91Ó°ÊÓ

Consider the two scalar potentials and two vector potentials of electrical $\mathrm{E}$ and magnetic potentials$\mathrm{B}$ .

Write the formula of curl term of electric field.

$\stackrel{\mathbf{→}}{\mathbf{E}}\mathbf{=}\mathbf{−}\mathbf{∇}\mathbf{V}\mathbf{−}\frac{\mathbf{∂}\stackrel{\mathbf{→}}{\mathbf{A}}}{\mathbf{∂}\mathbf{t}}$ …… (1)

Here,$\mathbf{∇}$ is derivative, $\mathit{V}$is voltage, $\mathit{A}$ is magnetic charge.

Write the formula of magnetic field should get a gradient and a time derivative.

role="math" localid="1658840585642" $\stackrel{\mathbf{→}}{\mathbf{B}}\mathbf{=}\mathbf{∇}\mathbf{×}\mathbf{A}$ …… (2)

Here, $\mathbf{∇}$ is derivative and $\mathbf{A}$ is magnetic charge.

Write the formula of Lorenz gauge of magnetic field.

role="math" localid="1658840685309" $\mathbf{(}{\mathbf{∇}}^{\mathbf{2}}{\stackrel{\mathbf{→}}{\mathbf{A}}}_{\mathbf{m}}\mathbf{−}\frac{\mathbf{1}}{{\mathbf{c}}^{\mathbf{2}}}\frac{{\mathbf{∂}}^{\mathbf{2}}{\mathbf{A}}_{\mathbf{m}}}{\mathbf{∂}{\mathbf{t}}^{\mathbf{2}}}\mathbf{)}\mathbf{−}\mathbf{∇}\mathbf{(}\mathbf{∇}\mathbf{â‹\dots }{\stackrel{\mathbf{→}}{\mathbf{A}}}_{\mathbf{m}}\mathbf{+}\frac{\mathbf{∂}{\mathbf{V}}_{\mathbf{m}}}{\mathbf{∂}\mathbf{t}}\mathbf{)}\mathbf{=}\mathbf{−}{\mathbf{μ}}_0{\stackrel{\mathbf{→}}{\mathbf{J}}}_m$ …… (3)

Here, $\mathbf{∇}$ is derivative, ${\mathbf{A}}_{\mathbf{m}}$ is magnetic charge of magnetic field, ${\mathbf{μ}}_{\mathbf{0}}$is permeability and ${\stackrel{\mathbf{→}}{\mathbf{J}}}_m$ is magnetic field density.

Write the formula of Lorenz gauge of electrical field.

role="math" localid="1658840830227" $\mathbf{(}\mathbf{−}{\mathbf{∇}}^{\mathbf{2}}{\stackrel{\mathbf{→}}{\mathbf{A}}}_{\mathbf{e}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{c}}^{\mathbf{2}}}\frac{{\mathbf{∂}}^{\mathbf{2}}{\stackrel{\mathbf{→}}{\mathbf{A}}}_{\mathbf{e}}}{\mathbf{∂}{\mathbf{t}}^{\mathbf{2}}}\mathbf{)}\mathbf{+}\mathbf{∇}\mathbf{(}\mathbf{∇}\mathbf{â‹\dots }{\stackrel{\mathbf{→}}{\mathbf{A}}}_{\mathbf{c}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{c}}^{\mathbf{2}}}\frac{\mathbf{∂}{\mathbf{V}}_{\mathbf{e}}}{\mathbf{∂}\mathbf{t}}\mathbf{)}\mathbf{=}{\mathbf{μ}}_0{\stackrel{\mathbf{→}}{\mathbf{J}}}_e$ …… (4)

Here,$\mathbf{∇}$ is derivative, ${\mathbf{A}}_{\mathbf{e}}$ is magnetic charge of electric field,${\mathbf{μ}}_0$ is permeability and ${\stackrel{\mathbf{→}}{\mathbf{J}}}_e$ is electric field density.

Before we add the magnetic charge, the fields are expressed in terms of potentials.

We anticipate that the formula should be entirely parallel when the magnetic charge is added, thus the electric field should receive a curl term and the magnetic field should receive a gradient and time derivative term:

Determine the curl term of electric field.

Substitute ${\mathrm{V}}_{\mathrm{c}}$ for $V$, $\frac{∂{\stackrel{→}{A}}_e}{∂t}−∇×{\stackrel{→}{A}}_m$ for $\frac{∂\stackrel{→}{A}}{∂t}$ into equation (1).

$\stackrel{→}{E}=−∇V_e−\frac{∂{\stackrel{→}{A}}_e}{∂t}−∇×{\stackrel{→}{A}}_m$

Substitute $−∇V_m−\frac{∂{\stackrel{→}{A}}_m}{∂t}+∇$ for $∇$ and ${\stackrel{→}{A}}_e$for $\stackrel{→}{A}$ into equation (2)

$\stackrel{→}{B}=−∇V_m−\frac{∂{\stackrel{→}{A}}_m}{∂t}+∇×{\stackrel{→}{A}}_e$

To make the subsequent formulations more symmetric, the sign of the curl term in the electric field was selected. Since the partial time derivative on the right has units of $\text{EL}/\text{T}=E\mathrm{Ï\dots }$, while B has units of $\text{E}/\mathrm{Ï\dots }$, the right equation is not dimensionally accurate because, if the left expression is accurate, then $A_m$ has units of $EL$, $L$ is length. To remedy this, we multiply it by $1/c^2$ in order to make it dimensionally proper.

Therefore, the value of curl term of electric field is $\stackrel{→}{E}=−∇V_e−\frac{∂{\stackrel{→}{A}}_e}{∂t}−∇×{\stackrel{→}{A}}_m$ and magnetic field should get a gradient and a time derivative is $\stackrel{→}{B}=−∇V_m−\frac{∂{\stackrel{→}{A}}_m}{∂t}+∇×{\stackrel{→}{A}}_e$.

We now begin by substituting the potential formulation in equation 7.44. (Formula 7.44.iii) results in

Determine the Lorenz gauge of magnetic field.

Substitute $0$ for $∇â‹\dots {\stackrel{→}{A}}_m+\frac{∂V_m}{∂t}$into equation (3).

${∇}^2{\stackrel{→}{A}}_m=−{\mathrm{μ}}_0{\stackrel{→}{J}}_m$

Determine the Lorenz gauge of electrical field.

Substitute $0$for$∇â‹\dots {\stackrel{→}{A}}_c+\frac{1}{c^2}\frac{∂V_e}{∂t}$ into equation (4).

${∇}^2{\stackrel{→}{A}}_e=−{\mathrm{μ}}_0{\stackrel{→}{J}}_e$

Therefore, the Lorenz gauge of magnetic and electrical field is ${∇}^2{\stackrel{→}{A}}_m=−{\mathrm{μ}}_0{\stackrel{→}{J}}_m$ and ${∇}^2{\stackrel{→}{A}}_e=−{\mathrm{μ}}_0{\stackrel{→}{J}}_e$.

Determine the electrical scalar potentials is:

$V_c=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}{∫}_V\frac{{\mathrm{ÒÏ}}_e({\stackrel{→}{r}}^{\text{'}},t_r)}{|{\stackrel{→}{r}}^{\text{'}}−\stackrel{→}{r}|}d{V}^{\text{'}}$

Now, determine the electrical vector potentials is:

${\stackrel{→}{A}}_e=\frac{{\mathrm{μ}}_0}{4\mathrm{π}}{∫}_V\frac{{\stackrel{→}{J}}_e({\stackrel{→}{r}}^{\text{'}},t_r)}{|{\stackrel{→}{r}}^{\text{'}}−\stackrel{→}{r}|}d{V}^{\text{'}}$

Determine the magnetic scalar potential is:

$V_m=\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}{∫}_V\frac{{\mathrm{ÒÏ}}_m({\stackrel{→}{r}}^{\text{'}},{\mathrm{â„“}}_r)}{|{\stackrel{→}{r}}^{\text{'}}−\stackrel{→}{r}|}d{V}^{\text{'}}$

Let's investigate. Attacking this with a Laplacian in the static case results in (because${∇}^2(1/|{\stackrel{→}{r}}^{\text{'}}−\stackrel{→}{r}|)=−4\mathrm{Ï€}\mathrm{δ}({\stackrel{→}{r}}^{\text{'}}−\stackrel{→}{r})$) , ${∇}^2{V}_{}=−{\mathrm{μ}}_0{\mathrm{ÒÏ}}_m\left(r\right)$which is merely (7.44.ii), making this look line.

Now, determine the magnetic vector potential is:

${\stackrel{→}{A}}_m=\frac{{\mathrm{μ}}_0}{4\mathrm{π}}{∫}_V\frac{{\stackrel{→}{J}}_m(\stackrel{→}{r^{\text{'}}},t_r)}{|{\stackrel{→}{r}}^{\text{'}}−\stackrel{→}{r}|}d{V}^{\text{'}}$

This may be attacked with a Laplacian (in static case), and the result is${∇}^2{\stackrel{→}{A}}_m=−{\mathrm{μ}}_0{\stackrel{→}{J}}_m$ , which is ${∇}^2{\stackrel{→}{A}}_m=−{\mathrm{μ}}_0{\stackrel{→}{J}}_m$ in static case.