# Ensemble average of polarization with applied field

This question is in regard to a dipole moment fluctuation formula seen in: MOLECULAR PHYSICS, 1983, VOL. 50, NO. 4, 841-858, on page 843.

For a system of polar liquid (water) at equilibrium, the ensemble average of the total dipole moment $$M$$ is: $$\left = \frac{\int dq M\exp(-\beta U(q))}{\int dq \exp(-\beta U(q))}$$ Where $$q$$ is a collective variable (for e.g. position and momentum) that dictates the Hamiltonian ($$U$$) of the system. The denominator is the partition function. Now consider a small external applied electric field $$E_o$$. The ensemble average of the total dipole moment is now: $$\left_E = \frac{\int dq M\exp(-\beta (U(q)-ME_o))}{\int dq \exp(-\beta (U(q)-ME_o))}$$ where the Hamiltonian has an extra term that accounts for the field-dipole interaction.

To simplify the above expression, literature often states "apply linearization on $$E_o$$" and that's where I'm confused. My interpretation of that is: $$\exp{(\beta ME_o) \approx 1 + \beta ME_o}$$ which makes $$\left_E = \frac{\int dq M\exp(-\beta U(q)) + M^2\beta ME_o \exp(-\beta U(q))}{\int dq \exp(-\beta (U(q)-ME_o))}$$ and I'm stuck here.

In the final expression shown in the paper, they arrived at: $$\left_E = \beta E_o\left$$ where the bracket around $$M^2$$ denotes average in equilibrium of no field, i.e. $$\left=\frac{\int dq M^2\exp(-\beta U(q))}{\int dq \exp(-\beta U(q))}$$.

Could someone help me explain how they arrived at their final expression?

I've found an answer. My misunderstanding was from the term "linearization". That term was meant for $$\left_E$$ and not $$E_o$$.
Starting from ensemble average of $$M$$ with a field $$E_o$$: $$\left_E = \frac{\int dq M\exp(-\beta (U(q)-ME_o))}{\int dq \exp(-\beta (U(q)-ME_o))}$$ The "linearization" here means to expand $$\left_E$$ to its first order in the Taylor series: $$\left_E \approx \left + \frac{\delta\left_E}{\delta E_o}$$ Applying product/quotient rule to the expression of $$\left_E$$, we'll find: $$\frac{\delta\left_E}{\delta E_o}= \frac {\int dq \beta M^2E_o\exp(-\beta (U(q)-ME_o))\int dq \exp(-\beta (U(q)-ME_o)) - \int dq M\beta E_o\exp(-\beta (U(q)-ME_o))\int dq M\exp(-\beta (U(q)-ME_o))} {\int dq \exp(-\beta (U(q)-ME_o))^2}$$
$$\frac{\delta\left_E}{\delta E_o}= \frac {\int dq \beta M^2E_o\exp(-\beta (U(q)-ME_o))} {\int dq \exp(-\beta (U(q)-ME_o))}- \frac {\int dq M\beta E_o\exp(-\beta (U(q)-ME_o))\int dq M\exp(-\beta (U(q)-ME_o))} {\int dq \exp(-\beta (U(q)-ME_o))^2}$$ Then by following the definition of ensemble average: $$\frac{\delta\left_E}{\delta E_o}= \beta E_o\left - \beta E_o\left\left$$
Finally, by also assuming $$\left=0$$ for an isotropic system with no field: $$\left_E = \left + \beta E_o (\left - \left\left) = \beta E_o\left$$.