Question 7.64: (a) Show that Maxwell’s equations with magnetic charge (Eq. ...

(a) Show that Maxwell’s equations with magnetic charge (Eq. 7.44) are invariant under the duality transformation

\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 {, } \\ \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}. \end{matrix} \right\}                                 (7.44)

\left.\begin{array}{l} E ^{\prime}= E \cos \alpha+c B \sin \alpha ,\\c B ^{\prime}=c B \cos \alpha- E \sin \alpha ,\\c q_{e}^{\prime}=c q_{e} \cos \alpha+q_{m} \sin \alpha ,\\q_{m}^{\prime}=q_{m} \cos \alpha-c q_{e} \sin \alpha.\end{array}\right\}                                 (7.68)

where c \equiv 1 / \sqrt{\epsilon_{0} \mu_{0}} and α is an arbitrary rotation angle in “E/B-space.” Charge and current densities transform in the same way as q_{e} \text { and } q_{n}. [This means, in particular, that if you know the fields produced by a configuration of electric charge, you can immediately (using \alpha=90^{\circ}) write down the fields produced by the corresponding arrangement of magnetic charge.] 

(b) Show that the force law (Prob. 7.38)

F =q_{e}( E + v \times B )+q_{m}\left( B -\frac{1}{c^{2}} v \times E \right)                                   (7.69)

is also invariant under the duality transformation.

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(a)   \nabla \cdot E ^{\prime}=( \nabla \cdot E ) \cos \alpha+c( \nabla \cdot B ) \sin \alpha=\frac{1}{\epsilon_{0}} \rho_{e} \cos \alpha+c \mu_{0} \rho_{m} \sin \alpha

 

=\frac{1}{\epsilon_{0}}\left(\rho_{e} \cos \alpha+c \mu_{0} \epsilon_{0} \rho_{m} \sin \alpha\right)=\frac{1}{\epsilon_{0}}\left(\rho_{e} \cos \alpha+\frac{1}{c} \rho_{m} \sin \alpha\right)=\frac{1}{\epsilon_{0}} \rho_{e}^{\prime}.

 

\nabla \cdot B ^{\prime}=( \nabla \cdot B ) \cos \alpha-\frac{1}{c}( \nabla \cdot E ) \sin \alpha=\mu_{0} \rho_{m} \cos \alpha-\frac{1}{c \epsilon_{0}} \rho_{e} \sin \alpha

 

=\mu_{0}\left(\rho_{m} \cos \alpha-\frac{1}{c \mu_{0} \epsilon_{0}} \rho_{e} \sin \alpha\right)=\mu_{0}\left(\rho_{m} \cos \alpha-c \rho_{e} \sin \alpha\right)=\mu_{0} \rho_{m}^{\prime} .

 

\nabla \times E ^{\prime}=( \nabla \times E ) \cos \alpha+c( \nabla \times B ) \sin \alpha=\left(-\mu_{0} J _{m}-\frac{\partial B }{\partial t}\right) \cos \alpha+c\left(\mu_{0} J _{e}+\mu_{0} \epsilon_{0} \frac{\partial E }{\partial t}\right) \sin \alpha

 

=-\mu_{0}\left( J _{m} \cos \alpha-c J _{e} \sin \alpha\right)-\frac{\partial}{\partial t}\left( B \cos \alpha-\frac{1}{c} E \sin \alpha\right)=-\mu_{0} J _{m}^{\prime}-\frac{\partial B ^{\prime}}{\partial t} .

 

\nabla \times B ^{\prime}=( \nabla \times B ) \cos \alpha-\frac{1}{c}( \nabla \times E ) \sin \alpha=\left(\mu_{0} J _{e}+\mu_{0} \epsilon_{0} \frac{\partial E }{\partial t}\right) \cos \alpha-\frac{1}{c}\left(-\mu_{0} J _{m}-\frac{\partial B }{\partial t}\right) \sin \alpha

 

=\mu_{0}\left( J _{e} \cos \alpha+\frac{1}{c} J _{m} \sin \alpha\right)+\mu_{0} \epsilon_{0} \frac{\partial}{\partial t}( E \cos \alpha+c B \sin \alpha)=\mu_{0} J _{e}^{\prime}+\mu_{0} \epsilon_{0} \frac{\partial E ^{\prime}}{\partial t}.

(b)   F ^{\prime}=q_{e}^{\prime}\left( E ^{\prime}+ v \times B ^{\prime}\right)+q_{m}^{\prime}\left( B ^{\prime}-\frac{1}{c^{2}} v \times E ^{\prime}\right)

 

=\left(q_{e} \cos \alpha+\frac{1}{c} q_{m} \sin \alpha\right)\left[( E \cos \alpha+c B \sin \alpha)+ v \times\left( B \cos \alpha-\frac{1}{c} E \sin \alpha\right)\right]

 

+\left(q_{m} \cos \alpha-c q_{e} \sin \alpha\right)\left[\left( B \cos \alpha-\frac{1}{c} E \sin \alpha\right)-\frac{1}{c^{2}} v \times( E \cos \alpha+c B \sin \alpha)\right]

 

=q_{e}\left[\left( E \cos ^{2} \alpha+c B \sin \alpha \cos \alpha-c B \sin \alpha \cos \alpha+ E \sin ^{2} \alpha\right)\right.

 

\left.+ v \times\left( B \cos ^{2} \alpha-\frac{1}{c} E \sin \alpha \cos \alpha+\frac{1}{c} E \sin \alpha \cos \alpha+ B \sin ^{2} \alpha\right)\right]

 

+q_{m}\left[\left(\frac{1}{c} E \sin \alpha \cos \alpha+ B \sin ^{2} \alpha+ B \cos ^{2} \alpha-\frac{1}{c} E \sin \alpha \cos \alpha\right)\right.

 

\left.+ v \times\left(\frac{1}{c} B \sin \alpha \cos \alpha-\frac{1}{c^{2}} E \sin ^{2} \alpha-\frac{1}{c^{2}} E \cos ^{2} \alpha-\frac{1}{c} B \sin \alpha \cos \alpha\right)\right]

 

=q_{e}( E + v \times B )+q_{m}\left( B -\frac{1}{c^{2}} v \times E \right)= F . \quad \text { qed }

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