Chapter 15, Problem 15.104P

(a)

To determine

The four component of $K$.

Answer to Problem 15.104P

The four component of $K$ is $\underset{\_}{K_1=q\left[\left(0\right)u_1+\left(B_3\right)u_2+\left(-B_2\right)u_3+\left(\frac{E_1}{c}\right)u_4\right]}$, $\underset{\_}{K_2=q\left[\left(-B_3\right)u_1+\left(0\right)u_2+\left(B_1\right)u_3+\left(\frac{E_2}{c}\right)u_4\right]}$, $\underset{\_}{K_3=q\left[\left(B_2\right)u_1+\left(-B_1\right)u_2+\left(0\right)u_3+\left(\frac{E_3}{c}\right)u_4\right]}$, and $\underset{\_}{K_4=q\left[\left(\frac{E_1}{c}\right)u_1+\left(\frac{E_2}{c}\right)u_2+\left(\frac{E_3}{c}\right)u_3+\left(0\right)u_4\right]}$.

Explanation of Solution

Write the expression for Lorentz force.

    $\text{F}=q\left(\text{E+v}\times \text{B}\right)$        (I)

Here, $\text{F}$ is the force, $q$ is the charge, $\text{E}$ is the electric field, $\text{v}$ is the velocity, and $\text{B}$ is the magnetic field.

Write the expression for $\text{v}\times \text{B}$.

    $\begin{array}{c}\text{v}\times \text{B}=\left|\begin{array}{ccc}i& j& k\\ v_1& v_2& v_3\\ B_1& B_2& b_3\end{array}\right|\\ =\left(v_2{B}_3-v_3{B}_2\right)i+\left(v_3{B}_1-v_1{B}_3\right)j+\left(v_1{B}_2-v_2{B}_1\right)k\end{array}$        (II)

Use the above equation in equation (I).

    $\text{F}=q\left(E_1+v_2{B}_3-v_3{B}_2\right)i+q\left(E_2+v_3{B}_1-v_1{B}_3\right)j+q\left(E_3+v_1{B}_2-v_2{B}_1\right)k$

Using the above equation, $\gamma \text{F}$ is given by

    $\gamma \text{F}=\gamma q\left(E_1+v_2{B}_3-v_3{B}_2\right)i+\gamma q\left(E_2+v_3{B}_1-v_1{B}_3\right)j+\gamma q\left(E_3+v_1{B}_2-v_2{B}_1\right)k$        (III)

Write the expression for $\text{v}\cdot \text{F}$.

    $\text{v}\cdot \text{F=}q\left(\text{v}\cdot \text{E+v}\cdot \left(\text{v}\times \text{B}\right)\right)$

The above equation can be written as

    $\begin{array}{c}\text{v}\cdot \text{F=}q\left(v_1{E}_1+v_2{E}_2+v_3{E}_3\right)\\ \frac{\gamma \text{v}\cdot \text{F}}{c}=\frac{\gamma q}{c}\left(v_1{E}_1+v_2{E}_2+v_3{E}_3\right)\end{array}$

Since $K=\left(\gamma \text{F},\frac{\gamma \text{v}\cdot \text{F}}{c}\right)$, and $u=\left(\gamma v,\gamma c\right)$, the equation for $K_1$ is given by

    $K_1=\gamma q\left(E_1+v_2{B}_3-v_3{B}_2\right)$

The above equation can be also written as

    $K_1=q\left[\left(0\right)\gamma v_1+\left(B_3\right)\left(\gamma v_2\right)+\left(-B_2\right)\left(\gamma v_3\right)+\left(\frac{E_1}{c}\right)(\gamma c)\right]$        (IV)

Since $u_1=\gamma v_1$, $u_2=\gamma v_2$, $u_3=\gamma v_3$, and $u_4=\gamma c$, the above equation is written as

    $K_1=q\left[\left(0\right)u_1+\left(B_3\right)u_2+\left(-B_2\right)u_3+\left(\frac{E_1}{c}\right)u_4\right]$        (V)

Write the expression for $K_2$.

    $K_2=\gamma q\left(E_2+v_3{B}_1-v_1{B}_3\right)$

Since $u_1=\gamma v_1$, $u_2=\gamma v_2$, $u_3=\gamma v_3$, and $u_4=\gamma c$, the above equation is written as

    $\begin{array}{c}{K}_2=q\left[\left(-B_3\right)\gamma v_1+\left(0\right)\gamma v_2+\left(B_1\right)\gamma v_3+\left(\frac{E_2}{c}\right)\gamma c\right]\\ =q\left[\left(-B_3\right)u_1+\left(0\right)u_2+\left(B_1\right)u_3+\left(\frac{E_2}{c}\right)u_4\right]\end{array}$        (VI)

Write the expression for $K_3$.

    $K_3=\gamma q\left(E_3+v_1{B}_2-v_2{B}_1\right)$

since $u_1=\gamma v_1$, $u_2=\gamma v_2$, $u_3=\gamma v_3$, and $u_4=\gamma c$, the above equation is written as

    $\begin{array}{c}{K}_3=q\left[\left(B_2\right)\left(\gamma v_1\right)+\left(-B_1\right)\left(\gamma v_2\right)+\left(0\right)\gamma v_3+\left(\frac{E_2}{c}\right)\gamma c\right]\\ =q\left[\left(B_2\right)u_1+\left(-B_1\right)u_2+\left(0\right)u_3+\left(\frac{E_2}{c}\right)u_4\right]\end{array}$

Write the expression for $K_4$.

    $K_4=\frac{\gamma q}{c}\left(v_1{E}_1+v_2{E}_2-v_3{E}_3\right)$

since $u_1=\gamma v_1$, $u_2=\gamma v_2$, $u_3=\gamma v_3$, and $u_4=\gamma c$, the above equation is written as

    $\begin{array}{c}{K}_4=\frac{\gamma q}{c}\left[\left(\frac{E_1}{c}\right)\left(\gamma v_1\right)+\left(\frac{E_2}{c}\right)\left(\gamma v_2\right)+\left(\frac{E_3}{c}\right)\left(\gamma v_3\right)+\left(0\right)\gamma c\right]\\ =q\left[\left(\frac{E_1}{c}\right)u_1+\left(\frac{E_2}{c}\right)u_2+\left(\frac{E_3}{c}\right)u_3+\left(0\right)u_4\right]\end{array}$

Conclusion:

Therefore, the four component of $K$ is $\underset{\_}{K_1=q\left[\left(0\right)u_1+\left(B_3\right)u_2+\left(-B_2\right)u_3+\left(\frac{E_1}{c}\right)u_4\right]}$, $\underset{\_}{K_2=q\left[\left(-B_3\right)u_1+\left(0\right)u_2+\left(B_1\right)u_3+\left(\frac{E_2}{c}\right)u_4\right]}$, $\underset{\_}{K_3=q\left[\left(B_2\right)u_1+\left(-B_1\right)u_2+\left(0\right)u_3+\left(\frac{E_3}{c}\right)u_4\right]}$, and $\underset{\_}{K_4=q\left[\left(\frac{E_1}{c}\right)u_1+\left(\frac{E_2}{c}\right)u_2+\left(\frac{E_3}{c}\right)u_3+\left(0\right)u_4\right]}$.

(b)

To determine

Using the results in subpart (a) calculate the matrix $FG$, and show that the tensor $F$ has the same form in Equation (15.143).

Answer to Problem 15.104P

The matrix $FG$ is given by $FG=\left[\begin{array}{cccc}0& B_3& -B_2& E_1/c\\ -B_3& 0& B_1& E_2/c\\ B_2& -B_1& 0& E_3/c\\ E_1/c& E_2/c& E_3/c& 0\end{array}\right]$, and it is shown that the tensor $F$ has the same form as Equation (15.143).

Explanation of Solution

Write the four component of $K$ in matrix form.

    $\left[\begin{array}{l}{K}_1\hfill \\ K_2\hfill \\ K_3\hfill \\ K_4\hfill \end{array}\right]=q\left[\begin{array}{cccc}0& B_3& -B_2& E_1/c\\ -B_3& 0& B_1& E_2/c\\ B_2& -B_1& 0& E_3/c\\ E_1/c& E_2/c& E_3/c& 0\end{array}\right]\left[\begin{array}{l}{u}_1\hfill \\ u_2\hfill \\ u_3\hfill \\ u_4\hfill \end{array}\right]$

The above equation is in the form $K=qFGu$, where $FG=\left[\begin{array}{cccc}0& B_3& -B_2& E_1/c\\ -B_3& 0& B_1& E_2/c\\ B_2& -B_1& 0& E_3/c\\ E_1/c& E_2/c& E_3/c& 0\end{array}\right]$.

Since $G$ is a unit vector, and $G^2=1$, $F$ is given by

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

From the above equation it is evident that $F$ has the same form as Equation (15.143).

Conclusion:

Therefore, The matrix $FG$ is given by $FG=\left[\begin{array}{cccc}0& B_3& -B_2& E_1/c\\ -B_3& 0& B_1& E_2/c\\ B_2& -B_1& 0& E_3/c\\ E_1/c& E_2/c& E_3/c& 0\end{array}\right]$, and it is shown that the tensor $F$ has the same form as Equation (15.143).