# Do we say that  phonon has effective mass through its dispersion relation?

The effective mass is proportional to the second derivative of the dispersion relation d2k/dE2. Do we say that phonon have effective mass through it ? Spin wave have.

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## 4 Answers

Phonon dispersions are generally indicated by a linear spectrum, so the second derivative is 0. Thus phonons are effectively massless, and have a velocity given by the first derivative.

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– user12445 Sep 25 '12 at 18:14

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@user12445: Optical modes are massive. – Ron Maimon Sep 26 '12 at 7:24

My opinion: Phonons are collective excitations of the crystal lattice vibration. They are massless Goldstone bosons resulting from the violation of the continuous translational symmetry of free space, by the crystal lattice. Phonons must carry momentum because they interact with electrons and change the momentum of the latter. An example is the electronic excitations in indirect gap semiconductors (I think this has been mentioned previously). Also, experiments of neutron scattering by a crystal indicate that phonons must have momentum. However, although phonons are massless, they have momentum by virtue of their wavelength. The effective mass equation 1/m = 1/h2d2E/dk2 arises from a semi-classical treatment of the motion of electrons and holes in the crystal lattice, and represents the effective mass of electrons and holes. I don’t think it applies to phonons. Similarly, it would be incorrect to conclude that phonons must travel at the speed of light because they are massless. This is because, unlike photons, phonons are not relativistically invariant objects - they don’t obey Dirac’s or Maxwell’s equations. They are excitations of a classical acoustic wave. I hope this contributes a little bit to this interesting discussion.

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This figure shows the phonon dispersions of ZnO. It is clear that while some phonons have very roughly linear dispersions many do not (especially close to zone center). The second derivative would be non zero in these regions. I hope this has added to the conversation.

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The effective mass of modes that are linear near the centre of the Brillouin zone will be zero since $\frac{d^{2}k}{d\omega^{2}}$ vanishes, but near extremes of the BZ this often isn't true and they will of course gain an effective mass.

Furthermore, optical modes are often quadratic near the BZ centre, which gives them an effective mass, and the same is generally true of surface acoustic modes.

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