91Ó°ÊÓ

Write the expression for the scalar potential.

$\mathbf{V}\mathbf{=}\mathbf{γ}\mathbf{(}\stackrel{\mathbf{¯}}{\mathbf{V}}\mathbf{+}\mathbf{v}{\stackrel{\mathbf{¯}}{\mathbf{A}}}_{\mathbf{x}}\mathbf{)}$ …… (1)

Here, $\mathrm{γ}$ is the Lorentz contraction, v is the velocity, and $\stackrel{\mathbf{¯}}{\mathbf{A}}$is the vector potential in $\stackrel{¯}{S}$frame.

Write the expression for the vector potential in$\stackrel{¯}{S}$ frame.

$\stackrel{\mathbf{¯}}{\mathbf{A}}=\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\frac{\stackrel{\mathbf{¯}}{\mathbf{m}}×\stackrel{\mathbf{¯}}{\widehat{\mathbf{r}}}}{{\stackrel{¯}{r}}^2}$ …..(2)

(a)

In$\stackrel{¯}{S}$frame, the electric potential$\left(\stackrel{¯}{V}\right)$is zero.

Substitute $\stackrel{¯}{V}=0$and ${\stackrel{\mathbf{¯}}{\mathbf{A}}}_x=\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\frac{{(\stackrel{\mathbf{¯}}{\mathbf{m}}×\stackrel{\mathbf{¯}}{\widehat{\mathbf{r}}})}_x}{{\stackrel{¯}{r}}^2}$in equation (1).

$\begin{array}{c}V=\mathrm{γ}\left(0+v\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\frac{{(\stackrel{\mathbf{¯}}{\mathbf{m}}×\stackrel{\mathbf{¯}}{\widehat{\mathbf{r}}})}_x}{{\stackrel{¯}{r}}^2}\right)\\ V=v\mathrm{γ}\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\frac{{(\stackrel{\mathbf{¯}}{\mathbf{m}}×\stackrel{\mathbf{¯}}{\widehat{\mathbf{r}}})}_x}{{\stackrel{¯}{r}}^2}\end{array}$ …(3)

Here,

$\begin{array}{c}{(\stackrel{\mathbf{¯}}{\mathbf{m}}×\stackrel{\mathbf{¯}}{\widehat{\mathbf{r}}})}_x=m_y\stackrel{¯}{z}=m_z\stackrel{¯}{y}\\ {(\stackrel{\mathbf{¯}}{\mathbf{m}}×\stackrel{\mathbf{¯}}{\widehat{\mathbf{r}}})}_x=m_{y}z−m_{z}y\end{array}$

Substitute ${(\stackrel{\mathbf{¯}}{\mathbf{m}}×\stackrel{\mathbf{¯}}{\widehat{\mathbf{r}}})}_x=m_{y}z−m_{z}y$in equation (3).

$V=v\mathrm{γ}\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\left(\frac{m_{y}z−m_{z}y}{{\stackrel{¯}{r}}^2}\right)$ ….. (4)

It is known that:

$\begin{array}{c}\stackrel{¯}{x}=\mathrm{γ}(x−vt)\\ \stackrel{¯}{x}=\mathrm{γ}{R}_x\end{array}$

The value of $\stackrel{¯}{y}$and $\stackrel{¯}{z}$will be,

$\begin{array}{c}\stackrel{¯}{y}=y\\ \stackrel{¯}{y}=R_y\\ \stackrel{¯}{z}=z\\ \stackrel{¯}{z}=R_z\end{array}$

From equation (4), calculate the value of ${\stackrel{¯}{r}}^2$.

$\begin{array}{c}{\stackrel{¯}{r}}^2={\mathrm{γ}}^2{R}_x^2+R_y^2+R_z^2\\ {\stackrel{¯}{r}}^2={\mathrm{γ}}^2(R_x^2+R_y^2+R_z^2)+(1−{\mathrm{γ}}^2)(R_y^2+R_z^2)\\ {\stackrel{¯}{r}}^2={\mathrm{γ}}^2\left(R^2−\frac{v^2}{c^2}{R}^2{\sin }^2\mathrm{θ}\right)\end{array}$

Substitute${\stackrel{¯}{r}}^2={\mathrm{γ}}^2\left(R^2−\frac{v^2}{c^2}{R}^2{\sin }^2\mathrm{θ}\right)$in equation (4).

$\begin{array}{c}V=v\mathrm{γ}\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\left(\frac{m_{y}z−m_{z}y}{r{\mathrm{γ}}^2\left(R^2−\frac{v^2}{c^2}{R}^2{\sin }^2\mathrm{θ}\right)}\right)\\ V=\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\frac{v\mathrm{γ}(m_{y}z−m_{z}y)}{{\mathrm{γ}}^3{R}^3{\left(1−\frac{v^2}{c^2}{\sin }^2\mathrm{θ}\right)}^{\frac{3}{2}}}\\ V=\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\frac{vâ‹\dots (\mathbf{m}×\mathbf{R})\left(1−\frac{v^2}{c^2}\right)}{R^3{\left(1−\frac{v^2}{c^2}{\sin }^2\mathrm{θ}\right)}^{\frac{3}{2}}}\end{array}$

Here,${\mathrm{μ}}_0=\frac{1}{{\mathrm{Î\mu }}_0{c}^2}$ and$\mathbf{v}â‹\dots (\mathbf{m}×\mathbf{R})=\mathbf{R}â‹\dots (\mathbf{v}×\mathbf{m})$.

Hence, the above equation becomes,

$V=\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\frac{\widehat{\mathbf{R}}â‹\dots (\mathbf{v}×\mathbf{m})\left(1−\frac{v^2}{c^2}\right)}{c^2{R}^3{\left(1−\frac{v^2}{c^2}{\sin }^2\mathrm{θ}\right)}^{\frac{3}{2}}}$ ……. (5)

Therefore, the scalar potential is $V=\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\frac{\widehat{\mathbf{R}}â‹\dots (\mathbf{v}×\mathbf{m})\left(1−\frac{v^2}{c^2}\right)}{c^2{R}^3{\left(1−\frac{v^2}{c^2}{\sin }^2\mathrm{θ}\right)}^{\frac{3}{2}}}$.

(b)

In the non-relativistic limit, write the equation for the scalar potential.

$\mathbf{v}=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{\widehat{\mathbf{R}}â‹\dots \mathbf{P}}{R^3}$ …… (6)

Compare the equations (5) and (6).

$\begin{array}{c}\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\frac{\widehat{\mathbf{R}}â‹\dots (\mathbf{v}×\mathbf{m})\left(1−\frac{v^2}{c^2}\right)}{c^2{R}^3{\left(1−\frac{v^2}{c^2}{\sin }^2\mathrm{θ}\right)}^{\frac{3}{2}}}=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{\widehat{\mathbf{R}}â‹\dots \mathbf{P}}{R^3}\\ \mathbf{p}=\frac{\mathbf{v}×\mathbf{m}}{c^2}\end{array}$

Therefore, the potential of an electric dipole is $\mathbf{p}=\frac{\mathbf{v}×\mathbf{m}}{c^2}$.