6
$\begingroup$

In Griffiths’ Introduction to Electrodynamics, he asks what changes would need to be made to Maxwell's equations to accommodate the existence of magnetic monopoles. Now, it is clear to me that the Gauss’s law and Ampere’s law must be left untouched. I also understand why the divergence of $\vec{B}$ should be $\alpha_{0}\rho_{m}$, where $\alpha_{0}$ is some constant to be determined experimentally and $\rho_{m}$ is the density of monopoles.

What I don’t understand is why the curl of $\vec{E}$ must be $\beta_{0}\vec{J}_{m}$, where $\beta_{0}$ is some constant.

The idea that moving electric monopoles produce a magnetic field is an experimental fact. Why must we assume that the same holds for magnetic monopoles as well? Is the symmetry of Maxwell's equations enough of a motivation for us to be sure that the electric field will now have a non-zero curl in the statics regime?

I get that such symmetries often motivate the discovery of such properties, but does it guarantee that the electric field curls?

$\endgroup$
10
$\begingroup$

With the standard caveat that all theory is predicated on experimental validation, I would strongly expect magnetic charge to obey a continuity equation. If you make the modifications you suggest without changing Faraday's law to include magnetic current, then you would find that $\frac{\partial \rho_m}{\partial t}=0$, implying that the total magnetic charge in any given region could never change.

Note that while Ampere's law is an experimental observation, the absence of the electrical current term similarly implies via Gauss' law for electric charge that the electric charge density of the universe is static, which would be intuitively upsetting were it not so easily demonstrated to be false.

$\endgroup$
4
  • $\begingroup$ Could you please clarify/elaborate the second paragraph, specifically "the absence of the electrical current term similarly implies via Gauss' law for electric charge that the electric charge density of the universe is static"? Your argument is not immediately clear to me. $\endgroup$ Sep 28 '20 at 6:57
  • $\begingroup$ @PhutureFysicist $\frac{\partial \rho}{\partial t} = \epsilon_0 \nabla \cdot \frac{\partial \mathbf E}{\partial t} = \epsilon_0 \nabla \cdot \left(\frac{1}{\epsilon_0\mu_0}\nabla\times\mathbf B - \frac{1}{\epsilon_0}\mathbf J\right) = -\nabla \cdot \mathbf J$. If $\mathbf J$ is absent from Ampere's law, then $\frac{\partial \rho}{\partial t} = 0$. $\endgroup$
    – J. Murray
    Sep 28 '20 at 7:01
  • $\begingroup$ More directly, the divergence of the curl is zero: $\nabla \cdot (\nabla \times \mathbf{E}) = 0$ for any vector-valued function $\mathbf{E}$. If we don't subtract $\mathbf{J}$, there would be a contradiction from magnetic Gauss's law: $\nabla \cdot (\nabla \times \mathbf{E}) = \nabla \cdot - \partial_t \mathbf{B} = - \partial_t (\nabla \cdot \mathbf{B}) = - 4 \pi \partial_t \rho$. This is basically @J.Murray's continuity argument in vector-calc speak. $\endgroup$
    – najkim
    Sep 28 '20 at 18:50
  • $\begingroup$ (Wikipedia insists on "Gauss's law".) $\endgroup$ Sep 28 '20 at 22:36

This site is temporarily in read only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .