Coupled Harmonic Oscillators – Equations of Motion and Analytical Solutions

classical-mechanicscoupled-oscillatorsharmonic-oscillatorhomework-and-exerciseslagrangian-formalism

We just started QFT, and I'm following our professor's notes but there is a passage I do not understand. We are speaking about a system of $N$ coupled harmonic oscillators $y_j(t)$ for $j = 1, …, N$ and we assume our lattice periodic: $j = j+ N$. The Lagrangian reads

$$L = \frac{m}{2}\sum_j \left(\dot{y_j}^2 -v^2\left(\frac{y_j – y_{j+1}}{\Delta}\right)^2\right)$$

Where $\Delta$ is the distance between atoms among $x$.

Now by Euler-Lagrange equations $$\frac{d}{dt} \frac{\partial L}{\partial \dot{y_j}} = \dfrac{\partial L}{\partial y_j}$$ I was able to get

$$\sum_j \ddot{y_j} = \left(-\frac{v^2}{\Delta^2}\right)\sum_j (y_j – y_{j+1}).$$

hence

$$\ddot{y_j} = \left(-\frac{v^2}{\Delta^2}\right)(y_j – y_{j+1}).$$

But my professor wrote

$$\ddot{y_j} = \left(-\frac{v^2}{\Delta^2}\right)(2y_j – y_{j+1} – y_{j-1}).$$

And here I got stuck.

I don't get if there is some "trick" or meaning for this writing, like something as

$$(y_j – y_{j+1}) = y_j – y_{j+1} – y_{j-1} + y_{j-1}.$$

and then rewriting it, bu I don't know.

Best Answer

I will write down explicitly the sum, so it is clear.

$$L=\frac{m}{2}\Big(\dot{y}_1+...+\dot{y}_i+...+\dot{y}_n -\upsilon^2\frac{(y_1-y_2)^2}{\Delta^2} -\upsilon^2\frac{(y_2-y_3)^2}{\Delta^2} -... -\upsilon^2\frac{(y_{i-1}-y_i)^2}{\Delta^2} -\upsilon^2\frac{(y_i-y_{i+1})^2}{\Delta^2} -... -\upsilon^2\frac{(y_n-y_{n+1})^2}{\Delta^2}\Big)$$ So for a certain $i$, there exist two terms in the Lagrangian, containing $y_i$.

Now, the Euler-Lagrange equations yield $$\frac{d}{dt}\Big(\frac{\partial L}{\partial \dot{y}_i}\Big)= m\ddot{y}_i\\ \Big(\frac{\partial L}{\partial y_i}\Big)= -\frac{\upsilon^2}{\Delta^2}\frac{\partial }{\partial y_i}\Big((y_{i-1}-y_i)^2- (y_{i}-y_{i+1})^2\Big)$$

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