# Interchanging $c\mathbf{B}$ with $\mathbf{E}$ in Ampere's law

I am reading Purcell's Electricity and Magnetism in the section about the displacement current in Maxwell's equations. He states that because the Lorentz-transformations of the electric and magnetic field are symmetric in $$\mathbf{E}$$ and $$c\mathbf{B}$$, we can interchange the roles of $$\mathbf{E}$$ and $$c\mathbf{B}$$ in Ampere's law so that $$\nabla \times \mathbf{B} = -\frac{1}{c^2}\frac{\partial \mathbf{E}}{\partial{t}}$$ I don't understand why this is true. Why can the roles of $$\mathbf{E}$$ and $$c\mathbf{B}$$ be interchanged?

• But that isn't what you get if you swap them. The LHS of Ampere's law would be $\nabla \times {\mathbf E}/c$ and the right hand side would have a current density. I think you mean the Maxwell-Faraday law. Jan 22, 2022 at 9:48

In Griffith's, "Problem 7.64 (a) Show that Maxwell's equations with magnetic charge (Eq. 7.44) are invariant under the duality transformation (7.68) $$\vec E' = \vec E \cos \alpha + c\vec B\sin\alpha$$ $$c\vec B' = c\vec B \cos \alpha — \vec E \sin \alpha,$$ $$cq'_e = cq_e \cos\alpha + q_m \sin \alpha$$ $$q'_m = q_m \cos \alpha — cq_e \sin \alpha,$$ where $$c = 1/\sqrt{\epsilon_0 \mu_0}$$ and $$\alpha$$ is an arbitrary rotation angle in "$$\vec E/\vec B$$-space." Charge and current densities transform in the same way as $$q_e$$ and $$q_m$$. [This means, in Chapter 7 Electrodynamics particular, that if you know the fields produced by a configuration of electric charge, you can immediately (using $$\alpha=90^\circ$$) write down the fields produced by the corresponding arrangement of magnetic charge.]
Try using $$\alpha=90^\circ$$.