Question

A proton has momentum with magnitude \(p_0\) when its speed is 0.400c. In terms of \(p_0\) , what is the magnitude of the proton's momentum when its speed is doubled to 0.800c?

Short answer

The new momentum is approximately 3.055 times \( p_0 \).

Step-by-step solution

Understand Momentum in Relativity

In relativistic physics, the momentum of a particle is given by the equation \( p = \frac{mv}{\sqrt{1 - \frac{v^2}{c^2}}} \), where \( m \) is the rest mass of the particle, \( v \) is the velocity, and \( c \) is the speed of light.

Express Original Momentum

The given problem states the proton has a momentum \( p_0 \) when its speed \( v_0 \) is 0.400c. Thus, \( p_0 = \frac{mv_0}{\sqrt{1 - \left(\frac{v_0}{c}\right)^2}} = \frac{m \cdot 0.4c}{\sqrt{1 - (0.4)^2}}.\)

Introduce New Velocity and Solve for New Momentum

We need to find the momentum when the velocity \( v \) is doubled to 0.800c. Using \( p = \frac{mv}{\sqrt{1 - \frac{v^2}{c^2}}} \), substitute \( v = 0.8c \) into the equation to get \( p = \frac{m \cdot 0.8c}{\sqrt{1 - (0.8)^2}}.\)

Relate the New Momentum to Original Momentum

Calculate the value of \( p \) in terms of \( p_0 \) by substituting \( m \) from \( p_0 \) and simplifying: \( p = \frac{0.8c \cdot p_0 \cdot \sqrt{1 - (0.4)^2}}{0.4c \cdot \sqrt{1 - (0.8)^2}}. \) Cancel terms and evaluate the expression.

Simplify the Expression

Compute \( \sqrt{1 - (0.4)^2} = \sqrt{0.84} \) and \( \sqrt{1 - (0.8)^2} = \sqrt{0.36} \). Substituting and simplifying gives \( p = p_0 \cdot \frac{2 \sqrt{0.84}}{\sqrt{0.36}} = p_0 \cdot \frac{2 \sqrt{0.84}}{0.6}. \) Further simplification results in \( p = p_0 \cdot \frac{2 \times 0.9165}{0.6} = p_0 \cdot 3.055.\)

Key concepts

Proton Speed

In classical mechanics, when we talk about speed, it's often a simple velocity value, such as 20 meters per second. However, when it comes to particles like protons moving at significant fractions of the speed of light, things become a bit more complex. In this context, we refer to this realm as "relativistic speed" because it requires the considerations of Einstein's theory of relativity.
The speed of light, represented by the symbol \( c \), is approximately 299,792,458 meters per second. When a proton moves at 0.400 times the speed of light \( (0.400c) \), it's still considerably slower than light, but fast enough that relativistic effects come into play. Relativistic effects include changes in momentum and time dilation, which aren't considered in classical physics.
Understanding the proton's speed in this context is crucial for correctly applying the relativistic momentum equation and predicting the behavior of subatomic particles in high-energy conditions.

Relativistic Physics

Relativistic physics fundamentally alters the way we understand motion and mass at high speeds. Albert Einstein's theory of relativity introduced concepts that deviate from traditional Newtonian physics, especially when objects approach the speed of light.
In the realm of relativistic physics:

• Mass and energy are interchangeable, linked by the equation \( E=mc^2 \).
• Time and space are intertwined, creating what is known as the space-time continuum.
• As an object's speed approaches the speed of light, its mass appears to increase, requiring more and more energy to continue accelerating.

In this exercise, understanding relativistic physics allows us to calculate the momentum of a proton traveling at significant fractions of the speed of light. These nuanced insights are vital for high-energy physics, such as in the operations of particle accelerators and studying cosmic phenomena.

Momentum Equation

The momentum equation in relativistic physics is quite different from the classical equation \( p = mv \) due to the effects of special relativity. The relativistic momentum equation is given by:
\[ p = \frac{mv}{\sqrt{1 - \frac{v^2}{c^2}}} \]
This formula accounts for the increase in momentum that occurs only at relativistic speeds, where \( v \) approaches \( c \).
When analyzing the problem of finding the momentum of a proton at a doubled speed of 0.800c, this equation helps us quantify how the interaction of velocity and relativistic effects influence momentum. By doubling the proton's speed, we see that its momentum increases not linearly but exponentially due to the relativistic factor \( \sqrt{1 - \frac{v^2}{c^2}} \).
Understanding how to manipulate and apply the relativistic momentum equation enables students to grasp the subtleties of motion at near-light speeds, crucial for any astrophysics or particle physics study.