Derivation of Lagrangian density for an infinite classical dielectric in interaction with the EM field

Question

I am tasked with reading and reproducing all the steps in J.J. Hopfield's 1958 paper "Theory of the Contribution of Excitons to the Complex Dielectric Constant of Crystals". Embarrassingly I am stuck on equation (3). Hopfield takes the Lagrangian density for an infinite classical dielectric in interaction with the electromagnetic field to be

Instead of taking this equation at face value, I would like to fully derive this Lagrangian density, but am unsure how to approach the problem. I have found many articles and sites tackling a similar problem, but never ending up with an expression as here. Any pointers on how to approach the problem or links to material that could be useful would be very much appreciated.

Answer 1

In this answer we work with units where the speed of light in vacuum is $c=1$, and Minkowski signature $(-,+,+,+)$.

The Lagrangian density (3) reads in cgs units

$$ \tag{3}{\cal L}(A_{\mu},{\bf P},{\bf M}) ~=~-\frac{1}{16\pi}F_{\mu\nu}F^{\mu\nu} +A_{\mu} J^{\mu}_b +\frac{1}{2\beta}\left(\frac{1}{\omega_0^2}\dot{\bf P}^2 -{\bf P}^2\right). $$

It consists of 3 terms:

1. A standard EM Maxwell term.

2. A source term with bounded 4-current $$ J^{\mu}_b~=~(\rho_b, {\bf J}_b) ~=~(-{\bf \nabla}\cdot {\bf P},{\bf \nabla}\times {\bf M}+ \dot{\bf P}),$$ where magnetization ${\bf M}={\bf 0}$. There is no free 4-current $J^{\mu}_f=0$. The source term reads $$ A_{\mu} J^{\mu}_b~\sim~{\bf E}\cdot{\bf P}+{\bf B}\cdot{\bf M}$$ modulo a total divergence term, which doesn't affect the Euler-Lagrange equations.

3. A harmonic oscillator with the polarization ${\bf P}$ as dynamical variable and with frequency $\omega_0$. The variation of the Lagrangian density (3) wrt. ${\bf P}$ leads to the constitutive equation $$ \tag{4}\frac{1}{\omega_0^2}\ddot{\bf P} +{\bf P}~=~ \beta {\bf E}, $$ cf. Ref. 1. Eq. (4) is the main reason to add the third term in the Lagrangian density (3). The harmonic oscillator of ${\bf P}$ is driven by the electric field ${\bf E}$ with a coupling constant $\beta$.

For completeness, let us mention that the variation of the Lagrangian density (3) wrt. the 4-gauge potential $A_{\mu}$ leads to Maxwell's equations (Gauss's and Ampere's law).

References:

1. J.J. Hopfield, Theory of the Contribution of Excitons to the Complex Dielectric Constant of Crystals, Phys. Rev. 112 (1958) 1555.

Answer 2

I will set constants like $c$ equal to one.

Then he starts with the normal relativisitc lagrangian, $\mathcal{L} = -\frac{1}{4}F^{\alpha \beta} F_{\alpha \beta} - A^\alpha J_\alpha$. Translating this into non-relativistic language, we get $\mathcal{L} = \frac{1}{2}(E^2 - B^2) - \phi \rho + \mathbf{A} \cdot \mathbf{J} $.

Now at this point he seems to assume no free charges or currents, and no magnetization. Therefore the microscopic charge density is the bound charge density $\rho_b = -\mathbf{\nabla} \cdot \mathbf{P}$, and the microscopic current density is related to changes in the polarization: $\mathbf{J} = \mathbf{J}_b = \partial_t \mathbf{P}$ Then $\mathcal{L} = \frac{1}{2}(E^2 - B^2) + \phi \mathbf{\nabla} \cdot \mathbf{P} + \mathbf{A} \cdot \partial_t \mathbf{P}$.

Recognizing $\mathbf{E} = -\partial_t \mathbf{A} - \mathbf{\nabla} \phi$ (this is off by a minus sign from what he says, I don't know why) and $\mathbf{B} = \mathbf{\nabla} \times \mathbf{A}$ we now have $\mathcal{L} = \frac{1}{2}((\partial_t \mathbf{A} + \mathbf{\nabla} \phi)^2 - (\mathbf{\nabla} \times \mathbf{A})^2) + \phi \mathbf{\nabla} \cdot \mathbf{P} + \mathbf{A} \cdot \partial_t \mathbf{P}$ This gives us the first two and last two terms of his expression up to factors of $4 \pi$ coming from I think gauss's law, which I have been ignoring.

Now I do not get his middle two terms. I have not read the paper. What are $\beta$ and $\omega_0$?