Problem: infinite diatomic crystal

Consider the following example of a discrete lattice consisting of a periodic repetition of two different masses m_1 and m_2 connected with a spring

Find the equation of motion for u_n and u_{n+1} by a force balance

Solution

\begin{align} m_1\ddot{u}_n &= k_s(u_{n-1} - u_n) - k_s(u_n - u_{n+1}) = k_s(u_{n-1} - 2 u_n + u_{n+1}) \\ m_2\ddot{u}_{n+1} &= k_s(u_n - u_{n+1}) - k_s(u_{n+1} - u_{n+2}) = k_s(u_{n} - 2 u_{n+1} + u_{n+2}) \end{align}

Deriving the steady state solution of a harmonic excitation, we assume that

\begin{align} u_n &=u_{0,n} e^{i\omega t} \\ u_{n+1}&=u_{0,n+1} e^{i\omega t}\\ \end{align}

In the following, we will skip the zero index of the complex amplitude u_{0,j} and just write u_{j} This results in the following equation of motion

(1) \begin{align} -\omega^2 m_1 u_n &= k_s(u_{n-1} - 2 u_n + u_{n+1}) \label{eq:wer} \\ -\omega^2 m_2 u_{n+1} &= k_s(u_{n} - 2 u_{n+1} + u_{n+2}) \label{eq:ert} \end{align}

In the next step, we apply the Floquet theorem to relate the displacements amplitudes u_j to a reference solution u_{ref,m_1} and u_{ref,m_2}

\begin{align} u_{n-1} &=u_{ref,m_2} e^{-ik2aL} \\ u_n &=u_{ref,m_1} e^{-ik2aL} \\ u_{n+1}&=u_{ref,m_2} e^{-ik(2a+2)L} \\ u_{n+2}&=u_{ref,m_1} e^{-ik(2a+2)L} \end{align}

The integer values a can be chosen arbitrarily as one can define any of the masses m_1 and m_2 as the reference mass. Inserting these relations into (1) yields

\begin{align} -\omega^2 m_1 u_{ref,m_1} e^{-i k 2a L} &= k_s\left(u_{ref,m_2}e^{-ik 2a L} - 2u_{ref,m_1}e^{-i k 2a L} + u_{ref,m_2}e^{-ik (2a+2) L}\right) \\ -\omega^2 m_2 u_{ref,m_2} e^{-i k (2a+2) L} &= k_s\left(u_{ref,m_1}e^{-ik 2a L} - 2u_{ref,m_2}e^{-i k (2a+2) L} + u_{ref,m_1}e^{-ik (2a+2) L}\right) \,\,. \end{align}

In the next step, we collect the terms depending on u_{ref,m_1} and u_{ref,m_2}

\begin{align} (2k - \omega^2 m_1) u_{ref,m_1}e^{-i k 2a L} - k_s\left(1 + e^{-ik2L}\right) u_{ref,m_2} e^{-i k 2a L}&= 0 \label{eq:EMphononic31}\\ (2k - \omega^2 m_2) u_{ref,m_2}e^{-i k (2a+2) L} - k_s\left(e^{ik2L} + 1\right) u_{ref,m_1}e^{-i k (2a+2) L} &= 0 \,\,. \label{eq:EMphononic32} \end{align}

The exponential terms depending on the value a cancel out

\begin{align} (2k - \omega^2 m_1) u_{ref,m_1} - k_s\left(1 + e^{-ik2L}\right) u_{ref,m_2} &= 0 \\ (2k - \omega^2 m_2) u_{ref,m_2} - k_s\left(e^{ ik2L} + 1\right) u_{ref,m_1} &= 0 \,\,. \end{align}

Writing everything in matrix notation yields

\begin{bmatrix} 2k_s-\omega^2 m_1 & - k_s\left(1 + e^{-ik2L}\right) \\ - k_s\left(e^{ ik2L} + 1\right) & 2k_s-\omega^2 m_2 \end{bmatrix} \begin{bmatrix} u_{ref,m_1}\\ u_{ref,m_2} \end{bmatrix} = \begin{bmatrix} 0\\ 0 \end{bmatrix} \,\,.

Looking for non-trivial solution with u_{ref,m_1}\neq 0 and u_{ref,m_2}\neq 0, the determinant of the matrix need to vanish. We use the relation 2cos(x)=e^{ix}+e^{-ix} and compute the eigenvalue problem with \lambda=\omega^2

\lambda^2 - 2k_s\left( \frac{1}{m_1} + \frac{1}{m_2} \right) \lambda + \frac{2k_s^2}{m_1 m_2}\left(1-\cos(2k L)\right) = 0
\lambda_{1,2} = k_s\left( \frac{1}{m_1} + \frac{1}{m_2} \right) \pm \sqrt{ k_s^2\left( \frac{1}{m_1} + \frac{1}{m_2} \right)^2 - \frac{2k_s^2}{m_1 m_2}\left(1-\cos(2k L)\right)}

For the circular frequency, there exist four solutions.

\begin{align} \omega_1(k) = \sqrt{\lambda_1}\quad \omega_2(k) = -\sqrt{\lambda_1}\quad \omega_3(k) = \sqrt{\lambda_2}\quad \omega_4(k) = -\sqrt{\lambda_2} \end{align}

These solution show for which combination of frequency and wave number the wave can propagate through the lattice.

\begin{align*} \omega_a&=\omega_3(\pi/2L)=\sqrt{\frac{2k_s}{m_2}} \quad \text{for }m_2 > m_1 \\ \omega_b&=\omega_1(\pi/2L)=\sqrt{\frac{2k_s}{m_1}} \quad \text{for }m_2 > m_1 \\ \omega_c&=\omega_1(0)=\sqrt{2k\left(\frac{1}{m_1} + \frac{1}{m_2}\right)} \end{align*}