Spatial separation between two events. For the passing reference frames of Fig. 37-25, events A and B occur with the following spacetime coordinates: according to the unprimed frame,(xA,tA)and role="math" localid="1663045013644" (xB,tB)according to the primed frame,(x'A,t'A) androle="math" localid="1663045027721" (x'B,t'B). In the umprimed frameΔt=tBtA=1.00 μsandrole="math" localid="1663045143133" Δx=xBxA=240 m.(a) Find an expression forin Δx'terms of the speed parameterβand the given data. GraphΔx'versusβfor two ranges ofβ: (b)0 to0.01and (c)0.1to 1. (d) At what value ofβisΔx'=0?

Short Answer

Expert verified

a) The expression for theΔx' is(240 m)(300 m)β1β2 .

(b) The graph betweenΔx' andβ for the value0 and 0.01.

(c)The graph between Δx'andβ for the value0.01 to 1.

(d) The value of βis 0.8.

Step by step solution

01

Write the given data from the question.

The time difference between the two events,

Δt=tBtAΔt=1 μs

The distance,

Δx=xBxAΔx=240 m

02

The formula to calculate the expression for Δt'and graphs between Δt' and speed parameter.

The expression to calculate the Lorentz factor is given as follows.

γ=11β2 …(i)

The equation to calculate the expression for theΔx' is given as follows.

Δx'=γ(ΔxβcΔt) …(ii)

Here, γis the Lorentz factor andc is the speed of light.

03

Calculate the expression for Δx'.

(a)

Calculate the expression forΔx'.

Calculate the expression for Δt'.

Substitute 11β2for γinto equation (ii).

Δx'=11β2(ΔxβcΔt)Δx'=ΔxβcΔt1β2

Substitute240 m forΔx ,3×108 m/s forc and1 μs forΔt into equation (i).

Δx'=240 mβ×3×108 m/s×1×106 s1β2Δx'=(240 m)(300 m)β1β2 …(iii)

Hence the expression for theΔx' is (240 m)(300 m)β1β2.

04

Draw the graph between Δx' and  β for the value 0 to 0.01 .

(b)

The graph betweenΔx' andβ is shown below.

05

Draw the graph between Δx' and β for the value 0.1 to 1.

(c)

The graph betweenΔx' and βis shown below.

06

Calculate the value of the β at which Δx'=0 .

Calculate the value of β.

Substitute0 forΔx' into equation (iii).

240300β1β2=0240300β=0300β=240β=0.8

Hence the value of βis0.8 .

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