91Ó°ÊÓ

The expression for the scaler potential from equation 10.49,

$V(r,t)=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{qc}{\sqrt{{(c^{2}t−râ‹\dots v)}^2+(c^2−v^2)(r^2−c^2{t}^2)}}$

The expression for the vector potential from equation 10.50,

$A(r,t)=\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\frac{qcv}{\sqrt{{(c^{2}t−râ‹\dots v)}^2+(c^2−v^2)(r^2−c^2{t}^2)}}$

The Lorentz conditions,

$∇â‹\dots A=−{\mathrm{μ}}_0{\mathrm{Î\mu }}_0\frac{∂V}{∂t}$

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

$\mathit{V}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{Ï€}{\mathbf{Î\mu }}_{\mathbf{0}}}\frac{\mathbf{q}\mathbf{c}}{\mathbf{(}\mathbf{r}\mathbf{c}\mathbf{-}\mathbf{r}\mathbf{×}\mathbf{v}\mathbf{)}}$

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

$\mathit{A}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\frac{{\mathbf{μ}}_{\mathbf{0}}}{\mathbf{4}\mathbf{π}}\frac{\mathbf{q}\mathbf{c}\mathbf{v}}{\mathbf{(}\mathbf{r}\mathbf{c}\mathbf{-}\mathbf{r}\mathbf{×}\mathbf{v}\mathbf{)}}$

Calculate the expression of$∇â‹\dots A$ .

$\begin{array}{l}∇â‹\dots A=∇â‹\dots \left[\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\frac{qcv}{\sqrt{{(c^{2}t−râ‹\dots v)}^2+(c^2−v^2)(r^2−c^2{t}^2)}}\right]\\ ∇â‹\dots A=\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}qcvâ‹\dots ∇{[{(c^{2}t−râ‹\dots v)}^2+(c^2−v^2)(r^2−c^2{t}^2)]}^{−\frac{1}{2}}\\ ∇â‹\dots A=\frac{{\mathrm{μ}}_{0}qc}{4\mathrm{Ï€}}vâ‹\dots \left\{\begin{array}{l}−\frac{1}{2}{[{(c^{2}t−râ‹\dots v)}^2+(c^2−v^2)(r^2−c^2{t}^2)]}^{−\frac{3}{2}}\\ +∇[{(c^{2}t−râ‹\dots v)}^2+(c^2−v^2)(r^2−c^2{t}^2)]\end{array}\right\}\\ ∇â‹\dots A=−\frac{{\mathrm{μ}}_{0}qc}{8\mathrm{Ï€}}\left\{\begin{array}{l}{[{(c^{2}t−râ‹\dots v)}^2+(c^2−v^2)(r^2−c^2{t}^2)]}^{−\frac{3}{2}}\\ +vâ‹\dots [−2(c^{2}t−râ‹\dots v)∇(\widehat{r}â‹\dots \widehat{v})+(c^2−v^2)∇\left(r^2\right)]\end{array}\right\}\end{array}$

Solve further as,

$\begin{array}{l}∇â‹\dots A=−\frac{{\mathrm{μ}}_{0}qc}{8\mathrm{Ï€}}\left\{\begin{array}{l}{[{(c^{2}t−râ‹\dots v)}^2+(c^2−v^2)(r^2−c^2{t}^2)]}^{−\frac{3}{2}}\\ +vâ‹\dots [−2(c^{2}t−râ‹\dots v)v+(c^2−v^2)2r]\end{array}\right\}\\ ∇â‹\dots A=−\frac{{\mathrm{μ}}_{0}qc}{4\mathrm{Ï€}}\left\{\begin{array}{l}{[{(c^{2}t−râ‹\dots v)}^2+(c^2−v^2)(r^2−c^2{t}^2)]}^{−\frac{3}{2}}\\ +[(c^{2}t−râ‹\dots v)v^2+(c^2−v^2)râ‹\dots v]\end{array}\right\}\\ ∇â‹\dots A=−\frac{{\mathrm{μ}}_{0}qc}{4\mathrm{Ï€}}\left\{\begin{array}{l}{[{(c^{2}t−râ‹\dots v)}^2+(c^2−v^2)(r^2−c^2{t}^2)]}^{−\frac{3}{2}}\\ +[c^{2}t{v}^2−(râ‹\dots v)v^2−c^2(râ‹\dots v)+v^2(râ‹\dots v)]\end{array}\right\}\end{array}$

Solve further as,

$∇â‹\dots A=−\frac{{\mathrm{μ}}_{0}qc}{4\mathrm{Ï€}}\left\{\begin{array}{l}{[{(c^{2}t−râ‹\dots v)}^2+(c^2−v^2)(r^2−c^2{t}^2)]}^{−\frac{3}{2}}\\ +\left[c^2(t{v}^2−râ‹\dots v)\right]\end{array}\right\}$

$∇â‹\dots A=−\frac{{\mathrm{μ}}_{0}qc}{4\mathrm{Ï€}}\frac{v^{2}t−râ‹\dots v}{{[{(c^{2}t−râ‹\dots v)}^2+(c^2−v^2)(r^2−c^2{t}^2)]}^{\frac{3}{2}}}$ ……. (1)

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

$\begin{array}{l}\frac{∂V}{∂t}=\frac{∂}{∂t}\left[\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{qc}{\sqrt{{(c^{2}t−râ‹\dots v)}^2+(c^2−v^2)(r^2−c^2{t}^2)}}\right]\\ \frac{∂V}{∂t}=\frac{qc}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{∂}{∂t}\left[\frac{1}{\sqrt{{(c^{2}t−râ‹\dots v)}^2+(c^2−v^2)(r^2−c^2{t}^2)}}\right]\\ \frac{∂V}{∂t}=\frac{qc}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}(−\frac{1}{2}){[{(c^{2}t−râ‹\dots v)}^2+(c^2−v^2)(r^2−c^2{t}^2)]}^{−\frac{3}{2}}+\frac{∂}{∂t}[{(c^{2}t−râ‹\dots v)}^2+(c^2−v^2)(r^2−c^2{t}^2)]\\ \frac{∂V}{∂t}=\frac{qc}{8\mathrm{Ï€}{\mathrm{Î\mu }}_0}{[{(c^{2}t−râ‹\dots v)}^2+(c^2−v^2)(r^2−c^2{t}^2)]}^{−\frac{3}{2}}+[2(c^{2}t−râ‹\dots v)c^2+(c^2−v^2)(−2c^{2}t)]\end{array}$

Solve further as,

$\begin{array}{l}\frac{∂V}{∂t}=\frac{−q{c}^3}{8\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{(−râ‹\dots v+v^{2}t)2}{{[{(c^{2}t−râ‹\dots v)}^2+(c^2−v^2)(r^2−c^2{t}^2)]}^{\frac{3}{2}}}\\ \frac{∂V}{∂t}=\frac{−q{c}^3}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{(v^{2}t−râ‹\dots v)}{{[{(c^{2}t−râ‹\dots v)}^2+(c^2−v^2)(r^2−c^2{t}^2)]}^{\frac{3}{2}}}\end{array}$

Multiply the above equation with $−{\mathrm{μ}}_0{\mathrm{Î\mu }}_0$.

$−{\mathrm{μ}}_0{\mathrm{Î\mu }}_0\frac{∂V}{∂t}=(−{\mathrm{μ}}_0{\mathrm{Î\mu }}_0)\frac{−q{c}^3}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{(v^{2}t−râ‹\dots v)}{{[{(c^{2}t−râ‹\dots v)}^2+(c^2−v^2)(r^2−c^2{t}^2)]}^{\frac{3}{2}}}$

$−{\mathrm{μ}}_0{\mathrm{Î\mu }}_0\frac{∂V}{∂t}=\frac{{\mathrm{μ}}_{o}q{c}^3}{4\mathrm{Ï€}}\frac{(v^{2}t−râ‹\dots v)}{{[{(c^{2}t−râ‹\dots v)}^2+(c^2−v^2)(r^2−c^2{t}^2)]}^{\frac{3}{2}}}$ ……. (2)

From the equation (1) and (2).

$∇â‹\dots A=−{\mathrm{μ}}_0{\mathrm{Î\mu }}_0\frac{∂V}{∂t}$

Hence, the Lorentz gauge conditions satisfied.