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An observer in frame \(S^{\prime}\) is moving to the right \((+x\) -direction at speed \(u=0.600 c\) away from a stationary observer in frame \(S .\) The observer in \(S^{\prime}\) measures the speed \(v^{\prime}\) of a particle moving to the right away from her. What speed \(v\) does the observer in \(S\) measure for the particle if \(v^{\prime}=0.400 c\) (b) \(v^{\prime}=0.900 c ;(\mathrm{c}) v^{\prime}=0.990 c ?\)

Short Answer

Expert verified
(a) 0.806c, (b) 0.974c, (c) 0.997c.

Step by step solution

01

Understand the problem

We have two reference frames, \(S\) and \(S'\). Frame \(S'\) is moving with velocity \(u = 0.600c\) relative to frame \(S\). We need to find the velocity \(v\) of a particle as measured in frame \(S\) when its velocity \(v'\) in frame \(S'\) is given for different values.
02

Set up the velocity transformation formula

We'll use the relativistic velocity addition formula to transform velocities between the two frames. The formula is given by:\[v = \frac{v' + u}{1 + \frac{v' u}{c^2}}\]
03

Solve for \(v\) when \(v' = 0.400c\)

Substitute \(v' = 0.400c\) and \(u = 0.600c\) into the velocity addition formula:\[v = \frac{0.400c + 0.600c}{1 + \frac{(0.400c)(0.600c)}{c^2}} = \frac{1.000c}{1 + 0.240} = \frac{1.000c}{1.240} \approx 0.806c\]Thus, when \(v' = 0.400c\), \(v \approx 0.806c\).
04

Solve for \(v\) when \(v' = 0.900c\)

Substitute \(v' = 0.900c\) and \(u = 0.600c\) into the velocity addition formula:\[v = \frac{0.900c + 0.600c}{1 + \frac{(0.900c)(0.600c)}{c^2}} = \frac{1.500c}{1 + 0.540} = \frac{1.500c}{1.540} \approx 0.974c\]Thus, when \(v' = 0.900c\), \(v \approx 0.974c\).
05

Solve for \(v\) when \(v' = 0.990c\)

Substitute \(v' = 0.990c\) and \(u = 0.600c\) into the velocity addition formula:\[v = \frac{0.990c + 0.600c}{1 + \frac{(0.990c)(0.600c)}{c^2}} = \frac{1.590c}{1 + 0.594} = \frac{1.590c}{1.594} \approx 0.997c\]Thus, when \(v' = 0.990c\), \(v \approx 0.997c\).

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Key Concepts

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

Understanding Reference Frames
Reference frames are crucial in understanding motion because they determine how we observe and measure velocity and position. In simple terms, a reference frame is like a coordinate system or a point of view from which you observe a situation. For example, the Earth itself is often used as a reference frame in daily life events.

In the context of relativity, we deal with inertial reference frames, which are frames that are not accelerating. In the exercise problem, there are two reference frames: frame \(S\) and frame \(S'\). Frame \(S'\), moving at a speed of \(0.600c\), is relative to the stationary frame \(S\). This setup allows us to make sense of how different observers (in \(S\) and \(S'\)) perceive the motion of objects (like a particle) differently.

The transformation of velocities between these frames is essential for understanding situations involving high-speed objects, such as those approaching the speed of light. By using the relativistic velocity addition formula, we can determine how the motion observed in one frame translates to another frame.
The Speed of Light Consistency
The speed of light, denoted as \(c\), is a fundamental constant of nature. It is incredibly fast, measuring approximately \(299,792,458\) meters per second. In the realm of physics, especially in the theory of relativity, the speed of light plays a rather special role: it is a universal speed limit.

Nothing can move faster than light in a vacuum, making it a crucial figure in relativistic equations, especially when handling high velocities comparable to \(c\). This is why it appears frequently in relativistic equations, such as the velocity addition formula in this exercise.

By incorporating \(c\), we are accounting for the effects predicted by Einstein's theory of relativity, which states that the laws of physics are the same for all observers, no matter how fast they are moving relative to each other. This principle is essential in predicting how different observers perceive time and space.
Introduction to Special Relativity
Special relativity is a groundbreaking theory developed by Albert Einstein, which fundamentally changed the way we understand space, time, and motion. This theory is built on two main postulates:
  • 1. The laws of physics are the same in all inertial reference frames.
  • 2. The speed of light in a vacuum is the same for all observers, regardless of the motion of the light source or observer.

These principles bring extraordinary consequences. For example, time dilation means that time can pass at different rates for observers in different states of motion. Length contraction tells us objects appear shorter in the direction of motion when they move closer to the speed of light.

In our exercise, the relativistic velocity addition formula illustrates these principles. It showcases how velocities add up when taking into account the constraints of special relativity. Rather than simply adding them algebraically, they are combined using the formula which ensures that no result exceeds \(c\), preserving the laws of physics consistent across all reference frames.

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Most popular questions from this chapter

A photon with energy \(E\) is emitted by an atom with mass \(m,\) which recoils in the opposite direction. (a) Assuming that the motion of the atom can be treated nonrelativistically, compute the recoil speed of the atom. (b) From the result of part (a), show that the recoil speed is much less than \(c\) whenever \(E\) is much less than the rest energy \(m c^{2}\) of the atom.

A person is standing at rest on level ground. How fast would she have to run to (a) double her total energy and (b) increase her total energy by a factor of 10?

Space Travel? Travel to the stars requires hundreds or thousands of years, even at the speed of light. Some people have suggested that we can get around this difficulty by accelerating the rocket (and its astronauts) to very high speeds so that they will age less due to time dilation. The fly in this ointment is that it takes a great deal of energy to do this. Suppose you want to go to the immense red giant Betelgeuse, which is about 500 light-years away. (A light- year is the distance that light travels in a year.) You plan to travel at constant speed in a 1000 -kg rocket ship (a little over a ton), which, in reality, is far too small for this purpose. In each case that follows, calculate the time for the trip, as measured by people on earth and by astronauts in the rocket ship, the energy needed in ioules, and the energy needed as a percentage of U.S. yearly use (which is 1.0 \(\times 10^{20} \mathrm{J} ) .\) For comparison, arrange your results in a table showing \(v_{\text { rocket }}, t_{\text { earth }}, t_{\text { rocket }}, E(\) in \(\mathrm{J}),\) and \(E\) (as \(\%\) of U.S. use). The rocket ship's speed is (a) \(0.50 \mathrm{c} ;\) (b) 0.99 \(\mathrm{c}\) ; (c) 0.9999 \(\mathrm{c} .\) On the basis of your results, does it seem likely that any government will invest in such high-speed space travel any time soon?

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?

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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