91Ó°ÊÓ

The formula used to find the time coordinate in some frames is $t^{\text{'}}=\frac{t-{\displaystyle \frac{\mathrm{β}x}{c}}}{\sqrt{1-{\left(\mathrm{β}\right)}^2}}$.

(a)

Here, the event 1 occurs at the origin at t=0 . So, the coordinate of the event 1 are .

$\left(x_1,t_1\right)=\left(0,0\right)$

Thus, the time of event 1 is 0s.

Here, the event occurs at x = 3.0 km and at $t=4.0\mathrm{μ}s$
. Given that $v=0.60c$and $\mathrm{β}=0.60$.

The distance coordinate of event 2 can be obtained as follows:

$$\begin{gathered} {x^{\text{'}}}_2=\frac{x-vt}{\sqrt{1-{\mathrm{β}}^2}} \\ =\frac{3000-0.60.3×{10}^8×4.0×{10}^{-6}}{\sqrt{1-{\left(0.60\right)}^2}} \\ =2.85×{10}^{3}m \end{gathered}$$

The time coordinate of event 2 can be obtained as follows:

$$\begin{gathered} {t^{\text{'}}}_2=\frac{t-{\displaystyle \frac{\mathrm{β}x}{c}}}{\sqrt{1-{\mathrm{β}}^2}} \\ =\frac{4.0×{10}^{-6}-{\displaystyle \frac{0.60×3000}{3.0×{10}^8}}}{\sqrt{1-{\left(0.60\right)}^2}} \\ =-2.5×{10}^{-6}s \end{gathered}$$

Thus, the time of the event 2 is $-2.5×{10}^{-6}s$.

(c)

Here, the two events in the frame occurs in the order of event 2 after 1. But in the frame the value of time coordinate${t^{\text{'}}}_2<0$ . So, event 1 occurs after 2.

Thus, the two observers see the two events in the reverse sequence.