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Phonons are obtaied by non-relativistic quantization of the lattice vibration. The dispersion relation is given by $\omega=c_s k$ where $c_s$ is the velocity of sound. What can we say about the mass of the phonon? I think it is not possible to compare this relation with the relativistic dispersion relation $E^2=p^2c^2+m^2c^4$ and conclude $m=0$. By mass, I do not mean the effective mass but the rest mass. Certainly, if the rest mass of phonon were zero it would have travelled with the velocity of light in vacuum.

I think in the non-relativistic approximation of the Einstein's energy momentum relation, the same $m$ appears in the non-relativistic kinetic energy $\frac{p^2}{2m}$. Therefore, we can still talk about rest mass in non-relativistic physics.

Moreover, phonon being a goldstone boson should have zero rest mass.

Edit: How does one define the rest mass of the phonon?

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Phonons are indeed massless, as you can see from their dispersion relation or from the fact that they are Goldstone bosons. The phonon dispersion relation that you wrote down tells us that we can excite a phonon mode, with some finite momentum, using an arbitrarily small amount of energy, hence they have no rest mass (in condensed matter language, they are not "gapped"). This does not mean that they travel with the speed of light; I guess one way to see that is that the lattice breaks Lorentz symmetry by giving us a preferred inertial frame. When formulating the theory of phonons we usually take the nonrelativistic limit $c \rightarrow \infty$ right from the start, so the speed of light never enters into any of the equations.

Phonons instead travel at the speed of sound $c_s$, which is the characteristic speed set by the lattice (if you compare the phonon dispersion to the relativistic dispersion relation you wrote down you see that $c_s$ replaces $c$, the speed of light).

Said differently, phonons are quasiparticles (=not true, elementary particles) that emerge in a theory with a lattice that breaks Lorentz symmetry, so your statement "if the rest mass of phonon was zero it would have traveled with the velocity of light in vacuum" does not apply to them.

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  • $\begingroup$ Nice, to-the-point answer. $\endgroup$ – Rococo May 19 '16 at 19:24
  • $\begingroup$ From the dispersion relation, it is not clear to me how did you conclude that phonons are massless (i.e., have zero rest mass). I guess, you're multiplying both sides of the dispersion relation $\omega=c_sk$ by $\hbar$ and arriving at $E=c_s p$. Then you compared it with $E^2=p^2c^2+m^2c_s^4$. But note that the relativistic dispersion relation is not in terms of the sound velocity $c_s$ but in terms of the velocity of light in vacuum $c$. The $m$ that appears in the non-relativistic Hamiltonian follows from the dispersion relation $E^2=p^2c^2+m^2c^4$ in the limit $\frac{p}{mc}\ll 1$. @VashVI $\endgroup$ – SRS Jun 30 '17 at 13:04
  • $\begingroup$ Indeed in the condensed matter parlance, phonons are gapless excitations. An arbitrarily small energy can excite a mode as you said and is also clear from the dispersion relation $\omega=c_sk$ or $E=c_s p$. But I'm reluctant to call the phonons to be massless i.e., having zero rest mass. On the other hand, massless relativistic quanta such as photon is gapless because $E=pc$ from them and also massless (have zero rest mass) because it can be compared with $E^2=p^2c_s^2+m^2c^4$. But this is not true for a non-relativistic quanta in condensed matter. It is gapless but not massless. $\endgroup$ – SRS Jun 30 '17 at 13:13
  • $\begingroup$ I guess it comes down to terminology. It seems natural to define rest mass as the total energy of a (quasi)particle when it's at rest, i.e. the "energy gap". Hence phonons, obeying a gapless dispersion relation, are massless. You seem to want to compare strictly to the relativistic dispersion relation, but as you know phonons are highly non-relativistic, they appear when you have some fixed lattice that breaks Lorentz symmetry. So in that case I'm not sure if there's a way to define the phonon rest mass that would satisfy you. $\endgroup$ – VashVI Sep 29 '17 at 6:52
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Phonons follow a wave equation, which is at least in first approximation simply a standard wave equation, the only difference to relativistic particles is that the speed of the waves is not c but the speed of sound $c_s$. But this does not change the mathematics of the equation, so that in general there may be phonons which follow a massless wave equations and phonons which follow one with something analogical to a mass term.

Acoustic and optical phonons are roughly analogical to massless and massive particles. For acoustic phonons, a wave with very long wavelength becomes simply a translation of the whole lattice, so that the energy becomes zero. This is, indeed, like a Goldstone boson related with translational symmetry.

For optical phonons, there is no such translational symmetry which forces the energy of waves with a long wavelength to go to zero. So they have non-zero energy even in the limit of infinite wavelength or momentum zero, similar to a mass term. Of course, this will not be exactly $E^2=p^2c_s^2 + m^2 c_s^4$, which is what one may obtain at best as an approximation near a minimum, but the most characteristic distinction remains: you need more than some minimal energy to create them.

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    $\begingroup$ @Schmelzer- From the relation $\omega=c_s k$ one can understand that they are gapless. But since this relation was derived on a non-relativistic footing, this relation does not contain c. And therefore, it is not clear to me how to compare it with the relativistic dispersion equation and "read off" the rest mass $m$. $\endgroup$ – SRS May 20 '16 at 9:23
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    $\begingroup$ The dispersion relation you cite holds only in the long wavelength limit ($k \rightarrow 0$) for acoustic phonons. The actual dispersion relation is more complicated and allows for both acoustic and optical phonons, the distinction being that acoustic phonons represent in-phase oscillations of the atoms in the lattice, while optical phonons arise from out-of-phase oscillations. $\endgroup$ – udrv May 20 '16 at 18:23
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    $\begingroup$ For example, for the simplest lattice that supports both optical and acoustic phonons the dispersion relation is essentially of the form $\omega^2 = A\left(1 \pm \sqrt{1 - B \sin^2(ka/2)} \right)$, where $a$ is the interatomic distance. The "-" branch gives acoustic phonons and behaves as $\omega \approx c_s k$ when $k\rightarrow 0$, while the "+" branch gives optical phonons and for $k\rightarrow 0$ yields $\omega \approx \omega_0 - (\hbar^2k^2/2\mu)$. $\endgroup$ – udrv May 20 '16 at 18:23
  • $\begingroup$ This looks very much like the nonrelativistic limit of a "relativistic" dispersion relation $E^2 = \mu^2 c_s^4 + p^2c_s^2$, but with a negative "effective mass" $\mu$. You can find a useful applet here: fermi.la.asu.edu/ccli/applets/phonon. Hope this helps. $\endgroup$ – udrv May 20 '16 at 18:27
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    $\begingroup$ Everybody knows that there are differences between particles and quasiparticles. If naming the parameter which is analogical to a mass "mass" confuses you, name it "quasimass", no problem. This is certainly not the place to argue about your confusion about dBB theory, if you want to discuss them, feel free to come to ilja-schmelzer.de/forum $\endgroup$ – Schmelzer Jun 2 '16 at 2:52

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