What is the physical meaning of the q in Fourier transforms of correlation functions?

Question

While studying polymer dynamics, one comes across a wide range of correlation functions. A lot of times, we are interested in the Fourier transforms of these functions, as the FT allows us to solve equations which come out of using these correlation functions.

Let's take the example of the RDF of a generic bead-spring polymer, where the location of monomer segment $i$ is denoted by $\mathbf{R}_i$. The RDF of the monomer beads is given by:

$$g(\mathbf{r})=\frac{1}{N}\sum_{n=1}^N \sum_{m=1}^N \langle \delta(\mathbf{r}-(\mathbf{R}_m - \mathbf{R}_n)) \rangle \tag{1} $$

The FT of $g(\mathbf{r})$ is given by: $$g(\mathbf{q}) = \int d\mathbf{r} e^{i\mathbf{q}\cdot\mathbf{r}}g(\mathbf{r})=\frac{1}{N}\sum _{n=1}^N \sum _{m=1}^N\langle \exp[i\mathbf{q}\cdot(\mathbf{R}_m-\mathbf{R}_n)] \rangle \tag{2}$$

The quantity $g(\mathbf{q})$ can be measured experimentally by SAXS.

One application they have in polymer physics of the above quantity is that they calculate the radius of gyration of a polymer, $R_g$, by using the above definition of $g(\mathbf{q})$, and Taylor expanding it at small $\mathbf{q}$. I don't understand the physical interpretation of what it means to be in the regime of small $\mathbf{q}$.

My question is, what exactly is $\mathbf{q}$ over here? The only time I have seen FT's being used is during functional analysis, or analysis of PDEs/ODEs.

Answer 1

Based on your previous questions, I assume this equation was from Section 1.2.1 of Doi's Polymer Physics.

While Doi doesn't explicitly state this anywhere that I can find, $\mathbf{q}$ is the label he uses for the momentum/wavevector. I'm more accustomed to seeing this denoted as $\mathbf{k}$, but as Anyon noted in the comments, $\mathbf{q}$ may be the more common notation when working with scattering. You should also watch out that $\mathbf{q} $ is also used in other places to denote the canonical position.

As for what it means to work in the small $\mathbf{q}$ limit, it helps to think of it as a wavevector, which is related proportional in magnitude to $\frac{1}{\lambda}$. So zero $\mathbf{q}$ corresponds to an infinite wavelength, which means we are essentially taking the limit where $g$ is changing very slowly over space.