1d atom chain continuum approximation

Question

Im reading "Condensed Matter Field Theory" by Atland and Simons. I have a fairly simple question. They are talking about a 1d chain of $N$ atoms, the Lagrangian for this system is straight forward: $$ L = \sum_{n=1}^{N} (\frac{m}{2}\dot{\phi}_n^2 - \frac{k_s}{2}(\phi_{n+1} - \phi_n)^2) $$ where $\phi_n$ is simply the displacement from equilibrium. He argues that iff $\phi_{n+1}-\phi_n << a \,\,\forall n$, where $a$ is the lattice spacing then: $$ \mathcal{L} = (\frac{m}{2}\dot{\phi}(x,t) - \frac{k_sa}{2}(\partial_x \phi(x,t))^2). $$ I have the following problem:

With this continuum business, one also identifies the integral as usual: $$ \int_0^{L=NA} f(x) dx= \sum_{n=0}^N f(an)a $$ Say, you choose $f(x)=x^2$, where $x$ would be the stretching of the spring like in the Lagrangian. Suppose we choose to integrate between two points $A$ and $B=A+a$ such that in the discrete case $\phi_{A+1}-\phi_{A}=0$ i.e. no stretching, but in the continuum only the endpoints are under no stretching, in between the continuum "continuation" would yield a non zero value of the integral. If this is true, the continuum approximation is adding non existing potential energy to the system. Why is my reasoning wrong or why is it okay to use this approximation.

Tl;dr: I basically need a detailed step by step derivation of going from the discrete Lagrangian of a 1d atomic chain to its continuum Lagrangian density.

Answer 1

I have to admit that I didn't follow your argument and about the $f(x)=x^2$ example (where does such a term exist in the Lagrangian? why $\phi_{A+a}-\phi_A=0$ here?) Specifically, note that we don't integrate over a function like $x^2$ but over the fields at a certain point $x$.

I do hope that I can elucidate the mapping from the lattice to the continuum.

We start with a lattice Lagrangian, as you give in your question

$$ L = \sum_n\left[ \frac{m}{2}\dot{\phi}^2_n-\frac{k_s}{2}\left(\phi_{n+1}-\phi_n\right)^2\right]$$ and now we want to replace everything with fields in the continuum limit. So we first need to define these fields. As sums are going to be replaced by integrals, it is very convenient to add the lattice spacing to the definitions of the fields themselves. So we define $\phi_n = \sqrt{a} \phi(na)$ (from now on $\phi$ followed with an argument in parenthesis is our new continuum-limit field). So we rewrite the Lagrangian as $$ L = \sum_n a \left[\frac{m}{2}\dot{\phi}^2(na)-\frac{k_s}{2}\left(\phi(na+a)-\phi(na)\right)^2 \right]$$ and we can readily transform the sum into an integral, with $dx = a$ $$ L = \int\! dx \left[\frac{m}{2}\dot{\phi}^2(x)-\frac{k_s}{2}\left(\phi(x+a)-\phi(x)\right)^2 \right]$$ now we just need to expand the difference, and keep the leading order (which is ok as long as this difference is indeed small - that is the functions are smooth enough) and we get $$ L = \int\! dx \left[\frac{m}{2}\dot{\phi}^2(x)-\frac{k_sa^2}{2}\left(\partial_x\phi(x)\right)^2 \right]$$ note that from dimensionality perspective, you need $a^2$ and not $a$ there (since $m/t^2$ has the same dimensions as $k_s$).