Chapter 7: Problem 1
The derivation of \((7.8)\) used the algebraic relation $$ (\gamma+1)^{2}\left(1-\kappa^{2}\right)^{2}=4 $$ where $$ \kappa=\frac{\beta \gamma}{\gamma+1} \quad \text { and } \quad\left(1-\beta^{2}\right) \gamma^{2}=1 $$ Show that this holds.
Short Answer
Expert verified
\( (\gamma+1)^{2}(1-\kappa^{2})^{2}=4 \) holds true.
Step by step solution
01
- Express \text{kappa} in terms of \text{beta} and \text{gamma}
Given \( \text{kappa} = \frac{\beta \gamma}{\gamma + 1} \), keep this expression for substitution.
02
- Work with the second given equation
Given \( (1 - \beta^{2}) \gamma^{2} = 1 \). Rearrange to express \( \beta^{2} \) in terms of \( \gamma \): \[ \beta^{2} = 1 - \frac{1}{\gamma^{2}} \].
03
- Substitute \text{beta squared} into the original relation
Substitute \( \beta^{2} = 1 - \frac{1}{\gamma^{2}} \) into the \( \text{kappa}^{2} \) term: \[ \kappa^{2} = \left( \frac{\beta \gamma}{\gamma + 1} \right)^{2} = \left( \frac{\gamma \sqrt{1 - \frac{1}{\gamma^{2}}}}{\gamma + 1} \right)^{2} = \left( \frac{\sqrt{\gamma^{2} - 1}}{\gamma + 1} \right)^{2} \].
04
- Simplify the \text{kappa squared} term
Simplify \( \kappa^{2} \): \[ \kappa^{2} = \frac{\gamma^{2} - 1}{(\gamma+1)^{2}} \].
05
- Substitute \text{kappa squared} into the given algebraic relation
Substitute \( \kappa^{2} \) back into the original equation \((\gamma + 1)^{2}(1 - \kappa^{2})^{2} = 4 \): \[ (\gamma + 1)^{2} \left( 1 - \frac{\gamma^{2} - 1}{(\gamma + 1)^{2}} \right)^{2} = 4 \].
06
- Simplify the expression inside the equation
Simplify the equation: \[ (\gamma + 1)^{2} \left( \frac{(\gamma+1)^{2}-(\gamma^{2}-1)}{(\gamma+1)^{2}} \right)^{2} = 4 \].
07
- Combine and further simplify
Simplify the fraction: \[ (\gamma + 1)^{2} \left( \frac{\gamma^{2} + 2\gamma + 1 - \gamma^{2} + 1}{(\gamma+1)^{2}} \right)^{2} = 4 \].This simplifies to \[ (\gamma+1)^{2} \left( \frac{2\gamma + 2}{(\gamma+1)^{2}} \right)^{2} = 4 \].
08
- Final Simplification
Simplify the final expression: \[ (\gamma + 1)^{2} \times \left( \frac{2(\gamma+1)}{(\gamma + 1)^{2}} \right)^{2} = 4 \].Further simplifying: \[ (\gamma + 1)^{2} \times \frac{4}{(\gamma + 1)^{2}} = 4 \].This simplifies to \[ 4 = 4 \], which confirms the given relation.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Lorentz Factor
In relativistic kinematics, the concept of the Lorentz factor is crucial. It's defined by the symbol \( \gamma \) and is given by the equation: \[ \gamma = \frac{1}{\sqrt{1 - \beta^2}} \]. This factor arises due to the effects of relativity when an object is moving close to the speed of light. It calculates how time, length, and relativistic mass change for objects moving at relativistic speeds.
When \( \beta \) (which is the velocity of the object divided by the speed of light) is close to zero, \( \gamma \) is approximately 1. However, as \( \beta \) approaches 1 (near the speed of light), \( \gamma \) increases significantly, indicating profound relativistic effects.
This factor simplifies calculations in special relativity, including time dilation and length contraction. For instance, the time dilation equation \[ \Delta t' = \gamma \Delta t \] shows that a moving clock runs slower than a stationary one by a factor of \( \gamma \).
Understanding the Lorentz factor is essential for solving problems in relativistic kinematics. It connects directly to how velocities and energies transform at high speeds.
When \( \beta \) (which is the velocity of the object divided by the speed of light) is close to zero, \( \gamma \) is approximately 1. However, as \( \beta \) approaches 1 (near the speed of light), \( \gamma \) increases significantly, indicating profound relativistic effects.
This factor simplifies calculations in special relativity, including time dilation and length contraction. For instance, the time dilation equation \[ \Delta t' = \gamma \Delta t \] shows that a moving clock runs slower than a stationary one by a factor of \( \gamma \).
Understanding the Lorentz factor is essential for solving problems in relativistic kinematics. It connects directly to how velocities and energies transform at high speeds.
Beta Parameter
In the realm of special relativity, the beta parameter, denoted by \( \ beta \), represents the ratio of an object's velocity (v) to the speed of light (c). It's defined as \[ \beta = \frac{v}{c} \]. This dimensionless quantity characterizes how an object's speed compares to the universal speed limit.
The value of \( \beta \) ranges from 0 to 1. A \( \beta \) value of 0 means the object is at rest, while a \( \beta \) value close to 1 indicates that the object is moving near the speed of light.
The beta parameter is significant in the equations of special relativity. For example, the relationship \[ 1 - \beta^2 \gamma^2 = 1 \] demonstrates how \( \beta \) impacts the Lorentz factor and vice versa.
This parameter simplifies complex relativistic equations, especially when combined with the Lorentz factor. It helps in expressing quantities like kinetic energy and momentum in a more straightforward manner.
Understanding \( \beta \) is essential for grasping how velocity affects relativistic transformations.
The value of \( \beta \) ranges from 0 to 1. A \( \beta \) value of 0 means the object is at rest, while a \( \beta \) value close to 1 indicates that the object is moving near the speed of light.
The beta parameter is significant in the equations of special relativity. For example, the relationship \[ 1 - \beta^2 \gamma^2 = 1 \] demonstrates how \( \beta \) impacts the Lorentz factor and vice versa.
This parameter simplifies complex relativistic equations, especially when combined with the Lorentz factor. It helps in expressing quantities like kinetic energy and momentum in a more straightforward manner.
Understanding \( \beta \) is essential for grasping how velocity affects relativistic transformations.
Algebraic Manipulation
Algebraic manipulation involves rearranging equations and expressions to simplify or solve them. This process is vital in solving relativistic kinematics problems.
Let's consider our given exercise. One of the steps involves substituting expressions to simplify the original equation. Given \[ (1 - \beta^2) \gamma^2 = 1 \], we can express \( \beta^2 \) as \[ \beta^2 = 1 - \frac{1}{\gamma^2} \].
Another key manipulation involves substituting \( \beta^2 \) into \( \kappa^2 \): \[ \kappa^2 = \left( \frac{\gamma \sqrt{1 - \frac{1}{\gamma^2}}}{\gamma+1} \right)^2 = \left( \frac{\sqrt{\gamma^2 - 1}}{\gamma + 1} \right)^2 \]. This is further simplified to \[ \kappa^2 = \frac{\gamma^2 - 1}{(\gamma+1)^2} \].
Steps like combining terms, factoring expressions, and simplifying fractions are all part of algebraic manipulation. These techniques are essential for solving and understanding complex scientific problems systematically. Whether one is solving for a specific variable or verifying an equation, algebraic manipulation is indispensable.
Let's consider our given exercise. One of the steps involves substituting expressions to simplify the original equation. Given \[ (1 - \beta^2) \gamma^2 = 1 \], we can express \( \beta^2 \) as \[ \beta^2 = 1 - \frac{1}{\gamma^2} \].
Another key manipulation involves substituting \( \beta^2 \) into \( \kappa^2 \): \[ \kappa^2 = \left( \frac{\gamma \sqrt{1 - \frac{1}{\gamma^2}}}{\gamma+1} \right)^2 = \left( \frac{\sqrt{\gamma^2 - 1}}{\gamma + 1} \right)^2 \]. This is further simplified to \[ \kappa^2 = \frac{\gamma^2 - 1}{(\gamma+1)^2} \].
Steps like combining terms, factoring expressions, and simplifying fractions are all part of algebraic manipulation. These techniques are essential for solving and understanding complex scientific problems systematically. Whether one is solving for a specific variable or verifying an equation, algebraic manipulation is indispensable.