# Why can the dispersion relation for a linear chain of atoms (connected by springs) be written as $\omega(k)=c_s \lvert k\rvert$?

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?

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.

• 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.
– qmd
May 28, 2016 at 13:48
• $$c_s$$ 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. May 28, 2016 at 14:06
• 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 $A$. May 28, 2016 at 14:10
• 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. May 28, 2016 at 14:21
• @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!
– qmd
May 28, 2016 at 14:33