Categories: Electromagnetism, Physics.

Clausius-Mossotti relation

The polarizability α\alpha of a small dielectric body (e.g. an atom or molecule) is defined to relate the electric field E\vb{E} applied to that body to the resulting dipole moment p\vb{p}:

p=αε0E\begin{aligned} \vb{p} = \alpha \varepsilon_0 \vb{E} \end{aligned}

If there are NN such bodies per unit volume, the polarization density P=ε0χeE\vb{P} = \varepsilon_0 \chi_e \vb{E} with P=Np\vb{P} = N \vb{p} suggests that χe=Nα\chi_e = N \alpha. However, this is an underestimation: each body’s induced dipole creates its own electric field, weakening the field felt by its neighbors. We need to include this somehow, but α\alpha is defined for a single dipole in a vacuum.

Let Eint\vb{E}_\mathrm{int} be the uniform internal field excluding the dipoles’ contributions, and E(r)\vb{E}(\vb{r}) the net field including them. Assume that the dipoles pi\vb{p}_i are arranged in a regular crystal lattice at sites Ri\vb{R}_i. Then E(r)\vb{E}(\vb{r}) is the sum of Eint\vb{E}_\mathrm{int} and all the dipoles’ fields:

E(r)=Eint+iEi(rRi)\begin{aligned} \vb{E}(\vb{r}) = \vb{E}_\mathrm{int} + \sum_{i} \vb{E}_i(\vb{r} - \vb{R}_i) \end{aligned}

Where the individual contribution Ei(r)\vb{E}_i(\vb{r}) of each dipole pi\vb{p}_i is as follows:

Ei(r)=14πε0(r^pir2)\begin{aligned} \vb{E}_i(\vb{r}) = - \frac{1}{4 \pi \varepsilon_0} \nabla \bigg( \frac{\vu{r} \cdot \vb{p}_i}{|\vb{r}|^2} \bigg) \end{aligned}

The atoms or molecules pi\vb{p}_i need not be perfect dipoles, as long as they approximate one when viewed from a distance much smaller than the crystal’s lattice constant. Clearly, in a multipole expansion of the true charge distribution ρi(r)\rho_i(\vb{r})’s electric potential Vi(r)V_i(\vb{r}), the dipole term will be dominant in that case, given by:

Vi(r)14πε01r2ρi(r)rcosθdr\begin{aligned} V_i(\vb{r}) \approx \frac{1}{4 \pi \varepsilon_0} \frac{1}{|\vb{r}|^2} \int_{-\infty}^\infty \rho_i(\vb{r}') \: |\vb{r}'| \cos{\theta} \dd{\vb{r}'} \end{aligned}

Where θ\theta is the angle between r\vb{r} and r\vb{r}', so this can be rewritten as a dot product with the unit vector r^\vu{r}, normalized from r\vb{r}:

Vi(r)=14πε01r2r^ ⁣ ⁣rρi(r)dr\begin{aligned} V_i(\vb{r}) = \frac{1}{4 \pi \varepsilon_0} \frac{1}{|\vb{r}|^2} \: \vu{r} \cdot \!\!\int_{-\infty}^\infty \vb{r}' \rho_i(\vb{r}') \dd{\vb{r}'} \end{aligned}

The integral is a more general definition of the dipole moment pi\vb{p}_i. You can convince yourself of this by defining ρi(r)\rho_i(\vb{r}) as two opposite charges +q+q and q-q respectively located at +di/2+\vb{d}_i/2 and di/2-\vb{d}_i/2; evaluating the integral then yields qdi=piq \vb{d}_i = \vb{p}_i. Therefore:

Vi(r)=14πε0r^pir2\begin{aligned} V_i(\vb{r}) = \frac{1}{4 \pi \varepsilon_0} \frac{\vu{r} \cdot \vb{p}_i}{|\vb{r}|^2} \end{aligned}

Then the corresponding electric field Ei\vb{E}_i is given by Vi- \nabla V_i as is well known.

The dipole p0\vb{p}_0 at r=0\vb{r} = 0 feels a net local field Eloc\vb{E}_\mathrm{loc}, given below. The crystal’s symmetry ensures that all its neighbors’ fields cancel out: since Ei(r)=Ei(r)\vb{E}_i(-\vb{r}) = -\vb{E}_i(\vb{r}), each dipole has a counterpart with the exact opposite field. We exclude the singular term E0(0)\vb{E}_0(0):

ElocE(0)=Eint+i0Ei(Ri)=Eint\begin{aligned} \vb{E}_\mathrm{loc} \equiv \vb{E}(0) = \vb{E}_\mathrm{int} + \sum_{i \neq 0} \vb{E}_i(-\vb{R}_i) = \vb{E}_\mathrm{int} \end{aligned}

Even if there is no regular lattice, this result still holds well enough, as long as the dipoles are uniformly distributed over a large volume.

So what was the point of including Ei(r)\vb{E}_i(\vb{r}) in the first place? Well, keep in mind that the sum over neighbors is nonzero for rRi\vb{r} \neq \vb{R}_i, which does affect the macroscopic field E\vb{E}, defined as:

E=Eloc+1ΩΩiEi(rRi)dr\begin{aligned} \vb{E} = \vb{E}_\mathrm{loc} + \frac{1}{\Omega} \int_\Omega \sum_i \vb{E}_i(\vb{r} - \vb{R}_i) \dd{\vb{r}} \end{aligned}

Where Ω\Omega is an arbitrary averaging volume, large enough to contain many dipoles, and also small enough to assume that the polarization is uniform within. Thanks to linearity and symmetry, we only need to evaluate this volume integral for a single dipole:

ΩEi(r)dr=14πε0Ω(r^pir2)dr=14πε0Ωr^pir2dn^\begin{aligned} \int_\Omega \vb{E}_i(\vb{r}) \dd{\vb{r}} &= - \frac{1}{4 \pi \varepsilon_0} \int_\Omega \nabla \bigg( \frac{\vu{r} \cdot \vb{p}_i}{|\vb{r}|^2} \bigg) \dd{\vb{r}} \\ &= - \frac{1}{4 \pi \varepsilon_0} \oint_{\partial \Omega} \frac{\vu{r} \cdot \vb{p}_i}{|\vb{r}|^2} \dd{\vu{n}} \end{aligned}

Where we have used the divergence theorem. Let us define Ω\Omega as a sphere with radius RR centered at r=0\vb{r} = 0. In spherical coordinates, the surface integral then becomes:

ΩEi(r)dr=14πε00π02πr^pir2r^r2sinθdφdθ\begin{aligned} \int_\Omega \vb{E}_i(\vb{r}) \dd{\vb{r}} &= - \frac{1}{4 \pi \varepsilon_0} \int_0^\pi \int_0^{2 \pi} \frac{\vu{r} \cdot \vb{p}_i}{|\vb{r}|^2} \: \vu{r} \: |\vb{r}|^2 \sin{\theta} \dd{\varphi} \dd{\theta} \end{aligned}

The radial coordinate disappears, so in fact the radius RR is irrelevant. We choose our coordinate system such that pi\vb{p}_i points in the positive zz-direction, i.e. pi=(0,0,pi)\vb{p}_i = (0, 0, |\vb{p}_i|), leaving:

ΩEi(r)dr=pi4πε00π02π(sinθcosφsinθsinφcosθ)sinθcosθdφdθ\begin{aligned} \int_\Omega \vb{E}_i(\vb{r}) \dd{\vb{r}} &= - \frac{|\vb{p}_i|}{4 \pi \varepsilon_0} \int_0^\pi \int_0^{2 \pi} \begin{pmatrix} \sin{\theta} \cos{\varphi} \\ \sin{\theta} \sin{\varphi} \\ \cos{\theta} \end{pmatrix} \sin{\theta} \cos{\theta} \dd{\varphi} \dd{\theta} \end{aligned}

Now, consider the following two straightforward indefinite integrals:

cosxsin2xdx=13sin3xcos2xsinxdx=13cos3x\begin{aligned} \int \cos{x} \sin^2{x} \dd{x} &= \:\:\,\, \frac{1}{3} \sin^3{x} \\ \int \cos^2{x} \sin{x} \dd{x} &= -\frac{1}{3} \cos^3{x} \end{aligned}

Applying these to our integral over θ\theta, the expression is reduced to just:

ΩEi(r)dr=pi12πε002π[sin3θcosφsin3θsinφcos3θ]0πdφ=pi6πε002π(001)dφ=pi3ε0\begin{aligned} \int_\Omega \vb{E}_i(\vb{r}) \dd{\vb{r}} &= \frac{|\vb{p}_i|}{12 \pi \varepsilon_0} \int_0^{2 \pi} \begin{bmatrix} \sin^3{\theta} \cos{\varphi} \\ \sin^3{\theta} \sin{\varphi} \\ \cos^3{\theta} \end{bmatrix}_0^\pi \dd{\varphi} \\ &= - \frac{|\vb{p}_i|}{6 \pi \varepsilon_0} \int_0^{2 \pi} \begin{pmatrix} 0 \\ 0 \\ \:1\: \end{pmatrix} \dd{\varphi} \\ &= - \frac{\vb{p}_i}{3 \varepsilon_0} \end{aligned}

Inserting this result into the macroscopic field E\vb{E}, and recognizing the second term as the linear polarization density P\vb{P}, we arrive at the key result:

E=Eloc13ε0ipiΩ    Eloc=E+13ε0P\begin{aligned} \vb{E} = \vb{E}_\mathrm{loc} - \frac{1}{3 \varepsilon_0} \sum_i \frac{\vb{p}_i}{\Omega} \qquad \implies \qquad \boxed{ \vb{E}_\mathrm{loc} = \vb{E} + \frac{1}{3 \varepsilon_0} \vb{P} } \end{aligned}

Whereas the individual dipoles are polarized by Eloc\vb{E}_\mathrm{loc}, the macroscopic polarization density is defined from E\vb{E}, so we now know that:

P=ε0χeE=ε0NαEloc=ε0Nα(E+13ε0P)\begin{aligned} \vb{P} = \varepsilon_0 \chi_e \vb{E} = \varepsilon_0 N \alpha \vb{E}_\mathrm{loc} = \varepsilon_0 N \alpha \bigg( \vb{E} + \frac{1}{3 \varepsilon_0} \vb{P} \bigg) \end{aligned}

Isolating this equation for P\vb{P}, and using that P=ε0χeE\vb{P} = \varepsilon_0 \chi_e \vb{E}, we get a more accurate expression for the electric susceptibility χe\chi_e:

χe=Nα1Nα/3\begin{aligned} \boxed{ \chi_e = \frac{N \alpha}{1 - N \alpha / 3} } \end{aligned}

Notice that for Nα1N \alpha \ll 1 this reduces to our earlier naive estimate: if each dipole’s neighbors are far away, they do not provide much shielding.

The corresponding macroscopic dielectric function εr\varepsilon_r is given by:

εr=1+χe=3+2Nα3Nα\begin{aligned} \boxed{ \varepsilon_r = 1 + \chi_e = \frac{3 + 2 N \alpha}{3 - N \alpha} } \end{aligned}

Experimentally, εr\varepsilon_r can be measured directly, but α\alpha cannot. By isolating εr\varepsilon_r for NαN \alpha, we finally arrive at the Clausius-Mossotti relation for calculating α\alpha:

13Nα=εr1εr+2\begin{aligned} \boxed{ \frac{1}{3} N \alpha = \frac{\varepsilon_r - 1}{\varepsilon_r + 2} } \end{aligned}

Since many theoretical models only calculate α\alpha (e.g. the Lorentz oscillator model), this result is useful for relating theory to experimental results.


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  3. A. Zangwill, Modern Electrodynamics, Cambridge.
  4. D.J. Griffiths, Introduction to Electrodynamics, 5th edition, Cambridge.