# Derivation of Perturbation Terms in Thermodynamic Perturbation Theory

In the "A critical evaluation of perturbation theories by Monte Carlo simulation of the first four perturbation terms in a Helmholtz energy expansion for the Lennard-Jones fluid" paper by T. van Westen and J. Gross the residual Helholtz energy is given by the expression

$$\begin{equation} A^{res} = A_0 + \int_0^1 \left< W_\lambda'(r^N)_{\lambda}\right> d\lambda \end{equation}$$ Here $$A_0$$ is a reference contribution term to the Helholtz Energy and $$\left<\right>_\lambda$$ denotes a statistical ensamble over a canonical fluid characterized by $$\lambda$$. $$\lambda$$ is a coupling character that can be gradually switched between the intermolecular potential of a reference fluid $$U_0$$($$\lambda = 0, W_{y = 0} = 0$$ and that of the desired fluid ($$\lambda = 1$$). In thermodnamic perturbation theory the decomposition of the intermolecular potential can be given by

$$\begin{equation} U_\lambda (r^N) = U_0(r^N) + W_\lambda(r^N) \end{equation}$$

In their paper they say that by expanding the ensamble average by a Taylor series about $$\lambda = 0$$ and letting the dimenionless Helholtz per particle be $$\tilde{a} = \beta A/N$$ where $$\beta = 1/kT$$ and defining $$\begin{equation} \tilde{a}^{res}= \displaystyle\sum_{i = 1}^{\infty}\tilde{a}_i \end{equation}$$ they find that

$$\begin{equation} \tilde{a_1} = \frac{1}{N}\left<\beta W'_\lambda \right>_0 \end{equation}$$

$$\begin{equation} \tilde{a_2} = \frac{1}{2!N}( \left< \beta W_\lambda''\right>_0 - \left< (\beta W'_\lambda - \left< \beta W_\lambda'\right>_0)^2\right>_0) \end{equation}$$

$$\begin{equation} \tilde{a_3} = \frac{1}{3!N} (\left_0 + 3(\left< \beta W_\lambda'\beta W_\lambda''\right>_0 - \left< \beta W_\lambda'\right>_0\left< \beta W_\lambda''\right>_0))+ \left< (\beta W_\lambda' - \left< \beta W_\lambda'\right>_0)^3\right>_0) \end{equation}$$.

How are the expressions for $$a_1$$,$$a_2$$ and $$a_3$$ derived? I am simply not able to derive these equations from the information. I do belive that it might be my mathematical understanding that is letting me down. Any help would be greatly appreciated!

• Still interested in help. – Lodin Ellingsen Oct 3 '19 at 9:29