Question

The parallel between rotations and Lorentz transformations is even more striking if we introduce the rapidity:

$\mathbf{\theta }\mathbf{=}{\mathbf{tanh}}^{\mathbf{-}\mathbf{1}}\left(\frac{v}{c}\right)$ (12.34)

(a) Express the Lorentz transformation matrix(Eq. 12.24) in terms of$\mathit{\theta }$, and compare it to the rotation matrix (Eq. 1.29).

In some respects, rapidity is a more natural way to describe motion than velocity. For one thing, it ranges fromrole="math" localid="1654511220255" $\mathbf{+}\mathbf{\infty }$ to $\mathbf{+}\mathbf{\infty }$, instead of -c to +c. More significantly, rapidities add, whereas velocities do not.

(b) Express the Einstein velocity addition law in terms of rapidity.

Short answer

(a) The Lorentz transformation matrix in terms of $\theta$is

$𝜦=\left[\begin{array}{cccc}\mathrm{cosh\theta }& -\mathrm{sinh\theta }& 0& 0\\ -\mathrm{sinh\theta }& \mathrm{cosh\theta }& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]$

(b) The Einstein velocity addition law in terms of rapidity is $\overline{\varphi }=\varphi -\theta$.

Step-by-step solution

Determine the Expression for the rapidity equation.

Write the equation for the rapidity.

$\mathit{\theta }\mathbf{=}{\mathbf{tanh}}^{\mathbf{-}\mathbf{1}}\left(\frac{v}{c}\right)$ …… (1)

Here, v is the velocity, and c is the speed of light.

Determine the Lorentz transformation matrix:

(a)

Rearrange the equation (1).

$\tanh \theta =\frac{\mathrm{v}}{\mathrm{c}}$ …… (2)

It is known that:

$\frac{\sinh \theta }{\cosh \theta }=\tanh \theta$

localid="1654672742690" ${\cosh }^2\theta -{\sin }^{2}h\theta =1$

Hence, equation (2) becomes,

$\frac{\sinh \theta }{\cosh \theta }=\frac{v}{c}$

Write the equation for the rotation matrix.

$\gamma =\frac{1}{\sqrt{1-{\left({\displaystyle \frac{v}{c}}\right)}^2}}$

Substitute$\frac{\sinh \theta }{\cosh \theta }$for $\frac{v}{c}$in the above equation.

$$\begin{gathered} \gamma =\frac{1}{\sqrt{1-{\left({\displaystyle \frac{\sinh \theta }{\cosh \theta }}\right)}^2}} \\ \gamma =\frac{1}{\sqrt{\left({\displaystyle \frac{{\cosh }^2\theta -{\sinh }^2\theta }{{\cosh }^2\theta }}\right)}} \\ \gamma =\sqrt{{\cosh }^2\theta } \\ \gamma =\cosh \theta \end{gathered}$$

Since it is known that:

$\beta =\frac{v}{c}=\tan \theta$

Hence, write the value of ${\gamma }_B$.

$$\begin{gathered} {\gamma }_B=\cosh \theta \times \tanh \theta \\ {\gamma }_B=\cosh \theta \times \frac{\sinh \theta }{\cosh \theta } \\ {\gamma }_B=\sinh \theta \end{gathered}$$

Write the matrix for Lorentz transformation with velocity v along the x-axis.

localid="1654670773676" $𝜦=\left[\begin{array}{cccc}\gamma & -\gamma \beta & 0& 0\\ -\gamma \beta & \gamma & 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]$

Substitute for and for in the above matrix.

$𝜦=\left[\begin{array}{cccc}\mathrm{cosh\theta }& -\mathrm{sinh\theta }& 0& 0\\ -\mathrm{sinh\theta }& \mathrm{cosh\theta }& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]$ …… (3)

Write the equation for the rotation matrix.

$R=\left(\begin{array}{cc}\cos \varphi & \sin \varphi \\ -\sin \varphi & \cos \varphi \end{array}\right)$

On comparing the equation (3) matrix with the rotation matrix, the matrix becomes,

localid="1654672813689" $R=\left(\begin{array}{ccc}\cos \varphi & \sin \varphi & 0\\ -\sin \varphi & \cos \varphi & 0\\ 0& 0& 1\end{array}\right)$

Therefore, the Lorentz transformation matrix in terms of $\theta$is

$𝜦=\left[\begin{array}{cccc}\cosh \theta & -\sin \theta & 0& 0\\ -\sinh \theta & \cosh \theta & 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]$

Determine the Einstein velocity addition law in terms of rapidity:

(b)

Write the expression for the velocity of a particle in the frame$\overline{s}$.

$\overline{u}=\frac{u-v}{1-{\displaystyle \frac{uv}{c^2}}}$

Divide by c in L.H.S and the numerator value.

$\frac{\overline{u}}{c}=\frac{{\displaystyle \frac{u}{c}}-{\displaystyle \frac{v}{c}}}{1-{\displaystyle \frac{uv}{c^2}}}$ …… (4)

Here, role="math" localid="1654671755426" $\frac{\overline{u}}{c}=\tanh \overline{\varphi }$and $\frac{u}{c}=\tanh \varphi$.

Substitute $\tanh \overline{\varphi }$for $\frac{\overline{u}}{c}$and $\frac{v}{c}$for $\tanh \theta$in equation (4).

role="math" localid="1654672215398" $$\begin{gathered} \tanh \overline{\varphi }=\frac{\tanh \varphi -\tanh \theta }{1-\tanh \varphi \tanh \theta } \\ \tanh \left(\varphi -\theta \right)=\frac{\tanh \varphi -\tanh \theta }{1-\tanh \varphi \tanh \theta } \end{gathered}$$

As$\tanh \left(\varphi -\theta \right)$ , solve the equation.

role="math" localid="1654672424139" $$\begin{gathered} \tanh \overline{\varphi }=\tanh \left(\varphi -\theta \right) \\ \overline{\varphi }=\varphi -\theta \end{gathered}$$

Therefore, the Einstein velocity addition law in terms of rapidity is $\overline{\varphi }=\varphi -\theta$.