 #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.
$$\frac {\partial f} {\partial \theta} = a_{\theta}(TT_{\theta})\theta + b_{\theta}\theta^3  tP = 0$$
$$ \frac {\partial f} {\partial P} = \alpha(TT_P)P t\theta$$
$$ P = t\theta \alpha (TT_P) $$
Then subbing this into the expression for θ gives me...
$$ \theta \left[ a_{\theta}(TT_{\theta}) + b_{\theta}\theta^2  t^2\alpha(TT_P) \right] = 0 $$
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 rewrite the expression for P in terms of ##T_{\theta}^{'}##.
$$\frac {\partial f} {\partial \theta} = a_{\theta}(TT_{\theta})\theta + b_{\theta}\theta^3  tP = 0$$
$$ \frac {\partial f} {\partial P} = \alpha(TT_P)P t\theta$$
$$ P = t\theta \alpha (TT_P) $$
Then subbing this into the expression for θ gives me...
$$ \theta \left[ a_{\theta}(TT_{\theta}) + b_{\theta}\theta^2  t^2\alpha(TT_P) \right] = 0 $$
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 rewrite the expression for P in terms of ##T_{\theta}^{'}##.
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