Question

(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.

Short answer

(a)

The value of Maxwell’s equations with magnetic charge are invariant under the duality transformation is $\nabla \times B\text{'}={\mu }_{0}J{\text{'}}_c+{\mu }_0{\epsilon }_0\frac{\partial E\text{'}}{\partial t}$.

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

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

The value of Maxwell’s equations with magnetic charge are invariant under the duality transformation islocalid="1657541223433" $\nabla \times B\text{'}={\mu }_0{J}_e+{\mu }_0{\epsilon }_0\frac{\partial E}{\partial t}$

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

Step-by-step solution

Write the given data from the question.

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

Determine the formula of Maxwell’s equations with magnetic charge are invariant under the duality transformation and value of force law.

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.

(a) Determine the value of Maxwell’s equations with magnetic charge are invariant under the duality transformation.

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

(b) Determine the value of force law.

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$F=q_e\left(E+\left(V\times B\right)\right)+q_m\left(B-\frac{1}{c^2}V\times E\right).$