What explains this asymmetry between the electric and magnetic fields if both electric and magnetic monopoles exist? Can't Maxwell's equations be formulated to be symmetric between the two in the presence of both monopoles? What leads to magnetic monopoles existing before the inflationary epoch and becoming infinitely diluted during inflation (if indeed that is the explanation to their absence), but the electric monopoles we know today only forming at the end of the electroweak era?
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1$\begingroup$ Please only ask one question per post $\endgroup$– BioPhysicistCommented Sep 22, 2020 at 8:07
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2 Answers
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Recall that electric-magnetic duality involves an $S$-transformation, namely the coupling constant undergoes $e\leftrightarrow 1/e$. This constant is very different from $1$ (in particular, $e\ll1$), which means that one side of the duality is weakly coupled while the other side is strongly coupled. Electric charges are very light – hence very easy to produce at low energies – while magnetic charges are very heavy, making them hard to produce today. The two types of excitations are dual, not symmetric.
Maxwell's equations are not symmetric under $E\leftrightarrow \pm B$. The transformation also involves $e\leftrightarrow 1/e$, which means this is not a symmetry, but a duality. The "equations of motion" on either side of the duality are the same, but the dynamics are different, because the objects have different properties (masses, charges, etc.).
answered Sep 25, 2020 at 14:38
AccidentalFourierTransformAccidentalFourierTransform
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$\begingroup$ Could you explain how a change in coupling constant and particle mass leads to being able to interpret magnetic monopoles as 0-dimensional topological defects and not electrons? $\endgroup$ Commented Oct 15, 2020 at 13:56
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$\begingroup$ @AnthonyKhodanian I didn't make that statement, and that is a new question, different from the one in the OP. The asymmetry is due to the fact that the duality is weak-strong, not due to monopoles being defects. But anyway, a monopole is a one-dimensional topological defect. It is a line (Dirac string), but the dependence on the path is topological: small deformations do not change expectation values. Again, this is a different question from the one in the OP, if you want to understand topological defects you should create a new thread. $\endgroup$ Commented Oct 16, 2020 at 15:32
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Also to answer the title question: what is a topological defect depends on the finite energy configurations of the fields. E&M without monopoles doesn't have nontrivial finite energy configurations (meaning you can smoothly deform them to zero). Also, E&M without electric charge but WITH monopoles doesn't have solutions with topological defects either. When you include both, the theory is complicated enough to give rise to a larger set of finite energy configurations, some of which are nontrivial.
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$\begingroup$ It's good answer. Could you specify "larger set of finite energy configurations"? Could you give some references? $\endgroup$– NikitaCommented Sep 29, 2020 at 21:03