0
$\begingroup$

The first and fourth Maxwell equations are often denotet in vaccum: $$ \nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0} $$ $$ \nabla \times \mathbf{B} = \mu_0\left( \mathbf{j}+\epsilon_0 \frac{\partial{\mathbf{E}}}{\partial{t}}\right) $$ and in matter: $$ \nabla \cdot \mathbf{D} = \rho_f $$ $$ \nabla \times \mathbf{H} = \mathbf{j}_f +\frac{\partial{\mathbf{D}}}{\partial{t}} $$

It is also often written, that the whole charge density $\rho = \rho_f+\rho_b$ is the sum of free and bound one. Similiarly $\mathbf{j} = \mathbf{j}_f+\mathbf{j}_b$, meaning the whole current density is the sum of the free and the bound one.

I ask then:

Why do the equations look like they do?

In vaccum there are no bound quantities, so using the "whole" quantities seems counterintuitiv. At the same time, I would expect bound quantities considered in matter. Is it just a problem of denotation? Something feels terribly off here.

$\endgroup$
0
$\begingroup$

The electric field $E$ is the field we apply, what we express with the first Maxwell equation is that its sources must come from the total density charge $\rho$. In a material, there will be some fixed charges, so the presence of $E$ will induce some dipoles, and this will make Polarization $P$ appear. Then polarization is related with the bound charge density $\rho_b$, while the rest of charges that are free to move we associate it with free density charge $\rho_f$ In vacuum, there is no bound density charge, so we have $\rho = \rho_f$.

In a presence of a material, we define displacement field $D$, or response field as $$ D= \epsilon_0 E + P $$ We normally consider the ideal situation for linear, homogeneous and isotropic material such that $P=\chi \epsilon_0 E$, where $\chi$ is electric susceptibility. In that way we could write $D=\epsilon_r \epsilon_0 E$, where $\epsilon_r = 1+ \chi$ is relative electric permitivity. $D$ is then taking into account the presence of free charges and bounded charges, althoug its sources are only the free charges. This can be seen easily, because the contribution of the Polarization is actually negative. It can be shown that $\nabla \cdot P =-\rho_b$, and as we know $\nabla \cdot E = \frac{\rho}{\epsilon_0}$. So if we take the divergence of the $D$ defined above and we use this results we have:

$$ \nabla \cdot D = \epsilon_0 \nabla \cdot E + \nabla \cdot P = \rho - \rho_b =\rho_f $$

since by definition $\rho = \rho_f + \rho_b$

$\endgroup$
  • $\begingroup$ Interesting. So one should fastidiously write "$\epsilon_0 \nabla\cdot E=\rho_f$" (vaccuum) and "$\epsilon_0 \nabla\cdot E=\rho$" (matter), right? $\endgroup$ – Josephus Dec 23 '16 at 23:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.