Dispersion on a chain of atoms

Take 15 minutes to prepare this exercise.

Then, if you lack ideas to begin, look at the given clue and start searching for the solution.

A detailed solution is then proposed to you.

If you have more questions, feel free to ask them on the forum.

We want to prove that  longitudinal elastic wave of an angular frequency \(\omega\) and a wave vector \(\vec k = k \vec u_x\) can propagate along  an infinite chain of atoms, of mass \(m\), of a spring constant \(K\) and rest positions :

\(x_0^n=na\)

In complex notation, the movement of the mass numbered \(n\) is :

\(\underline u_n (t)= Ae^{i(\omega t - kx_0^n)}\)

Question

Why doesn't \(A\) depend on \(n\) ?

Solution

The wave is neither stationary nor damped : all atoms are therefore the same amplitude for a progressive wave.

The system is generally translational invariant.

Thus, \(A\) is independent of the considered atom.

Question

Determining, as a function of \(k\), the possible values of the wave frequency \(\omega\) that can propagate on the chain.

What does we deduce from the nonlinearity of this relationship ?

Solution

The Newton's second law applied to the atom number \(n\) gives :

\(m{\ddot u_n}(t) = - K({u_n}(t) - {u_{n - 1}(t)}) + K({u_{n + 1}(t)} - {u_n}(t)) = - K(2{u_n}(t) - {u_{n - 1}(t)} - {u_{n + 1}(t)})\)

The solution proposed in this equation is injected and we obtain :

\(-m\omega ^2=K(e^{ika}+e^{-ika}-2)=-2K(1-coska)\)

Is :

\(\omega (k) = 2\sqrt {\frac{K}{m}} \left| {\sin \left( {\frac{{ka}}{2}} \right)} \right|\)

This dispersion relation is non-linear : there is dispersion.

Question

Calculate the phase \(v_{\varphi}(k)\) and group velocities \(v_g(k)\).

Give their limits for \(ka\to 0\) and \(ka \to \pi\) and comment.

Solution

The phase velocity is : (one located in the following in the case \(0<ka<\pi\))

\({v_\varphi } = \frac{\omega }{k} = 2\sqrt {\frac{K}{m}} \frac{{\sin \left( {\frac{{ka}}{2}} \right)}}{k}\)

And the group velocity :

\({v_g} = \frac{{d\omega }}{{dk}} = a\sqrt {\frac{K}{m}} \cos \left( {\frac{{ka}}{2}} \right)\)

For \(ka \to 0\) (thus big wavelengths) :

\({v_\varphi } \approx {v_g} = a\sqrt {\frac{K}{m}}\)

We find the velocity obtained in the approximation of continuous medium.

And for \(ka \to \pi\) :

\({v_\varphi } \approx \frac{{2a}}{\pi }\sqrt {\frac{K}{m}}\)

And :

\(v_g \approx 0\)

The wave doesn't pass anymore.