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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 $(1)-(7)$?

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    $\begingroup$ +1. It might be nice if you could use 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)". $\endgroup$ Jun 30 at 18:21
  • $\begingroup$ okay, i will do that. thanks for your input! @NikeDattani $\endgroup$
    – megamence
    Jun 30 at 20:24
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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.

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  • $\begingroup$ +1. Very well done for getting this done so quickly! I wonder, why did you use split instead of eqnarray? The disadvantage of split is that you can't label the equations, but I wonder if there may also be an advantage to it. $\endgroup$ Jun 30 at 18:17
  • $\begingroup$ Thanks - no reason beyond habit. I constantly mix up split and aligned, but I've never used eqnarray. Is there a reason to prefer it to aligned? $\endgroup$
    – wcw
    Jun 30 at 19:50
  • $\begingroup$ Man, these approximations come out of nowhere it feels like... thanks for answering! $\endgroup$
    – megamence
    Jun 30 at 20:40
  • $\begingroup$ Yeah, the explanation was definitely lacking, and the approximations do feel random written one after another like that. But the basic idea is simple enough: we want the linear response, so we series expand all nonlinear terms, organize the results, and drop anything quadratic and higher. $\endgroup$
    – wcw
    Jun 30 at 21:16

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