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(a) A meter stick is at rest in frame \(\mathcal{S}_{\mathrm{o}}\), which is traveling with speed \(V=0.8 c\) in the standard configuration relative to frame \(\mathcal{S}\). (a) The stick lies in the \(x_{\mathrm{o}} y_{\mathrm{o}}\) plane and makes an angle \(\theta_{\mathrm{o}}=60^{\circ}\) with the \(x_{\mathrm{o}}\) axis (as measured in \(\mathcal{S}_{\mathrm{o}}\) ). What is its length \(l\) as measured in \(\mathcal{S}\), and what is its angle \(\theta\) with the \(x\) axis? [Hint: It may help to think of the stick as the hypotenuse of a \(30-60-90\) triangle of plywood.] (b) What is \(l\) if \(\theta=60^{\circ} ?\) What is \(\theta_{\mathrm{o}}\) in this case?

Short Answer

Expert verified
(a) Length \(l = 0.866\,\text{m}\), angle \(\theta \approx 79.1^\circ\). (b) \(l = 1\,\text{m}\), \(\theta_o = 60^\circ\).

Step by step solution

01

Understanding the problem

The problem involves a meter stick at rest in frame \(\mathcal{S}_o\) which is moving relative to another frame \(\mathcal{S}\) with a speed \(V = 0.8c\). The stick makes an angle of 60° with the \(x_o\) axis. We need to find its length \(l\) and angle \(\theta\) in frame \(\mathcal{S}\), and consider what happens when the stick's angle \(\theta\) is 60° in this moving frame.
02

Analyze the initial setup in \(\mathcal{S}_o\)

In frame \(\mathcal{S}_o\), the stick is 1 meter long and oriented at an angle \(\theta_o = 60^\circ\). Since it's a 30-60-90 triangle, the lengths of the perpendicular sides can be calculated using trigonometric functions: the side along the \(x_o\)-axis is \(\cos 60^\circ\) and along the \(y_o\)-axis is \(\sin 60^\circ\), meaning the lengths are \(0.5 \, \text{m}\) and \(\sqrt{3}/2 \, \text{m}\) respectively.
03

Apply Length contraction in \(\mathcal{S}\)

In \(\mathcal{S}\), the length of the stick along the direction of motion (\(x\)-axis) experiences relativistic length contraction. The contracted length \(L_x\) is given by:\[ L_x = L_{x_o} \sqrt{1 - \left(\frac{V}{c}\right)^2} = 0.5 \times \sqrt{1 - 0.8^2} = 0.5 \times 0.6 = 0.3 \, \text{m} \]The length along the \(y\)-axis remains unchanged as length contraction only affects dimensions parallel to the motion.
04

Calculate the Length and Angle in \(\mathcal{S}\)

With \(L_x = 0.3 \, \text{m}\) and \(L_y = \sqrt{3}/2 \, \text{m}\), the actual length \(l\) in frame \(\mathcal{S}\) is the hypotenuse of the right triangle:\[ l = \sqrt{L_x^2 + L_y^2} = \sqrt{0.3^2 + (\sqrt{3}/2)^2} \approx 0.866 \, \text{m} \]The angle \(\theta\) with the \(x\)-axis is given by:\[ \tan \theta = \frac{L_y}{L_x} = \frac{\sqrt{3}/2}{0.3} \]\[ \theta = \arctan \left(\frac{\sqrt{3}}{0.6}\right) \approx 79.1^\circ\]
05

Transform Relative Conditions

For Part (b), if the angle \(\theta = 60^\circ\) in frame \(\mathcal{S}\), solve for \(l\):Knowing \(\tan\theta = \frac{L_y}{L_x}\), for \(\theta = 60^\circ\), \(\tan 60^\circ = \sqrt{3}\), leading to:\[ \frac{L_y}{L_x} = \sqrt{3} \rightarrow L_y = \sqrt{3}L_x \]Thus, with \(L_x = L_{x_o} \times 0.6\), calculate \(l = \sqrt{L_x^2 + L_y^2}\):\[ l = \sqrt{(L_{x_o} \times 0.6)^2 + (\sqrt{3}L_{x_o} \times 0.6)^2} \approx 1 \, \text{m} \]To find \(\theta_o\), if \(L_y = \sqrt{3} \times L_x = \text{initial }L_y\), the initial angle \(\theta_o\) remains \(60^\circ\) but is dependent on relative perception.

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Key Concepts

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

Length Contraction
One of the intriguing concepts in the realm of special relativity is length contraction. This phenomenon occurs because of the nature of space and time as described by Einstein's theory. When an object, such as a meter stick, moves at a high velocity relative to an observer, its length appears shorter to that observer when measured along the direction of motion.

This contraction is quantified by the Lorentz factor, given by the formula:
  • The contracted length, \(L'\), can be calculated using: \[ L' = L \sqrt{1 - \frac{V^2}{c^2}} \] where \(L\) is the original length in its rest frame, \(V\) is the velocity of the object, and \(c\) is the speed of light.
  • The contraction only affects the dimension parallel to the velocity.
In our specific problem, this contraction is evident as the meter stick travels with speed \(0.8c\). Calculations lead us to find that the contracted length along the motion is just \(0.3 \text{ m}\), proving the concept as real and calculable under given conditions.
Reference Frames
Understanding reference frames is crucial when diving into special relativity. Frames of reference dictate how we perceive motion and position. In the textbook problem, we are dealing primarily with two frames: \(\mathcal{S}_o\) (the rest frame of the stick) and \(\mathcal{S}\) (the moving observer's frame).

Here are key components to grasp about reference frames:
  • Each frame of reference can have different measurements for time and space due to their relative motion.
  • The concept ensures that physical phenomena remain consistent, yet they may appear differently to observers in relative motion.
  • Transforming between these frames requires us to apply the Lorentz transformation equations.
In the case at hand, the stick appears at different lengths and angles when viewed from these distinct frames of reference. It underscores how measurements are not absolute but rely on the observer's frame.
Lorentz Transformation
The Lorentz Transformation bridges two reference frames that are moving relative to one another. This transformation is fundamental in calculating how quantities like time, length, and angles change due to special relativity.

The equations involved in Lorentz transformations allow us to convert coordinates from one frame to another:
  • The change in time and space is accounted by: \[ x' = \frac{x - Vt}{\sqrt{1 - \frac{V^2}{c^2}}} \] \[ t' = \frac{t - \frac{Vx}{c^2}}{\sqrt{1 - \frac{V^2}{c^2}}} \]
  • For angles, like in our scenario, the transformation alters their perception:
  • Since the stick's perceived length and angle change, \(\tan \theta\) can be derived using the transformed lengths.
Understanding such transformations is vital in predicting how objects in motion are observed differently across frames. This allows us, for example, to see why the length contraction mentioned for the stick matters under the conditions given.

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Most popular questions from this chapter

Verify directly that \(x^{\prime} \cdot y^{\prime}=x \cdot y\) for any two four- vectors \(x\) and \(y,\) where \(x^{\prime}\) and \(y^{\prime}\) are related to \(x\) and \(y\) by the standard Lorentz boost along the \(x_{1}\) axis.

Like time dilation, length contraction cannot be seen directly by a single observer. To explain this claim, imagine a rod of proper length \(l_{\mathrm{o}}\) moving along the \(x\) axis of frame \(\mathcal{S}\) and an observer standing away from the \(x\) axis and to the right of the whole rod. Careful measurements of the rod's length at any one instant in frame \(\mathcal{S}\) would, of course, give the result \(l=l_{\mathrm{o}} / \gamma\). (a) Explain clearly why the light which reaches the observer's eye at any one time must have left the two ends \(A\) and \(B\) of the rod at different times. (b) Show that the observer would see (and a camera would record) a length more than \(l\). [It helps to imagine that the \(x\) axis is marked with a graduated scale.] ( \(\mathbf{c}\) ) Show that if the observer is standing close beside the track, he will see a length that is actually more than \(l_{\mathrm{o}}\); that is, the length contraction is distorted into an expansion.

The pion \(\left(\pi^{+} \text {or } \pi^{-}\right)\) is an unstable particle that decays with a proper half-life of \(1.8 \times 10^{-8}\) s. (This is the half-life measured in the pion's rest frame.) (a) What is the pion's half-life measured in a frame \(\mathcal{S}\) where it is traveling at \(0.8 c ?\) (b). If 32,000 pions are created at the same place, all traveling at this same speed, how many will remain after they have traveled down an evacuated pipe of length \(d=36 \mathrm{m} ?\) Remember that after \(n\) half-lives, \(2^{-n}\) of the original particles survive. (c) What would the answer have been if you had ignored time dilation? (Naturally it is the answer (b) that agrees with experiment.)

A positive pion decays at rest into a muon and neutrino, \(\pi^{+} \rightarrow \mu^{+}+\nu .\) The masses involved are \(m_{\pi}=140 \mathrm{MeV} / c^{2}, m_{\mu}=106 \mathrm{MeV} / c^{2},\) and \(m_{\nu}=0 .\) (There is now convincing evidence that \(m_{\nu}\) is not exactly zero, but it is small enough that you can take it to be zero for this problem.) Show that the speed of the outgoing muon has \(\beta=\left(m_{\pi}^{2}-m_{\mu}^{2}\right) /\left(m_{\pi}^{2}+m_{\mu}^{2}\right) .\) Evaluate this numerically. Do the same for the much rarer decay mode \(\pi^{+} \rightarrow \mathrm{e}^{+}+\nu,\left(m_{\mathrm{e}}=0.5 \mathrm{MeV} / c^{2}\right)\)

The relativistic kinetic energy of a particle is \(T=(\gamma-1) m c^{2} .\) Use the binomial series to express \(T\) as a series in powers of \(\beta=v / c .\) (a) Verify that the first term is just the non relativistic kinetic energy, and show that to lowest order in \(\beta\) the difference between the relativistic and non relativistic kinetic energies is \(3 \beta^{4} m c^{2} / 8 .\) (b) Use this result to find the maximum speed at which the non relativistic value is within \(1 \%\) of the correct relativistic value.

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