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Chapter 41: Problem 34

The insignia painted on the side of a spaceship is a circle with a line across it at \(45^{\circ}\) to the vertical. As the ship shoots past another ship in space, with a relative speed of \(0.95 \mathrm{c}\), the second ship observes the insignia. What angle does the observed line make to the vertical?

Short Answer

Expert verified
The observed angle is approximately 6.34 degrees to the vertical.

Step by step solution

01

Understanding the Problem

We have an insignia on a spaceship which is a circle with a line at a 45-degree angle to the vertical. This spaceship moves past another spaceship at a speed of 0.95 times the speed of light (\(c\)). The second ship observes this insignia at a different angle because of relativistic effects. Our task is to find this observed angle.
02

Consider Relevant Concept: Lorentz Transformation of Angles

Due to the relative motion of the spaceships at a significant fraction of the speed of light, relativistic effects come into play. Specifically, angles in the direction of motion undergo a transformation. This problem can be solved using the Lorentz Transformation for angles.
03

Apply the Lorentz Angle Transformation Formula

The angle transformation formula in special relativity is given by:\[\tan(\theta') = \frac{\sin(\theta)}{\gamma (\cos(\theta) - \frac{v}{c})}\]where \(\theta'\) is the observed angle, \(\theta\) is the original angle (45 degrees in this case), \(v\) is the velocity of the moving spaceship (0.95\(c\)), \(c\) is the speed of light, and \(\gamma = \frac{1}{\sqrt{1-v^2/c^2}}\) is the Lorentz factor.
04

Calculate the Lorentz Factor \(\gamma\)

First, calculate the Lorentz factor with the given speed:\[\gamma = \frac{1}{\sqrt{1 - (0.95)^2}}\]Computing this gives:\[\gamma \approx 3.202\]
05

Calculate the Original Trigonometric Values

The original angle \(\theta\) is 45 degrees: \[\sin(45^\circ) = \cos(45^\circ) = \frac{\sqrt{2}}{2}\]
06

Substitute Values into Tan Inverse Formula and Solve for \(\theta'\)

Now substitute the values into the angle transformation formula:\[\tan(\theta') = \frac{\sin(45^\circ)}{3.202 \left(\cos(45^\circ) - 0.95 \right)}\]Plug in the values:\[\tan(\theta') = \frac{\frac{\sqrt{2}}{2}}{3.202 \left(\frac{\sqrt{2}}{2} - 0.95 \right)}\]Solving this gives:\[\theta' \approx 6.34^\circ\]
07

Finalize the Solution

The angle \(\theta'\) observed by the second ship is approximately 6.34 degrees with respect to the vertical.

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Key Concepts

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

Relativistic Effects
In the realm of high-speed physics, relativistic effects become crucial when dealing with objects moving at speeds close to that of light. The exercise above highlights how such velocities can dramatically alter the apparent angles of observed phenomena.
Relativistic effects stem from Einstein's theory of relativity, which describes how time, space, and motion are interconnected. When an object travels at a significant fraction of the speed of light, measurements of distance, time, and even angles can differ markedly from those experienced at everyday speeds.
In the spaceship scenario, as one spaceship moves rapidly relative to the other, the line's angle on the insignia seems to shift from 45 degrees to approximately 6.34 degrees. This change is not due to any physical alteration but is instead a consequence of these relativistic effects.
These transformations can be quantified using specific mathematical tools, such as Lorentz transformations, to calculate how observed angles change with respect to the observer's motion.
Special Relativity
Special relativity, laid out by Albert Einstein in 1905, fundamentally changed our understanding of physics. It's based on two postulates: the laws of physics are identical in all inertial frames, and the speed of light is constant in a vacuum for all observers, regardless of their own speed.
Under special relativity, time and space are not independent entities but are intertwined into a four-dimensional construct known as spacetime. This means that velocity affects the perception of time and space.
  • Time Dilation: Moving clocks tick slower as seen from a stationary observer's frame, which explains anomalies like the twin paradox.
  • Length Contraction: Objects contract along the direction of motion when seen by a stationary observer.
  • Relativistic Simultaneity: Events that occur simultaneously in one frame may not be simultaneous in another.
This framework enables the calculation and understanding of how angles change as shown in the initial exercise, due to the high-speed observations involved.
Lorentz Factor
The Lorentz Factor, denoted by \( \gamma \), is a key component in the equations of special relativity. It is used for calculating relativistic effects, particularly noticeable at high velocities close to that of light.
It is defined as:\[\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\]where \( v \) is the velocity of the object and \( c \) is the speed of light. As the velocity approaches the speed of light, \( \gamma \) increases significantly, showcasing the dramatic effects relativity has at these scales.
  • For low speeds (much less than speed of light), \( \gamma \) approximates to 1, leading to negligible relativistic effects.
  • As \( v \) approaches \( c \), \( \gamma \) increases, emphasizing the increasing relativistic effects.
  • The Lorentz Factor plays a crucial role in transforming physical quantities like time, length, and as shown in the exercise, observed angles.
In calculating the observed angle \( \theta' \), \( \gamma \) was pivotal in accounting for how the angle appeared smaller due to the ship's high speed.

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Most popular questions from this chapter

The Sun radiates energy equally in all directions. At the position of the Earth \(\left(r=1.50 \times 10^{11} \mathrm{~m}\right)\), the irradiance of the Sun's radiation is \(1.4 \mathrm{~kW} / \mathrm{m}^{2}\). How much mass does the Sun lose per day because of the radiation? The area of a spherical shell centered on the Sun and passing through the Earth is $$ \text { Area }=4 \pi r^{2}=4 \pi\left(1.50 \times 10^{11} \mathrm{~m}\right)^{2}=2.83 \times 10^{23} \mathrm{~m}^{2} $$ Through each square meter of this area, the Sun radiates an energy per second of \(1.4 \mathrm{~kW} / \mathrm{m}^{2}\). Therefore, the Sun's total radiation per second is $$ \text { Energy } / \mathrm{s}=(\text { area })\left(1400 \mathrm{~W} / \mathrm{m}^{2}\right)=3.96 \times 10^{26} \mathrm{~W} $$ The energy radiated in one day \((86400 \mathrm{~s})\) is Energy/day \(=\left(3.96 \times 10^{26} \mathrm{~W}\right)(86400 \mathrm{~s} /\) day \() \times 3.42 \times 10^{31} \mathrm{~J} /\) day Because mass and energy are related through \(\Delta \mathrm{E}_{0}=\Delta m \mathrm{c}^{2}\), the mass loss per day is $$ \Delta m=\frac{\Delta E_{0}}{c^{2}}=\frac{3.42 \times 10^{31} \mathrm{~J}}{\left(2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)^{2}}=3.8 \times 10^{14} \mathrm{~kg} $$ For comparison, the Sun's mass is \(2 \times 10^{30} \mathrm{~kg}\).

Find the speed and momentum of a proton \(\left(m=1.67 \times 10^{-27} \mathrm{~kg}\right)\) that has been accelerated through a potential difference of 2000 MV. (We call this a \(2 \mathrm{GeV}\) proton.) Give your answers to three significant figures.

A person in a spaceship holds a meterstick as the ship shoots past the Earth with a speed \(u\) parallel to the Earth's surface. What does the person in the ship notice as the stick is rotated from parallel to perpendicular to the ship's motion? The stick behaves normally; it does not change its length, because it has no translational motion relative to the observer in the spaceship. However, an observer on Earth would measure the stick to be \((1 \mathrm{~m}) \sqrt{1-(v / \mathrm{c})^{2}}\) long when it is parallel to the ship's motion, and \(1 \mathrm{~m}\) long when it is perpendicular to the ship's motion.

A proton \(\left(\mathrm{m}=1.67 \times 10^{-27} \mathrm{~kg}\right)\) is accelerated to a kinetic energy of \(200 \mathrm{MeV}\). What is its speed at this energy?

A spacecraft moving at \(0.95 \mathrm{c}\) travels from the Earth to the star Alpha Centauri, which is \(4.5\) light years away. How long will the trip take according to \((a)\) Earth clocks and \((b)\) spacecraft clocks? (c) How far is it from Earth to the star according to spacecraft occupants? ( \(d\) ) What do they compute their speed to be? A light year is the distance light travels in 1 year, namely 1 light year \(=\left(2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)\left(3.16 \times 10^{7} \mathrm{~s}\right)=9.47 \times 10^{15} \mathrm{~m}\) Hence the distance to the star (according to earthlings) is $$ \begin{array}{l} d_{e}=(4.5)\left(9.47 \times 10^{15} \mathrm{~m}\right)=4.3 \times 10^{16} \mathrm{~m} \\ \text { (a) } \Delta t_{e}=\frac{d_{e}}{v}=\frac{4.3 \times 10^{16} \mathrm{~m}}{(0.95)\left(2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)}=1.5 \times 10^{8} \mathrm{~s} \end{array} $$ (b) Because clocks on the moving spacecraft run slower, $$ \Delta t_{\text {cnth }}=\Delta t_{\varepsilon} \sqrt{1-(v / \mathrm{c})^{2}}=\left(1.51 \times 10^{8} \mathrm{~s}\right)(0.312)=4.7 \times 10^{7} \mathrm{~s} $$ (c) For the spacecraft occupants, the Earth-star distance is moving past them with speed 0.95c. Therefore, that distance is shortened for them; they find it to be $$ d_{\text {cath }}=\left(4.3 \times 10^{16} \mathrm{~m}\right) \sqrt{1-(0.95)^{2}}=1.3 \times 10^{16} \mathrm{~m} $$ (d) For the spacecraft occupants, their relative speed is $$ v=\frac{d_{\text {cnft }}}{\Delta t_{\text {caff }}}=\frac{1.34 \times 10^{16} \mathrm{~m}}{4.71 \times 10^{7} \mathrm{~s}}=2.8 \times 10^{8} \mathrm{~m} / \mathrm{s} $$ which is 0.95c. Both Earth and spacecraft observers measure the same relative speed.
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