Does existence of magnetic monopole break covariant form of Maxwell’s equations for potentials?

Question

Absence of magnetic charges is reflected in one of Maxwell's fundamental equations: $$\operatorname{div} \vec B = 0. \tag1$$ This equation allows us to introducte concept of vector potential: $$\vec B = \operatorname{rot} \vec A.$$ Using this concept, it is possible to express Maxwell's equations in a graceful symmetric form: $$\nabla^2 \vec A - \frac{1}{c^2}\frac{\partial^2 \vec A}{\partial t^2} = - \frac{\vec j}{\epsilon_0 c^2}, \tag2$$ $$ \nabla^2 \phi -\frac{1}{c^2}\frac{\partial^2 \phi}{\partial t^2} = - \frac{\rho}{\epsilon_0}. \tag3 $$

Noticing, that vector $\vec A$ and scalar $\phi$ potentials, as well as electric current density $\vec j$ and charge density $\rho$, form a 4-vector in Minkovsky space-time. Therefore, Maxwell's equations can be expressed in a covariant form, using d'Alembertian: $$\nabla_{\mu}\nabla^{\mu} A_{\nu} = \frac{j_{\nu}}{\epsilon_0}. \tag4$$

If magnetic monopols exist, Maxwell's equation $(1)$ will look as: $$\operatorname{div} \vec B = \mu_0 c \rho_{\mathrm{magnet}}.$$

As the divergence of $\vec{B}$ isn't equal to zero, it impossible to introduce concept of vector potential. Thus, the equation in the form of $(4)$ will not be possible to express.

If $\nabla\cdot B\neq 0$, but $\nabla\times B=0$, you could assign a scalar potential. I guess.. — Manishearth, Mar 07 '12 at 17:29
possible duplicate of Effect of introducing magnetic charge on use of vector potential — Qmechanic, Mar 07 '12 at 17:31
But $\nabla \times \vec B \neq 0$ due to Maxwell's equation. — Sergio, Mar 07 '12 at 17:34
Not necessarily. $i=0$ does the trick. Basicallly, if there are loops then vector potential applies but not scalar. And the opposite if soirces/sinks of magnetic charge. But if both are together, then we have an issue. Which makes everything interesting.. — Manishearth, Mar 07 '12 at 17:45
Related question asked by OP on TP.SE: http://physics.stackexchange.com/q/27755/2451 — Qmechanic, Mar 09 '12 at 13:14
@Sergio May be not. See my question here: https://physics.stackexchange.com/questions/395167/physical-implication-of-textbfe-rightarrow-textbfb-textbfb-rightarrow — SRS, Mar 28 '18 at 09:40

score 10 · Accepted Answer · answered Mar 07 '12 at 18:06

10

Another option, besides modifying the potential $A_\mu = (A_i, \phi)$ in some way, is to introduce another 4-potential $C_\mu = (C_i, \psi)$. Then the electric and magnetic field are given by $$E = - \nabla \times C - \frac{\partial A}{\partial t} - \nabla \phi$$ $$B = \nabla \times A - \frac{\partial C}{\partial t} - \nabla \psi$$

More on this 2-potential approach can be found here: http://arxiv.org/abs/math-ph/0203043

answered Mar 07 '12 at 18:06

Tobias Diez

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In my opinion, the another 4-potential introduction is the the most natural way to solve this problem in contrast of Dirac's fibre approach. It is interesting to ask, how electromagnetic field quantization procedure will change as a result? – Sergio Mar 08 '12 at 19:39
I wonder if this is analogous to Hansen's potentials for a spacetime that is stationary but not static. – Aug 24 '13 at 03:31

score 4 · Answer 2 · answered Mar 07 '12 at 17:31

Yes, introducing a magnetic monopole into Maxwell’s equations means the existence of a vector potential that is defined everywhere and everywhere continuous is not possible anymore. In particular, this can be annoying because the vector potential representation is crucial to the prediction of the quantized value for a magnetic charge (both in the way it was historically developed, and in the way it is usually presented).

Even though the existence of the magnetic monopole hasn't been asserted, there has been quite a lot of research going on about how to go around this issue. Most formulations actually stick to some form of a vector potential, as it is the existing framework and is very convenient. This implies that such a monopole-compatible vector potential becomes a less straightforward beast. I'm not sure what is considered the best authority on this, but Wikipedia says on the issue (and it matches my own understanding):

In the mathematical theory of magnetic monopoles, A is allowed to be either undefined or multiple-valued in some places.

The topic was first approached by Dirac, and his position summarized here:

Dirac's reasoning shows that it is consistent in quantum mechanics to describe a magnetic monopole with the vector potential Equation 3, even though it has a "string" singularity

Dirac's approach is a bit outdated though. Nowadays, one would treat the subject using fibre-bundle theory. — Raskolnikov, Mar 07 '12 at 17:50
By the way, a relevant but somewhat old paper about it: TT Wu en CN Yang "Dirac Monopole without Strings Classical Lagrangian theory", Physical Review D Vol 14-2 p437-445, 1976. — Raskolnikov, Mar 09 '12 at 09:52

Does existence of magnetic monopole break covariant form of Maxwell’s equations for potentials?

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