- #1

JTFreitas

- 18

- 3

- Homework Statement
- The equation of motion of a linear chain of ##N## oscillators with equal masses can be written in terms of generalized coordinates:

$$ \ddot{q}_n(t) = \frac{\kappa}{m} (q_{n+1} + q_{n-1} - 2q_n) $$

We can expand the coordinate ##q_n## as

$$ q_{n}(t) = \sum_{k}a_{k}(t)u^{k}_{n}$$

Find the equation of motion in terms of ## \ddot{a}_k(t)##.

- Relevant Equations
- ##u^{k}_{n}## are a set of linearly independent basis functions:

$$u^{k}_{n} = \frac{e^{ikan}}{\sqrt{N}} $$

This way, they are orthonormal:

$$\sum_{n=1}^{N}(u^{k'}_{n})^{*} u^{k}_{n}= \delta_{kk'} $$

The discrete Fourier transform gives us the time-dependent coefficients:

$$ q_{n}(t) = \sum_{k}a_{k}(t)u^{k}_{n} \quad \leftrightarrow \quad a_{k}(t) = \sum_{n}(u^{k}_{n})^{*}q_{n}(t)$$

Forgot to mention earlier, but ##(u^{k}_{n})^{*}## is just the complex conjugate

Hello, I hope the equation formatting comes out right but I'll correct it if not.

So far, I have attempted to write ##\ddot{a}_k(t) = \sum_{n}(u^{k}_n)^*\ddot{q}_n(t) ##. Then I expand the right hand side with the original equation of motion, and I rewrite each coordinate according to its own decomposition:

$$q_n = \sum_{k}a_k u^{k}_{n}, \quad q_{n+1} = \sum_{k'}a_{k'} u^{k'}_{n+1}, \quad q_{n-1} = \sum_{k''}a_{k''} u^{k''}_{n-1} $$

This is an attempt to have something resembling the orthonormality relation in order to simplify things. This way, I get the expression

$$\ddot{a}_k(t) = \frac{\kappa}{m}\sum_{n}(u^{k}_n)^*\left(\sum_{k'}a_{k'} u^{k'}_{n+1} + \sum_{k''}a_{k''} u^{k''}_{n-1} - 2\sum_{k}a_k u^{k}_{n}\right)$$

Then I tried distributing the sum in n through the other sums:

$$\ddot{a}_k(t) = \frac{\kappa}{m}\left(\sum_{k'}a_{k'} \sum_{n}(u^{k}_n)^*u^{k'}_{n+1} + \sum_{k''}a_{k''} \sum_{n}(u^{k}_n)^*u^{k''}_{n-1} - 2\sum_{k}a_k \sum_{n}(u^{k}_n)^*u^{k}_{n}\right)$$

And I'm stuck here. I guess there is some simplifying step that makes each of the terms actually look like the orthonormality condition. We have it on the last term, and since ##k## is the same, we just have that equaling 1. But on the others, both the ##k## and ##n## are different, so I am unsure how to proceed.

So far, I have attempted to write ##\ddot{a}_k(t) = \sum_{n}(u^{k}_n)^*\ddot{q}_n(t) ##. Then I expand the right hand side with the original equation of motion, and I rewrite each coordinate according to its own decomposition:

$$q_n = \sum_{k}a_k u^{k}_{n}, \quad q_{n+1} = \sum_{k'}a_{k'} u^{k'}_{n+1}, \quad q_{n-1} = \sum_{k''}a_{k''} u^{k''}_{n-1} $$

This is an attempt to have something resembling the orthonormality relation in order to simplify things. This way, I get the expression

$$\ddot{a}_k(t) = \frac{\kappa}{m}\sum_{n}(u^{k}_n)^*\left(\sum_{k'}a_{k'} u^{k'}_{n+1} + \sum_{k''}a_{k''} u^{k''}_{n-1} - 2\sum_{k}a_k u^{k}_{n}\right)$$

Then I tried distributing the sum in n through the other sums:

$$\ddot{a}_k(t) = \frac{\kappa}{m}\left(\sum_{k'}a_{k'} \sum_{n}(u^{k}_n)^*u^{k'}_{n+1} + \sum_{k''}a_{k''} \sum_{n}(u^{k}_n)^*u^{k''}_{n-1} - 2\sum_{k}a_k \sum_{n}(u^{k}_n)^*u^{k}_{n}\right)$$

And I'm stuck here. I guess there is some simplifying step that makes each of the terms actually look like the orthonormality condition. We have it on the last term, and since ##k## is the same, we just have that equaling 1. But on the others, both the ##k## and ##n## are different, so I am unsure how to proceed.