Question

Confirm that the retarded potentials satisfy the Lorenz gauge condition.

$\mathbf{\nabla }\mathbf{\cdot }\mathbf{\left(}\frac{J}{r}\mathbf{\right)}\mathbf{=}\frac{1}{r}\mathbf{(}\mathbf{\nabla }\mathbf{\cdot }\mathbf{J}\mathbf{)}\mathbf{+}\frac{1}{2}\mathbf{(}{\mathbf{\nabla }}^{\mathbf{\text{'}}}\mathbf{\cdot }\mathbf{J}\mathbf{)}\mathbf{-}{\mathbf{\nabla }}^{\mathbf{\text{'}}}\mathbf{\cdot }\mathbf{\left(}\frac{J}{r}\mathbf{\right)}$

Where $\mathbf{\nabla }$denotes derivatives with respect to, and${\mathbf{\nabla }}^{\mathbf{\text{'}}}$ denotes derivatives with respect to${\mathbf{r}}^{\mathbf{\text{'}}}$. Next, noting that $\mathbf{J}\mathbf{(}{\mathbf{r}}^{\mathbf{\text{'}}}\mathbf{,}\mathbf{t}\mathbf{-}\mathbf{r}\mathbf{/}\mathbf{c}\mathbf{)}$depends on ${\mathrm{r}}^{\text{'}}$both explicitly and through, whereas it depends on r only through, confirm that

$\mathbf{\nabla }\mathbf{\cdot }\mathbf{J}\mathbf{=}\mathbf{-}\frac{1}{c}\dot{\mathbf{J}}\mathbf{\cdot }\mathbf{(}\mathbf{\nabla }\mathbf{r}\mathbf{)}$, ${\mathbf{\nabla }}^{\mathbf{\text{'}}}\mathbf{\cdot }\mathbf{J}\mathbf{=}\mathbf{-}\dot{\mathbf{\rho }}\mathbf{-}\frac{1}{c}\dot{\mathbf{J}}\mathbf{\cdot }\mathbf{\left(}{\mathbf{\nabla }}^{\mathbf{\text{'}}}\mathbf{r}\mathbf{\right)}$

Use this to calculate the divergence of$\mathbf{A}$ (Eq. 10.26).]

Short answer

The value of divergence of $\text{A}$ can be expressed as $-{\mu }_0{\epsilon }_0\frac{\partial }{\partial t}V$.

Step-by-step solution

Write the given data from the question

Consider a $\nabla$ denotes derivatives with respect to , and ${\nabla }^{\text{'}}$ denotes derivatives with respect to $r^{\text{'}}$. Next, noting that $\text{J}(-\text{r}/\text{c})$depends on $r^{\text{'}}$both explicitly and through , whereas it depends on r only through$r$ .

Determine the formula of divergence of A

Write the formula of divergence of.

$\mathbf{\nabla }\mathbf{\cdot }\mathbf{A}\mathbf{=}\frac{{\mathbf{\mu }}_0}{\mathbf{4}\mathbf{\pi }}\mathbf{\int }\mathbf{\nabla }\mathbf{\cdot }\mathbf{\left(}\frac{J}{r}\mathbf{\right)}$…… (1)

Here, ${\mathit{\mu }}_0$is the permeability of free space $\mathbf{\nabla }$denotes derivatives,$\mathrm{J}$ is surface density and $\mathrm{r}$ is resistance.

Determine the divergence of A

We know that from the product rule.

$\nabla \cdot \left(\frac{\text{J}}{\text{R}}\right)=\frac{1}{\text{r}}(\nabla \cdot \text{J})+\text{J}\cdot (\nabla \frac{1}{\text{r}})$ …… (2)

Then

localid="1658814372004" ${\nabla }^{\text{'}}\cdot \left(\frac{\text{J}}{\text{R}}\right)=\frac{1}{\text{r}}({\nabla }^{\text{'}}\cdot \text{J})+\text{J}\cdot \left({\nabla }^{\text{'}}\frac{1}{\text{r}}\right)$ …… (3)

As it knows that

$\text{r}=\text{r}-$$r^{\text{'}}$

Then,

$\nabla \left(\frac{1}{\text{r}}\right)=-{\nabla }^{\text{'}}\left(\frac{1}{\text{r}}\right)$

Substitute $-{\nabla }^{\text{'}}\left(\frac{1}{\text{r}}\right)$for$\nabla \left(\frac{1}{\text{r}}\right)$ in the equation (2).

$\nabla \cdot \left(\frac{\text{J}}{\text{r}}\right)=\frac{1}{\text{r}}(\nabla \cdot \text{J})-\text{J}\cdot \left({\nabla }^{\text{'}}\frac{1}{\text{r}}\right)$

From the equation (2)

$\text{J}\cdot \left({\nabla }^{\text{'}}\frac{1}{\text{r}}\right)={\nabla }^{\text{'}}\cdot \left(\frac{\text{J}}{\text{r}}\right)-\frac{1}{\text{r}}({\nabla }^{\text{'}}\cdot \text{J})$

Then, equation (1) reduced as follows:

$\begin{array}{c}\nabla \cdot \left(\frac{\text{J}}{\text{r}}\right)=\frac{1}{\text{r}}(\nabla \cdot \text{J})-({\nabla }^{\text{'}}\cdot \left(\frac{\text{J}}{\text{r}}\right)-\frac{1}{\text{r}}({\nabla }^{\text{'}}\cdot \text{J}))\\ =\frac{1}{\text{r}}(\nabla \cdot \text{J})-{\nabla }^{\text{'}}\left(\frac{\text{J}}{\text{r}}\right)+\frac{1}{\text{r}}({\nabla }^{\text{'}}\cdot \text{J})\end{array}$

We know that

$\begin{array}{c}\nabla \cdot \text{J}=\frac{\partial J_x}{\partial x}+\frac{\partial J_y}{\partial y}+\frac{\partial J_z}{\partial z}\\ =\frac{\partial J_x}{\partial t_r}\frac{\partial t_r}{\partial x}+\frac{\partial J_y}{\partial t_r}\frac{\partial t_r}{\partial y}+\frac{\partial J_z}{\partial t_r}\frac{\partial t_r}{\partial z}\end{array}$ …… (4)

But it is known that

$\begin{array}{l}\frac{\partial t_r}{\partial x}=\frac{-1}{c}\frac{\partial r}{\partial x}\\ \frac{\partial t_r}{\partial y}=\frac{-1}{c}\frac{\partial r}{\partial y}\\ \frac{\partial t_r}{\partial z}=\frac{-1}{c}\frac{\partial r}{\partial z}\end{array}$

Substitute $\frac{-1}{c}\frac{\partial r}{\partial x}$for$\frac{\partial t_r}{\partial x}$, $\frac{-1}{c}\frac{\partial r}{\partial y}$for$\frac{\partial t_r}{\partial y}$, and$\frac{-1}{c}\frac{\partial r}{\partial z}$ for $\frac{\partial t_r}{\partial z}$ in the equation (4).

$\nabla \cdot J=-\frac{1}{c}[\frac{\partial J_x}{\partial t_r}\cdot \frac{\partial r}{\partial x}+\frac{\partial J_y}{\partial t_r}\cdot \frac{\partial r}{\partial y}+\frac{\partial J_z}{\partial t_r}\cdot \frac{\partial r}{\partial z}]$

We know that

$\nabla \cdot r=\frac{\delta r}{\delta x}+\frac{\delta r}{\delta y}+\frac{\delta r}{\delta z}$

Then

$\nabla \cdot \text{J}=-\frac{1}{c}[\left(\frac{\partial J}{\partial t_r}\right)\cdot \left({\nabla }^{\text{'}}r\right)]$

Similarly

${\nabla }^{\text{'}}\cdot \text{J}=-\frac{\partial \rho }{\partial t}-\frac{1}{c}\frac{\partial \text{J}}{\partial t_r}\cdot \left({\nabla }^{\text{'}}r\right)$

Now substitute $\frac{-1}{c}\frac{\partial \text{J}}{\partial t_r}\cdot (\nabla r)$ for$(\nabla \cdot \text{J})$ and $-\frac{\partial \rho }{\partial t}-\frac{1}{c}\frac{\partial \text{J}}{\partial t_r}\cdot \left({\nabla }^{\text{'}}r\right)$for$({\nabla }^{\text{'}}\cdot \text{J})$ into the above equation.

$\begin{array}{c}\nabla \cdot \left(\frac{\text{J}}{\text{r}}\right)=\frac{1}{\text{r}}(\nabla \cdot \text{J})-{\nabla }^{\text{'}}\left(\frac{\text{J}}{\text{r}}\right)+\frac{1}{\text{r}}({\nabla }^{\text{'}}\cdot \text{J})\\ \nabla \cdot \left(\frac{\text{J}}{\text{r}}\right)=\frac{1}{\text{r}}(\nabla \cdot \text{J})+\frac{1}{\text{r}}\left({\nabla }^{\text{'}}\text{J}\right)-{\nabla }^{\text{'}}\frac{\text{J}}{\text{r}}\\ =\frac{1}{\text{r}}[\frac{-1}{c}\frac{\partial \text{J}}{\partial t_r}\cdot (\nabla \text{r})]+\frac{1}{\text{r}}[-\frac{\partial \rho }{\partial t}-\frac{1}{c}\frac{\partial \text{J}}{\partial t_r}\cdot \left({\nabla }^{\text{'}}\text{r}\right)]-{\nabla }^{\text{'}}\left(\frac{J}{\text{r}}\right)\end{array}$

Here, $\nabla \text{r}=-{\nabla }^{\text{'}}\text{r}$, then

$\nabla \cdot \left(\frac{\text{J}}{\text{r}}\right)=\frac{1}{\text{r}}\frac{\partial \rho }{\partial t}-{\nabla }^{\text{'}}\cdot \left(\frac{\text{J}}{\text{r}}\right)$

We know that vector potential is calculated by using the formulae

$\text{A}=\frac{{\mu }_0}{4\pi }\int \frac{I\left(t_r\right)}{r}dI$

Here, $I$is the current through the loop, $dI$is the length of the elementary part and $k$is the surface current.

$\text{A}(r,t)=\frac{{\mu }_0}{4\pi }\int \text{J}\left(\frac{r,t}{r}\right)d\tau$

Then

$\nabla \cdot \text{A}=\frac{{\mu }_0}{4\pi }\int \nabla \cdot \left(\frac{\text{J}}{\text{r}}\right)d\tau$

Substitute$\frac{\text{1}}{\text{r}}\frac{\partial \rho }{\partial t}-{\nabla }^{\text{'}}\cdot \left(\frac{\text{J}}{\text{r}}\right)$ for $\nabla \cdot \left(\frac{\text{J}}{\text{r}}\right)$ into the above equation.

$\begin{array}{c}\nabla \cdot \text{A}=\frac{{\mu }_0}{4\pi }[\frac{-\partial }{\partial t}\int \frac{\rho }{\text{r}}d\tau -\int {\nabla }^{\text{'}}\cdot \left(\frac{J}{r}\right)d\tau ]\\ =-{\mu }_0{\epsilon }_0\frac{\partial }{\partial t}[\frac{1}{4\pi {\epsilon }_0}\int \frac{\rho }{\text{r}}d\tau ]-\frac{{\mu }_0}{4\pi }\int {\rm N}\frac{1}{\text{r}}\cdot da\end{array}$

The$\text{J}=0$ across the surface at infinity, as far as we are aware. The last term then becomes zero, and the $\nabla \cdot \text{A}$equation follows the following reduction:

$\nabla \cdot \text{A}=-{\mu }_0{\epsilon }_0\frac{\partial }{\partial t}[\frac{1}{4\pi {\epsilon }_0}\int \frac{\rho }{r}d\tau ]$

Here, $V=\frac{1}{4\pi {\epsilon }_0}\int \frac{\rho }{\text{r}}d\tau$

Then, substitute $\frac{1}{4\pi {\epsilon }_0}\int \frac{\rho }{\text{r}}d\tau$ for $V$into the above equation.

$\begin{array}{l}\nabla \cdot \text{A}=-{\mu }_0{\epsilon }_0\frac{\partial }{\partial t}[\frac{1}{4\pi {\epsilon }_0}\int \frac{\rho }{\text{r}}d\tau ]\\ \nabla \cdot \text{A}=-{\mu }_0{\epsilon }_0\frac{\partial }{\partial t}V\end{array}$

Therefore, the value of divergence of$\text{A}$ can be expressed as$-{\mu }_0{\epsilon }_0\frac{\partial }{\partial t}V$.