Question 12.73: Generalize the laws of relativistic electrodynamics (Eqs. 12...

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

\frac{\partial F^{\mu \nu}}{\partial x^{\nu}}=\mu_{0} J^{\mu}, \quad \frac{\partial G^{\mu \nu}}{\partial x^{\nu}}=0                                           (12.127)

K^{\mu}=q \eta_{\nu} F^{\mu \nu}                                        (12.128)

The Blue Check Mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

Define the electric current 4-vector as before: J_{e}^{\mu}=\left(c \rho_{e}, J _{e}\right) , and the magnetic current analogously: J_{m}^{\mu}=\left(c \rho_{m}, J _{m}\right) . The fundamental laws are then

\partial_{\nu} F^{\mu \nu}=\mu_{0} J_{e}^{\mu}, \quad \partial_{\nu} G^{\mu \nu}=\frac{\mu_{0}}{c} J_{m}^{\mu}, \quad K^{\mu}=\left(q_{e} F^{\mu \nu}+\frac{q_{m}}{c} G^{\mu \nu}\right) \eta_{\nu}.

The first of these reproduces \nabla \cdot E =\left(1 / \epsilon_{0}\right) \rho_{e} \text { and } \nabla \times B =\mu_{0} J _{e}+\mu_{0} \epsilon_{0} \partial E / \partial t , just as before; the second yields \nabla \cdot B =\left(\mu_{0} / c\right)\left(c \rho_{m}\right)=\mu_{0} \rho_{m} \text { and }-(1 / c)[\partial B / \partial t+ \nabla \times E ]=\left(\mu_{0} / c\right) J _{m} \text {, or } \nabla \times E =-\mu_{0} J _{m}-\partial B / \partial t (generalizing Sec. 12.3.4). These are Maxwell’s equations with magnetic charge (Eq. 7.44). The third says

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

K^{1}=\frac{q_{e}}{\sqrt{1-u^{2} / c^{2}}}[ E +( u \times B )]_{x}+\frac{q_{m}}{c}\left[\frac{-c}{\sqrt{1-u^{2} / c^{2}}}\left(-B_{x}\right)+\frac{u_{y}}{\sqrt{1-u^{2} / c^{2}}}\left(-\frac{E_{z}}{c}\right)+\frac{u_{z}}{\sqrt{1-u^{2} / c^{2}}}\left(\frac{E_{y}}{c}\right)\right],

K =\frac{1}{\sqrt{1-u^{2} / c^{2}}}\left\{q_{e}[ E +( u \times B )]+q_{m}\left[ B -\frac{1}{c^{2}}( u \times E )\right]\right\} , or

F =q_{e}\left[ E +( u \times B ]+q_{m}\left[ B -\frac{1}{c^{2}}( u \times E )\right]\right.,

which is the generalized Lorentz force law (Eq. 7.69).

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

Related Answered Questions