(a) A μ = ( V / c , A x , A y , A z ) is a 4-vector (like x μ = ( c t , x , y , z ) ) , so (using Eq. 12.19): V = γ ( V ˉ + v A ˉ x ) \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) (a) A μ = ( V / c , A x , A y , A z ) is a 4-vector (like x μ = ( c t , x , y , z ) ) , so (using Eq. 12.19): V = γ ( V ˉ + v A ˉ x ) . But V ˉ = 0 \bar{V}=0 V ˉ = 0 , and
A ˉ x = μ 0 4 π ( m × r ‾ ) x r ˉ 3 \bar{A}_{x}=\frac{\mu_{0}}{4 \pi} \frac{( m \times \overline{ r })_{x}}{\bar{r}^{3}} A ˉ x = 4 π μ 0 r ˉ 3 ( m × r ) x
(i′) x = γ ( x ˉ + v t ˉ ) , (ii′) y = y ˉ , (iii′) z = z ˉ , (iv′) t = γ ( t ˉ + v c 2 x ˉ ) . } \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\} (i′) x = γ ( x ˉ + v t ˉ ) , (ii′) y = y ˉ , (iii′) z = z ˉ , (iv′) t = γ ( t ˉ + c 2 v x ˉ ) . ⎭ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎫ (12.19)
Now ( m × r ‾ ) x = m y z ˉ − m z y ˉ = m y z − m z y \text { Now }( m \times \overline{ r })_{x}=m_{y} \bar{z}-m_{z} \bar{y}=m_{y} z-m_{z} y Now ( m × r ) x = m y z ˉ − m z y ˉ = m y z − m z y . So
V = γ v μ 0 4 π ( m y z − m z y ) r ˉ 3 V=\gamma v \frac{\mu_{0}}{4 \pi} \frac{\left(m_{y} z-m_{z} y\right)}{\bar{r}^{3}} V = γ v 4 π μ 0 r ˉ 3 ( m y z − m z y ) Now x ˉ = γ ( x − v t ) = γ R x , y ˉ = y = R y , z ˉ = z = R z \bar{x}=\gamma(x-v t)=\gamma R_{x}, \bar{y}=y=R_{y}, \bar{z}=z=R_{z} x ˉ = γ ( x − v t ) = γ R x , y ˉ = y = R y , z ˉ = z = R z , where R is the vector (in S ) from the (instantaneous) location of the dipole to the point of observation. Thus
r ˉ 2 = γ 2 R x 2 + R y 2 + R z 2 = γ 2 ( R x 2 + R y 2 + R z 2 ) + ( 1 − γ 2 ) ( R y 2 + R z 2 ) = γ 2 ( R 2 − v 2 c 2 R 2 sin 2 θ ) \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) r ˉ 2 = γ 2 R x 2 + R y 2 + R z 2 = γ 2 ( R x 2 + R y 2 + R z 2 ) + ( 1 − γ 2 ) ( R y 2 + R z 2 ) = γ 2 ( R 2 − c 2 v 2 R 2 sin 2 θ ) (where θ is the angle between R and the x -axis, so that R y 2 + R z 2 = R 2 sin 2 θ ) \left.R_{y}^{2}+R_{z}^{2}=R^{2} \sin ^{2} \theta\right) R y 2 + R z 2 = R 2 sin 2 θ )
∴ V = μ 0 4 π v γ ( m y R z − m z R y ) γ 3 R 3 ( 1 − v 2 c 2 sin 2 θ ) 3 / 2 ; v ⋅ ( m × R ) = v ( m × R ) x = v ( m y R z − m z R y ) \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) ∴ V = 4 π μ 0 γ 3 R 3 ( 1 − c 2 v 2 s i n 2 θ ) 3 / 2 v γ ( m y R z − m z R y ) ; v ⋅ ( m × R ) = v ( m × R ) x = v ( m y R z − m z R y ) , so
V = μ 0 4 π v ⋅ ( m × R ) ( 1 − v 2 c 2 ) R 3 ( 1 − v 2 c 2 sin 2 θ ) 3 / 2 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}} V = 4 π μ 0 R 3 ( 1 − c 2 v 2 s i n 2 θ ) 3 / 2 v ⋅ ( m × R ) ( 1 − c 2 v 2 )
or, using μ 0 = 1 ϵ 0 c 2 and v ⋅ ( m × R ) = R ⋅ ( v × m ) : V = 1 4 π ϵ 0 R ^ ⋅ ( v × m ) ( 1 − v 2 c 2 ) c 2 R 2 ( 1 − v 2 c 2 sin 2 θ ) 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}} or, using μ 0 = ϵ 0 c 2 1 and v ⋅ ( m × R ) = R ⋅ ( v × m ) : V = 4 π ϵ 0 1 c 2 R 2 ( 1 − c 2 v 2 s i n 2 θ ) 3 / 2 R ⋅ ( v × m ) ( 1 − c 2 v 2 ) (b) In the nonrelativistic limit ( v 2 ≪ c 2 ) \left(v^{2} \ll c^{2}\right) ( v 2 ≪ c 2 )
V = 1 4 π ϵ 0 R ^ ⋅ ( v × m ) c 2 R 2 = 1 4 π ϵ 0 R ^ ⋅ p R 2 , with p = v × m c 2 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}} V = 4 π ϵ 0 1 c 2 R 2 R ⋅ ( v × m ) = 4 π ϵ 0 1 R 2 R ⋅ p , with p = c 2 v × m
which is the potential of an electric dipole.