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Chapter 38: Problem 60

A train moves at \(10 \mathrm{~km} / \mathrm{h}\) along the track. A passenger sprints toward the rear of the train at \(10 \mathrm{~km} / \mathrm{h}\) with respect to the train. Our knee-jerk motto says that the train clocks "run slow" with respect to clocks on the track, and the runner's watch "runs slow" with respect to train clocks. Therefore the runner's watch should "run doubly slow" with respect to clocks on the track. But the runner is at rest with respect to the track. What gives? (This example illustrates the danger of the simple knee-jerk motto "Moving clocks run slow.")

Short Answer

Expert verified
The runner is at rest relative to the track, so there is no additional time dilation. The runner's watch exhibits time dilation only relative to the train.

Step by step solution

01

Understand the Relative Motion

The train is moving at a speed of 10 km/h relative to the track. The passenger is sprinting towards the rear of the train at 10 km/h relative to the train. Since both are moving at the same speed in opposite directions, the net speed relative to the track is zero.
02

Determine the Runner's Speed Relative to the Track

Since the passenger is running towards the rear of the train at 10 km/h and the train itself is moving at 10 km/h in the opposite direction, these speeds cancel out. Therefore, the passenger's speed relative to the track is: \(0 = 10 \text{ km/h} - 10 \text{ km/h} \)The runner is at rest relative to the track.
03

Understand Time Dilation Concept

According to the theory of relativity, time dilation affects moving clocks. Each observer considers clocks in motion with respect to themselves as running slow. This example illustrates that the situation must be analyzed from a consistent frame of reference.
04

Analyze the Clocks Relative to the Train and Track

For an observer on the track, the train's clocks will appear to run slow due to the train's motion. For the runner, since they are at rest with respect to the track, there should be no additional time dilation aside from the train鈥檚 motion.
05

Clarify the Misconception

The knee-jerk motto 'Moving clocks run slow' suggests that the runner's watch 'runs doubly slow' because it is moving relative to both the train and the track. However, this is incorrect because the runner is not moving relative to the track. Thus, the runner's watch is not subjected to additional time dilation compared to the train's clocks, aside from that train's motion relative to the track.
06

Conclude the Analysis

The runner's watch does not 'run doubly slow' because, from the track's perspective, the runner is stationary relative to the track. Thus, the apparent time dilations do not compound. The correct understanding is that the runner's watch only exhibits time dilation relative to the train as both are in motion relative to each other.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Time Dilation
Time dilation is a fascinating consequence of the theory of relativity. It explains how time can run at different rates for observers moving relative to each other. When an object is moving close to the speed of light, time appears to slow down for it compared to a stationary observer.

For example, if you were travelling on a spaceship at a high speed, your clock would tick slower compared to people remaining on Earth. This effect doesn't become noticeable at everyday speeds like those of cars or trains. However, it becomes significant when approaching light speed.

In the exercise, this concept is crucial. The train passenger's watch does not 'run doubly slow' because the runner is stationary relative to the track. Hence, there is no compounded time dilation as initially suggested.
Theory of Relativity
Einstein鈥檚 theory of relativity revolutionized our understanding of space and time. It comprises two parts: special relativity and general relativity.

Special relativity, which is pertinent here, focuses on how space and time are linked for objects moving at a constant speed in a straight line. It introduces the concept that the laws of physics are the same for all non-accelerating observers, and that the speed of light within a vacuum is the same no matter the speed at which an observer travels.

A key point from special relativity relevant to our exercise is the concept of time dilation. Moving clocks run slow, meaning time slows down for objects in motion when observed from a stationary frame of reference. This theory helps explain the behavior of the clocks from the perspectives of the train, the runner, and an observer on the track.
Reference Frames
A reference frame is a perspective from which measurements such as position, velocity, and time are made. It can be thought of as a coordinate system or a viewpoint. In understanding relative motion, clearly identifying reference frames is crucial.

In the train and runner example, we use three reference frames:
  • The train's reference frame: from which the runner moves towards the rear at 10 km/h.
  • The track's reference frame: from which the train itself moves at 10 km/h.
  • The runner鈥檚 reference frame: considering the runner stationary relative to the track since the net speed is zero.
In each frame, velocities and the perception of time can vary, leading to different conclusions about the relative motion and time dilation.

The key to solving the problem is consistently applying the principles of special relativity within the correct reference frame.
Relative Velocity
Relative velocity is an essential concept in understanding motion in different frames of reference. It measures how fast an object is moving relative to another object.

In the exercise, the train moves at 10 km/h relative to the track. The passenger running towards the rear of the train also moves at 10 km/h but in the opposite direction to the train. When combining these velocities, we find the net speed of the runner relative to the track is zero.

This means that to an observer on the track, the runner is stationary. This is why the runner鈥檚 watch isn鈥檛 'doubly slow' - there is no additional motion to cause extra time dilation besides that caused by the train鈥檚 speed relative to the track.

By understanding relative velocity, you can determine how objects are moving concerning each other correctly and avoid misconceptions.

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Most popular questions from this chapter

The famous twin paradox is often introduced as follows: Two identical twins grow up together on Earth. When they reach adulthood, one twin zooms to a distant star and returns to find her stay-at-home sister much older than she is. Thus far no paradox. But Alexis Allen formulates the Twin Paradox for us: "The theory of special relativity tells us that all motion is relative. With respect to the traveling twin, the Earth-bound twin moves away and then returns. Therefore it is the Earthbound twin who should be younger than the 'traveling' twin. But when they meet again at the same place, it cannot possibly be that each twin is younger than the other twin. This Twin Paradox disproves relativity." The paradox is usually resolved by realizing that the traveling twin turns around. Everyone agrees which twin turns around, since the reversal of direction slams the poor traveler against the bulkhead of the decelerating starship, breaking her collarbone. The turnaround, evidenced by the broken collarbone, destroys the symmetry required for the paradox to hold. Good-bye Twin Paradox! Still, Alexis's father Cyril Allen has his doubts about this resolution of the paradox. "Your solution is extremely unsatisfying. It forces me to ask: What if the retro-rockets malfunction and will not fire at all to slow me down as I approach a distant star a thousand light-years from Earth? Then I cannot even stop at that star, much less turn around and head back to Earth. Instead, I continue moving away from Earth forever at the original constant speed. Does this mean that as I pass the distant star, one thousand light-years from Earth, it is no longer possible to say that I have aged less than my Earth-bound twin? But if not, then I would never have even gotten to the distant star at all during my hundred-year lifetime! Your resolution of the Twin Paradox is insufficient and unsatisfying." Write a half-page response to Cyril Allen, answering his objections politely but decisively.

(a) Find an equation for the unknown mass \(m\) of a particle if you know its momentum \(p\) and its kinetic energy \(K\). Show that this expression reduces to an expected result for nonrelativistic particle speeds. (b) Find the mass of a particle whose kinetic energy is \(K=55.0 \mathrm{MeV}\) and whose momentum is \(p=\) \(121 \mathrm{MeV} / \mathrm{c}\). Express your answer as a decimal fraction or multiple of the mass \(m_{\mathrm{e}}\) of the electron.

Evelvn Brown does not approve of our latticework of rods and clocks and the use of a light flash to synchronize them. (a) "I can synchronize my clocks in any way I choose!" she exclaims. Is she right? (b) Evelyn wants to synchronize two identical clocks, called Big Ben and Little Ben, which are at rest with respect to one another and separated by one million kilometers in their rest frame. She uses a third clock, identical in construction with the first two, that travels with constant velocity between them. As her moving clock passes Big Ben, it is set to read the same time as Big Ben. When the moving clock passes Little Ben, that outpost clock is set to read the same time as the traveling clock. "Now Big Ben and Little Ben are synchronized," says Evelyn Brown. Is Evelyn's method correct? (c) After Evelyn completes her synchronization of Little Ben by her method, how does the reading of Little Ben compare with the reading of a nearby clock on a latticework at rest with respect to Big Ben (and Little Ben) and synchronized by our standard method using a light flash? Evaluate in milliseconds any difference between the reading on Little Ben and the nearby lattice clock in the case that Evelyn's traveling clock moved at a constant velocity of 500000 kilometers per hour from Big Ben to Little Ben. (d) Evaluate the difference in the reading between the EvelynBrown- synchronized Little Ben and the nearby lattice clock when Evelyn's synchronizing traveling clock moves 1000 times as fast as the speed given in part (c).

The half-life of stationary muons is measured to be \(1.6\) microseconds. Half of any initial number of stationary muons decays in one half-life. Cosmic rays colliding with atoms in the upper atmosphere of the Earth create muons, some of which move downward toward the Earth's surface. The mean lifetime of high- speed muons in one such burst is measured to be 16 microseconds. (a) Find the speed of these muons relative to the Earth. (b) Moving at this speed, how far will the muons move in one half-life? (c) How far would this pulse move in one half-life if there were no relativistic time stretching? (d) In the relativistic case, how far will the pulse move in 10 half-lives? (e) An initial pulse consisting of \(10^{8}\) muons is created at a distance above the Earth's surface given in part (d). How many will remain at the Earth's surface? Assume that the pulse moves vertically downward and none are lost to collisions. (Ninety-nine percent of the Earth's atmosphere lies below \(40 \mathrm{~km}\) altitude.)

Jocelyn DeGuia takes off from Earth and moves toward the star Vega, which is 26 ly distant from Earth. Assume that Earth and Vega are relatively at rest and Jocelyn moves at \(v=0.99 c\) in the Earth-Vega frame. How much time will have elapsed on Earth (a) when Jocelyn reaches Vega and (b) when Earth observers receive a radio signal reporting that Jocelyn has arrived? (c) How much will Jocelyn age during her outward trip?
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