# Expanding ensemble averages

I am reading Statistical Physics of Particles by Kardar. I am struggling with problem 12d, in chapter 2, about semi-flexible polymers in two dimensions. The problem is as follows:

Configurations of a model polymer can be described by either a set of vectors $$\{ t_i \}$$ of length $$a$$ in two dimensions (for $$i=1,\ldots, N$$), or alternatively by angles $$\{ \phi _i \}$$ between successive vectors. The polymer is set at temperature $$T$$, and subject to energy $$H = -k \sum _{i=1}^{N-1} t_i \cdot t_{i+1} = -ka^2 \sum _{i=1}^{N-1} \cos \phi _i$$

The probability of a certain configuration is given by $$\exp (-H/kT)$$.

If the end of the polymer are pulled apart by a force $$F$$, the probabilities for polymer configurations are modified by the Boltzmann weight $$\exp (\mathbf{F}\cdot \mathbf{R}/kT)$$, by expanding this weight, or otherwise, show that $$\langle \mathbf{R} \rangle = K^{-1} \mathbf{F} + O\left( F^3 \right)$$

I know that $$\langle \mathbf{R} \rangle = \frac{\int \mathbf{R} \exp [-\beta H + \beta (\mathbf{F}\cdot \mathbf{R})] d\phi}{\int \exp [-\beta H + \beta (\mathbf{F}\cdot \mathbf{R})] d\phi}$$

From the previous parts of the problem, I know the $$\langle \mathbf{R} \rangle = 0$$ and $$\langle R^2 \rangle = a^2 N \coth \frac{1}{2\xi}$$, where $$\xi$$ is the persistence length.

My question is, how do I expand $$\langle \mathbf{R} \rangle$$? Can anyone give me a mathematically rigorous introduction to expanding such functions? I understand what Taylor expansions are, but how do I apply those principles to an object like $$\langle \mathbf{R} \rangle$$ described above?

I would appreciate any advice you have for me.

He is simply suggesting that you write for small $${\bf F}$$ $$\begin{equation} \exp(\beta({\bf F}\cdot {\bf R})) = 1 + \beta({\bf F}\cdot {\bf R}) +\frac{1}{2}\left [\beta({\bf F}\cdot {\bf R})\right]^2 +\frac{1}{3!}\left [\beta({\bf F}\cdot {\bf R})\right]^3+... \end{equation}$$ in your $$\langle {\bf R}\rangle$$ expression. You then have already calculated (or can easily calculate) the needed averages from your $$F=0$$ results.
Define the averages with no force using your notation $$\begin{equation} \langle O\rangle_0 = \frac{\int d\phi e^{-\beta H} O}{\int d\phi e^{-\beta H}} \,. \end{equation}$$ Then note that $$\langle R_\alpha \rangle_0 = 0$$, $$\langle R_\alpha R_\beta \rangle_0 = \delta_{\alpha\beta}\frac{1}{3}\langle R^2\rangle_0$$, $$\langle R_\alpha R_\beta R_\gamma\rangle_0 = 0$$, etc. along with your value of $$\langle R^2\rangle_0$$ from your previous results for the same problem, where $$R_\alpha$$ is the $$\alpha$$ component of $${\bf R}$$. You can now write your expression for $$\langle {\bf R}\rangle$$ as $$\begin{equation} \langle {\bf R}\rangle = \frac{\langle {\bf R} \exp(\beta({\bf F}\cdot {\bf R}))\rangle_0} {\langle \exp(\beta({\bf F}\cdot {\bf R}))\rangle_0} \,. \end{equation}$$ Expand the exponential as above. Only odd terms in $$F$$, survive in the numerator, and only even terms in the denominator with the 1 term dominant. The linear term in the numerator is the term you want. The cubic term in the numerator along with the expansion of the quadratic term in the denominator multiplying the linear term give the $$F^3$$ correction, but the analysis above shows that there are only odd terms in the full expansion, so you don't actually have to do any extra work.
• Thank you for your response, @user200143. I do not see how to use the averages calculated in from the $F=0$ results. If I employ the above expansion in both numerator and denominator of $\langle R \rangle$, how do I simplify that down to get the required expression? Jul 3, 2022 at 4:50