When we introduce electromagnetic field in Special Relativity, we add a term of −ecAidxi
If we now take divergence of both sides of this definition, we automatically get
∇⋅→H=0,
which is equivalent to inexistence of magnetic charges.
But suppose we've found a magnetic charge. What will change in our Lagrangian or in definition of electric and magnetic fields in this case to make ∇⋅→H=σ?
In this Phys.SE answer it's asserted that magnetic field would get an additional term "gradient of a scalar potential". Is this "a" scalar potential instead "the" A0 potential?
Answer
In the absence of magnetic monopoles, Maxwell's equations are
dF=0,d⋆F=Je,
where J is the 4-current 3-form due to electric charges (assuming a metric with signature (−,+,+,+)). For cohomological reasons, from the first equation one can asserts that there exists a 1-form A such that F=dA, and A is the interpreted as the 4-potential (ϕ,A) (up to the musical isomorphism between tangent and cotangent bundle to Minkowski spacetime). In the presence of magnetic monopoles (or charge, to even symmetrise terminology) the above equations would become
dF=Jm,d⋆F=Je,
where Jm is the 4-current for magnetic charges. Therefore in this extended theory of electrodynamics both the Faraday tensor F and its Hodge dual ⋆F (sometimes also denoted by G) figure in constitutive equations.
Since F is no longer a closed form, its expression must be modified by the introduction of a non-exact part, say C, so that
F=dA+C.
Since the equations are symmetric in F and ⋆F we can postulate there exist 1-forms B and D such that
⋆F=dB+D,
and assume that C depends on B, while D depends on A. But since ⋆⋆=−1 in special relativity, we conclude that
F=dA−⋆dB,
which can be related to the Helmholtz decomposition into polar and axial part for twice differentiable vector fields.
The Lorentz force for a particle with electric charge qe and magnetic charge qm would be K=ιu(qeF+qmG),
To make contact with the usual vector notation, observe that the Faraday tensor has the covariant matrix representation F=[0− ETE⋆H]
Reconstructing the Faraday tensor according to the prescription F=dA−⋆dB given above we then have, in terms of polar and axial parts F=(∇A0+∂A∂t−∇×B,∇×A+∇B0+∂B∂t),
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