Inertial frame S'moves at a speed0.60cof with respect to frame (Fig. 37-9). Further,x=x'=0att=t'=0.Two events are recorded. In frame S , event1occurs at the origin att=0and event 2 occurs on the x axis atx=3.0kmat t=4.0μs. According to the observerS' , 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

Expert verified

(a) According to the observer S', the time of the event 1 is0s .

(b) According to the observer S', the time of the event 2 is -2.5×10-6s.

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

Step by step solution

01

The time coordinate formula in frame

The formula used to find the time coordinate in some frames is t'=t-βxc1-β2.

02

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 .

x1,t1=0,0

Thus, the time of event 1 is 0s.

03

The time of event 2 

Here, the event occurs at x = 3.0 km and at t=4.0μs
. Given that v=0.60cand β=0.60.

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

x'2=x-vt1-β2=3000-0.60.3×108×4.0×10-61-0.602=2.85×103m

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

t'2=t-βxc1-β2=4.0×10-6-0.60×30003.0×1081-0.602=-2.5×10-6s

Thus, the time of the event 2 is -2.5×10-6s.

04

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 coordinatet'2<0 . So, event 1 occurs after 2.

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

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