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(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.

Short Answer

Expert verified
(a) The speed at which the momentum of a particle is twice as great from the nonrelativistic expression is \(v = c \sqrt{\frac{3}{4}}\). (b) The speed at which the force required to produce a given acceleration is twice as great as the force required to produce the same acceleration when the particle is at rest is \(v = c \sqrt{\frac{1}{3}}\)

Step by step solution

01

Understand and apply the relativistic momentum

Relativistic momentum is given by the equation \(p = \gamma m v \), where \(\gamma = \frac{1}{\sqrt{1 - (\frac{v}{c})^2}} \), m is the mass of the particle, v is the velocity and c is the speed of light. The nonrelativistic expression for momentum is \(p = m v \). We need to find the value of v when the relativistic momentum is twice the nonrelativistic momentum. That gives us the equation \(\gamma m v = 2m v\). Solving this gives \(\frac{1}{\sqrt{1 - (\frac{v}{c})^2}} = 2\). Simplifying this equation will yield the answer to part (a).
02

Simplify and solve the equation

Square both sides of the equation to get rid of the square root, we then have \(1 = 4 (1 - (\frac{v}{c})^2)\). Simplify this to find \(\frac{v}{c} = \sqrt{\frac{3}{4}}\). So, \(v = c \sqrt{\frac{3}{4}}\), which gives us the speed in terms of the speed of light for the first part.
03

Understanding and applying the concept of relativistic force

The relativistic concept of force is given as \(F = \gamma^3 m a\), where a is the acceleration. The nonrelativistic formula for force is \(F = m a\). Just as in the first part of the problem, you are looking for the speed at which the relativistic force is twice that of the nonrelativistic force. That provides the equation \(\gamma^3 m a = 2ma\). Proceed to solve this equation to find the speed for the second part.
04

Simplify and solve the equation

The mass and acceleration can be factored out from the equation leaving \(\gamma^3 = 2\). That reduces to \(\frac{1}{(1 - (\frac{v}{c})^2)^{3/2}} = 2\). Square both sides of the equation and simplify to find \(\frac{v}{c} = \sqrt{\frac{1}{3}}\). So, \(v = c \sqrt{\frac{1}{3}}\), which gives the speed in terms of the speed of light for the second part.

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Key Concepts

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

Relativistic Momentum
When we talk about motion at speeds close to that of light, we enter the realm of special relativity physics, a theory developed by Albert Einstein. In this domain, the momentum of an object does not simply follow the classical Newtonian mechanics equation, \( p = mv \), where \(p\) is the momentum, \(m\) is the mass, and \(v\) is the velocity of the object. Instead, we use the concept of relativistic momentum which incorporates the Lorentz factor, \( \gamma \), resulting in the equation \( p = \gamma mv \).

The reason for this change is that as objects move faster, nearing the speed of light (denoted as \(c\)), they gain relativistic mass, which means that their inertia increases, making it harder to increase their speed. To find out at which speed the momentum is twice that of the nonrelativistic value, we used the relativistic momentum formula and set it equal to twice the classical momentum \(2mv\) and solved for the velocity, \(v\). The resulting value, expressed in terms of the speed of light, shows the significant impact of relativity at high speeds.

Understanding this concept is crucial for advanced physics studies, including particle physics and cosmology, where velocities can approach the speed of light.
Lorentz Factor
The Lorentz factor or \( \gamma \) is a term that appears frequently in relativistic equations and plays a pivotal role in the theory of special relativity. It's given by the equation \( \gamma = \frac{1}{\sqrt{1 - (\frac{v}{c})^2}} \), where \(v\) is the velocity of the object and \(c\) is the speed of light. As the velocity \(v\) increases to significant fractions of the speed of light, \(\gamma\) increases dramatically, signifying that relativistic effects become more pronounced.

This factor is instrumental in understanding not just relativistic momentum but also time dilation and length contraction, two other important effects predicted by special relativity. For example, as the Lorentz factor increases, time ticks more slowly for an object in motion as compared to an observer at rest, which is a phenomenon known as time dilation.

In solving our exercise, squaring the Lorentz factor allowed us to simplify the equation and solve for \(v/c\), thereby finding the required velocity to observe the stated relativistic effects. This illustrates the significance of \(\gamma\) in translating between classical and relativistic formulas.
Special Relativity Physics
Special relativity physics is the framework we use to understand the physics of objects moving at or near the speed of light. This groundbreaking theory was developed by Einstein in 1905 and has since revolutionized our understanding of time, space, and motion. It's founded on two postulates: the laws of physics are the same in all inertial frames of reference, and the speed of light in a vacuum is the same for all observers, regardless of their relative motion.

Relativistic Force

In the context of this theory, the concept of force also gets a relativistic makeover. The classical equation \(F = ma\), where \(F\) is force and \(a\) is acceleration, is replaced by a more complex relationship involving the Lorentz factor—\(F = \gamma^3 ma\) when the force is applied in the direction of motion. This emphasis on the role of velocity relative to the speed of light determines how much harder it is to accelerate an object as it moves faster.

Our exercise demonstrated this by asking at what speed the magnitude of force required to produce a given acceleration is twice as great as when the particle is at rest. It exemplifies how the relativistic force equation deviates from classical intuition. Indeed, special relativity requires us to rethink many concepts we took for granted at low speeds, providing deeper insights into the nature of the universe.

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Most popular questions from this chapter

Calculate the magnitude of the force required to give a \(0.145 \mathrm{~kg}\) baseball an acceleration \(a=1.00 \mathrm{~m} / \mathrm{s}^{2}\) in the direction of the baseball's initial velocity when this velocity has a magnitude of (a) \(10.0 \mathrm{~m} / \mathrm{s} ;\) (b) \(0.900 c ;\) (c) \(0.990 c\) (d) Repeat parts (a), (b), and (c) if the force and acceleration are perpendicular to the velocity.

Spaceship \(A\) moves past the earth at \(0.80 c\) to the west. Spaceship \(B\) approaches \(A,\) moving to the east. Both spaceship crews measure their relative speed of approach to be \(0.98 c .\) What mass would the crews of both spaceships measure for the standard kilogram, kept at rest on the earth, (a) according to classical physics and (b) according to the special theory of relativity?

Quarks and gluons are fundamental particles that will be discussed in Chapter \(44 .\) A proton, which is a bound state of two up quarks and a down quark, has a rest mass of \(m_{\mathrm{p}}=1.67 \times 10^{-27} \mathrm{~kg}\). This is significantly greater than the sum of the rest mass of the up quarks, which is \(m_{\mathrm{u}}=4.12 \times 10^{-30} \mathrm{~kg}\) each, and the rest mass of the down quark, which is \(m_{\mathrm{d}}=8.59 \times 10^{-30} \mathrm{~kg} .\) Suppose we (incorrectly) model the rest energy of the proton \(m_{\mathrm{p}} c^{2}\) as derived from the kinetic energy of the three quarks, and we split that energy equally among them. (a) Estimate the Lorentz factor \(\gamma=\left(1-v^{2} / c^{2}\right)^{-1 / 2}\) for each of the up quarks using Eq. \((37.36) .\) (b) Similarly estimate the Lorentz factor \(\gamma\) for the down quark. (c) Are the corresponding speeds \(v_{\mathrm{u}}\) and \(v_{\mathrm{d}}\) greater than \(99 \%\) of the speed of light? (d) More realistically, the quarks are held together by massless gluons, which mediate the strong nuclear interaction. Suppose we model the proton as the three quarks, each with a speed of \(0.90 c,\) with the remainder of the proton rest energy supplied by gluons. In this case, estimate the percentage of the proton rest energy associated with gluons. (e) Model a quark as oscillating with an average speed of \(0.90 c\) across the diameter of a proton, \(1.7 \times 10^{-15} \mathrm{~m}\). Estimate the frequency of that motion.

A spaceship moving at constant speed \(u\) relative to us broad. casts a radio signal at constant frequency \(f_{0}\). As the spaceship approaches us, we receive a higher frequency \(f\); after it has passed, we receive a lower frequency. (a) As the spaceship passes by, so it is instantaneously moving neither toward nor away from us, show that the frequency we receive is not \(f_{0},\) and derive an expression for the frequency we do receive. Is the frequency we receive higher or lower than \(f_{0} ?\) (Hint: In this case, successive wave crests move the same distance to the observer and so they have the same transit time. Thus \(f\) equals \(1 / T .\) Use the dilation formula to relate the periods in the stationary and moving frames.) (b) A spaceship emits electromagnetic waves of frequency \(f_{0}=345 \mathrm{MHz}\) as measured in a frame moving with the ship. The spaceship is moving at a constant speed \(0.758 c\) relative to us. What frequency \(f\) do we receive when the spaceship is approaching us? When it is moving away? In each case what is the shift in frequency, \(f-f_{0} ?\) (c) Use the result of part (a) to calculate the frequency \(f\) and the frequency shift \(\left(f-f_{0}\right)\) we receive at the instant that the ship passes by us. How does the shift in frequency calculated here compare to the shifts calculated in part (b)?

(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 \(\left.m c^{2} .\right)\) (c) How do your answers in parts (a) and (b) compare?

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