is the reduced Gibbs free energy per column. To evaluate the inverse transform, we exploit the saddle point method and obtain (see Appendix for details)

$$Z(Y)\approx \exp \left[ -L\left( \rho _0\tan \theta +\frac{g(\rho _0)}{k_BT}%
\right) \right] ,  \label{eq:ZtoWapproximation}$$

where the saddle point ($\mu_0 =$ $-i\rho _0$) is defined implicitly by the stationarity condition $$-\frac{g^{\prime }(\rho _0)}{k_BT}=m\equiv \tan \theta .
\label{eq:saddlerho}$$ Here, prime (as in $g^{\prime }$) denotes a derivative with respect to $\rho $. This result can be regarded as applying a “torque” to the step to produce a rotation $\theta =\tan ^{-1}m$ from the minimum-energy, close-packed orientation.[@Leamy]

Taking the logarithm of Eq. (\[eq:ZtoWapproximation\]), we find the projected free energy per column $f(m)$ as a Legendre transform of the reduced Gibbs free energy per column $g(\rho _0)$: $$\frac{f(m)}{k_BT}\approx \rho _0m+\frac{g(\rho _0)}{k_BT}.  \label{eq:ftog}$$ Note that this expression is valid only for $L\gg 1$; for finite-sized systems, corrections are required. As standard for Legendre transforms,[@KSK] we have $$\frac{\ddot{f}(m)}{k_BT}=-\frac{k_BT}{g^{\prime \prime }(\rho _0)},$$ where $\ddot{f}\equiv \partial^2 f/\partial m^2$. Using $\beta (\theta)a=f(m )\cos
\theta $ and $m=\tan \theta $, with $a$ the lattice constant of the square (i.e., the column spacing, which is $(1/\sqrt{2})$ the conventional fcc lattice constant), we can rewrite the stiffness as 