91影视

The expression for the scaler potential from equation 10.49,

$V(r,t)=\frac{1}{4蟺{蔚}_0}\frac{qc}{\sqrt{{(c^{2}t鈭�r鈰�v)}^2+(c^2鈭�v^2)(r^2鈭�c^2{t}^2)}}$

The expression for the vector potential from equation 10.50,

$A(r,t)=\frac{{渭}_0}{4蟺}\frac{qcv}{\sqrt{{(c^{2}t鈭�r鈰�v)}^2+(c^2鈭�v^2)(r^2鈭�c^2{t}^2)}}$

The Lorentz conditions,

$鈭�鈰�A=鈭�{渭}_0{蔚}_0\frac{鈭�V}{鈭�t}$

The expression for the scaler potential for moving charge is given as follows.

$\mathit{V}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{蟺}{\mathbf{蔚}}_{\mathbf{0}}}\frac{\mathbf{q}\mathbf{c}}{\mathbf{(}\mathbf{r}\mathbf{c}\mathbf{-}\mathbf{r}\mathbf{脳}\mathbf{v}\mathbf{)}}$

The expression for the vector potential for moving charge is given as follows.

$\mathit{A}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\frac{{\mathbf{渭}}_{\mathbf{0}}}{\mathbf{4}\mathbf{蟺}}\frac{\mathbf{q}\mathbf{c}\mathbf{v}}{\mathbf{(}\mathbf{r}\mathbf{c}\mathbf{-}\mathbf{r}\mathbf{脳}\mathbf{v}\mathbf{)}}$

Calculate the expression of$鈭�鈰�A$ .

$\begin{array}{l}鈭�鈰�A=鈭�鈰�\left[\frac{{渭}_0}{4蟺}\frac{qcv}{\sqrt{{(c^{2}t鈭�r鈰�v)}^2+(c^2鈭�v^2)(r^2鈭�c^2{t}^2)}}\right]\\ 鈭�鈰�A=\frac{{渭}_0}{4蟺}qcv鈰�鈭�{[{(c^{2}t鈭�r鈰�v)}^2+(c^2鈭�v^2)(r^2鈭�c^2{t}^2)]}^{鈭�\frac{1}{2}}\\ 鈭�鈰�A=\frac{{渭}_{0}qc}{4蟺}v鈰�\left\{\begin{array}{l}鈭�\frac{1}{2}{[{(c^{2}t鈭�r鈰�v)}^2+(c^2鈭�v^2)(r^2鈭�c^2{t}^2)]}^{鈭�\frac{3}{2}}\\ +鈭�[{(c^{2}t鈭�r鈰�v)}^2+(c^2鈭�v^2)(r^2鈭�c^2{t}^2)]\end{array}\right\}\\ 鈭�鈰�A=鈭�\frac{{渭}_{0}qc}{8蟺}\left\{\begin{array}{l}{[{(c^{2}t鈭�r鈰�v)}^2+(c^2鈭�v^2)(r^2鈭�c^2{t}^2)]}^{鈭�\frac{3}{2}}\\ +v鈰�[鈭�2(c^{2}t鈭�r鈰�v)鈭�(\widehat{r}鈰�\widehat{v})+(c^2鈭�v^2)鈭�\left(r^2\right)]\end{array}\right\}\end{array}$

Solve further as,

$\begin{array}{l}鈭�鈰�A=鈭�\frac{{渭}_{0}qc}{8蟺}\left\{\begin{array}{l}{[{(c^{2}t鈭�r鈰�v)}^2+(c^2鈭�v^2)(r^2鈭�c^2{t}^2)]}^{鈭�\frac{3}{2}}\\ +v鈰�[鈭�2(c^{2}t鈭�r鈰�v)v+(c^2鈭�v^2)2r]\end{array}\right\}\\ 鈭�鈰�A=鈭�\frac{{渭}_{0}qc}{4蟺}\left\{\begin{array}{l}{[{(c^{2}t鈭�r鈰�v)}^2+(c^2鈭�v^2)(r^2鈭�c^2{t}^2)]}^{鈭�\frac{3}{2}}\\ +[(c^{2}t鈭�r鈰�v)v^2+(c^2鈭�v^2)r鈰�v]\end{array}\right\}\\ 鈭�鈰�A=鈭�\frac{{渭}_{0}qc}{4蟺}\left\{\begin{array}{l}{[{(c^{2}t鈭�r鈰�v)}^2+(c^2鈭�v^2)(r^2鈭�c^2{t}^2)]}^{鈭�\frac{3}{2}}\\ +[c^{2}t{v}^2鈭�(r鈰�v)v^2鈭�c^2(r鈰�v)+v^2(r鈰�v)]\end{array}\right\}\end{array}$

Solve further as,

$鈭�鈰�A=鈭�\frac{{渭}_{0}qc}{4蟺}\left\{\begin{array}{l}{[{(c^{2}t鈭�r鈰�v)}^2+(c^2鈭�v^2)(r^2鈭�c^2{t}^2)]}^{鈭�\frac{3}{2}}\\ +\left[c^2(t{v}^2鈭�r鈰�v)\right]\end{array}\right\}$

$鈭�鈰�A=鈭�\frac{{渭}_{0}qc}{4蟺}\frac{v^{2}t鈭�r鈰�v}{{[{(c^{2}t鈭�r鈰�v)}^2+(c^2鈭�v^2)(r^2鈭�c^2{t}^2)]}^{\frac{3}{2}}}$ 鈥︹��. (1)

Calculate the expression of $\frac{鈭�V}{鈭�t}$.

$\begin{array}{l}\frac{鈭�V}{鈭�t}=\frac{鈭�}{鈭�t}\left[\frac{1}{4蟺{蔚}_0}\frac{qc}{\sqrt{{(c^{2}t鈭�r鈰�v)}^2+(c^2鈭�v^2)(r^2鈭�c^2{t}^2)}}\right]\\ \frac{鈭�V}{鈭�t}=\frac{qc}{4蟺{蔚}_0}\frac{鈭�}{鈭�t}\left[\frac{1}{\sqrt{{(c^{2}t鈭�r鈰�v)}^2+(c^2鈭�v^2)(r^2鈭�c^2{t}^2)}}\right]\\ \frac{鈭�V}{鈭�t}=\frac{qc}{4蟺{蔚}_0}(鈭�\frac{1}{2}){[{(c^{2}t鈭�r鈰�v)}^2+(c^2鈭�v^2)(r^2鈭�c^2{t}^2)]}^{鈭�\frac{3}{2}}+\frac{鈭�}{鈭�t}[{(c^{2}t鈭�r鈰�v)}^2+(c^2鈭�v^2)(r^2鈭�c^2{t}^2)]\\ \frac{鈭�V}{鈭�t}=\frac{qc}{8蟺{蔚}_0}{[{(c^{2}t鈭�r鈰�v)}^2+(c^2鈭�v^2)(r^2鈭�c^2{t}^2)]}^{鈭�\frac{3}{2}}+[2(c^{2}t鈭�r鈰�v)c^2+(c^2鈭�v^2)(鈭�2c^{2}t)]\end{array}$

Solve further as,

$\begin{array}{l}\frac{鈭�V}{鈭�t}=\frac{鈭�q{c}^3}{8蟺{蔚}_0}\frac{(鈭�r鈰�v+v^{2}t)2}{{[{(c^{2}t鈭�r鈰�v)}^2+(c^2鈭�v^2)(r^2鈭�c^2{t}^2)]}^{\frac{3}{2}}}\\ \frac{鈭�V}{鈭�t}=\frac{鈭�q{c}^3}{4蟺{蔚}_0}\frac{(v^{2}t鈭�r鈰�v)}{{[{(c^{2}t鈭�r鈰�v)}^2+(c^2鈭�v^2)(r^2鈭�c^2{t}^2)]}^{\frac{3}{2}}}\end{array}$

Multiply the above equation with $鈭�{渭}_0{蔚}_0$.

$鈭�{渭}_0{蔚}_0\frac{鈭�V}{鈭�t}=(鈭�{渭}_0{蔚}_0)\frac{鈭�q{c}^3}{4蟺{蔚}_0}\frac{(v^{2}t鈭�r鈰�v)}{{[{(c^{2}t鈭�r鈰�v)}^2+(c^2鈭�v^2)(r^2鈭�c^2{t}^2)]}^{\frac{3}{2}}}$

$鈭�{渭}_0{蔚}_0\frac{鈭�V}{鈭�t}=\frac{{渭}_{o}q{c}^3}{4蟺}\frac{(v^{2}t鈭�r鈰�v)}{{[{(c^{2}t鈭�r鈰�v)}^2+(c^2鈭�v^2)(r^2鈭�c^2{t}^2)]}^{\frac{3}{2}}}$ 鈥︹��. (2)

From the equation (1) and (2).

$鈭�鈰�A=鈭�{渭}_0{蔚}_0\frac{鈭�V}{鈭�t}$

Hence, the Lorentz gauge conditions satisfied.