# Defining a local polarization field in a distribution of charge

I am currently building a theoretical model where charges of opposite signs are created by pairs and then diffuse and are drifted by an electrical field. I am taking this along a single so far, for simplicity. The object I'd like to arrive at the end is a local (density of) polarization, and not the global one (because the diffusion gives me two gaussians, spreading in time and drifting in opposites ways. But when I try to come up we've an expression for this polarization field, nothing pertinent comes out: if I look at one point and compute the barycenter of the charges from there, I get the same expression no manner where it is because of the global charge neutrality; and when I think about only computing charges in a small area about the point, I can't guarantee the neutrality of the charges inside this area (and so polarization is irrelevant in there). I am wondering if it's not because there is not more information than the one given by the separation of the two charge distribution barycenters.

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Gauss' law for a macroscopic system (for which polarization is well defined) is $$\nabla\cdot\mathbf{D} = 4 \pi \rho_f$$ where $\mathbf{D}$ is the electric displacement and $\rho_f$ is the density of free charges. The microscopic Gauss' law is given by $$\nabla\cdot\mathbf{E} = 4 \pi \rho$$ where $\mathbf{E}$ is the electric field and $\rho$ is the total charge density. The relationship between the electric displacement $\mathbf{D}$ and the electric field $\mathbf{E}$ defines the polarization $\mathbf{P}$, by $$\mathbf{D}=\mathbf{E}+4\pi\mathbf{P}$$ If we subtract the first equation from the second, we have \begin{aligned} \nabla\cdot(\mathbf{E}-\mathbf{D}) &= 4 \pi (\rho-\rho_f) \\ \nabla \cdot (-4 \pi \mathbf{P}) &= 4 \pi \rho_b \\ \nabla \cdot \mathbf{P} &= -\rho_b \end{aligned} So we have that the polarization divergence is established by $\rho_b$, the bound charge which contributes to the total charge $\rho$, but which is not free to move in response to electric fields, as are the charges included in $\rho_f$.
Since your model is attempting to describe charges displaced by an electric field, it seems to me that your system has no bound charges, and thus the concept of polarization cannot apply. You could still compute a dipole moment, $$\mathbf{p}(\mathbf{r})=\int_V \rho(\mathbf{r'}) (\mathbf{r'}-\mathbf{r})d^3\mathbf{r'},$$ but without some constraint (besides the large scale electric field) that prevents charges from moving, polarization is ill-defined.
 Yes, but as I said, the system being globally neutral, your dipole moment would be independent of $\vec{r}$: deriving it gives $-\int_V \rho(r')dr' = 0$. It is not a non-trivial local function. – Learning is a mess Feb 20 at 18:18 The integral of charge over all space is 0, but unless the charges are uniformly and continuously distributed, the dipole moment is not assured to be zero. Consider two opposite point charges at the center of a neutrally charged, conductive sphere. Certainly from outside the sphere, there are no detectable charges, and the system is perfectly neutral. However, inside the sphere, is the dipole moment 0? – KDN Feb 21 at 0:19 Or, to consider a system that is mathematically simpler, consider two charges $+q$ and $-q$ separated by a distance $d$, sitting between two conductive plates each distance $d/2$ away from the nearest charge (so the geometry looks like Plate|<--$d/2$-->$+q$<--$d$-->$-q$<--$d/2$-->|Plate ). You can use the method of images to solve this system quite simply. You will find that at the position of either of the real charges ($q+$ or $q-$, the dipole moment is $3 d q$ (up to a minus sign). And yet, the system is globally charge neutral, and there are no dipole moments outside the plates. – KDN Feb 21 at 0:35