Surface tension in equilibrium fluids I've been meaning to understand how exactly Eq 27 in this .pdf appears
https://michaelberryphysics.files.wordpress.com/2013/07/berry071.pdf
I've found other papers that mention it ( https://arxiv.org/abs/1603.05291v1  Eq 45.)
But no explicit proof either and they mention it's a common result
My reasoning so far involves:
$$\frac{\delta Z}{\delta A} = \frac{\delta Z}{\delta U} \frac{\delta U}{\delta r_{ij}} \frac{\delta r_{ij}}{\delta A} $$
$$\frac{\delta Z}{\delta U} = -Z/KT $$
$$\frac{\delta U}{\delta r_{ij}} \approx \phi'(r_{ij})$$
$$\frac{\delta r_{ij}}{\delta A} = \frac{x_{ij}^2-z_{ij}^2}{A r_{ij}}$$
So I guess that's where those terms come from. However I do not know how the subindices turn into only 1,2 and the pair correlation appears together with that 1/2 factor.
I'm guessing those three things are related or I might've done some calculation wrong, but I'm stuck here and any help would be appreciated.
 A: By straightforward calculation:
\begin{align}
\gamma &= \left(\frac{\partial F}{\partial A}\right)_{N,V,T} = -\frac{k_BT}{Z}\left(\frac{\partial Z}{\partial A}\right)_{N,V,T} \\
&= -\frac{k_BT}{Z}\left(\frac{\partial }{\partial A} \frac{1}{h^{3N}N!}\int\mathrm{d}\mathbf{r}^N\mathrm{d}\mathbf{p}^N\exp(-\beta\mathcal{H}(\mathbf{r}^N, \mathbf{p}^N))\right)_{N,V,T} \\
&= -\frac{k_BT}{Z}\left(\frac{\partial }{\partial A} \frac{1}{\Lambda^{3N}N!}\int\mathrm{d}\mathbf{r}^N \exp(-\beta U(\mathbf{r}^N))\right)_{N,V,T}
\end{align}
Now the Bogoliubov-Green trick, $\mathbf{r}_i = (\sqrt{A}x'_i, \sqrt{A}y'_i, \frac{V}{A}z'_i)$:
$$\int\mathrm{d}\mathbf{r}^N \exp(-\beta U(\mathbf{r}^N)) = V^N\int_0^1\mathrm{d}\mathbf{r}'^N \exp(-\beta U(\mathbf{r}^N))$$
so that
\begin{align}
\gamma &= -\frac{k_BT}{Z} \frac{V^N}{\Lambda^{3N}N!}\int_0^1\mathrm{d}\mathbf{r}'^N \left(\frac{\partial }{\partial A}\exp(-\beta U(\mathbf{r}^N))\right)_{N,V,T} \\
&= \frac{1}{Z} \frac{V^N}{\Lambda^{3N}N!}\int_0^1\mathrm{d}\mathbf{r}'^N \left(\frac{\partial U(\mathbf{r}^N)}{\partial A} \right)_{N,V,T}\exp(-\beta U(\mathbf{r}^N)) \\
&= \frac{1}{Z} \frac{1}{\Lambda^{3N}N!}\int \mathrm{d}\mathbf{r}^N \left(\frac{\partial U(\mathbf{r}^N)}{\partial A} \right)_{N,V,T}\exp(-\beta U(\mathbf{r}^N)) \\
&= \left\langle\frac{\partial U(\mathbf{r}^N)}{\partial A}\right\rangle_{N,V,T}
\end{align}
Taking then the potential as a pair potential, $U = \sum_{i<j}\varphi(r_{ij})$, we have 
\begin{align}
\gamma &= \left\langle\frac{\partial U(\mathbf{r}^N)}{\partial A}\right\rangle_{N,V,T} = \left\langle\frac{\partial \sum_{i<j}\varphi(r_{ij})}{\partial A}\right\rangle_{N,V,T} \\
&= \sum_{i<j}\left\langle\frac{\partial \varphi(r_{ij})}{\partial A}\right\rangle_{N,V,T} = \frac{1}{2}N(N-1) \left\langle\frac{\partial \varphi(r_{ij})}{\partial A}\right\rangle_{N,V,T}
\end{align}
for any $i, j$, so let's for notational convenience just set them as 1 and 2, respectively, and note that 
$$\left(\frac{\partial \varphi(r_{12})}{\partial A}\right)_{N,V,T} = \frac{\varphi'(r_{12})}{2Ar_{12}}(x_{12}^2+y_{12}^2-2z_{12}^2)$$
Putting this all together:
\begin{align}
\gamma &= \frac{1}{2}N(N-1) \frac{1}{Z} \frac{1}{\Lambda^{3N}N!}\int\mathrm{d}\mathbf{r}^N \frac{\varphi'(r_{12})}{2Ar_{12}}(x_{12}^2+y_{12}^2-2z_{12}^2)\exp(-\beta U(\mathbf{r}^N)) \\
&= \frac{1}{2Z} \frac{1}{\Lambda^{3N}(N-2)!}\int\mathrm{d}\mathbf{r}^N \frac{\varphi'(r_{12})}{2Ar_{12}}(x_{12}^2+y_{12}^2-2z_{12}^2)\exp(-\beta U(\mathbf{r}^N)) \\
&= \frac{1}{2} \int\mathrm{d}\mathbf{r}_1\mathrm{d}\mathbf{r}_2 \frac{\varphi'(r_{12})}{2Ar_{12}}(x_{12}^2+y_{12}^2-2z_{12}^2) \rho^{(2)}(\mathbf{r}_1, \mathbf{r}_2)
\end{align}
Where (see e.g. my answer here for further details)
$$\rho^{(2)}(\mathbf{r}_1, \mathbf{r}_2) = \frac{1}{\Lambda^{3N}Z(N-2)!}\int\mathrm{d}\mathbf{r}_3\cdots\mathrm{d}\mathbf{r}_N\mathrm{d}\mathbf{p}^N\exp(-\beta \mathcal{H}(\mathbf{r}^N, \mathbf{p}^N))$$
Noting the symmetry on the $x$-$y$ plane, we get the final expression:
$$
\gamma = \frac{1}{2A} \int\mathrm{d}\mathbf{r}_1\mathrm{d}\mathbf{r}_2 \rho^{(2)}(\mathbf{r}_1, \mathbf{r}_2)\varphi'(r_{12})\frac{x_{12}^2-z_{12}^2}{r_{12}}
$$
