I am reading Introduction to Polymer Physics by Doi and I am having trouble understanding a derivation by him on the concentration fluctuations in polymer solutions. I have outlined his method and have mentioned where I am having trouble following it:
Consider a mixture of two polymers A and B, having degrees of polymerization $N_A$ and $N_B$, respectively. Let $\phi_A, \phi_B$ be their respective volume fractions. So we can say that $\phi_A+\phi_B = 1$. Define a function based on the spatial coordinate $r$ such that $\phi_x(r)=1$ if polymer X is at point $r$, and $\phi _x (r) =0$ if polymer X is not a point $r$. Let $\langle...\rangle$ denote equilibrium ensemble average. Then, define \begin{align} \langle \phi _A(r) \rangle = \phi _A \quad(1) \end{align} Define $$\delta \phi _A(r)=\phi_A (r) - \phi _A\quad(2)$$ The fluctuation in concentration is defined by the correlation function $$S_{ab}(r-r')=\langle \delta \phi _a (r) \delta \phi _b (r')\rangle \quad(3)$$ where $a,b$ can be any of A and B. This can be proven by seeing that $\delta \phi _A (r)= - \delta \phi _B (r)$, so $$S_{AA}(r) = S_{BB}(r) = -S_{AB}(r) = -S_{BA}(r)$$ Therefore, the concentration fluctuations of an incompressible system are characterized by the single correlation function $S(r)=S_{AA}(r)$.
In order to evaluate $S_{ab}(r)$, we use the following relationship from linear response theory. Consider weak external potentials $u_A(r),u_B(r)$ which act respectively on A and B. The change in the system's potential energy is $$U_{\rm ext}=\int dr[u_A(r)\phi _A(r)+u_B(r)\phi_B(r)] \quad (4)$$ If the external field is small. the deviation $\overline{\delta \phi _a(r)}=\langle \phi _a(r) \rangle _{ext}-\phi _a$ can be written as a linear function of the external potential: $$\overline{\delta \phi_a(r)}=-\sum _{b}\int dr' \Gamma_{ab}(r-r')u_b(r')\quad (5)$$ where $\Gamma _{ab}(r)$ is called the response function. It is related to the correlation function S_{ab} is: $$\Gamma _{ab}(r) =\beta S_{ab}(r)\quad (6)$$
The tricky bit starts here. I want to now prove the above statement (6). I am going to reproduce exactly what Doi has:
To prove the above statement, let us write the intrinsic energy of the system as $U_0$, so that the equilibrium average $\overline{\delta \phi_a}$ in the presence of the external field can be expressed as \begin{align} \overline{\delta \phi _a} &=\frac{T_r\delta \phi _a \exp [-\beta(U_0+U_{ext})]}{T_r \exp [-\beta(U_0 +U_{ext})] } \\ &= \frac{T_r\delta \phi _a \exp [-\beta(U_0+U_{ext})]}{T_r \exp [-\beta(U_0)] }\cdot \frac{T_r \exp [-\beta U_0]}{T_r \exp [-\beta(U_0 +U_{ext})] } \\ &= \frac{\langle \delta \phi _a \exp (-\beta U_{ext})\rangle}{\langle \exp (-\beta U_{ext}\rangle} \quad (7) \end{align} where $\langle ... \rangle = \frac{T_r... \exp [-\beta(U_0)]}{T_r \exp [-\beta(U_0)] }$ denotes the equilibrium average when the external field is not applied. For weak external fields, $\exp(-\beta U_{ext})$ can be approximated as $1-\beta U_{ext}$. So from $(4),(7)$, $$\overline{\delta \phi_a (r)} = \langle \delta \phi _a (r) \rangle (1+\langle \beta U_{ext} \rangle )-\langle \delta \phi _a (r) \beta U_{ext}\rangle = -\beta \sum _b \int dr' \langle \delta \phi _a (r) \delta \phi _b (r') \rangle u_b (r')\quad (8)$$
I do not understand how they came up with the above equation. I apologize if I am not following some basic math, but I am really confused as to how Doi got the above equation $(8)$. If I approximate $$\exp(-\beta U_{ext})=1-\beta U_{ext}$$ If I plug it into $(7)$, I simply do not get the middle term in equation $(8)$. I get $$\overline{\delta \phi _a} = \frac{\langle \delta \phi_a (1-\beta U_{ext})\rangle }{\langle 1-\beta U_{ext} \rangle} = \frac{\langle\delta \phi_a \rangle -\langle \delta \phi _a \beta U_{ext}\rangle}{\langle 1-\beta U_{ext}\rangle} $$
How does he get to $(8)$ from $(1)-(7)$?
tag{1}
instead of\quad (1)
(see my latest edit to the answer), since this would make the formatting consistent with all the other questions on this site. Also, preferably you'd label all equations rather than just the ones you're referring to, because someone else might like to refer to them in a different question or answer or publication, like "using Eq. 9 in this post by megamance...". For the equations between your current (6) and (7), all we can do is say "in the first equation after (6)" or "in the second equation after (6)". $\endgroup$