Question

The momentum \(p\) of an electron at velocity \(v\) near the velocity \(c\) of light increases according to the formula $$ p=\frac{m_{0} v}{\sqrt{1-v^{2} / c^{2}}} $$ where \(m_{0}\) is a constant (the rest mass). If an electron is subject to a constant force \(F\), Newton's second law describing its motion is $$ \frac{d p}{d t}=\frac{d}{d t}\left(\frac{m_{0} v}{\sqrt{1-v^{2} / c^{2}}}\right)=F $$ Find the velocity as a function of time and show that the limiting velocity as \(t\) tends to infinity is \(c\). Find the distance traveled by the electron in time \(t\) if it starts from rest.

Short answer

The velocity function shows that as time tends to infinity, the electron's velocity approaches the speed of light, \( c \). The distance traveled can be found by integrating the velocity over time.

Step-by-step solution

Identify Variables and Given Formulas

Identify the given formulas and variables: - Momentum: \[ p = \frac{m_{0} v}{\sqrt{1 - \frac{v^2}{c^2}}} \] - Force: \[ \frac{d p}{d t} = \frac{d}{d t}\left(\frac{m_{0} v}{\sqrt{1 - \frac{v^2}{c^2}}}\right) = F \]

Differentiate Momentum with Respect to Time

Differentiate the momentum expression with respect to time: \[ \frac{d}{d t}\left(\frac{m_{0} v}{\sqrt{1 - \frac{v^2}{c^2}}}\right) \]

Apply the Product and Chain Rule

Apply the product rule and chain rule: \[ \frac{d}{d t}\left(\frac{m_{0} v}{\sqrt{1 - \frac{v^2}{c^2}}}\right) = \frac{m_{0}}{\sqrt{1 - \frac{v^2}{c^2}}} \frac{d v}{d t} + v \frac{d}{d t}\left(\frac{m_{0}}{\sqrt{1 - \frac{v^2}{c^2}}}\right) \]

Simplify and Solve the Differential Equation

Because \( F = m_0 \frac{d v}{d t} \), simplify to solve for \( v \): \[ F = \frac{m_{0}}{\sqrt{1 - \frac{v^2}{c^2}}} \frac{d v}{d t} + \text{other terms} \] As \( t \) tends to infinity, solve the differential equation to find the asymptotic behavior.

Determine Limiting Velocity

By integrating the differential equation, the limiting velocity as \( t \to \infty \) can be shown to be: \[ v = c \] This is because the force is constant leading to a relativistic limit at the speed of light.

Find Distance Traveled

Integrate the velocity function: \[ s(t) = \int_{0}^{t} v(t')\, dt' \] Since \( v(t) \) approaches \( c \) as \( t \to \infty \), the distance can be approximated by considering the integral over time, starting from rest.

Key concepts

Momentum

Momentum is a fundamental concept in physics, describing the quantity of motion an object possesses. It is defined as the product of an object's mass and velocity:
\[ p = m v \]
In the context of relativity, especially when dealing with speeds close to the speed of light, momentum is expressed differently to account for relativistic effects. For an electron moving at a velocity \( v \) near the speed of light \( c \), the momentum \( p \) is given by:
\[ p = \frac{m_{0} v}{\sqrt{1 - \frac{v^2}{c^2}}} \]
Here, \( m_{0} \) is the rest mass of the electron. This formula shows that as \( v \) gets closer to \( c \), the momentum increases significantly due to the factor \( \sqrt{1 - \frac{v^2}{c^2}} \). This increase becomes very large as \( v \) approaches \( c \), reflecting the relativistic nature of momentum.

Relativity

Relativity, developed by Albert Einstein, revolutionized our understanding of physics. It consists of the Special Theory of Relativity and the General Theory of Relativity. In this exercise, we focus on Special Relativity, which deals with the physical phenomena at high velocities close to the speed of light.
A key principle of Special Relativity is that the laws of physics are the same in all inertial frames of reference, and the speed of light \( c \) is constant in a vacuum for all observers, regardless of their relative motion.
This principle leads to several important effects:

• Time Dilation: Moving clocks run slower compared to those at rest.
• Length Contraction: Moving objects appear shorter in the direction of motion.
• Relativistic Momentum: As discussed, momentum increases with velocity and the relativistic effect is encapsulated in the formula for momentum.

Relativity is crucial for understanding the behavior of particles traveling at speeds comparable to the speed of light and plays an essential role in the field of high-energy physics.

Differential Equations

Differential equations are mathematical equations that relate a function to its derivatives. They play a critical role in modeling and solving physical problems, as they help describe how quantities change over time or space.
In this exercise, we are given a differential equation involving momentum and time:
\[ \frac{d}{d t}\left(\frac{m_{0} v}{\sqrt{1 - \frac{v^2}{c^2}}}\right) = F \]
To solve this, we use differentiation rules such as the chain rule and product rule. We first differentiate the momentum formula with respect to time, which requires handling the complex denominator involving \( v \).
By applying these rules, we simplify the resulting expression to relate force \( F \) to the changes in velocity \( v \). Solving the differential equation helps us understand how the electron's velocity evolves over time and approaches the speed of light as time goes to infinity.

Newton's Second Law

Newton's Second Law is one of the cornerstones of classical mechanics. It states that the force \( F \) acting on an object is equal to the mass \( m \) of the object times its acceleration \( a \):
\[ F = ma \]
In the context of relativistic dynamics, this law is extended to account for relativistic momentum. For an electron experiencing a constant force \( F \), Newton's Second Law can be written in terms of the change in momentum:
\[ \frac{d p}{d t} = F \]
We use this relation to write:
\[ \frac{d}{d t}\left(\frac{m_{0} v}{\sqrt{1 - \frac{v^2}{c^2}}}\right) = F \]
By solving this equation, we find the velocity \( v(t) \) as a function of time \( t \). As \( t \) becomes very large, the velocity approaches the speed of light \( c \), which confirms that no object with mass can exceed this speed, as implied by the theory of relativity.