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Chapter 37: Problem 36

(a) How much work must be done on a particle with mass \(m\) to accelerate it (a) from rest to a speed of 0.090\(c\) and (b) from a speed of 0.900\(c\) to a speed of 0.990\(c ?\) (Express the answers in terms of \(m c^{2} . )(\mathrm{c})\) How do your answers in parts \((\mathrm{a})\) and (b) compare?

Short Answer

Expert verified
(a) 0.004 mc^2, (b) 4.795 mc^2. Work for (b) is much greater due to relativistic effects.

Step by step solution

01

Identify the formula for relativistic work

The work done to accelerate a particle can be calculated by the change in kinetic energy, which is given by: \[ W = \Delta KE = \gamma mc^2 - m c^2 , \] where \( \gamma \) is the Lorentz factor, \( \gamma = \frac{1}{\sqrt{1-v^2/c^2}} \).
02

Calculate the work for part (a) (from rest to 0.090c)

First, find the Lorentz factor for speed \( v = 0.090c \): \[ \gamma_1 = \frac{1}{\sqrt{1-(0.090)^2}} = 1.004. \] The initial kinetic energy at rest is 0 (\(v=0\)). Thus, the work done is: \[ W_1 = (\gamma_1 - 1) mc^2 = (1.004 - 1) mc^2 = 0.004 mc^2. \]
03

Calculate the work for part (b) (from 0.900c to 0.990c)

First, find the Lorentz factors for both speeds: \[ \gamma_2 = \frac{1}{\sqrt{1-(0.900)^2}} = 2.294, \] \[ \gamma_3 = \frac{1}{\sqrt{1-(0.990)^2}} = 7.089. \] Then calculate the work done from 0.900c to 0.990c: \[ W_2 = (\gamma_3 - \gamma_2) mc^2 = (7.089 - 2.294) mc^2 = 4.795 mc^2. \]
04

Compare the work done in parts (a) and (b)

The work done in part (b) \((4.795 mc^2)\) is significantly greater than in part (a) \((0.004 mc^2)\). This difference highlights how much more energy is required to increase the speed of a particle at higher speeds close to the speed of light due to relativistic effects.

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Key Concepts

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

Work-Energy Principle
The work-energy principle is a fundamental concept in physics that connects the work done on an object to its change in kinetic energy.
In a relativistic context, this principle is expressed as the difference in the relativistic kinetic energy of a particle, which takes into account the effects of high velocities close to the speed of light.
It tells us how much work, or energy, we need to apply to accelerate an object from one speed to another.
  • The formula used is: \[ W = \Delta KE = \gamma mc^2 - mc^2 \] where \( W \) is the work done.
  • This formula incorporates the Lorentz factor \( \gamma \), which modifies the calculations based on the particle's velocity.
Understanding the work-energy principle with relativity helps us see why significant work is needed to speed up particles as they move faster.
Lorentz Factor
The Lorentz factor, denoted by \( \gamma \), is crucial in relativistic physics.
It arises from Einstein's theory of relativity and encapsulates the effect of speed on mass, time, and energy as objects move near the speed of light.
The factor is given by:
  • \( \gamma = \frac{1}{\sqrt{1-v^2/c^2}} \)
  • Here, \( v \) is the speed of the object, and \( c \) is the speed of light.
As the object's velocity \( v \) approaches the speed of light \( c \), \( \gamma \) dramatically increases, illustrating the increasingly larger "relativistic effects".
This means that more energy is needed to continue accelerating the particle.
Kinetic Energy
Kinetic energy in relativistic physics differs from its classical counterpart.
In the realm of high speeds, the classical formula, \( KE = \frac{1}{2}mv^2 \), is replaced by a relationship that incorporates the Lorentz factor:
  • Relativistic kinetic energy is: \[ KE = (\gamma - 1) mc^2 \]
  • This formula shows that at lower speeds, kinetic energy closely resembles classical predictions.
  • At speeds approaching \( c \), kinetic energy increases not linearly but with a more complex relationship, reflecting relativistic effects.
Understanding this is essential for comprehending why particle accelerators require massive amounts of energy to increase particles to near light speed.
Relativity
Relativity is the overarching theory developed by Albert Einstein that changes how we understand time, space, and energy.
It consists of two theories: Special Relativity and General Relativity. For the purpose of our problem, we are concerned with Special Relativity, which deals with particles moving at significant fractions of the speed of light.
  • It introduces ideas like time dilation, length contraction, and the relativity of simultaneity.
  • Special Relativity is key to understanding the work-energy problems in your physics homework, explaining why the energy required doesn't just scale linearly with speed.
  • These insights change the concept of energy itself, leading to new equations like \( E=mc^2 \) that links energy and mass.
This theory is fundamental for modern science and technology, influencing particle physics and cosmology.
Speed of Light
The speed of light, represented by \( c \), is a constant with a value of approximately 299,792,458 meters per second.
This is not just an important number; it is the universe's speed limit!
  • According to relativity, no object with mass can reach or exceed the speed of light.
  • The value of \( c \) deeply affects equations in relativistic physics, as seen in the Lorentz factor and energy formulas.
As particles approach this speed, their mass effectively increases, meaning more and more energy is needed to continue accelerating them.
This principle explains why the energy demands become enormous at high velocities, as seen in particle accelerators and nuclear physics.
Realizing the limit set by the speed of light helps in comprehending the mechanics of the universe as we accelerate particles in experiments.

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Most popular questions from this chapter

Determining the Masses of Stars. Many of the stars in the sky are actually binary stars, in which two stars orbit about their common center of mass. If the orbital speeds of the stars are high enough, the motion of the stars can be detected by the Doppler shifts of the light they empler. Stars for which this is the case are called spectroscopic binary stars. Figure \(\mathrm{P3} 7.75\) shows the simplest case of a spectroscopic binary star: two identical stars, each with mass \(m,\) orbiting their center of mass in a circle of radius \(R\) . The plane of the stars' orbits is edge-on to the line of sight of an observer on the earth. (a) The light produced by heated hydrogen gas in a laboratory on the earth has a frequency of \(4.568110 \times\) \(10^{14}\) Hz. In the light received from the stars by a telescope on the earth, hydrogen light is observed to vary in frequency between \(4.567710 \times 10^{14} \mathrm{Hz}\) and \(4.568910 \times 10^{14} \mathrm{Hz}\) . Determine whether the binary star system as a whole is moving toward or away from the earth, the speed of this motion, and the orbital speeds of the stars. Hint: The speeds involved are much less than \(c,\) so you may use the approximate result \(\Delta f / f=u / c\) given in Section 37.6 . (b) The light from each star in the binary system varies from its maximum frequency to its minimum frequency and back again in 11.0 days. Determine the orbital radius \(R\) and the mass \(m\) of each star. Give your answer for \(m\) in kilograms and as a multiple of the mass of the sun, \(1.99 \times 10^{30}\) kg. Compare the value of \(R\) to the distance from the earth to the sun, \(1.50 \times 10^{11}\) m. (This technique is actually used in astronomy to determine the masses of stars. In practice, the problem is more complicated because the two stars in a binary system are usually not identical, the orbits are usually not circular, and the plane of the orbits is usually tilted with respect to the line of sight from the earth.)

(a) At what speed is the momentum of a particle twice as great as the result obtained from the nonrelativistic expression \(m v ?\) Express your answer in terms of the speed of light. (b) A force is applied to a particle along its direction of motion. At what speed is the magnitude of force required to produce a given acceleration twice as great as the force required to produce the same acceleration when the particle is at rest? Express your answer in terms of the speed of light.

(a) Through what potential difference does an electron have to be accelerated, starting from rest, to achieve a speed of 0.980\(c ?\) (b) What is the kinetic energy of the electron at this speed? Express your answer in joules and in electron volts.

A source of electromagnetic radiation is moving in a radial direction relative to you. The frequency you measure is 1.25 times the frequency measured in the rest frame of the source. What is the speed of the source relative to you? Is the source moving toward you or away from you?

Electrons are accelerated through a potential difference of \(750 \mathrm{kV},\) so that their kinetic energy is \(7.50 \times 10^{5} \mathrm{eV}\) . (a) What is the ratio of the speed \(v\) of an electron having this energy to the speed of light, \(c ?\) (b) What would the speed be if it were computed from the principles of classical mechanics?
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