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  How to understand the emergent special relativity in the superfluid?

+ 8 like - 0 dislike

The superfluid vacuum theory was proposed to understand some features of the vacuum (aether) from the emergence point of view. Although made up of non-relativistic atoms, the low-energy excitations of superfluid, i.e. the phonons, are said to be "relativistic". I know that the superfluid phonons do have the linear dispersion $\omega = c k$, and the low-energy effective theory is Lorentz invariant. Also the superfluid is inviscid, so that the aether has no drag force. But I still doubt that the special relativity can really emerge in the superfluid. Because I can not show the following statement: The velocity of sound in the superfluid is the same for all inertial observers, regardless of their relative motion to the superfluid. I believe this should be a fundamental feature of relativity, but I do not see how to prove it. The velocity addition formula for sound is not the same as that for light. So I am very confused when people say that the phonons are relativistic. Any understanding of the emergent relativity in superfluid would be appreciated.

This post imported from StackExchange Physics at 2014-04-04 16:44 (UCT), posted by SE-user Everett You
asked May 6, 2013 in Theoretical Physics by Everett You (785 points) [ no revision ]

2 Answers

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The superfluid phonons have the linear dispersion $\omega=ck$ at low energies, and the low-energy effective theory is Lorentz invariant. But the superfluid phonons and the atoms that form the superfluid do not have the Lorentz symmetry. So the superfluid is an example of emergent Lorentz symmetry. The statement: The velocity of sound in the superfluid is the same for all inertial observers, regardless of their relative motion to the superfluid is valid if the clock and ruler are make by low energy phonons.

This post imported from StackExchange Physics at 2014-04-04 16:44 (UCT), posted by SE-user Xiao-Gang Wen

answered May 9, 2013 by Xiao-Gang Wen (3,339 points) [ revision history ]
edited Apr 4, 2014 by Xiao-Gang Wen
+ 0 like - 0 dislike

In general, a low-energy effective theory inherits continuous symmetries of the microscopic model. For this reason, the effective theory of superfluidity of non-relativistic particles should necessarily be invariant under Galilean relativity. Galilean invariance was used long ago by people like London, Landau and Popov for construction of the effective theory of superfluidity.

You are right, the low-energy degree of freedom in a superfluid- the U(1) Goldstone boson has a linear dispersion relation and unfortunately from that some people make a conclusion that the effective theory of superfluidity must be Lorentz invariant. Remember, however that the phonon is charge neutral collective degree of freedom and thus its linear dispersion is not in conflict with Galilean invariance. In other words from the form of the dispersion relation alone one can not conclude whether the theory is Lorentz or Galilean-invariant. What matters is the transformation property of the effective action!

I nice paper where the Galilean symmetry (and its local extension) of the effective theory of superfluidity is exploited is arXiv:cond-mat/0509786.

This post imported from StackExchange Physics at 2014-04-04 16:44 (UCT), posted by SE-user Sergej Moroz
answered Sep 21, 2013 by Sergej Moroz (15 points) [ no revision ]

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