Question

Derive Eq. 10.70. First show that $\frac{\partial {\mathbf{t}}_r}{\partial \mathbf{t}}=\frac{\mathrm{rc}}{\mathbf{r}\cdot \mathbf{u}}$

Short answer

The value of partial time derivative of the vector potential is$\frac{\partial \overrightarrow{\mathrm{A}}}{\partial \mathrm{t}}=\frac{\mathrm{qc}}{4{\mathrm{\pi \epsilon }}_0}\left[(\overrightarrow{\mathrm{l}}\cdot \overrightarrow{\mathrm{u}})\left(\frac{1}{\mathrm{c}}\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{\upsilon }}\right)+\frac{\mathrm{l}}{\mathrm{c}}\overrightarrow{\mathrm{\upsilon }}({\mathrm{c}}^2-{\mathrm{\upsilon }}^2+\overrightarrow{\mathrm{r}}\cdot \overrightarrow{\mathrm{a}})\right]$

Step-by-step solution

Write the given data from the question.

Consider the limitations of this site $l$ will be using letter $l$ as the distance from the charge to the point of interest.

Determine the formula of partial time derivative of the vector potential.

Write the formula of partial time derivative of the vector potential.

$\frac{\partial \overrightarrow{\mathbf{A}}}{\partial \mathbf{t}}=\frac{{\mathrm{\mu }}_0\mathbf{qc}}{\mathbf{4}\mathrm{\pi }}\mathbf{[}\frac{}{\overrightarrow{\mathbf{\iota }}\mathbf{\cdot }\overrightarrow{\mathbf{u}}}\frac{\mathbf{\partial }\overrightarrow{\mathbf{\upsilon }}}{\mathbf{\partial }\mathbf{t}}\mathbf{-}\frac{\overrightarrow{\mathbf{\upsilon }}}{{\mathbf{(}\overrightarrow{\mathbf{\iota }}\mathbf{\cdot }\overrightarrow{\mathbf{u}}\mathbf{)}}^{\mathbf{2}}}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{t}}\mathbf{(}\overrightarrow{\mathbf{\iota }}\mathbf{\cdot }\overrightarrow{\mathbf{u}}\mathbf{)}\mathbf{]}$ …… (1)

Here, ${\mathrm{\mu }}_0$ is permeability, $\mathbf{q}$ is test charge,$\overrightarrow{\mathbf{\iota }}$ is retarded time and $\overrightarrow{\mathbf{\upsilon }}$ is retarded time evaluated.

(a) Determine the value of partial time derivative of the vector potential.

The Lienard-Wiechert vector potential of the charge in motion is:

$\begin{array}{c}\overrightarrow{\mathrm{A}}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{\mathrm{qc}\overrightarrow{\mathrm{\upsilon }}}{\mathrm{\iota c}-\overrightarrow{\mathrm{\iota }}\cdot \overrightarrow{\mathrm{\upsilon }}}\\ =\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{\mathrm{qc}\overrightarrow{\mathrm{\upsilon }}}{\overrightarrow{\mathrm{\iota }}\cdot \overrightarrow{\mathrm{u}}}\end{array}$

All quantities evaluated at retarded time.

Consider the case:

$\begin{array}{c}\mathrm{\iota }=\mathrm{c}(\mathrm{t}-{\mathrm{t}}_{\mathrm{r}})\\ {\mathrm{\iota }}^2={\mathrm{c}}^2{(\mathrm{\iota }-{\mathrm{\iota }}_{\mathrm{r}})}^2\end{array}$

Solve for the partial time derivative as:

$\begin{array}{c}2\overrightarrow{\mathrm{\iota }}\cdot \frac{\partial \overrightarrow{\mathrm{\iota }}}{\partial \mathrm{t}}=2{\mathrm{c}}^2(\mathrm{\iota }-{\mathrm{\iota }}_{\mathrm{r}})\left(1-\frac{\partial {\mathrm{\iota }}_{\mathrm{r}}}{\partial \mathrm{t}}\right)\\ \frac{\partial {\mathrm{\iota }}_{\mathrm{r}}}{\partial \mathrm{t}}=1-\frac{\overrightarrow{\mathrm{\iota }}}{\mathrm{c}}\cdot \frac{\partial \overrightarrow{\mathrm{\iota }}}{\partial \mathrm{t}}\\ \frac{\partial {\mathrm{\iota }}_{\mathrm{r}}}{\partial \mathrm{t}}=\frac{\mathrm{c\iota }}{\overrightarrow{\mathrm{\iota }}\cdot (\mathrm{c}\widehat{\mathrm{\iota }}-\overrightarrow{\mathrm{\upsilon }})}\\ \frac{\partial {\mathrm{\iota }}_{\mathrm{r}}}{\partial \mathrm{t}}=\frac{\mathrm{c\iota }}{\overrightarrow{\mathrm{\iota }}\cdot \overrightarrow{\mathrm{u}}}\end{array}$

Since, $\overrightarrow{\mathrm{\iota }}=\overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{\omega }}\left({\mathrm{t}}_{\mathrm{r}}\right)$solve as:

$\begin{array}{c}\frac{\partial \overrightarrow{\mathrm{\iota }}}{\partial \mathrm{t}}=-\frac{\partial \overrightarrow{\mathrm{\omega }}}{\partial {\mathrm{t}}_{\mathrm{r}}}\frac{\partial {\mathrm{t}}_{\mathrm{r}}}{\partial \mathrm{t}}=-\overrightarrow{\mathrm{\upsilon }}\left({\mathrm{t}}_{\mathrm{r}}\right)\frac{\partial {\mathrm{t}}_{\mathrm{r}}}{\partial \mathrm{t}}\\ \frac{\partial {\mathrm{t}}_{\mathrm{r}}}{\partial \mathrm{t}}=1+\frac{\widehat{\mathrm{\iota }}}{\mathrm{c}}\cdot \overrightarrow{\mathrm{\upsilon }}\left({\mathrm{t}}_{\mathrm{r}}\right)\frac{\partial {\mathrm{t}}_{\mathrm{r}}}{\partial \mathrm{t}}\end{array}$

With this we can get the partial time derivative of the vector potential:

Determine the $\frac{\partial \overrightarrow{\upsilon }}{\partial t}$.

$\begin{array}{c}\frac{\partial \overrightarrow{\mathrm{\upsilon }}}{\partial \mathrm{t}}=\frac{\partial \overrightarrow{\mathrm{\upsilon }}}{\partial {\mathrm{t}}_{\mathrm{r}}}\frac{\partial {\mathrm{t}}_{\mathrm{r}}}{\partial \mathrm{t}}\\ =\overrightarrow{\mathrm{a}}\frac{\mathrm{c\iota }}{\overrightarrow{\mathrm{\iota }}\cdot \overrightarrow{\mathrm{u}}}\end{array}$

Determine$\frac{\partial }{\partial \mathrm{t}}(\overrightarrow{\mathrm{\iota }}\cdot \overrightarrow{\mathrm{u}})\frac{\partial \overrightarrow{\mathrm{\upsilon }}}{\partial \mathrm{t}}$.

$\begin{array}{c}\frac{\partial }{\partial \mathrm{t}}(\overrightarrow{\mathrm{\iota }}\cdot \overrightarrow{\mathrm{u}})=\mathrm{c}\frac{\partial \mathrm{\iota }}{\partial \mathrm{t}}-\frac{\partial \overrightarrow{\mathrm{\iota }}}{\partial \mathrm{t}}\cdot \overrightarrow{\mathrm{\upsilon }}-\frac{\partial \overrightarrow{\mathrm{\upsilon }}}{\partial \mathrm{t}}\cdot \overrightarrow{\mathrm{\iota }}\\ =\mathrm{c}\frac{\partial \mathrm{\iota }}{\partial \mathrm{t}}+{\mathrm{\upsilon }}^2\frac{\mathrm{c\iota }}{\overrightarrow{\mathrm{\iota }}\cdot \overrightarrow{\mathrm{u}}}-(\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{\iota }})\frac{\mathrm{c\iota }}{\overrightarrow{\mathrm{\iota }}\cdot \overrightarrow{\mathrm{u}}}\end{array}$

Determine the$\frac{\partial \mathrm{\iota }}{\partial \mathrm{t}}$.

$\begin{array}{c}\frac{\partial \mathrm{\iota }}{\partial \mathrm{t}}=\frac{\partial }{\partial \mathrm{t}}\left(\mathrm{c}(\mathrm{t}-{\mathrm{t}}_{\mathrm{r}})\right)\\ =\mathrm{c}\frac{\overrightarrow{\mathrm{\iota }}\cdot \overrightarrow{\mathrm{u}}-\mathrm{\iota c}}{\overrightarrow{\mathrm{\iota }}\cdot \overrightarrow{\mathrm{u}}}\end{array}$

Solve further as:

$\begin{array}{c}\frac{\partial }{\partial \mathrm{t}}(\overrightarrow{\mathrm{\iota }}\cdot \overrightarrow{\mathrm{u}})={\mathrm{c}}^2\frac{\overrightarrow{\mathrm{\iota }}\cdot \overrightarrow{\mathrm{u}}-\mathrm{\iota c}}{\overrightarrow{\mathrm{\iota }}\cdot \overrightarrow{\mathrm{u}}}+{\mathrm{\upsilon }}^2\frac{\mathrm{c\iota }}{\overrightarrow{\mathrm{I}}\cdot \overrightarrow{\mathrm{u}}}-(\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{\iota }})\frac{\mathrm{c\iota }}{\overrightarrow{\mathrm{I}}\cdot \overrightarrow{\mathrm{u}}}\\ =-\frac{1}{\overrightarrow{\mathrm{\iota }}\cdot \overrightarrow{\mathrm{u}}}(\mathrm{c\iota }({\mathrm{c}}^2-{\mathrm{\upsilon }}^2+\overrightarrow{\mathrm{\iota }}\cdot \overrightarrow{\mathrm{a}})-{\mathrm{c}}^2(\overrightarrow{\mathrm{\iota }}\cdot \overrightarrow{\mathrm{u}}))\end{array}$

Hence:

Determine the partial time derivative of the vector potential.

Substitute$\frac{\partial \overrightarrow{\mathrm{\upsilon }}}{\partial \mathrm{t}}$, $\frac{\partial \overrightarrow{\mathrm{\upsilon }}}{\partial \mathrm{t}}$, $\frac{\partial \mathrm{\iota }}{\partial \mathrm{t}}$ into equation (1).

$\begin{array}{c}\frac{\partial \overrightarrow{\mathrm{A}}}{\partial \mathrm{t}}=\frac{{\mathrm{\mu }}_0\mathrm{qc}}{4\mathrm{\pi }}\frac{1}{{(\overrightarrow{\mathrm{\iota }}\cdot \overrightarrow{\mathrm{u}})}^3}[\mathrm{\iota c}\overrightarrow{\mathrm{a}}(\overrightarrow{\mathrm{\iota }}\cdot \overrightarrow{\mathrm{u}})+\overrightarrow{\mathrm{\upsilon }}(\mathrm{cl}({\mathrm{c}}^2-{\mathrm{\upsilon }}^2+\overrightarrow{\mathrm{\iota }}\cdot \overrightarrow{\mathrm{a}})+{\mathrm{c}}^2(\overrightarrow{\mathrm{\iota }}\cdot \overrightarrow{\mathrm{u}}))]\\ =\frac{\mathrm{qc}}{4{\mathrm{\pi \epsilon }}_0}\left[(\overrightarrow{\mathrm{\iota }}\cdot \overrightarrow{\mathrm{u}})\left(\frac{1}{\mathrm{c}}\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{\upsilon }}\right)+\frac{\mathrm{\iota }}{\mathrm{c}}\overrightarrow{\mathrm{\upsilon }}({\mathrm{c}}^2-{\mathrm{\upsilon }}^2+\overrightarrow{\mathrm{r}}\cdot \overrightarrow{\mathrm{a}})\right]\end{array}$

Therefore, the value of partial time derivative of the vector potential is $\frac{\partial \overrightarrow{\mathrm{A}}}{\partial \mathrm{t}}=\frac{\mathrm{qc}}{4{\mathrm{\pi \epsilon }}_0}\left[(\overrightarrow{\mathrm{l}}\cdot \overrightarrow{\mathrm{u}})\left(\left(\frac{1}{\mathrm{c}}\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{\upsilon }}\right)\right)+\frac{\mathrm{l}}{\mathrm{c}}\overrightarrow{\mathrm{\upsilon }}({\mathrm{c}}^2-{\mathrm{\upsilon }}^2+\overrightarrow{\mathrm{r}}\cdot \overrightarrow{\mathrm{a}})\right]$.