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Chapter 37: Problem 6

As you pilot your space utility vehicle at a constant speed toward the moon, a race pilot fies past you in her spaceracer at a constant speed of 0.800\(c\) relative to you. At the instant the spaceracer passes you, both of you start timers at zero. (a) At the instant when you measure that the spaceracer has traveled \(1.20 \times 10^{8} \mathrm{m}\) past you, what does the race pilot read on her timer? (b) When the race pilot reads the value calculated in part (a) on her timer, what does she measure to be your distance from her? (c) At the instant when the race pilot reads the value calculated in part (a) on her timer, what do you read on yours?

Short Answer

Expert verified
(a) 0.300 s; (b) 7.20 脳 10^7 m; (c) 0.500 s

Step by step solution

01

Determine the Time in Your Frame

We know that the relative speed of the spaceracer is \(0.800c\) and it covers a distance of \(1.20 \times 10^{8}\) meters. By using the formula \(t = \frac{d}{v}\), where \(d\) is the distance and \(v\) is the velocity, we find the time:\[t = \frac{1.20 \times 10^{8} \, m}{0.800c} = \frac{1.20 \times 10^{8} \, m}{0.800 \times 3.00 \times 10^{8} \, m/s} = 0.500 \, s\]This is the time you measure in your frame.
02

Calculate the Time in the Race Pilot's Frame using Time Dilation

According to the special theory of relativity, time in a moving frame slows down by a factor of \(\gamma\), which is given by \(\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}\). Here, \(v = 0.800c\):\[\gamma = \frac{1}{\sqrt{1 - (0.800)^2}} = \frac{1}{\sqrt{0.36}} \approx 1.6667\]The time interval according to the race pilot is given by \(t' = \frac{t}{\gamma}\):\[t' = \frac{0.500}{1.6667} \approx 0.300 \, s\]The race pilot reads 0.300 seconds when you measure the race pilot has traveled \(1.20 \times 10^8 \, m\).
03

Calculate the Distance in the Race Pilot's Frame

To find out how far you appear to be from the race pilot in her frame, we need to use the Lorentz contraction formula. The real distance is reduced by a factor of \(\gamma\). Distance \(L'\) is calculated as:\[L' = \frac{L}{\gamma}\]The contracted length is:\[L' = \frac{1.20 \times 10^{8} \, m}{1.6667} \approx 7.20 \times 10^{7} \, m\]So, the race pilot measures a distance of \(7.20 \times 10^{7} \, m\) from you.
04

Confirm the Time in Your Frame Again

Since we want to understand what you read on your timer when the race pilot reads 0.300 seconds, we confirm that you measure 0.500 seconds as previously calculated in Step 1. Therefore, the time on your clock is 0.500 seconds.

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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 phenomenon predicted by Einstein's theory of Special Relativity. It describes how time appears to pass slower for an object in motion compared to one at rest. Imagine you're on a space mission, and you observe a race pilot zipping by in her speedy spaceracer. Due to her high velocity, time in her frame moves differently from yours.

When you see the spaceracer travelling at 0.800 times the speed of light, you may measure the time it takes her to travel a certain distance as 0.500 seconds. However, from her perspective within the race craft, her clock ticks more slowly. To compute the time dilation, we use the Lorentz factor, denoted by \( \gamma \), defined as \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \), where \( v \) is the velocity of the moving object and \( c \) is the speed of light. Here, \( \gamma \approx 1.6667 \).

Thus, when you observe that 0.500 seconds have passed, she measures her own passage of time as approximately 0.300 seconds. This apparent slowing down of time as perceived in a moving reference frame is the heart of time dilation.
Length Contraction
Length contraction is another intriguing aspect of Special Relativity. It refers to the shortening of objects in the direction of motion as perceived by an observer moving relative to the object. Suppose you're on your journey towards the Moon, and you watch the race pilot zoom past. From her frame, the distance you measure gets compressed.

To calculate this contracted length, we utilize the formula \( L' = \frac{L}{\gamma} \), where \( L \) represents the original length and \( L' \) is the contracted length. In this scenario, when you measure the spaceracer has travelled \( 1.20 \times 10^8 \) meters, she observes a contracted length of approximately \( 7.20 \times 10^7 \) meters. This is because her moving frame causes the length to shrink by the same factor \( \gamma \approx 1.6667 \) that affects time dilation.

So, if you ever race through the cosmos at near-light speeds, the universe itself appears to become more compact in the direction of your motion.
Special Relativity
Special Relativity, developed by Albert Einstein, revolutionized our understanding of space, time, and motion. This theory introduces two key ideas: time dilation and length contraction, which we explored earlier. It shows how at high velocities, the fabric of space and time behaves differently.

At the core of Special Relativity is the Lorentz Transformation, allowing us to calculate the changes in time and space coordinates as perceived by observers in different reference frames. However, it's essential to consider that these transformations are only significant at speeds approaching the speed of light.

The relativity effect implies there's no absolute frame of reference. Instead, the laws of physics remain consistent for all observers, irrespective of their constant velocities. For instance, in our space travel example, both you and the race pilot have valid viewpoints. Your perceptions just differ due to your relative motion. This concept challenges and enriches our classical views, highlighting the interconnectedness between time and space.

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Most popular questions from this chapter

The positive muon \(\left(\mu^{+}\right),\) an unstable particle, lives on average \(2.20 \times 10^{-6} \mathrm{s}\) (measured in its own frame of reference) before decaying. (a) If such a particle is moving, with respect to the laboratory, with a speed of \(0.900 c,\) what average lifetime is measured in the laboratory? (b) What average distance, measured in the laboratory, does the particle move before decaying?

Muons are unstable subatomic particles that decay to electrons with a mean lifetime of 2.2\(\mu \mathrm{s}\) . They are produced when cosmic rays bombard the upper atmosphere about 10 \(\mathrm{km}\) 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 -\mus lifetime? (b) According to your answer in part (a), it would seem that muons could never make it to the ground. But the \(2.2-\mu \mathrm{s}\) lifetime is measured in the frame of the muon, and muons are moving very fast. At a speed of \(0.999 c,\) 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 \(\mathrm{km}\) 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?

(a) Consider the Galilean transformation along the \(x\) -direction: \(x^{\prime}=x-\) vt and \(t^{\prime}=t .\) In frame \(S\) the wave equation for electromagnetic waves in a vacuum is $$\frac{\partial^{2} E(x, t)}{\partial x^{2}}-\frac{1}{c^{2}} \frac{\partial^{2} E(x, t)}{\partial t^{2}}=0$$ where \(E\) represents the electric field in the wave. Show that by using the Galilean transformation the wave equation in frame \(S^{\prime}\) is found to be $$\left(1-\frac{v^{2}}{c^{2}}\right) \frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial x^{\prime 2}}+\frac{2 v}{c^{2}} \frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial x^{\prime} \partial t^{\prime}}-\frac{1}{c^{2}} \frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial t^{\prime 2}}=0$$ 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 \(\partial / \partial x^{\prime}\) and \(\partial / \partial t^{\prime}\) by use of the chain rule.) (b) Repeat the analysis of part (a), but use the Lorentz coordinate transformations, Egs. \((37.21),\) and show that in frame \(S^{\prime}\) the wave equation has the same form as in frame \(S :\) $$\frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial x^{2}}-\frac{1}{c^{2}} \frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial t^{\prime 2}}=0$$ Explain why this shows that the speed of light in vacuum is \(c\) in both frames \(S\) and \(S^{\prime} .\)

One of the wavelengths of light emitted by hydrogen atoms under normal laboratory conditions is \(\lambda=656.3 \mathrm{nm},\) in the red portion of the electromagnetic spectrum. In the light emitted from a distant galaxy this same spectral line is observed to be Doppler- shifted to \(\lambda=953.4 \mathrm{nm},\) in the infrared portion of the spectrum. How fast are the emitting atoms moving relative to the earth? Are they approaching the earth or receding from it?

The starships of the Solar Federation are marked with the symbol of the federation, a circle, while starships of the Denebian Empire are marked with the empire's symbol, an ellipse whose major axis is 1.40 times longer than its minor axis \((a=1.40 b\) in Fig. P37.51). How fast, relative to an observer, does an empire ship have to travel for its marking to be confused with the marking of a federation ship?
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