Do Schwarzschild (zero angular momentum) and Kerr (rotating) black holes share the same information "loss" problem? The Penrose process in a Kerr black hole's ergosphere will spin it down, but only asymptotically toward zero angular momentum. All the fun is in the footnotes.
Your friendly, neighborhood typo-man has detected these:
"low density one on half" - one on should be on one;
"These flowing condensates doesn’t last very long" - doesn't should be don't.
I think "silent holes" would better than "dumb holes" since dumb has another meaning which was a pejorative descendant of the original meaning, but nobody asked me (nor should they).
Thanks for the interesting information. It sounds like condensed-matter physics might be the most fruitful research area at this point.
It doesn't really tell us anything we didn't know already, as they say "the entanglement confirms that there is an issue of information loss". And, like you say, it's not really a test of gravity as gravity might work completely differently.
JimV: Thanks, I've fixed that. I made a mistake with the prescheduling, this post was supposed to go out on Monday! I meant to read through it again this morning... Best,
"Since the rotational symmetry is essential for the red-shift in the gravitational potential"
Can you expand on this? Certainly, in general, rotational symmetry isn't necessary for a gravitational redshift. Of course, by the no-hair theorem, black holes are rotationally symmetric, at least asymptotically. But the quotation above seems rather strange. I'm sure I'm missing something, but I don't know what.
Irrelevant curiosity: Einstein="one stone" or "a stone", Steinhauer="stone beater". :-)
Phillip: What I meant is simply that you need 3 dimensions to get a 1/r potential, that's all. Yes, you can get the redshift otherwise... But it's not really clear to me from the paper. (Have you tried reading this thing?? It doesn't explain anything about nothing.) I had some exchange with the author, but I still don't know. I mean, just look at this density profile. Does this look like *any* particular curve? With some fantasy, it's kind of a step potential. Or maybe it has some 1/r drop. Or maybe not. And then there's the system size: the whole system is essentially just twice as large as the black hole interior. Then again, the wavelengths seem to be smaller than the black hole, which again doesn't make much sense. In summary, what I'm saying is that I don't understand how the scales involved here relate to actual black holes, and the paper doesn't illuminate this point. Best,
Phillip: No, rotational symmetry is not a requirement for redshift. Rotational symmetry in 3 dimensions is a requirement for a 1/r potential which gives rise to the correct redshift. A 1-dimensional system, or one with planar symmetry respectively, does not have a 1/r potential. Of course the condensate doesn't have a potential to begin with, but I don't see how the 1/r falloff is reproduced. Best, B.
I would like to understand this statement better,"Of course the condensate doesn't have a potential to begin with". To my my understanding, everything (matter,energy)participates in gravity, anything which exerts force would have potential. What am I missing? Does condensate not follow usual laws of mechanics?
kashyap: This is a misunderstanding, sorry. What I meant in my reply to Phillip is that the condensate merely mimics a potential gradient for the excitations. The condensate itself is of course in various potentials, gravitatational and electromagnetic, to hold it in place. Best,
I believe that this experiment is relevant for quantum gravity black holes because it is an example of a quantum system without singularities, i.e. fluid, which can reproduce an analogue of a black hole evaporation. Hence it is not difficult to imagine a quantum gravity theory based on a dynamical lattice (spacetime triangulation) which will be singularity free, where the phonons will be the gravitons and the elemantary particles. The effective theory for those phonons will be a QFT in a curved spacetime, so that one can have a black-hole background geometry and the Hawking evaporation as a semi-classical effect. The back-reaction effects than can be studied numerically or by looking at the analogue experiments.
I find these analog gravity papers a the same time very interesting, but also very frustrating. Most of the time they don't answer the most basic questions about the system that is being studied. For example: 1) What is the temperature of the fluid? 2) What is the Hawking temperature of the horizon? 3) If T_H<T, why would I expect to be able to detect Hawking radiation? More generally, why would quantum fluctuations (T_H\sim\hbar) be more important than ordinary, thermal fluctuations in the fluid?
In the most recent Steinhauser paper the temperature of the fluid is not stated, but in the earlier Nature physics paper it was estimated to be a T=1 nK. The expected horizon temperature (based on analog surface gravity) is also not stated. I tried to estimate it from the plots in the Nature Physics paper and got T_H=0.05 nK. This is much better than Unruh's water tanks (which are at 300K and T_H is also nK), but still not quite what you would want. The current paper has a measurement, T_H =1 nK. This seems like a strange accident, the Hawking temperature is equal to the ambient temperature of the fluid.
He measures a thermal spectrum, but in a regime where it basically looks like 1/w, which is a classical spectrum (and 1/f noise is a ubiquitous feature in many systems). He also looks for entanglement, but the measure of entanglement that is used in the paper is the structure factor, which is also a measure of purely classical correlations in fluids.
Yes, I agree with you. The paper is pretty terrible, I mean, it doesn't even state the dispersion relation or the metric, or what's plotted in the figures to begin with. That together with the absence of error estimates makes me think that it's a very preliminary note. Or at least I hope that there will be a somewhat more detailed paper in the future... Best,
"The high density region thus allowed the phonons to escape and corresponds to the outside of the horizon, whereas the low density region corresponds to the inside of the horizon." This looks very counterintuitive to me.High density region (like BH) should be inside the condensate or trap. Just outside horizon should be low density region. What am I missing?
The speed of sound in the fluid depends on the density. The higher the density, the higher the speed of sound. The higher the speed of sound, the easier it is for phonons to escape. Consequently, it's the low density (low speed of sound) region that causes a trapping. I don't know why you think this is unintuitive. Best,
Bee: My confusion is not about "The higher the density, the higher the speed of sound." That is high school physics.May be I should have been clearer. In BH high density regions are geometrically inside the BH.Light cannot escape from that region. Outside horizon there is very low density or vacuum.No matter how thin condensate is,it has probably higher density than outside (May be this is wrong assumption. May be outside air has higher density than condensate!!)
Hi Bee, thank-you for introducing me to the use of Bose-Einstein condensates in these so called "analogue gravity systems", as pioneered by Bill Unruh. Considering the sonic analogy, to call them "dumb" holes is entirely appropriate. Am I to understand that the fluid has zero viscosity, and that in this experimental setting it is in supersonic flow, accelerated by a laser? It was the bit about the potential "from a second laser" that I did not quite understand. I admit to having got a bit lost when I reached the technical section of Jeff's report. But I find the visual evidence of coupled phonons (albeit only the HF ones) departing in opposite directions from the "horizon" quite extraordinary and compelling.
Yes, the fluid has basically zero viscosity. This is good because the analogy is based on the zero viscosity approximation. The fluid is supersonic in one half, and subsonic in the other half. The supersonic region is the one that is "inside" the horizon. I didn't understand the part with the second laser either. All that I could extract is that it shines only on one half of the sample and that affects the number density - or else I am misreading the axis label of Figure 1b. As I complained above to somebody else already, the paper is miserably written, it isn't even explained what's shown in the Figures and the text is cryptic to say the least. Best,
I still don't understand your question. The density of the fluid has nothing to do with matter in an actual spacetime. The fluid is the analogue for the spacetime itself. Besides this, black holes are vacuum solutions, the density is zero. Best,
Bee: OK! I give up!! If the experiment is successful there will be long reviews of that and I might understand! If it does not work I do not have to know about it:-)
9:33 PM, November 09, 2015
Tl;dr: A new experiment demonstrates that Hawking radiation in a fluid is entangled, but only in the high frequency end. This result might be useful to solve the black hole information loss problem.
In August I went to Stephen Hawking’s public lecture in the fully packed Stockholm Opera. Hawking was wheeled onto the stage, placed in the spotlight, and delivered an entertaining presentation about black holes. The silence of the audience was interrupted only by laughter to Hawking’s well-placed jokes. It was a flawless performance with standing ovations.
In his lecture, Hawking expressed hope that he will win the Nobelprize for the discovery that black holes emit radiation. Now called “Hawking radiation,” this effect should have been detected at the LHC had black holes been produced there. But time has come, I think, for Hawking to update his slides. The ship to the promised land of micro black holes has long left the harbor, and it sunk – the LHC hasn’t seen black holes, has not, in fact, seen anything besides the Higgs.
But you don’t need black holes to see Hawking radiation. The radiation is a consequence of applying quantum field theory in a space- and time-dependent background, and you can use some other background to see the same effect. This can be done, for example, by measuring the propagation of quantum excitations in Bose-Einstein condensates. These condensates are clouds of about a billion or so ultra-cold atoms that form a fluid with basically zero viscosity. It’s as clean a system as it gets to see this effect. Handling and measuring the condensate is a big experimental challenge, but what wouldn’t you do to create a black hole in the lab?
The analogy between the propagation of excitations on background fluids and in a curved space-time background was first pointed out by Bill Unruh in the 1980s. Since then, many concrete examples have been found for condensed-matter systems that can be used as stand-ins for gravitational fields; they are summarily known as “analogue gravity system” – this is “analogue” as in “analogy,” not as opposed to “digital.”
In these analogue gravity systems, the quantum excitations are sound waves, and the corresponding quantum particles are called “phonons.” A horizon in such a space-time is created at the boundary of a region in which the velocity of the background fluid exceeds the speed of sound, thereby preventing the sound waves from escaping. Since these fluids trap sound rather than light, such gravitational analogues are also called “dumb,” rather than “black” holes.
Hawking radiation was detected in fluids a few years ago. But these measurements only confirmed the thermal spectrum of the radiation and not its most relevant property: the entanglement across the horizon. The entanglement of the Hawking radiation connects pairs of particles, one inside and one outside the horizon. It is a pure quantum effect: The state of either particle separately is unknown and unknowable. One only knows that their states are related, so that measuring one of the particles determines the measurement outcome of the other particle – this is Einstein’s “spooky action at a distance.”
The entanglement of Hawking radiation across the horizon is origin of the black hole information loss problem. In a real black hole, the inside partner of the entangled pair eventually falls into the singularity, where it gets irretrievably lost, leaving the state of its partner undetermined. In this process, information is destroyed, but this is incompatible with quantum mechanics. Thus, by combining gravity with quantum mechanics, one arrives at a result that cannot happen in quantum mechanics. It’s a classical proof by contradiction, and signals a paradox. This headache is believed to be remedied by the, still missing, theory of quantum gravity, but exactly what the remedy is nobody knows.
In a new experiment, Jeff Steinhauer from the Israel Institute of Technology measured the entanglement of the Hawking radiation in an analogue black hole; his results are available on the arxiv.
For this new experiment, the Bose Einstein condensate was trapped and put in motion with laser light, making it an effectively one-dimensional system in flow. In this trap, the condensate had low density on one half and a higher density in the other half, achieved by a potential step from a second laser. The speed of sound in such a condensate depends on the density, so that a higher density corresponds to a higher speed of sound. The high density region thus allowed the phonons to escape and corresponds to the outside of the horizon, whereas the low density region corresponds to the inside of the horizon.
The figure below shows the density profile of the condensate:
[Image]
Figure 1 from 1510.00621. The density profile of the condensate.
In this system then, Steinhauer measured correlations of fluctuations. These flowing condensates don’t last very long, so to get useful data, the same setting must be reproduced several thousand times. The analysis clearly shows a correlation between the excitations inside and outside the horizon, as can be seen in the figure below. The entanglement appears in the grey lines on the diagonal from top left to bottom right. I have marked the relevant feature with red arrows (ignore the green ones, they indicate matches between the measured angles and the theoretical prediction).
[Image]
When Steinhauer analyzed the dependence on the frequency, he found a correlation only in the high frequency end, not in the low frequency end. This is as intriguing as confusing. In a real black hole all frequencies should be entangled. But if the Hawking radiation was not entirely entangled across the horizon, that might allow information to escape. One has to be careful, however, to not read too much into this finding.
To begin with, let us be clear, this is not a gravitational system. It’s a system that shares some properties with the gravitational case. But when it comes to the quantum behavior of the background, that may or may not be a useful comparison. Even if it was, the condensate studied here is not rotationally symmetric, as a real black hole would be. Since the rotational symmetry is essential for the red-shift in the gravitational potential, I actually don’t know how to interpret the low frequencies. Possibly they correspond to a regime that real black holes just don’t have. And then the correlation might just have gotten lost in experimental uncertainties – limitations by finite system size, number of particles, noise, etc – on which the paper doesn’t have much detail.
The difference between the analogue gravity system, which is the condensate, and the real gravity system is that we do have a theory for the quantum properties of the condensate. If gravity was quantized in a similar way, then studies like the one done by Steinhauer, might indicate where Hawking’s calculation fails – for it must fail if the information paradox is to be solved. So I find this a very interesting development.
Will Hawking and Steinhauer get a Nobelprize for the discovery and detection of the thermality and entanglement of the radiation? I think this is very unlikely, for right now it isn’t clear whether this is even relevant for anything. Should this finding turn out to be key to developing a theory of quantum gravity however, that would be groundbreaking. And who knows, maybe Hawking will again be invited to Stockholm.
posted by Sabine Hossenfelder at 10:00 AM on Nov 1, 2015
"Dumb Holes Leak"
22 Comments -
Do Schwarzschild (zero angular momentum) and Kerr (rotating) black holes share the same information "loss" problem? The Penrose process in a Kerr black hole's ergosphere will spin it down, but only asymptotically toward zero angular momentum. All the fun is in the footnotes.
10:28 AM, November 01, 2015
Your friendly, neighborhood typo-man has detected these:
"low density one on half" - one on should be on one;
"These flowing condensates doesn’t last very long" - doesn't should be don't.
I think "silent holes" would better than "dumb holes" since dumb has another meaning which was a pejorative descendant of the original meaning, but nobody asked me (nor should they).
Thanks for the interesting information. It sounds like condensed-matter physics might be the most fruitful research area at this point.
2:40 PM, November 01, 2015
That was very interesting, thanks.
It doesn't really tell us anything we didn't know already, as they say "the
entanglement confirms that there is an issue of information loss". And, like you say, it's not really a test of gravity as gravity might work completely differently.
Cool experiment, though.
4:16 PM, November 01, 2015
JimV: Thanks, I've fixed that. I made a mistake with the prescheduling, this post was supposed to go out on Monday! I meant to read through it again this morning... Best,
B.
1:31 AM, November 02, 2015
Andrew: In the normal case the Hawking radiation is entangled at all frequencies. In this experiment it isn't.
1:32 AM, November 02, 2015
"Since the rotational symmetry is essential for the red-shift in the gravitational potential"
Can you expand on this? Certainly, in general, rotational symmetry isn't necessary for a gravitational redshift. Of course, by the no-hair theorem, black holes are rotationally symmetric, at least asymptotically. But the quotation above seems rather strange. I'm sure I'm missing something, but I don't know what.
Irrelevant curiosity: Einstein="one stone" or "a stone", Steinhauer="stone beater". :-)
2:39 AM, November 02, 2015
Phillip: What I meant is simply that you need 3 dimensions to get a 1/r potential, that's all. Yes, you can get the redshift otherwise... But it's not really clear to me from the paper. (Have you tried reading this thing?? It doesn't explain anything about nothing.) I had some exchange with the author, but I still don't know. I mean, just look at this density profile. Does this look like *any* particular curve? With some fantasy, it's kind of a step potential. Or maybe it has some 1/r drop. Or maybe not. And then there's the system size: the whole system is essentially just twice as large as the black hole interior. Then again, the wavelengths seem to be smaller than the black hole, which again doesn't make much sense. In summary, what I'm saying is that I don't understand how the scales involved here relate to actual black holes, and the paper doesn't illuminate this point. Best,
B.
2:48 AM, November 02, 2015
I don't see my question to which you are responding. But rotational symmetry is not a requirement for the redshift, right?
3:53 AM, November 02, 2015
Phillip: No, rotational symmetry is not a requirement for redshift. Rotational symmetry in 3 dimensions is a requirement for a 1/r potential which gives rise to the correct redshift. A 1-dimensional system, or one with planar symmetry respectively, does not have a 1/r potential. Of course the condensate doesn't have a potential to begin with, but I don't see how the 1/r falloff is reproduced. Best,
B.
4:14 AM, November 02, 2015
OK, got it now (and now my question is visible, before your response). The missing word is correct redshift.
4:38 AM, November 02, 2015
I would like to understand this statement better,"Of course the condensate doesn't have a potential to begin with".
To my my understanding, everything (matter,energy)participates in gravity, anything which exerts force would have potential. What am I missing? Does condensate not follow usual laws of mechanics?
7:34 AM, November 02, 2015
kashyap: This is a misunderstanding, sorry. What I meant in my reply to Phillip is that the condensate merely mimics a potential gradient for the excitations. The condensate itself is of course in various potentials, gravitatational and electromagnetic, to hold it in place. Best,
B.
8:31 AM, November 02, 2015
I believe that this experiment is relevant for quantum gravity black holes because it is an example of a quantum system without singularities, i.e. fluid, which can reproduce an analogue of a black hole evaporation. Hence it is not difficult to imagine a quantum gravity theory based on a dynamical lattice (spacetime triangulation) which will be singularity free, where the phonons will be the gravitons and the elemantary particles. The effective theory for those phonons will be a QFT in a curved spacetime, so that one can have a black-hole background geometry and the Hawking evaporation as a semi-classical effect. The back-reaction effects than can be studied numerically or by looking at the analogue experiments.
4:12 AM, November 04, 2015
I find these analog gravity papers a the same time very interesting, but also very frustrating. Most of the time they don't answer the most basic questions about the system that is being studied. For example: 1) What is the temperature of the fluid? 2) What is the Hawking temperature of the horizon? 3) If T_H<T, why would I expect to be able to detect Hawking radiation? More generally, why would quantum fluctuations (T_H\sim\hbar) be more important than ordinary, thermal fluctuations in the fluid?
In the most recent Steinhauser paper the temperature of the fluid is not stated, but in the earlier Nature physics paper it was estimated to be a T=1 nK. The expected horizon temperature (based on analog surface gravity) is also not stated. I tried to estimate it from the plots in the Nature Physics paper and got T_H=0.05 nK. This is much better than Unruh's water tanks (which are at 300K and T_H
is also nK), but still not quite what you would want. The current paper has a measurement, T_H =1 nK. This seems like a strange accident, the Hawking temperature is equal to the ambient temperature of the fluid.
He measures a thermal spectrum, but in a regime where it basically looks like 1/w, which is a classical spectrum (and 1/f noise is a ubiquitous feature in many systems). He also looks for entanglement, but the measure of entanglement that is used in the paper is the structure factor, which is also a measure of purely classical correlations in fluids.
12:04 AM, November 05, 2015
Thomas,
Yes, I agree with you. The paper is pretty terrible, I mean, it doesn't even state the dispersion relation or the metric, or what's plotted in the figures to begin with. That together with the absence of error estimates makes me think that it's a very preliminary note. Or at least I hope that there will be a somewhat more detailed paper in the future... Best,
B.
12:42 AM, November 05, 2015
"The high density region thus allowed the phonons to escape and corresponds to the outside of the horizon, whereas the low density region corresponds to the inside of the horizon."
This looks very counterintuitive to me.High density region (like BH) should be inside the condensate or trap. Just outside horizon should be low density region. What am I missing?
12:23 PM, November 05, 2015
kashyap:
The speed of sound in the fluid depends on the density. The higher the density, the higher the speed of sound. The higher the speed of sound, the easier it is for phonons to escape. Consequently, it's the low density (low speed of sound) region that causes a trapping. I don't know why you think this is unintuitive. Best,
B.
11:56 PM, November 05, 2015
Bee:
My confusion is not about "The higher the density, the higher the speed of sound." That is high school physics.May be I should have been clearer. In BH high density regions are geometrically inside the BH.Light cannot escape from that region. Outside horizon there is very low density or vacuum.No matter how thin condensate is,it has probably higher density than outside (May be this is wrong assumption. May be outside air has higher density than condensate!!)
9:28 AM, November 06, 2015
Hi Bee, thank-you for introducing me to the use of Bose-Einstein condensates in these so called "analogue gravity systems", as pioneered by Bill Unruh. Considering the sonic analogy, to call them "dumb" holes is entirely appropriate. Am I to understand that the fluid has zero viscosity, and that in this experimental setting it is in supersonic flow, accelerated by a laser? It was the bit about the potential "from a second laser" that I did not quite understand. I admit to having got a bit lost when I reached the technical section of Jeff's report. But I find the visual evidence of coupled phonons (albeit only the HF ones) departing in opposite directions from the "horizon" quite extraordinary and compelling.
6:56 AM, November 07, 2015
Rowan:
Yes, the fluid has basically zero viscosity. This is good because the analogy is based on the zero viscosity approximation. The fluid is supersonic in one half, and subsonic in the other half. The supersonic region is the one that is "inside" the horizon. I didn't understand the part with the second laser either. All that I could extract is that it shines only on one half of the sample and that affects the number density - or else I am misreading the axis label of Figure 1b. As I complained above to somebody else already, the paper is miserably written, it isn't even explained what's shown in the Figures and the text is cryptic to say the least. Best,
B.
10:36 AM, November 07, 2015
kashyap:
I still don't understand your question. The density of the fluid has nothing to do with matter in an actual spacetime. The fluid is the analogue for the spacetime itself. Besides this, black holes are vacuum solutions, the density is zero. Best,
B.
12:59 AM, November 09, 2015
Bee:
OK! I give up!! If the experiment is successful there will be long reviews of that and I might understand! If it does not work I do not have to know about it:-)
9:33 PM, November 09, 2015