91Ó°ÊÓ

Consider the distance between the source and the point of interest will be measured using the letter $l$ due to the site's limitations $I$.

Write the formula of (Lorenz gauge) potentials of a time-dependent ideal electric dipole at the origin.

$\mathbf{V}\mathbf{=}\frac{1}{\mathbf{4}{\mathbf{Ï€Î\mu }}_0}{∫}_V\frac{\mathrm{ÒÏ}({\stackrel{→}{r}}^{\text{'}},t_r)}{\mathrm{ι}}{\mathbf{dV}}^{\mathbf{\text{'}}}$ …… (1)

Here, role="math" localid="1658842513366" $\mathit{ÒÏ}$ is charge density, ${\mathbf{Î\mu }}_{\mathbf{0}}$ is permittivity, role="math" localid="1658842442385" ${\stackrel{\mathbf{→}}{\mathbf{r}}}^{\mathbf{\text{'}}}$is resistance, role="math" localid="1658842619960" ${\mathit{t}}_{\mathbf{r}}$ is retarded time and role="math" localid="1658842641287" $\mathit{ι}$ is letter at the distance from the source to the point of interest.

Write the formula of vector potentials of a time-dependent ideal electric dipole at the origin.

$\stackrel{\mathbf{→}}{\mathbf{A}}\mathbf{=}\frac{{\mathbf{μ}}_0}{\mathbf{4}\mathbf{π}}{∫}_V\frac{\stackrel{→}{J}({\stackrel{→}{r}}^{\text{'}},t_r)}{\mathrm{ι}}{\mathbf{dV}}^{\mathbf{\text{'}}}$ …… (2)

Here, ${\mathbf{μ}}_{\mathbf{0}}$is permeability, ${\stackrel{\mathbf{→}}{\mathbf{r}}}^{\mathbf{\text{'}}}$is resistance, ${\mathit{t}}_{\mathbf{r}}$ is retarded time and $\mathit{ι}$ is letter at the distance from the source to the point of interest.

Write the formula of magnetic field of a time-dependent ideal electric dipole at the origin.

$\stackrel{\mathbf{→}}{\mathbf{B}}\mathbf{=}\mathbf{∇}\mathbf{×}\stackrel{\mathbf{→}}{\mathbf{A}}$ …… (3)

Here, $\mathbf{∇}$ is derivative, $\stackrel{}{\mathbf{A}}$ is vector potential.

Write the formula of electric field of a time-dependent ideal electric dipole at the origin.

$\stackrel{\mathbf{→}}{\mathbf{E}}\mathbf{=}\mathbf{−}\mathbf{∇}\mathbf{V}\mathbf{−}\frac{\mathbf{∂}\stackrel{\mathbf{→}}{\mathbf{A}}}{\mathbf{∂}\mathbf{t}}$ …… (4)

Here, $\mathbf{∇}$ is derivative, $\mathbf{V}$ is velocity and $\mathit{A}$ is vector potential.

Determine the (Lorenz gauge) potentials of a time-dependent ideal electric dipole$\text{p}\left(\text{t}\right)$at the origin.

Substitute $\frac{1}{r}\left(1+\frac{\stackrel{→}{r}â‹\dots r^{\text{'}}}{r^2}\right)$for $\mathrm{ι}$and $t_0+\frac{\stackrel{→}{r}â‹\dots {\stackrel{→}{r}}^{\text{'}}}{r^2}$ for $t_r$ into equation (1).

Where, $t_0=t−r/c$ with this we can expand the potentials:

$\begin{array}{l}V=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}{∫}_V\frac{\mathrm{ÒÏ}({\stackrel{→}{r}}^{\text{'}},t_r)}{\mathrm{ι}}d{V}^{\text{'}}\\ ≈\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}{∫}_V(\mathrm{ÒÏ}({\stackrel{→}{r}}^{\text{'}},t_0)+\frac{\stackrel{→}{r}â‹\dots {\stackrel{→}{r}}^{\text{'}}}{rc}\stackrel{Ë™}{\mathrm{ÒÏ}}({\stackrel{→}{r}}^{\text{'}},t_0))(1+\frac{\stackrel{→}{r}â‹\dots {\stackrel{→}{r}}^{\text{'}}}{rc})\frac{1}{r}d{V}^{\text{'}}\\ ≈\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}{∫}_V(\mathrm{ÒÏ}({\stackrel{→}{r}}^{\text{'}},t_0)+\frac{\stackrel{→}{r}}{r^2}â‹\dots \left(\mathrm{ÒÏ}({\stackrel{→}{r}}^{\text{'}},t_0){\stackrel{→}{r}}^{\text{'}}\right)+\frac{\stackrel{→}{r}}{rc}â‹\dots \left(\stackrel{Ë™}{\mathrm{ÒÏ}}({\stackrel{→}{r}}^{\text{'}},t_0){\stackrel{→}{r}}^{\text{'}}\right))d{V}^{\text{'}}\end{array}$

Using the definition of a dipole moment, ignoring the fact that the first term is the total charge (zero), and leaving out the $o\left({r^{\text{'}}}^2\right)$terms:

Therefore, the value of (Lorenz gauge) potentials of a time-dependent ideal electric dipole at the origin $p\left(t\right)$is $V=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{\widehat{r}}{r^2}â‹\dots (\stackrel{→}{p}\left(t_0\right)+(r/c)\stackrel{Ë™}{\stackrel{→}{p}}\left(t_0\right))$.

Determine the vector potentials of a time-dependent ideal electric dipole $p\left(t\right)$at the origin.

Substitute $\frac{1}{r}\left(1+\frac{\stackrel{→}{r}â‹\dots r^{\text{'}}}{r^2}\right)$ for $\mathrm{ι}$and$t_0+\frac{\stackrel{→}{r}â‹\dots {\stackrel{→}{r}}^{\text{'}}}{r^2}$ for $t_r$ into equation (1).

role="math" localid="1658843272323" $\stackrel{→}{A}=\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}{∫}_V(\stackrel{→}{J}({\stackrel{→}{r}}^{\text{'}},t_0)+\frac{\stackrel{→}{r}â‹\dots {\stackrel{→}{r}}^{\text{'}}}{rc}\stackrel{→}{J}({\stackrel{→}{r}}^{\text{'}},t_0))\left(1+\frac{\stackrel{→}{r}â‹\dots {\stackrel{→}{r}}^{\text{'}}}{rc}\right)\frac{1}{r}d{V}^{\text{'}}$

However, according to the (eq. 5.31) we have:

${∫}_V\stackrel{→}{J}d{V}^{\text{'}}=\stackrel{˙}{\stackrel{→}{p}}$

And

$\begin{array}{c}\stackrel{→}{\mathrm{J}}\left({\stackrel{→}{\mathrm{r}}}^{\text{'}}\right)=\mathrm{ÒÏ}{\stackrel{→}{\mathrm{Ï\dots }}}^{\text{'}}\\ =\mathrm{ÒÏ}\stackrel{Ë™}{\stackrel{→}{\mathrm{r}}}\end{array}$

Therefore, all terms other than the first will have an $o\left({r^{\text{'}}}^2\right)$ or higher and may be eliminated. Due to this, just the initial phrase is preserved.

$\begin{array}{c}A≈\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}r}{∫}_V\stackrel{→}{J}({\stackrel{→}{r}}^{\text{'}},t_0)d{V}^{\text{'}}\\ =\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\frac{\stackrel{Ë™}{\stackrel{→}{\mathrm{ÒÏ}}}\left(t_0\right)}{r}\end{array}$

Therefore, the value of vector potentials of a time-dependent ideal electric dipole $\text{p}\left(\text{t}\right)$at the origin is $\stackrel{→}{A}≈\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\frac{\stackrel{˙}{\stackrel{→}{p}}\left(t_0\right)}{r}$.

Determine the magnetic field of a time-dependent ideal electric dipole $\text{p}\left(\text{t}\right)$ at the origin.

Substitute$\frac{{\mathrm{μ}}_0}{4\mathrm{π}}∇$ for $∇$ and $\frac{\stackrel{˙}{\stackrel{→}{p}}\left(t_0\right)}{r}$ for $\stackrel{→}{A}$into equation (3).

$\begin{array}{c}\stackrel{→}{\mathrm{B}}=\frac{{\mathrm{μ}}_0}{4\mathrm{π}}∇×\frac{\stackrel{˙}{\stackrel{→}{\mathrm{p}}}\left({\mathrm{t}}_0\right)}{\mathrm{r}}\\ =\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\left(\frac{1}{\mathrm{r}}∇×\stackrel{˙}{\stackrel{→}{\mathrm{p}}}\left({\mathrm{t}}_0\right)−\stackrel{˙}{\stackrel{→}{\mathrm{p}}}\left({\mathrm{t}}_0\right)×∇\frac{1}{\mathrm{r}}\right)\\ =\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\left(\frac{1}{\mathrm{r}}∇{\mathrm{t}}_0×\stackrel{¨}{\stackrel{→}{\mathrm{p}}}\left({\mathrm{t}}_0\right)+\frac{1}{{\mathrm{r}}^2}\stackrel{˙}{\stackrel{→}{\mathrm{p}}}\left({\mathrm{t}}_0\right)×\widehat{\mathrm{r}}\right)\\ =−\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\left(\frac{\widehat{\mathrm{r}}}{\mathrm{rc}}×\stackrel{¨}{\stackrel{→}{\mathrm{p}}}\left({\mathrm{t}}_0\right)+\frac{1}{{\mathrm{r}}^2}\stackrel{˙}{\stackrel{→}{\mathrm{p}}}\left({\mathrm{t}}_0\right)×\widehat{\mathrm{r}}\right)\end{array}$

Solve further as

$\stackrel{→}{B}=−\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\frac{1}{r^2}(\widehat{r}×\stackrel{˙}{\stackrel{→}{p}}\left(t_0\right)+(r/c)\widehat{r}×\stackrel{¨}{\stackrel{→}{p}}\left(t_0\right))$

Therefore, the value of magnetic field of a time-dependent ideal electric dipole $\text{p}\left(\text{t}\right)$ at the origin is $\stackrel{→}{B}=−\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\frac{1}{r^2}(\widehat{r}×\stackrel{˙}{\stackrel{→}{p}}\left(t_0\right)+(r/c)\widehat{r}×\stackrel{¨}{\stackrel{→}{p}}\left(t_0\right))$.

Determine the electric field of a time-dependent ideal electric dipole $p\left(t\right)$at the origin.

Substitute $\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\frac{\stackrel{¨}{\stackrel{→}{p}}\left(t_0\right)}{r}$ for $\frac{∂\stackrel{→}{A}}{∂t}$ and $\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}∇\left(\frac{\widehat{r}â‹\dots \stackrel{→}{p}\left(t_0\right)}{r^2}+\frac{\widehat{r}â‹\dots \stackrel{Ë™}{\stackrel{→}{p}}\left(t_0\right)}{r}\right)$for $V$ into equation (4).

$\stackrel{→}{E}=−\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}∇\left(\frac{\widehat{r}â‹\dots \stackrel{→}{p}\left(t_0\right)}{r^2}+\frac{\widehat{r}â‹\dots \stackrel{Ë™}{\stackrel{→}{p}}\left(t_0\right)}{r}\right)−\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\frac{\stackrel{¨}{\stackrel{→}{p}}\left(t_0\right)}{r}$

We may broaden the gradients of the scalar products using identity (ii) on page 21:

$\begin{array}{l}∇\frac{\widehat{r}â‹\dots \stackrel{→}{p}\left(t_0\right)}{r^2}=\frac{\widehat{r}}{r^2}×(∇×\stackrel{→}{p}\left(t_0\right))+\left(\frac{\widehat{r}}{r^2}â‹\dots ∇\right)\stackrel{→}{p}\left(t_0\right)+(\stackrel{→}{p}\left(t_0\right)â‹\dots ∇)\frac{\widehat{r}}{r^2}\\ ∇\frac{\widehat{r}â‹\dots \stackrel{→}{p}\left(t_0\right)}{r^2}=\frac{\widehat{r}}{rc}×(∇×\stackrel{→}{p}\left(t_0\right))+\left(\frac{\widehat{r}}{r^2}â‹\dots ∇\right)\stackrel{→}{p}\left(t_0\right)+(\stackrel{→}{p}\left(t_0\right)â‹\dots ∇)\frac{\widehat{r}}{rc}\end{array}$

Where we used the fact that $∇×(\widehat{r}/r^2)=0$. The terms are:

$\begin{array}{c}∇×\stackrel{→}{\mathrm{p}}\left({\mathrm{t}}_0\right)=−\frac{\widehat{\mathrm{r}}}{\mathrm{c}}×\stackrel{Ë™}{\stackrel{→}{\mathrm{p}}}\left({\mathrm{t}}_0\right)\\ ∇×\stackrel{Ë™}{\stackrel{→}{\mathrm{p}}}\left({\mathrm{t}}_0\right)=−\frac{\widehat{\mathrm{r}}}{\mathrm{c}}×\stackrel{¨}{\stackrel{→}{\mathrm{p}}}\left({\mathrm{t}}_0\right)\\ \left(\frac{\widehat{\mathrm{r}}}{{\mathrm{r}}^2}â‹\dots ∇\right)\stackrel{→}{\mathrm{p}}\left({\mathrm{t}}_0\right)=\left(\frac{\mathrm{x}}{{\mathrm{r}}^3}\frac{∂}{∂\mathrm{x}}+\frac{\mathrm{y}}{{\mathrm{r}}^3}\frac{∂}{∂\mathrm{y}}+\frac{\mathrm{z}}{{\mathrm{r}}^3}\right)\stackrel{→}{\mathrm{p}}\left({\mathrm{t}}_0\right)\\ =\frac{1}{{\mathrm{r}}^3}\left(\mathrm{x}\stackrel{Ë™}{\stackrel{→}{\mathrm{p}}}\left({\mathrm{t}}_0\right)\frac{∂{\mathrm{t}}_0}{∂\mathrm{x}}+\mathrm{y}\stackrel{Ë™}{\stackrel{→}{\mathrm{p}}}\left({\mathrm{t}}_0\right)\frac{∂{\mathrm{t}}_0}{∂\mathrm{y}}+\mathrm{z}\stackrel{Ë™}{\stackrel{→}{\mathrm{p}}}\left({\mathrm{t}}_0\right)\frac{∂{\mathrm{t}}_0}{∂\mathrm{z}}\right)\end{array}$

Solve further as

$\begin{array}{c}\left(\frac{\widehat{r}}{r^2}â‹\dots ∇\right)\stackrel{→}{p}\left(t_0\right)=\frac{\stackrel{Ë™}{\stackrel{→}{p}}\left(t_0\right)}{r^3}\stackrel{→}{r}â‹\dots ∇t_0\\ =−\frac{\stackrel{Ë™}{\stackrel{→}{p}}\left(t_0\right)}{r^{2}c}\end{array}$

Similarly:

$\left(\frac{\widehat{r}}{rc}â‹\dots ∇\right)\stackrel{Ë™}{\stackrel{→}{p}}\left(t_0\right)=−\frac{\stackrel{¨}{\stackrel{→}{p}}\left(t_0\right)}{c^{2}r}$

The final two terms are:

$\begin{array}{c}(\stackrel{→}{\mathrm{p}}\left({\mathrm{t}}_0\right)â‹\dots ∇)\frac{\widehat{\mathrm{r}}}{{\mathrm{r}}^2}=\left({\mathrm{p}}_{\mathrm{x}}\frac{∂}{∂\mathrm{x}}+{\mathrm{p}}_{\mathrm{y}}\frac{∂}{∂\mathrm{y}}+{\mathrm{p}}_{\mathrm{z}}\frac{∂}{∂\mathrm{z}}\right)\frac{\mathrm{x}\widehat{\mathrm{x}}+\mathrm{y}\widehat{\mathrm{y}}+\mathrm{z}\widehat{\mathrm{z}}}{{\mathrm{r}}^3}\\ ={\mathrm{p}}_{\mathrm{x}}\frac{{\mathrm{r}}^3\mathrm{x}−3{\mathrm{r}}^2−\mathrm{r}}{{\mathrm{r}}^6}+...\\ ={\mathrm{p}}_{\mathrm{x}}\frac{\mathrm{r}\widehat{\mathrm{x}}−3\mathrm{x}\widehat{\mathrm{r}}}{{\mathrm{r}}^4}+...\end{array}$

Summing the terms:

$\frac{1}{r^4}|p_x(r\widehat{x}−3x\widehat{r})+p_y(r\widehat{y}−3y\widehat{r})+p_z(r\widehat{z}−3z\widehat{r})|=\frac{1}{r^3}\left|\stackrel{→}{p}(t_0−3(\stackrel{→}{p}\left(t_0\right)â‹\dots \widehat{r})\widehat{r})\right|$

Similarly:

$\begin{array}{c}(\stackrel{Ë™}{\stackrel{→}{\mathrm{p}}}\left({\mathrm{t}}_0\right)â‹\dots ∇)\frac{\widehat{\mathrm{r}}}{\mathrm{rc}}=\left({\mathrm{p}}_{\mathrm{x}}\frac{∂}{∂\mathrm{x}}+{\mathrm{p}}_{\mathrm{y}}\frac{∂}{∂\mathrm{y}}+{\mathrm{p}}_{\mathrm{z}}\frac{∂}{∂\mathrm{z}}\right)\frac{\mathrm{x}\widehat{\mathrm{x}}+\mathrm{y}\widehat{\mathrm{y}}+\mathrm{z}\widehat{\mathrm{z}}}{{\mathrm{r}}^2\mathrm{c}}\\ ={\stackrel{Ë™}{\mathrm{p}}}_{\mathrm{x}}\frac{1}{\mathrm{c}}\frac{\widehat{\mathrm{x}}{\mathrm{r}}^2−2\mathrm{r}\frac{\mathrm{x}}{\mathrm{r}}\stackrel{→}{\mathrm{r}}}{{\mathrm{r}}^4}+...\\ ={\stackrel{Ë™}{\mathrm{p}}}_{\mathrm{x}}\frac{1}{\mathrm{c}}\frac{\mathrm{r}\widehat{\mathrm{x}}−2\mathrm{x}\widehat{\mathrm{r}}}{{\mathrm{r}}^3}+...\end{array}$

Summing the terms:

$\frac{1}{r^{3}c}[{\stackrel{Ë™}{p}}_x(\widehat{x}r−2x\widehat{r})+{\stackrel{Ë™}{p}}_y(\widehat{y}r−2y\widehat{r})+{\stackrel{Ë™}{p}}_z\left(z\widehat{r}\right)]=\frac{1}{r^{2}c}[\stackrel{Ë™}{\stackrel{→}{p}}\left(t_0\right)−2(\stackrel{Ë™}{\stackrel{→}{p}}\left(t_0\right)â‹\dots \widehat{r})\widehat{r}]$

Now, using the rule of BAC-CAB and (1) and (2) we have:

$\begin{array}{l}−\frac{\widehat{r}}{r^2}×\left(\frac{\widehat{r}}{c}×\stackrel{Ë™}{\stackrel{→}{p}}\left(t_0\right)\right)=−\frac{\widehat{r}}{c}\frac{\widehat{r}â‹\dots \stackrel{Ë™}{\stackrel{→}{p}}\left(t_0\right)}{r^2}+\stackrel{Ë™}{\stackrel{→}{p}}\left(t_0\right)\frac{1}{r^{2}c}\\ −\frac{\widehat{r}}{rc}×\left(\frac{\widehat{r}}{c}×\stackrel{Ë™}{\stackrel{→}{p}}\left(t_0\right)\right)=−\frac{\widehat{r}}{c}\frac{\widehat{r}â‹\dots \stackrel{Ë™}{\stackrel{→}{p}}\left(t_0\right)}{r^2}+\stackrel{Ë™}{\stackrel{→}{p}}\left(t_0\right)\frac{1}{r{c}^2}\end{array}$

Using this with (3), (4), (5) and (6) the gradient of potential is:

$∇V=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\left\{−\frac{\widehat{r}(\widehat{r}â‹\dots \stackrel{¨}{\stackrel{→}{p}}\left(t_0\right))}{c^{2}r}+\frac{1}{r^3}\left[(\stackrel{→}{p}+\frac{r}{c}\stackrel{Ë™}{\stackrel{→}{p}}\left(t_0\right))−3\widehat{r}(\widehat{r}â‹\dots (\stackrel{→}{p}−\frac{r}{c}\stackrel{Ë™}{\stackrel{→}{p}}\left(t_0\right)))\right]\right\}$

The electric field is finally, when the vector potential term is included and$c^2=1/{\mathrm{Î\mu }}_0{\mathrm{μ}}_0$ :

role="math" $\stackrel{→}{E}=−\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\left\{\frac{\stackrel{→}{p}\left(t_0\right)−\widehat{r}(\widehat{r}â‹\dots \stackrel{¨}{\stackrel{→}{p}}\left(t_0\right))}{r}+\frac{c^2}{r^3}\left[(\stackrel{→}{p}+\frac{r}{c}\stackrel{Ë™}{\stackrel{→}{p}}\left(t_0\right))−3\widehat{r}(\stackrel{→}{r}â‹\dots (\stackrel{→}{p}−\frac{r}{c}\stackrel{Ë™}{\stackrel{→}{p}}\left(t_0\right)))\right]\right\}$

Therefore, the value of electric field of a time-dependent ideal electric dipole$p\left(t\right)$ at the origin is $\stackrel{→}{E}=−\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\left\{\frac{\stackrel{→}{p}\left(t_0\right)−\widehat{r}(\widehat{r}â‹\dots \stackrel{¨}{\stackrel{→}{p}}\left(t_0\right))}{r}+\frac{c^2}{r^3}\left[\left(\stackrel{→}{p}+\frac{r}{c}\stackrel{Ë™}{\stackrel{→}{p}}\left(t_0\right)\right)−3\widehat{r}\left(\stackrel{→}{r}â‹\dots \left(\stackrel{→}{p}−\frac{r}{c}\stackrel{Ë™}{\stackrel{→}{p}}\left(t_0\right)\right)\right)\right]\right\}$