Problem 34 - University Physics with Modern Physics

Question

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

Step-by-Step Solution

Answer

(a) 0.0041mc², (b) 4.7946mc². More work is needed at higher speeds.

Step 1: Understanding Work in Relativistic Dynamics

Work done on a particle is equal to the change in its kinetic energy. In relativistic physics, kinetic energy, commonly denoted as \(K\), can be calculated for a particle of mass \(m\) and velocity \(v\) using the formula \(K = \( \gamma m c^2 - m c^2\) \), where \( \gamma = \frac{1}{\sqrt{1 - (v^2/c^2)}} \) is the Lorentz factor. It's important to note that \(c\) is the speed of light.

Step 2: Calculating Work from Rest to 0.090c

To find the work done to accelerate from rest to 0.090\(c\), calculate the change in kinetic energy, \(\Delta K\), which is \(K_{0.090c} - K_{0}\). The initial kinetic energy \( K_{0} = 0 \), so only the final kinetic energy is needed. Plugging into the formula for kinetic energy, \(K_{0.090c} = (\gamma_{0.090} m c^2) - (m c^2)\), where \(\gamma_{0.090} = \frac{1}{\sqrt{1 - (0.090)^2}}\). Calculate \(\gamma_{0.090}\), then evaluate \(K_{0.090c}\).

Step 3: Calculating Lorentz Factor and Kinetic Energy at 0.090c

Compute the Lorentz factor: \( \gamma_{0.090} = \frac{1}{\sqrt{1 - (0.090)^2}} \approx 1.0041\).Thus, \(K_{0.090c} = (1.0041 m c^2) - (m c^2) = 0.0041 m c^2 \).Therefore, the work done from rest to 0.090\(c\) is \(0.0041 m c^2\).

Step 4: Calculating Work from 0.900c to 0.990c

To compute the work from 0.900\(c\) to 0.990\(c\), calculate \(\Delta K = K_{0.990c} - K_{0.900c}\).Find the kinetic energy at each speed using the same formula with new velocities:- \(\gamma_{0.900} = \frac{1}{\sqrt{1 - (0.900)^2}} \approx 2.2942\)- \(\gamma_{0.990} = \frac{1}{\sqrt{1 - (0.990)^2}} \approx 7.0888\).Compute \(K_{0.900c} = 2.2942m c^2 - mc^2 = 1.2942 m c^2 \) and \(K_{0.990c}=7.0888m c^2 - mc^2 = 6.0888 m c^2 \).

Step 5: Calculating Change in Kinetic Energy from 0.900c to 0.990c

Subtract the kinetic energies: \(\Delta K = (6.0888 m c^2) - (1.2942 m c^2) = 4.7946 m c^2\).Thus, the work from 0.900\(c\) to 0.990\(c\) is \(4.7946 m c^2\).

Step 6: Comparing the Results

Now compare the two work amounts: from 0 to 0.090\(c\) requires \(0.0041 m c^2\) of work, while from 0.900\(c\) to 0.990\(c\) requires \(4.7946 m c^2\) of work. Notice that the latter requires significantly more work due to the increase in relative mass as the speed approaches the speed of light, demonstrating relativistic effects.

Key Concepts

Lorentz Factor

In relativistic dynamics, the Lorentz factor is a crucial concept used to understand how motion affects time, length, and energy. It's represented by the symbol \( \gamma \) (gamma) and is calculated by the formula \( \gamma = \frac{1}{\sqrt{1 - (v^2/c^2)}} \). Here, \( v \) is the velocity of the object, and \( c \) is the speed of light. The Lorentz factor shows how much time dilation and length contraction occur at high speeds. This means two things:

• As an object's speed approaches the speed of light, the Lorentz factor grows significantly, indicating that time and distances are perceived differently.
• At low speeds, the Lorentz factor is very close to 1, meaning that relativistic effects are negligible for everyday experiences.

This factor is essential for calculating relativistic kinetic energy and plays a major role in understanding how energies transform as objects speed up to velocities near the speed of light.

Relativistic Kinetic Energy

Kinetic energy in relativistic contexts differs from classical physics due to the effects highlighted by Einstein's theory of relativity. The formula used for relativistic kinetic energy is \( K = (\gamma - 1)mc^2 \), where \( m \) is the mass of the object and \( c \) is the speed of light.

• At low speeds, this expression simplifies, as the Lorentz factor \( \gamma \) approaches 1, making the overall energy similar to classical kinetic energy \( \frac{1}{2}mv^2 \).
• At high speeds, especially close to the speed of light, the relativistic kinetic energy increases rapidly because \( \gamma \) becomes much larger.

This increase in energy showcases the reality that much more work is required to continue accelerating a mass as it nears light speed compared to when it's at lower speeds. This is because, effectively, the object's mass seems to increase due to relativistic effects, culminating in a larger energy requirement for the same amount of acceleration.

Work-Energy Principle

The work-energy principle is a foundational concept in physics, which states that the work done on an object is equal to the change in its kinetic energy. In the realm of relativistic mechanics, this principle still holds, but the calculations incorporate the effects of relativity.

• The change in kinetic energy \( \Delta K \) is used to determine how much work is done when changing the velocity of an object, specifically from speed \( v_1 \) to \( v_2 \).
• This principle remains pivotal as it helps bridge the gap between force-based dynamics and energy-based computations.
• For a particle accelerating to high speeds, greater amounts of work are observed due to the relativistic effects, which means the changes in kinetic energy are larger and require recalibrated calculations.

Thus, applying the work-energy principle in a relativistic context provides insights into how much energy (or work) is needed for significant velocity changes, crucial for understanding motion in high-energy environments, such as particles moving in accelerators.