Why aren't Faraday's law of induction and Maxwell-Ampere's law symmetric? [duplicate]

Question

I don't see Faraday's law of induction and Maxwell-Ampere's law are totally symmetric in the sense that Maxwell-Ampere's law has a factor of $ϵ_0μ_0$: \begin{align} \nabla\times\mathbf E&=-\frac{\partial\mathbf B}{\partial t} \\ \nabla\times\mathbf B&=\color{blue}{\mu_0\varepsilon_0}\frac{\partial\mathbf E}{\partial t} \end{align} Can anyone explain me why.

Answer 1

Faraday's law and Ampere's law are manifestly asymmetric, as Faraday's law is a homogeneous equation and Ampere's law is not. Therefore, I assume you mean in the absence of a current density, in which case the equations become (in SI units) $$\nabla \times \vec E = -\frac{\partial \vec B}{\partial t}$$ and $$\nabla \times \vec B = \epsilon_0 \mu_0 \frac{\partial \vec E}{\partial t}$$

There is a necessary asymmetry present because $\vec E$ and $\vec B$ have different dimensions, as do $\nabla$ and $\frac{\partial}{\partial t}$.

In the gaussian unit system, the electric field and magnetic field have the same units, so the equations become $$ \nabla \times \vec E = -\frac{1}{c} \frac{\partial \vec B}{\partial t}$$ $$ \nabla \times \vec B = \frac{1}{c} \frac{\partial \vec E}{\partial t}$$

We can do even better and use natural units where $c=1$, in which case we have $$ \nabla \times \vec E = -\frac{\partial \vec B}{\partial t}$$ $$ \nabla \times \vec B = \frac{\partial \vec E}{\partial t}$$

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The lesson here is that it's important to be able to see through the outward appearance of an equation to its inner structure. Often times, constants can be "absorbed" by redefining certain quantities and/or changing unit systems. This is one aspect of a larger concept called nondimensionalization, and it's an important skill to develop as time goes on.