Three dimensional classical continuum limit, wave equation

Question

Many textbooks of classical mechanics or classical field theory mention that a three dimensional "string" (the continuum limit of a lattice) leads to/can be described by the 3 dimensional wave equation

$\square \psi =$ some inhomogeneity.

However, all of the textbooks I've read don't really show that but just say that is either a logical/plausible extension of the 1D string and it's wave equation

$(\frac{1}{c^2} \frac{d^2}{dt^2} - \frac{d^2}{dx^2}) \psi =$ some inhomogeneity,

which one can get with the continuum limit of a chain of harmonic oscillators) or that one can easily derive it with analogous methods - without actually showing how that is done.

So, my question is: How do you get the three dimensional wave equation as a continuum limit of a lattice of harmonic oscillators?

Answer 1

Take for example a series of coupled harmonic oscillators. In discrete case, you have the system of equations: $$ \ddot u_i +\omega^2\sum_{\langle i,j\rangle}u_i-u_j = 0 $$ with $\omega$ in dependent of the site (homogeneity). In the continuum limit, you want to assimilate the discrete $u_i$ by a field $\phi$ given by: $$ \phi(x) = u_i $$ with $x=ia$ and $a$ the lattice spacing. The tricky step is to appropriately renormalise the coupling constant $\omega_0$ in order to have an interesting continuum limit (for statistical or quantum fields, this is more technical). With the benefit of hindsight, the correct scaling is: $$ \omega_0^2 = \frac{c^2}{a^2} $$ which you can anticipate from dimensional analysis alone. In this case, using the $a\to0$ limit: $$ \begin{align} \frac{1}{a^2}\sum_{\langle i,j\rangle}u_i-u_j &= \frac{1}{a^2}\sum_{\langle i,j\rangle}\phi(ia)-\phi(ja) \\ &\to -\Delta \phi(x) \end{align} $$

For example in 1D, this would give: $$ \begin{align} \sum_{\langle i,j\rangle}\phi(ia)-\phi(ja) &= 2\phi(x)-\phi(x+a)-\phi(x-a) \\ &= 2\phi(x)-\phi(x)-\partial_x\phi(x)a-\frac{\partial_x^2\phi(x)}{2}a^2-\phi(x)+\partial_x\phi(x)a-\frac{\partial_x^2\phi(x)}{2}a^2\\ \\ &= -\partial_x^2\phi(x)a^2 \end{align} $$

You thus recover the PDE in the continuum limit: $$ \partial_t^2\phi-\frac{1}{c^2}\Delta \phi = 0 $$

Btw, it turns out that the continuum limit can also be seen as the long wavelength limit of the discrete lattice model as well.

Hope this helps.