# Linear chain of oscillators and normal coordinates

JTFreitas
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.

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What is $a$ in the definition of $u^k_n$?

JTFreitas
What is $a$ in the definition of $u^k_n$?
In terms of a clear definition, ##a_k## is just the coefficient of expansion of ##q(t)## as a discrete Fourier transform, and it carries the time dependence.

In essence, I introduce the new definition of ##q_n(t)## because my initial equation is coupled (I have ##q_n, q_{n-1}, q_{n+1} ## in a single equation), expanding them in terms of these functions allows me to decouple the equation (essentially having all my different #a_k# terms becoming a single term.) In fact, I do not have a clear definition of what ##a_k## equals, because my goal is to have a differential equation in terms of ##a_k##, so that I can find what it is equal to.

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This is an attempt to have something resembling the orthonormality relation in order to simplify things. This way, I get the expression
$$\ddot{a}_k(t) = \frac{\kappa}{m}\sum_{n}(u^{k}_n)^*\left(\sum_{k'}a_{k'} u^{k'}_{n+1} + \sum_{k''}a_{k''} u^{k''}_{n-1} - 2\sum_{k}a_k u^{k}_{n}\right)$$
You need to change the dummy summation index on the last sum on the right to ##k'''## in order to distinguish it from the fixed ##k## that appears on the left side and in the factor ##(u^{k}_n)^*##.
Then I tried distributing the sum in n through the other sums:
$$\ddot{a}_k(t) = \frac{\kappa}{m}\left(\sum_{k'}a_{k'} \sum_{n}(u^{k}_n)^*u^{k'}_{n+1} + \sum_{k''}a_{k''} \sum_{n}(u^{k}_n)^*u^{k''}_{n-1} - 2\sum_{k}a_k \sum_{n}(u^{k}_n)^*u^{k}_{n}\right)$$
Can you express ##u^{k'}_{n+1}## in terms of ##u^{k'}_{n}##? Similarly for ##u^{k''}_{n-1}##.

JTFreitas
You need to change the dummy summation index on the last sum on the right to ##k'''## in order to distinguish it from the fixed ##k## that appears on the left side and in the factor ##(u^{k}_n)^*##.

Can you express ##u^{k'}_{n+1}## in terms of ##u^{k'}_{n}##? Similarly for ##u^{k''}_{n-1}##.

Is it strictly necessary to change the last index? The original differential equation was for ##q_n##, so the corresponding ##k## ought to be the same, or am I misunderstanding something?

And with regards to re-expressing the basis functions, I am wondering if maybe we can change the index of ##u^{k'}_{n+1}## by multiplying out some phase factor?

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$$\ddot{a}_k(t) = \frac{\kappa}{m}\sum_{n}(u^{k}_n)^*\left(\sum_{k'}a_{k'} u^{k'}_{n+1} + \sum_{k''}a_{k''} u^{k''}_{n-1} - 2\sum_{k}a_k u^{k}_{n}\right)$$
Is it strictly necessary to change the last index? The original differential equation was for #q_n#, so the corresponding ##k## ought to be the same, or am I misunderstanding something?
It's necessary to distinguish the summation index in the last sum from the fixed value of ##k## on the left side.

If you want to use ##k## as the summation index in the last sum, then you need to change the ##k## index on the left side and also in the factor ##(u^{k}_{n})^{*}## to some other symbol, say ##l##. So, you could write $$\ddot{a}_l(t) = \frac{\kappa}{m}\sum_{n}(u^{l}_n)^*\left(\sum_{k'}a_{k'} u^{k'}_{n+1} + \sum_{k''}a_{k''} u^{k''}_{n-1} - 2\sum_{k}a_k u^{k}_{n}\right)$$ If you are having trouble seeing why the ##l## index must be distinguished from the summation index ##k## in the last sum, then we can go through it more carefully.

And with regards to re-expressing the basis functions, I am wondering if maybe we can change the index of ##u^{k'}_{n+1}## by multiplying out some phase factor?
Yes. Using ##u^{k}_{n} = \frac{1}{\sqrt{N}} e^{ikan} ##, you should be able to write ##u^{k+1}_{n}## as a phase factor times ##u^{k}_{n}##. [Correction: I should have written that you should be able to write ##u^{k}_{n+1}## as a phase factor times ##u^{k}_{n}##.]

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• JTFreitas
JTFreitas
It's necessary to distinguish the summation index in the last sum from the fixed value of ##k## on the left side.

If you want to use ##k## as the summation index in the last sum, then you need to change the ##k## index on the left side and also in the factor ##(u^{k}_{n})^{*}## to some other symbol, say ##l##. So, you could write $$\ddot{a}_l(t) = \frac{\kappa}{m}\sum_{n}(u^{l}_n)^*\left(\sum_{k'}a_{k'} u^{k'}_{n+1} + \sum_{k''}a_{k''} u^{k''}_{n-1} - 2\sum_{k}a_k u^{k}_{n}\right)$$ If you are having trouble seeing why the ##l## index must be distinguished from the summation index ##k## in the last sum, then we can go through it more carefully.

Yes. Using ##u^{k}_{n} = \frac{1}{\sqrt{N}} e^{ikan} ##, you should be able to write ##u^{k+1}_{n}## as a phase factor times ##u^{k}_{n}##.

Okay, got it. Possibly you mean the ##u^{k}_{n+1}## as a phase factor times ##u^{k}_{n}##?

Oh, seeing how you altered the LHS index to ##l##, I think I understand why we need to change the summation index. The k-index corresponds to the k-th function of the n-th particle coordinate, but when I expand the n-th coordinate on the right hand side, that index is a dummy index, so it does not correspond to the k-th index too, so it has to be different. Is this right?

I was trying to follow some steps in my textbook and at some point the equation looked like this, so I was not too sure about the indices.

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Possibly you mean the ##u^{k}_{n+1}## as a phase factor times ##u^{k}_{n}##?
Yes, thanks.

Oh, seeing how you altered the LHS index to ##l##, I think I understand why we need to change the summation index. The k-index corresponds to the k-th function of the n-th particle coordinate, but when I expand the n-th coordinate on the right hand side, that index is a dummy index, so it does not correspond to the k-th index too, so it has to be different. Is this right?
Yes, I think that's right.

• JTFreitas