What sort of coupling of oscillators yields a potential of $\lambda q(t)Q(t)$?

Question

I'm trying to work through Galley's paper, The classical mechanics of non-conservative systems. It starts with a toy problem to demonstrate the need for his full approach. I'm certain understanding this toy problem is not essential for the paper, but the particulars leave me wondering if he is describing an actual physically-realizable system, or if he is choosing an unrealistic toy problem in order to make the mathematics easier.

An illustrative example. To demonstrate the shortcoming of Hamilton's principle, consider a harmonic oscillator with amplitude $q(t)$, mass $m$, and frequency $\omega$ coupled with strength $\lambda$ to another harmonic oscillator with amplitude $Q(t)$, mass $M$, and frequency, $\Omega$. The action for this system is

$$S[q,Q] = \int_{t_i}^{t_f}dt\left\{\frac m 2(\dot q^2 - \omega^2q^2) + \lambda qQ + \frac M 2(\dot Q^2 - \Omega^2Q^2)\right\}.\tag{1}$$

The two individual oscillator terms make sense, as the usual $T-U$ construction. However, I am trying to figure out what to do with $\lambda qQ$. If I try to get a force by taking the gradient, I get something that doesn't appear to be remotely linear. It feels strange that one might start with the simplest systems one can work with (harmonic oscillators), and then couple them with a rather exotic contraption.

Am I missing an obvious way to create a simple system with this potential term, or is this really just a rather unrealistic way to couple the systems for the sole purpose of making the later mathematics easier.

Answer 1

I think the problem of coupled pendulums or any coupled harmonic oscillators are based on a mutual force proportional to the relative displacement of the oscillators

$$ m\,\frac{d^2q}{dt^2} = - k(q - Q) $$

$$ m\,\frac{d^2Q}{dt^2} = - k(Q - q). $$

You can check that the potential that gives the right force of this system is

$$ U(q,Q) = \frac 12 k (q-Q)^2 $$

The full Hamiltonian is

$$ H = \frac 12 m \dot q^2 + \frac 12 m \dot Q^2 + \frac 12 (q-Q)^2 $$

If you open the binomial in the potential, we can write

$$ H = \left (\frac 12 \dot q^2 + \frac 12 kq^2\right ) + \left(\frac 12 m\dot Q^2 + \frac 12 kQ^2 \right) + kqQ $$

which is a linear term, like the one you have. The Masses could be different, with result in different independent frequencies.

Answer 2

The Langrange function is

$$\mathcal L=\frac m2 \left( {{\dot q}}^{2}-{\omega}^{2}{q}^{2} \right) +\lambda\,q\,Q+ \frac M2 \left( {{\dot Q}}^{2}-{\Omega}^{2}{Q}^{2} \right) $$

substitute $~\omega^2=\frac \lambda m~,\Omega^2=\frac \lambda M~$ you obtain

$$\mathcal L=\underbrace{\frac m2{{\dot q}}^{2}+\frac M2{{\dot Q}}^{2}}_{T}\underbrace{-\frac 12\,\lambda\,{q}^{2}+ \lambda\,q\,Q-\frac 12\,\lambda\,{Q}^{2}}_{-\frac \lambda 2\,(q-Q)^2=-U} $$

$~\lambda \quad [N/m]~$ is the spring constant

$$m\,\ddot q+\lambda(q-Q)=0\quad, \omega^2=\frac \lambda m\\ M\,\ddot Q-\lambda(q-Q)=0\quad, \Omega^2=\frac \lambda M$$