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Chapter 37: Problem 14

A rocket ship flies past the earth at 91.0% of the speed of light. Inside, an astronaut who is undergoing a physical examination is having his height measured while he is lying down parallel to the direction in which the ship is moving. (a) If his height is measured to be 2.00 m by his doctor inside the ship, what height would a person watching this from the earth measure? (b) If the earth-based person had measured 2.00 m, what would the doctor in the spaceship have measured for the astronaut's height? Is this a reasonable height? (c) Suppose the astronaut in part (a) gets up after the examination and stands with his body perpendicular to the direction of motion. What would the doctor in the rocket and the observer on earth measure for his height now?

Short Answer

Expert verified
(a) Earth observer measures 0.83 m. (b) Doctor measures 4.83 m. (c) Both measure 2.00 m.

Step by step solution

01

Identify the Scenario

This problem revolves around the relativistic effect known as length contraction, which affects measurements of length when objects move close to the speed of light.
02

Understand Length Contraction Formula

The formula for length contraction is given by \(L = L_0 \sqrt{1 - \frac{v^2}{c^2}}\), where \(L_0\) is the proper length (length measured in the object's rest frame), \(v\) is the velocity of the moving object, and \(c\) is the speed of light.
03

Calculate Earth's Observed Length for Part (a)

Since the height measured inside the ship (proper length \(L_0\)) is 2.00 m, and the rocket moves at 91.0% the speed of light \((v = 0.91c)\), the observer on Earth measures a contracted length \(L = 2.00 \times \sqrt{1 - (0.91)^2}\). Calculate \(L = 2.00 \times \sqrt{1 - 0.8281} = 2.00 \times \sqrt{0.1719}\) which approximately equals \(0.83 \text{ m}\).
04

Calculate Ship's Observed Length for Part (b)

If the earth-based observer measures the height as 2.00 m, this is the contracted length for that observer. Therefore, the doctor in the ship measures the longer, proper length \(L_0 = 2.00 / \sqrt{1 - 0.8281}\). Calculate \(L_0 = 2.00 / 0.4144\), which approximately equals \(4.83 \text{ m}\). While this is a mathematically calculated height, it is not reasonable for a human.
05

Evaluate Height Perpendicular to Motion for Part (c)

When the astronaut stands up perpendicular to the direction of motion, no length contraction occurs in that dimension. Therefore, both the doctor and the Earth observer measure the height as 2.00 m, the same as if the astronaut were at rest.

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Key Concepts

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

Special Relativity
Special relativity is a theory introduced by Albert Einstein. It describes how the laws of physics work when objects move at high speeds, close to the speed of light. A key aspect is that the speed of light, denoted as "c", is constant and the same for all observers, regardless of their motion. As objects approach this speed, time and space behave differently from our everyday experiences.
Einstein's theory leads to intriguing phenomena such as time dilation and length contraction. These mean that time appears to move slower, and lengths appear shorter for objects moving at near-light speeds, compared to what observers at rest would measure. Understanding these effects is crucial for many problems and applications in physics.
Proper Length
Proper length, denoted as \(L_0\), refers to the length of an object measured by an observer who is at rest relative to the object. This measurement is the longest possible measurement of an object. For example, the astronaut's height of 2.00 m inside the spaceship is his proper length.
Proper length is essential to distinguishing between what an observer on Earth and what one on a rocket moving at high speeds would measure. It is the baseline measurement used in the length contraction formula to find how much shorter the length will appear to an observer moving relative to the object.
Relativistic Effects
Relativistic effects become significant when objects travel at speeds close to the speed of light. One of the main effects is length contraction. To an outside observer, such as someone on Earth observing a passing spaceship, objects moving at such speeds appear shorter in the direction of motion.
Using the length contraction formula \(L = L_0 \sqrt{1 - \frac{v^2}{c^2}}\), we can calculate the contracted length. For instance, if the spaceship speed is 91% of the speed of light \((v = 0.91c)\), the Earth observer sees the astronaut's height contracted from 2.00 m to about 0.83 m. Relativistic effects only appear at velocities that are substantial fractions of light speed.
Speed of Light
The speed of light is a fundamental constant in physics, represented by "c" and approximately equal to \(3 \times 10^8\) meters per second.
Nothing with mass can match or exceed the speed of light, according to current physical understanding. This speed acts as a cosmic speed limit, dictating how light and information travel across the universe. It deeply influences concepts in special relativity, such as length contraction and time dilation, demonstrating how crucial "c" is in determining the behavior of objects at high speeds.

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Most popular questions from this chapter

(a) Consider the Galilean transformation along the \(x\)-direction : \(x' = x - vt\) and \(t' = t\). In frame \(S\) the wave equation for electromagnetic waves in a vacuum is $${\partial^2E(x, t)\over \partial x^2} - {1 \over c^2} {\partial^2E(x, t)\over \partial t^2} = 0$$ where \(E\) represents the electric field in the wave. Show that by using the Galilean transformation the wave equation in frame \(S'\) is found to be $$(1 - {v^2 \over c^2} ) {\partial^2E(x', t') \over \partial{x'^2}} + {2v \over c^2} {\partial^2E(x', t') \over \partial{x'} \partial{t'}} - {1 \over c^2} {\partial^2E(x', t') \over \partial{t'^2}} = 0$$ This has a different form than the wave equation in \(S\). Hence the Galilean transformation \(violates\) the first relativity postulate that all physical laws have the same form in all inertial reference frames. (\(Hint\): Express the derivatives \(\partial/\partial{x}\) and \(\partial/\partial{t}\) in terms of \(\partial/\partial{x'}\) and \(\partial/\partial{t'}\) by use of the chain rule.) (b) Repeat the analysis of part (a), but use the Lorentz coordinate transformations, Eqs. (37.21), and show that in frame \(S'\) the wave equation has the same form as in frame \(S\): $${\partial^2E(x', t') \over \partial{x'}^2} - {1 \over c^2} {\partial^2E(x', t') \over \partial{t'}^2} = 0$$ Explain why this shows that the speed of light in vacuum is c in both frames \(S\) and \(S'\).

A muon is created 55.0 km above the surface of the earth (as measured in the earth's frame). The average lifetime of a muon, measured in its own rest frame, is 2.20 \(\mu\)s, and the muon we are considering has this lifetime. In the frame of the muon, the earth is moving toward the muon with a speed of 0.9860\(c\). (a) In the muon's frame, what is its initial height above the surface of the earth? (b) In the muon's frame, how much closer does the earth get during the lifetime of the muon? What fraction is this of the muon's original height, as measured in the muon's frame? (c) In the earth's frame, what is the lifetime of the muon? In the earth's frame, how far does the muon travel during its lifetime? What fraction is this of the muon's original height in the earth's frame?

A spaceship flies past Mars with a speed of 0.985c relative to the surface of the planet. When the spaceship is directly overhead, a signal light on the Martian surface blinks on and then off. An observer on Mars measures that the signal light was on for 75.0 ms. (a) Does the observer on Mars or the pilot on the spaceship measure the proper time? (b) What is the duration of the light pulse measured by the pilot of the spaceship?

Muons are unstable subatomic particles that decay to electrons with a mean lifetime of 2.2 \(\mu\)s. They are produced when cosmic rays bombard the upper atmosphere about 10 km above the earth's surface, and they travel very close to the speed of light. The problem we want to address is why we see any of them at the earth's surface. (a) What is the greatest distance a muon could travel during its 2.2 - \(\mu\)s lifetime? (b) According to your answer in part (a), it would seem that muons could never make it to the ground. But the 2.2- \(\mu\)s lifetime is measured in the frame of the muon, and muons are moving very fast. At a speed of 0.999c, what is the mean lifetime of a muon as measured by an observer at rest on the earth? How far would the muon travel in this time? Does this result explain why we find muons in cosmic rays? (c) From the point of view of the muon, it still lives for only 2.2 \(\mu\)s, so how does it make it to the ground? What is the thickness of the 10 km of atmosphere through which the muon must travel, as measured by the muon? Is it now clear how the muon is able to reach the ground?

Everyday Time Dilation. Two atomic clocks are carefully synchronized. One remains in New York, and the other is loaded on an airliner that travels at an average speed of 250 m/s and then returns to New York. When the plane returns, the elapsed time on the clock that stayed behind is 4.00 h. By how much will the readings of the two clocks differ, and which clock will show the shorter elapsed time? (\(Hint\): Since \(u \ll c\), you can simplify \(\sqrt{1 - u^2/c^2}\) by a binomial expansion.)
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