where $\{\Delta\}$ is the set of all $\Delta_i$ each of which ranges over all integers. From $Z(Y)$ we can find the orientation dependence of the free energy $F(Y) = -k_B T \ln Z(Y)$, the [*projected*]{} free energy per length $f(m) = F(Y)/L$, and the line tension (or free energy per length) $\beta(\theta) = f(m)\cos \theta$ (since the step length is $L/ \cos \theta$); thence, we can find the stiffness $\tilde{\beta}(\theta) =\beta (\theta )+
\partial^2\beta (\theta )/\partial \theta ^2$.

For future reference, note that the process of extracting an atom from the step-edge and replacing it alongside the edge, discussed in the penultimate paragraph of the introduction, creates two pairs of $\Delta \! =\! +1$ and $\Delta \! =\! -1$, costing $4H$ according to Eq. (\[eq:interfaceEnergy\]) and removing a net of 2 NN bonds, so that $H \! =\! -\epsilon_1/2k_BT\! =\! \varepsilon_k/k_BT$. Similarly, we compare the energies of two NN atoms, abutting \[the lower side of\] a step edge ($\{\Delta_{i}\} \! =\! 0$) at $i_0$ and either parallel or perpendicular to the edge. In the first case, $\Delta_{i_0} \! =\! +1$ and $\Delta_{i_0+2} \! =\! -1$, with an added energy of $2H + 2(U\! +\!D)$ according to Eq. (\[eq:interfaceEnergy\]). In the 