Chapter 35: Problem 8
Consider a particle moving with a speed less than \(0.5 c\). If the speed of the particle is doubled, by what factor will the momentum increase? a) less than 2 b) equal to 2 c) greater than 2
Short Answer
Expert verified
Is it (a) less than 2, (b) equal to 2, or (c) greater than 2?
Answer: (c) greater than 2
Step by step solution
01
Write down the relativistic momentum formula
The relativistic momentum formula is given by:
$$p = \gamma m v$$
where
\(p\) is the momentum
\(\gamma\) is the relativistic factor, defined as \(\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\)
\(m\) is the mass of the particle
\(v\) is the speed of the particle
\(c\) is the speed of light
02
Calculate the initial relativistic momentum
Let's denote the initial speed of the particle as \(v_i\) (which is less than 0.5c). The initial momentum (\(p_i\)) can be calculated as follows:
$$p_i = \gamma_i m v_i$$
where \(\gamma_i\) is the initial relativistic factor.
03
Calculate the final relativistic momentum
Since the speed is doubled, the final speed of the particle is \(v_f = 2v_i\). We can now calculate the final momentum (\(p_f\)) using the relativistic momentum formula:
$$p_f = \gamma_f m(2v_i)$$
where \(\gamma_f\) is the final relativistic factor.
04
Compare the initial and final momenta
The factor by which the momentum will increase is the ratio of the final momentum to the initial momentum:
$$\text{Factor} = \frac{p_f}{p_i} = \frac{\gamma_f m(2v_i)}{\gamma_i m v_i}$$
Notice that the mass \(m\) and the initial speed \(v_i\) cancel out in the equation above, leaving us with:
$$\text{Factor} = \frac{2\gamma_f}{\gamma_i}$$
05
Determine which statement is true
Now we can compare the factors to see which statement is true:
- If the factor is less than 2, the momentum has increased by a factor less than 2. (option a)
- If the factor is equal to 2, the momentum has increased by a factor of 2. (option b)
- If the factor is greater than 2, the momentum has increased by a factor greater than 2. (option c)
We know that when the speed of a particle increases, the relativistic factor \(\gamma\) will always increase. Therefore, \(\gamma_f > \gamma_i\). As a result, we can conclude that:
$$\text{Factor} = \frac{2\gamma_f}{\gamma_i} > \frac{2\gamma_i}{\gamma_i} = 2$$
So the correct answer is (c) the momentum will increase by a factor greater than 2.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Special Relativity
Special relativity, a fundamental theory developed by Albert Einstein, revolutionized our understanding of space, time, and energy. This theory is based on two key postulates: the laws of physics are the same for all observers in uniform motion relative to one another, and the speed of light in a vacuum, denoted by the symbol \( c \), is the same for all observers, regardless of their motion or the motion of the source of light.
One of the most intriguing aspects of special relativity is time dilation, where time itself is perceived to slow down for objects moving at speeds close to the speed of light relative to an observer. This concept is crucial when discussing the relativistic momentum formula \( p = \gamma m v \), as the factor \( \gamma \), known as the Lorentz factor, adjusts the classical momentum to incorporate the effects of high velocities. At velocities much lower than the speed of light, relativistic effects are negligible and classical physics prevails. However, as the velocity approaches \( c \), the effects become significant, and the differences between classical and relativistic momentum grow.
When explaining special relativity in the context of momentum, it's vital to convey the non-intuitive aspects of the theory in an approachable manner. As speeds increase towards the speed of light, not only does time change, but mass effectively increases, meaning the momentum of an object will also increase disproportionately compared to its increase in speed.
One of the most intriguing aspects of special relativity is time dilation, where time itself is perceived to slow down for objects moving at speeds close to the speed of light relative to an observer. This concept is crucial when discussing the relativistic momentum formula \( p = \gamma m v \), as the factor \( \gamma \), known as the Lorentz factor, adjusts the classical momentum to incorporate the effects of high velocities. At velocities much lower than the speed of light, relativistic effects are negligible and classical physics prevails. However, as the velocity approaches \( c \), the effects become significant, and the differences between classical and relativistic momentum grow.
When explaining special relativity in the context of momentum, it's vital to convey the non-intuitive aspects of the theory in an approachable manner. As speeds increase towards the speed of light, not only does time change, but mass effectively increases, meaning the momentum of an object will also increase disproportionately compared to its increase in speed.
Momentum Increase Factor
The momentum increase factor in the context of special relativity reflects how the momentum of an object changes when its velocity is altered. Traditionally, in classical mechanics, the momentum of a moving object with a constant mass is directly proportional to its velocity. However, under the influence of special relativity, this relationship becomes more complex.
Momentum in special relativity is given by the equation \( p = \gamma m v \), where\( m \) represents mass, and \( v \) represents velocity. Here, the Lorentz factor \( \gamma \) is critical, as it changes as a function of velocity relative to the speed of light. As velocity increases, \( \gamma \) increases non-linearly, meaning that for velocities significantly lower than \( c \), the \( \gamma \) value is close to 1, and momentum behaves classically. In contrast, as an object's velocity approaches the speed of light, \( \gamma \) increases sharply, leading to a momentum that increases to a much greater extent than the velocity.
The exercise provided demonstrates an instance where doubling the velocity does not merely double the momentum. Instead, the Lorentz factor becomes greater as velocity increases, leading to a scenario where the momentum has increased by a factor greater than 2. It is an excellent example to highlight the non-linear nature of relativistic momentum compared to classical momentum.
Momentum in special relativity is given by the equation \( p = \gamma m v \), where\( m \) represents mass, and \( v \) represents velocity. Here, the Lorentz factor \( \gamma \) is critical, as it changes as a function of velocity relative to the speed of light. As velocity increases, \( \gamma \) increases non-linearly, meaning that for velocities significantly lower than \( c \), the \( \gamma \) value is close to 1, and momentum behaves classically. In contrast, as an object's velocity approaches the speed of light, \( \gamma \) increases sharply, leading to a momentum that increases to a much greater extent than the velocity.
The exercise provided demonstrates an instance where doubling the velocity does not merely double the momentum. Instead, the Lorentz factor becomes greater as velocity increases, leading to a scenario where the momentum has increased by a factor greater than 2. It is an excellent example to highlight the non-linear nature of relativistic momentum compared to classical momentum.
Speed of Light
The speed of light, typically denoted by \( c \), is one of the most important and intriguing constants in physics. At approximately \( 299,792,458 \) meters per second in a vacuum, this ultimate speed limit has profound implications for how we understand the universe. In the context of special relativity, the concept of the speed of light being the same for all observers, regardless of their relative motion, leads to many non-intuitive effects, such as time dilation and length contraction.
The speed of light is also central to the equation for relativistic momentum, as seen in the exercise. Because \( c \) appears in the denominator of the Lorentz factor \( \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \), when \( v \) significantly increases, the value of \( \gamma \) heavily influences both time and momentum. This is why an increase in the velocity of a particle, even if below \( c \), can lead to relativistic effects that dramatically alter its momentum.
Understanding the speed of light not only aids in grasping theoretical concepts but also applies to real-world technologies, such as GPS and particle accelerators, where relativistic effects must be taken into account. When discussing the speed of light in educational content, it's crucial to reinforce its role as a universal constant, one that shapes the very fabric of space-time and underlies the relativistic phenomena we observe in high-speed particles.
The speed of light is also central to the equation for relativistic momentum, as seen in the exercise. Because \( c \) appears in the denominator of the Lorentz factor \( \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \), when \( v \) significantly increases, the value of \( \gamma \) heavily influences both time and momentum. This is why an increase in the velocity of a particle, even if below \( c \), can lead to relativistic effects that dramatically alter its momentum.
Understanding the speed of light not only aids in grasping theoretical concepts but also applies to real-world technologies, such as GPS and particle accelerators, where relativistic effects must be taken into account. When discussing the speed of light in educational content, it's crucial to reinforce its role as a universal constant, one that shapes the very fabric of space-time and underlies the relativistic phenomena we observe in high-speed particles.
