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Landau–Peierls instability

Consider a one-dimensionally ordered crystal in 3D space. The density function is then given by ${\displaystyle \rho =\rho ($z)} . Since this is a 1D system, only the displacement ${\displaystyle u}$ along the ${\displaystyle z}$-direction due to thermal fluctuations can smooth out the density function; dispalcement in the other two directions are irrelevant. The net change in the free energy due to the fluectuations is given by

${\displaystyle {\mathcal {F}}=\int (F-F_{0})dV}$

where ${\displaystyle F_{0}}$ is the free energy without flcutuations. Note that ${\displaystyle {\mathcal {F}}}$ cannot depend on ${\displaystyle u}$ or be a linear function of ${\displaystyle \nabla u}$ because the first case corresponds to a simple uniform translation and the second case is unstable. Thus, ${\displaystyle {\mathcal {F}}}$ must be quadratic in the derivatives of ${\displaystyle u}$. These are given by^[3]

${\displaystyle {\mathcal {F}}={\frac {C}{2}}\int dV\left[\left({\frac {\partial u}{\partial z}}\right)^{2}+\lambda _{1}{\frac {\partial u}{\partial z}}\left({\frac {\partial ^{2}u}{\partial x^{2}}}+{\frac {\partial ^{2}u}{\partial y^{2}}}\right)+\lambda _{2}\left({\frac {\partial ^{2}u}{\partial x^{2}}}+{\frac {\partial ^{2}u}{\partial y^{2}}}\right)^{2}\right]}$

where ${\displaystyle C}$, ${\displaystyle \lambda _{1}}$ and ${\displaystyle \lambda _{2}}$ are material constants; in smectics, where the symmetry ${\displaystyle z\mapsto -z}$ must be obeyed, the second term has to be set zero, i.e., ${\displaystyle \lambda _{1}=0}$. In the Fourier space (in a unit volume), this is just

${\displaystyle {\mathcal {F}}={\frac {C}{2}}\int {\frac {d^{3}k}{(2\pi )^{3}}}(k_{z}^{2}+\lambda _{1}k_{z}\kappa ^{2}+\lambda _{2}\kappa ^{4})|{\hat {u}}($k)|^{2},\quad \kappa ^{2}=k_{x}^{2}+k_{y}^{2}.}

From the equipartition theorem, we can deduce that^[4]

${\displaystyle |{\hat {u}}($k)|^{2}={\frac {k_{B}T}{C(k_{z}^{2}+\lambda _{1}k_{z}\kappa ^{2}+\lambda _{2}\kappa ^{4})}}.}

The mean square displacement is thus given by

${\displaystyle \langle u^{2}($r)\rangle ={\frac {k_{B}T}{(2\pi )^{3}C}}\int _{1/L}^{k_{c}}{\frac {d^{3}k}{k_{z}^{2}+\lambda _{1}k_{z}\kappa ^{2}+\lambda _{2}\kappa ^{4}}}}

where the integral is cut off at a large wavenumber that is comparable to the linear dimension of the element undergoing deformation. In the thermodynamic limit, ${\displaystyle L\to \infty }$, the integral diverges logarthmically. This means that the an element at a particular point is displaced through very large distances and therefore smoothes out the function ${\displaystyle \rho ($x)} , leaving ${\displaystyle \rho =}$constant as the only solution and destroying the 1D ordering.

See also[edit]

• Peierls transition

References[edit]