618 lines
66 KiB
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618 lines
66 KiB
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ESSENTIAL
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MODERN PHYSICS
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Study Guide Workbook
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(with Full Solutions)
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Chris McMullen, Ph.D.
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Essential Modern Physics Study Guide Workbook Chris McMullen, Ph.D.
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monkeyphysicsblog.wordpress.com improveyourmathfluency.com chrismcmullen.com
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Copyright © 2019 Chris McMullen, Ph.D. All rights reserved.
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Zishka Publishing ISBN: 978-1-941691-28-1 Textbooks > Science > Physics Study Guides > Workbooks> Science
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CONTENTS
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Introduction 1 Special Relativity Concepts 2 Time Dilation and Length Contraction 3 The Lorentz Transformation 4 Relative Velocity 5 Relativistic Momentum and Energy 6 Blackbody Radiation 7 The Photoelectric Effect 8 The Compton Effect 9 Bohr’s Model 10 The de Broglie Relation 11 Heisenberg’s Uncertainty Principle 12 Differential Equations 13 Schrödinger’s Equation 14 Probabilities and Expectation Values About the Author
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INTRODUCTION
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The goal of this study guide workbook is to provide practice and help carrying out essential problemsolving strategies that are standard in modern physics. The aim here is not to overwhelm the student with comprehensive coverage of every type of problem, but to focus on the main strategies and techniques with which most physics students struggle. This workbook is not intended to serve as a substitute for lectures or for a textbook, but is rather intended to serve as a valuable supplement. Each chapter includes a concise review of the essential information, a handy outline of the problem-solving strategies, and examples which show step-bystep how to carry out the procedure. This is not intended to teach the material, but is designed to serve as a time-saving review for students who have already been exposed to the material in class or in a textbook. Students who would like more examples or a more thorough introduction to the material should review their lecture notes or read their textbooks. Every problem in this study guide workbook applies the same strategy which is solved step-bystep in at least one example within the chapter. Study the examples and then follow them closely in order to complete the exercises. Many of the exercises are broken down into parts to help guide the student through the exercises. Each exercise tabulates the corresponding answers on the same page. Students can find full solutions at the end of each chapter. The prerequisites for using this workbook include first-year physics (including energy, waves, and electricity and magnetism) and calculus (including derivatives and integrals). Although Schrödinger’s equation in quantum mechanics is a differential equation, students do not need previous exposure to differential equations: This workbook provides a concise introduction to basic differential equations in Chapter 12, and shows how to apply these methods in the remaining chapters.
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1 SPECIAL RELATIVITY CONCEPTS
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Relevant Terminology
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Galilean relativity – our experience with the relative motion of objects traveling at speeds much slower than the speed of light.
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Special relativity – the physics of the relative motion of objects where at least one object is traveling at a very high speed (compared to the speed of light).
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Ether – a hypothetical substance once believed to fill space; it was believed to serve as a medium for the transmission of light waves.
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Photon – a single particle in a beam of light.
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Interferometer – a device involving the interference of two beams of light, which was used by Michelson and Morley to measure the speed of light relative to the earth.
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Time dilation – the phenomenon whereby time appears to travel more slowly for objects moving fast (close to light speed) relative to other observers.
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Length contraction – the phenomenon whereby objects moving fast (close to light speed) appear shorter relative to other observers.
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Simultaneity – when two events occur at the exact same moment relative to an observer, the events are said to be simultaneous for that observer.
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Momentum – mass times velocity.
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Inertia – the natural tendency of any object to maintain constant momentum.
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Mass – a measure of inertia.
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Vacuum – a region of space completely devoid of matter (it doesn’t even contain air).
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Inertial reference frame – a frame that travels with constant velocity.
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Galilean Relativity
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In our everyday experience with objects that travel much slower than the speed of light, relative velocities appear to obey the formula for vector addition. Suppose that one observer (designated R) is at rest while a second observer (designated M) is moving with speed v relative to the first observer. Suppose also that each observer sets up a coordinate system with the x-axis along the relative velocity
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If each observer measures the velocity of an object, the x-components of the velocities that they measure (uR and uM) are related by the following vector addition formula, provided that neither
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observer nor the object are traveling close to light speed.
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uM = uR – v
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This equation is actually pretty simple: It’s just subtraction. The challenge is to remember what the notation means (M stands for moving, while R stands for rest) so that you can apply the equation correctly. We will explore this equation further in the examples that follow.
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The Mysterious Ether
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Physicists once believed in a hypothetical substance called the ether, which was believed to permeate all of space. At the time, all other waves besides light were known to travel in a medium. You can see ripples travel along the surface of water. Sound waves create alternating regions of compression (high pressure) and rarefaction (low pressure) in a medium such as air, water, wood, or metal. Light was also known to be a wave, yet sunlight can travel through space (a near-perfect vacuum). Since all other waves required a medium in which to propagate, the concept of the ether could explain the transmission of light through space.
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It turns out that the ether hypothesis is incorrect, as demonstrated by the Michelson-Morley experiment. Light can travel through a perfect vacuum (without an ether).
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The Earth, Light, and the Hypothetical Ether
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The result of the Michelson-Morley experiment—that the ether doesn’t exist—came as a big shock to the physics community at the time. To understand why, we must explore the ether as it had been believed to exist. The ether was believed to permeate all of space. The reference frame of the ether was believed to serve as an absolute reference frame. That is, the speed of light was only believed to
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travel c = 2.9979 × 108 m/s in a reference frame that was at rest relative to the ether. Furthermore, it was believed that the vector addition equation of Galilean relativity applied to objects moving any speed, including light itself. (Like the ether, this also proved to be incorrect.)
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The earth orbits the sun and therefore must be moving relative to the hypothetical ether (as the direction of earth’s velocity is constantly changing). From the point of view of the ether, the earth is moving relative to the ether with instantaneous speed v. From the point of view of earthlings, the earth seems to be stationary and we would instead interpret the ether to be moving with speed v. (You should have experience with this. If you are sitting in a bus that is moving, objects outside of the bus appear to be moving relative to you.) From the perspective of “stationary” earthlings, the ether is seen as an “ether wind” (when you run through air that is originally still, the air seems to pass by you like a sort of wind).
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Imagine that you get in a motorboat and travel along the surface of a river. The boat would travel with a speed of 30 m/s on a still pond, but there is a river current of 10 m/s. When the boat is headed downstream, it would be traveling 30 + 10 = 40 m/s relative to the land. When the boat is headed upstream, it would be traveling 30 – 10 = 20 m/s relative to the land. When the boat is headed crossstream, apply the Pythagorean theorem to determine that the boat travels
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relative to the land.
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The same principle as the motorboat and river was believed to apply to the earth traveling through the hypothetical ether. Imagine that you shine a beam of light from earth and wish to measure the speed of
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light. According to the ether hypothesis, the light would travel with speed c = 2.9979 × 108 m/s relative to the ether. The earth travels with speed v = 30,000 m/s relative to the sun. As the earth orbits the sun, the direction of its velocity constantly changes. The speed of light would equal c + v when the light is heading “downwind” (when the earth happens to be traveling along the ether wind), the speed of light would equal c – v when the light is heading “upwind,” and the speed of light would
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equal
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when the light is heading “across wind.” However, when Michelson and Morley investigated this, no such changes in the speed of light were detected. The laws of Galilean relativity do not apply to light or to objects moving fast (close to light speed). The ether does not exist. The speed of light turned out to be a universal constant (independent of the motion of the observer or source).
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Why c + v, c – v, and
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Seemed to Make Sense
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Imagine a father and son playing catch on a boat that is at rest. They throw the ball with a speed of 20 m/s east and west. They continue to play catch the same way even when the boat starts traveling 50 m/s to the east. When the ball travels 20 m/s east relative to the boat, it is traveling 50 + 20 = 70 m/s to the east relative to the land. When the ball travels 20 m/s to the west relative to the boat, it is traveling 50 – 20 = 30 m/s to the east relative to the land. This is the way relative velocities work at speeds that are small compared to the speed of light. It agrees with everyday experience.
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Now imagine a spaceship traveling 0.7c (or 70% the speed of light) relative to earth. The spaceship suddenly turns on its headlights. How fast is the beam of light traveling relative to earth? Based on our experience with low speeds (Galilean relativity), it’s intuitive to expect the beam of light to be traveling 0.7c + c = 1.7c relative to the earth. But that’s not what happens. It turns out that the beam of
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light travels c = 2.9979 × 108 m/s relative to the ship and that the beam of light also travels c =
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2.9979 × 108 m/s relative to earth, with no contradiction! Although c + v seemed to make sense, it turned out to contradict experiments.
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The Michelson-Morley Experiment
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Michelson and Morley used an interferometer that was designed to detect small changes in the speed of light. When an incident beam of light reached a glass slab, it split into two: One beam reflected from the surface of the glass towards mirror M1, and a second beam refracted through the
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glass towards mirror M2. After reflecting off mirrors M1 and M2, the beams met back up at the glass,
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and light from each beam reached a telescope, where an interference pattern was viewed. The entire apparatus could be rotated.
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The two beams would travel in different directions relative to the ether, and thus the two beams would have different speeds relative to the earth if the ether hypothesis were true. For example, if the beam heading toward M1 were traveling upwind with speed c – v (in which case it would then travel
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downwind with speed c + v after reflection), then the beam heading toward M1 would be traveling
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crosswind with speed
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Thus, the two beams would return to the glass at different times, creating an interference pattern when viewed through the telescope (since the beams would be slightly out of phase due to the time lag).
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However, the Michelson-Morley experiment failed to detect any change in the speed of light as the apparatus was rotated. It turns out that light (which is an electromagnetic wave) does not require a medium (such as the ether) in order to propagate. Light can travel through a perfect vacuum. There is no preferred or absolute reference frame for measuring the speed of light. It turns out that the speed of light is a universal constant, independent of the motion of the source or the observer.
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The Problem
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The Michelson-Morley experiment contradicted the expectations of the ether hypothesis. How could the speed of light be the same in each beam of the interferometer, regardless of the orientation of the apparatus?
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If a boat is traveling north, it can launch a cannonball farther to the north (relative to the land) than it could if the boat were at rest. This is the principle of vector addition applied in Galilean relativity, which agrees with human experience (with objects traveling much slower than the speed of light).
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Imagine a spaceship traveling one-half the speed of light (0.5c) relative to earth. Also imagine a beam of light traveling parallel to the spaceship. An observer on earth measures the speed of the beam of
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light to be c = 2.9979 × 108 m/s. What will an observer on the spaceship measure the speed of the beam of light to be? According to the Michelson-Morley experiment, the answer is the same: c =
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2.9979 × 108 m/s. (Note that the answer isn’t 0.5c.)
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Your experience with relative motion at low speeds is much different. If you’re sitting in a bus traveling 40 m/s and a car passes you traveling 50 m/s, each second the car gets 10 m further ahead of the bus, so the car seems to be traveling 10 m/s relative to you. However, if you’re in a spaceship and proceed to measure the speed of light, the Michelson-Morley experiment shows that you will get c =
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2.9979 × 108 m/s regardless of how fast the ship is traveling. Even if the spaceship travels 0.99c
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relative to earth, an observer on the ship still measures the speed of light to be c = 2.9979 × 108 m/s (and not 0.01c).
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Albert Einstein introduced his theory of special relativity to resolve this seeming paradox, but it came with some interesting consequences: time dilation and length contraction. The underlying issue is that time, space, and light behave much differently than our everyday experience with low-speed motion suggests. The two different observers (in the spaceship and on earth) actually measure distance and time differently due to time dilation and length contraction. If there is a meterstick on the spaceship and an observer on earth proceeds to measure the length of that meterstick as the spaceship travels very fast (close to light speed), the observer on earth measures the meterstick to be significantly less than one meter long. Similarly, if there is a pendulum on the spaceship that oscillates with a period of exactly one second relative to an observer on the spaceship, an observer on earth measures the period of that same pendulum to last significantly longer than one second. When the relative speed between two observers is in the neighborhood of the speed of light, the observers significantly disagree on such basic notions as what meters and seconds are! As bizarre as this may seem, Einstein’s theory of special relativity not only explains the Michelson-Morley experiment, it also agrees with countless other scientific tests.
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Einstein’s Theory of Special Relativity
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The theory of special relativity applies to objects that move with constant velocity (meaning that they move with constant speed and also travel in a straight line). Einstein developed his theory of special relativity from two fundamental postulates:
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1. The laws of physics are the same in any inertial reference frame. (An inertial reference frame is any coordinate system that has constant velocity. Recall that constant velocity means both constant speed and traveling in a straight line.) This means that any physics experiment will yield the same results in any laboratory that travels with constant velocity, whether the laboratory is at rest on earth or moving in a spaceship with a velocity of 0.8c relative to the earth.
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2. Any observer in an inertial reference frame would measure the speed of light to be c = 2.9979
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× 108 m/s, regardless of the velocity of the observer and also regardless of the velocity of the light source.
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One consequence of special relativity is that there isn’t any preferred (or absolute) reference frame (such as an ether). Any inertial reference frame is equally as good as any other.
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What does it mean to be at rest? You can be at rest relative to earth, but earth is revolving around the sun. Even if an object is at rest relative to the sun, the sun is traveling through space relative to other stars. Since the laws of physics are the same in all inertial reference frames, it would be impossible to find a particular reference frame that you could say is truly at “rest.” As long as you’re moving with constant velocity, you’re entitled to consider yourself to be at “rest” and to consider everything else moving relative to you. Even if you’re in a bus that is traveling 30 m/s west past a station, you may consider yourself to be at rest and may consider the station as traveling 30 m/s to the east (opposite to you). A woman at the station may consider herself to be at rest and consider you to be traveling 30 m/s to the west. You are both entitled to be correct when working out physics with your own inertial reference frames. According to Einstein, it’s all relative.
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As we will explore mathematically in Chapters 2-5, the theory of special relativity comes with a few seemingly strange consequences. In particular, even seemingly fundamental concepts like length and time are relative, and different observers in different inertial reference frames measure length and time differently:
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• Two events that appear to occur simultaneously in one inertial reference frame may appear to occur at different times in another inertial reference frame.
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• Length contraction: When an object is moving relative to an inertial reference frame, the object appears shorter (along the direction of motion) than it does relative to an inertial reference frame that is at rest relative to the object.
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• Time dilation: Time passes more slowly on a clock in a moving inertial reference frame than it does for an inertial reference frame that is at rest relative to the clock.
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Simultaneity Is Relative
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One consequence of the postulates of special relativity is that whether or not two events appear to occur simultaneously (at the same time) depends on the inertial reference frame from which the events are observed. If two events are observed to occur simultaneously in one inertial reference frame, they may not be observed to occur simultaneously in another inertial reference frame.
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For example, consider the spaceship traveling close to the speed of light in the illustration above. The spaceship passes very close to a space station that is orbiting the earth. The space station is traveling so much slower than the speed of light that it is practically at rest relative to the very fast spaceship. One astronaut is stationed at point Q on the space station, while another astronaut is stationed at point P on the spaceship. There are lights on the space station at the points labeled 1 and 2. These two lights remain off almost all of the time. The lights are programmed to flash quickly at the exact instant that the two ends of the spaceship happen to be (momentarily) positioned directly above the lights. At this exact instant, point A is directly above point 1, point B is directly above point 2, and points P and Q are exactly midway between points 1 and 2 (and are thus midway between points A and B).
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Observer Q is at rest relative to the space station and is exactly midway between points 1 and 2, such that the light emitted by each point during the flash travels the same distance to reach observer Q. Thus, observer Q on the space station sees the two lights flash simultaneously. In contrast, observer P is traveling close to the speed of light relative to points 1 and 2. Light from point 2 reaches observer P before light from point 1. Observer P doesn’t see the two lights flash simultaneously. Which observer is correct? According to special relativity, both are correct. Whether you are practically at rest (like the space station) or traveling close to the speed of light relative to the earth (like the spaceship), the laws of physics are the same. There isn’t a preferred inertial reference frame that makes observations more “correct.”
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Time Dilation
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Time passes more slowly on a clock in a moving inertial reference frame than it does for an inertial reference frame that is at rest relative to the clock. This is known as time dilation.
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You can see how time dilation is a direct consequence of the postulates of special relativity by considering the illustration above. The spaceship is traveling to the right with a speed that is close to the speed of light. The spaceship passes by a space station that is orbiting the earth. The space station is traveling so much slower than the speed of light that it is practically at rest relative to the very fast spaceship. As the spaceship is passing the space station, an astronaut inside of the spaceship turns on a flashlight, shining a beam of light straight upward in the diagram (perpendicular to the direction that the spaceship is traveling). The spaceship has transparent walls such that an observer inside of the space station can see the beam of the flashlight shining inside of the spaceship as it passes the space station.
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Relative to the astronaut inside of the spaceship (observer P), the flashlight beam appears to travel straight upward, as shown in the left diagram above. Relative to an observer inside of the space station (observer Q), the flashlight beam appears to travel diagonally up and to the right, as shown in the right diagram above. (Of course, the photons in the beam of light have inertia, which is the natural tendency of all objects to travel with constant momentum. If you ride in an airplane traveling 500 mph and throw a ball straight upwards, you will catch the ball in your hand because of inertia. The ball certainly won’t smack the back of the airplane mid-flight. If you’ve forgotten about inertia, it may help to review an introductory physics textbook.)
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Observer P sees the beam of light take a shorter path (distance LP, which is straight upward), whereas
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observer Q sees the beam of light take a longer path (distance LQ, which is diagonal). According to
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the second postulate of special relativity, both observers must measure the speed of light to be the same value. Either observer takes the distance (L) traveled and divides by the corresponding time (t) measured to determine the speed of light. The subscripts P and Q indicate which observer makes the measurement.
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Since LQ > LP, in order for both observers to measure the same value for the speed of light, it follows
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that tQ > tP. Since observer P is at rest relative to the flashlight, while observer Q is moving relative
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to the flashlight, the inequality tQ > tP means that time travels more slowly for clocks (and observers
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and objects) that are moving relative to an event. This is called time dilation. We will explore the mathematics of time dilation further in Chapter 2.
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The effect is mutual. Note that observer Q is moving relative to P, but also that observer P is moving relative to Q. Therefore, time dilation depends on your perspective:
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• Relative to observer P on the spaceship, time passes more slowly for observer Q who is on the space station. This is what we found in our example since we let observer P shine the flashlight inside of the spaceship.
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• Relative to observer Q on the space station, time passes more slowly for observer P who is on the spaceship. If we had let observer Q shine the flashlight inside of the space station (instead of letting observer P shine the flashlight inside of the spaceship), we would have found that the time measured by observer P would have been longer.
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Length Contraction
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When an object is moving relative to an inertial reference frame, the object appears shorter (along the direction of motion) than it does relative to an inertial reference frame that is at rest relative to the object. This is known as length contraction.
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You can see how length contraction comes about by considering the illustration above. The spaceship is traveling from earth’s sun to Alpha Centauri with a speed that is close to the speed of light. Observer S is an astronaut aboard the spaceship, while observer E is stationed on earth. Each observer proceeds to measure the distance from the sun to Alpha Centauri (L) and the time of the trip (t).
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Observer E on earth measures the distance to be LE and the time to be tE, such that the speed of the
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spaceship is v = LE/tE. Since the earth is “moving” relative to the spaceship, time is dilated for
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observer E, meaning that tE > tS. Observer S is at “rest” relative to the spaceship, and sees Alpha
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Centauri getting closer to the spaceship with the same speed v. (The perspective is differentwhether earth is at rest and the spaceship is moving, or whether the spaceship is at rest and the stars are moving—but the speed is the same.) Observer S measures the distance to be LS and the time to be
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tS such that v = LS/tS.
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Since observer S is at “rest” relative to the spaceship, time passes “normally” for observer S, such that tS < tE. If tS < tE, how can v = LS/tS and v = LE/tE both result in the same speed? The answer is
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that LS < LE. This means that the distance between the stars appears shorter for observer S than it
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does for observer E. Note that observer S is moving relative to the distance between the stars. Therefore, distance is shorter (along the direction of motion) relative to an inertial reference frame that it is moving relative to the distance than it is for a reference frame that is at rest relative to the distance. This is known as length contraction. We will explore the mathematics of length contraction further in Chapter 2.
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The effect is mutual. Note that observer S is moving relative to E, but also that observer E is moving relative to S. Therefore, length contraction depends on your perspective:
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• Relative to observer S on the spaceship, distances measured along the direction of the spaceship’s motion (such as the distance between the sun and Alpha Centauri) appear shorter than they do for observer E on earth. This is what we found in our example since observer S is moving relative to the distance between the two stars.
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• Relative to observer E on earth, the length of the spaceship is shorter than it is for observer S on the spaceship. If we had considered measurements of the length of the spaceship (instead of the distance between the stars) in our example, this is what we would have found.
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• Relative to observer S on the spaceship traveling from the sun to Alpha Centauri, a second spaceship parked on earth would appear shorter (along the line connecting the two stars) than it is for observer E on earth.
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Note that the effects of time dilation and length contraction that we discussed in this example involve two different perspectives:
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• For observer E on earth, v = LE/tE, we noted that time was dilated (tE > tS) for observer E
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because the earth is “moving” relative to the spaceship.
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• For observer S on the spaceship, v = LS/tS, we noted that length was contracted (LS < LE) for
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observer S because the spaceship is “moving” relative to the two stars.
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Relativistic Mass
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As the relative speed between two inertial reference frames gets closer to the speed of light, the effects of special relativity—including time dilation and length contraction—become more pronounced. In the limit that the relative speed approaches the speed of light, time slows down to a complete stop and length contracts to zero. However, you can’t actually reach this limit. An object that has mass can be accelerated to nearly light speed (like 0.99c or 0.999c), but can never reach the speed of light exactly.
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In Galilean relativity, it would be very easy to accelerate an object faster than the speed of light. According to Newton’s second law of motion, the net force acting on an object equals the object’s mass times its acceleration:
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(This equation applies to objects that have constant mass, but that’s not a problem for Galilean relativity, where ordinarily an object’s mass isn’t expected to change while it accelerates.) The mass
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of an electron is 9.1 × 10–31 kg. If you applied a force of just 1 N (one Newton) to an electron, according to Galilean relativity, the electron would experience an acceleration of
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What does this acceleration mean? It means that starting from rest, after a just 1 s (one second), an
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electron would have a speed of v = 1.1 × 1030 m/s (since acceleration describes the rate at which velocity increases). That’s way, way faster than the speed of light in vacuum, which is c = 2.9979 ×
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108 m/s.
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In the laboratory, it doesn’t happen that way. Although it is very easy to accelerate electrons to very high speeds, once the speed of an electron reaches the neighborhood of the speed of light in vacuum, it becomes increasingly harder to accelerate the electron. We can accelerate electrons up to 0.9c or even 0.99c, but trying to reach 0.99999c is extremely difficult. Why? According to Einstein’s theory of relativity, mass isn’t constant: The faster an electron travels, the more mass the electron has (relative to the laboratory). We call this relativistic mass.
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Recall that mass is a measure of inertia in the following sense: The more mass an object has, the more difficult it is to overcome the object’s inertia in order to accelerate the object. The relativistic mass of an object describes the object’s inertia. As the object travels closer to the speed of light (relative to an inertial reference frame), the greater its relativistic mass (and relativistic inertia): It becomes harder and harder to accelerate the object.
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There are two types of mass:
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• An object’s rest mass tells you how difficult it is to accelerate the object relative to an observer for which the observer and object are both at rest.
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• An object’s relativistic mass tells you how difficult it is to accelerate the object relative to an observer for which the object is traveling close to the speed of light.
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Proper Time, Proper Length, and Rest Mass
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When you compare measurements of time, distance, or mass made by observers in different inertial reference frames, it’s important to be able to determine which of the measurements will be larger and which will be smaller. If you can identify the proper time, the proper length, and the rest mass properly, this will help with your comparisons.
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• The proper time corresponds to a time interval measured by a clock that is at rest relative to the events. Observers who are moving relative to the events measure a greater time interval due to time dilation.
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• The proper length corresponds a distance measured by a tape measure (or other device used for measuring distance) that is at rest. Observers who are moving relative to the distance measure a shorter distance due to length contraction.
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• The rest mass corresponds to the mass of an object measured by an observer who is at rest relative to the object. Observers who are moving relative to the object measure a larger mass called relativistic mass.
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Note: In some relativity questions, the proper time and proper length are not measured by the same observer: Proper time and proper length may come from different perspectives (that is, two different inertial reference frames).
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||
Symbols and SI Units
|
||
|
||
|
||
|
||
|
||
Constants
|
||
|
||
|
||
|
||
|
||
Strategy for Solving Galilean Relativity Problems
|
||
To solve a problem involving Galilean relativity (which applies to problems with relative speeds that are small compared to the speed of light in vacuum), follow these steps:
|
||
• Setup a coordinate system with the x-axis along the direction of the relative velocity,
|
||
• Galilean relativity involves vector subtraction. For one-dimensional problems, simply subtract according to the following equation:
|
||
uM = uR – v
|
||
•• uR is the velocity of an object as measured by an observer at rest (called R).
|
||
•• uM is the velocity of an object as measured by an observer (called M) that is moving relative
|
||
to the observer that is at rest.
|
||
•• v is the relative speed between the two observers.
|
||
|
||
|
||
|
||
|
||
Strategy for Solving Conceptual Special Relativity Problems
|
||
To solve a conceptual problem involving time dilation, length contraction, or relativistic mass, follow these steps. (For mathematical problems, see Chapters 2-5.)
|
||
• It may help to draw a diagram and label both objects and observers.
|
||
• The laws of special relativity apply to inertial observers—observers who travel with (approximately) constant velocity (which means that speed and direction are both constant). Any inertial observer is free to claim to be at rest, and can make equally valid arguments based on this claim.
|
||
• When applying time dilation, identify the proper time: The proper time is measured by an observer whose clock is at rest relative to the events. Any observer who is moving relative to the events will measure a longer time due to time dilation.
|
||
• When applying length contraction, identify the proper length: The proper length is measured by an observer who is at rest relative to the distance being measured. Any observer who is moving relative to the distance will measure a shorter distance due to length contraction.
|
||
• When working with relativistic mass, identify the rest mass: The rest mass is measured by an observer who is at rest relative to the object. Any observer who is moving relative to the object will measure a greater mass called relativistic mass.
|
||
|
||
|
||
Example: A monkey is riding on a boat that is traveling 12 m/s to the east along a river. As the boat passes a boy who is standing on the bank of the river, the monkey throws a banana 18 m/s to the east relative to the monkey. How fast is the banana moving relative to the boy?
|
||
Setup a coordinate system with +x directed to the east. Identify the given information:
|
||
• The relative speed between the monkey and the boy is v = 12 m/s.
|
||
• The monkey is the moving observer. The velocity of the banana relative to the monkey is uM =
|
||
18 m/s to the east.
|
||
• The boy is at rest. The velocity of the banana relative to the boy is uR.
|
||
Since these speeds are small compared to the speed of light, we may apply the equation for Galilean relativity:
|
||
The banana is moving 30 m/s relative to the boy.
|
||
Note: It is standard in physics to neglect air resistance unless stated otherwise in a problem.
|
||
|
||
|
||
Example: A monkey is riding on a boat that is traveling 12 m/s to the east along a river. As the boat passes a boy who is standing on the bank of the river, the monkey throws a banana 18 m/s to the west relative to the monkey. How fast is the banana moving relative to the boy?
|
||
Compare these two examples carefully. What is different? This time the monkey throws the banana to the west, which is opposite to the boat’s motion. Setup a coordinate system with +x directed to the east. Identify the given information:
|
||
• The relative speed between the monkey and the boy is v = 12 m/s.
|
||
• The monkey is the moving observer. The velocity of the banana relative to the monkey is uM =
|
||
–18 m/s. Velocity includes direction: For one-dimensional problems, minus signs distinguish between forward and backward.
|
||
• The boy is at rest. The velocity of the banana relative to the boy is uR.
|
||
Since these speeds are small compared to the speed of light, we may apply the equation for Galilean relativity:
|
||
The banana is moving 6 m/s relative to the boy. Since uR is negative, the banana is heading west
|
||
(along –x) relative to the boy.
|
||
|
||
|
||
Example: A monkey drives a blue car 30 m/s to the north. On the same street, another monkey drives a red car 20 m/s to the south. What is the velocity of the red car relative to the monkey in the blue car?
|
||
Don’t overthink it. This problem is simpler, since there isn’t a banana (or other object) moving relative to both observers. We’re just trying to determine the relative velocity
|
||
between the two monkeys. You can reason this out as follows:
|
||
• In one second, the blue car will travel 30 m north relative to the ground.
|
||
• In one second, the red car will travel 20 m south relative to the ground.
|
||
• Thus, in one second, the red car will be 20 + 30 = 50 m south of the blue car.
|
||
• Since the red car will be 50 m further south of the blue car each second, the velocity of the red car is –50 m/s relative to the blue car, meaning 50 m/s to the south.
|
||
|
||
|
||
Example: One astronaut is in a space station that is orbiting the earth. Another astronaut is in a spaceship. The space station is practically at rest relative to the very fast spaceship, and the spaceship is traveling close to the speed of light relative to the earth. The spaceship and space station both have transparent walls such that either astronaut can make observations (with the aid of a telescope) of what’s going inside of the other space craft.
|
||
(A) How does the astronaut inside of the spaceship appear to age relative to the astronaut inside of the space station?
|
||
Identify the proper time. The astronaut on the spaceship measures the proper time because this astronaut is at rest relative to himself (or herself). An observer who is moving relative to the spaceship will measure a longer time due to time dilation. Therefore, relative to the astronaut on the space station, the astronaut on the spaceship appears to age more slowly than the astronaut on the space station. (Let’s not worry about make-up, genetic differences in aging, etc. Let’s treat the astronauts equally, as if they are identical twins.)
|
||
(B) Both astronauts measure the length of the spaceship. Compare their measurements.
|
||
Identify the proper length. The astronaut on the spaceship is at “rest” relative to the distance being measured. An observer who is moving relative to the spaceship will measure a shorter distance due to length contraction. Therefore, the spaceship appears to be shorter relative to the astronaut inside of the space station than it does relative to the astronaut inside of the spaceship.
|
||
(C) Both astronauts measure the mass of the spaceship. Compare their measurements.
|
||
The astronaut in the spaceship measures the spaceship’s rest mass, whereas the astronaut in the space station measures the spaceship’s relativistic mass, which is larger.
|
||
|
||
|
||
|
||
|
||
Chapter 1 Problems
|
||
1. A monkey is standing on the top of a train that is traveling 36 m/s to the south. A girl is standing on the ground beside the railroad tracks. As the train passes the girl, the monkey throws an apple with a speed of 12 m/s relative to the train.
|
||
(A) If the monkey throws the apple to the south, what is the speed of the apple relative to the girl?
|
||
(B) If the monkey throws the apple to the north, what is the speed of the apple relative to the girl?
|
||
Want help? Check the solution at the end of the chapter.
|
||
Answers: 1. (A) 48 m/s (B) 24 m/s
|
||
|
||
|
||
2. A monkey is riding inside of a train car that is traveling to the west with constant velocity. A girl is standing on the ground beside the railroad tracks. The monkey paints a red X on the floor of the train car (that is, the X is on the bottom of the train car, not on the ground). As the train car is passing the girl, the monkey holds a banana directly over the X and releases the banana. The train car has large windows (with no curtains), such that the girl is able to watch the banana fall and see where it lands.
|
||
(A) Where does the banana land? Explain your answer.
|
||
(B) What path does the banana take relative to the monkey?
|
||
(C) What path does the banana take relative to the girl?
|
||
Want help? Check the solution at the end of the chapter.
|
||
Answers: 2. (A) on the X
|
||
(B) straight line (C) parabola
|
||
|
||
|
||
3. On a still pond, a monkey rides a boat 32 m/s to the east. This monkey throws an orange 20 m/s to the east relative to himself. The monkey’s father rides a boat 24 m/s to the east, the monkey’s mother is standing still on a pier, and the monkey’s uncle rides a boat 10 m/s to the west.
|
||
(A) Find the velocity of the monkey relative to his father.
|
||
(B) Find the velocity of the monkey relative to his mother.
|
||
(C) Find the velocity of the monkey relative to his uncle.
|
||
(D) Find the velocity of the monkey’s father relative to the monkey.
|
||
(E) Find the velocity of the monkey’s mother relative to the monkey.
|
||
(F) Find the velocity of the monkey’s uncle relative to the monkey.
|
||
(G) Find the velocity of the monkey’s father relative to the monkey’s mother.
|
||
(H) Find the velocity of the monkey’s mother relative to the monkey’s father.
|
||
(I) Find the velocity of the monkey’s father relative to the monkey’s uncle.
|
||
(J) Find the velocity of the monkey’s uncle relative to the monkey’s father.
|
||
(K) Find the velocity of the monkey’s mother relative to the monkey’s uncle.
|
||
(L) Find the velocity of the monkey’s uncle relative to the monkey’s mother.
|
||
(M) Find the velocity of the orange relative to the monkey.
|
||
(N) Find the velocity of the orange relative to the monkey’s father.
|
||
(O) Find the velocity of the orange relative to the monkey’s mother.
|
||
(P) Find the velocity of the orange relative to the monkey’s uncle.
|
||
Want help? Check the solution at the end of the chapter.
|
||
Answers: 3. (A) 8 m/s E (B) 32 m/s E
|
||
(C) 42 m/s E (D) 8 m/s W (E) 32 m/s W
|
||
(F) 42 m/s W (G) 24 m/s E (H) 24 m/s W
|
||
(I) 34 m/s E (J) 34 m/s W (K) 10 m/s E
|
||
(L) 10 m/s W (M) 20 m/s E (N) 28 m/s E
|
||
|
||
|
||
(O) 52 m/s E (P) 62 m/s E
|
||
|
||
|
||
4. Two identical spaceships travel with a relative speed close to the speed of light. As shown below, the spaceships travel in opposite directions. At the exact moment that the spaceships pass one another, scientists aboard each ship create identical clones of a chimpanzee. In this way, the two chimpanzees are effectively identical twins. Their names are Marco and Polo.
|
||
(A) Scientists aboard Marco’s ship have one photograph of Marco on his 30th birthday, and compare it to a photograph taken with a telescope of exactly how Polo looked after 30 years according to calendars kept by Marco’s scientists (after accounting for the time it took for light to reach Marco’s ship from Polo’s ship). How do the photographs compare?
|
||
(B) Scientists aboard Polo’s ship have one photograph of Polo on his 30th birthday, and compare it to a photograph taken with a telescope of exactly how Marco looked after 30 years according to calendars kept by Polo’s scientists (after accounting for the time it took for light to reach Polo’s ship from Marco’s ship). How do the photographs compare?
|
||
(C) Which team of scientists is correct? Explain your answers to parts (A) and (B).
|
||
(D) Which ship appears longer according to measurements made by Marco’s scientists?
|
||
(E) Which ship appears longer according to measurements made by Polo’s scientists?
|
||
(F) Which ship has more mass according to measurements made by Marco’s scientists?
|
||
(G) Which ship has more mass according to measurements made by Polo’s scientists?
|
||
(H) Which ship appears taller according to measurements made by Marco’s scientists?
|
||
Want help? Check the solution at the end of the chapter.
|
||
Answers: 4. (A) Polo looks younger than 30
|
||
(B) Marco looks younger than 30
|
||
(C) both! (D) Marco’s (E) Polo’s
|
||
(F) Polo’s (G) Marco’s (H) same
|
||
|
||
|
||
5. As illustrated below, when a spaceship is parked at rest inside of a space garage, the ship is too long to fit inside of the garage.
|
||
As illustrated below, when the same spaceship is traveling close to the speed of light, it can momentarily fit inside of the same space garage according to observers stationed in the space garage. (This space garage is special: There are front and back doors which sense the presence of the spaceship, and which open or close almost instantly. Of course, the spaceship will only be inside of the space garage with both doors shut for a tiny fraction of a second relative to observers stationed in the space garage.)
|
||
(A) Explain how the spaceship is able to fit inside of the space garage with both doors closed simultaneously (for a tiny fraction of a second) from the point of view of observers stationed in the space garage.
|
||
(B) From the point of view of astronauts inside of the spaceship, is the spaceship able to fit inside of the space garage with both doors closed simultaneously? Explain.
|
||
Want help? Check the solution at the end of the chapter.
|
||
Answers: 5. (A) length contraction
|
||
(B) no; the doors do not appear to be closed simultaneously relative to observers in the spaceship
|
||
|
||
|
||
6. When cosmic rays interact with atoms high in earth’s atmosphere, particles called muons can be produced. Muons are particles similar to electrons, except that they have about 200 times as much mass and are unstable. Muons decay very quickly: A muon produced at rest only lasts a couple of microseconds before decaying into other particles. Based on the short average lifetime of a muon (as measured in the muon’s rest frame), almost none of the muons produced in the upper atmosphere should reach earth’s surface, yet a very large number of these muons are detected at earth’s surface.
|
||
(A) From the point of view of scientists stationed on the ground, explain how a large number of muons are able to reach earth’s surface even though their average lifetime (as measured in the muon’s rest frame) is too short for them to survive that long.
|
||
(B) Now explain this from the reference frame of the muons (instead of the reference frame of scientists stationed on the ground).
|
||
Want help? Check the solution at the end of the chapter.
|
||
Answers: 6. (A) time dilation
|
||
(B) length contraction
|
||
|
||
|
||
7. For each question below, state whether it is theoretically possible or impossible according to special relativity. Explain your answers.
|
||
(A) A chimpanzee could travel to a star that is 10,000 light-years away in her natural lifetime. Note: A light-year is the distance that light travels in one year.
|
||
(B) A chimpanzee could go on a space trip and appear younger than her own daughter when she returns to earth.
|
||
(C) A chimpanzee could go on a space trip and appear younger than she was when she left the earth.
|
||
Want help? Check the solution at the end of the chapter.
|
||
Answers: 7. (A) possible
|
||
(B) possible (C) impossible
|
||
|
||
|
||
|
||
|
||
Solutions to Chapter 1
|
||
1. Setup a coordinate system with +x directed to the south. Identify the given information:
|
||
• The relative speed between the monkey and the girl is v = 36 m/s.
|
||
• The monkey is the moving observer. The velocity of the apple relative to the monkey is uM.
|
||
This value will be different in parts (A) and (B) since the apple is thrown in a different direction in each part.
|
||
• The girl is at rest. The velocity of the apple relative to the girl is uR.
|
||
Since these speeds are small compared to the speed of light, we may apply the equation for Galilean relativity. As usual, we neglect air resistance unless stated otherwise.
|
||
uM = uR – v
|
||
(A) Since the apple is thrown south and we chose +x to point south, uM = 12 m/s.
|
||
The apple is moving 48 m/s relative to the girl.
|
||
(B) Since the apple is thrown north and we chose +x to point south, uM = –12 m/s.
|
||
The apple is moving 24 m/s relative to the girl.
|
||
|
||
|
||
2. (A) The banana lands directly on the X. Why? Because the banana has inertia. Recall from Newton’s laws of motion (which are taught in first-year physics) that inertia is the natural tendency of all objects to maintain constant velocity. According to Newton’s second law of motion, a net external force is needed to accelerate an object (and thus change the object’s velocity), since
|
||
When the monkey releases the banana, a net gravitational force acts downward, causing the banana to accelerate downward. However, there are no forces acting horizontally, so the banana doesn’t accelerate horizontally. The banana maintains a constant horizontal component of velocity (vx), while
|
||
the vertical component of velocity (vy) changes. That’s why, in projectile motion, horizontally we
|
||
have
|
||
and vertically we have
|
||
Horizontally, vx is constant (because ax = 0), whereas vertically there is uniform acceleration (ay =
|
||
–g = –9.81 m/s2 is constant).
|
||
You can verify this by riding in an airplane. If you throw an eraser straight upward relative to you while sitting inside of the airplane, you will catch the eraser because it has inertia. The eraser surely won’t land behind you, even if the airplane is traveling 300 m/s and if the eraser is in the air for half a second (in which case the airplane travels 150 m horizontally).
|
||
(B) Relative to the monkey, the banana appears to fall in a straight line downward, no different than if the train had been parked when the monkey released the banana. The laws of physics are the same in any inertial reference frame, meaning that the result of dropping a banana from rest will be the same whether the train is at rest or moving with constant velocity.
|
||
(C) Relative to the girl, the banana follows the arc of a parabola, beginning with a horizontal tangent. The same path would result if the girl had the banana and threw it horizontally. The banana follows the path of a projectile, which is parabolic.
|
||
From the monkey’s point of view, the monkey claims that the train is at “rest” (but that the girl and ground are moving). According to the monkey, vx = 0 and the banana falls straight downward.
|
||
From the girl’s point of view, the girl claims that she is at “rest” (but that the train is moving). According to the girl, vx isn’t 0. Relative to the girl,
|
||
|
||
|
||
and
|
||
which can be combined to make
|
||
which is the equation of a parabola. To derive this equation, you need to combine the following equations:
|
||
|
||
|
||
3. Setup a coordinate system with +x directed to the east.
|
||
|
||
|
||
|
||
|
||
4. (A) Marco appears to age normally relative to the scientists on Marco’s ship. Because the proper time for Polo’s aging process is measured by Polo’s team, Marco’s team will measure Polo’s aging process to occur more slowly due to time dilation. Therefore, when Marco’s team compares their photographs, Polo will appear younger.
|
||
(B) Polo appears to age normally relative to the scientists on Polo’s ship. Because the proper time for Marco’s aging process is measured by Marco’s team, Polo’s team will measure Marco’s aging process to occur more slowly due to time dilation. Therefore, when Polo’s team compares their photographs, Marco will appear younger.
|
||
(C) Both teams are correct. Marco’s team believes that Polo appears younger when Marco celebrates
|
||
his 30th birthday, and Polo’s team believes that Marco appears younger when Polo celebrates his 30th birthday, and both teams are correct because both teams are inertial observers (since both spaceships travel with constant velocity). According to special relativity, there is no preferred reference frame; and the laws of physics are the same for all inertial observers. (For the chimpanzees to actually meet up, note that one would have to accelerate.)
|
||
(D) Marco’s ship appears normal relative to the scientists on Marco’s ship. Because the proper length for Polo’s ship is measured by Polo’s team, Marco’s team will measure the length of Polo’s ship to be shorter than normal due to length contraction; Marco’s appears longer.
|
||
(E) Polo’s ship appears normal relative to the scientists on Polo’s ship. Because the proper length for Marco’s ship is measured by Marco’s team, Polo’s team will measure the length of Marco’s ship to be shorter than normal due to length contraction; Polo’s appears longer.
|
||
(F) Marco’s team measures Marco’s ship to have its rest mass. When Marco’s team measures the mass of Polo’s ship, Polo’s ship’s relativistic mass appears greater than its rest mass. (Let’s assume that both teams of scientists have the same combined rest mass.)
|
||
(G) Polo’s team measures Polo’s ship to have its rest mass. When Polo’s team measures the mass of Marco’s ship, Marco’s ship’s relativistic mass appears greater than its rest mass.
|
||
(H) The height is the same for each. (Length contraction occurs only along the direction of the velocity. The height, which is perpendicular to the velocity, is unaffected.)
|
||
|
||
|
||
5. (A) Since the proper length of the spaceship is measured by astronauts inside of the ship, astronauts stationed in the space garage measure the length of the spaceship to be shorter than normal due to length contraction. The concept of length contraction explains how the ship is short enough to fit in the garage while traveling close to the speed of light.
|
||
(B) The spaceship has its usual length relative to astronauts inside of the spaceship. You can’t explain this in terms of length contraction because length contraction would make the space garage appear shorter than usual relative to astronauts inside of the spaceship. The answer has to do with the fact that different inertial observers often disagree on whether or not two events occur simultaneously; this is one of those times. Whereas astronauts stationed in the garage momentarily see the spaceship fit inside of the garage with both doors temporarily closed, astronauts aboard the spaceship don’t see the two doors closed at the same instant. Rather, astronauts aboard the spaceship see the door on the right side of our diagram close first (while the back of the spaceship sticks out on the left side of our diagram). They then see the front door open. The spaceship continues a short ways until the back of the spaceship is safely inside of the garage. At this point, the door on the left side of our diagram quickly closes (while the front of the spaceship sticks out on the right side of our diagram).
|
||
|
||
|
||
6. (A) Since the proper time would be measured by a reference frame traveling with the muons themselves, scientists on earth’s surface measure a longer lifetime due to time dilation. That is, because the muons travel close to the speed of light relative to the earth, their average lifetime is much longer than it would be if they were at rest, which allows them to travel a longer distance before they decay.
|
||
(B) From the reference frame of the muons, their lifetime is normal (they measure the proper time). Instead, the muons “see” length contraction: Since scientists on earth’s surface measure the proper length for the muons’ trip, a reference frame attached to the muons would measure a shorter distance. That is, because the muons travel close to the speed of light (although from their perspective, the muons are at rest and the earth is traveling close to the speed of light towards the muons), the distance between earth’s surface and where they are produced in earth’s atmosphere is much shorter than would be if they were at rest, which allows them to reach earth’s surface in less time. It’s instructive to compare how these different perspectives lead to equivalent conclusions through two quite different effects.
|
||
|
||
|
||
7. (A) Theoretically, it is possible, provided that the chimpanzee travels in a spaceship with a speed that is close enough to the speed of light (that’s the hard part). Relative to observers on earth, the chimpanzees would age much more slowly than normal through time dilation. (However, the initial part of the trip requires acceleration, and the end of the trip involves deceleration, both of which involve general relativity, which is a step beyond special relativity.)
|
||
(B) Theoretically, it is possible, for the same reason as part A, provided that the chimpanzee travels in a spaceship that is close enough to the speed of light. If the chimpanzee ages slowly enough due to time dilation, she could appear younger than her daughter when she returns to earth. However, whatever the age difference is between the chimpanzee and her daughter, at least that number of years must pass on earth for this to be possible. For example, if the chimpanzee is 30 years old and her daughter is 12 years old, at least 18 years must pass on earth during the trip (plus additional years depending on how fast the ship travels; we’ll explore the mathematics involved in time dilation in Chapter 2).
|
||
(C) This is theoretically impossible (without plastic surgery or age defying medicine). While time can slow down due to time dilation, in special relativity time can’t go backwards.
|
||
|
||
|
||
|
||
|
||
2 TIME DILATION AND LENGTH CONTRACTION
|
||
Relevant Terminology
|
||
Time dilation – the phenomenon whereby time appears to travel more slowly for objects moving fast (close to light speed) relative to other observers.
|
||
Length contraction – the phenomenon whereby objects moving fast (close to light speed) appear shorter relative to other observers.
|
||
Proper time – a time interval measured by a clock that is at rest relative to the events. An observer who is moving relative to the events measures a greater time interval due to time dilation.
|
||
Proper length – a distance measured by an observer who is at rest relative to the distance. An observer who is moving relative to the distance measures a shorter distance due to length contraction.
|
||
Inertial reference frame – a frame that travels with constant velocity.
|
||
|
||
|
||
|
||
|
||
Time Dilation
|
||
Time passes more slowly on a clock in a moving inertial reference frame than it does for an inertial reference frame that is at rest relative to the clock. This is known as time dilation.
|
||
In the diagram above, the spaceship is traveling to the right with a speed that is close to the speed of light. The spaceship passes by a space station that is practically at rest relative to the very fast spaceship. As the spaceship is passing the space station, an astronaut inside of the spaceship turns on a flashlight, shining a beam of light straight upward in the diagram (perpendicular to the direction that the spaceship is traveling).
|
||
Relative to the astronaut inside of the spaceship (observer P), the flashlight beam appears to travel straight upward (left diagram). Relative to an observer inside of the space station (observer Q), the flashlight beam appears to travel diagonally (right diagram).
|
||
In the illustration above:
|
||
• LP is the distance that the light travels according to observer P.
|
||
• LQ is the distance that the light travels according to observer Q.
|
||
• LS is the distance that the spaceship travels horizontally during this time.
|
||
These three distances are related by the Pythagorean theorem:
|
||
According to the second postulate of special relativity, both observers must measure the speed of light to be the same value. Either observer takes the distance (L) traveled and divides by the corresponding time (t) measured to determine the speed of light (c). The subscripts P and Q indicate
|
||
|
||
|
||
which observer makes the measurement.
|
||
Observer Q can also measure the speed (v) of the spaceship by dividing the horizontal distance (LS)
|
||
traveled by the corresponding time (tQ).
|
||
Multiply each equation by the corresponding time.
|
||
LP = c tP
|
||
LQ = c tQ
|
||
LS = v tQ
|
||
Substitute these expressions into the equation from the Pythagorean theorem.
|
||
Solve for tQ.
|
||
Divide the numerator and denominator each by c2.
|
||
|
||
|
||
|
||
|
||
|
||
|
||
Length Contraction
|
||
When an object is moving relative to an inertial reference frame, the object appears shorter (along the direction of motion) than it does relative to an inertial reference frame that is at rest relative to the object. This is known as length contraction.
|
||
In the diagram above, a spaceship is traveling from earth’s sun to Alpha Centauri with a speed that is close to the speed of light. Observer S is an astronaut aboard the spaceship, while observer E is stationed on earth.
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Observer E on earth measures the distance to be LE and the time to be tE, such that the speed of the
|
||
spaceship is
|
||
v = LE / tE
|
||
Observer S on the spaceship measures the distance to be LS and the time to be tS such that
|
||
v = LS / tS
|
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According to observer E, the ship is “moving” while the stars are at rest, whereas according to observer S, the earth and stars are “moving” while the spaceship is at “rest,” but either way the speed of the spaceship is the same:
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Solve for LS.
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||
Since the spaceship’s clock is at “rest” relative to the journey, observer S measures the proper time and the passage of time aboard the spaceship appears dilated relative to observer E. Use the time dilation equation with tS as the proper time:
|
||
|
||
|
||
Note: When applying the above equation to other problems, the proper length will often not be measured by an observer on earth. That just happened to be the case in this example. In a given problem, you must apply the concept of proper length to determine which observer measures the proper length.
|
||
Note that the effects of time dilation and length contraction that we discussed in this example involve
|
||
|
||
|
||
two different perspectives:
|
||
• For observer E on earth, v = LE / tE, we noted that time was dilated (tE > tS) for observer E
|
||
because the earth is “moving” relative to the spaceship.
|
||
• For observer S on the spaceship, v = LS / tS, we noted that length was contracted (LS < LE) for
|
||
observer S because the spaceship is “moving” relative to the two stars.
|
||
|
||
|
||
|
||
|
||
Time Dilation and Length Contraction Equations
|
||
In the equations below, t0 represents the proper time (measured by an observer who is at rest relative
|
||
to the events) and L0 represents the proper length (measured by an observer who is at rest relative to
|
||
the distance). Note that t0 and L0 are not necessarily measured by the same observer in a problem (in
|
||
fact, these were measured by different observers when we derived the equation for length contraction in the previous section).
|
||
|
||
|
||
|
||
|
||
Symbols and SI Units
|
||
|
||
|
||
|
||
|
||
Constants
|
||
|
||
|
||
|
||
|
||
Strategy for Solving Time Dilation and Length Contraction
|
||
Problems
|
||
To solve a problem involving time dilation or length contraction, follow these steps:
|
||
• It may help to draw a diagram and label both objects and observers.
|
||
• When applying time dilation, identify the proper time (t0), which is measured by an observer
|
||
whose clock is at rest relative to the events.
|
||
• When applying length contraction, identify the proper length (L0), which is measured by an
|
||
observer who is at rest relative to the distance being measured.
|
||
Note that t0 and L0 are not necessarily measured by the same observer in a problem.
|
||
|
||
|
||
Example: A chimpanzee astronaut sleeps for 8 hours according to a spaceship’s clock while traveling at 0.5c (half the speed of light) relative to the earth. For how much time does the chimpanzee appear to be sleeping relative to an observer on earth?
|
||
Which observer measures the proper time? The spaceship’s clock is at rest relative to the chimpanzee. Therefore, the chimpanzee measures the proper time: t0 = 8 hr. The observer on earth
|
||
measures a greater time, td, due to time dilation. The relative speed is v = 0.5c. Use the time dilation
|
||
equation.
|
||
|
||
|
||
Example: A spaceship has a length of 20 m when it is parked near the surface of the earth. When a chimpanzee astronaut in the spaceship travels at 0.8c relative to the earth, what length does the spaceship appear to have relative to an observer on earth?
|
||
Which observer measures the proper length? The chimpanzee is at rest relative to the length of the spaceship. Therefore, the chimpanzee measures the proper length: L0 = 20 m. The observer on earth
|
||
measures a shorter distance, Lc, due to length contraction. The relative speed is v = 0.8c. Use the
|
||
length contraction equation.
|
||
|
||
|
||
|
||
|
||
Chapter 2 Problems
|
||
1. A chimpanzee astronaut travels in a spaceship at 0.6c relative to the earth. According to the chimpanzee, the spaceship is 30 m long and the trip takes 12 years.
|
||
(A) How long is the spaceship relative to an observer on earth?
|
||
(B) How long does the trip take relative to an observer on earth?
|
||
(C) How far does the spaceship travel according to an observer on earth?
|
||
Note: One light-year (ly) is the distance that light travels in one year.
|
||
(D) How far does the spaceship travel according to the chimpanzee?
|
||
(E) Explain your answers to parts C and D.
|
||
Want help? Check the solution at the end of the chapter.
|
||
Answers: 1. (A) 24 m (B) 15 years
|
||
(C) 9.0 ly (D) 7.2 ly
|
||
(E) time dilation vs. length contraction
|
||
|
||
|
||
2. How fast must a spaceship travel relative to another observer in order to appear half as long as it really is?
|
||
Want help? Check the solution at the end of the chapter.
|
||
Answer:
|
||
|
||
|
||
3. Muons that are produced at rest have an average lifetime of
|
||
A beam of muons is produced that travels 0.99c relative to the earth.
|
||
(A) How far does classical physics expect the muons to travel on average? (For this question, pretend that the muons don’t follow the laws of special relativity.)
|
||
(B) Relative to observers on earth, how far will the muons actually travel on average?
|
||
(C) Relative to a reference frame attached to the muons, how far do the muons travel on average?
|
||
(D) Explain your answers to parts B and C.
|
||
Want help? Check the solution at the end of the chapter.
|
||
Answers: 3. (A) 0.65 km
|
||
(B) 4.6 km (C) 0.65 km
|
||
(D) time dilation vs. length contraction
|
||
|
||
|
||
4. A chimpanzee astronaut travels in a spaceship at (12/13)c relative to the earth. According to observers stationed on earth, the trip takes 26 years.
|
||
(A) How far does the spaceship travel relative to the earth?
|
||
(B) How far does the spaceship travel relative to the chimpanzee?
|
||
(C) How long does the trip take relative to the chimpanzee?
|
||
Want help? Check the solution at the end of the chapter.
|
||
Answers: 4. (A) 24 ly (B) (120/13) ly = 9.2 ly (C) 10 yr
|
||
|
||
|
||
|
||
|
||
Solutions to Chapter 2
|
||
1. The relative speed is v = 0.6c.
|
||
(A) Regarding the length of the spaceship, the chimpanzee measures the proper length since the chimpanzee is at rest relative to the spaceship: L0 = 30 m. Use the equation for length contraction to
|
||
determine what an observer on earth measures.
|
||
Recall that the way to divide by a fraction is to multiply by its reciprocal.
|
||
(B) Regarding the time of the trip, the chimpanzee measures the proper time since the ship’s clock is at “rest” relative to the journey: t0 = 12 yr. Use the equation for time dilation to determine what an
|
||
observer on earth measures.
|
||
(C) Multiply the speed of the ship by the time of the trip as measured by an earth observer.
|
||
|
||
|
||
Note that one lightyear (ly) equals c times 1 yr. That is, the speed of light times one year equals the
|
||
distance that light travels in one year. (If you prefer meters, use c = 2.9979 × 108 m/s and convert 1 yr
|
||
to 31,536,000 seconds to get 8.5 × 1016 m.) Unlike part A, the observer on earth measures the proper length for this distance (which is at rest relative to earth).
|
||
(D) Multiply the speed of the ship by the time of the trip as measured by the chimpanzee.
|
||
Since the observer on earth measures the proper length of the trip (since the starting and ending points of the journey—which are likely the sun and a nearby star—aren’t moving relative to the earth), the chimpanzee measures a shorter length due to length contraction. We could have obtained the same answer from the length contraction equation (using d for the distance of the trip, so as not to confuse it with the L that we used for the length of the ship in part A). It’s instructive to compare part D with part A, since in part A the chimpanzee measured the proper length, whereas in part D the observer on earth measures the proper length.
|
||
|
||
|
||
|
||
|
||
(A) Multiply the speed of the muons by the average lifetime.
|
||
(B) The average lifetime will be dilated relative to observers on earth.
|
||
(C) In the muons’ reference frame, the lifetime is normal, but the distance that they travel (on average) is contracted.
|
||
|
||
|
||
(A) Multiply the speed of the ship by the time of the trip as measured by an earth observer. |