How do I write the Lagrangian for a system with 2 different locations of oscillation?

Question

I have a system where there is a particle placed in each of the minima of the potential $$U(x)=\beta(x^2-\alpha^2)^2.$$ The particles are also connected by a massless spring where the equilibrium length is the distance from one minimum to another.

The potential about small oscillations at $\pm\alpha$ is $U(x)=4\alpha^2 \beta (x\mp \alpha)^2$.

When writing down the Lagrangian, can I pick either of the two potentials or does it matter which one I pick? I'm going off the assumption that there must be only 1 Lagrangian for this system, not 2.

Answer 1

Just write down the Lagrangian for the full $U(x)$ and spring system. For the second paragraph, I would advise giving those different names $U_{appr}$. You can write down the equations of motion without approximation. But solving them is a separate matter.

So

$$ \mathcal{L} = \frac{1}{2} m_1 \dot{x}_1^2 + \frac{1}{2} m_2 \dot{x}_2^2 - \beta (x_1^2 -a^2)^2 - \beta (x_2^2 -a^2)^2 - \frac{1}{2} k (x_2 - x_1 - 2a)^2\\ \frac{d}{dt} \frac{d\mathcal{L}}{d\dot{x_1}} - \frac{d\mathcal{L}}{dx_1} = 0\\ \frac{d}{dt} \frac{d\mathcal{L}}{d\dot{x_2}} - \frac{d\mathcal{L}}{dx_2} = 0\\ $$

You get the equations of motion for each of them. You should look for a solution that has

$$ x_1 \approx -a + \epsilon y_1 + O(\epsilon^2)\\ x_2 \approx a + \epsilon y_2 + O(\epsilon^2 )\\ $$

Plug this approximation in to the equations of motion and keep only up to order $\epsilon$. Try solving that instead.

Answer 2

Revised Answer

This problem is simple enough that you can determine the frequencies of the normal modes quite easily. The Lagrange Method is not necessary.

The 2 minima are mirror images of each other. There will be 2 normal modes : the particles moving anti-symmetrically in the same direction, and symmetrically in opposite directions. In both cases you can ignore the other particle : in the anti-symmetric case because the spring remains relaxed, in the symmetric case because the spring acts as though fixed at the line of symmetry.

Any other small oscillation is a linear combination of these 2 modes.

For small displacements $z=x-a$ from equilibrium, the potential $U(x)$ is parabolic - ie $U(z) \approx \frac12kz^2$. This applies for all potentials, including the spring - exact in this case.

For small displacements from equilibrium at the local minimum, the potential well is equivalent to a spring of constant $k_1$. The restoring force on the particle is $F(z)=-\frac{dU(z)}{dz}$ to 1st order in extension $z$, evaluated at $z=0$. Compare the result with Hooke's Law $F=-k_1 z$ to find the spring constant $k_1$. Then you can immediately write down the angular frequency and period of small oscillations.

In the 2nd case you have 2 springs in parallel; the combined spring constant is $k_2=k_1+2K$, where $K$ is the constant for the given spring. Note that for a given tension in a spring the extension at either end is half the total extension. You are effectively cutting the spring in half. It takes twice the force to produce the same extension in half the spring, so the spring constant is doubled, not halved.