Question

An ideal magnetic dipole moment m is located at the origin of an inertial system $\overline{\mathbf{S}}$ that moves with speed v in the x direction with respect to inertial system S. In$\overline{\mathbf{S}}$ the vector potential is

$\overline{\mathbf{A}}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}}{\mathbf{4}\mathbf{\pi }}\frac{\overline{\mathbf{m}}\mathbf{\times }\overline{\widehat{\mathbf{r}}}}{{\overline{\mathbf{r}}}^{\mathbf{2}}}$

(Eq. 5.85), and the scalar potential$\overline{\mathbf{V}}$ is zero.

(a) Find the scalar potential V in S.

(b) In the nonrelativistic limit, show that the scalar potential in S is that of an ideal electric dipole of magnitude

$\mathbf{p}\mathbf{=}\frac{\mathbf{v}\mathbf{\times }\mathbf{m}}{{\mathbf{c}}^{\mathbf{2}}}$

located at$\overline{\mathbf{O}}$ .

Short answer

(a) The scalar potential is $V=\frac{{\mu }_0}{4\pi }\frac{\widehat{\mathbf{R}}\cdot (\mathbf{v}\times \mathbf{m})\left(1-\frac{v^2}{c^2}\right)}{c^2{R}^3{\left(1-\frac{v^2}{c^2}{\sin }^2\theta \right)}^{\frac{3}{2}}}$.

(b) The potential of an electric dipole is$\mathbf{p}=\frac{\mathbf{v}\times \mathbf{m}}{c^2}$ .

Step-by-step solution

Expression for the scalar potential:

Write the expression for the scalar potential.

$\mathbf{V}\mathbf{=}\mathbf{\gamma }\mathbf{(}\overline{\mathbf{V}}\mathbf{+}\mathbf{v}{\overline{\mathbf{A}}}_{\mathbf{x}}\mathbf{)}$ …… (1)

Here, $\gamma$ is the Lorentz contraction, v is the velocity, and $\overline{\mathbf{A}}$is the vector potential in $\overline{S}$frame.

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

$\overline{\mathbf{A}}=\frac{{\mu }_0}{4\pi }\frac{\overline{\mathbf{m}}\times \overline{\widehat{\mathbf{r}}}}{{\overline{r}}^2}$ …..(2)

Determine the scalar potential V in S:

(a)

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

Substitute $\overline{V}=0$and ${\overline{\mathbf{A}}}_x=\frac{{\mu }_0}{4\pi }\frac{{(\overline{\mathbf{m}}\times \overline{\widehat{\mathbf{r}}})}_x}{{\overline{r}}^2}$in equation (1).

$\begin{array}{c}V=\gamma \left(0+v\frac{{\mu }_0}{4\pi }\frac{{(\overline{\mathbf{m}}\times \overline{\widehat{\mathbf{r}}})}_x}{{\overline{r}}^2}\right)\\ V=v\gamma \frac{{\mu }_0}{4\pi }\frac{{(\overline{\mathbf{m}}\times \overline{\widehat{\mathbf{r}}})}_x}{{\overline{r}}^2}\end{array}$ …(3)

Here,

$\begin{array}{c}{(\overline{\mathbf{m}}\times \overline{\widehat{\mathbf{r}}})}_x=m_y\overline{z}=m_z\overline{y}\\ {(\overline{\mathbf{m}}\times \overline{\widehat{\mathbf{r}}})}_x=m_{y}z-m_{z}y\end{array}$

Substitute ${(\overline{\mathbf{m}}\times \overline{\widehat{\mathbf{r}}})}_x=m_{y}z-m_{z}y$in equation (3).

$V=v\gamma \frac{{\mu }_0}{4\pi }\left(\frac{m_{y}z-m_{z}y}{{\overline{r}}^2}\right)$ ….. (4)

It is known that:

$\begin{array}{c}\overline{x}=\gamma (x-vt)\\ \overline{x}=\gamma R_x\end{array}$

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

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

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

$\begin{array}{c}{\overline{r}}^2={\gamma }^2{R}_x^2+R_y^2+R_z^2\\ {\overline{r}}^2={\gamma }^2(R_x^2+R_y^2+R_z^2)+(1-{\gamma }^2)(R_y^2+R_z^2)\\ {\overline{r}}^2={\gamma }^2\left(R^2-\frac{v^2}{c^2}{R}^2{\sin }^2\theta \right)\end{array}$

Substitute${\overline{r}}^2={\gamma }^2\left(R^2-\frac{v^2}{c^2}{R}^2{\sin }^2\theta \right)$in equation (4).

$\begin{array}{c}V=v\gamma \frac{{\mu }_0}{4\pi }\left(\frac{m_{y}z-m_{z}y}{r{\gamma }^2\left(R^2-\frac{v^2}{c^2}{R}^2{\sin }^2\theta \right)}\right)\\ V=\frac{{\mu }_0}{4\pi }\frac{v\gamma (m_{y}z-m_{z}y)}{{\gamma }^3{R}^3{\left(1-\frac{v^2}{c^2}{\sin }^2\theta \right)}^{\frac{3}{2}}}\\ V=\frac{{\mu }_0}{4\pi }\frac{v\cdot (\mathbf{m}\times \mathbf{R})\left(1-\frac{v^2}{c^2}\right)}{R^3{\left(1-\frac{v^2}{c^2}{\sin }^2\theta \right)}^{\frac{3}{2}}}\end{array}$

Here,${\mu }_0=\frac{1}{{\epsilon }_0{c}^2}$ and$\mathbf{v}\cdot (\mathbf{m}\times \mathbf{R})=\mathbf{R}\cdot (\mathbf{v}\times \mathbf{m})$.

Hence, the above equation becomes,

$V=\frac{{\mu }_0}{4\pi }\frac{\widehat{\mathbf{R}}\cdot (\mathbf{v}\times \mathbf{m})\left(1-\frac{v^2}{c^2}\right)}{c^2{R}^3{\left(1-\frac{v^2}{c^2}{\sin }^2\theta \right)}^{\frac{3}{2}}}$ ……. (5)

Therefore, the scalar potential is $V=\frac{{\mu }_0}{4\pi }\frac{\widehat{\mathbf{R}}\cdot (\mathbf{v}\times \mathbf{m})\left(1-\frac{v^2}{c^2}\right)}{c^2{R}^3{\left(1-\frac{v^2}{c^2}{\sin }^2\theta \right)}^{\frac{3}{2}}}$.

Show that p=v×mc2:

(b)

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

$\mathbf{v}=\frac{1}{4\pi {\epsilon }_0}\frac{\widehat{\mathbf{R}}\cdot \mathbf{P}}{R^3}$ …… (6)

Compare the equations (5) and (6).

$\begin{array}{c}\frac{{\mu }_0}{4\pi }\frac{\widehat{\mathbf{R}}\cdot (\mathbf{v}\times \mathbf{m})\left(1-\frac{v^2}{c^2}\right)}{c^2{R}^3{\left(1-\frac{v^2}{c^2}{\sin }^2\theta \right)}^{\frac{3}{2}}}=\frac{1}{4\pi {\epsilon }_0}\frac{\widehat{\mathbf{R}}\cdot \mathbf{P}}{R^3}\\ \mathbf{p}=\frac{\mathbf{v}\times \mathbf{m}}{c^2}\end{array}$

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