If there is a “holy grail” to be found in modern astrophysics, it probably has something to do with finding out what’s going on inside of black holes. Since no light escapes from their event horizons, studying their insides directly is impossible. As if that wasn’t bad enough, our best theories tend to break down inside the event horizon, limiting our ability to study them even theoretically with present models.
Despite all that, there are ways to get at the behavior of black holes. A recent line of work is approaching the problem in a different way—by analogy. Rather than trying to observe real black holes or trying to simulate them mathematically, researchers are constructing analogs of black holes. These constructions can be observed in a lab, right here on Earth.
Of course, scientists have no way of creating an actual gravitational singularity on a table-top, so they had to rely on the next best thing. The essence of a black hole is that it has an event horizon—a point of no return from which no light can escape. By analogy, in a fluid, there can be a point of no return for sound waves. If, for example, the fluid is moving faster than the speed of sound, no sound can outrun the fluid to escape in the opposite direction. That’s the basic idea behind a new experiment published in the journal Nature Physics (abstract) —an experiment that apparently makes a Hawking radiation laser out of a sonic black hole.
[Additional Coverage]: http://www.universetoday.com/115307/hawking-radiation-replicated-in-a-laboratory/
(Score: 0) by Anonymous Coward on Wednesday October 29 2014, @07:12AM
good explanation of hawking radiation in the article. I knew more about virtual particles but didn't realize hawking radiation had something to do with them
(Score: 2) by aristarchus on Wednesday October 29 2014, @07:28AM
If it is sonic, it is not really Hawking Radiation, then, is it? Speed of sound as an absolute constant substituting for the speed of light? Well, an interesting idea, but not sure how far the analogy will hold. Could be torn apart by an actual event horizon.
(Score: 2) by davester666 on Wednesday October 29 2014, @06:15PM
That's fine. As long as they don't actually try to create a real black hole here.
"It'll be really small, so it won't be a problem to contain it. Trust us. Nothing can go wrong."
(Score: 1) by boristhespider on Wednesday October 29 2014, @08:44PM
"If it is sonic, it is not really Hawking Radiation, then, is it?"
Yes, yes it is. Hawking radiation is radiation from a horizon. As Steinhauer has done, and has been theorised for about 30 years now, if you can set up a system that possesses a horizon, and can produce a thermal bath of quantum particles, you produce Hawking radiation. In a dilute Bose-Einstein condensate, we've got the setup we need -- an inviscid, irrotational, barotropic fluid (which can be pushed to move faster than the local speed of sound); and zero thermal sources of noise which would otherwise drown the phonons; a dispersion relation that in the longer-wavelength limit has the correct form (unlike superfluid Helium, which otherwise satisifies this; superfluid Helium exhibits the roton dip, where energy rather than going into phonons is pushed into rotons).
Basically, if you analyse the propagation of sound waves in *any* irrotational, barotropic, inviscid system then you find that they propagate on an effective spacetime -- effectively, sound waves in this kind of system follow the same paths (in the right coordinate system) as light rays do in a gravitational system. This is exact. It also means if you force the medium to move faster than the local sound speed, you form a horizon -- which is easy to visualise. Imagine a river rolling down an infinite hill, moving faster and faster. At some point it will start moving faster than the speed of sound in the water. Any soundwaves in the river are going to be swept along with it, and at the point that the river is moving faster than the sound speed, *no sound waves can get out*. That obviously resembles a horizon, but the beautiful thing is that if you do the maths properly, it really *is* a horizon. If you express the evolution equation for the (phase of the) sound waves in a geometric language, you find it maps out the null geodesics in what's known as Painleve-Gullstrand coordinates, which are a representation of Schwazschild spacetime where the event horizon stays regular. That means that those sound waves *propagate along null geodesics of a Schwazschild spacetime*. This is true in a classical system just as much as a quantum system.
This kind of study - pioneered by Bill Unruh - has shown not least that every single system that exhibits a horizon will exhibit Hawking radiation. The connection with black holes is not spurious, but it is slightly misleading. The emission of Hawking radiation is associated with the horizon, with the kinematics and not with the dynamics of the spacetime. (Beautifully, it also means that any geometric theory of gravity will produce Hawking radiation. Even if GR is wrong, which it probably is, we'll see Hawking radiation anyway, pending some extraordinarily unlikely revolution in our understanding of gravity.) There's a set of coordinates that are particularly germane here -- Rindler coordinates. Rindler coordinates are those attached to observers undergoing permanent acceleration away from some origin, and if you look at the spacetime diagram for Rindler "spacetime" (https://people.math.osu.edu/gerlach.1/perf_cont_interf_oct97_abstrandfig_la/img31.gif) you can immediately see resemblances to Schwarzschild spacetime (http://upload.wikimedia.org/wikipedia/commons/6/6b/KruskalKoords.gif). Turns out those resemblances are quite strong -- those diagonal lines that look like the Schwarzschild horizons *are* horizons, and an observer in Rindler coordinates will observe the emission of Hawking radiation. (This is basically the Unruh effect.) And yet Rindler coordinates are nothing more or less than bog-standard Minkowski spacetime.
The only thing is that you need to have a system where we can actually do this, and no classical system qualifies. They're all viscous, not barotropic, and tend to have quite a bit of vorticity. However, in a quantum system we can actually set up an analogue black hole spacetime. Fascinatingly, if you then start writing down the (effective) quantum field theory of the phonons in this spacetime in the vicinity of the horizon, you get a "squeezer" Hamiltonian -- pair production. if you look at this more closely, you find that you do, indeed, have Hawking radiation.
Basically all you really need is a system with quantum particles that have about the right dispersion relationship, and you need a setup with a horizon of some form, and you *will* get Hawking radiation. Somewhat surprisingly, the physics is the same.
So long answer to a short question: yes, it is Hawking radiation.
(Score: 2) by hubie on Wednesday October 29 2014, @09:46PM
Excellent response, thank you.
Is that true in general, or do you need to have an inner boundary (the "white hole") to establish the proper dispersion relation to get the lasing effect as described in Steinhauer's paper? Is the lasing effect required to produce Hawking radiation, or is it just an effect to make it easier to detect?
(Score: 1) by boristhespider on Wednesday October 29 2014, @10:12PM
Certainly if you want to do it with a fluid system, you inevitably get the white hole if you want the black hole -- even on a classical level that's clear, since there's no physical way we can get something that moves faster and faster than the speed of sound and never slows back down again. (Failing anything else, we get a hard boundary at the edge of the fluid where the speed of sound is identically zero, so it *has* to pass back through the speed of sound, even if in a thin layer.) The dispersion relationship is more a factor of the actual type of fluid you're considering -- for superfluid Helium, for instance, it breaks down at the roton dip no matter what games you might play with the background setup -- but it's certainly true you need to ensure that if your dispersion relationship is only valid in a certain range of wavelengths that you can actually *contain* those wavelengths...
The details of Steinhauer's setup I'm afraid I can't comment on. I did my Masters on acoustic holes, but I've been twelve years out of this field so while I'm not strictly a layman in the area, especially since I've been in gravitational physics much of the rest of the time, I'm also not quite an expert.
(Score: 1) by boristhespider on Wednesday October 29 2014, @10:15PM
Also, sorry, I didn't quite answer your question. To get Hawking radiation all you need (in theory) is the presence of a horizon, and a source of (quasi-)particles. In reality, you're not going to have much luck in a lab setting up a black hole horizon without the white hole. If nothing else, the sheer focus of sound waves at the centre of your acoustic hole would obliterate your system pretty quickly.
(Score: 2) by hubie on Friday October 31 2014, @02:00AM
Thanks again. I did my research in cosmic rays, so I never had to worry about the particles until they got to our solar system. :)
(Score: 2) by aristarchus on Thursday October 30 2014, @07:20AM
Maybe I have not been keeping up with physics: What the hell is a phonon? A sound "particle"? Prey tell! And a roton must be a rotating sound particle. Yes, this does keep the analogy intact. But what the?
(Score: 0) by Anonymous Coward on Thursday October 30 2014, @01:03PM
Exactly. A phonon is a quasiparticle describing the basic excitation of a mechanical wave (that is, a sound wave).
(Score: 0) by Anonymous Coward on Thursday October 30 2014, @04:02PM
Sorry, I should have been a bit clearer. As the AC above says, you're exactly right -- a phonon is a "quasiparticle", basically a quantum of sound, in roughly the same way a photon is a quantum of radiation. So far as I know (and I'm very far from being an expert in superfluid Helium) the roton was originally thought to be linked with vortices but may or may not actually be so. In any event, it's either a different form of quasiparticle that arises specifically in superfluid He, or it's a different aspect of the phonons in superfluid He. (And the distinction between those is a bit arbitrary anyway.)
-- boristhespider
(Score: 2) by wonkey_monkey on Wednesday October 29 2014, @07:53AM
Is this the same thing as when you run the tap into the kitchen sink and you get a circle of smooth water flowing out in a circle, with a turbulent edge? I heard that was somehow very slightly analogous to a black hole.
systemd is Roko's Basilisk
(Score: 1) by boristhespider on Wednesday October 29 2014, @08:25PM
No. This is a direct analogue. So long as you stay within the regime of the approximation, this is an exact analogy of a gravitational black hole. To get the setup working you need a fluid that is irrotational, inviscid and barotropic -- which the water swirling around the plug in your sink is absolutely not. You also need to ensure there are no other sources of noise, literally, since the analogue hole is mapped out by phonons, which are quantised sound waves, and there must be *no* other sources of noise, something which is only possible in a quantum fluid such as the Bose-Einstein condensate used here. BECs are practically the only fluid we've got where you can play this trick, since you also need the right dispersion relationship and other relatively easily-produced quantum fluids, such as liquid Helium, have totally knackered dispersion relations.
If anyone's interested, I made a few posts on the other site when this was discussed there (as AC since I was at work):
http://science.slashdot.org/comments.pl?sid=5820855&cid=48130377 [slashdot.org]
http://science.slashdot.org/comments.pl?sid=5820855&cid=48130453 [slashdot.org]
http://science.slashdot.org/comments.pl?sid=5820855&cid=48130667 [slashdot.org]
http://science.slashdot.org/comments.pl?sid=5820855&cid=48133843 [slashdot.org]
This one wasn't me: http://science.slashdot.org/comments.pl?sid=5820855&cid=48220377 [slashdot.org] but it's an interesting post.
(Score: 1) by boristhespider on Wednesday October 29 2014, @10:23PM
(I'd like to quantify the word "exact" slightly. It's an exact analogue of a gravitational hole, so long as we're working on a background level, which is applicable for Hawking radiation until the point at which the backreaction of the loss of energy to Hawking radiation on the hole becomes significant. That's certainly important for the analogue holes, because it could fairly strongly limit the time for which the analogue hole is a good approximation before the radiation has depleted the condensate to the level that the approximation breaks down. The same does happen for a gravitational hole but because the scales are so different it takes a hell of a lot longer unless we're looking at femtometre scale holes.
As soon as we look at perturbations on the hole, the analogy breaks down absolutely. So we can't, for instance, model the merging of two Schwarzschild holes by flinging two acoustic holes at one another. This is because while we've mimicked the kinematics of the background spacetime -- the geometry, basically -- the *dynamics* of the two systems are wildly different. An acoustic hole is obeying non-relativistic quantum physics or, if you're working with a speculative fluid, it's obeying absolutely Newtonian physics. A gravitational hole is obeying the Einstein equations, and these are obviously rather different. So while the analogy is perfect for a particular regime, we have to be careful that we don't leave that regime.)
(Score: 2) by mrchew1982 on Wednesday October 29 2014, @02:53PM
Hōōin Kyōma might lend them the Phone Wave if they ask nicely and disavow any relationship with CERN.