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Chapter 37: Problem 8

At \(x=x^{\prime}=0\) and \(t=t^{\prime}=0\) a clock ticks aboard an extremely fast spaceship moving past us in the \(+x\) -direction with a Lorentz factor of 100 so \(v \approx c .\) The captain hears the clock tick again \(1.00 \mathrm{~s}\) later. Where and when do we measure the second tick to occur?

Short Answer

Expert verified
The observer measures the second tick to occur at approximately \(100s\) and at a distance of \(100c*s\) in the +x direction.

Step by step solution

01

Understand the Lorentz Transformation

The Lorentz Transformation is a coordinate transformation between two coordinate systems which move at constant velocity relative to each other. They will be used as follows:\n\nFor time: \(t = \gamma (t^{\prime} + \frac{vx^{\prime}}{c^2})\)\n\nFor position: \(x = \gamma(x^{\prime} + vt^{\prime})\)\n\nwhere:\n\(\gamma\) is the Lorentz factor,\n\(t^{\prime}\) is the time in the moving frame,\n\(t\) is the time in the non-moving frame,\n \(x^{\prime}\) is the distance in the moving frame,\n \(x\) is the distance in the non-moving frame, and \n\(v\) is the velocity.
02

Plug In Given Values

The problem has given us \(t^{\prime} = 1.00 \, s\) (the time in the spacecraft), \(x^{\prime} = 0\) (because the event takes place in the spacecraft), and \(\gamma = 100\). We can now plug these values into the above equations.\n\nSo, \(t = \gamma (t^{\prime} + \frac{vx^{\prime}}{c^2}) = 100(1.00s + \frac{v*0}{c^2}) = 100s\)\n\nand \(x = \gamma(x^{\prime} + vt^{\prime})= 100(0 + v*1.00s)\)
03

Approximate \(v 鈮� c \)

Since the spacecraft is moving extremely fast and \(v 鈮� c \), then \(x = 100(0 + c*1.00s) = 100c*s\)

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Key Concepts

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

Lorentz factor
The Lorentz factor, denoted by the Greek letter \( \gamma \), is a critical component in the realm of special relativity by Albert Einstein. It becomes significant when objects move at velocities close to the speed of light. The Lorentz factor is given by the formula:
  • \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \)
where \( v \) is the velocity of the moving object, and \( c \) is the speed of light. The bigger the Lorentz factor, the greater the relativistic effects on time and space. As per the given exercise, the Lorentz factor is 100, indicating very high speed close to that of light. This shows that time, as observed from stationary observers, stretches, and space distances warp when measured against the fast-moving frame.
time dilation
Time dilation is a fascinating outcome of Einstein's theory of special relativity. It describes how time appears to slow down for objects moving at high velocities relative to an observer at rest. This becomes apparent through the Lorentz transformations, allowing us to see how time in one frame compares to another. In simpler terms, when a clock moves at speeds approaching the speed of light, it ticks slower compared to a clock that is not moving or moving at lower speeds.
  • In the exercise, the clock aboard the spaceship ticks again 1 second later in the moving frame.
  • However, to an observer at rest, this same 1 second took 100 seconds to pass due to the Lorentz factor of 100.
This phenomenon unravels the vast reaches of time differences experienced at different velocities.
special relativity
Special relativity surges into our understanding with Einstein's groundbreaking equations. It links the concepts of space and time into one unified continuum, spacetime. Unlike our ordinary experiences, special relativity shows us that space and time are intertwined and relative.Some fundamental ideas in special relativity include:
  • The speed of light (\( c \)) is constant in all inertial frames of reference.
  • Nothing can travel faster than light within our universe.
  • The laws of physics are the same in all inertial frames of reference.
Incorporating the Lorentz transformations allows us to calculate observations like time dilation and length contraction, based on the velocity of moving objects. For example, in the given problem, we witness both time dilation and changes in spatial measurements when a spaceship moves at velocities close to light speed.

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Most popular questions from this chapter

A baseball coach uses a radar device to measure the speed of an approaching pitched baseball. This device sends out electromagnetic waves with frequency \(f_{0}\) and then measures the shift in frequency \(\Delta f\) of the waves reflected from the moving baseball. If the fractional frequency shift produced by a baseball is \(\Delta f / f_{0}=2.86 \times 10^{-7},\) what is the baseball's speed in \(\mathrm{km} / \mathrm{h} ?\) (Hint: Are the waves Doppler- shifted a second time when reflected off the ball?)

(a) How much work must be done on a particle with mass \(m\) to accelerate it (a) from rest to a speed of \(0.090 c\) and (b) from a speed of \(0.900 c\) to a speed of \(0.990 c ?\) (Express the answers in terms of \(\left.m c^{2} .\right)\) (c) How do your answers in parts (a) and (b) compare?

What is the speed of a particle whose kinetic energy is equal to (a) its rest energy and (b) five times its rest energy?

Electromagnetic radiation from a star is observed with an earth-based telescope. The star is moving away from the earth at a speed of \(0.520 c\). If the radiation has a frequency of \(8.64 \times 10^{14} \mathrm{~Hz}\) in the rest frame of the star, what is the frequency measured by an observer on earth?

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)?
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