Chapter 15, Problem 15.34P

To determine

The value of ${\Lambda }_{\text{B}}(\theta )$ and check the result with the motion of the spatial origin of the frame

Answer to Problem 15.34P

${\Lambda }_B(\theta )$ is verified by finding the motion of the spatial origin of frame $S$ as observed in $S^{\prime }$.

Explanation of Solution

The four vector ${\Lambda }_R(\theta )$ is given by,

    ${\Lambda }_R(\theta )=\left[\begin{array}{cccc}\cos \theta & \sin \theta & 0& 0\\ -\sin \theta & \cos \theta & 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]$        (I)

Similarly, ${\Lambda }_R(-\theta )$ is given by,

    ${\Lambda }_R(-\theta )=\left[\begin{array}{cccc}\cos (-\theta )& \sin (-\theta )& 0& 0\\ -\sin \theta & \cos \theta & 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]$

    $\begin{array}{c}{\Lambda }_R(\theta )=\left[\begin{array}{cccc}\cos \theta & \sin \theta & 0& 0\\ -\sin \theta & \cos \theta & 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]\\ ={\left[{\Lambda }_R\left(\theta \right)\right]}^{-}\end{array}$        (II)

The value of ${\Lambda }_R\left(\theta \right)\cdot {\Lambda }_R\left(-\theta \right)$ is given by,

$\begin{array}{c}{\Lambda }_R\left(\theta \right)\cdot {\Lambda }_R\left(-\theta \right)=\left[\begin{array}{cccc}{\cos }^2\theta +{\sin }^2\theta & -\cos \theta \sin \theta +\cos \theta \sin \theta & 0& 0\\ -\cos \theta \sin \theta +\cos \theta \sin \theta & {\sin }^2\theta +{\cos }^2\theta & 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]\\ =\left[\begin{array}{llll}1\hfill & 0\hfill & 0\hfill & 0\hfill \\ 0\hfill & 1\hfill & 0\hfill & 0\hfill \\ 0\hfill & 0\hfill & 1\hfill & 0\hfill \\ 0\hfill & 0\hfill & 0\hfill & 1\hfill \end{array}\right]\end{array}$        (IV)

Equation (IV) shows that ${\Lambda }_R(\theta )$ is orthogonal.

The matrix ${\Lambda }_B(\theta )$ can be found with the orthogonal property of the vector ${\Lambda }_R(\theta )$ as,

    ${\Lambda }_B(\theta )={\Lambda }_R(-\theta ){\Lambda }_B(0){\Lambda }_R(\theta )$        (V)

The value of ${\Lambda }_B(0)=\left[\begin{array}{cccc}\gamma & 0& 0& -\gamma \beta \\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ -\gamma \beta & 0& 0& \gamma \end{array}\right]$        (VI)

Use equation (I), (II), and (VI) in equation (V) to find ${\Lambda }_B(\theta )$.

$\begin{array}{c}{\Lambda }_B(\theta )=\left[\begin{array}{cccc}\cos \theta & -\sin \theta & 0& 0\\ \sin \theta & \cos \theta & 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]\left[\begin{array}{cccc}\gamma & 0& 0& -\gamma \beta \\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ -\gamma \beta & 0& 0& \gamma \end{array}\right]\left[\begin{array}{cccc}\cos \theta & \sin \theta & 0& 0\\ -\sin \theta & \cos \theta & 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]\\ =\{\left[\begin{array}{cccc}\gamma \cos \theta +0+0& 0-\sin \theta +0+0& 0+0+0+0& -\gamma \beta \cos \theta +0+0+0\\ \gamma \sin \theta +0+0+0& 0+\cos \theta +0+0& 0+0+0+0& -\gamma \beta \sin \theta +0+0+0\\ 0+0+0+0& 0+0+0+0& 0+0+1+0& 0+0+0+0\\ 0+0+0-\gamma \beta & 0+0+0+0& 0+0+0+0& 0+0+0+\gamma \end{array}\right]\\ \left[\begin{array}{cccc}\cos \theta & \sin \theta & 0& 0\\ -\sin \theta & \cos \theta & 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]\end{array}$

$\begin{array}{c}=\left[\begin{array}{cccc}\gamma \cos \theta & -\sin \theta & 0& -\gamma \beta \cos \theta \\ \gamma \sin \theta & \cos \theta & 0& -\gamma \beta \sin \theta \\ 0& 0& 1& 0\\ -\gamma \beta & 0& 0& \gamma \end{array}\right]\left[\begin{array}{cccc}\cos \theta & \sin \theta & 0& 0\\ -\sin \theta & \cos \theta & 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]\\ =\left[\begin{array}{cccc}\gamma {\cos }^2\theta +{\sin }^2\theta +0+0& \gamma \cos \theta \sin \theta -\cos \theta \sin \theta +0+0& 0+0+0+0& 0+0+0-\gamma \beta \cos \theta \\ \gamma \sin \theta \cos \theta -\sin \theta \cos \theta +0+0& \gamma {\sin }^2\theta +{\cos }^2\theta +0+0& 0+0+0+0& 0+0+0-\gamma \beta \sin \theta \\ 0+0+0+0& 0+0+0+0& 0+0+1+0& 0+0+0+0\\ -\gamma \beta \cos \theta +0+0+0& -\gamma \beta \sin \theta +0+0+0& 0+0+0+0& 0+0+0+\gamma \end{array}\right]\end{array}$

  $=\left[\begin{array}{cccc}{\gamma }^2{\cos }^2\theta +{\sin }^2\theta & (\cos \theta \sin \theta )(\gamma -1)& 0& -\gamma \beta \cos \theta \\ (\cos \theta \sin \theta )(\gamma -1)& \gamma {\sin }^2\theta +{\cos }^2\theta & 0& -\gamma \beta \sin \theta \\ 0& 0& 1& 0\\ -\gamma \beta \cos \theta & -\gamma \beta \sin \theta & 0& \gamma \end{array}\right]$        (V)

Now considering the spatial origin frame $S$ is given by,

$S=\left[\begin{array}{c}0\\ 0\\ 0\\ x_4\end{array}\right]$        (VI)

The motion of spatial origin in frame $S$ as observed in $S^{\prime }$ is given by,

$S^{\prime }={\Lambda }_B(\theta )S$        (VII)

$\begin{array}{c}\left[\begin{array}{c}{\dot{x}}_1^{*}\\ x_2^{\prime }\\ x_3^{\prime }\\ x_4^{\prime }\end{array}\right]=\left[\begin{array}{cccc}{\gamma }^2{\cos }^2\theta +{\sin }^2\theta & \cos \theta \sin \theta (\gamma -1)& 0& -\gamma \beta \cos \theta \\ \cos \theta \sin \theta (\gamma -1)& \gamma {\sin }^2\theta +{\cos }^2\theta & 0& -\gamma \beta \sin \theta \\ 0& 0& 1& 0\\ -\gamma \beta \cos \theta & -\gamma \beta \sin \theta & 0& \gamma \end{array}\right]\left[\begin{array}{l}0\hfill \\ 0\hfill \\ 0\hfill \\ x_4\hfill \end{array}\right]\\ =\left[\begin{array}{c}0+0+0-\gamma \beta x_4\cos \theta \\ 0+0+0-\gamma \beta x_4\sin \theta \\ 0+0+0+0\\ 0+0+0+\gamma x_4\end{array}\right]=\left[\begin{array}{c}-\gamma \beta x_4\cos \theta \\ -\gamma \beta x_4\sin \theta \\ 0\\ \gamma x_4\end{array}\right]\end{array}$

Thus it can be expressed as,

$\begin{array}{c}{x}_1{}^{\prime }=-\frac{v}{c}\times ct\cos \theta \\ =-\gamma vt\cos \theta \\ =-\gamma \left(0-vt\cos \theta \right)\\ =\gamma \left(x_1-vt\cos \theta \right)\end{array}$        (IX)

Here, $v$ is the velocity with which the frame $S^{\prime }$ is moving relative to $S$ at an angle $\theta$. Then the $x$ component of velocity is,

$\begin{array}{l}{v}_x=v\cos \theta \\ v_y=v\sin \theta \end{array}$        (X)

Using equation (X),

$x^{\prime }=\gamma \left(x-v_{x}t\right)$        (XI)

Similarly,

$\begin{array}{l}{y}^{\prime }=\gamma \left(y-v_{y}t\right)\\ z^{\prime }=z\end{array}$

Conclusion

Therefore, ${\Lambda }_B(\theta )$ is verified by finding the motion of the spatial origin of frame $S$ as observed in $S^{\prime }$.