Question

The coordinates of event Aare $\mathbf{(}{\mathbf{x}}_{\mathbf{A}}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{}{\mathit{t}}_{\mathbf{A}\mathbf{,}}$ and the coordinates of event B are$\left(x_B,0,0\right)\mathbf{,}\mathbf{}{\mathit{t}}_{\mathbf{A}}$. Assuming the displacement between them is spacelike, find the velocity of the system in which they are simultaneous.

Short answer

The velocity of the system is $v=\left(\frac{t_B-t_A}{x_B-x_A}\right)c^2$.

Step-by-step solution

Expression for the Lorentz transformation equation:

Write the expression for the Lorentz transformation equation.

$\overline{\mathbf{t}\mathbf{}}\mathbf{=}\mathit{Y}\left(t-\frac{v}{c^2}x\right)$ …… (1)

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

Determine the velocity of the system:

Write equation (1) in terms of $∆$.

$∆\overline{t}=\gamma \left(∆t-\frac{v}{c^2}∆x\right)$

Here, $∆\overline{t}=0$

Hence, the equation becomes,

$$\begin{gathered} \gamma \left(∆t-\frac{v}{c^2}∆x\right)=0 \\ ∆t-\frac{v}{c^2}∆x=0 \\ ∆t-\frac{v}{c^2}∆x \\ v=\frac{c^2∆t}{∆x} \end{gathered}$$

As the coordinates of event A and B are given as $\left(x_A,0,0\right),t_A$and $\left(x_B,0,0\right),t_B$respectively, the velocity of the system will be,

$$\begin{gathered} v=\frac{c^2\left(t_B-t_A\right)}{\left(x_B-x_A\right)} \\ v=\left(\frac{t_B-t_A}{x_B-x_A}\right)c^2 \end{gathered}$$

Therefore, the velocity of the system is $v=\left(\frac{t_B-t_A}{x_B-x_A}\right)c^2$.