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.
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 fiel
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.
Mimicking a theory, not a phenomenon (Score:5, Insightful)
Re: (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 fiel
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.