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Chapter 37: Q20P (page 1146)

As in Fig. 37-9, reference frame S'passes reference Sframe with a certain velocity. Events 1 and 2 are to have a certain temporal separation ∆t'according to the S'observer. However, their spatial separation ∆x'according to that observer has not been set yet. Figure 37-24 gives their temporal separation ∆x'according to the observer as a function of for a range of values. The vertical axis scale is set by ∆ta=6μs. What is ∆t'?

Short Answer

Expert verified

The value of ∆t'is 6.3×10-7s.

Step by step solution

01

The Lorentz transformation

For two events the value of ∆tis given by ∆t=∆t'+β∆x'c1-β2or∆t=11-β2.∆t'+11-β2.β∆x'c .

02

The calculation

Here, in the above equation the coefficient of ∆x'is considered as the slope of the graph. Here, the slope of the graph is 6-2μs400m=0.01.

So, the expression for∆t'=∆t1-β2 becomes . From the graph, we can interpret the value ofβ as 0.949. So, the value of ∆t'can be calculated as follows:

∆t'=∆t1-β2=2×10-6×1-0.9492=6.3×10-7s

Thus, the value of∆t' is 6.3×10-7s.

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Most popular questions from this chapter

Continuation of Problem 65. Let reference frame C in Fig. 37-31 move past reference frame D (not shown). (a) Show thatMAD=MABMBCMCD

(b) Now put this general result to work: Three particles move parallel to a single axis on which an observer is stationed. Let plus and minus signs indicate the directions of motion along that axis. Particle A moves past particle B atβAB=+0.20 . Particle B moves past particle C at βBC=−0.40. Particle C moves past observer D atβCD=+0.60 . What is the velocity of particle A relative to observer D? (The solution technique here is much faster than using Eq. 37-29.)

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