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2

The answer to your question is, to a minute degree, is yes, since the total energy of something measures its source strength / coupling to gravity. So, if an system has energy $\Delta E$ added to it, it will weigh more - its gravitational mass will increase by $\Delta E/c^2$. Practically, the effect is almost unmeasurable; a magnet's stored magnetic energy ...

3

Because by expanding the sinus term into a taylor expansion, you get $\sin(x)\approx x - \frac{x^3}{6} +\cdots$ So, for small values of k you are allowed to take just the linear term.

4

@GerryHarp's answer contains the gist of the main idea, but there is a point that doesn't quite make sense: There is no such thing as bare ions crystal in solid state physics. To answer your last question: The "bare phonon frequencies" definitely do not refer to phonon frequencies in the absence of electrons. In fact, an ion crystal without electrons is ...

4

Lets start with the bare ions. You set up a linear lattice of protons where the boundaries are held in place by some means. The protons want to get as far apart from each other as possible. So the minimum energy state has the protons spaced on the line with equal distances between them. So yes, the ions form a regular lattice without the electrons. Now ...

5

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 ...

19

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 ...

1

As you mentioned, a phonon is a quantized sound wave. Let us ask then what is the difference between a quantum and classical sound wave in a crystal? At room temperature, for most solids, the specific heat is pretty close to $3Nk$. This is the classical Dulong-Petit answer. However, at low temperature, the specific heat deviates away from the Dulong-Petit ...

1

Light, passing through a transparent medium, is scattered when it interacts with that medium's spatial & temporal variations producing the medium's refractive index. This is called Brillouin scattering and can be described as interaction of a photon with a phonon representing the medium's compressional deformation. Although mostly studied in the context ...

4

This is what I thought at first: You need a restoring force in order to have a phonon: after all, phonons result from the quantization of the lattice energy written as a sum of harmonic oscillators. The hamiltonian of a gas cannot be written in this way: there are no restoring forces (you cannot "pull" a gas), so there can be no phonons. As Rococo wrote, ...

0

I'm tempted to answer in the affirmative, depending upon BOUNDARY CONDITIONS. Without getting into it too far Suppose the ideal gas is contained within (for simplicity) a cubic box of length L. One has that the wavenumbers $$k_{n}=\frac{n\pi}{2L}$$ are then quantized, such that there are phonon-like modes of propagation. One will then obtain something ...

9

The only mention of this subject I can recall seeing is an aside in Xiao-Gang Wen's book, Quantum Field Theory of Many-Body Systems. Footnote on page 86: A sound wave in air does not correspond to any discrete quasiparticle. This is because the sound wave is not a fluctuation of any quantum ground state. Thus, it does not correspond to any excitation ...

0

My intuition is that ideal gasses must have some form of quantized quasiparticle/collective excitation (I'll just call it a quasiparticle), but it usually isn't (or never is) a useful concept. (Just like a basketball at room temperature has quantized energy levels, but it's not useful to think of them.) Whether you call that quasiparticle a "phonon" is a ...

1

The low-energy optical branch appears to cross the acoustic branch, but that doesn't mean that hybridization has occurred. In fact, it's an indication that hybridization has not occurred. As far as I can tell from the diagram, the modes in question will be completely independent. I can't imagine any physical consequence. Update after comment ...

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