out the formulas for u, S, g, and $\overleftrightarrow{\mathbf{T}}$in the presence of magnetic charge. [Hint: Start with the generalized Maxwell equations (7.44) and Lorentz force law (Eq. 8.44), and follow the derivations in Sections 8.1.2, 8.2.2, and 8.2.3.]

Write the expression for Maxwell’s equation with a magnetic charge.

$\mathbf{\nabla }\mathbf{·}\mathit{F}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{\epsilon }}_{\mathbf{0}}}{\mathit{\mu }}_{\mathbf{0}}$ …… (1)

Consider the second equation as:

$\mathbf{\nabla }\mathbf{·}\mathit{B}\mathbf{=}\mathit{p}{\mathit{\mu }}_{\mathbf{0}\mathbf{m}}$ …… (2)

Consider the third equation as:

$\mathbf{\nabla }\mathbf{·}\mathit{E}\mathbf{=}\mathbf{-}{\mathit{\mu }}_{\mathbf{0}}{\mathit{J}}_{\mathbf{m}}\mathbf{-}\frac{\mathbf{\partial }\mathbf{B}}{\mathbf{\partial }\mathbf{t}}$ …… (3)

Consider the fourth equation as:

$\mathbf{\nabla }\mathbf{\times }\mathit{B}\mathbf{=}\mathbf{-}{\mathit{\mu }}_{\mathbf{0}}{\mathit{J}}_{\mathbf{e}}\mathbf{+}{\mathit{\mu }}_{\mathbf{0}}{\mathit{\epsilon }}_{\mathbf{0}}\frac{\mathbf{\partial }\mathbf{E}}{\mathbf{\partial }\mathbf{t}}$ …… (4)

Here, E is the electric field, ${\epsilon }_0$is the permittivity of free space, ${\rho }_e$is the electric charge density, $J_e$is the electric current density, J_m is the magnetic current density, B is the magnetic field, t is the time and ${\mu }_0$is the permeability of free space.

Write the expression for the Lorentz force.

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

Here, q is the charge, c is the speed of light, and v is the speed.

Multiply by dl in equation (5) to calculate the work done by the electromagnetic force in the interval dt.

$\mathit{F}\mathbf{·}\mathit{d}\mathit{l}\mathbf{=}\left[q_e\left(E+v\times B\right)+q_m\left(B-\frac{1}{c^2}v\times E\right)\right]\mathbf{·}\mathit{d}\mathit{t}$ ……. (6)

Write the expression for displacement in terms of velocity and time.

dl = vdt

Substitute the known values in equation (6).

$$\begin{gathered} \mathit{F}\mathbf{·}\mathit{d}\mathit{l}\mathbf{=}\mathbf{[}{\mathbf{q}}_{\mathbf{e}}\left(E+v\times B\right)\mathbf{+}{\mathbf{q}}_{\mathbf{m}}\left(B-\frac{1}{c^2}v\times E\right)\mathbf{]}\mathbf{·}\mathit{v}\mathit{d}\mathit{t} \\ \mathit{F}\mathbf{·}\mathit{d}\mathit{l}\mathbf{=}\mathbf{[}{\mathbf{q}}_{\mathbf{e}}\mathbf{(}\mathbf{E}\mathbf{+}\mathbf{v}\mathbf{\times }\mathbf{B}\mathbf{)}\mathbf{·}\mathbf{vdt}\mathbf{+}{\mathbf{q}}_{\mathbf{m}}\mathbf{(}\mathbf{B}\mathbf{-}\frac{\mathbf{1}}{{\mathbf{c}}^{\mathbf{2}}}\mathbf{v}\mathbf{\times }\mathbf{E}\mathbf{)}\mathbf{·}\mathbf{vdt}\mathbf{]} \\ \mathit{F}\mathbf{·}\mathit{d}\mathit{l}\mathbf{=}\mathbf{[}\mathbf{(}\mathbf{(}{\mathbf{q}}_{\mathbf{e}}\mathbf{E}\mathbf{·}\mathbf{vdt}\mathbf{)}\mathbf{+}{\mathbf{q}}_{\mathbf{e}}\mathbf{(}\mathbf{v}\mathbf{\times }\mathbf{B}\mathbf{)}\mathbf{·}\mathbf{vdt}\mathbf{)}\mathbf{+}\mathbf{(}{\mathbf{q}}_{\mathbf{m}}\mathbf{B}\mathbf{·}\mathbf{vdt}\mathbf{-}\frac{{\mathbf{q}}_{\mathbf{m}}}{{\mathbf{c}}^{\mathbf{2}}}\mathbf{(}\mathbf{v}\mathbf{\times }\mathbf{E}\mathbf{)}\mathbf{·}\mathbf{vdt}\mathbf{)}\mathbf{]} \\ \mathit{F}\mathbf{·}\mathit{d}\mathit{l}\mathbf{=}\mathbf{[}\left(q_{e}E·\mathrm{vdt}\right)\mathbf{+}\mathbf{(}{\mathbf{q}}_{\mathbf{m}}\mathbf{B}\mathbf{·}\mathbf{vdt}\mathbf{)}\mathbf{]} \end{gathered}$$ ……. (7)

Since,

Substitute the known values in equation (7).

$\frac{\mathbf{d}\mathbf{W}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{\int }\left[E·J_e+B·J_m\right]\mathit{d}\mathit{\zeta }$

Use equations (3) and (4) to remove ${\mathit{J}}_{\mathbf{e}}$and ${\mathit{J}}_{\mathbf{m}}$in the above equation.

$$\begin{gathered} \frac{\mathbf{d}\mathbf{W}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{\int }\left[E·\left(\frac{\nabla \times B}{{\mu }_0}-{\epsilon }_0\frac{\partial E}{\partial \partial }\right)+B·\left(-\frac{∆\times E}{{\mu }_0}-\frac{1}{{\mu }_0}\frac{\partial B}{\partial t}\right)\right] \\ \frac{\mathbf{d}\mathbf{W}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{\int }\left[\left(\frac{E·\left(\nabla \times B\right)}{{\mu }_0}-{\epsilon }_{0}E·\frac{\partial E}{\partial \partial }\right)+\left(-\frac{B·\left(∆\times E\right)}{{\mu }_0}-\frac{1}{{\mu }_0}\frac{\partial B}{\partial t}\right)\right] \end{gathered}$$ ……. (9)

Since,

$\mathbf{\nabla }\mathbf{·}\left(E\times B\right)\mathbf{=}\left(\nabla \times E\right)\mathbf{-}\mathit{E}\mathbf{·}\left(\nabla \times B\right)$

On further solving equation (8),

$$\begin{gathered} \frac{\mathbf{d}\mathbf{W}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{\int }\mathbf{-}\frac{\mathbf{1}}{{\mathbf{\mu }}_{\mathbf{0}}}\mathbf{\nabla }\mathbf{·}\left(E\times B\right)\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{t}}\left({\epsilon }_0{E}^2+\frac{1}{{\mu }_0}{B}^2\right) \\ \frac{\mathbf{d}\mathbf{W}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{-}\frac{\mathbf{d}}{\mathbf{d}\mathbf{t}}\mathbf{\int }\frac{\mathbf{1}}{\mathbf{2}}\left({\epsilon }_0{E}^2+\frac{1}{{\mu }_0}{B}^2\right)\mathit{d}\mathit{\zeta }\mathbf{-}\frac{\mathbf{1}}{{\mathbf{\mu }}_{\mathbf{0}}}\boxed{\xcancel{}}\mathbf{}\mathbf{\int }\left(E\times B\right)\mathbf{·}\mathit{d}\mathit{a} \end{gathered}$$

Hence, the energy stored in the field will be,

$\mathit{u}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\left({\epsilon }_0{E}^2+\frac{1}{{\mu }_0}{B}^2\right)$

So, the Poynting vector will be,

$\mathit{S}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{\mu }}_{\mathbf{0}}}\left(E\times B\right)$

Write the expression for electromagnetic momentum density.

$\mathit{g}\mathbf{=}{\mathit{\mu }}_{\mathbf{0}}{\mathit{\epsilon }}_{\mathbf{0}}\mathit{S}$

Substitute the known values in the above equation.

$$\begin{gathered} \mathit{g}\mathbf{=}{\mathit{\mu }}_{\mathbf{0}}{\mathit{\epsilon }}_{\mathbf{0}}\mathbf{\left(}\frac{\mathbf{1}}{{\mathbf{\mu }}_{\mathbf{0}}}\left(E\times B\right)\mathbf{\right)} \\ \mathit{g}\mathbf{=}{\mathit{\epsilon }}_{\mathbf{0}}\mathbf{(}\mathbf{E}\mathbf{\times }\mathbf{B}\mathbf{)} \end{gathered}$$

Write the expression for the total electromagnetic force on the charges for volume.

$\mathit{F}\mathbf{=}\mathbf{\int }\left[\left(E+V\times B\right){\rho }_e+\left(B-\frac{1}{c^2}v\times E\right){\rho }_m\right]\mathit{d}\mathit{\zeta }$

Write the expression for force per unit volume.

$$\begin{gathered} \frac{\mathbf{F}}{\mathbf{d}\mathbf{\zeta }}\mathbf{=}\mathbf{\int }\mathbf{[}\left(E+V\times B\right){\mathbf{\rho }}_{\mathbf{e}}\mathbf{+}\left(B-\frac{1}{c^2}v\times E\right){\mathbf{\rho }}_{\mathbf{m}}\mathbf{]} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{f}\mathbf{=}\mathbf{[}\mathbf{(}{\mathbf{\rho }}_{\mathbf{e}}\mathbf{E}\mathbf{+}{\mathbf{\rho }}_{\mathbf{e}}\mathbf{v}\mathbf{\times }\mathbf{B}\mathbf{)}\mathbf{+}\mathbf{(}{\mathbf{\rho }}_{\mathbf{m}}\mathbf{B}\mathbf{-}\frac{\mathbf{1}}{{\mathbf{c}}^{\mathbf{2}}}{\mathbf{\rho }}_{\mathbf{m}}\mathbf{v}\mathbf{\times }\mathbf{E}\mathbf{)}\mathbf{]} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{f}\mathbf{=}\mathbf{[}\mathbf{(}{\mathbf{\rho }}_{\mathbf{e}}\mathbf{E}\mathbf{+}{\mathbf{J}}_{\mathbf{e}}\mathbf{\times }\mathbf{B}\mathbf{)}\mathbf{+}\mathbf{(}{\mathbf{\rho }}_{\mathbf{m}}\mathbf{B}\mathbf{-}\frac{\mathbf{1}}{{\mathbf{c}}^{\mathbf{2}}}{\mathbf{J}}_{\mathbf{m}}\mathbf{\times }\mathbf{E}\mathbf{)}\mathbf{]} \end{gathered}$$

From equations (1), (2), (3) and (4),

$\mathit{f}\mathbf{=}{\mathit{\epsilon }}_{\mathbf{0}}\left[\left(\nabla ·E\right)E-\frac{1}{2}\nabla \left(E^2\right)+\left(E·\nabla \right)E\right]\mathbf{+}\frac{\mathbf{1}}{{\mathbf{\mu }}_{\mathbf{0}}}\left[\left(\nabla ·B\right)B-\frac{1}{2}\nabla \left(B^2\right)+\left(B·\nabla \right)B\right]\mathbf{-}{\mathit{\epsilon }}_{\mathbf{0}}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{t}}\left(E\times B\right)$

So, the stress tensor will be,

${\mathit{T}}_{\mathbf{ij}}\mathbf{=}{\mathit{\epsilon }}_{\mathbf{0}}\mathbf{(}{\mathbf{E}}_{\mathbf{i}}{\mathbf{E}}_{\mathbf{j}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{\delta }}_{\mathbf{ij}}{\mathbf{E}}^{\mathbf{2}}\mathbf{)}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{\mu }}_{\mathbf{0}}}\mathbf{(}{\mathbf{B}}_{\mathbf{i}}{\mathbf{B}}_{\mathbf{j}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{\delta }}_{\mathbf{ij}}{\mathbf{B}}^{\mathbf{2}}\mathbf{)}$

Therefore, the energy stored in the field is $\mathit{u}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{(}{\mathbf{\epsilon }}_{\mathbf{0}}{\mathbf{E}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{\mu }}_{\mathbf{0}}}{\mathbf{B}}^{\mathbf{2}}\mathbf{)}$, the Poynting vector is $\mathit{S}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{\mu }}_{\mathbf{0}}}\mathbf{(}\mathbf{E}\mathbf{\times }\mathbf{B}\mathbf{)}$, the electromagnetic momentum density is $\mathit{g}\mathbf{=}{\mathit{\epsilon }}_{\mathbf{0}}\mathbf{(}\mathbf{E}\mathbf{\times }\mathbf{B}\mathbf{)}$, and the stress tensor is ${\mathit{T}}_{\mathbf{ij}}\mathbf{=}{\mathit{\epsilon }}_{\mathbf{0}}\mathbf{(}{\mathbf{E}}_{\mathbf{i}}{\mathbf{E}}_{\mathbf{j}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{\delta }}_{\mathbf{ij}}{\mathbf{E}}^{\mathbf{2}}\mathbf{)}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{\mu }}_{\mathbf{0}}}\mathbf{(}{\mathbf{B}}_{\mathbf{i}}{\mathbf{B}}_{\mathbf{j}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{\delta }}_{\mathbf{ij}}{\mathbf{B}}^{\mathbf{2}}\mathbf{)}$. Thus, the formulas for the momentum density and the stress are unchanged.