Problem 55 - Fundamentals of Physics

Question

A certain particle of mass \(m\) has momentum of magnitude \(m c\). What are (a) \(\beta\), (b) \(\gamma\), and (c) the ratio \(K / E_{0}\) ?

Step-by-Step Solution

Answer

(a) \( \beta = \frac{1}{\sqrt{2}} \); (b) \( \gamma = \sqrt{2} \); (c) \( \frac{K}{E_{0}} = \sqrt{2} - 1 \).

Step 1: Understand the Relation between Momentum and Velocity

The momentum of a particle is given by the relativistic expression: \( p = \gamma m v \), where \( \gamma = \frac{1}{\sqrt{1-\beta^2}} \) and \( \beta = \frac{v}{c} \). Since the magnitude of the momentum is given as \( p = mc \), we have:\[ mc = \gamma m v \]Solving for \( \beta \), this simplifies to \( c = \gamma v \). Therefore, it can be expressed as \( v = \frac{c}{\gamma} \).

Step 2: Solve for \( \beta \)

From \( \beta = \frac{v}{c} \) and using \( v = \frac{c}{\gamma} \) from the previous step, we have:\[ \beta = \frac{c}{c \gamma} = \frac{1}{\gamma} \]Since \( \gamma \) is greater than 1 for any moving particle, calculate \( \beta \). We'll solve for \( \gamma \) next to get \( \beta \).

Step 3: Solve for \( \gamma \)

From Step 1, \( p = mc = \gamma m v \) simplifies to:\[ \gamma = \frac{mc}{mv} \]\[ \gamma = \frac{c}{v} \]Because \( v = \frac{c}{\gamma} \), \( \beta = \frac{1}{\gamma} \), we equate:\[ \gamma = \frac{1}{\sqrt{1-\left(\frac{1}{\gamma}\right)^2}} \]Solve for \( \gamma \gamma^2 = 2 \Rightarrow \gamma = \sqrt{2} \).

Step 4: Calculate \( \beta \) with \( \gamma \)

Now that \( \gamma = \sqrt{2} \), substitute into \( \beta = \frac{1}{\gamma} \):\[ \beta = \frac{1}{\sqrt{2}} \]

Step 5: Find the Ratio \( \frac{K}{E_{0}} \)

The kinetic energy \( K \) is given by \( K = (\gamma - 1) E_0 \), where \( E_0 = mc^2 \) is the rest energy.\[ K = (\sqrt{2} - 1) mc^2 \]\( \frac{K}{E_{0}} = \sqrt{2} - 1 \) simplifies to: \( \frac{K}{E_{0}} = \sqrt{2} - 1 \).

Key Concepts

Momentum

In the realm of relativistic mechanics, momentum is a bit different than what we encounter in classical physics. Instead of the simple formula \( p = mv \) we use in classical mechanics, relativistic momentum is given by the equation \( p = \gamma mv \). Here, \( \gamma \) is the Lorentz factor, which modifies the momentum calculation based on the velocity of the object.

Momentum in this context considers the effects of special relativity, especially significant as objects approach the speed of light \( c \). The momentum increases more sharply than in non-relativistic scenarios due to the factor \( \gamma \), making calculations crucially dependent on the speed relative to the speed of light.

• Relativistic momentum increases rapidly as velocity approaches \( c \).
• It's essential for calculations involving subatomic particles traveling at high velocities.
• Represents a conserved quantity, just like in classical mechanics, but modified by relativistic effects.

Overall, understanding relativistic momentum requires an appreciation of how velocities close to light speed lead to significant effects that differ from everyday experiences.

Lorentz Factor

The Lorentz factor, denoted by \( \gamma \), plays a pivotal role in understanding how relativistic effects change our perceptions of motion and energy. It is defined as \( \gamma = \frac{1}{\sqrt{1-\beta^2}} \), where \( \beta \) is the ratio of the object's velocity to the speed of light (\( \beta = \frac{v}{c} \)).

This factor becomes particularly crucial when an object's velocity approaches the speed of light. As \( \beta \) increases, \( \gamma \) increases dramatically, leading to significant changes in momentum and kinetic energy calculations. In the problem given, solving for \( \gamma \) helped determine the unknowns in the exercise.

• The Lorentz Factor account for relativistic effects at high velocities.
• As a result, it influences both time dilation and length contraction phenomena.
• \( \gamma \) is greater than or equal to 1. It equals 1 when the velocity is zero.

In essence, the Lorentz Factor is a gateway to understanding the profound consequences of moving at speeds approaching light.

Kinetic Energy

Relativistic kinetic energy is given by the equation \( K = (\gamma - 1)E_0 \), where \( E_0 = mc^2 \) is the rest energy of the particle. This connection to rest energy signifies a departure from the classical kinetic energy formula \( K = \frac{1}{2}mv^2 \).

As velocity nears the speed of light, kinetic energy increases sharply because of the \( \gamma \) factor. This makes particle accelerators, such as those used in high-energy physics experiments, consider these relativistic effects. The exercise asked for the ratio \( \frac{K}{E_0} \), showcasing how these values differ from classical approximations.

• Relativistic kinetic energy can not be simply added like classical kinetic energy because of \( \gamma \)'s influence.
• At low speeds, relativistic kinetic energy approximates classical kinetic energy.
• The rest energy \( E_0 \) is a significant component, showing energy stored due to mass.

Relativistic kinetic energy widens our understanding and showcases the need for new formulas as we reach extremes in velocity.