Chapter 15, Problem 15.102P

(a)

To determine

Using the standard boost along $x_1$ axis, calculate the corresponding boost along the $x_3$ axis.

Answer to Problem 15.102P

The corresponding boost along the $x_3$ axis is $\underset{\_}{{E^{\prime }}_1=\gamma \left(E_1-\beta c{B}_2\right)}$, $\underset{\_}{{E^{\prime }}_2=\gamma \left(E_2+\beta c{B}_1\right)}$, $\underset{\_}{{E^{\prime }}_3=E_3}$, $\underset{\_}{{B^{\prime }}_1=\gamma \left(B_1+\beta \frac{E_2}{c}\right)}$, $\underset{\_}{{B^{\prime }}_2=\gamma \left(B_2-\beta \frac{E_1}{c}\right)}$, $\underset{\_}{{B^{\prime }}_3=B_3}$.

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.

To compute the corresponding boost along $x_3$ axis, change the coordinates from $1\to 3,2\to 1,\text{and 3}\to \text{2}$.

The corresponding boost along $x_3$ axis for ${E^{\prime }}_1$ is given by

    ${E^{\prime }}_1=\gamma \left(E_1-\beta c{B}_2\right)$

The corresponding boost along $x_3$ axis for ${E^{\prime }}_2$ is given by

    ${E^{\prime }}_2=\gamma \left(E_2+\beta c{B}_1\right)$

The corresponding boost along $x_3$ axis for ${E^{\prime }}_3$ is given by

    ${E^{\prime }}_3=E_3$

The corresponding boost along $x_3$ axis for ${B^{\prime }}_1$ is given by

    ${B^{\prime }}_1=\gamma \left(B_1+\beta \frac{E_2}{c}\right)$

The corresponding boost along $x_3$ axis for ${B^{\prime }}_2$ is given by

    ${B^{\prime }}_2=\gamma \left(B_2-\beta \frac{E_1}{c}\right)$

The corresponding boost along $x_3$ axis for ${B^{\prime }}_3$ is given by

    ${B^{\prime }}_3=B_3$

Conclusion:

Therefore, the corresponding boost along the $x_3$ axis is $\underset{\_}{{E^{\prime }}_1=\gamma \left(E_1-\beta c{B}_2\right)}$, $\underset{\_}{{E^{\prime }}_2=\gamma \left(E_2+\beta c{B}_1\right)}$, $\underset{\_}{{E^{\prime }}_3=E_3}$, $\underset{\_}{{B^{\prime }}_1=\gamma \left(B_1+\beta \frac{E_2}{c}\right)}$, $\underset{\_}{{B^{\prime }}_2=\gamma \left(B_2-\beta \frac{E_1}{c}\right)}$, $\underset{\_}{{B^{\prime }}_3=B_3}$.

(b)

To determine

The inverse of transformation obtained in subpart (a), and verify the results for the fields of a moving line charge.

Answer to Problem 15.102P

The inverse of transformation obtained in subpart (a) are $\underset{\_}{E_1=\gamma \left({E^{\prime }}_1+\beta c{B^{\prime }}_2\right)}$, $\underset{\_}{E_2=\gamma \left({E^{\prime }}_2-\beta c{B^{\prime }}_1\right)}$, $\underset{\_}{E_3={E^{\prime }}_3}$, $\underset{\_}{B_1=\gamma \left({B^{\prime }}_1-\beta \frac{{E^{\prime }}_2}{c}\right)}$, $\underset{\_}{B_2=\gamma \left({B^{\prime }}_2+\beta \frac{{E^{\prime }}_1}{c}\right)}$, and $\underset{\_}{B_3={B^{\prime }}_3}$, and the results are verified.

Explanation of Solution

Write the expression for inverse transformation of ${E^{\prime }}_1$.

    $E_1=\frac{{E^{\prime }}_1}{\gamma }+\beta c{B}_2$        (VII)

Here, $E_1$ is the inverse transformation of ${E^{\prime }}_1$.

Use $B_2=\left(\frac{{B^{\prime }}_2}{\gamma }+\beta \frac{E_1}{c}\right)$ in the above equation.

    $\begin{array}{c}{E}_1=\frac{{E^{\prime }}_1}{\gamma }+\beta c\left(\frac{{B^{\prime }}_2}{\gamma }+\beta \frac{E_1}{c}\right)\\ E_1=\frac{{E^{\prime }}_1}{\gamma }+\left(\frac{\beta c{B^{\prime }}_2}{\gamma }+{\beta }^{2}c\right)\\ \left(1-{\beta }^2\right)E_1=\frac{{E^{\prime }}_1}{\gamma }+\beta c\frac{{B^{\prime }}_2}{\gamma }\end{array}$

Rearrange the above equation.

    $\begin{array}{c}\frac{E_1}{{\gamma }^2}=\frac{{E^{\prime }}_1}{\gamma }+\beta c\frac{B^{\prime }}{\gamma }\\ E_1=\gamma \left(E^{\prime }+\beta c{B}^{\prime }\right)\end{array}$        (VIII)

Write the expression for inverse transformation of ${E^{\prime }}_2$.

    $E_2=\frac{{E^{\prime }}_2}{\gamma }-\beta c{B}_1$        (IX)

Here, $E_2$ is the inverse transformation of ${E^{\prime }}_2$.

Use $B_1=\left(\frac{{B^{\prime }}_1}{\gamma }+\beta \frac{E_2}{c}\right)$ in the above equation.

    $\begin{array}{c}{E}_2=\frac{{E^{\prime }}_2}{\gamma }-\beta c\left(\frac{{B^{\prime }}_1}{\gamma }+\beta \frac{E_2}{c}\right)\\ =\frac{{E^{\prime }}_2}{\gamma }-\frac{\beta c{B^{\prime }}_1}{\gamma }+{\beta }^2{E}_2\\ \left(1-{\beta }^2\right)E_2=\frac{{E^{\prime }}_2}{\gamma }-\frac{\beta c}{\gamma }{B}^{\prime }\\ E_2=\gamma \left({E^{\prime }}_2-\beta c{B}^{\prime }\right)\end{array}$        (X)

Write the expression for inverse transformation of ${E^{\prime }}_3$.

    $E_3={E^{\prime }}_3$        (XI)

Here, $E_3$ is the inverse transformation of ${E^{\prime }}_3$.

Write the expression for inverse transformation of ${B^{\prime }}_1$.

    $B_1=\frac{{B^{\prime }}_1}{\gamma }-\frac{\beta }{c}\gamma \left({E^{\prime }}_2-\beta c{B^{\prime }}_1\right)$

Here, $B_1$ is the inverse transformation of ${B^{\prime }}_1$.

The above equation can be written as

    $\begin{array}{l}{B}_1=\frac{{B^{\prime }}_1}{\gamma }-\frac{\beta }{c}\gamma {E^{\prime }}_2+{\beta }^2\gamma {B^{\prime }}_1\hfill \\ B_1=\gamma \left[{B^{\prime }}_1\left(\frac{1}{{\gamma }^2}+{\beta }^2\right)-\beta \frac{{E^{\prime }}_2}{c}\right]\hfill \end{array}$

Rearrange the above equation.

    $\begin{array}{c}{B}_1=\gamma \left[\left(1-{\beta }^2+{\beta }^2\right){B^{\prime }}_1-\beta \frac{{E^{\prime }}_2}{c}\right]\\ =\gamma \left[{B^{\prime }}_1-\beta \frac{{E^{\prime }}_2}{c}\right]\end{array}$        (XII)

Write the expression for inverse transformation of ${B^{\prime }}_2$.

    $B_2=\frac{{B^{\prime }}_2}{\gamma }+\frac{\beta }{c}\gamma \left({E^{\prime }}_1+\beta c{B^{\prime }}_2\right)$        (XIII)

Here, $B_2$ is the inverse transformation of ${B^{\prime }}_2$.

The above equation can be written as

    $\begin{array}{c}{B}_2=\gamma \left[{B^{\prime }}_2\left(\frac{1}{{\gamma }^2}+{\beta }^2\right)+\beta \frac{{E^{\prime }}_1}{c}\right]\\ =\gamma \left[{B^{\prime }}_2+\beta \frac{{E^{\prime }}_1}{c}\right]\end{array}$        (XIV)

Write the expression for inverse transformation of ${B^{\prime }}_3$.

    $B_3={B^{\prime }}_3$        (XV)

From equation (15.148), $E^{\prime }$ is given by

    $E^{\prime }=\frac{2k{\lambda }^{\prime }}{{{\rho }^{\prime }}^2}\left(x^{\prime },y^{\prime },0\right)$        (XVI)

From the above equation ${E^{\prime }}_1$ is given by

    ${E^{\prime }}_1=\frac{2k{\lambda }^{\prime }}{{{\rho }^{\prime }}^2}{x}^{\prime }$        (XVII)

From the equation (XVI) ${E^{\prime }}_2$ is given by

    ${E^{\prime }}_2=\frac{2k{\lambda }^{\prime }}{{{\rho }^{\prime }}^2}y\text{'}$        (XVIII)

From the equation (XVI) ${E^{\prime }}_3$ is given by

    ${E^{\prime }}_3=0$        (XIX)

Since $B^{\prime }=0$, ${B^{\prime }}_1$ is given by

    ${B^{\prime }}_1=0$        (XX)

Since $B^{\prime }=0$, ${B^{\prime }}_2$ is given by

    ${B^{\prime }}_2=0$        (XXI)

Since $B^{\prime }=0$, ${B^{\prime }}_3$ is given by

    ${B^{\prime }}_3=0$        (XXII)

From equation (XVII), write $E_1$.

    $\begin{array}{c}{E}_1=\gamma \left[\frac{2k{\lambda }^{\prime }}{{{\rho }^{\prime }}^2}{x}^{\prime }+0\right]\\ =\gamma \frac{2k{\lambda }^{\prime }}{{{\rho }^{\prime }}^2}{x}^{\prime }\end{array}$        (XXIII)

Since we have considered, the Lorentz boost along the 3 direction as $x^{\prime }=x$, $y^{\prime }=y$, and $z^{\prime }=0$. Thus ${\rho }^{\prime }=\rho$.

Use the above condition in equation (XXIII).

    $\begin{array}{c}{E}_1=\gamma \frac{2k{\lambda }^{\prime }}{{\rho }^2}x\\ =\frac{2k\lambda }{{\rho }^2}x\end{array}$        (XXIV)

Similarly $E_2$ is given by

    $E_2=\frac{2k\lambda }{{\rho }^2}y$        (XXV)

Where $\lambda =\gamma {\lambda }^{\prime }$.

Similarly $E_3$ is given by

    $E_3=0$        (XXVI)

Using equation (XXIV), (XXV), and (XXVI), write the expression for $E$.

    $E=\frac{\gamma k{\lambda }^{\prime }}{{\rho }^2}\left(x,y,0\right)$

Since ${\lambda }^{\prime }=\frac{\lambda }{\gamma }$, the above equation is written as

    $\begin{array}{c}E=\frac{k\lambda }{{\rho }^2}\left(x,y,0\right)\\ =\frac{k\lambda }{\rho }\widehat{\rho }\end{array}$

From the equation, it is evident that equation (15.151) is verified.

Write the expression for $B_1$.

    $B_1=\gamma \left[0-\frac{\beta }{c}\frac{2k{\lambda }^{\prime }}{{{\rho }^{\prime }}^2}{y}^{\prime }\right]$

The above equation can be written as

    $B_1=-\gamma \beta \frac{2k{\lambda }^{\prime }}{c{\rho }^2}$        (XXVII)

Write the expression for $B_2$.

    $B_2=\gamma \left[0+\frac{\beta }{c}\frac{2k{\lambda }^{\prime }}{{{\rho }^{\prime }}^2}{x}^{\prime }\right]$

The above equation can be written as

    $B_2=\gamma \beta \frac{2k{\lambda }^{\prime }}{c{\rho }^2}x$        (XXVIII)

Write the expression for $B_3$.

  $B_3=0$        (XXIX)

Using equation (XXVII), (XXVIII), and (XXIX), write the equation for $B$.

    $B=\gamma \beta \frac{2k{\lambda }^{\prime }}{c{\rho }^2}\left(-y,x,0\right)$        (XXX)

In the above equation $\frac{k}{c^2}$, is rearranged as

      $\begin{array}{c}\frac{k}{c^2}=\frac{1}{4\pi {\epsilon }_0{c}^2}\\ =\frac{1}{4\pi {\epsilon }_0\frac{1}{{\epsilon }_0{\mu }_0}}\\ =\frac{{\mu }_0}{4\pi }\end{array}$

Use the above condition in equation (XXX).

    $\text{B}=\frac{2{\mu }_0}{4\pi }\frac{\lambda v}{{\rho }^2}\left(-y,x,0\right)$

Since $\left(\frac{-y}{\rho },\frac{x}{\rho },0\right)$ is taken as $\widehat{\phi }$, the above equation is written as

    $B=\frac{{\mu }_0}{2\pi }\frac{\lambda v}{\rho }\widehat{\phi }$

From the above equation it is evident that equation (15.152) is verified.

Conclusion:

Therefore, the inverse of transformation obtained in subpart (a) are $\underset{\_}{E_1=\gamma \left({E^{\prime }}_1+\beta c{B^{\prime }}_2\right)}$, $\underset{\_}{E_2=\gamma \left({E^{\prime }}_2-\beta c{B^{\prime }}_1\right)}$, $\underset{\_}{E_3={E^{\prime }}_3}$, $\underset{\_}{B_1=\gamma \left({B^{\prime }}_1-\beta \frac{{E^{\prime }}_2}{c}\right)}$, $\underset{\_}{B_2=\gamma \left({B^{\prime }}_2+\beta \frac{{E^{\prime }}_1}{c}\right)}$, and $\underset{\_}{B_3={B^{\prime }}_3}$, and the results are verified.