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Chapter 35: Problem 18

Prove that in all cases, two sub-light-speed velocities "added" relativistically will always yield a sub-light-speed velocity. Consider motion in one spatial dimension only.

Short Answer

Expert verified
Question: Prove that the total sub-light-speed velocity obtained from adding two sub-light-speed velocities relativistically is always less than the speed of light. Answer: By using the relativistic velocity addition formula and analyzing the denominators, we demonstrated that the total velocity w is always sub-light-speed, meaning it is always less than the speed of light.

Step by step solution

01

Review the relativistic velocity addition formula

The relativistic addition of velocities formula is given as: $$ w = \frac{v + u}{1 + \frac{v \cdot u}{c^2}} $$ where w is the total velocity, v and u are the two sub-light-speed velocities being added, and c is the speed of light.
02

Define the given conditions

In our given problem, the velocities are sub-light-speed, which means \(0 \leq v < c\) and \(0 \leq u < c\). Our goal is to prove that the total velocity w is also sub-light-speed, so \(0 \leq w < c\).
03

Analyze the denominator of the formula

Let's analyze the denominator of the fraction \(\frac{v \cdot u}{c^2}\). Since both v and u are sub-light-speed, their product is positive and smaller than \(c^2\). Therefore, the fraction is positive and smaller than 1. $$ 0 < \frac{v \cdot u}{c^2} < 1 $$ Adding 1 to this inequality, $$ 1 < 1 + \frac{v \cdot u}{c^2} < 2 $$
04

Show that the total velocity is sub-light-speed

Now, we'll use the formula for relativistic velocity addition to prove that the total velocity w is sub-light-speed: $$ w = \frac{v + u}{1 + \frac{v \cdot u}{c^2}} $$ In the worst-case scenario (maximum), v and u are both close to c. Thus, \((v + u)\) is close to \(2c\), and the denominator from step 3 is less than 2. Then, the maximum value of w can be obtained when the numerator is the largest and the denominator is the smallest: $$ w_{max} = \frac{2c}{1 + \frac{v \cdot u}{c^2}} $$ Since the denominator is always greater than 1, it follows that the fraction is less than 2, so \(w_{max} < 2c\). Therefore, the total velocity w is always sub-light-speed, as required to prove.

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Most popular questions from this chapter

At rest, a rocket has an overall length of \(L .\) A garage at rest (built for the rocket by the lowest bidder) is only \(L / 2\) in length. Luckily, the garage has both a front door and a back door, so that when the rocket flies at a speed of \(v=0.866 c\), the rocket fits entirely into the garage. However, according to the rocket pilot, the rocket has length \(L\) and the garage has length \(L / 4\). How does the rocket pilot observe that the rocket does not fit into the garage?

An astronaut in a spaceship flying toward Earth's Equator at half the speed of light observes Earth to be an oblong solid, wider and taller than it appears deep, rotating around its long axis. A second astronaut flying toward Earth's North Pole at half the speed of light observes Earth to be a similar shape but rotating about its short axis. Why does this not present a contradiction?

Consider two clocks carried by observers in a reference frame moving at speed \(v\) in the positive \(x\) -direction relative to ours. Assume that the two reference frames have parallel axes, and that their origins coincide when clocks at that point in both frames read zero. Suppose the clocks are separated by distance \(l\) in the \(x^{\prime}-\) direction in their own reference frame; for instance, \(x^{\prime}=0\) for one clock and \(x^{\prime}=I\) for the other, with \(y^{\prime}=z^{\prime}=0\) for both. Determine the readings \(t^{\prime}\) on both clocks as functions of the time coordinate \(t\) in our reference frame.

Use the relativistic velocity addition to reconfirm that the speed of light with respect to any inertial reference frame is \(c\). Assume one-dimensional motion along a common \(x\) -axis.

Using relativistic expressions, compare the momentum of two electrons, one moving at \(2.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}\) and the other moving at \(2.00 \cdot 10^{3} \mathrm{~m} / \mathrm{s}\). What is the percentage difference between classical momentum values and these values?
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