Because of the scale of the experimental setup, it is quite obvious that no gravitational effects are involved. Hence, there is no possibility for this experiment to recreate phenomena at the intersection of quantum mechanics and general relativity. What the Steinbauer does is he replicates a particular model of the black hole. If his setup works, fine, but it doesn't prove a single thing about how black holes behave - because he did not create one.
by Anonymous Coward writes:
on Monday October 13, 2014 @11:10AM (#48130453)
It tells us how horizons behave. The production of Hawking radiation in a gravitational black hole relies (and relies only) on the presence of a horizon. In an acoustic hole, we've got a horizon for phonons, rather than for photons, but it's still a horizon. The actual structure of the geometry in the simplest cases is Schwarzschild, but one can play some interesting games to get a more complicated setup which is more usable - and in any event, it also exhibits a horizon. Therefore, while the effective field theory describing the phonons holds, and while the system exhibits a horizon, any observation of Hawking radiation will directly test the processes by which we believe Hawking radiation is produced in actual gravitational holes. It's based on basically the same physics. What it *won't* tell us is anything about the backreaction of Hawking radiation on a gravitational hole, because while the kinematics of an analogue hole are the same as a gravitational hole (at an unperturbed level), the dynamics are completely different. Further, the system will eventually produce enough Hawking radiation that the condensate will be depleted to an extent that the analogy is no longer valid even at a background level. However, while the analogy is valid -- and that can be quantified and therefore controlled -- we can still exploit it. And when the analogy isn't valid we're still learning useful things about the behaviour of supercold fluids.
by Anonymous Coward writes:
on Monday October 13, 2014 @11:31AM (#48130667)
When it comes to the derivation of Hawking radiation, surprisingly, yes. It relies on there being a quantum vacuum for some type of particle (in gravitational radiation, photons; in the analogue case, phonons), and it relies on there being a horizon (in the gravitational case this is an event horizon; in the analogue case it's an acoustic horizon). It also relies on the analogue medium producing phonons of the right form -- a form where (in the appropriate regime) the phonons have a quantum theory that acts like photons do. In particular, you have to have a medium where the phonons have a "squeezer" Hamiltonian in the vicinity of the acoustic hole, meaning that pair production will happen. The derivation of Hawking radiation from this point takes the same form in both a gravitational hole and an acoustic hole. Of course, when the conditions change, as they inevitably will, the analogy breaks down but in its regimes of validity there's no issue, and we can quantify the extent to which the analogy is holding.
The production of Hawking radiation in a gravitational black hole relies (and relies only) on the presence of a horizon.
Does it? Because from what I've understood, it's caused by virtual particles getting sufficient energy from interaction with a field to become real particles, and even horizon is simply the boundary above which particles so produced can escape. If so, then any strong enough field should produce Hawking radiation - for example, a strong enough electric field would produce a stream of elect
It was pity stayed his hand.
"Pity I don't have any more bullets," thought Frito.
-- _Bored_of_the_Rings_, a Harvard Lampoon parody of Tolkein
Mimicking a theory, not a phenomenon (Score:5, Insightful)
Re:Mimicking a theory, not a phenomenon (Score:5, Informative)
It tells us how horizons behave. The production of Hawking radiation in a gravitational black hole relies (and relies only) on the presence of a horizon. In an acoustic hole, we've got a horizon for phonons, rather than for photons, but it's still a horizon. The actual structure of the geometry in the simplest cases is Schwarzschild, but one can play some interesting games to get a more complicated setup which is more usable - and in any event, it also exhibits a horizon. Therefore, while the effective field theory describing the phonons holds, and while the system exhibits a horizon, any observation of Hawking radiation will directly test the processes by which we believe Hawking radiation is produced in actual gravitational holes. It's based on basically the same physics. What it *won't* tell us is anything about the backreaction of Hawking radiation on a gravitational hole, because while the kinematics of an analogue hole are the same as a gravitational hole (at an unperturbed level), the dynamics are completely different. Further, the system will eventually produce enough Hawking radiation that the condensate will be depleted to an extent that the analogy is no longer valid even at a background level. However, while the analogy is valid -- and that can be quantified and therefore controlled -- we can still exploit it. And when the analogy isn't valid we're still learning useful things about the behaviour of supercold fluids.
Re: (Score:3)
Re:Mimicking a theory, not a phenomenon (Score:5, Informative)
When it comes to the derivation of Hawking radiation, surprisingly, yes. It relies on there being a quantum vacuum for some type of particle (in gravitational radiation, photons; in the analogue case, phonons), and it relies on there being a horizon (in the gravitational case this is an event horizon; in the analogue case it's an acoustic horizon). It also relies on the analogue medium producing phonons of the right form -- a form where (in the appropriate regime) the phonons have a quantum theory that acts like photons do. In particular, you have to have a medium where the phonons have a "squeezer" Hamiltonian in the vicinity of the acoustic hole, meaning that pair production will happen. The derivation of Hawking radiation from this point takes the same form in both a gravitational hole and an acoustic hole. Of course, when the conditions change, as they inevitably will, the analogy breaks down but in its regimes of validity there's no issue, and we can quantify the extent to which the analogy is holding.
Re: (Score:2)
Does it? Because from what I've understood, it's caused by virtual particles getting sufficient energy from interaction with a field to become real particles, and even horizon is simply the boundary above which particles so produced can escape. If so, then any strong enough field should produce Hawking radiation - for example, a strong enough electric field would produce a stream of elect