# Lagrangian of 2D square lattice of point masses connected by springs

Zee's QFT book mentions the Lagrangian of a square 2D horizontal lattice of point masses, connected by springs, and considering only vertical displacements $q_{i}$, as

$L = \frac{1}{2} \sum\limits_{a}(m \dot{q}_{a}^{2}) - \frac{1}{2} \sum\limits_{ab}(k_{ab}q_{a}q_{b}) - \frac{1}{2} \sum\limits_{abc}(g_{abc}q_{a}q_{b}q_{c}) - ...$

I have done elementary exercises in Lagrangian Mechanics, using $\frac{1}{2}k(l-l_{0})^{2}$ as the potential energy of the springs, but, after naively trying to derive (with which I mean "to build") that Lagrangian by myself, I suspect I must be missing some kind of additional cross-contributions to the potential energy (and I have no idea where did that triple products $q_{a}q_{b}q_{c}$ emerged from...).

I know this is the abc of solid state physics, that gives rise to the phonons and other interesting stuff, but I am almost completely ignorant in that area. Can anybody at least put me in the right track on how to derive (i.e. to build, departing from some given assumptions) that Lagrangian?

NOTE: In other words, say you want to build that Lagrangian, considering only vertical movements of the masses. The kinetic energy term is obvious, but for the potential energy, is it enough to naively sum $\frac{1}{2}k(l-l_{0})^{2}$ of all the springs? (of course written as a function of the $q_{i}$). Or, perhaps, is there any additional contribution to the potential energy that comes from the fact that the springs are somehow having some influence on each other?

EDIT with some remarks:

Remark 1: A somewhat similar approach to what I am looking for, can be found for a linear chain of atoms, here (Ben Simons, Notes on Quantum Condensed Matter Field Theory, chapter 1)

Remark 2: Thanks very much for correcting my misuse of the english word "derive". Ok, a Lagrangian is not derived. When I say "to derive a Lagrangian" I want to mean "to build a Lagrangian departing from some assumptions" like is the usual approach. For example, I can build the Lagrangian from a double pendulum from the assumptions that the masses of the rods can be neglected and there is no friction, and so I simply add the kinetic energy of the two masses and subtract their gravitational potential energy.

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There isn't really anything to be "derived". The first term is the kinetic term of all the point masses. The remaining terms are the most generic interactions among the point masses you can write down. –  Olaf Dec 13 '12 at 17:36
Apart from the quadratic term, there are only two other terms: The kinetic energy (which you should also have in your classical Lagrangian from which you started your derivation) and the third-order term $g_{abc}q_aq_bq_c$, which you cannot recover from a HO, but only from higher-order contributions, which essentially correspond to multi-particle interactions. –  Claudius Dec 13 '12 at 19:20
Dear Mephisto, it's hard to understand what you mean by a "derivation". The Lagrangian is a defining expression of a theory. One could perhaps "derive" the Lagrangian from a picture of the situation but you didn't give us any. But if the picture is what most of us vaguely imagine, it's obvious that you and Zee wrote the correct Lagrangian. The quadratic and cubic terms are the most general quadratic and cubic terms one may have. In some long-distance approximation, all the 4th-order and higher terms maybe shown to be irrelevant - which is why they're called "irrelevant", too. –  Luboš Motl Dec 13 '12 at 19:26
That potential is defined as the series entering the Lagrangian :). That sounds a bit silly, but it really is nothing more than having $L = T-V$ and $V$ some function of all the $q$'s. Assuming $V$ is analytic we might as well use its Taylor series. Keeping only the lowest order terms produces Zee's potential. –  Olaf Dec 13 '12 at 19:30
In other words, if I would ask you to write down the Lagrangian of a particle hanging vertically from a spring, you wouldn't answer "Oh, it is just the kinetic energy $\frac{1}{2}m\dot{y}^{2}$ minus a generic expansion to account for the potential energy". You would answer "It is $\frac{1}{2}m\dot{y}^{2}-\frac{1}{2}k(y-l_{0})^{2} +mgy$". Now I ask you to write the Lagrangian from a horizontal lattice of identical point masses connected by identical springs that are allowed to move only vertically. How would you approach the problem? –  Mephisto Dec 13 '12 at 23:20

I'm not sure if a Condensed Matter book is going to give you what you want: as pointed out by commenters, you cannot derive a Lagrangian, you can only justify it because it represents the correct physics. But here is a simple interpretation of the 3rd order term. For small deformations, Hooke's law holds and the restoring force $F_{a}=−k_{ab}q_b$. (For isotropic systems this reduces to the familiar $F_a = -kq_a$.) But for larger deformations (beyond the proportionality limit) you get non-linear corrections to the spring constant $\delta k_{ab} \sim g_{abc} q_c$, where $g_{abc}$ is some material-dependent constant. So the constant $g_{abc}$ quantifies how much the stress acting on the springs alters their springiness.
@Mephisto Nowhere does it say that $a\neq b \neq c$. Indeed, later on in the book (p. second page of chapter I.3) he takes the sum over nearest neighbours only. Then only the terms $\sim g_{aba}q_aq_b q_a$ survive and my interpretation is precisely what is going on. To introduce couplings between the springs, you have to add more springs. The whole point of the example is that all interactions between the particles of the mattress can be thought of metaphorically as non-linear springs. –  Mark Mitchison Dec 14 '12 at 21:18