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In the twin paradox example, Alice boards a spaceship that flies to a space station 3.25 light-years away and then returns with a speed of \(0.65 c\). a) Calculate the total distance Alice traveled during the trip, as measured by Alice. b) With the aforementioned total distance, calculate the total time duration for the trip, as measured by Alice.

Short Answer

Expert verified
Answer: The total distance traveled by Alice in her reference frame is 9.7274 light-years, and the total time duration for the trip as measured by Alice is 6.6837 years.

Step by step solution

01

a) Calculate the total distance traveled by Alice

First, let's calculate the distance Alice travels as measured by her. Because the speed of light is constant in all reference frames, we know that the distance Alice travels in one reference frame (i.e., her own frame) can be related to the distance traveled as measured from another frame (i.e., the Earth's frame) using the Lorentz transformations. The trip can be divided into two parts - going to the space station and returning back to Earth. Each leg of the trip has the same distance as measured by Earth, which is 3.25 light-years. Thus, the total distance traveled as measured by Earth is\(2\times 3.25\) light-years. Now, we need to use the Lorentz transformation to find the distance traveled by Alice in her reference frame. The Lorentz transformation equation for distance is: $$x' = \gamma (x - vt)$$ Where \(x'\) is the distance in Alice's reference frame, \(x\) is the distance in Earth's reference frame, \(v\) is the velocity of the spaceship, \(t\) is the time in Earth's reference frame, and \(\gamma\) is the Lorentz factor, given by: $$\gamma = \frac{1}{\sqrt{1 - (v/c)^2}}$$ We know that \(x = 3.25\,\text{light-years}\), \(v = 0.65c\), and \(c\) is the speed of light. However, since we are dealing with distances, we can't just directly apply time dilation. Instead, we need to find out the time dilation factor and then relate that to the distance traveled. Let's first find the Lorentz factor, \(\gamma\): $$\gamma = \frac{1}{\sqrt{1 - (0.65c/c)^2}} = \frac{1}{\sqrt{1 - 0.65^2}} = 1.4966$$ Since we are only interested in distance traveled and the time it took for the moving spaceship, we can simplify the Lorentz transformation equation for distance as: $$x' = \gamma x$$ Now, we can calculate the distance traveled by Alice to the space station in her reference frame: $$x'_{1} = 1.4966 \times 3.25 = 4.8637\,\text{light-years}$$ The total distance Alice travels, as measured by her, is twice this value (as she travels the same distance back to Earth): $$x'_{\text{total}} = 2 \times 4.8637 = 9.7274\,\text{light-years}$$ So, the total distance traveled by Alice in her reference frame is \(9.7274\,\text{light-years}\).
02

b) Calculate the total time duration for the trip as measured by Alice

Now, we need to find out the total time duration for the trip as measured by Alice. As mentioned before, we need to look at the time dilation factor since the time experienced by Alice (moving frame) will be different from that experienced by Earth (stationary frame). The time dilation equation is: $$t' = \frac{t}{\gamma}$$ Where \(t'\) is the time as measured by Alice, \(t\) is the time in Earth's reference frame, and \(\gamma\) is the Lorentz factor. The time taken in each leg (trip to the space station and the return trip) can be found by dividing the distance by the velocity: $$t = \frac{x}{v} = \frac{3.25\,\text{light-years}}{0.65c} = 5\,\text{years}$$ Then the total time as measured by Earth for Alice's trip would be: $$t_{\text{total}} = 2 \times 5 = 10\,\text{years}$$ Now, we can find the total time duration for the trip as experienced by Alice using the time dilation equation: $$t'_{\text{total}} = \frac{t_{\text{total}}}{\gamma} = \frac{10\,\text{years}}{1.4966} = 6.6837\,\text{years}$$ So, the total time duration for the trip as measured by Alice is about \(6.6837\) years.

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Key Concepts

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

Lorentz Transformation
The Lorentz transformation is the cornerstone of Albert Einstein's theory of special relativity. It provides the mathematical formulas that describe how measurements of space and time by two observers moving relative to each other are related. For instance, if a spaceship travels at a significant fraction of the speed of light, we can use the Lorentz transformation to compare the distance and time measurements from the perspective of an observer on Earth to those of an observer on the spaceship.

Specifically in our exercise, the Lorentz transformation shows us how Alice, the traveling twin, perceives distance, despite the fact that her velocity is a substantial fraction (\(0.65 c\)) of the speed of light. In the step-by-step solution you saw how the transformation was applied to calculate the distance as measured by Alice. This concept reinforces that distances and times are not absolute but are relative to the observer's state of motion.
Time Dilation
The phenomenon of time dilation is one of the most fascinating predictions of special relativity. It tells us that time is not experienced at the same rate for observers in different reference frames, especially when they are moving at speeds close to the speed of light relative to each other. Time dilation arises naturally from the Lorentz transformation equations and has been confirmed by numerous experiments.

For Alice in the exercise, time dilation means that although 10 years pass on Earth during her journey, she experiences only approximately 6.6837 years. This is because the Lorentz factor (\( \frac{1}{\text{sqrt}{1 - (v/c)^2}} \) shrinks the time interval that Alice experiences in her reference frame. Time dilation doesn't just apply to hypothetical space journeys; it also has practical implications for satellite technology and GPS systems.
Reference Frames
In physics, a reference frame is essentially a perspective from which an observer measures distances, times, and other physical quantities. It's like choosing a viewpoint in a painting - the scene can look different depending on where you stand. In special relativity, reference frames are especially important because they help define how different observers might measure different 'realities'.

In our twin paradox example, there are two key reference frames: the Earth frame (which is considered an inertial frame of reference because it is stationary or moving at a constant velocity) and the spaceship frame (which is the non-inertial frame due to acceleration when turning around at the space station). The exercise shows that measurements of distances and times differ between these frames due to the relative velocity of the spaceship to Earth.
Special Relativity
Special relativity revolutionized our understanding of space, time, and energy. It applies to all physical phenomena in the absence of gravity and postulates that the laws of physics are the same for all observers in inertial frames of reference, and that the speed of light in a vacuum is constant, regardless of the state of motion of the source or observer.

The twin paradox, an exercise we're discussing, is a famous thought experiment in special relativity. It highlights how different the concepts of space and time are from our everyday experiences. Thanks to special relativity, we understand that Alice's time is dilated (runs slower) compared to her twin on Earth. Through this astounding theory, time travel becomes a part of scientific reality, not just science fiction, as moving close to the speed of light causes significant changes in the passage of time.

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Most popular questions from this chapter

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.

In mechanics, one often uses the model of a perfectly rigid body to model and determine the motion of physical objects (see, for example, Chapter 10 on rotation). Explain how this model contradicts Einstein's special theory of relativity.

Consider motion in one spatial dimension. For any velocity \(v,\) define parameter \(\theta\) via the relation \(v=c \tanh \theta\) where \(c\) is the vacuum speed of light. This quantity is variously called the velocity parameter or the rapidity corresponding to velocity \(v\). a) Prove that for two velocities, which add according to the Lorentzian rule, the corresponding velocity parameters simply add algebraically, that is, like Galilean velocities. b) Consider two reference frames in motion at speed \(v\) in the \(x\) -direction relative to one another, with axes parallel and origins coinciding when clocks at the origin in both frames read zero. Write the Lorentz transformation between the two coordinate systems entirely in terms of the velocity parameter corresponding to \(v\), and the coordinates.

You shouldn't invoke time dilation due to your relative motion with respect to the rest of the world as an excuse for being late to class. While it is true that relative to those at rest in the classroom, your time runs more slowly, the difference is likely to be negligible. Suppose over the weekend you drove from your college in the Midwest to New York City and back, a round trip of \(2200 .\) miles, driving for 20.0 hours each direction. By what amount, at most, would your watch differ from your professor's watch?

How much work is required to accelerate a proton from rest up to a speed of \(0.997 c ?\)

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