Finding transition temperature of Landau ferroelectric

  • #1
baseballfan_ny
91
23
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}^{'}##.
 
Last edited:

Answers and Replies

  • #2
anuttarasammyak
Gold Member
1,843
944
It is a cubic equation of ##T'_{\theta}## whose solution we know by formula.
 

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