tag:blogger.com,1999:blog-22973357.post5621511389890372761..comments2023-09-27T07:44:19.769-04:00Comments on Sabine Hossenfelder: Backreaction: Dumb Holes LeakSabine Hossenfelderhttp://www.blogger.com/profile/06151209308084588985noreply@blogger.comBlogger22125tag:blogger.com,1999:blog-22973357.post-35090781166386710312015-11-09T21:33:07.221-05:002015-11-09T21:33:07.221-05:00Bee:
OK! I give up!! If the experiment is successf...Bee:<br />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:-)kashyap vasavadahttps://www.blogger.com/profile/10732897306667764590noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-9716290122456651592015-11-09T00:59:27.584-05:002015-11-09T00:59:27.584-05:00kashyap:
I still don't understand your questi...kashyap:<br /><br />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,<br /><br />B.Sabine Hossenfelderhttps://www.blogger.com/profile/06151209308084588985noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-66396764448180504632015-11-07T10:36:51.426-05:002015-11-07T10:36:51.426-05:00Rowan:
Yes, the fluid has basically zero viscosit...Rowan:<br /><br />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 Sabine Hossenfelderhttps://www.blogger.com/profile/06151209308084588985noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-20408278800155527672015-11-07T06:56:02.646-05:002015-11-07T06:56:02.646-05:00Hi Bee, thank-you for introducing me to the use of...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? ItRowanhttps://www.blogger.com/profile/01366046281549156033noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-24953111701698354162015-11-06T09:28:57.665-05:002015-11-06T09:28:57.665-05:00Bee:
My confusion is not about "The higher th...Bee:<br />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 bekashyap vasavadahttps://www.blogger.com/profile/10732897306667764590noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-86394647509355191172015-11-05T23:56:11.931-05:002015-11-05T23:56:11.931-05:00kashyap:
The speed of sound in the fluid depends ...kashyap:<br /><br />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,<br /><br />B.Sabine Hossenfelderhttps://www.blogger.com/profile/06151209308084588985noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-2590382898282635592015-11-05T12:23:00.196-05:002015-11-05T12:23:00.196-05:00"The high density region thus allowed the pho..."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."<br />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?kashyap vasavadahttps://www.blogger.com/profile/10732897306667764590noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-35875404220329398592015-11-05T00:42:39.203-05:002015-11-05T00:42:39.203-05:00Thomas,
Yes, I agree with you. The paper is prett...Thomas,<br /><br />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,<br /><Sabine Hossenfelderhttps://www.blogger.com/profile/06151209308084588985noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-79447540499518625382015-11-05T00:04:21.158-05:002015-11-05T00:04:21.158-05:00I find these analog gravity papers a the same time...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, whyThomashttps://www.blogger.com/profile/06838983632065478955noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-53063121309011261132015-11-04T04:12:59.315-05:002015-11-04T04:12:59.315-05:00I believe that this experiment is relevant for qua...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 A. Mikovichttps://www.blogger.com/profile/03175906801121515444noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-20635813429151974862015-11-02T08:31:45.893-05:002015-11-02T08:31:45.893-05:00kashyap: This is a misunderstanding, sorry. What I...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,<br /><br />B. Sabine Hossenfelderhttps://www.blogger.com/profile/06151209308084588985noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-21063120646317239592015-11-02T07:34:07.386-05:002015-11-02T07:34:07.386-05:00I would like to understand this statement better,&...I would like to understand this statement better,"Of course the condensate doesn't have a potential to begin with".<br />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 vasavadahttps://www.blogger.com/profile/10732897306667764590noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-61142435365989661782015-11-02T04:38:20.138-05:002015-11-02T04:38:20.138-05:00OK, got it now (and now my question is visible, be...OK, got it now (and now my question is visible, before your response). The missing word is <i>correct</i> redshift.<br />Phillip Helbighttps://www.blogger.com/profile/12067585245603436809noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-54411171364626491222015-11-02T04:14:12.164-05:002015-11-02T04:14:12.164-05:00Phillip: No, rotational symmetry is not a requirem...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 Sabine Hossenfelderhttps://www.blogger.com/profile/06151209308084588985noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-38434946525698757712015-11-02T03:53:36.819-05:002015-11-02T03:53:36.819-05:00I don't see my question to which you are respo...I don't see my question to which you are responding. But rotational symmetry is not a requirement for the redshift, right?<br /><br />Phillip Helbighttps://www.blogger.com/profile/12067585245603436809noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-29209724461420372412015-11-02T02:48:47.935-05:002015-11-02T02:48:47.935-05:00Phillip: What I meant is simply that you need 3 di...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 Sabine Hossenfelderhttps://www.blogger.com/profile/06151209308084588985noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-46649873405162470032015-11-02T02:39:46.459-05:002015-11-02T02:39:46.459-05:00"Since the rotational symmetry is essential f...<i>"Since the rotational symmetry is essential for the red-shift in the gravitational potential"</i><br /><br />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 Phillip Helbighttps://www.blogger.com/profile/12067585245603436809noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-59379113764509697402015-11-02T01:32:12.039-05:002015-11-02T01:32:12.039-05:00Andrew: In the normal case the Hawking radiation i...Andrew: In the normal case the Hawking radiation is entangled at all frequencies. In this experiment it isn't.Sabine Hossenfelderhttps://www.blogger.com/profile/06151209308084588985noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-80674139489749334112015-11-02T01:31:32.728-05:002015-11-02T01:31:32.728-05:00JimV: Thanks, I've fixed that. I made a mistak...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,<br /><br />B.Sabine Hossenfelderhttps://www.blogger.com/profile/06151209308084588985noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-20794197261945243052015-11-01T16:16:12.676-05:002015-11-01T16:16:12.676-05:00That was very interesting, thanks.
It doesn't...That was very interesting, thanks.<br /><br />It doesn't really tell us anything we didn't know already, as they say "the <br />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.<br /><br />Cool experiment, though.Andrew Thomashttps://www.blogger.com/profile/03852211910001840777noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-25261710953955523442015-11-01T14:40:03.293-05:002015-11-01T14:40:03.293-05:00Your friendly, neighborhood typo-man has detected ...Your friendly, neighborhood typo-man has detected these:<br /><br />"low density one on half" - one on should be on one;<br /><br />"These flowing condensates doesn’t last very long" - doesn't should be don't.<br /><br />I think "silent holes" would better than "dumb holes" since dumb has another meaning which was a pejorative descendant of the JimVhttps://www.blogger.com/profile/10198704789965278981noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-75723315627377987182015-11-01T10:28:16.241-05:002015-11-01T10:28:16.241-05:00Do Schwarzschild (zero angular momentum) and Kerr ...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.Uncle Alhttps://www.blogger.com/profile/05056804084187606211noreply@blogger.com