Question: Temporal separation between two events. Events and occur with the following spacetime coordinates in the reference frames of Fig. 37-25: according to the unprimed frame,andaccording to the primed frame, (xA,tA)and(xB,tB). In the unprimed frame t=tB-tA=1.00μsand Δx=xB-xA=240m. (a) Find an expression for Δt'in terms of the speed parameterβand the given data. Graph Δt' versus βfor the following two ranges of β: (b) 0 to 0.01and 0.1 (c) 0.1 to 1. (d) At what value of βis Δt'minimum and (e) what is that minimum? (f) Can one of these events cause the other? Explain.

Short Answer

Expert verified

Answer

(a) The expression forΔt'is1-0.80β1-β2μs.

(b) The graph betweenΔt'andβfor the value 0 and 0.01

.

(c)The graph betweenΔt' andβ for the value 0.01 to 1.

(d) The value of the βis 0.8.

(e) The minimum value ofΔt' is0.6μs .

(f) The event Acan cause event B .

Step by step solution

01

Write the given data from the question.

The time difference between the two events,

Δt=tB-tAΔt=1μs

The distance,

Δx=xB-xAΔx=240m

02

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

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

γ=11-β2 …(i)

The equation to calculate the expression for theis given as follows.

Δt'=γ(Δt-βΔxc) …(ii)

Here,βis the speed parameter, γis the Lorentz factor and c is the speed of light.

03

Calculate the expression forΔt'  .

(a)

Calculate the expression for Δt'.

Substitute11-β2forγinto equation (ii).

Δt'=11-β2Δt-βΔxc

Substitute 240mforx,1μsforΔtand3×108m/sforcr into above equation.

Δt'=11-β21×10-6s-β×240m3×108Δt'=11-β21×10-6s-80×10-8sβΔt'=11-β21×10-6s-0.80×10-6sβΔt'=1-0.80β1-β2 …(iii)

Hence the expression forΔt' is,1-0.80β1-β2s .

04

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

(b)

The graph between Δt'andβis shown below.

05

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

(c)

The graph betweenΔt' and βis shown below

06

Calculate the value of β at which Δt' is minimum.

(d)

To calculate the value of the take the derivative of the equation (iii),

dΔt'dβ=-0.81-β2-1-0.8β-2β121-β21-β2dΔt'dβ=β1-0.8β-0.81-β21-β23/2dΔt'dβ=β-0.81-β23/2

Substitute 0 for dΔt'dβinto above equation.

0=β-0.81-β23/2β-0.8=0β=0.8

Hence the value of theβ is 0.8 .

07

Calculate the minimum value of Δt' at β=0.8

(e)

Calculate the minimum value of Δt'.

Substitute 0.8 for βinto equation (iii).

Δt'=1-0.80×0.81-0.82Δt'=1-0.640.6Δt'=0.360.6Δt'=0.6μs

Hence the minimum value ofΔt' isrole="math" localid="1663062673499" 0.6μs .

08

8: Determine that can one of these events cause another event

Calculate the velocity,

v=ΔxΔt

Substitute 240mforΔxand1μsforΔtinto above equation.

v=240m1×10-6sv=240×106m/sv=2.4×108m/s

Since the velocity 2.4×108m/sis less than the speed of the light therefore event A can cause the event B.

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