Given some solid, how can we model sound waves moving through it?

Let us assume that this is a simple solid, made of identical atoms spaced evenly. We know that atoms in a solid lattice are held together by binding forces, for example ionic, covalent and metallic bonds. The Mie potential shown on the right is generally a good model for binding forces in solids.

Let us model the binding forces using springs. We model all atoms as connected to only their nearest neighbours by identical springs. This model is accurate in the small displacement limit, that is the low temperature limit.

Let's explore this model. We will start considering the simple case of a one dimensional lattice, building up the theory. Here we will develop the concept of the first Brillouin Zone and how energy moves through the solid in the form of waves. Then we will look at a two dimensional lattice and consider transverse and longitudinal waves passing through the solid. We will continue to three dimensional solids for completeness.

We start with the one dimensional case: imagine having a chain of identical atoms. We then approximate their interactions to only be with nearest neighbours. We do this by saying that neighbouring atoms are connected by springs, all of which are identical.

The solution to this model is most easily dealt with when the infinite lattice approximation is made. This is where we approximate the solid size as much larger than the maximum displacement of individual atoms. This is physically acceptable as in reality, atom movements in solids are much smaller than the size of the solids. The most general solution is $$ u_n(t) = Re \sum_{k}\tilde{u}_k e^{i(nka-\omega_kt)} .$$ This is a sum of an arbitrary number of waves, each with their own wavevector. $u_n$ is the displacement of the $n^{th}$ atom from its equilibrium position and $k$ is the wavevector of the incoming wave. (Derivation is available at the end of this suite under the 'Derivation' button.)

Here we simulate for a single $k$: $$ u_n(t) = u_k \cos(nka-\omega_kt)$$ For any given $k$: $$ \omega_k = 2 \omega_D \left| \sin \left( \frac{k a}{2} \right) \right|$$ where $$ \omega_D = \sqrt{\frac{\kappa}{m}}$$ $\kappa$ is the spring constant, $m$ is the mass of a single atom, $a$ is the atomic spacing and $\omega_D$ is also known as the Debye frequency. Here, we assume $a = \omega_D = 1$ and $\tilde{u}_k$ has zero phase for simplicity.

In general, the energy of an acoustic wave of angular frequency $\omega_k$ is $$ E(k) = \hbar \omega_k. $$

$u_k$ is the amplitude of the response of the individual atoms to wave mode $k$. Notice that the overall motion is unaffected if $2n \pi$ for any integer $n$ is added to $k$. Mathematically, this is due to the factor $e^{i2n\pi} = 1$ in the general solution. Physically, it makes sense due to the given trigonometric dispersion relation for $\omega_k$ - $k + 2n\pi$ has the same energy as $k$. (Try adding integer multiple of 2 to the $d$ value!)

For the upcoming visualisations, we will ignore this redundancy by taking $-\pi \lt ka \leqslant \pi $, known as the first Brillouin Zone. This applies to components in the higher dimensions too.

Also notice that when $k \to \pm \pi$ (i.e. $d \to \pm 1$), there are two standing waves. Neighbouring atoms move in anti-phase. (Reduce $u_k$ to see this effect better.)

Now consider an infinitely wide sheet of identical atoms. The most general solution is: $$ \textbf{u}_n(t) = Re \sum_{\textbf{k}}\tilde{\textbf{u}}_\textbf{k} e^{i(\textbf{k} \cdot \textbf{R}_n - \omega_\textbf{k} t)} $$ where $\textbf{u}_n$ is the displacement of the $n^{th}$ atom and $\textbf{R}_n$ is the lattice site that atom, and $\textbf{k}$ is the wavevector of each incoming wave.

We simulate for a single $\mathbf{k}$: $$ \textbf{u}_n(t) = \textbf{u}_\textbf{k} \cos(\textbf{k} \cdot \textbf{R}_n - \omega_k t)$$ For any given $\mathbf{k}$: $$ \omega^2_{\mathbf{k}} = 4 \omega^2_D \left[ \sin^2 \left(\frac{k_x a}{2} \right) + \sin^2 \left(\frac{k_y a}{2} \right) \right]$$ where $$ \omega_D = \sqrt{\frac{\kappa}{m}}$$ $\kappa$ is the spring constant, $m$ is the mass of a single atom, $a$ is the atomic spacing and $\omega_D$ is also known as the Debye frequency. We also assume $a = \omega_D = 1$ and $\tilde{\textbf{u}}_\textbf{k}$ has zero phase for simplicity.

We also trace the wave, which travels at phase velocity $\mathbf{v}=\frac{\mathbf{k}}{w}$.

$\textbf{u}_k$ represents the response of the individual atoms to wave $\textbf{k}$. Notice that $|\textbf{k} \cdot \textbf{u}_k|$ and $|\textbf{k} \times \textbf{u}_k|$ determine the type of wave: longitudinal, transverse or a combination of both. Try making $|\textbf{k} \cdot \textbf{u}_k| = 0 $ or $|\textbf{k} \times \textbf{u}_k| = 0 $. What are the types of wave you see in these respective cases?

Finally, we look at a three dimensional infinite crystal lattice. This simulation is meant for completeness. The colour coding here is to track layers of atoms. If this looks confusing for you, you may wish to return to the two dimensional case as it encompasses all the essentials of physics involved.

A reminder of the general solution: $$ \textbf{u}_n(t) = Re \sum_{\textbf{k}}\tilde{\textbf{u}}_\textbf{k} e^{i(\textbf{k} \cdot \textbf{R}_n - \omega_\textbf{k} t)} $$ For a single $\mathbf{k}$: $$ \textbf{u}_n(t) = \textbf{u}_k \cos(\textbf{k} \cdot \textbf{R}_n-\omega_k t)$$ For any given $\mathbf{k}$: $$ \omega^2_{\mathbf{k}} = $$ $$ 4 \omega^2_D \left[ \sin^2 \left(\frac{k_x a}{2} \right) + \sin^2 \left(\frac{k_y a}{2} \right) + \sin^2 \left(\frac{k_z a}{2} \right) \right]$$ where all the symbols have the same meanings as before.

The types of waves available are still longitudinal, transverse or a combination of both. This is indicated by $|\textbf{k} \cdot \textbf{u}_k|$ and $|\textbf{k} \times \textbf{u}_k|$.

Derivation