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A spaceship at rest in a certain reference frame \(S\) is given a speed increment of \(0.50 c .\) Relative to its new rest frame, it is then given a further \(0.50 c\) increment. This process is continued until its speed with respect to its original frame \(S\) exceeds \(0.999 c .\) How many increments does this process require?

Short Answer

Expert verified
5 increments are needed to exceed 0.999c.

Step by step solution

01

Understanding of Relativistic Velocity Addition

In special relativity, velocities do not simply add like in classical physics. When a spaceship already moving at a velocity \(u\) is given an additional speed \(v\), its final velocity \(v'\) is given by the formula: \[ v' = \frac{u + v}{1 + \frac{uv}{c^2}} \]. This formula accounts for effects of relativity and ensures velocities do not exceed the speed of light \(c\).
02

Initial Setup

Initially, the spaceship is at rest in frame \(S\) (\(u = 0\)) and receives a first speed increment of \(0.50c\). Thus, velocity after the first increment \(v_1\) is given by: \[ v_1 = \frac{0 + 0.50c}{1 + \frac{0 \times 0.50c}{c^2}} = 0.50c \].
03

Calculating Second Increment

The spaceship now sees itself at rest in its new reference frame. It receives another \(0.50c\) increment relative to its new rest frame. Applying the velocity addition formula with \(u = 0.50c\) and \(v = 0.50c\): \[ v_2 = \frac{0.50c + 0.50c}{1 + \frac{0.50c \times 0.50c}{c^2}} = \frac{c}{1 + 0.25} = \frac{0.8c}{1.25} = 0.8c \].
04

Subsequent Increments

Continue using the formula for each increment. For the third speed increment relative to its new rest frame, \(u = 0.8c\) and \(v = 0.50c\), therefore: \[ v_3 = \frac{0.8c + 0.50c}{1 + \frac{0.8c \times 0.50c}{c^2}} = \frac{1.3c}{1.4} = 0.9286c \].
05

Checking the Final Condition

The fourth increment needs to be calculated since \(0.9286c < 0.999c\). Applying the relativistic formula again with \(u = 0.9286c\) and \(v = 0.50c\): \[ v_4 = \frac{0.9286c + 0.50c}{1 + \frac{0.9286c \times 0.50c}{c^2}} = 0.9825c \].
06

Final Increment Calculation

The velocity \(0.9825c\) is still less than \(0.999c\), so one more increment is necessary. For the fifth increment: \[ v_5 = \frac{0.9825c + 0.50c}{1 + \frac{0.9825c \times 0.50c}{c^2}} > 0.999c \]. This results in a speed exceeding \(0.999c\), satisfying the condition.

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Key Concepts

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

Special Relativity
Special relativity is a theory in physics introduced by Albert Einstein in 1905. It revolutionized how we understand motion, space, and time, especially when objects move at speeds close to the speed of light. Usually, when we think about velocity, we imagine simply adding the speeds together. However, special relativity shows us that velocities need to be calculated using a specific formula when approaching the speed of light.

One of the most critical aspects of special relativity is the concept that the speed of light, denoted as \(c\), is the maximum speed anything can travel through the universe. This means that no matter how you add velocities, the speed of light cannot be surpassed. This is where the relativistic velocity addition formula comes into play. It ensures that even when two speeds are combined, the result remains under the light speed limit.

Special relativity also introduces the concept of time dilation and length contraction. These phenomena occur because, as an object’s speed approaches the speed of light, time for the object appears to slow down relative to a stationary observer, and the object itself seems compressed along the direction of motion. This doesn't mean that time actually slows down for the object, but it’s perceived that way from a different frame of reference.
Speed of Light
The speed of light is a fundamental constant of nature and is approximately \(299,792,458\) meters per second or about \(300,000\) kilometers per second. It is the highest speed at which information or matter can travel through space. In physics, we denote this speed with the symbol \(c\).

The speed of light plays a crucial role in the theory of relativity. One of the cornerstone principles of special relativity is that no object with mass can reach or exceed the speed of light. As objects gain speed, their relativistic mass increases, requiring more and more energy to continue accelerating them. Thus, as a massive object nears the speed of light, it would require infinite energy to actually reach \(c\), effectively making it impossible.

Furthermore, the speed of light is invariant, meaning it is the same in all reference frames, regardless of the relative motion of source and observer. This universality is a pivotal aspect of special relativity, leading to some non-intuitive consequences such as time dilation and length contraction.
Reference Frame
A reference frame is the perspective from which an observer measures and observes physical events. It includes both a coordinate system and a clock to compare various phenomena. In the context of relativity, reference frames are crucial because they affect the measurements of time and space.

In our problem, the spaceship starts at rest in a reference frame, say frame \(S\). When it gains a speed increment of \(0.50c\), any further measurement for that spaceship should be taken from its new reference frame, where it's now at rest. This shifting of reference frames is a vital part of relativistic calculations, as it ensures the velocity addition formula is applied correctly.

In non-relativistic physics, different reference frames might measure different speeds, but they would not fundamentally alter time or spatial measurements. However, in relativity, moving to a different frame can lead to changes in the perceived rates of time passage or spatial distances due to effects like time dilation. Therefore, understanding how reference frames influence observations is key to understanding problems in special relativity.

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Most popular questions from this chapter

A spaceship of rest length \(130 \mathrm{~m}\) races past a timing station at a speed of \(0.740 c .\) (a) What is the length of the spaceship as measured by the timing station? (b) What time interval will the station clock record between the passage of the front and back ends of the ship?

If we intercept an electron having total energy 1533 \(\mathrm{MeV}\) that came from Vega, which is 26 ly from us, how far in lightyears was the trip in the rest frame of the electron?

An astronaut exercising on a treadmill maintains a pulse rate of 150 per minute. If he exercises for \(1.00 \mathrm{~h}\) as measured by a clock on his spaceship, with a stride length of \(1.00 \mathrm{~m} / \mathrm{s}\), while the ship travels with a speed of \(0.900 c\) relative to a ground station, what are (a) the pulse rate and (b) the distance walked as measured by someone at the ground station?

The car-in-the-garage problem. Carman has just purchased the world's longest stretch limo, which has a proper length of \(L_{c}=30.5 \mathrm{~m}\). In Fig. \(37-32 a\), it is shown parked in front of a garage with a proper length of \(L_{g}=6.00 \mathrm{~m} .\) The garage has a front door (shown open) and a back door (shown closed). The limo is obviously longer than the garage. Still, Garageman, who owns the garage and knows something about relativistic length contraction, makes a bet with Carman that the limo can fit in the garage with both doors closed. Carman, who dropped his physics course before reaching special relativity, says such a thing, even in principle, is impossible. To analyze Garageman's scheme, an \(x_{c}\) axis is attached to the limo, with \(x_{c}=0\) at the rear bumper, and an \(x_{g}\) axis is attached to the garage, with \(x_{g}=0\) at the (now open) front door. Then Carman is to drive the limo directly toward the front door at a velocity of \(0.9980 \mathrm{c}\) (which is, of course, both technically and financially impossible). Carman is stationary in the \(x_{c}\) reference frame; Garageman is stationary in the \(x_{g}\) reference frame. There are two events to consider. Event \(1:\) When the rear bumper clears the front door, the front door is closed. Let the time of this event be zero to both Carman and Garageman: \(t_{g 1}=t_{c 1}=0\). The event occurs at \(x_{c}=x_{g}=0\). Figure \(37-32 b\) shows event 1 according to the \(x_{g}\) reference frame. Event 2 : When the front bumper reaches the back door, that door opens. Figure \(37-32 c\) shows event 2 according to the \(\bar{x}_{8}\) reference frame. According to Garageman, (a) what is the length of the limo, and what are the spacetime coordinates (b) \(x_{g^{2}}\) and (c) \(t_{g 2}\) of event \(2 ?\) (d) For how long is the limo temporarily "trapped" inside the garage with both doors shut? Now consider the situation from the \(x_{c}\) reference frame, in which the garage comes racing past the limo at a velocity of \(-0.9980 c\). According to Carman, (e) what is the length of the passing garage, what are the spacetime coordinates (f) \(x_{c 2}\) and (g) \(t_{c 2}\) of event \(2,(\mathrm{~h})\) is the limo ever in the garage with both doors shut, and (i) which event occurs first? (j) Sketch events 1 and 2 as seen by Carman. (k) Are the events causally related; that is, does one of them cause the other? (I) Finally, who wins the bet?

What is the momentum in MeV/c of an electron with a kinetic energy of \(2.00 \mathrm{MeV} ?\)

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