Chapter 15, Problem 15.101P

(a)

To determine

To prove that $\text{E}\cdot \text{B}$ and $E^2-c^2{B}^2$ is invariant under any Lorentz transformation.

Answer to Problem 15.101P

It is proved that $\text{E}\cdot \text{B}$ and $E^2-c^2{B}^2$ is invariant under any Lorentz transformation.

Explanation of Solution

Write the expression for transformed electric field using standard boost.

    ${E^{\prime }}_1=E_1$        (I)

Here, ${E^{\prime }}_1$ is the transformed electric field along $x$ direction, and $E_1$ is the original electric field component along $x$ direction.

Write the expression for transformed electric field using standard boost.

    ${E^{\prime }}_2=\gamma \left(E_2-\beta B_3\right)$        (II)

Here, ${E^{\prime }}_2$ is the transformed electric field along $y$ direction, $B_3$ is the original magnetic field component along $z$ direction and $E_2$ is the original electric field component along $y$ direction.

Write the expression for transformed electric field using standard boost.

    ${E^{\prime }}_3=\gamma \left(E_3+\beta c{B}_2\right)$        (III)

Here, ${E^{\prime }}_3$ the transformed electric field component along $z$ direction, $E_3$ is the original electric field component along $z$ direction, and $B_2$ is the original magnetic field component along $y$ direction.

Write the expression for transformed of magnetic field using standard boost.

    ${B^{\prime }}_1=B_1$        (IV)

Here, ${B^{\prime }}_1$ is the transformed of magnetic field component along $x$ direction, and $B_1$ is the original magnetic field along $x$ direction.

Write the expression for transformed of magnetic field using standard boost.

    ${B^{\prime }}_2=\gamma \left(B_2+\frac{\beta E_3}{c}\right)$        (V)

Here, ${B^{\prime }}_2$ is the transformed of magnetic field component along $y$ direction, $c$ is the speed of light and $B_2$ is the original magnetic field along $y$ direction.

Write the expression for transformed of magnetic field using standard boost.

    ${B^{\prime }}_3=\gamma \left(B_3-\beta E_2/c\right)$        (VI)

Here, ${B^{\prime }}_1$ is the transformed of magnetic field component along $x$ direction, and $B_1$ is the original magnetic field along $x$ direction.

Write the equation for $E^{\prime }\cdot B^{\prime }$.

    $\begin{array}{c}{E}^{\prime }\cdot B^{\prime }=\left(i{E^{\prime }}_1+j{E^{\prime }}_2+k{E^{\prime }}_3\right)\cdot \left(i{B^{\prime }}_1+j{B^{\prime }}_2+k{B^{\prime }}_3\right)\\ ={E^{\prime }}_1{B^{\prime }}_1+{E^{\prime }}_2{B^{\prime }}_2+{E^{\prime }}_3{B^{\prime }}_3\end{array}$        (VII)

Use equation (I), (II), (III), (IV), (V) and (VI) in the above equation.

    $\begin{array}{c}{E}^{\prime }\cdot B^{\prime }=E_1{B}_1+\gamma \left(E_2-\beta c{B}_3\right)\gamma \left(B_2+\beta E_3/c\right)+\gamma \left(E_3+\beta c{B}_2\right)\gamma \left(B_3-\beta E_2/c\right)\\ =E_1{B}_1+\gamma \left(E_2-\beta c{B}_3\right)\gamma \left(B_2+\beta E_3/c\right)+\gamma \left(E_3+\beta c{B}_2\right)\gamma \left(B_3-\beta E_2/c\right)\\ =E_1{B}_1+{\gamma }^2\left[E_2{B}_2+\frac{\beta E_2{E}_3}{c}-\beta c{B}_2{B}_3-{\beta }^2{E}_3{B}_3+E_3{B}_3-\frac{\beta E_3{E}_2}{c}+\beta c{B}_2{B}_3-{\beta }^2{E}_2{B}_2\right]\\ =E_1{B}_1+{\gamma }^2\left[E_2{B}_2-{\beta }^2{E}_2{B}_2+E_3{B}_3-{\beta }^2{E}_3{B}_3\right]\end{array}$

Rearrange the above equation.

    $\begin{array}{c}{E}^{\prime }\cdot B^{\prime }=E_1{B}_1+\frac{1}{1-{\beta }^2}\left[\left(1-{\beta }^2\right)E_2{B}_2+\left(1-{\beta }^2\right)E_3{B}_3\right]\\ =E_1{B}_1+E_2{B}_2+E_3{B}_3\end{array}$

The above equation can be written as

    $\begin{array}{c}{E}^{\prime }\cdot B^{\prime }=\left(i{E}_1+j{E}_2+k{E}_3\right)\cdot \left(i{B}_1+j{B}_2+k{B}_3\right)\\ =\text{E}\cdot \text{B}\end{array}$        (VIII)

Write the equation for ${E^{\prime }}^2-c^2{B^{\prime }}^2$.

    $\begin{array}{c}{E^{\prime }}^2-c^2{B^{\prime }}^2=\left({E^{\prime }}_1^2+{E^{\prime }}_2^2+{E^{\prime }}_3^2\right)-\left(c^2{B^{\prime }}_1^2+c^2{B^{\prime }}_2^2+c^2{B^{\prime }}_3^2\right)\\ =E_1^2+{\left[\gamma \left(E_2-\beta c{B}_3\right)\right]}^2+{\left[\gamma \left(E_3+\beta c{B}_2\right)\right]}^2-c^2{B}_1^2-c^2{\left[\gamma \left(B_2+\beta E_3/c\right)\right]}^2-c^2{\left[\gamma \left(B_3-\beta E_2/c\right)\right]}^2\\ =E_1^2+{\gamma }^2\left(E_2^2+{\beta }^2{c}^2{B}_3^2-2\beta c{E}_2{B}_3\right)+{\gamma }^2\left(E_3^2+{\beta }^2{c}^2{B}_2^2+2\beta c{E}_3{B}_2\right)\\ -c^2{B}_1^2-c^2{\gamma }^2\left(B_2^2+\frac{{\beta }^2{E}_3^2}{c^2}+\frac{2\beta B_2{E}_3}{c}\right)-c^2{\gamma }^2\left(B_3^2+\frac{{\beta }^2{E}_2^2}{c^2}-\frac{2\beta E_2{B}_3}{c}\right)\\ =E_1^2+E_2^2\left({\gamma }^2-{\gamma }^2{\beta }^2\right)+E_3^2\left(\gamma -{\gamma }^2{\beta }^2\right)-c^2{B}_1^2-c^2{B}_2^2\left(\gamma -{\gamma }^2{\beta }^2\right)\\ +E_2{B}_3\left(-2\beta {\gamma }^{2}c+2\beta {\gamma }^{2}c\right)+E_3{B}_2\left(2\beta {\gamma }^{2}c-2\beta {\gamma }^{2}c\right)\end{array}$

Rearrange the above equation.

    ${E^{\prime }}^2-c^2{B^{\prime }}^2=E_1^2+E_2^2{\gamma }^2\left(1-{\beta }^2\right)-E_3^2{\gamma }^2\left(1-{\beta }^2\right)-c^2{B}_1^2-c^2{B}_2^2{\gamma }^2\left(1-{\beta }^2\right)-c^2{B}_3^2{\gamma }^2\left(1-{\beta }^2\right)$        (IX)

Since ${\gamma }^2\left(1-{\beta }^2\right)=1$

Use the above condition in equation (IX).

    $\begin{array}{c}{E^{\prime }}^2-c^2{B^{\prime }}^2=E_1^2+E_2^2+E_3^2-c^2\left(B_1^2+B_2^2+B_3^2\right)\\ =E^2-c^2{B}^2\end{array}$

Conclusion:

Therefore, it is proved that $\text{E}\cdot \text{B}$ and $E^2-c^2{B}^2$ is invariant under any Lorentz transformation.

(b)

To determine

To prove that if $\text{E}$ and $\text{B}$ are perpendicular in frame $S$, then they are perpendicular in all the frame $S$.

Answer to Problem 15.101P

It is proved that if $\text{E}$ and $\text{B}$ are perpendicular in frame $S$, then they are perpendicular in all the frame $S$.

Explanation of Solution

When electric field $\text{E}$  and magnetic field $\text{B}$ are perpendicular to each other, their dot product is equal to zero $\text{E}\cdot \text{B}=0$.

Since $\text{E}\cdot \text{B=}{\text{E}}^{\prime }\cdot {\text{B}}^{\prime }$, the dot product of electric field vector and magnetic field vector is also zero ${\text{E}}^{\prime }\cdot {\text{B}}^{\prime }=0$. Thus $\text{E}$ and $\text{B}$ are perpendicular in frame $S$.

Conclusion:

Therefore, it is proved that if $\text{E}$ and $\text{B}$ are perpendicular in frame $S$, then they are perpendicular in all the frame $S$.

(c)

To determine

To prove that if $E>cB$ exist in a frame $S$, then there exist no frame such that $E=0$.

Answer to Problem 15.101P

It is proved that if $E>cB$ exist in a frame $S$, then there exist no frame such that $E=0$.

Explanation of Solution

Consider that in frame $S$, $E>cB$.

Write the relation between $E$ and $B$ in $S$ frame.

    $E>cB$

Square the above equation on both the sides.

    $\begin{array}{c}{E}^2>c^2{B}^2\\ E^2-c^2{B}^2>0\end{array}$        (X)

But $E^2-c^2{B}^2={E^{\prime }}^2-{c^{\prime }}^2{B^{\prime }}^2$

Use the above condition in equation (X).

    $\begin{array}{c}{E^{\prime }}^2-c^2{B^{\prime }}^2>0\\ {E^{\prime }}^2>c^2{B^{\prime }}^2\end{array}$        (XI)

Since $c^2{B^{\prime }}^2$ is never less than zero, therefore the above equation is reduced to

    $c^2{B^{\prime }}^2\ge 0$

From the above equation it is clear that ${E^{\prime }}^2>0$, or $E^{\prime }>0$, and since $E^{\prime }$ is not equal to zero no frame exist having $E=0$.

Conclusion:

Therefore, it is proved that if $E>cB$ exist in a frame $S$, then there exist no frame such that $E=0$