Landau–Peierls instability

Landau–Peierls instability refers to the phenomenon in which the mean square displacements due to thermal fluctuatuions diverge in the thermodynamic limit and is named after Lev Landau (1937) and Rudolf Peierls (1934). This instability prevails in one-dimensional ordering of atoms/molecules in 3D space such as 1D crystals and smectics and also in two-dimensional ordering in 2D space such as a monomolecular adsorbed filsms at the interface between two isotrophic phaes. The divergence is logarthmic, which is rather slow and therefore it is possible to realize substances in practice that are subject to Landau–Peierls instability.

Mathematical description
Consider a one-dimensionally ordered crystal in 3D space. The density function is then given by $$\rho=\rho(z)$$. Since this is a 1D system, only the displacement $$u$$ along the $$z$$-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


 * $$\mathcal F = \int(F-F_0) dV$$

where $$F_0$$ is the free energy without flcutuations. Note that $$\mathcal F$$ cannot depend on $$u$$ or be a linear function of $$\nabla u$$ because the first case corresponds to a simple uniform translation and the second case is unstable. Thus, $$\mathcal F$$ must be quadratic in the derivatives of $$u$$. These are given by


 * $$\mathcal F = \frac{C}{2}\int dV\left[\left(\frac{\partial u}{\partial z}\right)^2 + \lambda_1 \frac{\partial u}{\partial z}\left(\frac{\partial^2 u}{\partial x^2}+ \frac{\partial^2 u}{\partial y^2}\right) + \lambda_2 \left(\frac{\partial^2 u}{\partial x^2}+ \frac{\partial^2 u}{\partial y^2}\right)^2\right] $$

where $$C$$, $$\lambda_1$$ and $$\lambda_2$$ are material constants; in smectics, where the symmetry $$z\mapsto -z$$ must be obeyed, the second term has to be set zero, i.e., $$\lambda_1=0$$. In the Fourier space (in a unit volume), this is just


 * $$\mathcal F = \frac{C}{2}\int \frac{d^3k}{(2\pi)^3} (k_z^2 + \lambda_1 k_z \kappa^2 + \lambda_2 \kappa^4)|\hat u(k)|^2, \quad \kappa^2 = k_x^2 + k_y^2.$$

From the equipartition theorem, we can deduce that


 * $$|\hat u(k)|^2 = \frac{k_B T}{C(k_z^2 + \lambda_1 k_z \kappa^2 + \lambda_2 \kappa^4)}.$$

The mean square displacement is thus given by


 * $$\langle u^2(r)\rangle = \frac{k_B T}{(2\pi)^3 C} \int_{1/L}^{k_c} \frac{d^3k}{k_z^2 + \lambda_1 k_z \kappa^2 + \lambda_2 \kappa^4}$$

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, $$L\to \infty$$, 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 $$\rho(x)$$, leaving $$\rho=$$constant as the only solution and destroying the 1D ordering.