Charge and current densities, polarization and magnetization

The charge and current density for $N$ charges $q_n$ at positions ${\bf r}_n$ are

$$\rho({\bf r}) = \sum_nq_n \delta({\bf r}_n-{\bf r}),\quad {\bf j}({\bf r}) = \sum_nq_n \dot{{\bf r}}_n \delta({\bf r}_n-{\bf r})$$ (3.1)

These can be expressed in terms of polarization fields $P$ (electric polarization) and $\mathbf{M}$ (magnetic polarization or magnetization) via

$$\rho=-\div\P,\quad {\bf j}= \frac{d\P}{dt}+\mathbf{\nabla \times} \mathbf{M}.$$ (3.2)

In some traditional formulations of electromagnetism, one distinguishes between `bound' and `free' charges which, however, from a fundamental point of view is a little bit artificial. The above definition of $P$ and $\mathbf{M}$ thus refers to the total charge and current charge densities without such separation.

The interesting thing now is the fact that $P$ and $\mathbf{M}$ are not uniquely defined by Eq. (VII.3.2). They are arbitrary in very much the same way as the potentials $\phi$ and $\mathbf{A}$ are arbitrary. Note that only the longitudinal part of $P$ is uniquely determined from Eq. (VII.3.2) and given by the charge density,

$$\rho = -\div\P\leadsto i{\bf k}\P({\bf k}) = \rho({\bf k})$$

$$\leadsto \P_\Vert({\bf k}) \equiv [{\bf k} {\bf P}({\bf k})]{\bf k}/k^2 = -i \rho({\bf k}){\bf k}/k^2.$$ (3.3)

One can transform $P$ and $\mathbf{M}$ according to

$$\P\to \P' = \P+ \mathbf{\nabla \times} {\bf U},\quad \mathbf{M}\to \mathbf{M}' = \mathbf{M}- \frac{d{\bf U}}{dt} + \nabla u,$$ (3.4)

with $\P'$ and $\mathbf{M}'$ still fulfilling Eq. (VII.3.2). Following Woolley, this arbitraryness is related to charge conservation (I think this interpretation can be carried over to QM and be linked to U(1) invariance of QED). In the classical theory it is the divergence operator playing the central role ( $\varepsilon_0 \div\mathbf{E}= \rho$, Gauss' law) and the central object in this discussion therefore is the Greens function ${\bf g}({\bf r},{\bf r}')$ in

$${\bf\nabla}_{\bf r}\cdot {\bf g}({\bf r},{\bf r}')=-\delta({\bf r}-{\bf r}').$$ (3.5)

Fourier transformation yields

$$-i {\bf k} {\bf g}({\bf k}) = -1\leadsto {\bf g}_\Vert({\bf k}) \equiv [{\bf k} {\bf g}({\bf k})]{\bf k}/k^2 = -i {\bf k}/k^2$$

$${\bf g}_\perp({\bf k})$$ arbitrary. (3.6)

In other words, all the arbitraryness (all the fuss about gauge invariance) sits in the transversal part of the Greens function ${\bf g}({\bf r},{\bf r}')$ of the divergence operator.

The polarization is now expressed by ${\bf g}({\bf r},{\bf r}')$ that solves $\rho=-\div\P$, i.e.

$$\P({\bf r}) = \int d{\bf r}' {\bf g}({\bf r},{\bf r}') \rho({\bf r}').$$ (3.7)