36.8.1 Using Relativity

Relativity is phrased in terms of events; hence relativity problems are solved by interpreting the problem statement in terms of specific events.

Problem-Solving Strategy 36.1
Relativity

MODEL Frame the problem in terms of events, things that happen at a specific place and time.

VISUALIZE A pictorial representation defines the reference frames.

  • Sketch the reference frames, showing their motion relative to each other.

  • Show events. Identify objects that are moving with respect to the reference frames.

  • Identify any proper time intervals and proper lengths. These are measured in an object’s rest frame.

SOLVE The mathematical representation is based on the Lorentz transformations, but not every problem requires the full transformation equations.

  • Problems about time intervals can often be solved using time dilation: Δt=γ Δτ.

  • Problems about distances can often be solved using length contraction: L=ℓ/γ.

REVIEW Are the results consistent with Galilean relativity when v≪c?

EXAMPLE 36.7 Ryan and Priya revisited

Priya is standing in the center of a long, flat railroad car that has firecrackers tied to both ends. The car moves past Ryan, who is standing on the ground, with velocity v=0.8c. Flashes from the exploding firecrackers reach him simultaneously 1.0 μs after the instant that Priya passes him, and he later finds burn marks on the track 300 m to either side of where he had been standing.

  1. According to Ryan, what is the distance between the two explosions, and at what times do the explosions occur relative to the time that Priya passes him?

  2. According to Priya, what is the distance between the two explosions, and at what times do the explosions occur relative to the time that Ryan passes her?

MODEL Let the explosion on Ryan’s right, the direction in which Priya is moving, be event R. The explosion on his left is event L.

VISUALIZE Priya and Ryan are in inertial reference frames. As FIGURE 36.30 shows, Priya’s frame S′ is moving with v=0.8c relative to Ryan’s frame S. We’ve defined the reference frames such that Priya and Ryan are at the origins. The instant they pass, by definition, is t=t′=0 s. The two events are shown in Ryan’s reference frame.

FIGURE 36.30 A pictorial representation of the reference frames and events.

The figure illustrates a man on the ground observing a woman on a rail car moving to the right, with explosion events at its ends.

SOLVE

  1. The two burn marks tell Ryan that the distance between the explosions was L=600 m. Light travels at c=300 m/μs, and the burn marks are 300 m on either side of him, so Ryan can determine that each explosion took place 1.0 μs before he saw the flash. But this was the instant of time that Priya passed him, so Ryan concludes that the explosions were simultaneous with each other and with Priya’s passing him. The spacetime coordinates of the two events in frame S are (xR, tR)=(300 m, 0 μs) and (xL, tL)=(−300 m, 0 μs).

  2. We already know, from our qualitative analysis in Section 36.5, that the explosions are not simultaneous in Priya’s reference frame. Event R happens before event L in S′, but we don’t know how they compare to the time at which Ryan passes Priya. We can now use the Lorentz transformations to relate the spacetime coordinates of these events as measured by Ryan to the spacetime coordinates as measured by Priya. Using v=0.8c, we find that γ is

    γ=11−v2/c2=11−0.82=1.667

    For event L, the Lorentz transformations are

    xL′=1.667((−300 m)−(0.8c)(0 μs))=−500 mtL′=1.667((0 μs)−(0.8c)(−300 m)/c2)=1.33 μs

    And for event R,

    xR′=1.667​​​(300 m)−(0.8c)(0 μs)​​​=500 m tR′=1.667​​​(0 μs)−(0.8c)(300 m)/c2​=−1.33 μs

    According to Priya, the two explosions occur 1000 m apart. Furthermore, the first explosion, on the right, occurs 1.33 μs before Ryan passes her at t′=0 s. The second, on the left, occurs 1.33 μs after Ryan goes by.

REVIEW Events that are simultaneous in frame S are not simultaneous in frame S′. The results of the Lorentz transformations agree with our earlier qualitative analysis.

A follow-up discussion of Example 36.7 is worthwhile. Because Ryan moves at speed v=0.8c=240 m/μs relative to Priya, he moves 320 m during the 1.33 μs between the first explosion and the instant he passes Priya, then another 320 m before the second explosion. Gathering this information together, FIGURE 36.31 shows the sequence of events in Priya’s reference frame.

FIGURE 36.31 The sequence of events as seen in Priya’s reference frame.

The figure consists of three parts that illustrate the sequence of explosion events as seen in Priya's frame of reference.
FIGURE 36.31 Full Alternative Text

The firecrackers define the ends of the railroad car, so the 1000 m distance between the explosions in Priya’s frame is the car’s length L′ in frame S′. The car is at rest in frame S′, hence length L′ is the proper length: ℓ=1000 m. Ryan is measuring the length of a moving object, so he should see the car length contracted to

L=1−β2ℓ=ℓγ=1000 m1.667=600 m

And, indeed, that is exactly the distance Ryan measured between the burn marks.

Finally, we can calculate the spacetime interval s between the two events. According to Ryan,

s2=c2(Δt2)−(Δx)2=c2(0 μs)2−(600 m)2=−(600 m)2

Priya computes the spacetime interval to be

s2=c2(Δt′)2−(Δx′)2=c2(2.67 μs)2−(1000 m)2=−(600 m)2

Their calculations of the spacetime interval agree, showing that s really is an invariant, but notice that s itself is an imaginary number.