= 1$, and applying , to ${{\mathcal V}}\circ R$ we have that $$\mu_{V_\varphi} = T^{-1}\sharp\mu_{{{\mathcal V}}\circ R},$$ where $$\mathcal V\circ R(T(x)) = V_\varphi(x) - \log(1+|x|^2).$$ This implies that $\mu_{V_\varphi}$ has compact support since $N$ does not belong to the support of $\mu_{{{\mathcal V}}\circ R} $. Moreover, using again to evaluate ${{\mathcal V}}(R T(x))$ we find for any $x$ such that $RT(x)\neq N$, i.e. $x\neq -d/c$, $$\begin{aligned}
&V_\varphi(x)  = V(T^{-1}RT(x)) - \log(1+|T^{-1}RT(x)|^2)  + \log(1+|x|^2),\\
& V_\varphi(-d/c) = {{\mathcal V}}(N) + \log(1+|d/c|^2)=\log(1+|d/c|^2) +\liminf_{|x|\to +\infty} \{V(x) - \log(1+|x|^2)\}.\end{aligned}$$ Finally we find that $$\label{vp} V_\varphi(x)  = V(\varphi(x)) - \log(1+|\varphi(x)|^2) + \log(1+|x|^2), \quad V_\varphi(-d/c) = \liminf_{y\to-d/c} V_\varphi(y).$$

Now we rewrite the discrete energy by changing variables, to find that, writing $w_{n,V}$ instead of $w_n$ to clarify the dependence on $V$, $$\label{disc1}w_{n,V}(x_1,\dots,x_n) = -\sum_{i\neq j}^n\log|\varphi(y_i) - \varphi(y_j)|+ n\sum_{i=1}^nV(\varphi(y_i)),$$ where $x_i = \varphi(y_i)$. Now we use the identity (see [@Hardy]) $$\|T(x)-T(y)\|=\frac{|x-y|}{\sqrt{1+|x|^2}\sqrt{1+|y|^2}}$$ applied to $\varphi( x)$, $\varphi(y)$ together with the fact that $\varphi = T^{-1} R T$ and that $R$ is a rotation to get $$\|T(x)-T(y)\|=\frac{|\varphi(x)-\varphi(y)|}{\sqrt{1+|\varphi(x)|^2}\sqrt{1+|\varphi(y)|^2}}.$$ The two together imply that $$\log|\varphi(x) - \varphi(y)| = \log|x-y| + \frac12 \log (1+|\varphi(x)|^2) + \frac12 \log (1+|\varphi(y)|^2) - \frac12 \log (1+|x|^2)- \frac12 \log (1+|y|^2).$$ Replacing in shows that $$\label{chvar}w_{n,V}(x_1,\dots, x_n) = 