Without magnetic monopoles, the Maxwell equations are these (I'm dropping vector notations and all constants, for simplicity): \begin{align} \nabla \cdot E &= \rho_{elec}, \tag{1} \\[12pt] \nabla \times E &= -\, \frac{\partial B}{\partial t}, \tag{2} \\[12pt] \nabla \cdot B &= 0, \tag{3} \\[12pt] \nabla \times B &= J_{elec} + \frac{\partial E}{\partial t}. \tag{4} \end{align} These equations can be presented into an explicitely relativistic form (much simpler): \begin{gather} \partial_a F^{ab} = J_{elec}^b, \tag{5} \\[12pt] \partial_a {}^{\star}F^{ab} = 0, \tag{6} \end{gather} where ${}^{\star}F^{ab} = \frac{1}{2} \, \varepsilon^{abcd} F_{cd}$ is the Hodge dual of $F_{ab}$. Equations (2), (3) are equivalent to (6). They imply the existence of a gauge potential : $A^a = \{ \, \phi, A^i\}$ such that \begin{align} B &= \nabla \times A, \tag{7} \\[12pt] E &= -\, \nabla \phi - \frac{\partial A}{\partial t}, \tag{8} \end{align} or \begin{equation} F_{ab} = \partial_a A_b - \partial_b A_a. \tag{9} \end{equation} Thus (2), (3) and (6) are automatically and trivially satisfied. This was the starting point of all gauge theories.

Now, introducing magnetic monopoles imposes that equations (2), (3) and (6) must be modified: \begin{align} \nabla \cdot E &= \rho_{elec}, \tag{10} \\[12pt] \nabla \times E &= J_{magn} -\, \frac{\partial B}{\partial t}, \tag{11} \\[12pt] \nabla \cdot B &= \rho_{magn}, \tag{12} \\[12pt] \nabla \times B &= J_{elec} + \frac{\partial E}{\partial t}, \tag{13} \end{align} or equivalently \begin{gather} \partial_a F^{ab} = J_{elec}^b, \tag{14} \\[12pt] \partial_a {}^{\star}F^{ab} = J_{magn}^b. \tag{15} \end{gather} The Maxwell equations are now more symetric with magnetic monopoles.

But then, how do you introduce the classical gauge potentials, since $B =\nabla \times A$ cannot be valid anymore ($\nabla \cdot (\nabla \times A) \equiv 0$, unless you play "artificial" tricks with the space topology)? The relativistic expression (9) cannot be introduced anymore. And because of the new symetry, why not introduce an expression like $E = \nabla \times A$ instead?

To me, introducing magnetic monopoles appears to destroy the gauge potential theory. I don't see how the gauge potentials could be justified if there are magnetic monopoles. Playing tricks with topology feels like a patch to the theory, and I'm having hard times in accepting that gauge potentials must be pushed on the topology side.

  • 2
    $\begingroup$ Well, if you don't approve of cutting out Dirac strings, usually this is handled by promoting $A$ to a connection on a fiber bundle. For example, see here. $\endgroup$ – knzhou Dec 3 '18 at 12:15
  • 1
    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/22018/2451 and links therein. $\endgroup$ – Qmechanic Dec 3 '18 at 13:07
  • 1
    $\begingroup$ @knzhou Let's hope magnetic monopoles are never found, for I am not sure if my eyes can handle a connection on a fibre bundle! $\endgroup$ – my2cts Dec 3 '18 at 13:23
  • $\begingroup$ How could we define an electromagnetic Lagrangian without the gauge potentials ? AFAIK, we need the potentials to define and apply a variational principle. The magnetic monopole appears to destroy the variational principle! $\endgroup$ – Cham Dec 3 '18 at 18:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.