Linear chain of oscillators and normal coordinates

• JTFreitas
In summary, the conversation discusses an attempt to simplify a coupled equation through the use of a discrete Fourier transform. The main goal is to decouple the equation and express it in terms of a coefficient of expansion, denoted as ##a_k##. The conversation also touches upon the need to distinguish the summation index in the last term of the right hand side from the fixed value of ##k## on the left side. It is suggested to use a different index, such as ##l##, in order to avoid confusion. Additionally, it is discussed how the basis functions can be re-expressed using a phase factor.
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.

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

pasmith said:
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.

JTFreitas said:
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}##.

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

1. What is a linear chain of oscillators?

A linear chain of oscillators is a theoretical model used in physics to understand the behavior of a chain of particles connected by springs. Each particle in the chain is considered to be an oscillator, meaning it has the ability to vibrate or oscillate back and forth.

2. What are normal coordinates in the context of a linear chain of oscillators?

Normal coordinates are a set of coordinates used to describe the motion of a linear chain of oscillators. They are chosen in such a way that the equations of motion for each oscillator are decoupled, making it easier to analyze the system.

3. How are the normal coordinates related to the normal modes of vibration in a linear chain of oscillators?

The normal coordinates are directly related to the normal modes of vibration in a linear chain of oscillators. Each normal coordinate corresponds to a specific normal mode, which is a pattern of vibration where all particles in the chain oscillate with the same frequency and phase.

4. How does the number of normal modes in a linear chain of oscillators relate to the number of particles in the chain?

In a linear chain of oscillators, the number of normal modes is equal to the number of particles minus one. This means that for a chain of N particles, there will be N-1 normal modes. This relationship is known as the equipartition theorem.

5. Can a linear chain of oscillators have an infinite number of normal modes?

Yes, a linear chain of oscillators can have an infinite number of normal modes. This is because the chain can have an infinite number of particles, and each particle can contribute to a different normal mode. However, in most practical applications, only a finite number of normal modes are considered.

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