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.

 

[pasmith]

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.

 

[TSny]

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?

 

[TSny]

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

 

[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.

 

[TSny]

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.