Solving for the deviation in the polymer volume fraction under the random phase approximation

Question

I am reading through Introduction to Polymer Physics by Doi, and I am studying the concentration fluctuation in polymer solutions, specifically the random phase approximation. For the most part, I understand the logic, until they try to solve for the volume fractions from the following three equations: \begin{align} \overline{\delta \phi _A (r)} &= -\beta \int dr' S_{AA}^0 (r-r')[u_A(r')+w_A(r')+V(r')]\label{1}\tag{1}\\ \overline{\delta \phi _B (r)} &= -\beta \int dr' S_{BB}^0 (r-r')[u_B(r')+w_B(r')+V(r')]\label{2}\tag{2}\\ \end{align} Using the constraint relation $\phi _A(r) + \phi_B(r) = 1$, we have $$\overline{\delta \phi _A (r)} + \overline{\delta \phi_B (r)}=0\label{3}\tag{3}$$ The objective is to solve the above 3 equations. We also have the following definitions: \begin{align} w_A(r) &= -z[\epsilon_{AA}\overline{\phi _A(r)}+\epsilon _{AB} \overline{\phi _B (r)}] \label{4}\tag{4}\\ w_B(r) &= -z[\epsilon _{BA}\overline{\phi_A(r)}+\epsilon _{BB}\overline{\phi_B (r)}]\label{5}\tag{5} \end{align} They use Fourier transform to solve these equations and I don't quite understand the process. This is how Doi puts it forward: To solve the above equations, we will use the Fourier transform defined as follows: \begin{align} \overline{\phi _{Aq}}=\int dr e^{iq\cdot r}\overline{\delta \phi _A (r)}=\int dr e^{iq\cdot r}\overline{\phi _A (r)}\label{6}\tag{6} \end{align} (And then suddenly) the above equations take the form \begin{align} \overline{\phi_{A}} = -\beta S_{AA}^0[u_A-z(\epsilon _{AA}\overline{\phi_A}+\epsilon_{AB} \overline{\phi _B}) +V)]\label{7}\tag{7} \\ \overline{\phi_{B}} = -\beta S_{BB}^0[u_B-z(\epsilon _{BA}\overline{\phi_A}+\epsilon_{BB} \overline{\phi _B}) +V)]\label{8}\tag{8} \end{align}

My question is, how did he get to the above equations from the Fourier transforms? I am confused about the math here, if anything.

Answer 1

Doi makes this slightly more confusing because just after these equations, he writes:

Here for simplicity we have dropped the subscript q.

So all of the pieces of your Eqs \eqref{7} and \eqref{8} are actually the Fourier transforms of the pieces of your \eqref{1} and \eqref{2}, so they should all have q subscripts. Its also common to write Fourier transformed functions as $\tilde{f}(q)$ to emphasize that it is not the same functional form as $f(r)$.

The reason the integral disappears is actually due to a general result called the Convolution Theorem. A convolution of two functions $f$ and $g$ is defined as $$f(r)*g(r)\equiv\int f(r-r')g(r')dr'=\int f(r')g(r-r')dr'\tag{1}$$ The Convolution theorem states that $$\mathcal{F}[f(r)*g(r)]=\tilde{f}(q)\cdot \tilde{g}(q)\tag{2}$$ where $\mathcal{F}$ is the Fourier transform. This means convolutions in real space become products in the reciprocal space (and vice versa).

Because your \eqref{1} and \eqref{2} have the form of a convolution, their Fourier transforms are just the products of the Fourier transform of $S_{AA/BB}$ with the Fourier transform of each potential term.