Problem 76 - Fundamentals of Physics

Question

The total energy of a proton passing through a laboratory apparatus is \(10.611 \mathrm{~nJ} .\) What is its speed parameter \(\beta ?\) Use the proton mass given in Appendix B under "Best Value," not the commonly remembered rounded number.

Step-by-Step Solution

Answer

\( \beta = \sqrt{1 - \frac{E_0^2}{E^2}} \), after solving for \( \gamma \).

Step 1: Identify the given information

The total energy of a proton is given as 10.611 nJ (nanojoules). We need to calculate the speed parameter \( \beta \), which is the ratio of the proton's velocity \( v \) to the speed of light \( c \). The proton's rest mass from the appendix is \( m_p = 1.6726219 \times 10^{-27} \) kg.

Step 2: Convert the energy into joules

Since nanojoules are given, first convert the total energy from nanojoules to joules: \[ 10.611 \, \text{nJ} = 10.611 \times 10^{-9} \, \text{J} \]. This conversion is necessary for using it in the equations related to relativistic energies.

Step 3: Use the relativistic energy formula

The relativistic energy \( E \) is given by: \[ E = \gamma m_p c^2 \], where \( \gamma \) is the Lorentz factor defined as \( \gamma = \frac{1}{\sqrt{1-\beta^2}} \), \( c \) is the speed of light \( 3 \times 10^8 \, \text{m/s} \). We are given \( E = 10.611 \times 10^{-9} \, \text{J}\) and need to solve for \( \beta \).

Step 4: Calculate the rest energy

Calculate the rest energy of the proton using: \[ E_0 = m_p c^2 = 1.6726219 \times 10^{-27} \, \text{kg} \times (3 \times 10^8 \, \text{m/s})^2 \]. Simplify this to find \( E_0 \).

Step 5: Solve for the Lorentz factor \( \gamma \)

From the equation \( E = \gamma E_0 \), solve for \( \gamma \): \[ \gamma = \frac{E}{E_0} \]. Substitute the values of \( E \) and \( E_0 \) to find \( \gamma \).

Step 6: Determine \( \beta \) using \( \gamma \)

Use the formula relating \( \gamma \) and \( \beta \): \[ \gamma = \frac{1}{\sqrt{1-\beta^2}} \]. Solve for \( \beta \): \[ \beta = \sqrt{1 - \frac{1}{\gamma^2}} \]. Substitute the value of \( \gamma \) from the previous step to find \( \beta \).

Key Concepts

Lorentz factor

The Lorentz factor, often symbolized by the Greek letter \( \gamma \), is a crucial concept in Einstein's theory of relativity. It accounts for the effects of relativistic speeds on time, length, and energy as an object approaches the speed of light. When calculating relativistic energies, as was needed in the exercise problem, the Lorentz factor comes in handy. It is defined as:\[ \gamma = \frac{1}{\sqrt{1 - \beta^2}} \]Where \( \beta \) is the dimensionless speed parameter, the ratio of a particle’s velocity \( v \) to the speed of light \( c \). The Lorentz factor establishes how much time dilates or how much mass increases as the speed of an object nears the speed of light.

• For speeds much less than the speed of light, \( \gamma \) is approximately 1, meaning relativistic effects are negligible.
• As \( \beta \) approaches 1 (meaning the speed approaches the speed of light), \( \gamma \) increases dramatically, reflecting more pronounced relativistic effects.

In our exercise, by finding \( \gamma \), we can determine the relativistic speed (\( \beta \)) of the proton from its given energy, allowing us insight into its motion through the apparatus.

Proton mass

Protons are subatomic particles found in the nucleus of an atom, and their mass is an essential factor when dealing with nuclear physics and relativistic equations. The mass of a proton is a fundamental constant used to determine energies involved in particle physics. In our exercise, we specifically use the precise mass of the proton provided in a physics appendix, totaling \( m_p = 1.6726219 \times 10^{-27} \) kg.The rest mass energy of any particle, including the proton, can be calculated using the formula:\[ E_0 = m_p c^2 \]Where:

• \( E_0 \) is the rest mass energy.
• \( m_p \) is the rest mass of the particle, in this scenario, the proton.
• \( c \) is the speed of light.

This equation highlights the energy that is intrinsic to the mass of a particle alone, without taking into account any kinetic energy due to motion. In the exercise, calculating the rest energy \( E_0 \) is a key step in finding the proton's Lorentz factor and subsequent speed parameter \( \beta \).

Speed of light

The speed of light, denoted as \( c \), is one of the most fundamental constants in physics, holding significant importance in both classical and modern physics realms. It is precisely \( 299,792,458 \) meters per second, although often approximated in calculations as \( 3 \times 10^8 \) m/s for simplicity. Given the speed of light's pivotal role in the universe's structure, it acts as a cosmic speed limit. Nothing with mass can exceed this speed, leading to profound implications for relativistic physics.In the exercise, the speed of light plays a critical role within the relativistic energy equations, such as:\[ E = \gamma m_p c^2 \]and helps in translating a particle's relativistic effects into everyday physics understanding.

• All electromagnetic waves travel at this speed in a vacuum, including light.
• As objects approach this speed, time dilation increases, meaning time slows down relative to an observer not moving at these speeds.

Understanding the significance of the speed of light allows for the comprehension of phenomena such as time travel, energy-mass equivalence, and various cosmic events. In solving our exercise, the constant \( c \) provides the foundation for relating energy changes to speed and is vital for calculating \( \beta \), further linking velocities in our relativistic framework.