Chapter 15, Problem 15.105P

To determine

The matrix $F^{\prime }$, and to verify the transformation equations (15.146) for the electromagnetic fields.

Answer to Problem 15.105P

The matrix $F^{\prime }=\left[\begin{array}{cccc}0& \gamma \left(B_3-\beta E_2/c\right)& -\gamma \left(B_2+\beta E_3/c\right)& -E_1/c\\ -\gamma \left(B_3-\beta E_2/c\right)& 0& B_1& -\frac{\gamma \left(E_2-\beta c{B}_3\right)}{c}\\ \gamma \left(B_2+\beta E_3/c\right)& -B_1& 0& -\frac{\gamma \left(E_3+\beta c{B}_2\right)}{c}\\ E_1/c& \frac{\gamma \left(E_2-\beta c{B}_3\right)}{c}& \frac{\gamma \left(E_3+\beta c{B}_3\right)}{c}& 0\end{array}\right]$, and the transformation equations are verified.

Explanation of Solution

Write the matrix form of $F$.

    $F=\left[\begin{array}{cccc}0& B_3& -B_2& -E_1/c\\ -B_3& 0& B_1& -E_2/c\\ B_2& -B_1& 0& -E_3/c\\ E_1/c& E_2/c& E_3/c& 0\end{array}\right]$        (I)

Here, $F$ is the four tensor.

Write the matrix form of standard Lorentz boost $\Lambda$.

    $\Lambda =\left[\begin{array}{cccc}\gamma & 0& 0& -\gamma \beta \\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ -\gamma \beta & 0& 0& \gamma \end{array}\right]$        (II)

Here, $\Lambda$ is the matrix form of standard Lorentz boost.

From the above equation $\tilde{\Lambda }$

    $\tilde{\Lambda }=\left[\begin{array}{cccc}\gamma & 0& 0& -\gamma \beta \\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ -\gamma \beta & 0& 0& \gamma \end{array}\right]$        (III)

Since $F^{\prime }=\Lambda F\tilde{\Lambda }$, from equation (I), (II), and (III) the matrix representation of $F^{\prime }$ is given as

    $\begin{array}{c}{F}^{\prime }=\left[\begin{array}{cccc}\gamma & 0& 0& -\gamma \beta \\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ -\gamma \beta & 0& 0& \gamma \end{array}\right]\left[\begin{array}{cccc}0& B_3& -B_2& -E_1/c\\ -B_3& 0& B_1& -E_2/c\\ B_2& -B_1& 0& -E_3/c\\ E_1/c& E_2/c& E_3/c& 0\end{array}\right]\left[\begin{array}{cccc}\gamma & 0& 0& -\gamma \beta \\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ -\gamma \beta & 0& 0& \gamma \end{array}\right]\\ =\left[\begin{array}{cccc}0+0+0-\gamma \beta E_1/c& \gamma B_3+0+0-\gamma \beta E_2/c& -\gamma B_2+0+0-\gamma \beta E_3/c& -\gamma E_1/c+0+0+0\\ 0-B_3+0+0& 0+0+0+0& 0+B_1+0+0& 0-E_2/c+0+0\\ 0+0+B_2+0& 0+0-B_1+0& 0+0+0+0& 0+0-E_3/c+0\\ 0+0+0+\gamma E_1/c& -\gamma \beta B_3+0+0+\gamma E_2/c& \gamma \beta B_2+0+0+\gamma E_3/c& \gamma \beta E_1/c+0+0+0\end{array}\right]\left[\begin{array}{cccc}\gamma & 0& 0& -\gamma \beta \\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ -\gamma \beta & 0& 0& \gamma \end{array}\right]\\ =\left[\begin{array}{cccc}-\gamma \beta E_1/c& \gamma B_3-\gamma \beta E_2/c& -\gamma B_2-\gamma \beta E_3/c& -\gamma E_1/c\\ -B_3& 0& B_1& -E_2/c\\ B_2& -B_1& 0& -E_3/c\\ \gamma E_1/c& -\gamma \beta B_3+\gamma E_2/c& \gamma \beta B_2+\gamma E_3/c& \gamma \beta E_1/c\end{array}\right]\left[\begin{array}{cccc}\gamma & 0& 0& -\gamma \beta \\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ -\gamma \beta & 0& 0& \gamma \end{array}\right]\end{array}$

The above equation can be written as

    $\begin{array}{c}{F}^{\prime }=\left[\begin{array}{cccc}-{\gamma }^2\beta E_1/c+0+0+{\gamma }^2\beta E_1/c& 0+\gamma B_3-\gamma \beta E_2/c+0+0& 0+0-\gamma B_2-\gamma \beta E_3/c+0& {\gamma }^2{\beta }^2{E}_1/c+0+0-{\gamma }^2{E}_1/c\\ -\gamma B_3+0+0+\gamma \beta E_2/c& 0+0+0+0& 0+0+B_1+0& \gamma \beta B_3+0+0-\gamma E_2/c\\ \gamma B_1+0+0+\gamma \beta E_3/c& 0-B_1+0+0& 0+0+0+0& -\gamma \beta B_2+0+0-\gamma E_3/c\\ {\gamma }^2{E}_1/c+0+0-{\gamma }^2{\beta }^2{E}_1/c& 0-\gamma \beta B_3+\gamma E_2/c+0+0& 0+0+\gamma {\beta }_2+\gamma E_3/c+0& -{\gamma }^2\beta E_1/c+0+0{\gamma }^2\beta E_1/c\end{array}\right]\\ =\left[\begin{array}{cccc}0& \gamma \left(B_3-\beta E_2/c\right)& \gamma \left(-B_2-\beta E_3/c\right)& {\gamma }^2\left({\beta }^2-1\right)E_1/c\\ \gamma \left(-B_3+\beta E_2/c\right)& 0& B_1& \gamma \left(\beta B_3-E_2/c\right)\\ \gamma \left(B_2+\beta E_3/c\right)& -B_1& 0& \gamma \left(-\beta B_2-E_3/c\right)\\ {\gamma }^2\left(1-{\beta }^2\right)E_1/c& \gamma \left(-\beta B_3+E_2/c\right)& \gamma \left(\beta B_2+E_3/c\right)& 0\end{array}\right]\end{array}$        (IV)

The above can be also written as

    $F^{\prime }=\left[\begin{array}{cccc}0& {B^{\prime }}_3& -{B^{\prime }}_2& -{E^{\prime }}_1/c\\ -{B^{\prime }}_3& 0& {B^{\prime }}_1& -{E^{\prime }}_2/c\\ {B^{\prime }}_2& -{B^{\prime }}_1& 0& -{E^{\prime }}_3/c\\ {E^{\prime }}_1/c& {E^{\prime }}_2/c& {E^{\prime }}_3/c& 0\end{array}\right]$        (V)

By comparing (IV) and (V), write the expression for ${B^{\prime }}_3$.

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

By comparing (IV) and (V), write the expression for ${B^{\prime }}_2$.

    $\begin{array}{c}-{B^{\prime }}_2=-\gamma \left(B_2+\beta E_3/c\right)\\ {B^{\prime }}_2=\gamma \left(B_2+\beta E_3/c\right)\end{array}$        (VII)

By comparing (IV) and (V), write the expression for ${B^{\prime }}_1$.

    ${B^{\prime }}_1=B_1$        (VIII)

By comparing (IV) and (V), write the expression for ${E^{\prime }}_1$.

    $\begin{array}{c}-{E^{\prime }}_1/c=E^{\prime }/c\\ {E^{\prime }}_1=E_1\end{array}$        (IX)

By comparing (IV) and (V), write the expression for ${E^{\prime }}_2$.

    $\begin{array}{c}-{E^{\prime }}_2/c=-\gamma \left(E_2-\beta c{B}_3\right)\\ {E^{\prime }}_2=\gamma \left(E_2-\beta c{B}_3\right)\end{array}$        (X)

By comparing (IV) and (V), write the expression for ${E^{\prime }}_3$.

    $\begin{array}{c}-{E^{\prime }}_3/c=-\gamma \left(E_3+\beta c{B}_2\right)\\ {E^{\prime }}_3=\gamma \left(E_3+\beta c{B}_2\right)\end{array}$        (XI)

The results obtained in equation (VI), (VII), (VIII), (IX), and (X) is same as the transformation equation in Equation (15-146).

Conclusion:

Therefore, the matrix $F^{\prime }$ is derived as

$F^{\prime }=\left[\begin{array}{cccc}0& \gamma \left(B_3-\beta E_2/c\right)& -\gamma \left(B_2+\beta E_3/c\right)& -E_1/c\\ -\gamma \left(B_3-\beta E_2/c\right)& 0& B_1& -\frac{\gamma \left(E_2-\beta c{B}_3\right)}{c}\\ \gamma \left(B_2+\beta E_3/c\right)& -B_1& 0& -\frac{\gamma \left(E_3+\beta c{B}_2\right)}{c}\\ E_1/c& \frac{\gamma \left(E_2-\beta c{B}_3\right)}{c}& \frac{\gamma \left(E_3+\beta c{B}_3\right)}{c}& 0\end{array}\right]$, and the transformation equations are verified.