 #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:
$$ \ddot{q}_n(t) = \frac{\kappa}{m} (q_{n+1} + q_{n1}  2q_n) $$
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 timedependent 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:
$$q_n = \sum_{k}a_k u^{k}_{n}, \quad q_{n+1} = \sum_{k'}a_{k'} u^{k'}_{n+1}, \quad q_{n1} = \sum_{k''}a_{k''} u^{k''}_{n1} $$
This is an attempt to have something resembling the orthonormality relation in order to simplify things. This way, I get the expression
$$\ddot{a}_k(t) = \frac{\kappa}{m}\sum_{n}(u^{k}_n)^*\left(\sum_{k'}a_{k'} u^{k'}_{n+1} + \sum_{k''}a_{k''} u^{k''}_{n1}  2\sum_{k}a_k u^{k}_{n}\right)$$
Then I tried distributing the sum in n through the other sums:
$$\ddot{a}_k(t) = \frac{\kappa}{m}\left(\sum_{k'}a_{k'} \sum_{n}(u^{k}_n)^*u^{k'}_{n+1} + \sum_{k''}a_{k''} \sum_{n}(u^{k}_n)^*u^{k''}_{n1}  2\sum_{k}a_k \sum_{n}(u^{k}_n)^*u^{k}_{n}\right)$$
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:
$$q_n = \sum_{k}a_k u^{k}_{n}, \quad q_{n+1} = \sum_{k'}a_{k'} u^{k'}_{n+1}, \quad q_{n1} = \sum_{k''}a_{k''} u^{k''}_{n1} $$
This is an attempt to have something resembling the orthonormality relation in order to simplify things. This way, I get the expression
$$\ddot{a}_k(t) = \frac{\kappa}{m}\sum_{n}(u^{k}_n)^*\left(\sum_{k'}a_{k'} u^{k'}_{n+1} + \sum_{k''}a_{k''} u^{k''}_{n1}  2\sum_{k}a_k u^{k}_{n}\right)$$
Then I tried distributing the sum in n through the other sums:
$$\ddot{a}_k(t) = \frac{\kappa}{m}\left(\sum_{k'}a_{k'} \sum_{n}(u^{k}_n)^*u^{k'}_{n+1} + \sum_{k''}a_{k''} \sum_{n}(u^{k}_n)^*u^{k''}_{n1}  2\sum_{k}a_k \sum_{n}(u^{k}_n)^*u^{k}_{n}\right)$$
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.