Maxwell’s equations with magnetic charge read:
\text { (i) } \nabla \cdot E =\frac{1}{\epsilon_{0}} \rho_{e} , \text { (iii) } \nabla \times E =-\mu_{0} J _{m}-\frac{\partial B }{\partial t} ,
\text { (ii) } \nabla \cdot B =\mu_{0} \rho_{m} , \text { (iv) } \nabla \times B =\mu_{0} J _{e}+\mu_{0} \epsilon_{0} \frac{\partial E }{\partial t} ,
where \rho_{e} is the electric charge density, \rho_{m} is the magnetic charge density, J _{e} is the electric current density, and J _{m} is the magnetic current density.
If there are only electric charges and currents, then the usual potential formulation applies (Section 10.1.1), with E =- \nabla V_{e}-\partial A _{e} / \partial t, B = \nabla \times A _{e} . If there are only magnetic charges and currents,
\text { (i) } \nabla \cdot E =0 , \text { (iii) } \nabla \times E =-\mu_{0} J _{m}-\frac{\partial B }{\partial t} ,
\text { (ii) } \nabla \cdot B =\mu_{0} \rho_{m} , \text { (iv) } \nabla \times B =\mu_{0} \epsilon_{0} \frac{\partial E }{\partial t} .
This time equation (i) says that E can be expressed as the curl of a vector potential: E =- \nabla \times A _{m} , and plugging this into (iv) yields \nabla \times\left( B +\mu_{0} \epsilon_{0} \partial A _{m} / \partial t\right)= 0 , which tells us that the quantity in parentheses can be represented as the gradient of a scalar: B =-\nabla V_{m}-\mu_{0} \epsilon_{0} \partial A _{m} / \partial t . [The signs of V_{m} \text { and } A _{m} are arbitrary, but I think this choice yields the most symmetrical formulation.] Putting these into (ii) and (iii) yields
\square^{2} V_{m}=-\mu_{0} \rho_{m}-\mu_{0} \epsilon_{0} \frac{\partial}{\partial t}\left( \nabla \cdot A _{m}+\frac{\partial V_{m}}{\partial t}\right), \quad \square^{2} A _{m}=-\mu_{0} J _{m}+ \nabla \left( \nabla \cdot A _{m}+\frac{\partial V_{m}}{\partial t}\right) .
The “Lorenz gauge” for the magnetic potentials, \nabla \cdot A _{m}=-\partial V_{m} / \partial t , kills the terms in parentheses.
In general, if there are both electric and magnetic charges and currents present, the total fields are the sums (by the superposition principle):
E =-\nabla V_{e}-\frac{\partial A _{e}}{\partial t}-\nabla \times A _{m}, \quad B =-\nabla V_{m}-\mu_{0} \epsilon_{0} \frac{\partial A _{m}}{\partial t}+\nabla \times A _{e} .
If we choose to work in the Lorenz gauge:
\nabla \cdot A _{e}=-\mu_{0} \epsilon_{0} \frac{\partial V_{e}}{\partial t}, \quad \nabla \cdot A _{m}=-\frac{\partial V_{m}}{\partial t} ,
Maxwell’s equations become (in terms of the potentials)
\square^{2} V_{e}=-\frac{1}{\epsilon_{0}} \rho_{e}, \quad \square^{2} V_{m}=-\mu_{0} \rho_{m}, \quad \square^{2} A _{e}=-\mu_{0} J _{e}, \quad \square^{2} A _{m}=-\mu_{0} J _{m} ,
and the retarded solutions are
V_{e}( r , t)=\frac{1}{4 \pi \epsilon_{0}} \int \frac{\rho_{e}\left( r ^{\prime}, t_{r}\right)}{ᴫ} d \tau^{\prime}, \quad V_{m}( r , t)=\frac{\mu_{0}}{4 \pi} \int \frac{\rho_{m}\left( r ^{\prime}, t_{r}\right)}{ ᴫ } d \tau^{\prime} ,
A _{e}( r , t)=\frac{\mu_{0}}{4 \pi} \int \frac{ J _{e}\left( r ^{\prime}, t_{r}\right)}{ ᴫ } d \tau^{\prime}, \quad A _{m}( r , t)=\frac{\mu_{0}}{4 \pi} \int \frac{ J _{m}\left( r ^{\prime}, t_{r}\right)}{ᴫ} d \tau^{\prime} .