Derivation on correlation function and response functions in polymer physics

Question

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 $(4),(7)$, $$\overline{\delta \phi_a (r)} = \langle \delta \phi _a (r) \rangle (1+\langle \beta U_{ext} \rangle )-\langle \delta \phi _a (r) \beta U_{ext}\rangle = -\beta \sum _b \int dr' \langle \delta \phi _a (r) \delta \phi _b (r') \rangle u_b (r')\quad (8)$$

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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 $(1)-(7)$?

Answer 1

It looks Doi makes some extra simplifications (beyond expanding the exponential) that are valid when the external field is weak.

Let's start with what you wrote and make one simplification, $$ \begin{eqnarray} \overline{\delta \phi _a} &= \frac{\langle\delta \phi_a \rangle -\langle \delta \phi _a \beta U_{ext}\rangle}{\langle 1-\beta U_{ext}\rangle} \tag{1}\\ &= \frac{\langle\delta \phi_a \rangle -\langle \delta \phi _a \beta U_{ext}\rangle}{1 - \langle \beta U_{ext}\rangle}.\tag{2} \\ \end{eqnarray} $$

Then, since we want the linear response, we continue approximating on the basis that $U_{ext}$ is small. To expand the denominator, we use $$ \frac{1}{1-x} \approx 1 + x,\tag{3} $$ which is valid for small $x$. After dropping any terms with two or more factors of $U_{ext}$, we get $$ \overline{\delta \phi _a} = \langle\delta \phi_a \rangle \left[1 + \langle \beta U_{ext}\rangle\right] -\langle \delta \phi _a \beta U_{ext}\rangle , \tag{4}$$

which is Doi's expression.