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In the text I am reading, it is made clear that it is incorrect to assume that the electric displacement $\textbf{D}$ is identical to the electric field with the exception that it is raised from the free charge instead of total charge. A similar case is made with $\textbf{H}$. For the electric displacement, we know that $\nabla \times \textbf{D}=\nabla \times \textbf{P}$. We also know that $\nabla \cdot \textbf{H}=\nabla \cdot \textbf{M}$. So what would the physical interpretation of $\nabla \times \textbf{P}$ and $\nabla \cdot \textbf{M}$ be? Furthermore, are there any materials in which $\nabla \times \textbf{P}$ and $\nabla \cdot \textbf{M}$ are zero?

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Electric displacement D is not the same as E by definition. In vacuum both quantities are equal but in matter D differs from E due to the presence of polarization P. We know that in electrostatics curl E is zero everywhere but curl D may not be since it depends on curl P. Curl D depends on distribution of polarization in matter. Curl P would be zero in a material with uniform P (except on the surface). Divergence M is nonzero at ends of a bar magnet but zero inside assuming a uniform M. Both curl P and divergence M impose certain boundary conditions on B and H and are useful in determining electric and magnetic fields.

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  • $\begingroup$ A uniform polarization would be if all the eletric dipoles line up in the presence of an electric field correct? $\endgroup$ – Oscar Flores Dec 10 '14 at 20:47
  • $\begingroup$ The presence of electric field is not the essential requirement. For instance in materials called ferroelectrics we get polarization at macroscopic level even without applying an electric field by lowering the temperature. Yes a uniform polarization means all the electric dipoles are pointing in the same direction. $\endgroup$ – SAKhan Jan 11 '15 at 20:26

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