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.
Answer
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.
This makes perfect sense: the presence of excitations of the field (mattress) alters the way the excitations move, i.e. a self-interaction of the field (or once you quantise, quanta of the field interacting with each other). In particular, you can see that if two wavepackets collide, the increased amplitude of the deformation will alter the effective spring constant at the impact point, resulting in scattering effects. In the absence of the non-harmonic terms the wavepackets would just pass right through each other.
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