# Is sound a Nambu-Goldstone mode?

The usual sound exists in solids, liquids, and gases, as a long-wavelength excitation with linear dispersion. Can its presence be attributed to the spontaneous breaking of some symmetry? In other words, is it a Goldstone mode of some symmetry?

• You can view a massless klein gordon field as a goldstone boson for a spontaneously broken shift symmetry (since the vev is not invariant under shifts of the value of phi). I'm guessing it's the same for sound waves, the background density breaks the shift symmetry andsoundwaves could be viewed as the associated goldstone mode – Andrew Apr 22 '15 at 22:10

Yes sound is a goldstone mode. Consider, for example, an ideal gas with particles at positions $\mathbf{x}_i$. There is a symmetry where we can displace each particle by some displacement $\mathbf{u}$. Of course this symmetry breaks spontaneously. By definition, we only observe $\mathbf{u}=\mathbf{0}$.

The goldstone modes corresponding to this symmetry are modes where $\mathbf{u}$ is nonzero and varies spatially with some wavevector $\mathbf{k}$. That is, each particle gets displaced according to $\mathbf{x}_i \to \mathbf{x}_i + \mathbf{u} \cos(\mathbf{k} \cdot \mathbf{x}_i)$. This displacment will cause a sinusoidal variation in density, and therefore sinusoidal varation in pressure, which is what sound is.

Notice that the energy of the mode goes to zero as $\mathbf{k}$ goes to zero, since the $\mathbf{k}=\mathbf{0}$ limit is just a uniform shift, which requires zero energy. That is the idea behind goldstone modes. This same logic applies in liquids and solids as well.

• Thanks to Andrew and NowIGetToLearnWhatAHeadIs but I still am not convinced (read: do not understand). Both a liquid and a gas have a continuous shift symmetry, while a solid has a discrete shift symmetry. In particular, this translational symmetry is not broken in an ideal gas, contrary to what seems to be implied in the answer. It is obvious that the translation of all particles in the system, as explained in the answer is a zero-energy excitation. But it is not obvious how the system looks without this symmetry, or how this symmetry is spontaneously broken. – jarm Apr 23 '15 at 8:01
• I think you are saying that you do see how sound in solids is a goldstone mode, because there is a continuous symmetry that breaks and gives you a discrete symmetry. Now when I was talking about liquids and gases, I was referring to a particle based model. This model has a continuous symmetry that is broken, and the state with a broken symmetry has no remaining symmetry because particles in a liquid and gas are disordered. – Brian Moths Apr 23 '15 at 19:28
• Now you were probably looking at it from a field-theoretic point of view, where the fluid is represented by density, pressure, and velocity fields. In that case the ground state does have translational symmetry (zero velocity, uniform density and pressure), so there does not seem to be a broken symmetry, and so you would be led to say that sound is not a goldstone mode. That is an interesting point and I didn't think about that. – Brian Moths Apr 23 '15 at 19:31

It depends on what sound you are talking about. Yes, in crystalline solids, sound is nothing but propagating phonons, in which case it is the Goldstone modes corresponding to broken translational symmetry.

In fluids, sound is a pressure-density wave and there it is not a Goldstone mode. It instead arises because of conservation laws governing the conservation of momentum and of mass (non-relativistically and in the absence of reactions) along with the presence of inertia.