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I am confused by most discussions of analog Hawking radiation in fluids (see, for example, the recent experimental result of Weinfurtner et al. Phys. Rev. Lett. 106, 021302 (2011), arXiv:1008.1911). The starting point of these discussions is the observation that the equation of motion for fluctuations around stationary solutions of the Euler equation have the same mathematical structure as the wave equation in curved space (there is a fluid metric $g_{ij}$ determined by the background flow). This background metric can have sonic horizons. The sonic horizons can be characterized by an associated surface gravity $\kappa$, and analog Hawking temperature $T_H \sim \kappa\hbar/c_s$.

My main questions is this: Why would $T_H$ be relevant? Corrections to the Euler flow are not determined by quantizing small oscillations around the classical flow. Instead, hydrodynamics is an effective theory, and corrections arise from higher order terms in the derivative expansion (the Navier-Stokes, Burnett, super-Burnett terms), and from thermal fluctuations. Thermal fluctuations are governed by a linearized hydro theory with Langevin forces, but the strength of the noise terms is governed by the physical temperature, not by Planck's constant.

A practical question is: In practice $T_H$ is very small (because it is proportional to $\hbar$). How can you claim to measure thermal radiation at a temperature $T_H << T$?

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Great question! –  user566 Jan 6 '12 at 22:08
    
@Thomas, I went ahead and merged both accounts, to avoid confusion, hope that is OK with you. –  user566 Jan 6 '12 at 23:12
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@Thomas: I used to be puzzled by this a bit when I was doing stuff on it. I think I eventually convinced myself that it's because the thermal state you get is actually "classical", in that e.g. as a Wigner distribution it's always positive. That turns out to be the weird thing about Unruh/Hawking radiation --- the form is actually not very quantum at all. –  genneth Jan 10 '12 at 13:28
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If you have radiation at a rate proportional to $\hbar$, how can it not be a quantum effect? –  Thomas Jan 12 '12 at 4:09
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@FrédéricGrosshans: and, of course, blackbody radiation was how quantum mechanics was originally discovered in the first place. –  Jerry Schirmer May 23 '12 at 14:07
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2 Answers

I remember attending a seminar by Unruh a few months ago and the same question arised. As far as I remember, he enfasized that in these hydrodynamic analogs of black holes, the flow is not quantized, it is a classical fluid, and everything is classical and that the dumb hole behaves like a quantum amplifier emitting quantum noise from the Horizon. Calculations indicate that the spectrum of these outgoing phonons must be "thermal" (like in black hole case) and a small temperature can be associated to it. I would not call this temperature "quantum", even if it is proportional to $\hbar$ since it can be completely determined from classical attributes of the fluid (speed of flow, adiabatic properties of the fluid, etc) which can be measured classically. The appearance of $\hbar$ is supposed to be a consequence of quantum amplification of the "phonon field" near the Horizon.

Looking at the experimental paper I don't see anywhere a direct measurement of temperature. They don't stick a thermometer in the water stream (for obvious reasons). They just verify that the dumb hole seems to have a "thermal character" while acting on surface waves. They just measure the amplitude of ingoing and outgoing waves and find that their squared ratios follows the Boltzmann distribution in the frequency which agrees with Hawking radiation.

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The conditions for the existence of the Hawking effect are described in classical terms, i.e you need

1) A Lorentz signature metric

2) A horizon (given, for example, by space flowing into a BH faster than the speed of light, or fluid flowing downstream faster than the speed of sound)

3) Surface gravity at the horizon

Those conditions are then applied to a quantum field which satisfies a wave equation (e.g. Klein Gordon field on spacetime). The standard analysis proceeds by treating photon paths as null geodesics (eikonal approximation). The acoustic analogue of this is that sound waves (the quantum excitations of which are phonons) follow null geodesics in the Lorentzian acoustic metric. So my guess is that the reason for the presence of Planck's constant in the Hawking temperature expression in the acoustic case is that they're treating Hawking radiation in the phonon (quantum) field.

Indeed, Visser says

an acoustic event horizon will emit Hawking radiation in the form of a thermal bath of phonons at a temperature

$$kT_H=\frac{\hbar g_H}{2\pi c} $$

Here $g_H$ is the acoustic surface gravity and $c$ is the speed of sound

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