Question 10.33: Develop the potential formulation for electrodynamics with m...

Develop the potential formulation for electrodynamics with magnetic charge (Eq. 7.44). [Hint: You’ll need two scalar potentials and two vector potentials. Use the Lorenz gauge. Find the retarded potentials (generalizing Eqs. 10.26), and give the formulas for E and B in terms of the potentials (generalizing Eqs. 10.2 and 10.3).]

\left. \begin{matrix} \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} \text {. } \end{matrix} \right\}                               (7.44)

V( r , t)=\frac{1}{4 \pi \epsilon_{0}} \int \frac{\rho\left( r ^{\prime}, t_{r}\right)}{ᴫ} d \tau^{\prime}, \quad A ( r , t)=\frac{\mu_{0}}{4 \pi} \int \frac{ J \left( r ^{\prime}, t_{r}\right)}{ᴫ} d \tau^{\prime}                                     (10.26)

B = × A                                                    (10.2)

E =-\nabla V-\frac{\partial A }{\partial t}                                    (10.3)

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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} .

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