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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 like:

$$k_{nml}=\sqrt{k_{n}^{2}+k_{m}^{2}+k_{l}^{2}}=\frac{\pi}{2L}\sqrt{n^{2}+m^{2}+l^{2}} $$

with momenta:

$$p_{nml}=\hslash k_{nml}$$

and other various properties depending upon your dispersion relation. Actually without boundary conditions, I would say no phonon modes. which is very interesting when you think about it. In any practical application there are always boundary conditions, especially with sound.

In a sense then, normal modes of sound are phonons.

EDIT: a quick search of this yields the same thing: https://en.m.wikipedia.org/wiki/Gas_in_a_box It is known as the Thomas-Fermi approximation and is used for massive or massless non or weakly interacting particles

It's also worth noting that, for phonons in a crystal, it is in actuality, the boundary of the whole crystal that cause quantization of phonon wavevectors http://users.physik.fu-berlin.de/~pascual/Vorlesung/SS09/slides/EPIV-09SS-SolSt_K3-Lattice%20vibrations.pdf (page 2) For example, in an infinite crystal, the allowed wavevectors would be continuous, and momentum wouldn't be quantized. Any good solid state text explains this, I like Kittel's intro to solid state physics

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 like:

$$k_{nml}=\sqrt{k_{n}^{2}+k_{m}^{2}+k_{l}^{2}}=\frac{\pi}{2L}\sqrt{n^{2}+m^{2}+l^{2}} $$

with momenta:

$$p_{nml}=\hslash k_{nml}$$

and other various properties depending upon your dispersion relation. Actually without boundary conditions, I would say no phonon modes. which is very interesting when you think about it. In any practical application there are always boundary conditions, especially with sound.

In a sense then, normal modes of sound are phonons.

EDIT: a quick search of this yields the same thing: https://en.m.wikipedia.org/wiki/Gas_in_a_box It is known as the Thomas-Fermi approximation and is used for massive or massless non or weakly interacting particles

It's also worth noting that, for phonons in a crystal, it is in actuality, the boundary of the whole crystal that cause quantization of phonon wavevectors http://users.physik.fu-berlin.de/~pascual/Vorlesung/SS09/slides/EPIV-09SS-SolSt_K3-Lattice%20vibrations.pdf (page 2) For example, in an infinite crystal, the allowed wavevectors would be continuous, and momentum wouldn't be quantized at all! Any good solid state text explains this, I like Kittel's intro to solid state physics

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 like:

$$k_{nml}=\sqrt{k_{n}^{2}+k_{m}^{2}+k_{l}^{2}}=\frac{\pi}{2L}\sqrt{n^{2}+m^{2}+l^{2}} $$

with momenta:

$$p_{nml}=\hslash k_{nml}$$

and other various properties depending upon your dispersion relation. Actually without boundary conditions, I would say no phonon modes. which is very interesting when you think about it. In any practical application there are always boundary conditions, especially with sound.

In a sense then, normal modes of sound are phonons.

EDIT: Not sure why I was downvoted, a quick search of this yields the same thing: https://en.m.wikipedia.org/wiki/Gas_in_a_box It is known as the Thomas-Fermi approximation and is used for massive or massless non or weakly interacting particles

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 like:

$$k_{nml}=\sqrt{k_{n}^{2}+k_{m}^{2}+k_{l}^{2}}=\frac{\pi}{2L}\sqrt{n^{2}+m^{2}+l^{2}} $$

with momenta:

$$p_{nml}=\hslash k_{nml}$$

and other various properties depending upon your dispersion relation. Actually without boundary conditions, I would say no phonon modes. which is very interesting when you think about it. In any practical application there are always boundary conditions, especially with sound.

In a sense then, normal modes of sound are phonons.

EDIT: a quick search of this yields the same thing: https://en.m.wikipedia.org/wiki/Gas_in_a_box It is known as the Thomas-Fermi approximation and is used for massive or massless non or weakly interacting particles

It's also worth noting that, for phonons in a crystal, it is in actuality, the boundary of the whole crystal that cause quantization of phonon wavevectors http://users.physik.fu-berlin.de/~pascual/Vorlesung/SS09/slides/EPIV-09SS-SolSt_K3-Lattice%20vibrations.pdf (page 2) For example, in an infinite crystal, the allowed wavevectors would be continuous, and momentum wouldn't be quantized. Any good solid state text explains this, I like Kittel's intro to solid state physics

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 like:

$$k_{nml}=\sqrt{k_{n}^{2}+k_{m}^{2}+k_{l}^{2}}=\frac{\pi}{2L}\sqrt{n^{2}+m^{2}+l^{2}} $$

with momenta:

$$p_{nml}=\hslash k_{nml}$$

and other various properties depending upon your dispersion relation. Actually without boundary conditions, I would say no phonon modes. which is very interesting when you think about it. In any practical application there are always boundary conditions, especially with sound.

In a sense then, normal modes of sound are phonons.

Not sure why I was downvoted, a quick search of this yields the same thing: https://en.m.wikipedia.org/wiki/Gas_in_a_box It is known as the Thomas-Fermi approximation and is used for massive or massless non or weakly interacting particles

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 like:

$$k_{nml}=\sqrt{k_{n}^{2}+k_{m}^{2}+k_{l}^{2}}=\frac{\pi}{2L}\sqrt{n^{2}+m^{2}+l^{2}} $$

with momenta:

$$p_{nml}=\hslash k_{nml}$$

and other various properties depending upon your dispersion relation. Actually without boundary conditions, I would say no phonon modes. which is very interesting when you think about it. In any practical application there are always boundary conditions, especially with sound.

In a sense then, normal modes of sound are phonons.

EDIT: Not sure why I was downvoted, a quick search of this yields the same thing: https://en.m.wikipedia.org/wiki/Gas_in_a_box It is known as the Thomas-Fermi approximation and is used for massive or massless non or weakly interacting particles