Temperature dependence of dipolar polarization

Dipolar (orientational) polarization occurs in materials with permanent molecular dipoles $p_0$ (e.g. H₂O, HCl). Without a field the dipoles point randomly, so $P = 0$. An applied field tries to align them, while thermal agitation tries to keep them random. The result depends on temperature.

 E = 0                E -->
  \  /  |  -           ->  /  ->  \
  -  \  /  |           ->  ->  /  ->
  random, P = 0        partly aligned, P > 0

Energy of a dipole in a field

A dipole at angle $\theta$ to the field $E$ has potential energy

$$U = -p_0E\cos\theta$$

and its component along the field is $p_0\cos\theta$.

Average dipole moment (Boltzmann statistics)

The probability of a dipole having orientation $\theta$ is proportional to $e^{-U/kT}$. The number of dipoles in the solid angle $d\Omega = 2\pi\sin\theta\,d\theta$ is proportional to $e^{p_0E\cos\theta/kT}\sin\theta\,d\theta$. So

$$\langle p_0\cos\theta \rangle = p_0\frac{\int_0^{\pi}\cos\theta\, e^{a\cos\theta}\sin\theta\, d\theta}{\int_0^{\pi}e^{a\cos\theta}\sin\theta\, d\theta}, \qquad a = \frac{p_0E}{kT}$$

Put $x = \cos\theta$, $dx = -\sin\theta\,d\theta$:

$$\langle p_0\cos\theta \rangle = p_0\frac{\int_{-1}^{1}x e^{ax}dx}{\int_{-1}^{1}e^{ax}dx} = p_0\left[\coth a - \frac{1}{a}\right] = p_0 L(a)$$

$L(a)$ is the Langevin function.

Usual condition (weak field, normal temperature)

Normally $p_0E \ll kT$ (e.g. $p_0 \sim 10^{-30}$ C m, $E = 10^{6}$ V/m gives $p_0E \approx 10^{-24}$ J, while $kT \approx 4 \times 10^{-21}$ J at 300 K), so $a \ll 1$ and $L(a) \approx a/3$. Then

$$\langle p \rangle = \frac{p_0^2 E}{3kT}$$

so the dipolar polarizability and polarization are

$$\alpha_d = \frac{p_0^2}{3kT}, \qquad P_d = N\frac{p_0^2}{3kT}E$$

Conclusion

$\alpha_d \propto 1/T$: as temperature rises, thermal agitation disorders the dipoles more and dipolar polarization falls. This is the Debye relation. Including electronic polarization, the total polarizability is

$$\alpha = \alpha_e + \frac{p_0^2}{3kT}$$

A plot of $\alpha$ (or $\varepsilon_r - 1$) against $1/T$ is a straight line: its slope gives $p_0$ and its intercept gives $\alpha_e$. For non-polar materials the slope is zero, which shows that only dipolar polarization depends on temperature.