Amplitudes of Fourier expansion of a vector as the generalized coordinates

Question

On page 14, under the subtopic "Constraints", when discussing about generalized coordinates, Goldstein says the following:

All sorts of quantities may be impressed to serve as generalized coordinates. Thus, the amplitudes in a Fourier expansion of ${\bf r}_j$ may be used as generalized coordinates, or we may find it convenient to employ quantities with the dimensions of energy or angular momentum.

I understand that generalized coordinates need not be orthogonal position vectors. But what does Fourier expansion of a vector even mean? A vector has 3 coordinates, now the amplitudes of Fourier expansion are infinite. How can they be used as generalized coordinates?

Answer 1

To stay within finite (as opposed to infinite) degrees of freedom (which is the topic of the book), Goldstein might have in mind a complex discrete (as opposed to continuous) Fourier transform $$ {\bf r}_j(t)~=~\sum_{k=1}^N e^{2\pi i jk/N }{\bf q}_j(t)$$ (or a real version thereof). Here the amplitudes ${\bf q}_j(t)$ play the role of generalized coordinates. This is for instance useful for finding normal modes in lattice models.

Answer 2

A response from Physicsforums:

As I already said, that's not enough context to make sense about this statement. I'd tend to think, it doesn't make sense here at all since for point mechanics you have a finite number of independent degrees of freedom in configuration space, and it doesn't make sense to use a Fourier transformation of some kind to relabel them.

This changes for the case of fields, where you have a continuous number of field-degrees of freedom, which you may either label in terms of $\vec{x}$ or of $\vec{k}$, where $$\phi(t,\vec{x})=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{k}}{(2 \pi)^3} \tilde{\phi}(t,\vec{k}) \exp(-\mathrm{i} \vec{k} \cdot \vec{x}).$$

-vanhees71