Meaning of $E$, $ D$, $B$ and $H$-fields

Question

The Lorentz force is

$$ F = q(E + v \times B) $$

where $E$ is denoted the electric field and $B$ is denoted the magnetic field. Further, Maxwell's equations are

$$\begin{array}{rcl} \nabla \cdot D &=& \rho\\ \nabla \times H - \frac{\partial D}{\partial t} &=& J\\ \nabla \times E + \frac{\partial B}{\partial t} &=& 0\\ \nabla \cdot B &=& 0 \end{array}$$

The question now is, there seems to be a difference in that it is $E$ that shows up in the Lorentz force, while the first Maxwell equations say something about $D$, while $B$ is present both in the Lorentz force and in the last Maxwell equation.

What is the reason for this discrepancy, or apparent asymmetry between how the electric and magnetic fields are treated? This question is closely related to the discussion in Asymmetry definitions electric $\chi_e$ and magnetic susceptibility $\chi_m$ but that discussion failed to raise how the field appears in the Lorentz force and there is also no clear answer to the question.

Answer 1

The Lorentz force formula for EM force on charged particle is valid only in vacuum, where $\mathbf D=\epsilon_0\mathbf E$ and $\mathbf B = \mu_0 \mathbf H$. It can be written in different ways:

$$ \mathbf F = q\mathbf E + q\mathbf v\times \mathbf B\,\,\,(*) $$ or $$ \mathbf F = q\mathbf E + q\mathbf v\times \mu_0 \mathbf H $$ or

$$ \mathbf F = q\mathbf D/\epsilon_0 + q\mathbf v\times \mathbf B $$

or

$$ \mathbf F = q\mathbf D/\epsilon_0 + q\mathbf v\times \mu_0 \mathbf H. $$

So the "asymmetry" you observe is due to particular choice of $(*)$ as the convention from the above 4 possibilities.