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Frederick Lindemann's heuristic associates melting with vibrational displacement reaching a specified fraction of the nearest-neighbor distance $d$. The displacement convention must be stated: one-coordinate variance and full three-dimensional variance differ by a factor of three. With the high-temperature Einstein mean-square displacement, the displayed convention gives $T_m=c_L^2{d}^{2}m{k}_B{T}_E^2/{\hslash }^2$. Neither $c_L$ nor the independent-oscillator approximation is universal.

For one Cartesian displacement, the harmonic potential is $m_{\mathrm{Ar}}{\omega }_E^2{x}^2/2$. At $T\gg T_E$, the equipartition theorem assigns $k_{B}T/2$ to this quadratic potential term. HenceThis is a one-coordinate mean squared displacement, not the total three-dimensional displacement variance, which is three times larger in the isotropic model.

For comparison, the exact Einstein mean-square displacement isUsing $\coth y\simeq 1/y$ recovers the requested high-temperature result. Its nonzero zero-temperature limit is the displacement contribution of the zero-point energy.

Square the stated Lindemann melting criterion and insert the high-temperature displacement:Thus, with $m_{\mathrm{Ar}}=(0.040\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1})/N_A$,Here $T_m/T_E\simeq 1.66$, so the use of the high-temperature approximation is only moderately justified. It is an estimate, not a controlled $T/T_E\to \infty$ limit. Retaining the exact Einstein mean-square displacement in the same heuristic criterion gives about $97\mathrm{K}$, indicating the size of that particular correction. The empirical threshold $0.05d$ and the independent-oscillator model introduce additional uncertainty.