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

$$\begin{gathered} \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{E}\mathbf{\text{'}}\mathbf{=}\mathit{E}\mathit{c}\mathit{o}\mathit{s}\mathit{\alpha }\mathbf{+}\mathit{c}\mathit{B}\mathit{s}\mathit{i}\mathit{n}\mathit{\alpha }\mathbf{,} \\ \mathit{c}\mathit{B}\mathbf{\text{'}}\mathbf{=}\mathit{c}\mathit{B}\mathit{c}\mathit{o}\mathit{s}\mathit{\alpha }\mathbf{-}\mathit{E}\mathit{s}\mathit{i}\mathit{n}\mathit{\alpha }\mathbf{,} \\ \mathit{c}\mathit{q}{\mathbf{\text{'}}}_{\mathbf{e}}\mathbf{=}\mathit{c}{\mathit{q}}_{\mathbf{e}}\mathit{c}\mathit{o}\mathit{s}\mathit{\alpha }\mathbf{+}{\mathit{q}}_{\mathbf{m}}\mathit{s}\mathit{i}\mathit{n}\mathit{\alpha }\mathbf{,} \\ \mathit{q}{\mathbf{\text{'}}}_{\mathbf{m}}\mathbf{=}{\mathit{q}}_{\mathbf{m}}\mathit{c}\mathit{o}\mathit{s}\mathit{\alpha }\mathbf{-}\mathit{c}{\mathit{q}}_{\mathbf{e}}\mathit{s}\mathit{i}\mathit{n}\mathit{\alpha }\mathbf{,} \end{gathered}$$

Where $\mathit{c}\mathbf{\equiv }\mathbf{1}\mathbf{/}\sqrt{{\mathbf{\epsilon }}_{\mathbf{0}}{\mathbf{\mu }}_{\mathbf{0}}}$and $\mathit{\alpha }$is an arbitrary rotation angle in “E/B-space.” Charge and current densities transform in the same way as ${\mathit{q}}_{\mathbf{e}}$and ${\mathit{q}}_{\mathbf{m}}$. [This means, in particular, that if you know the fields produced by a configuration of electric charge, you can immediately (using $\mathit{\alpha }\mathbf{=}\mathbf{90}\mathbf{°}$) write down the fields produced by the corresponding arrangement of magnetic charge.]

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

$\mathbf{F}\mathbf{=}{\mathbf{q}}_{\mathbf{e}}\left(E+V\times B\right)\mathbf{+}{\mathbf{q}}_{\mathbf{m}}\mathbf{}\left(B-\frac{1}{c^2}V\times E\right)$

is also invariant under the duality transformation.

We have to show that

(a)$\nabla ·E\text{'}=\frac{{\rho }_e}{{\epsilon }_0}$

(b) $\nabla ·B\text{'}={\mu }_0{\rho }_m$

(c) $\nabla \times E\text{'}=-{\mu }_0{J}_m-\frac{\partial B}{\partial t}$

(d) $\nabla \times B\text{'}={\mu }_0{J}_e+{\mu }_0{\epsilon }_0\frac{\partial E}{\partial t}$

Write the formula ofMaxwell’s equations with magnetic charge is invariant under the duality transformation.

$\mathbf{\nabla }\mathbf{‐}\mathit{E}\mathbf{\text{'}}$ …… (1)

Here, $\mathit{E}\mathbf{\text{'}}$is electrical field.

Write the formula of Maxwell’s equations with magnetic charge is invariant under the duality transformation.

localid="1657626502486" $\mathbf{∆}\mathbf{.}\mathbf{B}\mathbf{\text{'}}$ …… (2)

Here,$\mathbf{B}\mathbf{\text{'}}$is magnetic field.

Write the formula of Maxwell’s equations with magnetic charge is invariant under the duality transformation.

localid="1657626718393" $\mathbf{\nabla }\mathbf{\times }\mathit{E}\mathbf{\text{'}}$ …… (3)

Write the formula of Maxwell’s equations with magnetic charge is invariant under the duality transformation.

$\mathbf{\nabla }\mathbf{\times }\mathbf{B}\mathbf{\text{'}}$ …… (4)

Here,$\mathbf{B}\mathbf{\text{'}}$ is magnetic field.

Write the formula of force law.

$\mathbf{F}\mathbf{=}{\mathbf{q}}_{\mathbf{e}}\mathbf{(}\mathbf{E}\mathbf{\times }\mathbf{V}\mathbf{\times }\mathbf{B}\mathbf{)}\mathbf{+}{\mathbf{q}}_{\mathbf{m}}\left(B-\frac{1}{C^2}V\times E\right)$ …… (5)

Here,${\mathbf{q}}_{\mathbf{e}}$ is electric charge,${\mathbf{q}}_{\mathbf{m}}$ is magnetic charge,$\mathbf{B}$ is magnetic field and$\mathbf{v}$ is voltage.

Consider given equation as:

$$\begin{gathered} E\text{'}=Ecos\alpha +cBsin\alpha , \\ cB\text{'}=cBcos\alpha -Esin\alpha , \\ cq{\text{'}}_e=c{q}_{e}cos\alpha +q_{m}sin\alpha , \\ q{\text{'}}_m=q_{m}cos\alpha -c{q}_{e}sin\alpha , \end{gathered}$$

Determine the value of Maxwell’s equation with magnetic charge is invariant under the duality transformation.

Substitute $\mathrm{Ecos}\alpha +c\mathrm{Bsin}\alpha$for $\mathrm{E}\text{'}$into equation (1).

localid="1657686378180" $$\begin{gathered} \nabla .\mathrm{E}\text{'}=\nabla .\left[\mathrm{Ecos}\alpha +c\mathrm{Bsin}\alpha \right] \\ =\left(\nabla .\mathrm{E}\right)\cos \alpha +c\left(\nabla .B\right)\sin \alpha \\ =\frac{1}{{\epsilon }_0}\left({\rho }_e\right)\cos \alpha +c{\mu }_0{\rho }_m\sin \alpha \\ =\frac{1}{{\epsilon }_0}\left[{\rho }_e\cos \alpha +\frac{1}{c}{\rho }_m\sin \alpha \right] \end{gathered}$$

Solve further as,

$$\begin{gathered} \nabla .E=\frac{1}{{\epsilon }_0}\left[{\rho }_e\cos \alpha +\frac{1}{c}{\rho }_m\sin \alpha \right] \\ =\frac{1}{{\epsilon }_0}{\rho }_e^{\text{'}} \end{gathered}$$

Therefore, the value of Maxwell’s equations with magnetic charge are invariant under the duality transformation is$\nabla \times B\text{'}={\mu }_0{J}_c^{\text{'}}+{\mu }_0{\epsilon }_0\frac{\partial E\text{'}}{\partial t}.$

Determine the value of Maxwell’s equation with magnetic charge is invariant under the duality transformation.

Substitute$B\cos \alpha -\frac{E}{c}\sin \alpha$for$B\text{'}$into equation (2).

$$\begin{gathered} \nabla .B\text{'}=B\cos \alpha -\frac{E}{c}\sin \alpha \\ =\left(\nabla .B\right)\cos \alpha -\frac{\left(\nabla \times E\right)}{c}\sin \alpha \\ ={\mu }_0{\rho }_m\cos \alpha -\frac{{\rho }_e}{{\epsilon }_{0}c}\sin \alpha \\ ={\mu }_0\left(f_m\cos \alpha -c{\rho }_e\sin \alpha \right) \end{gathered}$$

Solve further as,

$\nabla .B\text{'}={\mu }_0{\rho }_m^{\text{'}}$

Therefore, the value of Maxwell’s equations with magnetic charge are invariant under the duality transformation is$\nabla .B\text{'}={\mu }_0{\rho }_m.$

Determine the value of Maxwell’s equation with magnetic charge is invariant under the duality transformation.

Substitute$E\cos \alpha +cBs\mathrm{in}\alpha$for$E\text{'}$into equation (3).

$$\begin{gathered} \nabla \times E\text{'}=E\cos \alpha +cB\sin \alpha \\ =\left(\nabla \times E\right)\cos \alpha +c\left(\nabla \times B\right)\sin \alpha \\ =\left(-{\mu }_0{J}_m-\frac{\partial B}{\partial t}\right)\cos \alpha +c\left({\mu }_0{J}_e+{\mu }_0{\epsilon }_0\frac{\partial E}{\partial t}\right)\sin \alpha \\ =-{\mu }_0\left(J_m\cos \alpha -c{J}_e\sin \alpha \right)-\frac{\partial }{\partial t}\left(B\cos \alpha -\frac{E}{c}\sin \alpha \right) \end{gathered}$$

Solve further as,

$\nabla \times E\text{'}=-{\mu }_0{J}_m^{\text{'}}-\frac{\partial B\text{'}}{\partial t}$

Therefore, the The value of Maxwell’s equations with magnetic charge are invariant under the duality transformation is$\nabla \times E\text{'}=-{\mu }_0{J}_m^{\text{'}}-\frac{\partial B\text{'}}{\partial t}.$

Determine the value of Maxwell’s equation with magnetic charge is invariant under the duality transformation.

Substitute$B\cos \alpha -\frac{1}{c}E\sin \alpha$for$B\text{'}$into equation (4).

$$\begin{gathered} \nabla \times B\text{'}=\nabla \times \left(B\cos \alpha -\frac{1}{c}E\sin \alpha \right) \\ =\left(\nabla \times B\right)\cos \alpha -\frac{1}{c}\left(\nabla \times E\right)\sin \alpha \\ =\left[{\mu }_0{J}_{e+}{\mu }_0{\epsilon }_0\frac{\partial E}{\partial t}\right]\cos \alpha -\frac{1}{c}\left[-{\mu }_0{J}_m-\frac{\partial B}{\partial t}\right]\sin \alpha \\ ={\mu }_0\left(J_e\cos \alpha +\frac{1}{c}{J}_m\sin \alpha \right)+{\mu }_0{\epsilon }_0\frac{\partial }{\partial t}\left(E\cos \alpha +cB\sin \alpha \right) \end{gathered}$$

Solve further as:

$\nabla \times B\text{'}={\mu }_0{J}_e^{\text{'}}+{\mu }_0{\epsilon }_0\frac{\partial E\text{'}}{\partial t}$

Therefore, the value of Maxwell’s equations with magnetic charge are invariant under the duality transformation is$\nabla \times B\text{'}={\mu }_0{J}_e^{\text{'}}+{\mu }_0{\epsilon }_0\frac{\partial E\text{'}}{\partial t}$

Here, the force law for a monopole$q_m$traveling across the electric and magnetic fields E and B at velocity $v$ is

Determine the force law.

localid="1657692041980" $$\begin{gathered} F=q_e\left(E+\left(V\times B\right)\right)+q_m\left(B-\frac{1}{c^2}V\times E\right) \\ ThenF\text{'}=q_e^{\text{'}}(E\text{'}+(V\times B\text{'}\left)\right)+q_m^{\text{'}}\left(B\text{'}-\frac{1}{c^2}V\times E\text{'}\right) \\ Substitute\left(q_e\cos \alpha +\frac{1}{c}{q}_m\sin \alpha \right)for{q}_e^{\text{'}},(E\cos \alpha +c\sin \alpha )forE\text{'}, \\ \left(B\cos \alpha -\frac{1}{c}E\right)forB\text{'},\left(q_m\cos \alpha -c{q}_e\sin \alpha \right)for{q}_m^{\text{'}},\left(B\cos \alpha -\frac{1}{c}E\sin \alpha \right)forB \\ and\left(cE\cos \alpha +cB\sin \alpha \right)forE\text{'}intoequation\left(5\right). \end{gathered}$$

F'=
$$\begin{gathered} \left(q_e\cos \alpha +\frac{1}{c}{q}_m\sin \alpha \right)\left[\left(E\cos \alpha +cB\sin \alpha \right)+v\times \left(B\cos \alpha -\frac{1}{c}E\sin \alpha \right)\right]+(q_m\cos \alpha -c{q}_e\sin \alpha )\left(Bc\right) \\ =\left\{=q_e\left[\begin{array}{c}(E{\cos }^2\alpha +cB\sin \alpha \cos \alpha -cB\sin \alpha \cos \alpha -cB\sin \alpha \cos \alpha +E{\sin }^2\alpha +\\ v\times \left(B{\cos }^2\alpha -\frac{1}{c}E\sin \alpha \cos \alpha +\frac{1}{c}E\sin \alpha \cos \alpha +B{\sin }^2\alpha \right)\end{array}\right] \\ +q_m\left[\begin{array}{c}\left(\frac{1}{c}E\sin \alpha \cos \alpha +B{\sin }^2\alpha +B{\cos }^2\alpha -\frac{1}{c}E\sin \alpha \cos \alpha \right)+v\\ \times \left(\frac{1}{c}B\sin \alpha \cos \alpha -\frac{E}{c^2}{\sin }^2\alpha -\frac{E}{c^2}{\cos }^2\alpha -\frac{B}{c}\sin \alpha \cos \alpha \right)\end{array}\right]\right\} \\ =q_e\left(E\times (V\times B)\right)+q_m\left(B-\frac{1}{c^2}v\times E\right) \\ =F \end{gathered}$$

Therefore, the value of force law is$F=q_e\left(E+\left(V\times B\right)\right)+q_m\left(B-\frac{1}{c^2}V\times E\right).$