# Homework Help: Normal coordinates (small oscillations)

1. Feb 1, 2006

### Chen

Hello,

I solved the problem of small oscillations for a 3-atom molecule, such as CO2, which is modeled as 3 masses connected by 2 springs. Both springs have a constant k, the outer masses are m and the middle one is M.

There are 3 modes of oscillations, and one of them is of course $$\omega$$ = 0, i.e it is a rigid translation of the molecule as a whole. I've also found the normal coordinates for each mode, and for this particular one I found:

$$Q = \frac{1}{\sqrt{2m+M}} (mq_1 + Mq_2 + mq_3)$$

Where qi is the "real" coordinates of each molecule. This seems pretty logical, right? Because basically I foud that the normal coordinates for the $$\omega$$ = 0 mode is exactly the coordinate of the center of mass (after normalization).

However, the exact same problem was solved in Goldstein's "Classical mechanics" (3rd ed.), and a different normal coordinate was found there. It was:

$$Q = \frac{1}{\sqrt{2m+M}} (\sqrt{m}q_1 + \sqrt{M}q_2 + \sqrt{m}q_3)$$

Which is not what I found, nor do I understand its meaning. My friend thinks that both answers are correct, and the difference is just in normalization; I don't agree, one of these answers must be wrong. I'd think that my answer is correct, but since the other one is taken for the book, I'm not so sure.

Thanks,
Chen

2. Feb 1, 2006

### Integral

Staff Emeritus
Dimensionally speaking only one of those solutions can be correct. Yours differs by $\sqrt {m}$. I doubt there is a dimensional error in Goldstein.

Actually it is clear that yours is incorrect.

3. Feb 1, 2006

### Chen

Well, suppose that in my solution I divide by (2m+M) and not its square root, so the dimensions are not a problem. It still makes more sense than the solution in Goldstein, no? Because it's the coordinate of the center of mass. So unless I don't understand the meaning of a normal coordinate, I'd think it looks pretty right.

I double- and triple- checked my solution many times. I diagonalized V (potential energy) over T (kinetic energy), and found that the diagonalizing matrix A, such that $$A^{-1}VA = I$$ is:

$$A = \left(\begin{array}{ccc}\frac{1}{\sqrt{2m+M}}&\frac{1}{\sqrt{2m}}&\frac{M}{\sqrt{2mM(2m+M)}}\\\frac{1}{\sqrt{2m+M}}&0&\frac{-2m}{\sqrt{2mM(2m+M)}}\\\frac{1}{\sqrt{2m+M}}&\frac{-1}{\sqrt{2m}}&\frac{M}{\sqrt{2mM(2m+M)}}\end{array}\right)$$

Which is orthogonal over T, i.e $$A^tTA = I$$. This is also the matrix that appears in Goldstein. Therefore, the normal coordinates are:

$$\vec{Q} = A^{-1}\vec{q} = A^tT\vec{q}$$

Where T is the trivial matrix, with simply m, M, m on the diagonal.

Now you see how I came to my solution?

I don't suppose you could explain why that is?

Thanks,
Chen

Last edited: Feb 1, 2006
4. Feb 2, 2006

Anyone?

Chen