Update to clarify The dispersion relation for an one-dimensional chain of atoms each of mass $m$ and attached to each other by identical springs of force constant $K$ (which for a continuous medium will become the Bulk modulus representing its elastic property) is given by $$\omega(k)=2\sqrt{\frac{K}{m}}|\sin(\frac{ka}{2})|.\tag{1}$$ Here $a$ denotes the equilibrium spacing between the atoms.

We note that the dispersion relation (1) is not linear. For long wavelength modes, Eq.(1) becomes $$\omega=\sqrt{\frac{K}{m}}(|k|a),\tag{2}$$ and it is a standard trick to read off the velocity of sound from (2) using the formula $c_s=\omega/k$. For a reference, see Ashcroft and Mermin, Eq. 22.29 and 22.31.


$\bullet$ Does it mean that the small wavelength modes satisfying Eq.(1) but not Eq.(2) cannot be the carriers of the sound wave?

$\bullet$ Why instead of linearizing (1), calculate the group velocity $\frac{d\omega}{dk}$ and attribute that to be the velocity of sound?

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    $\begingroup$ So maybe you can briefly clarify how you see how a spring constant $K$ and a mass $m$ can possibly enter into the physics of sound wave? $\endgroup$ Feb 1, 2017 at 14:19
  • $\begingroup$ I would really like to know the reason for the downvotes. $\endgroup$
    – SRS
    Feb 1, 2017 at 22:47
  • $\begingroup$ I'm not the one who downvoted so I can't answer but your question is hard to follow: for a 1d chain of atoms the wave would be transverse, but sound is longitudinal so the two situations don't match much, and my original comment stands: how exactly do you see the connection between atoms connected by springs and sound waves? $\endgroup$ Feb 1, 2017 at 23:04
  • $\begingroup$ I just realized I was thinking about your chain incorrectly (of course it can be transverse) yet the question remains: what is the analogue of $K$ in sound waves? $\endgroup$ Feb 1, 2017 at 23:49
  • $\begingroup$ @ZeroTheHero The atoms move in the direction of the propagation of the wave. Why would the wave be transverse? Moreover, it is the standard trick to linearize the dispersion relation for the monatomic chain, and read off the velocity of sound from it. $\endgroup$
    – SRS
    Feb 2, 2017 at 5:01

2 Answers 2


Your question seems to have the premise: "The speed of sound is constant by definition, therefore, if high-frequency phonons move significantly faster or slower than low-frequency phonons, those high-frequency phonons must not be sound waves".

Well, I disagree with the premise. The speed of sound is frequency dependent. I have seen this stated, explicitly and implicitly, in words and figures, over and over in tons of different sources. I don't think it's controversal.

  • $\begingroup$ Then why in Ashchroft and Mermin's book they linearize (1), and define the velocity of sound as a constant? @Steve Byrnes $\endgroup$
    – SRS
    Feb 6, 2017 at 16:50
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    $\begingroup$ Are you referring to (22.31)? "This [linear relation] is the type of behavior we are accustomed to in the cases of light waves and ordinary sound waves." If so, I think you're reading too much into this quote. In the "ordinary sound waves" in air that we experience in everyday life, sound dispersion is not usually noticeable, so we are "accustomed to" thinking about a constant (freq-independent) speed of sound ... even though in reality we all know that it is never exactly constant. $\endgroup$ Feb 6, 2017 at 19:05

It seems you have answered your own question since, for small arguments, $\sin(ka/2)\approx ka/2$. Hence, the physics of your system is such that the dispersion relation you give is precisely some long wavelength limit.

Dispersion relations are rarely exactly linear but usually linear only in some limit. For instance, the dispersion relation for a piano string is of the form $$ \frac{\omega^2}{k^2}=\frac{T_0}{\rho_0}+\alpha k^2 $$ where $T_0$ is the tension in the string, $\rho_0$ is the mass density and $\alpha$ is a small positive constant that would be $0$ if the string were perfectly flexible. One recovers the more usual linear dispersion relation by neglecting the $\alpha k^2$.

Therefore, the dispersion relation you give is only the dominant part of the full dispersion relation, rather than the exact relation.

If you consider sound in a gas, the physics of course is different since atoms in a chain are fixed around a position whereas molecules in a gas are not. If you consider sound in a solid, typical values of $a$ (which would be the interspacing between atoms on the chain) will be very small and values of $k=2\pi/\lambda$ would be of the order of the inverse length of the chain, i.e. also small. i.e. the longest wavelengths you can fit in your chain will be the order of the chain itself. Both arguments point to $ka/2$ being typically quite small and so justify the approximation $\sin(ka/2)\approx $ka/2$.


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