The calculation I sent for Comprehensive exam was only valid for large values of \(R_2-R_1\) or \(R_2\) (as we fixed R1=1). The creutz ratio does not provide the spatial potential but the potential difference of two points.
The following points has to be noted.
- The creutz ratio \(\chi(R_1,R_{2})=-\lim_{z2 \to \infty} ln(\frac{W(R_1,Z_1)W(R_2,Z_2)}{W(R_1,Z_2)W(R_2,Z_1)})\) reduces to difference of potential if we use \(W(R,Z)= e^{-V_s(R)Z+\beta Z(Z+R)}\), where \(V_s(R)= \sigma R+C-\alpha / R\).
- $\(\chi(R_1,R_2)=(z_1-z_2)\left[ \sigma R_1 - \frac{\alpha}{R_1} + c -(\sigma R_2 - \frac{\alpha}{R_2} + c ) \right\)$
- \(V_s(R_2)-V_s(R_1) = \lim_{z2 \to \infty} ln(\frac{W(R_1,Z_1)W(R_2,Z_2)}{W(R_1,Z_2)W(R_2,Z_1)}) \) reduces to V_s(R_2-R_1) only for the terms where \(V_s(R)=\sigma R +C\) which is true for large values of R.
- Now if we like to plot the spatial potential and extract the coulombic portion, we have to plot \(V_s(R_2)-V_s(R_1)\) vs \(R_2\) as \(R_1=1\). Since R_2 is fixed, V(R_1) is also fixed, and the coulombic portion is not changed by shift in potential by a constant. But for different temperatures \(V_s(R_{1})\) may be different and the consntant could be different.
- Now the challenge is to find \(V_s(R_{1})\)