Question 12.64: An ideal magnetic dipole moment m is located at the origin o...

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

\overline{ A }=\frac{\mu_{0}}{4 \pi} \frac{\overline{ m } \times \overline{\hat{ r }}}{\bar{r}^{2}},

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

A _{ dip }( r )=\frac{\mu_{0}}{4 \pi} \frac{ m \times \hat{ r }}{r^{2}}                                     (5.85)

(a) Find the scalar potential V in S. [Answer:

\left.\frac{1}{4 \pi \epsilon_{0}} \frac{\hat{ R } \cdot( v \times m )}{c^{2} R^{2}} \frac{\left(1-v^{2} / c^{2}\right)}{\left.1-\left(v^{2} / c^{2}\right) \sin ^{2} \theta\right)^{3 / 2}}\right]

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

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

located at \overline{ O }.

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\text { (a) } \left.A^{\mu}=\left(V / c, A_{x}, A_{y}, A_{z}\right) \text { is a 4-vector (like } x^{\mu}=(c t, x, y, z)\right) \text {, so (using Eq. 12.19): } V=\gamma\left(\bar{V}+v \bar{A}_{x}\right) . But \bar{V}=0 , and

\bar{A}_{x}=\frac{\mu_{0}}{4 \pi} \frac{( m \times \overline{ r })_{x}}{\bar{r}^{3}}

 

\left. \begin{matrix} \text { (i′) } x=\gamma(\bar{x}+v \bar{t}), \\ \text { (ii′) } y=\bar{y}, \\ \text { (iii′) } z=\bar{z} \text {, } \\ \text { (iv′) } t=\gamma\left(\bar{t}+\frac{v}{c^{2}} \bar{x}\right) \text {. } \end{matrix} \right\}                            (12.19)

\text { Now }( m \times \overline{ r })_{x}=m_{y} \bar{z}-m_{z} \bar{y}=m_{y} z-m_{z} y . So

V=\gamma v \frac{\mu_{0}}{4 \pi} \frac{\left(m_{y} z-m_{z} y\right)}{\bar{r}^{3}}

Now \bar{x}=\gamma(x-v t)=\gamma R_{x}, \bar{y}=y=R_{y}, \bar{z}=z=R_{z} , where R is the vector (in S) from the (instantaneous) location of the dipole to the point of observation. Thus

\bar{r}^{2}=\gamma^{2} R_{x}^{2}+R_{y}^{2}+R_{z}^{2}=\gamma^{2}\left(R_{x}^{2}+R_{y}^{2}+R_{z}^{2}\right)+\left(1-\gamma^{2}\right)\left(R_{y}^{2}+R_{z}^{2}\right)=\gamma^{2}\left(R_{2}-\frac{v^{2}}{c^{2}} R^{2} \sin ^{2} \theta\right)

(where θ is the angle between R and the x-axis, so that \left.R_{y}^{2}+R_{z}^{2}=R^{2} \sin ^{2} \theta\right).

\therefore V=\frac{\mu_{0}}{4 \pi} \frac{v \gamma\left(m_{y} R_{z}-m_{z} R_{y}\right)}{\gamma^{3} R^{3}\left(1-\frac{v^{2}}{c^{2}} \sin ^{2} \theta\right)^{3 / 2}} ; v \cdot( m \times R )=v( m \times R )_{x}=v\left(m_{y} R_{z}-m_{z} R_{y}\right) , so

V=\frac{\mu_{0}}{4 \pi} \frac{ v \cdot( m \times R )\left(1-\frac{v^{2}}{c^{2}}\right)}{R^{3}\left(1-\frac{v^{2}}{c^{2}} \sin ^{2} \theta\right)^{3 / 2}},

\text { or, using } \mu_{0}=\frac{1}{\epsilon_{0} c^{2}} \text { and } v \cdot( m \times R )= R \cdot( v \times m ): \quad V=\frac{1}{4 \pi \epsilon_{0}} \frac{\widehat{ R } \cdot( v \times m )\left(1-\frac{v^{2}}{c^{2}}\right)}{c^{2} R^{2}\left(1-\frac{v^{2}}{c^{2}} \sin ^{2} \theta\right)^{3 / 2}}

(b) In the nonrelativistic limit \left(v^{2} \ll c^{2}\right):

V=\frac{1}{4 \pi \epsilon_{0}} \frac{\widehat{ R } \cdot( v \times m )}{c^{2} R^{2}}=\frac{1}{4 \pi \epsilon_{0}} \frac{\widehat{ R } \cdot p }{R^{2}}, \quad \text { with } \quad p =\frac{ v \times m }{c^{2}},

which is the potential of an electric dipole.

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