 Solution

Under uniform geometric scaling, write ${\mathbf{r}}_i^{\prime }=\lambda {\mathbf{r}}_i$. Each volume element contributes a factor ${\lambda }^3$, while the inverse-sixth-power separation contributes ${\lambda }^{-6}$. The double integral therefore scales as ${\lambda }^3{\lambda }^3{\lambda }^{-6}=1$, andThis assumes unchanged material Hamaker constant, pairwise-additive nonretarded interactions and the same intervening medium. It is a simultaneous scaling of both body sizes and their separation, not merely moving fixed bodies apart.