PDA

View Full Version : Need clarification on Black Hole Concepts



mmaayeh
2010-Aug-09, 06:05 PM
I just got a chance to listen to a very good podcast by the Silicon Valley Astronomy lecture series:

Dr. Alex Filippenko (University of California, Berkeley): "Hearts of Darkness: Black Holes in Space"
May 19, 2010

After listening to the lecture, I was trying to get my head around some concepts I thought I understood and new concepts I don't understand. But, it was fun and enjoyable to listen to so, I need to listen to it again.

However, I have some questions on the following concepts discussed:


Hawking Radiation and negative mass/energy. I am not sure if I miss understood the lecture but, it sounded like that negative mass is considered when one of the virtual particles fall into the black hole and the other is ejected. The balance is negative energy because one virtual particle is expelled? Is this the correct concept or is it literally "negative energy" falling in?
Membrane paradox -- can some one clarify this point. I am not sure I understand this at all. What is this membrane and what is the nature of this membrane.
Matter falling towards the horizon seems to be stuck there and never falling in. So, relative to an outside observer watching the one falling towards the horizon then, is the clock of the one falling into the event horizon slowing down to be essentially stopped so that they never fall in or they literally falling in?
Also, there was a concept that the space-time is highly curved so that it is rotated into another dimension. That was confusing for me as was the concept that space-time curves into one of the existing 3 dimensions or other dimensions. If other dimensions then, is this an indication of multiple dimensions other than our 3 dimensions and one of time. Or, everything is curved into a single dimension as everything falls into the singularity. Also, the concept of Membrane paradox was introduced again but, not sure how this plays into it.

grant hutchison
2010-Aug-09, 06:53 PM
1. The negative energy is, in effect, the result of "book-keeping" by a distant observer, who sees energy come out and the black hole lose mass. If you fell into the hole alongside the infalling particle, it would look entirely normal, with a normal rest-mass.
2. I don't know what the membrane paradox is; do you mean "information paradox" or "membrane paradigm"?
3. Stuff falling in never crosses the event horizon in the coordinates of a distant observer, but does cross the event horizon according to its own coordinates. Although an infaller sees normal time elapse on his own clock, an external observer has no time on his own worldline which is simultaneous with any of the infaller's time below the event horizon.
4. I think what you're remembering is that spacetime is so strongly curved below the event horizon that the time dimension is swapped with the radial spacial dimension. No extra dimensions are required.

Grant Hutchison

mugaliens
2010-Aug-10, 09:10 AM
...that spacetime is so strongly curved below the event horizon that the time dimension is swapped with the radial spacial dimension. No extra dimensions are required.

Grant Hutchison

Swapped? How so, Grant?

WaxRubiks
2010-Aug-10, 09:19 AM
how do you get the transfer from time dimension to spacial dimension? Is that gradual or an immediate jump?

caveman1917
2010-Aug-10, 10:07 AM
Swapped? How so, Grant?

This (http://members.cox.net/jhaldenwang/black_hole.htm) may help. Basically it has to do with an extra minus sign appearing from the EH inwards.

One could think of it this way, what is the difference between a time dimension and a spatial dimension?
A spatial dimension is bidirectional (one can go forwards and backwards), but a time dimension is unidirectional.
Once inside the EH the radial spatial dimension takes on that role, one can only go forwards along with it, it has switched from a bidirectional to a unidirectional nature.
The sidewalks moves so fast (>c) that it's impossible to 'run' fast enough against it, one is inevitably dragged along with it.

mmaayeh
2010-Aug-10, 10:19 AM
1. The negative energy is, in effect, the result of "book-keeping" by a distant observer, who sees energy come out and the black hole lose mass. If you fell into the hole alongside the infalling particle, it would look entirely normal, with a normal rest-mass.
Grant Hutchison
Oh okay, I thought it was more of these lines.



2. I don't know what the membrane paradox is; do you mean "information paradox" or "membrane paradigm"?
Grant Hutchison

Yes, I meant membrane paradigm and not "paradox". I just wondered what the nature and behavior of "membrane paradigm".

grant hutchison
2010-Aug-10, 10:22 AM
how do you get the transfer from time dimension to spacial dimension? Is that gradual or an immediate jump?If you plot the axes in suitable coordinates, the infaller's time and space axes close together like the blades of a pair of scissors, align when at the event horizon, and cross over inside the event horizon.
Of course the infaller notices nothing unusual, but it means that the Schwarzschild radial direction is "timelike" for those inside the horizon. The singularity no longer lies in a direction one can point at, but forms a barrier across future time. One cannot manoeuvre away from it: the best an infaller can do (sometimes, depending on his initial trajectory) is to manoeuvre on to a longer trajectory that arrives at the singularity a little later.

Grant Hutchison

mmaayeh
2010-Aug-10, 10:22 AM
This (http://members.cox.net/jhaldenwang/black_hole.htm) may help. Basically it has to do with an extra minus sign appearing from the EH inwards.
One could think of it this way, what is the difference between a time dimension and a spatial dimension?
A spatial dimension is bidirectional (one can go forwards and backwards), but a time dimension is unidirectional.
Once inside the EH the radial spatial dimension takes on that role, one can only go forwards along with it, it has switched from a bidirectional to a unidirectional nature.
The sidewalks moves so fast (>c) that it's impossible to 'run' fast enough against it, one is inevitably dragged along with it.

So, let me see if I understand this too. So, does this mean that the spatial dimension will behave like a time dimension, in some sense? Or, do I have this backwards? That it is all space and no-time within the EH? Or, how should we conceptualize this?

caveman1917
2010-Aug-10, 10:32 AM
So, let me see if I understand this too. So, does this mean that the spatial dimension will behave like a time dimension, in some sense? Or, do I have this backwards? That it is all space and no-time within the EH? Or, how should we conceptualize this?

Define: a = (1-2m/r). The EH is at (r = 2m).

The Swarzschild metric around a black hole is this (considering only radially infalling particles):
ds˛ = a*dt˛ - (1/a)*dr˛. Note that the only fundamental difference between the two dimensions is the sign.

Outside the EH: a > 0 (r > 2m)
Inside the EH: a < 0 (r < 2m)

But as a switches sign - so does t and r 'switch roles'.

But the metric is only a conceptualization from the perspective of an outside observer - how he would make sense of what's happening to our infaller. Extrapolating from theory - he can never know for sure. To the infaller everything seems normal - the only thing he notices is that his feet suddenly start to pull.

grant hutchison
2010-Aug-10, 10:34 AM
Yes, I meant membrane paradigm and not "paradox". I just wondered what the nature and behavior of "membrane paradigm".Kip Thorne used "paradigm" in the way Kuhn did, to suggest a particular way of thinking about something.
Because of the way spacetime behaves at the event horizon of a black hole, it turns out to be physically useful and informative to think of the event horizon as if it were a sort of electrically conducting membrane; one can even assign physical properties to this "membrane", such as electrical resistance. This appearance of membrane-like properties only occurs for stationary distant observers, however. An infalling observer encounters nothing unusual.

Grant Hutchison

grant hutchison
2010-Aug-10, 10:39 AM
So, let me see if I understand this too. So, does this mean that the spatial dimension will behave like a time dimension, in some sense?Yes, one of the infaller's spacial axes becomes "timelike" for an external observer plotting Schwarzschild coordinates.
This gross mismatch between the infaller's coordinates and those used by an external observer is why an external observer is never simultaneous with anything that happens inside the event horizon: he simply can't extend his own simultaneity across the event horizon and into the region of reversed coordinates.

Grant Hutchison

mmaayeh
2010-Aug-10, 11:00 AM
Define: a = (1-2m/r). The EH is at (r = 2m).

The Swarzschild metric around a black hole is this (considering only radially infalling particles):
ds˛ = a*dt˛ - (1/a)*dr˛. Note that the only fundamental difference between the two dimensions is the sign.

Outside the EH: a > 0 (r > 2m)
Inside the EH: a < 0 (r < 2m)

But as a switches sign - so does t and r 'switch roles'.

But the metric is only a conceptualization from the perspective of an outside observer - how he would make sense of what's happening to our infaller. Extrapolating from theory - he can never know for sure. To the infaller everything seems normal - the only thing he notices is that his feet suddenly start to pull.

So, this is just a change in reference with respect to an outside observer only and physically time is still time and space is still space with the infaller. But, what is interesting in the equation above is when r=2m where the metric goes to infinite. Not sure how to ask this but, what kind of nature must the EH have in order to have a metric at infinite?

mmaayeh
2010-Aug-10, 11:06 AM
Yes, one of the infaller's spacial axes becomes "timelike" for an external observer plotting Schwarzschild coordinates.
This gross mismatch between the infaller's coordinates and those used by an external observer is why an external observer is never simultaneous with anything that happens inside the event horizon: he simply can't extend his own simultaneity across the event horizon and into the region of reversed coordinates.

Grant Hutchison

As you and Caveman pointed out that the reference for the outside observer is flipped after an infaller crosses the EH. Which in the end truly means that the infaller is cut off from the rest of the universe.

mmaayeh
2010-Aug-10, 11:16 AM
Kip Thorne used "paradigm" in the way Kuhn did, to suggest a particular way of thinking about something.
Because of the way spacetime behaves at the event horizon of a black hole, it turns out to be physically useful and informative to think of the event horizon as if it were a sort of electrically conducting membrane; one can even assign physical properties to this "membrane", such as electrical resistance. This appearance of membrane-like properties only occurs for stationary distant observers, however. An infalling observer encounters nothing unusual.

Grant Hutchison

So, this is a construct more or less and not an actual physical phenomenon, as indicated in this thread.

caveman1917
2010-Aug-10, 11:19 AM
So, this is just a change in reference with respect to an outside observer only and physically time is still time and space is still space with the infaller.

Yes. But it is a physical reality to the outside observer, by virtue of him extending his coordinates - this is how he conceptualizes what is 'really physically' going on.
Every observer is his own judge as to his own physical reality. This is more like a disagreement between two observers. However this disagreement is so fundamental that it could never be allowed, and that's where the EH horizon conveniently saves the day again. Our outside observer will never 'observe' himself being wrong about this, so there is no physical contradiction popping up in his view, his perspective - to him - is as physical as the infaller's perspective is to the infaller.


But, what is interesting in the equation above is when r=2m where the metric goes to infinite. Not sure how to ask this but, what kind of nature must the EH have in order to have a metric at infinite?

Yes indeed. It reaches a singularity at the EH.
For this it's important to see that there are two fundamentally different kinds of singularities, physical singularities and coordinate singularities.
Only the former are 'real'.
What happens with the latter is that in some chosen coordinate system (swarzschild metric) a singularity pops up. But choosing a different coordinate system the singularity goes away. This means the singularity was a weakness of the chosen coordinate system, not a 'true' singularity. For that to be there should exist no allowed reference frame in which the singularity doesn't exist - such as is the case with the singularity at the center of a black hole.

It was thought for a while the swarzschild singularity was real, it took many years for a coordinate system to be found that didn't contain a singularity at r = 2m.

grant hutchison
2010-Aug-10, 01:15 PM
So, this is a construct more or less and not an actual physical phenomenon, as indicated in this thread.It's a construct that gives the right answers to certain physical questions, and to some extent it can be explored by "real" observers. If we lower an observer quasi-statically towards the event horizon, he encounters a hot sea of particles just above the event horizon, with a temperature that rises to infinity as the event horizon is approached ever more closely. This particle sea is analogous to the Unruh radiation encountered by a continuously accelerating observer in flat space. If we calculate the spectrum of particles which reach infinity, it gives us Hawking radiation. And (so I understand) the electrical properties of this plasma match those of the "membrane" used in the membrane paradigm. So there is an actual physical phenomenon associated with the paradigm; it's just not observable by all observers: which is a recurring theme in black hole physics.

Grant Hutchison

Ken G
2010-Aug-10, 04:14 PM
Indeed, the forces required to keep that observer quasi-static as he/she approaches the event horizon would be spectacular. Since the free-fall observer doesn't see the bath of Unruh radiation, it suggests that it is not the event horizon that is "responsible" for the "membrane", but rather the spectacular force. The membrane sounds quite similar to what is called the "Rindler horizon" of an accelerating observer in free space with no black hole around. So those do not actually appear to be properties of black holes themselves-- rather, the black hole just presents a curvature of spacetime that when we coordinatize using special-relativity-inspired approaches, the concept of "staying in the same place" stands in for what we normally think of as "having a spectacular acceleration."

But how that connects with Hawking radiation is mysterious to me. For one thing, Hawking radiation of a normal black hole is so minuscule that there's no chance of observing it in any way I know of. Indeed, I'm not sure there's another example of a widely held physical theory that is so far from ever having a shred of direct experimental support-- it's all just faith in how general relativity and quantum mechanics will work when they overlap, even though the two theories have only ever been checked independently. I believe that Unruh radiation has been tested, but as we've seen, the black hole is not playing any fundamental role in the Unruh radiation, it's the force on the particle that is doing that. Yet somehow, this connects to Hawking radiation, and that is something that should be happening to the black hole.

grant hutchison
2010-Aug-10, 04:35 PM
Taylor and Wheeler make the mathematical connection between Unruh, the acceleration required to stay stationary is the Schwarzschild metric, and the Hawking radiation detected by a distant observer. The Unruh temperature climbs towards infinity as one approaches the event horizon quasistatically, but the radiation from each layer is gravitationally red-shifted by an amount which makes each successive layer of Unruh radiation "look" like the Hawking temperature to a distant stationary observer.

But I realize you're talking about the deeper significance of being able to obtain Hawking from Unruh in this way. It does seem like we're being told something important by the fact that we can "get at" Hawking radiation by using an approach that never crosses the event horizon.

Grant Hutchison

mmaayeh
2010-Aug-10, 04:50 PM
Yes. But it is a physical reality to the outside observer, by virtue of him extending his coordinates - this is how he conceptualizes what is 'really physically' going on.
Every observer is his own judge as to his own physical reality. This is more like a disagreement between two observers. However this disagreement is so fundamental that it could never be allowed, and that's where the EH horizon conveniently saves the day again. Our outside observer will never 'observe' himself being wrong about this, so there is no physical contradiction popping up in his view, his perspective - to him - is as physical as the infaller's perspective is to the infaller.



Yes indeed. It reaches a singularity at the EH.
For this it's important to see that there are two fundamentally different kinds of singularities, physical singularities and coordinate singularities.
Only the former are 'real'.
What happens with the latter is that in some chosen coordinate system (swarzschild metric) a singularity pops up. But choosing a different coordinate system the singularity goes away. This means the singularity was a weakness of the chosen coordinate system, not a 'true' singularity. For that to be there should exist no allowed reference frame in which the singularity doesn't exist - such as is the case with the singularity at the center of a black hole.

It was thought for a while the swarzschild singularity was real, it took many years for a coordinate system to be found that didn't contain a singularity at r = 2m.

Man that is amazing. So, would you say that the singularity at the center of a black hole is a physical singularity or it is due to a coordinate singularity. I imagine you are talking about a physical singularity at this point.

Also, I am sure it has been done but, what would happen if you place the center of the coordinate at the EH? Would you see a different kind of nature when looking out outside of the black hole vs. inside the black hole (if that is possible?)?

Ken G
2010-Aug-10, 05:36 PM
But I realize you're talking about the deeper significance of being able to obtain Hawking from Unruh in this way. It does seem like we're being told something important by the fact that we can "get at" Hawking radiation by using an approach that never crosses the event horizon.
Exactly, it makes it seem like the Hawking radiation, at any distance from the black hole, is nothing but the Unruh radiation stemming from the proper acceleration of the observer. The latter is something the observer is doing, arbitrarily trying to stay "stationary" in space curved by the black hole, but doesn't seem to have anything directly to do with the black hole. How does such a thing make a real black hole evaporate, and why would it even be there were there no observers with forces on them to witness this Unruh radiation?

It must be that Unruh radiation is not exactly created by the acceleration of the observer, but rather, it is in some sense "always there" but simply not visible until you accelerate. In other words, the radiation is part of the physical manifestation of the spacetime, but that manifestation is fully frame dependent-- it's not just that some space looks like some time or vice versa if you change the observer motion, there must be a whole suite of physical effects that are similarly swapped around, including a background of thermal radiation. As usual, the "story" we tell for "why that radiation is there" can vary, but the physical outcomes due to its presence must be the same.

publius
2010-Aug-10, 09:07 PM
Remember the difference between Rindler and a black hole is the presence of real curvature. While I don't understand it, that is the key to the difference I think. As the mass of black hole increases, the invariant curvature at the horizon (this goes simply as the magnitude of the tidal force, ~1/r^3) decreases, and as the black hole gets bigger, the Hawking radiation decreases and the time to "evaporate" increases. Thus in the limit of large mass, the black hole horizon is like a Rindler horizon. Indeed, the EH would be a larger and large sphere and begin to look like then flat plane of Rindler.

But for a small black hole, the curvature gets significant at the EH and things are different. The "real" aspect of Hawking must have to do with the invariant curvature.

The way I understand it, and this is not a simple thing so take my understanding with a grain of salt, is that Unruh radiation is due to the coordinate change of going to accelerating coordinates. The effect of that coordinate change on the QFT equations is to change how the vacuum state "looks". What is an empty, zero particle state, now has particles in it. This is not a change of the vacuum, only how you see it.

Add real curvature to the mix, and something more happens, something that might be considered real. But again, this is beyond my understanding.

-Richard

caveman1917
2010-Aug-11, 02:06 AM
Exactly, it makes it seem like the Hawking radiation, at any distance from the black hole, is nothing but the Unruh radiation stemming from the proper acceleration of the observer. The latter is something the observer is doing, arbitrarily trying to stay "stationary" in space curved by the black hole, but doesn't seem to have anything directly to do with the black hole. How does such a thing make a real black hole evaporate, and why would it even be there were there no observers with forces on them to witness this Unruh radiation?

I didn't do the math on this, but intuition would suggest something in the way that it might be directly related to the black hole.

We commonly think of a black hole as a singularity surrounded by an EH.
We are causally disconnected by the EH from anything in between, what really happens is that we say that our theory predicts that - in order to have invariant curvature of such magnitude at this spherical radius - there must be an invariant singularity of such mass at the center.
So disregarding this extrapolation (which may nevertheless very well be correct) we end up with defining a black hole as some specific curvature outside an EH.


As the mass of black hole increases, the invariant curvature at the horizon (this goes simply as the magnitude of the tidal force, ~1/r^3) decreases, and as the black hole gets bigger, the Hawking radiation decreases and the time to "evaporate" increases.

We may then consider Hawking radiation as an effect of the curvature of spacetime at (or very near) the EH.
So we can define Hawking radiation as expressing (curvature => radiation).

To the Rindler observer. The curvature he sees is only a coordinate effect. However to him it is real.
But also, introducing this curvature - he has 'done away' with him being accelerating, the two perspectives are equivalent - and so should they. Either accelerating in flat spacetime or interial in curved spacetime. Considering Unruh radiation as the effect of acceleration and using the equivalence we can also define Unruh radiation as expressing (curvature => radiation).

Since both observers are intertial and experience the same conditions - any physical result must be the same.

The dichotomy goes away if we consider it one thing: Hawking-Unruh radiation, expressing that curvature means spacetime radiates.
We take them as different because we started of from two different extremal conditions, interial with high invariant curvature on onehand, and accelerating in flat conditions on the other hand.

What does it mean when we say invariant? It only says that if i experience something - so does the rest. The relevant curvature is the total from the invariant part (which must be there by definition) and the part the observer himself introduces by his own coordinate choice - such is the experience from the relevant observer.

This would however mean that either the EH of a Rindler observer would slowly recede, or that Hawking radiation actually doesn't decrease the mass of the black hole.

Ken G
2010-Aug-11, 05:33 AM
The effect of that coordinate change on the QFT equations is to change how the vacuum state "looks". What is an empty, zero particle state, now has particles in it. This is not a change of the vacuum, only how you see it.

Add real curvature to the mix, and something more happens, something that might be considered real. But again, this is beyond my understanding.
Yes, it seems that "Bogoliubov transformations" are the key to see how the vacuum looks different for different observers, but the difference if the curvature is "real" is that you need to do a Bogoliubov transformation to "stay in the same place" in some meaningful global sense. Hawking evaporation must be related to that global state.

Strange
2010-Aug-11, 12:37 PM
I naively (as the math is waaay beyond me) assumed that the Rindler/event horizons and Unruh/Hawking radiation were equivalent in the same way that acceleration and gravity are equivalent. As is so often the case, it seems it ain't that simple...

Ken G
2010-Aug-11, 02:47 PM
I naively (as the math is waaay beyond me) assumed that the Rindler/event horizons and Unruh/Hawking radiation were equivalent in the same way that acceleration and gravity are equivalent. As is so often the case, it seems it ain't that simple...Yes, that's what we're struggling with too-- it's clear that the equivalence principle is at play here, but there appears to be something else as well. Unruh radiation is just another way to look at a vacuum, it is in some sense "already in the vacuum", so it can't change the vacuum. But Hawking radiation is normally said to change the black hole-- it can evaporate it (for pathologically small black holes that have never been found). It sounds like there is something fundamentally unstable about setting a black hole down in a vacuum-- there is something "already in" the vacuum that will "cause" the black hole to evaporate, and one way to see that is to look at it from a perspective where it appears to be Unruh radiation that makes the black hole go away.

It also makes one wonder if there is a perspective where the black hole does not evaporate at all, just as there are perspectives where the Unruh radiation is not seen. That would have to mean Hawking was wrong to surrender the famous bet-- if some perspective has the black hole not evaporating, then the "information" cannot come out.

Strange
2010-Aug-11, 05:36 PM
Yes, that's what we're struggling with too

Well that is oddly reassuring :)


It also makes one wonder if there is a perspective where the black hole does not evaporate at all

Would someone in free fall towards the black hole still see the Hawking radiation?

loglo
2010-Aug-11, 07:58 PM
Yes, that's what we're struggling with too-- it's clear that the equivalence principle is at play here, but there appears to be something else as well. Unruh radiation is just another way to look at a vacuum, it is in some sense "already in the vacuum", so it can't change the vacuum. But Hawking radiation is normally said to change the black hole-- it can evaporate it (for pathologically small black holes that have never been found). It sounds like there is something fundamentally unstable about setting a black hole down in a vacuum-- there is something "already in" the vacuum that will "cause" the black hole to evaporate, and one way to see that is to look at it from a perspective where it appears to be Unruh radiation that makes the black hole go away.

It also makes one wonder if there is a perspective where the black hole does not evaporate at all, just as there are perspectives where the Unruh radiation is not seen. That would have to mean Hawking was wrong to surrender the famous bet-- if some perspective has the black hole not evaporating, then the "information" cannot come out.

From what I can tell the perspective Hawking is taking is that the non-evaporating (eternal) black hole is one of the states being summed over in the path integral but it is not an "observable" since he is "taking measurements from infinity" like a particle scattering experiment.

In Hawking's own words (http://math.ucr.edu/home/baez/week207.html):-

" Information is lost in topologically nontrivial metrics, like the eternal black hole. On the other hand, information is preserved in topologically trivial metrics. The confusion and paradox arose because people thought classically, in terms of a single topology for spacetime. It was either R4, or a black hole. But the Feynman sum over histories allows it to be both at once. One can not tell which topology contributed the observation, any more than one can tell which slit the electron went through, in the two slits experiment. All that observation at infinity can determine is that there is a unitary mapping from initial states to final, and that information is not lost.

My work with Hartle showed the radiation could be thought of as tunnelling out from inside the black hole. It was therefore not unreasonable to suppose that it could carry information out of the black hole. This explains how a black hole can form, and then give out the information about what is inside it, while remaining topologically trivial."

I took this from the old Baez page linked above that was created shortly after Hawking's announcement as it was the only place where I could see any analysis. I see his paper (http://prd.aps.org/abstract/PRD/v72/i8/e084013) has 140 citations so it must have made an impact of some sort. Does he anyone know what came of it all? Is it really settled?

One thing I missed when I read this last was the reference to work with Hartle showing how Hawking radiation could be treated as particles tunnelling out of the black hole. Does anyone know which paper(s) he is referring to specifically?

mugaliens
2010-Aug-11, 09:35 PM
Hey, loglo - outstanding question!


One thing I missed when I read this last was the reference to work with Hartle showing how Hawking radiation could be treated as particles tunnelling out of the black hole. Does anyone know which paper(s) he is referring to specifically?

I was going to say, "I do not agree with Hawking's quote, "My work with Hartle showed the radiation could be thought of as tunnelling out from inside the black hole. It was therefore not unreasonable to suppose that it could carry information out of the black hole. This explains how a black hole can form, and then give out the information about what is inside it, while remaining topologically trivial.""

Then, I thought about it.

caveman1917
2010-Aug-11, 10:14 PM
Would someone in free fall towards the black hole still see the Hawking radiation?

Yes, just as well as anyone inertial from very far away would see it. Hawking radiation is the effect of invariant curvature (and thus an invariant EH) around a BH, it is the same for all inertial observers.

Rindler observers will also create an EH - it is variant (due to their acceleration) - but to them it is just as real as sitting inertially outside a black hole.

grant hutchison
2010-Aug-11, 10:36 PM
Interesting, though, that the infaller and the hoverer will have different explanations for where the Hawking radiation comes from. The infaller sees virtual pairs being promoted to reality by tidal gravity; one member of the pair then escapes to become Hawking radiation at infinity. The hoverer near the horizon inhabits a sea of real particles, some of which evaporate to become Hawking radiation at infinity.

Grant Hutchison

publius
2010-Aug-11, 11:32 PM
Rindler observers will also create an EH - it is variant (due to their acceleration) - but to them it is just as real as sitting inertially outside a black hole.

Actually the Rindler horizon is an invariant -- it's just that is only "important" for the particular Rindler "family" in aligns with. To inertial (free falling) observers, the Rindler horizon is a plane moving at light speed, chasing the Rindler family, but never quite catching it. Inertial observers see it get asympotically close, but never make it. To the Rindler family, it stays a constant distant behind them. If an observer changes his acceleration, the "old" horizon can be thought as now moving, and a "new" horizon, another null surface takes its place.

Note the null surface part. You'll note that ds on the horizon is zero, whether it's a Rindler horizon or a Schwarzschild horizon. That shows it is "moving at light speed" even though it is stationary for the Rindler and Schwarzschild families. And note that inertial observers confirm this "moving at light speed" thing. Inertial observers in flat space-time see the Rindler horizon as a plane moving at light speed and ds is obviously zero for that obvious null path.

Different observers coordinatize that horizon in different ways, but it is the same set of events in space-time. It's just a particular set is more significant in certain ways to different observers.

caveman1917
2010-Aug-11, 11:47 PM
To the Rindler family, it [the EH] stays a constant distant behind them.
added [the EH]

Does Unruh radiation make that distance increase for Rindler observers in the same sense Hawking radiation (supposedly) makes the EH recede for Swarzschild observers?

publius
2010-Aug-12, 01:27 AM
added [the EH]

Does Unruh radiation make that distance increase for Rindler observers in the same sense Hawking radiation (supposedly) makes the EH recede for Swarzschild observers?

Good question. I don't really know the answer. But let's make sure we frame the question properly. If our Rindler observers are just pure mathematical observers, in the sense of riding some abstract reference frame, the answer is no. It has to be no. Minkowski space-time is just that. It doesn't matter what reference frame we observe it in. Now, how about real accelerating observers. Do they interact somehow with the Unruh radiation and how does that affect things? That's the part I don't know.

But we can sort of approach this from an Equivalence Principle point of view. This gets complex and I don't understand it enough to be really talking about, but I'll plow on. Newtonian gravity: and infinite plane sheet of mass makes a uniform gravitational field, constant g throughout all the half space about it, and a constant g in the opposite direction on the other side. In GR, such a thing ain't such a simple animal -- indeed the concept of a uniform gravitational field in GR is something of a will-o-the-wisp. But Rindler can be constructed from such a thing, albeit in a very strange way. To an inertial observer, the free faller, the mass distribution is an infinite plane, but one that is, <ahem> moving at light speed towards the observer, then bounces back at x = 0. :lol: That sucker is actually something different from flat space-time, but the half space above it will match up with the Rindler observers, who only see "half" (or is it a quarter, I forget) the space-time anyway . IOW, the infinite plane is right there at the Rindler horizon. Note space-time will be flat to the inertial observer because the infinite sheet is coming at light speed. It's causal effects get there no sooner than it does. After it "hits", who knows what happens. Again, we need a real expert on this to explain how it works.

So, the effect of all that for our purposes, is the amount of mass energy behind the RIndler horizon is infinite, and thus no finite flux of radiation is going to reduce it. :)

-Richard

publius
2010-Aug-12, 02:09 AM
Here is a recent paper related to this subject, the notion of the uniform field in General Relativity:

http://arxiv.org/abs/1003.3022

Note the name Rindler is never mentioned, but that's the constant proper acceleration metric. Note the other one defining constant acceleration -- that metric is flat only locally. And note the last part where they start on the gravitational side. They come up with a space-time that is indeed singular on a spatial plane -- the infinite mass sheet so to speak, although they don't describe it as such (because probably such a description is not the most accurate in the world. :) ) and that agrees with Rindler for those "above it", and is flat above it.

I encourage you all to read that paper. The authors approach things from a true GR way of seeing the world. What do we mean by "acceleration". Are you sure it means what you think it means? Let's make sure it does. And you see they come to the conclusion that what is meant by "uniform gravitational field" in a relativistic sense cannot be the same thing as we think it means from the classical limit. IOW, Rindler it is and a uniform gravitional field is indeed a *flat* space-time, with the varying proper acceleration along the Rindler ruler (the rocket and rope thing) with the exception of that plane singularity for a "real" mass presence.

-Richard

Ken G
2010-Aug-12, 05:35 AM
So, the effect of all that for our purposes, is the amount of mass energy behind the RIndler horizon is infinite, and thus no finite flux of radiation is going to reduce it.I don't know the answer to caveman1917's question either, but I think there might be a problem with this one. Doesn't the infinite plane of mass that a Rindler observer might perceive is hanging out at the Rindler horizon a finite surface mass density whose value depends on the Rindler proper acceleration? So it's only infinite by virtue of having infinite lateral extent. But then, so would the Unruh radiation-- it would be per square area, and per square area the mass is finite. Is it possible that the Rindler observer does indeed recede from the horizon, such that both his distance from it, and its apparent mass density, depends not only on his proper acceleration, but also the history of how long he has maintained it? After all, Unruh radiation is not a purely GR result, so might not appear in a purely GR Rindler analysis-- you have to connect the GR solutions to a quantum mechanical treatment of the vacuum.

Another question I have is, if an accelerated observer sees a thermal bath, cannot that bath cause, say, excitations in atoms carried by the accelerated observer? What interpretation would an inertial observer give to such a transition-- can accelerated atoms spontaneously excite without requiring there be any photons present?

publius
2010-Aug-12, 06:16 AM
Ken,

I just don't know. The Rindler horizon is a null surface and the mass sheet is thus moving at light speed (I think) -- the sheet is right on the horizon. Thus the energy density of the sheet would indeed be infinite, per square area. But I just don't know for sure. Note in that paper they talk of a singularity on the plane. (You may be on to something about the history thing -- Rindler and these relations are about eternal acceleration)

And I'm baffled as well about the interaction of an accelerating observer with the radiation flux he sees. And this question is not "GR" as there's no space-time curvature, no "real gravity", afoot at all! Indeed, if we take the POV of an inertial observer this is purely SR and flat space-time and how QFT plays out there. QFT is supposed to be fully SR relativistic. If that is so, then there has to be something whereby the inertial observer would predict whatever interactions the accelerating observer sees as caused by the Unruh flux. Unruh should live entirely in the QFT-SR theory realm.


-Richard

Ken G
2010-Aug-12, 06:47 AM
I just don't know. The Rindler horizon is a null surface and the mass sheet is thus moving at light speed (I think) -- the sheet is right on the horizon. Thus the energy density of the sheet would indeed be infinite, per square area. Not to the Rindler observer though, right? To them, should it not be more or less the Newtonian mass density that corresponds to their proper acceleration? I think you made a good point about how things look to the inertial observer in the equivalent-mass-sheet formulation-- the mass sheet is moving at c, and the spacetime has not yet "felt" its presence, but that's all from the inertial perspective-- the accelerating observer who is hovering in an equivalence-principle balance with the perceived gravity seems more straightforward to think about.
Note in that paper they talk of a singularity on the plane. (You may be on to something about the history thing -- Rindler and these relations are about eternal acceleration)That raises another question about Unruh radiation-- the places I've seen it talked about seemed to imply that you perceive a thermal bath in the vacuum as soon as you have proper acceleration, not that you need to have maintained that proper acceleration for some time (like, forever!). In special relativity, the standard coordinates can be attributed a physical meaning only for inertial observers who have always been inertial, not for those who are instantly comoving with such an eternally inertial observer (that's the "Andromeda paradox" business). It sounds like you are saying that Rindler coordinates can only be attributed physical meaning (and have the equivalence principle applied to them, for example) for observers who have always had that same proper acceleration. If so, that might present difficulties in using the equivalence principle-- if it leads to the kinds of paradoxes we are grappling with, maybe it is impossible to maintain a fixed proper acceleration forever.

Given how tricky this all is, perhaps we first need the "simpler" question answered:

And I'm baffled as well about the interaction of an accelerating observer with the radiation flux he sees. And this question is not "GR" as there's no space-time curvature, no "real gravity", afoot at all! Indeed, if we take the POV of an inertial observer this is purely SR and flat space-time and how QFT plays out there. QFT is supposed to be fully SR relativistic. If that is so, then there has to be something whereby the inertial observer would predict whatever interactions the accelerating observer sees as caused by the Unruh flux. Unruh should live entirely in the QFT-SR theory realm.Yes, I don't get that either. In fact, I think this might be an area of expert dispute. The Wiki makes some bizarre statements that experts agree on the "Unruh effect", which is a thermodynamic equilibrium at the Unruh temperature (the "warm bath" for an accelerating observer), but they don't agree on "Unruh radiation", which sounds like they are talking about emissions by the accelerating observer, not the bath of radiation that such an observer sees. They certainly don't give us any answer for the seemingly invariant excitations that the Unruh effect should be causing in atoms carried by the accelerating observer.

The Wiki seems a bit at sea here, and clearer answers from other sources are hard to come by, but it seems hard to conclude anything else but that atoms that are accelerating for an inertial observer must appear to that observer to be maintaining an equilibrium excitation by what seems like an equilibrium with virtual particles. They must look like atoms that get excited for no apparent reason, and then undergo radiative de-excitations without creating real photons. Perhaps that also dovetails with Hawking radiation-- an atom suspended just above the event horizon would also appear, to the inertial observer at infinity, to be excited for no reason, and not make real photons when it de-excites. However, they might then conclude that the atom was being excited by virtual photons coming from the vicinity of the EH, and when they de-excite, they create virtual photons, but those virtual photons do get "promoted" into real photons after that, by the "tidal gravity".

I guess what I'm saying is, it sounds like inertial observers will always interpret both Unruh radiation and Hawking radiation as virtual photons, which can alter the excitation of real atoms without there being any real photons around (the energy must come from whatever is accelerating the atoms). The difference is, the presence of a real event horizon (again as perceived by the inertial observer) in the black-hole case allows the virtual photons to be "promoted" into real photons, by virtue of the fact that negative-energy particles are allowed on the "other side" of a real event horizon. That would be why Hawking radiation can yield evaporation but Unruh radiation cannot-- in the latter case there's no real event horizon for the inertial observer to perceive, so the virtual photons can't move any real mass around by sidestepping the interdiction against negative energy particles. (The observer with proper acceleration thinks there's a horizon there in either case, so they do see evaporation as the virtual particles are being promoted to real particles either way, but maybe in the Rindler case it is allowed for the accelerating observer to think the mass sheet is evaporating while the inertial observer sees that there really was never any mass sheet there in the first place).

publius
2010-Aug-12, 07:03 AM
Ken,

You know, now I'm just stumped. :lol: I always assumed that Unruh was "instant" as well.

And second, I got to reading and I'm just not sure anymore about the "infinite mass sheet" business and Rindler. I think there's something there but I'm just no longer sure if such a thing works like I thought it does. There's always trouble when you play with infinity. So the Equivalence Principle insipired question of does the Rindler observer think his horizon is receding or "negative mass" crossing it, I'm now just going to punt and say I don't know.

Yes, we'll worry about the answer to the "simpler" question first.


-Richard

Ken G
2010-Aug-12, 07:34 AM
I edited my previous remark to include a speculation about the role of virtual particles here. I think you were right in your answer to caveman1917, that an observer with constant proper acceleration does not perceive any change in the mass sheet or their distance from it. So they might say that there is a net flux of virtual particles that are being promoted to real particles by virtue of having negative-energy partners going to "minus infinity" on the other side of the Rindler horizon. That doesn't evaporate the mass sheet, because unlike with the black hole, there are now "three places" of interest-- the horizon, the observer, and "minus infinity"-- there's a noncompact topology.

ETA: Nope, I'm wrong-- sending negative energy photons off to minus infinity would still reduce the gravity of the Rindler mass sheet, as soon as the positive energy photons passed by the Rindler observer. So that would still evaporate the mass sheet and cause the Rindler observer to separate from the Rindler horizon, which doesn't make any sense if they are maintaining fixed proper acceleration. I'm stumped. I cannot reconcile the views of an accelerating observer, and an inertial one at the same point, whose motion is only instantaneously known, and only one sees Unruh radiation, versus the same scenario with real tidal gravity of a black hole, and both see Hawking radiation. For example, I can't tell if you have two observers, both hovering just outside an EH of a real black hole, and one is suddenly dropped into free fall, does the one still hovering see both Unruh radiation and Hawking radiation, while the free faller sees only Hawking? And how does the hoverer know that the Hawking radiation will evaporate the black hole, but the Unruh radiation won't?

publius
2010-Aug-12, 07:58 PM
Something else just hit me: the stress-energy tensor. If the accelerating sees a "thermal bath", then by Jimminy that means he's seeing a non-zero stress-energy tensor all around him. The vacuum is not empty. The stress-energy tensor must be invariant, which would imply the *inertial observer* must see the same invariant (although it will resolve differently in different coordinates). That means we're no longer talking empty space-time!

And that makes sense, as the quantum vacuum is full of something (but we know how that works out trying to reconcile that with the cosmological constant). So I was wrong about Unruh being the domain of plain SR and QFT, it is very much in the realm of merging GR with QFT! Unruh is a result of QFT, and thus, even for inertial observers, we're not talking empty space-time. It can't be empty, because it would have to be empty for all observers, due to the invariance of the stress-energy tensor. If something is there for one observer, it's there for all.

The questions we have are not going to be resolved until we get a working merger of QFT and GR. Searching around, I see abstracts about the Unruh and Hawking stress-energy tensors, which apparently are just various approximate derivations (good guesses!). AFAIK, no workable solution for an evaporating black hole space-time has ever been worked out and probably won't until we have a quantum theory of gravity or whatever is required to marry GR and QFT.

publius
2010-Aug-13, 04:03 AM
Ken,

In digging around, I stumbled on this post from a thread in Physics Forums:

http://www.physicsforums.com/showpost.php?p=887929&postcount=53



Your point about the consistency required between the views different observers take for the Rindler spacetime Unruh effect is a good one. In this Unruh effect, suppose the quantum field in is the vacuum of the inertial observer. Then, the expectation values of the components of the stress energy tensor for this observer are all zero. Since the stress-energy tensor is a tensor, the values of the components of the stress energy tensor are all zero in every coordinate system, including the coordinate system of the accelerated observer.

What is happening here? How does the non-accelerated observer feel a temperature? To explain this, I'll quote a passage from Birrell and Davies about how an idealized accelerated particle detector reponds to the inertial vacuum.

"The explanation comes from a consideration of the agency that brings about the acceleration of the detector in the first place. As the detector accelerates, its coupling to the field causes the emission of quanta, which produces a resistance against the accelerating force. The work done by the external force to overcome this resistance supplies the missing energy that feeds into the field via the quanta emitted from the detector, and also into the detector which simulaneously makes upward transitions. But as far as the detector is concerned, the net affect is the absorption of thermally distributed quanta."

publius
2010-Aug-13, 04:21 AM
So apparently we have this:

To an inertial observer we have a quantum field thingy in the vacuum state -- the expectation values of all components of the stress-energy tensor are zero. They thus must be zero for all observers. The "Boggy-Boo" transforms are just a coordinate effect. They transform virtual particles to real particles and vice-versa but they do not change the expectation values of the invariant stress-energy tensor.

An atom is accelerated. The Rindler frame of that atom sees a radiation bath and is possibly excited by that bath. The inertial observer watching that sees the coupling of that atom to the field as causing resistance to its acceleration. It must supply energy to overcome that resistance and that is what excites the atom.

-Richard

Ken G
2010-Aug-13, 04:48 AM
Well, at least we can forgive ourselves for not knowing the answers. What I don't get is why we are expected to take Hawking radiation as gospel-- it will never be observed, and it doesn't emerge from a consistent theory that can answer even simple questions like what is the difference between Hawking and Unruh radiation.

publius
2010-Aug-13, 05:13 AM
Well, at least we can forgive ourselves for not knowing the answers.

You were right on track in post #37, at least according to the Birrell and Davies book quoted in that PF post:

http://www.amazon.com/Quantum-Cambridge-Monographs-Mathematical-Physics/dp/0521278589

To the inertial observer, the accelerated system "gets hot" because of the work done by accelerating against the quantum field. To the Rindler observer accelerating with it, it gets hot by an equilibrium with the Unruh radiation. To the inertial observer this hot system is emitting real photons.

Now, as to how Hawking radiation becomes really real and evaporates a black hole, I can only wave my hands and say "real curvature" I suppose. :lol:

-Richard

caveman1917
2010-Aug-13, 09:21 AM
So it seems the Rindler observer and the "Rindler-inertial" observer perspectives can be made consistent.

That still leaves the main question hanging about the Hawking-Unruh connection though.
Too bad we'll probably need to wait for a quantum gravity theory for that.

Strange
2010-Aug-13, 01:18 PM
That still leaves the main question hanging about the Hawking-Unruh connection though.
Too bad we'll probably need to wait for a quantum gravity theory for that.

And then I will understand even less than I do now! (Unless some of the ATM crowd are right and it is all really simple, if we could only see it :))

Thanks to publius and Ken G for an enlightening discussion.

publius
2010-Aug-14, 03:59 AM
The physicists in all of us should be screaming "Unruh(Hawking)? Let's measure it!". Well, like Ken was saying, the prospects for measuring Hawking radiation seem nil and will probably remain so for the foreseable future, but Unruh may be a different story. The Unruh temperature is proportional to the acceleration, so acclerate something enough to get a meaningful temperature. Well, 1 earth g produces a Unruh temperature of a mere 10^(-20)K, so all we need is 10^24g or so to get hot enough. :lol:

Obviously one is not going to accelerate a particle detector at that rate without destroying it (and probably a lot of other equipment as well including blowing up your own fool self, probably), but there may be an indirect way. The same effect that produces Unruh should alter the frequency of accelerated electrons in a measurable way if the acceleration is around 10^26g. There are proposals to do that, but apparently that is still out of reach, but maybe doable. The trouble is the vacuum polarization limits the E field and thus the acceleration of electrons. 10^28g is apparently the maximum before the E field starts popping out electron-positron pairs.

But it won't be a direct measurment, only an effect related to the Unruh effect.

Then there is this:

http://en.wikipedia.org/wiki/Sokolov-Ternov_effect

which claims to be related to Unruh as well.

These guys say an acclerated detector will NOT radiate, although apparently will still "get hot":

http://arxiv.org/abs/quant-ph/0509151

This guy says he has observed the Unruh emission:

http://arxiv.org/abs/cond-mat/0510743


-Richard