Landau–de Gennes theory

Remarks

A Q-tensor is uniaxial if two eigenvalues coincide; then $Q = s \big(n \otimes n - \frac13 I\big)$ for a unit vector $n$ and a scalar order parameter $s$. Example: for uniaxial $Q$ the eigenvalues are $2s/3, -s/3, -s/3$, so $\tr(Q^2) = 2s^2/3$ and $\tr(Q^3) = 2s^3/9$, and $$f_B = \frac{a}{3} s^2 - \frac{2b}{27} s^3 + \frac{c}{9} s^4.$$ The isotropic state $s = 0$ and a nematic state $s > 0$ have equal energy when the bracket in $f_B = s^2 \big(\frac{a}{3} - \frac{2b}{27} s + \frac{c}{9} s^2\big)$ has a double root, that is, at $a = b^2/(27c)$, with $s = b/(3c)$. So $s$ jumps from $0$ to $b/(3c)$: the isotropic–nematic transition is first order, at a temperature above $T^*$. Non-example: the matrix $\mathrm{diag}(1, 0, 0)$ is symmetric but has trace $1$, so it is not a Q-tensor. For uniaxial $Q$ with constant $s$, the elastic term reduces to the one-constant Oseen–Frank energy with $K = 2 L s^2$.