Question

(a) Show that $\mathbf{(}\mathbf{E}\mathbf{\cdot }\mathbf{B}\mathbf{)}$is relativistically invariant.

(b) Show that $\mathbf{(}{\mathbf{E}}^{\mathbf{2}}\mathbf{-}{\mathbf{c}}^{\mathbf{2}}{\mathbf{B}}^{\mathbf{2}}\mathbf{)}$is relativistically invariant.

(c) Suppose that in one inertial system$\mathbf{B}\mathbf{=}\mathbf{0}$but $\mathbf{E}\mathbf{\ne }\mathbf{0}$(at some point P). Is it possible to find another system in which the electric field is zero atP?

Short answer

(a)$(\mathbf{E}\cdot \mathbf{B})$ is relativistically invariant.

(b) $({\mathrm{E}}^2-{\mathrm{c}}^2{\mathrm{B}}^2)$is relativistically invariant.

(c) It is not possible to find another system in which the electric field is zero at P.

Step-by-step solution

Expression for the set of transformation rules for an electric and magnetic field:

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

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

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

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

Show that (E⋅B) is relativistically invariant:

(a)

Consider the dot product of an electric and magnetic field.

$\overline{E}\cdot \overline{B}={\overline{E}}_x{\overline{B}}_x+{\overline{E}}_y{\overline{B}}_y+{\overline{E}}_z{\overline{B}}_z$

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

$\begin{array}{c}\overline{E}\cdot \overline{B}=E_x{B}_x+\left(\gamma (E_y-v{B}_z)\right)(\gamma \left(\left(B_y+\frac{v}{c^2}{E}_z\right)\right)+\left(\gamma (E_z+v{B}_y)\right)\left(\gamma \left(B_z-\frac{v}{c^2}{E}_y\right)\right)\\ \overline{E}\cdot \overline{B}=E_x{B}_x+{\gamma }^2(E_y-v{B}_z)\left(B_y+\frac{v}{c^2}{E}_z\right)+{\gamma }^2(E_z+v{B}_z)\left(B_z-\frac{v}{c^2}{E}_y\right)\\ \overline{E}\cdot \overline{B}=E_x{B}_x+{\gamma }^2\left[\begin{array}{l}{E}_y{B}_y+\frac{v}{c^2}{E}_y{E}_z-v{B}_y{B}_z-\frac{v^2}{c^2}{E}_z{B}_z+\\ E_z{B}_z-\frac{v}{c^2}{E}_y{E}_z+v{B}_y{B}_z-\frac{v^2}{c^2}{E}_y{B}_y\end{array}\right]\\ \overline{E}\cdot \overline{B}=E_x{B}_x+{\gamma }^2\left[E_y{B}_y\left(1-\frac{v^2}{c^2}\right)+E_z{B}_z\left(1-\frac{v^2}{c^2}\right)\right]\end{array}$

On further solving,

$\begin{array}{c}\overline{E}\cdot \overline{B}=E_x{B}_x+{\gamma }^2\left(1-\frac{v^2}{c^2}\right)(E_y{B}_y+E_z{B}_z)\\ \overline{E}\cdot \overline{B}=E_x{B}_x+{\gamma }^2\left(\frac{1}{{\gamma }^2}\right)(E_y{B}_y+E_z{B}_z)\\ \overline{E}\cdot \overline{B}=E_x{B}_x+E_y{B}_y+E_z{B}_z\\ \overline{E}\cdot \overline{B}=\overline{E}\cdot \overline{B}\end{array}$

Therefore,$(\mathbf{E}\cdot \mathbf{B})$is relativistically invariant.

Show that (E2-c2B2) is relativistically invariant:

(b)

Consider the equation,

${\overline{E}}^2-c^2{\overline{B}}^2=\left({\overline{E}}_x^2+{\overline{E}}_y^2+{\overline{E}}_z^2\right)-c^2\left({\overline{B}}_x^2+{\overline{B}}_y^2+{\overline{B}}_z^2\right)$

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

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

On further solving,

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

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

Step 4:

(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.