1
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$
    – ProfRob
    Commented Jan 22, 2022 at 9:48

1 Answer 1

1
$\begingroup$

In Purcell Ch 6, he says "The equations in the box confront us with an astonishing fact, their symmetry with respect to E and B. If the printer had mistakenly interchanged E's with B's, and y's with z's, the equations would come out exactly the same!"--- so, you have swap y and z as well.

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 $\alpha=90^\circ$.

The fancier term is "Hodge duality". https://en.wikipedia.org/wiki/Hodge_star_operator

In 3-dimensions, this duality allows one to associate an axial vector with a bivector (as in the cross-product). Note: 3-2=1... a sort of complimentary dimension.

In 4-dimensions, this duality allows a 2-form (like the electromagnetic field tensor) to be mapped to its dual (4-2)-form.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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