Question

Generalize the laws of relativistic electrodynamics (Eqs. 12.127 and 12.128) to include magnetic charge. [Refer to Sect. 7.3.4.]

Short answer

The generalized law of relativistic electrodynamics to include magnetic charge is

$K=\frac{1}{\sqrt{1-{\displaystyle \frac{u^2}{c^2}}}}{\left[q_e\left[E+\left(\mu \times B\right)\right]\right]}_x+q_m\left[\left(B-\frac{1}{c^2}\left(u\times E\right)\right)\right]$

Step-by-step solution

Expression for Maxwell’s equation:

Write the three Maxwell’s equations.

$$\begin{gathered} {\mathbf{\partial }}_{\mathbf{v}}{\mathit{F}}^{\mathbf{m}\mathbf{v}}\mathbf{=}{\mathit{m}}_{\mathbf{0}}{\mathbf{J}}_{\mathbf{e}}^{\mathbf{m}} \\ {\mathbf{\partial }}_{\mathbf{v}}{\mathit{G}}^{\mathbf{m}\mathbf{v}}\mathbf{=}\frac{{\mathbf{m}}_{\mathbf{0}}}{\mathbf{c}}{\mathbf{J}}_{\mathbf{e}}^{\mathbf{m}} \\ {\mathbf{K}}^{\mathbf{m}}\mathbf{=}\left(q_e{F}^{\mathrm{mv}}+\frac{q_m}{c}{G}^{\mathrm{mv}}\right){\mathbf{h}}_{\mathbf{v}} \end{gathered}$$

Write all the above three equations with a magnetic charge.

localid="1657699942558" $$\begin{gathered} \nabla .E=\frac{{\rho }_e}{{\epsilon }_0} \\ \nabla \times B={\mu }_0{J}_e+{\mu }_0{\epsilon }_0\frac{\partial E}{\partial t} \\ \nabla .B=\left(\frac{{\mu }_0}{c}\right)\left(c{\rho }_m\right)={\mu }_0{\rho }_m \\ -\frac{1}{c}\left(\frac{\partial b}{\partial t}+\nabla \times E\right)=\frac{{\mu }_0}{c}{J}_m \end{gathered}$$

Determine the laws of relativistic electrodynamics to include magnetic charge:

Write the equation for the Minkowski force on a charge q.

$K^m=q{h}_v{F}^{mv}$

Let $\mu =1$:

$$\begin{gathered} K^{1}q{\eta }_v{F}^{1v}=q\left(-{\eta }^0{F}^{10}+{\eta }^1{F}^{11}+{\eta }^2{F}^{12}+{\eta }^3{F}^{13}\right) \\ {K}^1=q\left[\frac{-c}{\sqrt{1-{\displaystyle \frac{u^2}{c^2}}}}\left(-B_x\right)+\frac{u_y}{\sqrt{1-{\displaystyle \frac{u^2}{c^2}}}}\left(-\frac{E_x}{c}\right)+\frac{u_z}{\sqrt{1-{\displaystyle \frac{u^2}{c^2}}}}\left(-\frac{E_x}{c}\right)\right] \\ K=\frac{1}{\sqrt{1-{\displaystyle \frac{u^2}{c^2}}}}{\left[q_e\left[E+\left(\mu \times B\right)\right]\right]}_x+q_m\left[\left(B-\frac{1}{c^2}\left(u\times E\right)\right)\right] \end{gathered}$$

Therefore, the generalized law of relativistic electrodynamics to include magnetic charge is$K=\frac{1}{\sqrt{1-{\displaystyle \frac{u^2}{c^2}}}}{\left[q_e\left[E+\left(\mu \times B\right)\right]\right]}_x+q_m\left[\left(B-\frac{1}{c^2}\left(u\times E\right)\right)\right]$.