analyze the situation using accelerating reference frames (per the equivalence principle).

If Hawking radiation emanates from the black hole's horizon then there must be some equivalent from the Rindler horizon of the accelerating reference frame. This is known as Unruh radiation. Unruh radiation is not seen by an inertial observer. So, by the equivalence principle, I would expect that an observer free-falling across the EH of a supermassive BH would also not detect any Hawking radiation.

What I am less certain of is if an observer were not free falling across the EH, but passed the EH while accelerating (but not enough to "hover", or even accelerating towards the singularity).

I just read something on a similar note elsewhere. It also made me wonder if a free falling observer may not observe Hawlking radiation, however I'm not sure that makes sense either.

One problem with using Supermassive Black Holes is that you have to be careful not to end up ignoring the effect you are looking for. The in falling observer may not see Hawking radiation because it is too little.

I'm just starting to find some better search times to find stuff on google on this topic. The below link looks interesting, but I suspect that even if I was subscribed, I would need to learn a lot before having much chance of even getting an intuitive understanding of what it is talking about.
Hawking radiation as perceived by different observers.

The physical image that emerges from these analyses is rather rich and compelling. Among many other results, we find that generic freely-falling observers do not perceive vacuum when crossing the horizon, but an effective temperature a few times larger than the one that they perceived when it started to free-fall. We explain this phenomenon as due to a diverging Doppler effect at horizon crossing.

 

[Dale]

The in falling observer may not see Hawking radiation because it is too little.

That is a good point. The temperature of a black hole is lower the more massive it is, so a supermassive BH would be very cold, perhaps undetectably cold.

However, IMO, the equivalence principle trumps Hawking radiation, and you can always make the EH of a Schwarzschild BH equivalent to a Rindler horizon for a sufficiently small region of spacetime. However, this is a very handwavy argument that I recognize as being somewhat weak, so I won't do more than merely present it.

I understand the equivalence principle much more than I understand Hawking radiation, so I rely on it more and tend to be skeptical of Hawking radiation where it seems to conflict. I.e. I would need to see some rigorous proof before believing that a free-falling observer sees Hawking radiation at all, let alone the Hawking-like radiation you are discussing.

 

[lukesfn]

I understand the equivalence principle much more than I understand Hawking radiation, so I rely on it more and tend to be skeptical of Hawking radiation where it seems to conflict. I.e. I would need to see some rigorous proof before believing that a free-falling observer sees Hawking radiation at all, let alone the Hawking-like radiation you are discussing.

I'd like to know if a free falling observer is expected to see Hawking radiation or not. Reading much else where, I have seen a lot of conflicting information about this.

I noticed a similar thread here on physicsforums. however, there seems to be a lot of contradictory answers.

Consider a small black hole, radiating a large amount of hawking radiation, and two observers, at a distance where gravity is small, one observer hovering, one free falling, both with the same instantaneous velocity. The hovering observer, would not need to accelerate much to stay hovering due to the low gravity, so would not see much Unruh radiation. The free falling observer sees Unruh radiation. But, because it is a small BH it should be radiating a lot, so one might expect the hovering observer to see a large amount of hawking radiation, not a small amount, and have trouble understanding why the free falling observer sees none, leading to think that curvature of space is very important to take into consideration with hawking radiation.

I can think of some holes in this line of thinking, such as, perhaps a freefaller observes hawlking radiation out side the horizon, but not at the horizon, or maybe there the hawking radiation smoothly changes direction at the horizon, but those ideas bring up a lot of paradoxes such as, shouldn't the hawking radiation cool down the further it gets from a BH, not increase?

hmm. the way Wikipedia tries to describe it is interesting.

Hawking radiation is required by the Unruh effect and the equivalence principle applied to black hole horizons. Close to the event horizon of a black hole, a local observer must accelerate to keep from falling in. An accelerating observer sees a thermal bath of particles that pop out of the local acceleration horizon, turn around, and free-fall back in. The condition of local thermal equilibrium implies that the consistent extension of this local thermal bath has a finite temperature at infinity, which implies that some of these particles emitted by the horizon are not reabsorbed and become outgoing Hawking radiation

Can a distant observer use the equivalence principle and claim that they are very quickly accelerating away from the EV and make calculations based off that about what they would observe at the EV, even though, they are actually accelerating very slowly, or even free falling, and there is a large amount of curved space between them and the EV? They would have to then compensate for what ever effect to pass back though curved space to them to be observed though.

 

[lukesfn]

I've been doing some extra reading.

These all link directly to pdf files. The first has a good summery about observations of halwking radiation from different observes in the introduction. The others confirm the some of same by different methods. I don't pretend to have tried to understand the maths, so I have to trust the conclusions.

(1) Taking the Temperature of a Black Hole

(2) Hawking radiation as seen by an in-falling observer

(3) Hawking radiation as perceived by didifferent
observers

From these, the temperature observed of hawking radiation seems to depend on the observers location, velocity, and acceleration. Whether or not a free falling observer sees hawking radiation or not, depends on there.

According to [2], the only a free falling observer with zero instantaneous radial velocity at the horizon will see no hawking radiation. I'm a little confused what zero radial velocity means in this context. I guess it is a I guess it means that if we imagine space moving past the EV at the speed of light, the observer would be not moving relative that space.

If read quite a few claims in other places that a free falling observer sees no hawking radiation, but it sounds like the standard lore may have been over generalized.

Anyway, I think I will keep reading more of these papers.

I'm not quite sure what a zero radial velocity observer sees from just out side the EV, and just inside yet.

I'm wondering if all observers see hawking radiation out side the BH EV, but only some inside do.

 

[Naty1]

I saved this post from another discussion:

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

These consideration resolve an apparent paradox concerning the Hawking effect. The proper time for a freely-falling observer to reach the event horizon is finite, yet the free-fall time as measured at infinity is infinite. Ignoring back-reaction, the black hole will emit an infinite amount of radiation during the time that the falling observer is seen, from a distance to reach the event horizon. Hence it would appear that, in the falling frame, the observer should encounter an infinite amount of radiation in a finite time, and so be destroyed. On the other hand, 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.

The paradox is resolved when a careful distinction is made between particle number and energy density. When the observer approaches the horizon, the notion of a well-defined particle number loses its meaning at the wavelengths of interest in the Hawking radiation; the observer is 'inside' the particles. We need not, therefore, worry about the observer encountering an infinite number of particles. On the other hand, energy does have a local significance. In this case, however, although the Hawking flux does diverge as the horizon is approached, so does the static vacuum polarization, and the latter is negative. The falling observer cannot distinguish operationally between the energy flux due to oncoming Hawking radiation and that due to the fact that he is sweeping through the cloud of vacuum polarization. The net result is to cancel the divergence on the event horizon, and yield a finite result, ...

 

[Naty1]

I'm wondering if all observers see hawking radiation out side the BH EV,

no way things will ever be THAT straightforward!...Observations are generally 'relative'.

If read quite a few claims in other places that a free falling observer sees no hawking radiation, but it sounds like the standard lore may have been over generalized.

That's my understanding for a radially free falling observer...Apparently for those observers on an arcing trajectory, some radiation might be observed. But the above quote [prior post] suggests not. A poster in another discussion said the energy of observers also plays a role.

I have been looking for the following statement for sometime...a prior post of mine...

This confirms that Hawking radiation as described by 'particle separation' at the horizon is not mathematically based...just one way to think about what is happening. [Penrose is a collaborator of Hawking on many BH/GR issues.]

Baez: is right about the particle antiparticle pair production NOT following directly from the math...According to Roger Penrose in THE ROAD TO REALITY, page 836, such a view was Hawking’s "intuitive" explanation. Penrose explains that the production of real particles with positive and negative energy can be explained in terms the Killing vector K ...

Kip Thorne, BLACK HOLES AND TIME WARPS, PGS 435-440

Hawking concluded that a black hole behaves precisely as though its horizon has a finite temperature. There are several different ways to picture black hole evaporation...all acknowledge vacuum fluctuations as the ultimate source of the emitted radiation...the simplist is that...tidal gravity pulls the virtual photons apart and the one that escapes carries away the energy that the hole's tidal gravity gave it.

As soon as one is separated from its virtual partner, it becomes "real" (observable)...but only to an accelerating (stationary) observer outside the horizon.

Since black holes are still considerably colder than our universe, they radiate nothing but do absorb energy (radiation) from the universe and slowly grow. So nobody has ever seen Hawking radiation.

If we could hang a thermometer on a long string and dangle it just outside the horizon of a black hole, it would register incredibly high perhaps approaching infinite temperature. On the other hand a free falling observer falling towards the black hole registers no such increase in temperature and passes harmlessly through the mathematical (theoretical) horizon without any immediate effect except increasing gravitational acceleration and tidal effects...So thermal and quantum radiation become two sides of the same coin near a horizon.

I don't personally any longer believe any of Thorne's statements are completely correct...but who am I to contradict him.

As an example, in a discussion in these forums I recall we concluded that a blackbody radiates as long as it is above absolute zero...and a BH is a black body...so it does radiate, but not evaporate in our life times, because it absorbs way more than it emits...

A key to all understanding all this are posts above discussing global and local constructs. As is always the case in relativity, different mathematical perspectives [different models] offer different [complementary] effects.

 

[PeterDonis]

The hovering observer, would not need to accelerate much to stay hovering due to the low gravity, so would not see much Unruh radiation. The free falling observer sees Unruh radiation.

This is backwards, isn't it? To first order, at least (see further comments in a moment), the accelerated observer should see Unruh radiation and the free-fall observer should not.

However, just to inject some more confusion into this thread , Wald, in his book Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics (published in the 1990's) has an interesting discussion of the Unruh effect. Consider an accelerated detector that detects a quantum of Unruh radiation, by which we mean it absorbs the quantum and thereby gains some energy. This means that the detector changes state: but the state change will also be detectable by a free-falling observer who is passing the detector. To the free-falling observer, this process will look like the detector is *emitting* a particle and its state is changing due to "back reaction" from the emission.

In other words, if Unruh radiation is actually detected, the detection induces a state change in the quantum field that can be detected by all observers, though the interpretation of what the state change means will be different for different observers. The accelerated observer will see the state change as a decrease in particle number, while the free-falling observer will see it as an *increase* in particle number. (This is possible because the particle number operator is observer-dependent.) I haven't seen much discussion of this point in other literature, although it seems obvious once stated. Similar remarks would, of course, apply to Hawking radiation.

 

[lukesfn]

I've been more carefully examining the articles I posted earlier, in particular the following.

[3] Hawking radiation as perceived by different
observers

I'm reconsidering the following comment by me.

According to [3], the only a free falling observer with zero instantaneous radial velocity at the horizon will see no hawking radiation. I'm a little confused what zero radial velocity means in this context. I guess it is a I guess it means that if we imagine space moving past the EV at the speed of light, the observer would be not moving relative that space.

(I think may have accidently written "according to [2]" rathen then [3] earlier, although [2] may say similar as well)
I now think that radial zero velocity in this context means, the object is instantaneously stationary relative to the horizon, which would mean this free falling observer is actually moving at the speed of light, so is some special kind of hypothetical classical observer.

If I understand correctly, this would mean that according to [3], the only observer to see no hawking radiation (or equivalent) is a hypothetical observer moving at the speed of light. It seems that some people might have assumed that because this observer sees no hawking radiation a slower moving observer would see a blue shift of nothing, which is still nothing, however, according to [3], the blue shift is infinite, and when you take the limit of the infinite multiplied by zero, you actually get a finite value.

From [3], the temperature seen by a free-faller at radius r and dropped from radius r₀ is equal to
1/(4m) (1 + 2m/r) (1 - 2m/r)^1/2
×{(1 - 2m/r₀)^1/2 + (2m/r×(1-r/r₀))^1/2 }^1/2
×{(1 - 2m/r₀)^1/2 - (2m/r×(1-r/r₀))^1/2 }^-1/2
Some constants such as G and c have apparently been set as 1 in this paper to arrive at this equation. Before dropped from r₀, the object is being accelerated so that it is hovering r₀. This determines the starting velocity, which becomes infinite at the horizon (at the horizon r₀ = 2m).

However, taking find the limit for both r₀ = 2m, and r = 2m, we find zero temperature at the horizon. This corresponds to the case mentioned above where the hypothetical observer is moving at the speed of light. More interestingly, r₀ > 2m, and r = 2m, meaning the observer is moving less then the speed of light at the horizon, they will see a finite temperature.

Interestingly, this formula can also be extended inside the horizon. Some complex numbers might appear, but the cancel out. It would appear that the in-falling observer with this forumula sees hawking radiation increasing all the way to the singularity approaching infinity.

This leads back to my first post which speculates about consequences of perceived hawking radiation by an observer inside the horizon, which there seems to be some mathematical models to show they may very well perceive.

I made a comment in the first post about there being an event horizon between a free falling observer inside the EV and the center, but I now realize that although the observer would perceive the horizon when they look in front of them, and see hawking radiation coming from it, the location of the horizon would actually be behind them and the hawking radiation moving inwards. I already expect the hawking radiation would be moving inwards, however, this new perspective of an in falling observer seeing hawking radiation coming from a horizon that is behind them them into the singularity makes it harder to imagine how any process this radiation could could lead to mass leaving the BH.

I think this new perspective might help resolved my original question, however the concept of perceiving mass transferred out of the BH but towards it at the same time is quite a strange one, and I think I need to follow the mathematics a bit further.

This confirms that Hawking radiation as described by 'particle separation' at the horizon is not mathematically based...just one way to think about what is happening.

I haven't brought up any discussion of the exact mechanism that causes the radiation because I am not sure it is important to this discussion. Not yet anyway.

These consideration resolve an apparent paradox concerning the Hawking effect. The proper time for a freely-falling observer to reach the event horizon is finite, yet the free-fall time as measured at infinity is infinite. Ignoring back-reaction, the black hole will emit an infinite amount of radiation during the time that the falling observer is seen, from a distance to reach the event horizon. Hence it would appear that, in the falling frame, the observer should encounter an infinite amount of radiation in a finite time, and so be destroyed. On the other hand, 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.

I've also considered the similar case of where the BH actually evaporates, which makes a bit more sense since it how can an infinite amount of radiation come from a shrinking BH, however, there is still a puzzle where you might think that the distant observer would see the BH evaporate before the falling observer hits the horizon, so you might expect them to be destroyed by a very large amount of hawking radiation. But they would observe the hawking radiation to arpoach an infinite wavelength at towards horizon and the observer to approach zero wave length towards the horizon, which is not a good mathematical state to try to predict the outcome. I think that my new understanding from reading more [3], now might explain how the infinity slightly wins over the zero and settles at a finite radiation.

This is backwards, isn't it? To first order, at least (see further comments in a moment), the accelerated observer should see Unruh radiation and the free-fall observer should not.

Sorry. I managed to leave out an important a "no". "the free falling observer sees no Unruh radiation". You can see a couple sentences below I commented how I had trouble understanding why they see don't see radiation radiation.

Below would be the fixed version, which might be a little more clear.

Consider a small black hole, radiating a large amount of hawking radiation, and two observers, at a distance where gravity is small, one observer hovering, one free falling, both with the same instantaneous velocity. The hovering observer, would not need to accelerate much to stay hovering due to the low gravity, so would not see much Unruh radiation. The free falling observer sees NO Unruh radiation. But, because it is a small BH it should be radiating a lot, so one might expect the hovering observer to see a large amount of hawking radiation, not a small amount, and have trouble understanding why the free falling observer sees none, leading to think that curvature of space is very important to take into consideration with hawking radiation.

To the free-falling observer, this process will look like the detector is *emitting* a particle and its state is changing due to "back reaction" from the emission.

It definitely is confusing to wonder how the free falling observer sees there detector emitting energy while and outside observer sees it receiving energy. I've also read somewhere (lost the reference sorry) that the falling observer will see there detector gain mass, but not see any radiation. However, I'm starting to suspect both of these cases might be non sense.

 

[Naty1]

Above I posted:

Since black holes are still considerably colder than our universe, they radiate nothing but do absorb energy (radiation) from the universe and slowly grow.

I think a black body actually emits regardless of the surrounding temperature...better to say therefore that a BH will always absorb more than it emits when colder than it's surroundings.

 

[Naty1]

Wald, in his book Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics ...

In other words, if Unruh radiation is actually detected, the detection induces a state change in the quantum field that can be detected by all observers... . (This is possible because the particle number operator is observer-dependent.)...

My understanding was that the two observers in fact detect different quantum fields...that accelerated detectors will register different particle counts than an inertial detector...in other words, there is no unique observer independent vacuum energy...hence particle number observations vary are observer dependent as you said].

Ok..I checked my notes and found a long discussion on this very same issue started by tom.stoer August 2012, here:

https://www.physicsforums.com/showthread.php?t=625633

no need to go over it again here...I can't find any 'summary' of my own conclusions but here are some pieces of likely interest...

So it seems that we are no longer talking about one unique quantum state with frame dependent interpretations but that we have two truly different quantum states, two different "realities".

Vanadium: Let’s imagine…..I have a "perfect detector" and as soon as it sees a particle, it puts up a flag. All observers agree that this flag goes up - or not- irrespective of frame. The rocket takes off, and after a moment, the flag goes up.

If you ask the rocket traveler what happened, she will say that the rocket started accelerating, they saw Unruh radiation, and the detector sent up a flag. If you ask the stay-at-home traveler, she will say that the rocket started accelerating, began to radiate, and the detector measured this radiation and sent up a flag.

This is in one of Unruh's papers:., "suppose ….a rocket had no changing multipoles and thus will not radiate". It turns out that in that case the detector will not respond to radiation. If you want the detector to be able to detect, it needs some sort of multipole to respond to the radiation, and that very multipole begins to emit. So the answer is that what one observer characterizes as absorption, the other characterizes as emission.

and Bill_k observed:

...In QFT, we build the theory in terms of Poincare unirreps. As soon as you transform to an accelerating FoR, you can't expect the Poincare-based theory to remain good in all aspects. Consider a simple uniform-acceleration case. The accelerated observer is no longer in Minkowski spacetime, but rather in Rindler spacetime, with extra horizons, etc...

I'll come back later and try reading that prior thread and see if I can reach any summary conclusions to post here...gotta go...

 

[PeterDonis]

My understanding was that the two observers in fact detect different quantum fields

That's not the way I understood what I read in Wald, but I have not read a lot of the literature on this. My understanding of Wald is that the underlying quantum field is the same, but the particle number operator is observer-dependent: the operator realized by the accelerated detector is a different operator from the one realized by the inertial detector. So they can detect different particle numbers from the same underlying quantum field.

When each observer detects a change in particle number, they are detecting the same underlying state change in the quantum field; but the inertial observer interprets the state change as an increase in particle number (an emission from the accelerated detector), while the accelerated observer interprets it as a decrease in particle number (an absorption by the accelerated detector).

So I agree with the rest of what you said, but I'm not sure that the above quote is consistent with it. However, as I said, I have not read a lot of the literature on this, and it certainly might be more complicated than I think it is.

https://www.physicsforums.com/showthread.php?t=625633

Interesting quotes, I'll read through the thread.

 

[PeterDonis]

I'll read through the thread.

I took a quick read through, and I would have responded to tom.stoer similarly to the way Ilmrak did: in QFT, "particles" are no longer fundamental entities, quantum fields are. "Particle number" is no longer a fundamental element of reality, i.e., it's not a property of a quantum state per se; it's only a property of a particular operator applied to a quantum state, and different observers will have different particle number operators, just as they have different operators for other observables, such as momentum or energy. So particle number being observer-dependent is really no more mysterious than momentum or energy being observer-dependent.

 

[anorlunda]

There will still be another event horizon between the observer and the center of the BH

The math does not predict one.

Wait. I thought that the definition of the Schwarzschild radius was where the GR radius of curvature of space matched the actual radius. Let S be the Schwarzschild Radius. Assume the BH mass is concentrated in the singularity with radius Y. Then for every radius R, where Y<R<S, the GR radius of curvature should be even less. Therefore I expect every such interior radius R to also be an event horizon.

Another way of saying it is that light, or information, can never travel radially outward in the interior volume enclosed by the EH. Not just at the EH be everywhere inside the BH volume. Is that correct?

 

[PeterDonis]

I thought that the definition of the Schwarzschild radius was where the GR radius of curvature of space matched the actual radius.

No, it isn't. It's where the radial coordinate r = 2M, where M is the mass of the hole, but r is not the "radius of curvature of space" or the "actual radius" from the center at r = 0. (The latter concept doesn't make sense in any case, because the singularity at r = 0 is to the future of the horizon; inside the horizon, the direction of decreasing r is timelike, not spacelike.)

Assume the BH mass is concentrated in the singularity with radius Y.

The singularity has no radius; it's at r = 0. Also, the BH mass is not concentrated there; once the object that originally collapsed to form the hole has collapsed to r = 0 and formed a singularity, the stress-energy inside that object is no longer anywhere in the spacetime.

Then for every radius R, where Y<R<S, the GR radius of curvature should be even less.

No, the radius of curvature of spacetime gets *larger* as you go to smaller values of the radial coordinate r from the horizon. At the singularity, r = 0, the radius of curvature of spacetime diverges to infinity.

light, or information, can never travel radially outward in the interior volume enclosed by the EH. Not just at the EH be everywhere inside the BH volume. Is that correct?

Yes, it's correct, but it doesn't imply the other things you said.

 

[Naty1]

PeterDonis:

I had posted:

My understanding was that the two observers in fact detect different quantum fields.

and in fact even in the Horava-Lifgarbagez gravity paper they say:

The Unruh effect is a statement that the Minkowski vacuum appears as a thermal state with an indefinite number of particles when viewed by a uniformly accelerated observer.

and that's pretty much all I meant except for the 'indefinite' word...not sure what that is...not important...I was a bit surprised to see Wald make the particle distinction...but then that's why he writes physics books and I don't!

PeterDonis posted:

I took a quick read through, and I would have responded to tom.stoer similarly to the way Ilmrak did: in QFT, "particles" are no longer fundamental entities, quantum fields are. ... So particle number being observer-dependent is really no more mysterious than momentum or energy being observer-dependent.

That sure is one way to look at it...but that stuff is mysterious enough! [see excerpts below]

I did like Ilmrak's comments ...

Some additional points along those lines from the earlier discussion:

...Interpretation of ‘reality’ [observations or measurements] is precisely the purpose of a group of transformations between reference frames. Without it, one observer's "reality" cannot be meaningfully compared with another's...

In QFT... you can't expect the Poincare-based theory to remain good in all aspects. ... The accelerated observer is no longer in Minkowski spacetime, but rather in Rindler spacetime, with extra horizons, etc.

Ticks [measurements] are caused by some property of the object observed, such as charge, momentum, position, etc. Is the sea property of a field or of a particle? The ticks do not answer these question; an answer depends on the physical theory you are using to interpret the ticks...

I see I lost interest in that earlier thread when it got focused on ' reality'..philosophy...maybe Zapper closed it for a time...But I see the last post in that old thread refers to the Rovelli paper "What is a particle"...and that also led to similar discussions as this one...
It's a very worthwhile read for those interested in particle issues.

This was interesting, from:

" Horava-Lifgarbagez gravity, absolute time, and objective particles in curved space"

The absolute time involved in this theory allows to define an objective notion of particles associated with quantization of fields in classical gravitational backgrounds. The Unruh effect and other observer-dependent notions of particles in curved space are interpreted as effects caused by interaction between the objective vacuum and the measuring apparatus made up of objective particles...

[traditional particles in curved space-time] the notion of [particle] frequency depends on the choice of the time coordinate. On the other hand, the principle of covariance with respect to arbitrary spacetime coordinate transformations, which plays a crucial role in general relativity, implies that there is not any natural choice of the time coordinate. Consequently, there is no any natural definition of particles either.

...it is not clear which definition of particles, if any, corresponds to the particles that
contribute to the covariantly transforming energy-momentum tensor that determines
the gravitational field...

They go on to note that with their Horava-Lifgarbagez absolute time and a Schwarzschild black Hole:

...exact metric (7) contains two horizons at the outer horizon r+ ≈ 2M is close to the Schwarzschild horizon, while the inner horizon r− ≈ 0 is close to the singularity. ... the singularity on the horizon is not merely a coordinate singularity. This means that, contrary to the usual interpretation of Hawking radiation, all observers should agree that the black hole radiates, including the freely falling observers near the horizon...

So it is interesting to see that with an 'absolute time' everybody observes the same particles.

 

[lukesfn]

I've been trying to avoid the thread moving in this direction because the discussion can quickly become divergent. Although it may have some relevance, it may not be important to my original question. I've read that the mathematics for leaking horizons uses mathematics of fields and doesn't need a particle interpretation, and similar effects are expected in different analog systems with very different underlying physics.

I don't know enough about QFT for this discussion to be very helpful for me yet. I would love to learn more. Currently, I think a lot of confusion comes from words been given a specific definition when used in a theory which can be interpreted differently when considering the common speak English meaning. I don't know much about QFT, but I am suspicious that the the disagreement could actually be between what is a virtual particle and a real particle, or weather we really need to consider particles at all.

From wikipedia:

Although Unruh's prediction that an accelerating detector would see a thermal bath is not controversial, the interpretation of the transitions in the detector in the non-accelerating frame are. It is widely, although not universally, believed that each transition in the detector is accompanied by the emission of a particle, and that this particle will propagate to infinity and be seen as Unruh radiation.

If you look from the point of view of non-accelerated observer, perhaps we could consider the particle the accelerating observer receives is virtual which would make the partner in the pair it came from appear real, then it would be interpreted this as an emission of radiation from the observer. Anyway, thinking about it this way suddenly make me realize how the energy would add up even if one observer sees a particle being emitted and one received. Last time I asked, I didn't consider how the direction of momentum might work. It suddenly makes sense that a non accelerated observer could see thermal radiation that would appear to come from the accelerated observer. However, in the case of a BH, both observers see a horizon, so they could both see hawking radiation coming from it.

Getting back to my original question and looking at what happens to hawking radiation as you approach the singularity.

I will return to the formula I gave earlier. Under certain assumptions, this gives observed hawking radiation for a free faller where m is mass of the BH, r is the observers radius, and /r₀ is the radius the object was dropped from. (Before being dropped, the object was hovering at a constant radius)
1/(4m) (1 + 2m/r) (1 - 2m/r)^1/2
×{(1 - 2m/r₀)^1/2 + (2m/r×(1-r/r₀))^1/2 }^1/2
×{(1 - 2m/r₀)^1/2 - (2m/r×(1-r/r₀))^1/2 }^-1/2

Now if we simplify things by taking the limit /r₀→∞
1/(4m) (1 + 2m/r) (1 - 2m/r)^1/2
×{1 + (2m/r)^1/2 }^1/2
×{1 - (2m/r)^1/2 }^-1/2

In the approximation in the paper it came from, this formula now gives a temperature of hawking radiation for an observer free falling from ∞. This formula can be continued all the way to the singularity. (Note that I don't think that the paper ever tried to extend this formula inside the horizon, I am doing that on my on initiative) Anyway, towards the singularity, the observed temperature tends towards ∞. This implies that at some finite radius, the observer will see a radiation with energy greater then the mass of the BH. Therefore, I've shown that the speculation in my original post has some mathematical basis. I still find it quite an interesting question.

I guess that for a large BH, the wave length of this radiation near the singularity would be too short to make a lot of sense due to the Planck length. However, the radiation is being emitted in all directions so there should be a larger radius where the observer would expect to pass radiation emitted in all directions that would be greater then the BH. If we step back even further, the formula also shows that the free faller will be seeing the BH evaporate the entire way in towards the singularity, so the mass and energy levels might get smaller as we go in, changing the formula.

So I have a paradox, how can any singularity form, if this formula for observed hawking radiation seems to imply that a free falling observer appear to see the singularity evaporate before they get there?

 

[PeterDonis]

how can any singularity form, if this formula for observed hawking radiation seems to imply that a free falling observer appear to see the singularity evaporate before they get there?

I think the general opinion among physicists is that quantum effects do prevent a singularity from forming. (Note that "general opinion" does not mean "unanimous"--I think there are still plenty of physicists who don't share it.) The mechanism you're describing could, I suppose, be one way that could happen, although in what I have read on the subject (which is not a lot), I have not really seen any discussion of mechanisms by which quantum effects would prevent a singularity from forming, just general arguments about how they have to do so because unitarity has to be preserved.

However, even if quantum effects do prevent a singularity from forming, that does not mean they will prevent a horizon from forming. I don't see how the mechanism you describe could prevent the horizon from forming, since even by the formulas you refer to, the Hawking radiation observable by a free-falling observer at the horizon of any black hole of astronomical significance (stellar mass or larger) will be too faint to measure--certainly not sufficient to have any significant effect on the formation of the horizon.

 

[lukesfn]

I think the general opinion among physicists is that quantum effects do prevent a singularity from forming. (Note that "general opinion" does not mean "unanimous"--I think there are still plenty of physicists who don't share it.) The mechanism you're describing could, I suppose, be one way that could happen, although in what I have read on the subject (which is not a lot), I have not really seen any discussion of mechanisms by which quantum effects would prevent a singularity from forming, just general arguments about how they have to do so because unitarity has to be preserved.

I've also had the impression that there is a general opinion is that some kind of quantum effect probably occurs that prevents a singularity from being formed, however, all the discussion I've heard about this points to the need for new unifying physics to begin to explain how this might happen. I haven't heard about anybody investigating how semiclassical gravity analysis might prevent a singularity from forming. I would think people must have done plenty of analysis though.

I found a thread on another forum where somebody else asked the exact same question as me, but the following discussion was short, and unfortunately didn't lead anywhere helpful.

However, even if quantum effects do prevent a singularity from forming, that does not mean they will prevent a horizon from forming.

100% Agreed. As I understand it, a horizon exist whenever a large enough mass is contained with in a small enough radius.

 

[lukesfn]

I never found a good answer to this question, and I have kept wondering what it means that an observer falling into a BH, at some point before the singularity, should observer a temperature greater then the energy of the BH.

I've had a couple thoughts though. One is that I am not sure about how the temperature relates to energy. I guess the intensity would also be important. I'm not sure if the formulas showed what happened to the intensity, but I don't suppose it drops away, intuitively, I would expect the intensity to increase as well.

Another important question is, what happens to the BH relativistic mass (Or energy) in the frame of the infaller? Does the energy of the BH increase so that the level of radiation observed coming from type BH never exceed that of the BH? That might make some kind of sense, however, I am pretty confused about how the maths might work out in that case.

Either-way, I guess some pretty weird things are going to start happening at those energy levels.

 

[dauto]

Let me add another voice to the discussion. The free falling observers never see any Hawking radiation. Not before they cross the Horizon, not while they cross the Horizon, not after they cross the horizon. That's a simple consequence of the equivalence principle valid for any B-hole no matter its mass.

 

[PeterDonis]

an observer falling into a BH, at some point before the singularity, should observer a temperature greater then the energy of the BH.

Do you have any references where you've seen this discussed? I haven't seen any calculations of the behavior of quantum fields close to the singularity, so I'm not sure what the basis is for the temperature behavior you're referring to.

what happens to the BH relativistic mass (Or energy) in the frame of the infaller?

This isn't really a meaningful question; to an infaller that's falling through the horizon, the BH isn't an "object" that the concept of "relativistic mass" or energy can be applied to; that sort of concept can only be meaningfully applied to the BH by an observer very far away, to whom the BH just looks like an ordinary gravitating object. There is an $M$ parameter in the metric, but that parameter doesn't depend on the observer's state of motion.