fact that atoms close to $i$ partially shield it from the solvent. $G_\mathrm{pol}$ represents the electrostatic energy required to solvate a pre-assembled group of molecules and thus this term is added to the vacuum electrostatic energy. The various implementations for the GB term differ in how $\alpha_i$ is computed.

For illustrative purposes, let us consider the model of Hawkins [*et al.*]{} [@hawkins95; @hawkins96; @tsui00; @tsui01]. (With minor modifications, the technique is applicable to other GB models.) We express $\alpha_i$ as [@hawkins95 eq. (10)] $$\label{alpha}
\frac1{\alpha_i} = \frac1{\rho_i} -
 \sum_{j\ne i} \Delta_{ij},$$ where $\rho_i$ is the radius of atom $i$, $$\label{deltaij}
\Delta_{ij}
  = \int_{\rho_i}^\infty \frac{dr}{r^2} H_{ij}(r; r_{ij}, \rho_j)$$ is the reduction in the effective inverse Born radius of atom $i$ due to atom $j$. Here $H_{ij}$ is the fraction of the area of a sphere of radius $r$ centered on the $i$th atom eclipsed by a $j$th atom and is given by [@hawkins95 eq. (12)] $$H_{ij}=\left\{
\begin{array}{l@{\hspace{1em}}l}
\displaystyle
\frac{\rho_j^2-(r_{ij}-r)^2}{4r_{ij}r}, &
\mbox{for $\abs{r_{ij}-\rho_j} \le r \le r_{ij}+\rho_j$},\\
1, & \mbox{for $r < \rho_j - r_{ij}$},\\
0, & \mbox{otherwise ($r \gtrless r_{ij} \pm \rho_j$)}.\\
\end{array}
\right.$$ Evaluating the integral in eq. (\[deltaij\]) then gives $$\Delta_{ij} = 
\left\{
\begin{array}{l@{\hspace{1em}}l}
0, & \mbox{for $ \rho_i 