Curl of the dipole moment vector - why is it not always zero

[zezima1]

I'm confused about what polarization of a dielectric does to its electrical properties. It is clear to me that polarization causes every little atom to get a tiny dipole moment. A measure of the polarization is therefore P = dipole moment per unit volume. However, what is really a dipole moment of an atom? Is it just the potential connected to the field of the polarized atom?
If so I need some help: My book defines what is called the electric displacement, which is used for a new form of Gauss' law which holds for dielectrics. I won't bother to write that here, since you have probably seen that many times. Well, we can then ask the question: Is the path integral independent of your path in the presence of dielectrics. And the general answer to that is for some reason no, because the curl of P (the dipole moment per unit volume) is not always zero. Now I don't really understand why the path integral cannot be independent of the path. I won't pretend to understand why the curl of P is not zero since I'm not sure what P really is - is it a field or something else? Well in either way: No matter what we are dealing with electrostatics right? Because this is all stationary charges and so are the charges formed by the polarization. And since no magnetic forces are present we have: nabla x sum of all E fields = 0. So the curl of the total field, even counting the polarization must be zero musn't it? So what is it that makes a work integral still be dependent on the path? My teacher says it happens between two different materials, but I just don't see why the curl of the total field (or what ever this electric displacement) represents.

 

[gabbagabbahey]

I'm confused about what polarization of a dielectric does to its electrical properties. It is clear to me that polarization causes every little atom to get a tiny dipole moment.

Really? That shouldn't be clear to you at all. Polarization is a macroscopic (average) quantity. All kinds of crazy stuff can (and does!) go on at the microscopic level, but the polarization (Defined as the dipole moment per unit volume) is really an average quantity (usually averaged over a region containing thousands of atoms, just like density!).

A measure of the polarization is therefore P = dipole moment per unit volume. However, what is really a dipole moment of an atom? Is it just the potential connected to the field of the polarized atom?

Dipole moment of a distribution is a quantity that is independent of where you are measuring it from. It is defined as \mathbf{p}=\int\mathbf{r}'\rho(\mathbf{r}')d^3r' (the integral is over the charge distribution and independent of which field point you are measuring from!). The dipole moment of an atom is just this quantity, evaluated for an atom. The dipole moment of any distibution is just this quantity, evaluated for that distribution.

The dipole moment is useful, because it is easy to express the dipole contribution of the potential of the distribution in terms of it

V_{\text{dip}}=\frac{1}{4\pi\epsilon_0}\mathbf{p} \cdot \frac{\hat{\mathbf{r}}}{r^2}

This contribution to the potential usually dominates at large distances from the distribution, when the net charge is zero (which is quite a common scenario!).

If so I need some help: My book defines what is called the electric displacement, which is used for a new form of Gauss' law which holds for dielectrics. I won't bother to write that here, since you have probably seen that many times. Well, we can then ask the question: Is the path integral independent of your path in the presence of dielectrics. And the general answer to that is for some reason no, because the curl of P (the dipole moment per unit volume) is not always zero. Now I don't really understand why the path integral cannot be independent of the path. I won't pretend to understand why the curl of P is not zero since I'm not sure what P really is - is it a field or something else? Well in either way: No matter what we are dealing with electrostatics right? Because this is all stationary charges and so are the charges formed by the polarization. And since no magnetic forces are present we have: nabla x sum of all E fields = 0. So the curl of the total field, even counting the polarization must be zero musn't it? So what is it that makes a work integral still be dependent on the path? My teacher says it happens between two different materials, but I just don't see why the curl of the total field (or what ever this electric displacement) represents.

The electric displacement is just a useful quantity for calculations (and in some cases, a direcly measurable quantity), it is not an electrostatic field.

The polarization may or may not have zero curl. It depends on the material and the source of the polarization.

For example, for a nice linear dielectric placed in an electric field, a significant portion of the material's dipoles will align with the external field (or in the opposite direction) and the resulting polarization (dipole moment per unit volume) may be directly proportional to the external field (\mathbf{P}=\alpha\mathbf{E}). In such a case, it's curl will vanish since \mathbf{\nabla}\times\mathbf{P}=\alpha(\mathbf{\nabla}\times\mathbf{E})=0.

However, there are many many materials for which the polarization will not be directly proportional to the external field (and even some called electrets which carry their own permanant polarizations, without an external field being applied!), and so the curl of the polarization may not vanish.

There is no reason to assume that the polarization will in general have zero curl. It is not an electrostatic field (it doesn't even have the same units!).