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

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

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.

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 $\frac{\partial \overrightarrow{\mathrm{A}}}{\partial \mathrm{t}}=\frac{\mathrm{qc}}{4{\mathrm{\pi \epsilon }}_0}\left[(\overrightarrow{\mathrm{l}}\cdot \overrightarrow{\mathrm{u}})\left(\left(\frac{1}{\mathrm{c}}\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{\upsilon }}\right)\right)+\frac{\mathrm{l}}{\mathrm{c}}\overrightarrow{\mathrm{\upsilon }}({\mathrm{c}}^2-{\mathrm{\upsilon }}^2+\overrightarrow{\mathrm{r}}\cdot \overrightarrow{\mathrm{a}})\right]$.