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Chapter 5: Problem 1

How fast must a particle move before its kinetic energy equals its rest energy? [0.866 c]

Short Answer

Expert verified
Answer: A particle's kinetic energy equals its rest energy at a velocity of approximately 0.866c, where c is the speed of light.

Step by step solution

01

Understanding Kinetic and Rest Energy

Kinetic energy (KE) is the energy of a particle due to its motion, while rest energy is the energy a particle has when it's at rest (not moving). The relativistic kinetic energy formula is given by: KE = (γ - 1) mc^2 where m is the mass of the particle, c is the speed of light, and γ is the Lorentz factor, given by: γ = 1 / sqrt(1 - (v^2/c^2)) The rest energy (E) of a particle is given by: E = mc^2 We want to find the velocity (v) at which the kinetic energy equals the rest energy.
02

Set Kinetic Energy Equal to Rest Energy

To find the velocity at which the kinetic energy equals the rest energy, we set the relativistic kinetic energy formula equal to the rest energy formula: (γ - 1) mc^2 = mc^2
03

Simplify the Equation

Notice that mc^2 is a common factor on both sides of the equation. Divide both sides by mc^2 to cancel it: (γ - 1) = 1
04

Solve for the Lorentz Factor (γ)

Now, we need to solve for γ: γ = 1 + 1 γ = 2
05

Find the Velocity (v)

We have the value of γ. Let's plug it back into the Lorentz factor formula and solve for the velocity: 2 = 1 / sqrt(1 - (v^2/c^2)) Square both sides of the equation: 4 = 1 / (1 - (v^2/c^2)) Now, multiply both sides by (1 - (v^2/c^2)) to eliminate the denominator: 4(1 - (v^2/c^2)) = 1 Distribute the 4 on the left side: 4 - 4(v^2/c^2) = 1 Subtract 4 from both sides: -4(v^2/c^2) = -3 Divide both sides by -4: (v^2/c^2) = 3/4 Multiply both sides by c^2: v^2 = (3/4)c^2 Take the square root of both sides: v = sqrt((3/4)c^2) And finally, simplify the equation: v ≈ 0.866c So, a particle must move at approximately 0.866 times the speed of light before its kinetic energy equals its rest energy.

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Key Concepts

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

Rest Energy
When we speak of the energy of an object at rest, we're talking about its inherent energy—energy that it possesses because of its mass, regardless of its position or motion. In physics, this is often referred to as 'rest energy' and is calculated using the famous equation from Einstein's theory of relativity: \( E = mc^2 \).

The 'm' in this equation represents the mass of the object, while 'c' is the speed of light in a vacuum, approximately \(3 \times 10^8 \) meters per second. The speed of light squared is an enormous number, signifying that even a small amount of mass corresponds to a huge amount of energy. Understanding rest energy is crucial, as it forms the basis of the principle of mass-energy equivalence, a core concept in modern physics which indicates that mass can be converted into energy and vice versa.

For any particle, the rest energy is often the starting point for understanding more complex forms of energy when the particle is in motion, such as kinetic energy in relativistic contexts.
Lorentz Factor
Relativity introduces us to speeds that are a significant fraction of the speed of light, and these velocities do not adhere to classical mechanics. Here, the Lorentz factor (\( \gamma \)) becomes an essential concept. The Lorentz factor describes how time, length, and relativistic mass change for an object while it is moving.

The formula for the Lorentz factor is \( \gamma = 1 / \sqrt{1 - (v^2/c^2)} \), where 'v' is the velocity of the object and 'c' is again the speed of light. As the speed of an object approaches the speed of light, the Lorentz factor increases dramatically. At speeds much less than the speed of light, the Lorentz factor is close to 1, meaning relativistic effects are negligible—Newton's classical mechanics holds true. However, as the speed reaches a significant fraction of light speed, relativistic effects cannot be ignored, and the object's kinetic energy increases more than what would be predicted by classical energy formulas.
Speed of Light
The speed of light in a vacuum, symbolized by 'c', is a fundamental constant of nature and is approximately \(299,792,458 \) meters per second. This speed is the ultimate speed limit in the universe, and according to Einstein's theory of relativity, nothing can travel faster than light.

In the context of relativistic kinetic energy, the speed of light is the threshold that separates Newtonian mechanics from relativistic mechanics. It functions as a conversion factor between mass and energy and plays a significant role in determining the movement and interaction of particles at high speeds. Since the speed of light is such a large number, when squared in equations like Einstein's mass-energy equivalence formula, it illustrates the massive amounts of energy contained within a small mass. Moreover, as an object approaches the speed of light, its kinetic energy increases without bound as a result of the relativistic effects—spotlighting the intricate link between velocity, energy, and mass in the realm of high-speed physics.

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Most popular questions from this chapter

If one neutron and one pi-meson are to ernerge from the collision of a photon with a stationary proton, find the threshold frequency of the photon in terms of the rest mass \(n\) of a proton o neutron (here assumed equal) and that, \(m\), of a pi-meson. [Answer \(\left.c^{2}\left(m^{2}+2 m n\right) / 2 h n .\right]\)

If \(\Delta x^{*}=(c \Delta t, \Delta \mathbf{r})\) is the four-vector join of two events on the worldline of a uniformly moving particle (or photon), prove that the frequency vector of its de Broglie wave is given by $$ N^{\mu}=\frac{v}{c \Delta t} \Delta x^{\mu}, $$ whence \(v / \Delta t\) is invariant. Compare with Exercise III \((7)\).

A rocket propels itself rectilinearly by giving portions of its mass a constant fbackward) velocity \(U\) relative to its instantaneous rest frame. It continues to do so until it attains a velocity \(V\) relative to its initial rest frame. Prove that the ratio of the initial to the final rest mass of the rocket is given by $$ \frac{M_{\mathrm{i}}}{M_{\mathrm{f}}}=\left(\frac{c+V}{c-V}\right)^{\frac{c}{2 U}} $$

taking \(h=6.63 \times 10^{-27} \mathrm{crgs}\) and \(c=300000 \mathrm{~km} \mathrm{~s}^{-1}\), calculate how many photons of wavelength \(5 \times 10^{-5} \mathrm{~cm}\) (in the yellowgreen of the visible spectrum) must fall per second on a blackened plate to produce a force of one dyne. [Answer: \(7.3 \times 10^{21}\). Hint: force equals momentum absorbed per unit time.]

Prove that, in relativistic as in Newtonian mechanics, the time rate of change of the angular momentum \(h=\mathbf{r} \times p\) of a particle about an origin \(\mathrm{O}\) is equal to the couple \(\mathrm{r} \times \mathrm{f}\) of the applied force about \(\mathrm{O}\). If 1 \(^{\mu v}\) is the particle's four- angular momentum, and if we define the fourcouple \(G^{\mu v}=x^{\mu} F^{v}-x^{v} F^{\mu}\), where \(x^{\mu}\) and \(F^{\mu}\) are the four-vectors corresponding to \(\mathbf{r}\) and \(\mathbf{f}\), prove that \((\mathrm{d} / \mathrm{d} \tau) L^{\mu \nu}=G^{\mu \nu}\) and that the spacc-space part of this equation corresponds to the above threevector result.
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