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Chapter 37: Problem 41

Compute the kinctic energy of a proton (mass \(1.67 \times$$10^{-27} \mathrm{kg}\) using both the nonrelativistic and relativistic expressions, and compute the ratio of the two results (relativistic divided by nonrelativistic) for speeds of (a) \(8.00 \times 10^{7} \mathrm{m} / \mathrm{s}\) and (b) \(2.85 \times 10^{8} \mathrm{m} / \mathrm{s}\) .

Short Answer

Expert verified
For speed (a), the ratio is close to 1. For speed (b), the ratio is significantly greater than 1.

Step by step solution

01

Define Nonrelativistic Kinetic Energy

The formula for nonrelativistic kinetic energy is given by \( KE_{nr} = \frac{1}{2}mv^2 \). Here, we have the mass of the proton \( m = 1.67 \times 10^{-27} \; \text{kg} \) and the velocity \( v \). We will calculate \( KE_{nr} \) for both speeds.
02

Calculate Nonrelativistic Kinetic Energy for Speed (a)

Substitute \( v = 8.00 \times 10^7 \; \text{m/s} \) into the nonrelativistic formula: \[KE_{nr, (a)} = \frac{1}{2} \times 1.67 \times 10^{-27} \times (8.00 \times 10^7)^2\] Calculate to find \( KE_{nr, (a)} \).
03

Calculate Nonrelativistic Kinetic Energy for Speed (b)

Substitute \( v = 2.85 \times 10^8 \; \text{m/s} \) into the nonrelativistic formula: \[KE_{nr, (b)} = \frac{1}{2} \times 1.67 \times 10^{-27} \times (2.85 \times 10^8)^2\] Calculate to find \( KE_{nr, (b)} \).
04

Define Relativistic Kinetic Energy

The formula for relativistic kinetic energy is \( KE_r = (\gamma - 1)mc^2 \), where \( \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \) and \( c = 3 \times 10^8 \; \text{m/s} \). We will calculate \( KE_r \) for both speeds.
05

Calculate Relativistic Kinetic Energy for Speed (a)

First, find \( \gamma \) for \( v = 8.00 \times 10^7 \; \text{m/s} \):\[\gamma = \frac{1}{\sqrt{1-\left(\frac{8.00 \times 10^7}{3 \times 10^8}\right)^2}}\]Then, calculate \( KE_r \):\[KE_{r, (a)} = (\gamma - 1) \times 1.67 \times 10^{-27} \times (3 \times 10^8)^2\]
06

Calculate Relativistic Kinetic Energy for Speed (b)

First, find \( \gamma \) for \( v = 2.85 \times 10^8 \; \text{m/s} \):\[\gamma = \frac{1}{\sqrt{1-\left(\frac{2.85 \times 10^8}{3 \times 10^8}\right)^2}}\]Then, calculate \( KE_r \):\[KE_{r, (b)} = (\gamma - 1) \times 1.67 \times 10^{-27} \times (3 \times 10^8)^2\]
07

Calculate the Ratio of Relativistic to Nonrelativistic Kinetic Energy for Speed (a)

The ratio \( R_{(a)} \) is given by:\[R_{(a)} = \frac{KE_{r, (a)}}{KE_{nr, (a)}}\]Calculate \( R_{(a)} \) using the values obtained from the previous steps.
08

Calculate the Ratio of Relativistic to Nonrelativistic Kinetic Energy for Speed (b)

The ratio \( R_{(b)} \) is given by:\[R_{(b)} = \frac{KE_{r, (b)}}{KE_{nr, (b)}}\]Calculate \( R_{(b)} \) using the values obtained from the previous steps.

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Key Concepts

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

Nonrelativistic Kinetic Energy
The concept of nonrelativistic kinetic energy is tied closely to what we typically learn in basic physics about energy in motion. Simply put, kinetic energy is the energy an object has due to its motion. For slow-moving objects, especially those moving at much less than the speed of light, the classical or nonrelativistic formula for kinetic energy works well: \[ KE_{nr} = \frac{1}{2}mv^2 \] Here, \( m \) represents mass, and \( v \) is the velocity of the object. In our scenario, we're calculating the kinetic energy of a proton, whose mass is \( 1.67 \times 10^{-27} \mathrm{kg} \). This formula gives a good estimate of kinetic energy when the velocity is a small fraction of the speed of light, which is common in everyday situations.
Relativistic Kinetic Energy
As speeds approach that of light, Einstein鈥檚 theory of special relativity shows that classical physics doesn't quite hold up. The relativistic kinetic energy accounts for the increased mass of an object as it moves ever faster. The formula is:\[ KE_r = (\gamma - 1)mc^2 \] Here, \( c \) is the speed of light, and \( \gamma \) (gamma) is the Lorentz factor, calculated as:\[ \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \] This factor becomes significant at speeds close to the speed of light, reflecting the dependency of a particle鈥檚 energy on its velocity. This equation is crucial for understanding high-speed particles in systems like particle accelerators.
Proton Mass
When dealing with particles like protons, their mass is a foundational element in any kinetic energy calculation. The proton is one of the fundamental constituents of atoms, having a mass of approximately \( 1.67 \times 10^{-27} \mathrm{kg} \). Though seemingly negligible, this mass is crucial in scientific calculations, especially in physics experiments and theoretical calculations. Protons, being part of the atomic nucleus, interact under forces including the strong nuclear force, making their mass essential in both classical and modern physics calculations.
Speed of Light
In physics, the speed of light (\( c \), approximately \( 3 \times 10^8 \mathrm{m/s} \)) is a significant constant that plays an essential role in the structure of the universe. It is not just a measure of the fastest speed at which information or matter can travel, but a fundamental component of Einstein's mass-energy equivalence and special relativity. Understanding its role is key when considering relativistic effects, where velocities approaching this constant hugely influence the behavior and energy of moving objects, as seen in our relativistic energy formulas.
Special Relativity
The groundbreaking theory of special relativity was introduced by Albert Einstein in 1905. It revolutionized how we understand time, space, and energy. This theory demonstrates how the laws of physics are identical for all observers in uniform motion relative to one another, introducing the concept that the speed of light is constant in vacuum. It reshapes how we calculate energy and momentum for objects moving at high velocities, leading to a more accurate understanding of kinetic energy beyond classical Newtonian mechanics. This understanding allows scientists to predict the behavior of particles moving close to the speed of light, an essential aspect of modern physics.

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Most popular questions from this chapter

In high-energy physics, new particles can be created by collisions of fast- moving projectile particles with stationary particles. Some of the kinetic energy of the incident particle is used to create the mass of the new particle. A proton-proton collision can result in the creation of a negative kaon \(\left(\mathrm{K}^{-}\right)\) and a positive \(\mathrm{kaon}\left(\mathrm{K}^{+}\right) :\) $$p+p \rightarrow p+p+\mathbf{K}^{-}+\mathbf{K}^{+}$$ (a) Calculate the minimum kinetic energy of the incident proton that will allow this reaction to occur if the second (target) proton is initially at rest. The rest energy of each kaon is 493.7 \(\mathrm{MeV}\) , and the rest energy of each proton is 938.3 MeV. (Hint: It is useful here to work in the frame in which the total momentum is zero. See Problem 8.100 , but note that here the Lorentz transformation must be used to relate the velocities in the laboratory frame to those in the zero-total-momentum frame.) (b) How does this calculated minimum kinetic energy compare with the total rest mass energy of the created kaons? (c) Suppose that instead the two protons are both in motion with velocities of equal magnitude and opposite direction. Find the minimum combined kimetic energy of the two protons that will allow the reaction to occur. How does this calculated minimum kinetic energy compare with the total rest mass energy of the created kaons? (This example shows that when colliding beams of particles are used instead of a stationary target, the energy requirements for producing new particles are reduced substantially.)

A particle with mass \(m\) accelerated from rest by a constant force \(F\) will, according to Newtonian mechanics, continue to accelerate without bound; that is, as \(t \rightarrow \infty, v \rightarrow \infty .\) Show that according to relativistic mechanics, the particle's speed approaches \(c\) as \(t \rightarrow \infty\) . I Note: Auseful integralis \(\int\left(1-x^{2}\right)^{-3 / 2} d x=x / \sqrt{1-x^{2}} \cdot 1\)

(a) Consider the Galilean transformation along the \(x\) -direction: \(x^{\prime}=x-v t\) and \(t^{\prime}=t\) . In frame \(S\) the wave equation for electromagnetic waves in a vacuum is $$\frac{\partial^{2} E(x, t)}{\partial x^{2}}-\frac{1}{c^{2}} \frac{\partial^{2} E(x, t)}{\partial t^{2}}=0$$ where \(E\) represents the electric field in the wave. Show that by using the Galilean transformation the wave equation in frame \(S^{\prime}\) is found to be $$\left(1-\frac{v^{2}}{c^{2}}\right) \frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial x^{\prime 2}}+\frac{2 v}{c^{2}} \frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial x^{\prime} \partial t^{\prime}}-\frac{1}{c^{2}} \frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial t^{2}}=0$$ This has a different form than the wave equation in \(S\) . Hence the Galiean transformation violates the first relativity postulate that all physical laws have the same form in all inertial reference frames. (Hint: Express the derivatives \(\partial / \partial x\) and \(\partial / \partial t\) in terms of part (a), but use the Lorentz coordinate transformations, Eqs. (37.21), and show that in frame \(S^{\prime}\) the wave equation has the same form as in frame \(S\) : $$\frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial x^{2}}-\frac{1}{c^{2}} \frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial t^{\prime 2}}=0$$ Explain why this shows that the speed of light in vacuum is \(c\) in both frames \(S\) and \(S^{\prime} .\)

Show that when the source of electromagnetic waves moves away from us at 0.600 \(\mathrm{c}\) , the frequency we measure is half the value measured in the rest frame of the source.

A proton (rest mass \(1.67 \times 10^{ \times 27} \mathrm{kg} )\) has total energy that is 4.00 times its rest energy. What are (a) the kinetic energy of the proton; (b) the magnitude of the momentum of the proton; (c) the speed of the proton?
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