Question

Inertial frame ${\mathit{S}}^{\mathbf{\text{'}}}$moves at a speed$\mathbf{0}\mathbf{.}\mathbf{60}\mathit{c}$of with respect to frame (Fig. 37-9). Further,$\mathit{x}\mathbf{=}{\mathit{x}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{0}$at$\mathit{t}\mathbf{=}{\mathit{t}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{0}$.Two events are recorded. In frame S , event$1$occurs at the origin at$\mathit{t}\mathbf{=}\mathbf{0}$and event 2 occurs on the $\mathit{x}$ axis at$\mathit{x}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\mathit{k}\mathit{m}$at $\mathit{t}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathit{\mu }\mathit{s}$. According to the observer$S^{\text{'}}$ , what is the time of (a) event 1 and (b) event 2? Do the two observers see the same sequence or the reverse?

Short answer

(a) According to the observer $S^{\text{'}}$, the time of the event 1 is$0s$ .

(b) According to the observer $S^{\text{'}}$, the time of the event 2 is $-2.5\times {10}^{-6}s$.

(c) The two observers see the two events in the reverse order.

Step-by-step solution

The time coordinate formula in frame

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

The time of event 1

(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.

The time of event 2 

Here, the event occurs at x = 3.0 km and at $t=4.0\mu s$
. Given that $v=0.60c$and $\beta =0.60$.

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

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

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

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

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

The sequence of the event 

(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.