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Chapter 37: Problem 4

A spaceship flies past Mars with a speed of \(0.985 c\) 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 \(75.0 \mu\) s. (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?

Short Answer

Expert verified
(a) The observer on Mars measures the proper time because the event (blinking light) is happening at the same position for him. (b) Use the time dilation formula to find the duration of the light pulse as measured by the spaceship pilot.

Step by step solution

01

Determine the observer measuring proper time

The proper time is the time interval that is measured by an observer who perceives the start and end of an event happening in the exact same position. Here, the event is the signal light turning on and off. This event occurs at a fixed location on Mars surface. Hence, the observer on Mars measures the proper time.
02

Apply the time dilation formula

The formula for time dilation is given by \( \Delta t = \frac{\Delta t_0}{\sqrt{1-\left(\frac{v}{c}\right)^2}} \), where \( \Delta t \) is the dilated time (time measured by the moving observer), \( \Delta t_0 \) is the proper time, \( v \) is the relative speed and \( c \) the speed of light. In this case, \( \Delta t_0 \) is the time measured by Mars observer (75.0 碌s), \( v \) is the spaceship speed (0.985 c), and \( c \) is the speed of light. Now, we just need to insert these values into the equation and solve it.
03

Compute the time as measured by spaceship pilot

By substituting the given values into the time dilation formula, we calculate the time interval (duration of the light pulse) as measured by the spaceship pilot.

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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 concept in the realm of special relativity, introduced by Albert Einstein. It's all about how time appears to stretch or dilate when observed from a moving frame of reference. This might sound odd, but imagine you're on a spaceship zipping past a planet. Time for you, relative to someone standing still on the planet, would actually be experienced differently.
An important aspect of time dilation is the time dilation formula:\[ \Delta t = \frac{\Delta t_0}{\sqrt{1-\left(\frac{v}{c}\right)^2}} \]
Here, \( \Delta t \) is the dilated time鈥攚hat the person on the spaceship would measure. \( \Delta t_0 \) is the proper time,\( v \) is the relative speed of the moving object, and \( c \) is the speed of light. In simpler terms, this equation tells us how much time stretches by when something is moving incredibly fast. So, as the speed of the spaceship increases, the time dilation effect becomes more pronounced.
Proper Time
Proper time is the time measured by an observer for whom the events (such as a light blinking on and off) occur at the same point in space. This is crucial because it provides a baseline measurement, unaffected by high speeds or other factors.
In our exercise, the observer standing on Mars is the one who measures the proper time, since the signal light from Mars blinks at a fixed location relative to them. This observer is stationary with respect to the event, so their measurement of time (75.0 microseconds) is the proper time in this context.
Understanding proper time helps us unravel complex situations where multiple frames of reference are involved. It gives us a grounded perspective on the event without the influence of motion.
Relative Speed
Relative speed is the speed of one object as observed from another moving object. It鈥檚 a crucial element in understanding phenomena in special relativity, because many effects like time dilation depend on this relative motion.
In our example, the spaceship travels at a speed of \(0.985 c\) relative to Mars. This high relative speed is significant because it puts the theory of special relativity into action. The closer the speed of the spaceship approaches the speed of light, the more pronounced the relativistic effects will become.
Relative speed allows us to compare the motions of two objects effectively. It's this relative nature of speeds that determines how much time dilation we observe. When solving problems, understanding this concept helps clarify how different observers, moving at different speeds, perceive time and space differently.

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Most popular questions from this chapter

A spacecraft of the Trade Federation flies past the planet Coruscant at a speed of \(0.600 c\). A scientist on Coruscant measures the length of the moving spacecraft to be \(74.0 \mathrm{~m}\). The spacecraft later lands on Coruscant, and the same scientist measures the length of the now stationary spacecraft. What value does she get?

If a muon is traveling at \(0.999 c,\) what are its momentum and kinetic energy? (The mass of such a muon at rest in the laboratory is 207 times the electron mass.)

A spaceship moving at constant speed \(u\) relative to us broad. casts a radio signal at constant frequency \(f_{0}\). As the spaceship approaches us, we receive a higher frequency \(f\); after it has passed, we receive a lower frequency. (a) As the spaceship passes by, so it is instantaneously moving neither toward nor away from us, show that the frequency we receive is not \(f_{0},\) and derive an expression for the frequency we do receive. Is the frequency we receive higher or lower than \(f_{0} ?\) (Hint: In this case, successive wave crests move the same distance to the observer and so they have the same transit time. Thus \(f\) equals \(1 / T .\) Use the dilation formula to relate the periods in the stationary and moving frames.) (b) A spaceship emits electromagnetic waves of frequency \(f_{0}=345 \mathrm{MHz}\) as measured in a frame moving with the ship. The spaceship is moving at a constant speed \(0.758 c\) relative to us. What frequency \(f\) do we receive when the spaceship is approaching us? When it is moving away? In each case what is the shift in frequency, \(f-f_{0} ?\) (c) Use the result of part (a) to calculate the frequency \(f\) and the frequency shift \(\left(f-f_{0}\right)\) we receive at the instant that the ship passes by us. How does the shift in frequency calculated here compare to the shifts calculated in part (b)?

Quarks and gluons are fundamental particles that will be discussed in Chapter \(44 .\) A proton, which is a bound state of two up quarks and a down quark, has a rest mass of \(m_{\mathrm{p}}=1.67 \times 10^{-27} \mathrm{~kg}\). This is significantly greater than the sum of the rest mass of the up quarks, which is \(m_{\mathrm{u}}=4.12 \times 10^{-30} \mathrm{~kg}\) each, and the rest mass of the down quark, which is \(m_{\mathrm{d}}=8.59 \times 10^{-30} \mathrm{~kg} .\) Suppose we (incorrectly) model the rest energy of the proton \(m_{\mathrm{p}} c^{2}\) as derived from the kinetic energy of the three quarks, and we split that energy equally among them. (a) Estimate the Lorentz factor \(\gamma=\left(1-v^{2} / c^{2}\right)^{-1 / 2}\) for each of the up quarks using Eq. \((37.36) .\) (b) Similarly estimate the Lorentz factor \(\gamma\) for the down quark. (c) Are the corresponding speeds \(v_{\mathrm{u}}\) and \(v_{\mathrm{d}}\) greater than \(99 \%\) of the speed of light? (d) More realistically, the quarks are held together by massless gluons, which mediate the strong nuclear interaction. Suppose we model the proton as the three quarks, each with a speed of \(0.90 c,\) with the remainder of the proton rest energy supplied by gluons. In this case, estimate the percentage of the proton rest energy associated with gluons. (e) Model a quark as oscillating with an average speed of \(0.90 c\) across the diameter of a proton, \(1.7 \times 10^{-15} \mathrm{~m}\). Estimate the frequency of that motion.

A nuclear bomb containing \(12.0 \mathrm{~kg}\) of plutonium explodes. The sum of the rest masses of the products of the explosion is less than the original rest mass by one part in \(10^{4}\). (a) How much energy is released in the explosion? (b) If the explosion takes place in \(4.00 \mu \mathrm{s}\) what is the average power developed by the bomb? (c) What mass of water could the released energy lift to a height of \(1.00 \mathrm{~km} ?\)
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