A permanent dipole of moment in a local field at temperature has orientational energy

PHYSICS · Multi-part · Fermi University Physics · Electricity, Magnetism & Electromagnetism · Electric fields in matter · Electric fields in matter

The Boundary-Value Problem, Ch Molecular polarisability, permanent dipoles and the Clausius-Mossotti relation

Problem

A permanent dipole of moment pp in a local field ElocE_{\rm loc} at temperature TT has orientational energy U(θ)=−pEloccos⁡θU(\theta) = -pE_{\rm loc}\cos\theta, with θ\theta the angle between p\mathbf{p} and Eloc\mathbf{E}_{\rm loc}. Let x≡pEloc/kBTx \equiv pE_{\rm loc}/k_BT. ((a)) Starting from the Boltzmann weight excos⁡θe^{x\cos\theta} averaged over solid angle, show that the mean dipole moment along the field is ⟨pz⟩=pL(x)\langle p_z\rangle = pL(x) with L(x)=coth⁡x−1/xL(x) = \coth x - 1/x (the Langevin function), and that for x≪1x \ll 1 this reduces to ⟨pz⟩=p2Eloc/3kBT\langle p_z\rangle = p^2E_{\rm loc}/3k_BT. ((b)) A liquid of number density NvN_v contains molecules with electronic polarisability α\alpha and permanent dipole moment pp. Using pavg=(α+p2/3kBT)Eloc\mathbf{p}_{\rm avg} = (\alpha + p^2/3k_BT)\mathbf{E}_{\rm loc}, P=Nvpavg\mathbf{P} = N_v\mathbf{p}_{\rm avg}, and the Lorentz local field Eloc=E+P/3ε0\mathbf{E}_{\rm loc} = \mathbf{E} + \mathbf{P}/3\varepsilon_0, derive the polar form of the Clausius-Mossotti relation, εr−1εr+2=Nv3ε0(α+p23kBT).\frac{\varepsilon_r - 1}{\varepsilon_r + 2} = \frac{N_v}{3\varepsilon_0}\left(\alpha + \frac{p^2}{3k_BT}\right). ((c)) Show that this relation gives an infinite εr\varepsilon_r (a spontaneous, self-sustaining polarisation with no applied field) once the right-hand side reaches 11, and solve for the temperature TcT_c at which this happens, Tc=p2Nv9ε0kB(1−Nvα3ε0).T_c = \frac{p^2N_v}{9\varepsilon_0k_B\left(1 - \dfrac{N_v\alpha}{3\varepsilon_0}\right)}. Evaluate TcT_c for liquid water at its actual density, taking Nv=3.34×1028 m−3N_v = 3.34\times10^{28}\,\mathrm{m^{-3}}, α/4πε0=1.5×10−30 m3\alpha/4\pi\varepsilon_0 = 1.5\times10^{-30}\,\mathrm{m^3} and p=6.2×10−30 C mp = 6.2\times10^{-30}\,\mathrm{C\,m}, and comment on what this predicted value means physically, given that water at room temperature is not spontaneously polarised.

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