Finding transition temperature of Landau ferroelectric

[baseballfan_ny]

Homework Statement

In terms of Landau theory, a “proper ferroelectric” is completely analogous to a ferromagnet, where the order parameter is the polarization $P$ (dipole moment per volume). We can write a free energy for the ferroelectric phase transition as $F_P = \frac 1 2 \alpha (T - T_P)P^2 + \frac 1 4 g_4 P^4 + ...$.
For a molecular crystal a structural phase transition can occur where the molecular axis tilts by an angle $\theta$ with respect to the crystal axis. The associated free energy is $F_{\theta} = \frac 1 2 a_{\theta}(T - T_{\theta}) \theta^2 + \frac 1 4 b_{\theta} \theta^4 + ...$.
Under certain circumstances (the molecule is chiral, i.e., lacks inversion symmetry), the dipole moment on the molecule can couple to the molecular tilt, with a coupling term $F_{P \theta} = -t \theta P$ where t is the coupling constant.

$T_{\theta} > T_P$ and a tilt phase transition will occur first on cooling. A net polarization can exist due to coupling.

a. In the tilted phase $T_P < T < T_{\theta}$, determine $\theta$ and $P$ in terms of $\alpha$, $a_{\theta}$, $b_{\theta}$, $T_P$, and $T_{\theta}$. You can neglect the $P^4$ term.
b. Because of the coupling, the tilt transition does not occur at $T_{\theta}$, but some other temp $T_{\theta}^{'}$. Determine $T_{\theta}^{'}$ and the critical exponent $b$ for order parameter $\theta$.

Relevant Equations

$\frac {\partial F} {\partial \xi} = 0$

So for part a, I separately minimized F wrt $\theta$ and $P$ and got the following.

$$\frac {\partial f} {\partial \theta} = a_{\theta}(T-T_{\theta})\theta + b_{\theta}\theta^3 - tP = 0$$
$$ \frac {\partial f} {\partial P} = \alpha(T-T_P)P -t\theta$$
$$ P = t\theta \alpha (T-T_P) $$

Then subbing this into the expression for θ gives me...

$$ \theta \left[ a_{\theta}(T-T_{\theta}) + b_{\theta}\theta^2 - t^2\alpha(T-T_P) \right] = 0 $$

which I can solve for $\theta = 0$ or $\theta = \sqrt {\frac {t^2} { \alpha (T - T_P) } }- a_{\theta}(T - T_{\theta})$. And plugging this in gives $P = \pm \frac {t} {\alpha (T - T_P)} \sqrt {\frac {t^2} { \alpha (T - T_P) } }- a_{\theta}(T - T_{\theta})$.

Now my problem is with b, because I don't think I have the proper approach of finding the temperature at which the phase transition occurs, $T_{\theta}^{'}$. My idea was that maybe $T_{\theta}^{'}$ is the temperature $T$ when $\theta$ goes to 0.

That is
$$\pm \sqrt {\frac {t^2} { \alpha (T_{\theta}^{'} - T_P) } }- a_{\theta}(T_{\theta}^{'} - T_{\theta}) = 0 $$
$$ \frac {t^2} { \alpha (T_{\theta}^{'} - T_P) } = (T_{\theta}^{'} - T_{\theta})(T_{\theta}^{'} - T_P) $$

which seems a bit difficult to solve for $T_{\theta}^{'}$. The hint I have from class is that I should somehow re-write the expression for P in terms of $T_{\theta}^{'}$.

 

[anuttarasammyak]

It is a cubic equation of $T'_{\theta}$ whose solution we know by formula.