Asymmetry definitions electric $\chi_e$ and magnetic susceptibility $\chi_m$

Question

Could anybody explain the asymmetry in the definitions of the electric susceptibility $\chi_e$ and the magnetic susceptibility $\chi_m$: $$D=\epsilon_0 E+P\qquad\qquad P=\epsilon_0\chi_e E$$ $$B=\mu_0 H+\mu_0 M\qquad\qquad M=\chi_m H$$ The following definition would make more sense to me: $$D=\epsilon_0 E+P\qquad\qquad P=\epsilon_0\chi_e E$$ $$B=\mu_0 H+M\qquad\qquad M=\color{red}{\mu_0}\chi_m H$$ Edit: The electric and magnetic parts appear in the Maxwell equations in a very similar way: $$\nabla\times E+\frac{\partial B}{\partial t}=0$$ $$\nabla\times H-\frac{\partial D}{\partial t}=j$$ $$\nabla\cdot D=\rho$$ $$\nabla\cdot B=0$$

Answer 1

It's ultimately a matter of definition but note there is an asymmetry in how the defining factors $\epsilon_0$ and $\mu_0$ occur. From Gauss's law $$ \oint \vec E\cdot d\vec S =\frac{q_{encl}}{\epsilon_0}\qquad \qquad \vec\nabla \cdot \vec E=\frac{\rho}{\epsilon_0} $$ with $\epsilon_0$ entering in the denominator. From Ampere's law $$ \oint \vec B\cdot d\vec \ell = \mu_0 I_{encl}\qquad \qquad \vec\nabla\times \vec B=\mu_0\vec J $$ with $\mu_0$ now appearing in the numerator. As a result the fields $\vec B$ and $\vec E$ which are defined by "elementary" integral laws are related to $\vec H$ and $\vec D$ by multiplicative or dividing factors rather than by two multiplicative or two dividing factors. Note that the magnetization $\vec M$ is defined in terms of the field $\vec H$ whereas the polarization $\vec P$ is in terms of the "fundamental field" $\vec E$.