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On the german wikipedia site (right under "Akustische Moden"), the dispersion relation for a linear chain of atoms (connected by springs):

$$\omega(k)=2 \sqrt{\frac{K}{M}} \left \vert \sin{\frac{ka}{2}}\right \vert$$

is approximated as:

$$ \omega (k)\approx c_s \lvert k\rvert$$

for small $k$. ($c_s$ is the speed of sound). Why are we allowed to do that?

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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.

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    $\begingroup$ Thanks. That would give me: $$a \sqrt{\frac{K}{M}} \lvert k\rvert$$ though. I don't see how $a \sqrt{\frac{K}{M}}$ can be approximated as $c_s$. $a$ is the distance between two atoms in the lattice, $K$ the spring konstant and $M$ the mass. $\endgroup$
    – qmd
    May 28, 2016 at 13:48
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    $\begingroup$ $$c_s$$ is just the speed of sound, which connects (in your linear dispersion relation) the frequency with the momentum. So, this is just the speed of sound by definition. $\endgroup$ May 28, 2016 at 14:06
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    $\begingroup$ To elaborate upon what qmd, QuantumMechanics said, the speed of sound is defined either the "phase velocity" $\omega/k$, or the "group velocity", $d\omega / d k$. Whenever $\omega=Ak$ for some $A$, one may see that the group velocity and the phase velocity coincide and both are equal to $A$. $\endgroup$ May 28, 2016 at 14:10
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    $\begingroup$ Yes, $\omega=Ak$ holds if and only if the group velocity is equal to the phase velocity, and this is also equivalent to having no dispersion - i.e. no dependence of either velocity on $k$ or equivalently on $\omega$. You can replace $a\sqrt{K/M}$ by $c_s$ e.g. using the delete key several times (on a PC), or using a rubber on a pencil, and then writing $c_s$ to the place you emptied, or by writing the equation once again with $c_s$ instead of the previous form of the coefficient. The point is that they are equal by the definition of $c_s$ so you are allowed to do that. $\endgroup$ May 28, 2016 at 14:21
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    $\begingroup$ @LubošMotl Thanks for the humerous explanation (I think I needed that slap in the face). ;) It makes sense now. I just didn't see how $a \sqrt{\frac{K}{M}}$ could be a velocity but the unit analysis: $$[m]\sqrt{\frac{[kg]}{[kg][s^2]}}=\frac{[m]}{[s]}$$ checks out. Also, by just looking at the definition of the dispersion relation $\omega=v_p k$ it makes total sense why $v_p=c_s$ by definition. I don't know why I couldn't see that before. Thank you Lubos and thank you QuantumMechanics! $\endgroup$
    – qmd
    May 28, 2016 at 14:33

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