91Ó°ÊÓ

Write the expression for the set of transformation rules for an electric field.

$\begin{array}{c}{\stackrel{\mathbf{¯}}{\mathbf{E}}}_{\mathbf{x}}\mathbf{=}{\mathbf{E}}_{\mathbf{x}}\\ {\stackrel{\mathbf{¯}}{\mathbf{E}}}_{\mathbf{y}}\mathbf{=}\mathbf{γ}\mathbf{(}{\mathbf{E}}_{\mathbf{y}}\mathbf{-}{\mathbf{vB}}_{\mathbf{z}}\mathbf{)}\\ {\stackrel{\mathbf{¯}}{\mathbf{E}}}_{\mathbf{z}}\mathbf{=}\mathbf{γ}\mathbf{(}{\mathbf{E}}_{\mathbf{z}}\mathbf{+}{\mathbf{vB}}_{\mathbf{y}}\mathbf{)}\end{array}$

Write the expression for the set of transformation rules for the magnetic field.

$\begin{array}{c}{\stackrel{\mathbf{¯}}{\mathbf{B}}}_{\mathbf{x}}\mathbf{=}{\mathbf{B}}_{\mathbf{x}}\\ {\stackrel{\mathbf{¯}}{\mathbf{B}}}_{\mathbf{y}}\mathbf{=}\mathbf{γ}\left(B_y+\frac{v}{c^2}{E}_z\right)\\ {\stackrel{\mathbf{¯}}{\mathbf{B}}}_{\mathbf{z}}\mathbf{=}\mathbf{γ}\left(B_z-\frac{v}{c^2}{E}_y\right)\end{array}$

(a)

Consider the dot product of an electric and magnetic field.

$\stackrel{¯}{E}â‹\dots \stackrel{¯}{B}={\stackrel{¯}{E}}_x{\stackrel{¯}{B}}_x+{\stackrel{¯}{E}}_y{\stackrel{¯}{B}}_y+{\stackrel{¯}{E}}_z{\stackrel{¯}{B}}_z$

Substitute ${\stackrel{¯}{E}}_x=E_x$, ${\stackrel{¯}{B}}_x=B_x$,${\stackrel{¯}{E}}_y=\mathrm{γ}(E_y−v{B}_z)$, ${\stackrel{¯}{B}}_y=\mathrm{γ}\left(B_y+\frac{v}{c^2}{E}_z\right)$, ${\stackrel{¯}{E}}_z=\mathrm{γ}(E_z+v{B}_y)$and ${\stackrel{¯}{B}}_z=\mathrm{γ}\left(B_z−\frac{v}{c^2}{E}_y\right)$in the above expression.

$\begin{array}{c}\stackrel{¯}{E}â‹\dots \stackrel{¯}{B}=E_x{B}_x+\left(\mathrm{γ}(E_y−v{B}_z)\right)(\mathrm{γ}\left(\left(B_y+\frac{v}{c^2}{E}_z\right)\right)+\left(\mathrm{γ}(E_z+v{B}_y)\right)\left(\mathrm{γ}\left(B_z−\frac{v}{c^2}{E}_y\right)\right)\\ \stackrel{¯}{E}â‹\dots \stackrel{¯}{B}=E_x{B}_x+{\mathrm{γ}}^2(E_y−v{B}_z)\left(B_y+\frac{v}{c^2}{E}_z\right)+{\mathrm{γ}}^2(E_z+v{B}_z)\left(B_z−\frac{v}{c^2}{E}_y\right)\\ \stackrel{¯}{E}â‹\dots \stackrel{¯}{B}=E_x{B}_x+{\mathrm{γ}}^2\left[\begin{array}{l}{E}_y{B}_y+\frac{v}{c^2}{E}_y{E}_z−v{B}_y{B}_z−\frac{v^2}{c^2}{E}_z{B}_z+\\ E_z{B}_z−\frac{v}{c^2}{E}_y{E}_z+v{B}_y{B}_z−\frac{v^2}{c^2}{E}_y{B}_y\end{array}\right]\\ \stackrel{¯}{E}â‹\dots \stackrel{¯}{B}=E_x{B}_x+{\mathrm{γ}}^2\left[E_y{B}_y\left(1−\frac{v^2}{c^2}\right)+E_z{B}_z\left(1−\frac{v^2}{c^2}\right)\right]\end{array}$

On further solving,

$\begin{array}{c}\stackrel{¯}{E}â‹\dots \stackrel{¯}{B}=E_x{B}_x+{\mathrm{γ}}^2\left(1−\frac{v^2}{c^2}\right)(E_y{B}_y+E_z{B}_z)\\ \stackrel{¯}{E}â‹\dots \stackrel{¯}{B}=E_x{B}_x+{\mathrm{γ}}^2\left(\frac{1}{{\mathrm{γ}}^2}\right)(E_y{B}_y+E_z{B}_z)\\ \stackrel{¯}{E}â‹\dots \stackrel{¯}{B}=E_x{B}_x+E_y{B}_y+E_z{B}_z\\ \stackrel{¯}{E}â‹\dots \stackrel{¯}{B}=\stackrel{¯}{E}â‹\dots \stackrel{¯}{B}\end{array}$

Therefore,$(\mathbf{E}â‹\dots \mathbf{B})$is relativistically invariant.

(b)

Consider the equation,

${\stackrel{¯}{E}}^2−c^2{\stackrel{¯}{B}}^2=\left({\stackrel{¯}{E}}_x^2+{\stackrel{¯}{E}}_y^2+{\stackrel{¯}{E}}_z^2\right)−c^2\left({\stackrel{¯}{B}}_x^2+{\stackrel{¯}{B}}_y^2+{\stackrel{¯}{B}}_z^2\right)$

Substitute ${\stackrel{¯}{E}}_x=E_x$, ${\stackrel{¯}{B}}_x=B_x$, ${\stackrel{¯}{E}}_y=\mathrm{γ}(E_y−v{B}_z)$, ${\stackrel{¯}{B}}_y=\mathrm{γ}\left(B_y+\frac{v}{c^2}{E}_z\right)$, ${\stackrel{¯}{E}}_z=\mathrm{γ}(E_z+v{B}_y)$and ${\stackrel{¯}{B}}_z=\mathrm{γ}\left(B_z−\frac{v}{c^2}{E}_y\right)$in the above expression.

$\begin{array}{c}{\stackrel{¯}{E}}^2−c^2{\stackrel{¯}{B}}^2=\left\{\begin{array}{l}\left[E_x^2+{\mathrm{γ}}^2{(E_y−v{B}_z)}^2+{\mathrm{γ}}^2{(E_z−v{B}_y)}^2\right]−\\ c^2\left[B_x^2+{\mathrm{γ}}^2{\left(B_y+\frac{v}{c^2}{E}_z\right)}^2+{\mathrm{γ}}^2{\left(B_z−\frac{v}{c^2}{E}_y\right)}^2\right]\end{array}\right\}\\ {\stackrel{¯}{E}}^2−c^2{\stackrel{¯}{B}}^2=\left[\begin{array}{l}{E}_x^2+{\mathrm{γ}}^2(E_y^2−2E_{y}v{B}_z+v^2{B}_z^2+E_z^2+2E_{z}v{B}_y+v^2{B}_y^2−c^2{B}_y^2−2c^2\frac{v}{c^2}{B}_y{E}_z)−\\ c^2\frac{v^2}{c^4}{E}_z^2−c^2{B}_z^2+2c^2\frac{v^2}{c^2}{B}_z{E}_y−c^2\frac{v^2}{c^4}{E}_y^2−c^2{B}_x^2\end{array}\right]\\ {\stackrel{¯}{E}}^2−c^2{\stackrel{¯}{B}}^2=E_x^2−c^2{B}_x^2+{\mathrm{γ}}^2\left[\begin{array}{l}{E}_y^2\left(1−\frac{v^2}{c^2}\right)+E_z^2\left(1−\frac{v^2}{c^2}\right)−\\ c^2{B}_y^2\left(1−\frac{v^2}{c^2}\right)−c^2{B}_z^2\left(1−\frac{v^2}{c^2}\right)\end{array}\right]\end{array}$

On further solving,

$\begin{array}{c}{\stackrel{¯}{E}}^2−c^2{\stackrel{¯}{B}}^2=\left(E_x^2+E_y^2+E_z^2\right)−c^2\left(B_x^2+B_y^2+B_z^2\right)\\ {\stackrel{¯}{E}}^2−c^2{\stackrel{¯}{B}}^2=E^2−c^2{B}^2\end{array}$

Therefore, $(E^2−c^2{B}^2)$is relativistically invariant.

(c)

Based on the given problem, if $\mathbf{B}=0$then, the equation ${\mathrm{E}}^2−{\mathrm{c}}^2{\mathrm{B}}^2$will also be equal to zero.

As the equation ${\mathrm{E}}^2−{\mathrm{c}}^2{\mathrm{B}}^2$is already proved as relativistically invariant, the equation must be true in any reference frame.

Therefore, it is not possible to find another system in which the electric field is zero at P.