91影视

Write the equation for the rapidity.

$\mathit{胃}\mathbf{=}{\mathbf{tanh}}^{\mathbf{-}\mathbf{1}}\left(\frac{v}{c}\right)$ 鈥︹�� (1)

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

(a)

Rearrange the equation (1).

$\tanh 胃=\frac{\mathrm{v}}{\mathrm{c}}$ 鈥︹�� (2)

It is known that:

$\frac{\sinh 胃}{\cosh 胃}=\tanh 胃$

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

Hence, equation (2) becomes,

$\frac{\sinh 胃}{\cosh 胃}=\frac{v}{c}$

Write the equation for the rotation matrix.

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

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

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

Since it is known that:

$尾=\frac{v}{c}=\tan 胃$

Hence, write the value of ${纬}_B$.

$$\begin{gathered} {纬}_B=\cosh 胃脳\tanh 胃 \\ {纬}_B=\cosh 胃脳\frac{\sinh 胃}{\cosh 胃} \\ {纬}_B=\sinh 胃 \end{gathered}$$

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

localid="1654670773676" $饾湨=\left[\begin{array}{cccc}纬& -纬尾& 0& 0\\ -纬尾& 纬& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]$

Substitute for and for in the above matrix.

$饾湨=\left[\begin{array}{cccc}\mathrm{肠辞蝉丑胃}& -\mathrm{蝉颈苍丑胃}& 0& 0\\ -\mathrm{蝉颈苍丑胃}& \mathrm{肠辞蝉丑胃}& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]$ 鈥︹�� (3)

Write the equation for the rotation matrix.

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

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

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

Therefore, the Lorentz transformation matrix in terms of $胃$is

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

(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.

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

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

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

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

As$\tanh \left(蠒-胃\right)$ , solve the equation.

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

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