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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<W_\lambda''' \right>_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!

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

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