because the Casimir repulsion is stronger for teflon. If, instead of a pairwise calculation, we perform an exact three-body calculation of the $h_e$ values for these radii at the equilibrium sphere–sphere separation $d_e = 92.8$ nm, we find that the $h_e$ values change by $< 1$%. Conversely, if we keep $h_e$ fixed and compute the three-body change in $d_e$ (compared to $h\to\infty$), again we find that the change is $<1$%. As mentioned above, the small size of the Si sphere makes it unsurprising that the Si sphere does not change the equilibrium $h_e$ of the much larger teflon sphere. Furthermore, the sensitivity of the sphere–sphere force $F_d$ to the teflon $h$ is equal to the sensitivity of the teflon sphere–plate force $F_h$ to $d$, thanks to the equivalence $\partial
F_d/\partial h = -\partial^2 U/\partial d \partial h = \partial F_h
/\partial d$ where $U$ is the energy. Therefore, one would also not expect the finite value of $h_e$ for the Si sphere to modify the equilibrium $d_e$. Size asymmetry alone, however, does not explain why the finite $h_e$ of the *teflon* sphere does not affect the sphere-plate interactions of the Si sphere. Even if the Si sphere were of infinitesimal 