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Chapter 37: Problem 8

An alien spacecraft is flying overhead at a great distance as you stand in your backyard. You see its searchlight blink on for 0.150 s. The first officer on the spacecraft measures that the searchlight is on for 12.0 ms. (a) Which of these two measured times is the proper time? (b) What is the speed of the spacecraft relative to the earth, expressed as a fraction of the speed of light \(c\)?

Short Answer

Expert verified
(a) The proper time is 12.0 ms. (b) The speed is approximately 99.68% of the speed of light.

Step by step solution

01

Identify Proper Time

Proper time is the time measured by an observer who is at rest relative to the event being measured. In this case, the proper time is the time measured by the first officer on the spacecraft, as the officer is at rest relative to the searchlight.
02

Use Time Dilation Formula

The time dilation formula is given by \( t = \frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}} \), where \( t \) is the time measured by an observer moving relative to the event, \( t_0 \) is the proper time, \( v \) is the velocity of the moving observer, and \( c \) is the speed of light.
03

Convert Measurements to Same Units

Convert the proper time \( t_0 = 12.0 \) ms to seconds: \( 12.0 \text{ ms} = 0.012 \text{ s} \). The observer on Earth measures \( t = 0.150 \text{ s} \).
04

Solve for Velocity Fraction

Rearrange the time dilation formula to solve for \( \frac{v}{c} \): 1. \( t = \frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}} \) 2. Solve for \( \frac{v^2}{c^2} \), giving \( 1-\frac{v^2}{c^2} = \left(\frac{t_0}{t}\right)^2 \)3. \( \frac{v^2}{c^2} = 1-\left(\frac{t_0}{t}\right)^2 \)4. \( \frac{v}{c} = \sqrt{1-\left(\frac{0.012}{0.150}\right)^2} \)
05

Calculate and Simplify

Calculate \( \left(\frac{0.012}{0.150}\right)^2 = 0.0064 \). Therefore, \( \frac{v}{c} = \sqrt{1 - 0.0064} = \sqrt{0.9936} \). This gives \( \frac{v}{c} \approx 0.9968 \).

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Key Concepts

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

Proper Time
In the realm of physics, specifically in the context of special relativity, **proper time** becomes an essential concept. Proper time is the time interval measured by an observer who is at rest relative to the process being observed. It is the true duration of an event from the perspective of someone or something that is not moving relative to the event.

This concept is crucial when dealing with time dilation because the proper time represents the "true" time for events. In our exercise, the officer on the spacecraft measures the searchlight blink duration as 12.0 ms. This is considered the proper time because the officer and the light event are both in the same reference frame—the spacecraft itself. In contrast, the observer on Earth measures a different time, 0.150 s, due to the relativistic effects of time dilation since the spacecraft is moving relative to Earth.
Velocity of Spacecraft
In understanding how fast the spacecraft is moving relative to Earth, we need to delve into the **velocity of the spacecraft** in terms of the speed of light. The formula that governs this relationship is derived from time dilation, which takes into account the effects of an object's speed as it approaches the speed of light.

The important thing to remember is:
  • The observer on Earth perceives a longer time for the searchlight blink due to the relative motion of the spacecraft.
  • This time dilation occurs because the spacecraft is moving at a significant fraction of the speed of light (\( c \)).
  • By rearranging the time dilation formula, we solved for the ratio of the spacecraft's velocity to the speed of light.
In our example, after using the time dilation formula and substituting the known values, we calculated the velocity of the spacecraft as a fraction of the speed of light, approximately 0.9968. This indicates the spacecraft is traveling at almost the speed of light itself, leading to the significant time difference measured by the observers.
Speed of Light
The **speed of light** in a vacuum is a fundamental constant of nature, symbolized by \( c \). It holds a value of approximately 299,792,458 meters per second. This isn't just any random measurement—it defines the ultimate speed limit of the universe.

Key points about the speed of light:
  • Nothing can travel faster than this limit.
  • In our exercise, the speed of light serves as a reference to determine how close the spacecraft's velocity gets to this universal ceiling.
  • The time dilation effects become significant as velocities approach the speed of light, as seen with the spacecraft.
Understanding the speed of light helps in imagining how time can "stretch" and "shrink" depending on the observer's frame of reference and velocity. As the velocity approaches \( c \), time dilation becomes noticeable, which is exactly what we observe in the time difference between the spacecraft and Earth in the exercise.

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Most popular questions from this chapter

A pursuit spacecraft from the planet Tatooine is attempting to catch up with a Trade Federation cruiser. As measured by an observer on Tatooine, the cruiser is traveling away from the planet with a speed of 0.600c. The pursuit ship is traveling at a speed of 0.800c relative to Tatooine, in the same direction as the cruiser. (a) For the pursuit ship to catch the cruiser, should the velocity of the cruiser relative to the pursuit ship be directed toward or away from the pursuit ship? (b) What is the speed of the cruiser relative to the pursuit ship?

(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 \(S'\).

In certain radioactive beta decay processes, the beta particle (an electron) leaves the atomic nucleus with a speed of 99.95\(\%\) the speed of light relative to the decaying nucleus. If this nucleus is moving at 75.00\(\%\) the speed of light in the laboratory reference frame, find the speed of the emitted electron relative to the laboratory reference frame if the electron is emitted (a) in the same direction that the nucleus is moving and (b) in the opposite direction from the nucleus's velocity. (c) In each case in parts (a) and (b), find the kinetic energy of the electron as measured in (i) the laboratory frame and (ii) the reference frame of the decaying nucleus.

An unstable particle is created in the upper atmosphere from a cosmic ray and travels straight down toward the surface of the earth with a speed of 0.99540c relative to the earth. A scientist at rest on the earth's surface measures that the particle is created at an altitude of 45.0 km. (a) As measured by the scientist, how much time does it take the particle to travel the 45.0 km to the surface of the earth? (b) Use the length-contraction formula to calculate the distance from where the particle is created to the surface of the earth as measured in the particle's frame. (c) In the particle's frame, how much time does it take the particle to travel from where it is created to the surface of the earth? Calculate this time both by the time dilation formula and from the distance calculated in part (b). Do the two results agree?

(a) How fast must you be approaching a red traffic (\(\lambda=\) 675 nm) for it to appear yellow (\(\lambda=\) 575 nm)? Express your answer in terms of the speed of light. (b) If you used this as a reason not to get a ticket for running a red light, how much of a fine would you get for speeding? Assume that the fine is $1.00 for each kilometer per hour that your speed exceeds the posted limit of 90 km/h.
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