It is well known that the low-energy physics of many non-relativistic condensed matter systems can be described by field theories that display emergent Lorentz symmetry. The heuristic way to figure out the effective "speed of light" (really more like a speed of sound) is to expand the dispersion relation $\epsilon(p) = \epsilon(\hbar k) = \hbar\, \omega(k) = \hbar c k + o(k^2)$ to linear order in $k$, and the coefficient of the linear term gives the speed of sound.

But what about a condensed-matter system whose low-energy physics is described by a field theory consisting of several massless (or low-mass) weakly interacting fields not related by symmetry? It seems to me that performing this procedure separately for each field would generically yield a different linear-order coefficient for each dispersion relation, and therefore a different effective speed of sound, for each field. Is this the case?

If so, then each field would be individually Lorentz-invariant when considered in isolation, but the interacting theory as a whole would not be, due to the different fields' different "speeds of light". And yet, while I've often seen lattice condensed-matter systems coarse-grained into field theories with multiple fields with different masses and coupling constants, I can't recall ever seeing one where the different fields have different speeds of sound.

  • 1
    $\begingroup$ The simplest example I can think of is graphene. The speed of sound for the dirac fermions is much larger than the speed of sound of the phonon field. $\endgroup$
    – KF Gauss
    Apr 18, 2019 at 6:26
  • 1
    $\begingroup$ Generically if you have fermions of different bands, then each one has their own fermi velocity, this is the case for metals like iron. $\endgroup$
    – KF Gauss
    Apr 18, 2019 at 6:30
  • 1
    $\begingroup$ I totally agree with what you're saying here. I imagine the reason for the discrepancy is the usual theorist's bias of just omitting (or perhaps not even knowing about) any details that would spoil the nice symmetries. $\endgroup$
    – knzhou
    Apr 21, 2019 at 10:31


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.