In fact, when $\epsilon_2/\epsilon_1$ is 1/6, so that $-\epsilon_2/2k_BT = (\epsilon_2/\epsilon_1)(\varepsilon_k/k_BT) \approx (1/6)4.64$, the model-predicted value of $k_B T/ \tilde{\beta}a$ has decreased to less than half its $\epsilon_2 \!=\! 0$ value (viz. by a factor of 0.46, vs. 0.63 if Eq. (\[eq:ZVP\]) is used), so about 3/2 the experimental ratio. If $\epsilon_2/\epsilon_1$ increases even further, $k_B T/ \tilde{\beta}a$ further decreases and develops positive curvature, causing an endpoint local minimum to appear at $\theta = 45^\circ$. We can determine when this occurs by expanding Eq. (\[eq:nnnstiffness\]) about $m=1$: $$\begin{aligned}
  \label{eq:mexpand}
  \frac{k_B T}{\tilde{\beta}a} =
%\frac{e^{\frac{\epsilon_2}{2k_BT}}}{\sqrt{2}} + \left( %\frac{e^{-\frac{\epsilon_2}{2k_BT}}}{8 \sqrt{2}} \! -\!\frac{3 %e^{\frac{\epsilon_2}{2k_BT}}}{4 \sqrt{2}}  \right) (m\! -\!1)^2 \! +\! \ldots
\frac{e^{-D}}{\sqrt{2}} + \left( \frac{e^{D}}{8 \sqrt{2}} \! -\!\frac{3 e^{-D}}{4 \sqrt{2}}  \right) (m\! -\!1)^2 \! +\! \ldots\end{aligned}$$ Setting the coefficient of $(m-1)^2$ to zero gives $-2D = \epsilon_2/k_BT = -\ln(6) \approx -1.8$, which corresponds to a value of $k_B T/\tilde{\beta}a=\sqrt{3}/6  \approx 0.29$, about 2/5 the value at $\epsilon_2 \! =\! 0$. For $T=320$K and $\varepsilon_k = 128$ meV, this corresponds to $\epsilon_2/\epsilon_1 \approx 0.2$. However, for the NNN interaction alone to account for the factor-of-4 discrepancy between model/theory and experiment reported by Dieluweit et al.[@dieluweit02], Fig. \[fig:nnnplot\] shows that $\epsilon_2/\epsilon_1 \approx 0.3$ would be required.

