Chapter 7: Problem 5
A particle with momentum \(6.0 \mathrm{GeV} / c\) has a total energy of \(11.2 \mathrm{GeV}\). Determine the mass of the particle and its speed.
Short Answer
Expert verified
The particle's mass is approximately 9.46 GeV/c², and its speed is about 0.537c.
Step by step solution
01
Identify the Known Variables
The momentum of the particle is given as \( p = 6.0 \, \text{GeV}/c \), and the total energy \( E = 11.2 \, \text{GeV}\). We need to find the mass \( m \) and speed \( v \) of the particle.
02
Use the Energy-Momentum Relation
The energy-momentum relation is \( E^2 = (pc)^2 + (m_0c^2)^2 \). Substitute \( E = 11.2 \, \text{GeV} \) and \( p = 6.0 \, \text{GeV}/c \) into the equation:\[(11.2)^2 = (6)^2 + (m_0c^2)^2\]Solve for \( (m_0c^2) \).
03
Solve for the Rest Mass Energy \((m_0c^2)\)
Calculate \((11.2)^2 = 125.44\) and \((6)^2 = 36\). Therefore, \[125.44 = 36 + (m_0c^2)^2\]\[(m_0c^2)^2 = 125.44 - 36 = 89.44\]\[m_0c^2 = \sqrt{89.44} \approx 9.46 \, \text{GeV}\]
04
Find the Mass \(m_0\)
Since \( m_0c^2 = 9.46 \, \text{GeV} \) and \( c = 1 \, \text{GeV}/(\text{GeV}/c)\), solve for \( m_0 \):\[m_0 = \frac{9.46 \, \text{GeV}}{c^2} = 9.46 \, \text{GeV}/c^2\]
05
Calculate the Speed \(v\) Using Lorentz Factor
Use the formula \( E = \gamma m_0c^2 \) where \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \). We have\[\gamma = \frac{11.2}{9.46}\]Calculate \( \gamma \approx 1.184\).
06
Solve for the Speed \(v\)
We know \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} = 1.184 \). Solve for \( v/c \):\[\sqrt{1 - \frac{v^2}{c^2}} = \frac{1}{1.184}\]\[1 - \frac{v^2}{c^2} = \left(\frac{1}{1.184}\right)^2\]\[\frac{v^2}{c^2} = 1 - \left(\frac{1}{1.184}\right)^2\]Calculate \(v/c \approx 0.537 \).
07
Conclusion
The mass of the particle is \( m_0 \approx 9.46 \, \text{GeV}/c^2 \) and its speed is \( v \approx 0.537c \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Energy-Momentum Relation
The energy-momentum relation is a key concept in relativistic dynamics. It connects a particle's energy, momentum, and mass through the equation:
Rest mass energy is represented by \( m_0c^2 \), while \( c \) is the speed of light.
This relation tells us that energy depends not just on mass but also on the momentum a particle carries.
When solving for the rest mass, we rearrange the formula:
- \( E^2 = (pc)^2 + (m_0c^2)^2 \)
Rest mass energy is represented by \( m_0c^2 \), while \( c \) is the speed of light.
This relation tells us that energy depends not just on mass but also on the momentum a particle carries.
When solving for the rest mass, we rearrange the formula:
- Start with \( E^2 - (pc)^2 = (m_0c^2)^2 \)
Lorentz Factor
An important concept while dealing with particles moving at significant fractions of the speed of light is the Lorentz factor \( \gamma \). It is a measure of how much quantities like time, length, and mass transform between two reference frames:
For our particle, using its energy and rest mass energy, we derived \( \gamma = \frac{11.2}{9.46} \approx 1.184 \).
This significant value of \( \gamma \) shows the relativistic effects are non-negligible. We can then find the speed, \( v \), using \( \gamma \). By rearranging the Lorentz factor formula, we calculated \( v \approx 0.537c \).
- \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \)
For our particle, using its energy and rest mass energy, we derived \( \gamma = \frac{11.2}{9.46} \approx 1.184 \).
This significant value of \( \gamma \) shows the relativistic effects are non-negligible. We can then find the speed, \( v \), using \( \gamma \). By rearranging the Lorentz factor formula, we calculated \( v \approx 0.537c \).
Rest Mass Energy
Rest mass energy is a concept that arises from the equation \( E_0 = m_0c^2 \), discovered by Einstein.
This energy is inherent to the particle even when it is at rest, meaning there is no loss of energy when a particle is not moving.
It signifies the energy "contained" due to its mass alone. In our problem, the rest mass energy is calculated as \( 9.46 \, \text{GeV} \).
This energy is inherent to the particle even when it is at rest, meaning there is no loss of energy when a particle is not moving.
It signifies the energy "contained" due to its mass alone. In our problem, the rest mass energy is calculated as \( 9.46 \, \text{GeV} \).
- This shows the potential energy stored within the mass itself, transformable into kinetic or other energy forms when the particle moves.
- For any moving particle, observing its rest mass energy helps understand both intrinsic energy and relativistic transformations in different frames.