Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A mad physicist claims to have observed the decay of a particle of mass \(M\) into two identical particles of mass \(m,\) with \(M<2 m .\) In response to the objections that this violates conservation of energy, he replies that if \(M\) was traveling fast enough it could easily have energy greater than \(2 m c^{2}\) and hence could decay into the two particles of mass \(m\). Show that he is wrong. [He has forgotten that both energy and momentum are conserved. You can analyse this problem in terms of these two conservation laws, but it is much simpler to go to the rest frame of \(M .]\)

Short Answer

Expert verified
The decay violates energy conservation since \( Mc^2 < 2mc^2 \).

Step by step solution

01

Conservation of Energy in the Rest Frame

In the rest frame of the particle with mass \(M\), the total energy is simply its rest energy, which is given by \(E = Mc^2\).
02

Condition for Decay in the Rest Frame

To decay into two particles of mass \(m\), the initial particle’s energy \(Mc^2\) must be at least equal to \(2mc^2\). This is due to the energy conservation law which requires the total energy before and after the decay to be equal.
03

Analyze Mass Condition

Given the problem states \(M < 2m\), the condition \(Mc^2 \geq 2mc^2\) cannot be satisfied. Thus, even in an abstract situation where it could have additional energy due to motion, in the rest frame, energy conservation is violated.
04

Conservation of Momentum Consideration

Apart from energy, momentum must also be conserved. In the rest frame, initial momentum is zero. The momentum of the decay products could only sum to zero if each has opposite momentum, implying each has non-zero velocity, which necessitates each having energy greater than \(mc^2\) in the rest frame. This total energy must exceed \(2mc^2\) in the lab frame, which again contradicts \(Mc^2\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rest Frame
The concept of the rest frame is essential in understanding problems of particle decay. The rest frame of a particle refers to a reference frame in which the particle is at rest, meaning it has no kinetic energy. This frame allows us to focus entirely on the particle's rest energy, which is the energy due to its mass alone. It simplifies analysis, as we do not need to consider additional energy due to motion.

To illustrate, if we have a particle of mass \( M \), its rest energy in the rest frame is given by \( E = Mc^2 \). This energy serves as a baseline for any decay processes we might observe. In analyzing particle decay, the rest frame helps us apply conservation laws straightforwardly by concluding that all the initial energy comes from the rest mass. By focusing only on this rest energy, it becomes clear if supposed decay processes are possible or not within the given constraints of conservation laws.
Momentum Conservation
Momentum conservation is a fundamental principle in physics, stating that the total momentum of a closed system remains constant over time. In particle decay scenarios, this law plays a pivotal role in determining outcome feasibility.
  • Initial momentum in the rest frame is zero, as the particle is at rest.
  • For momentum to be conserved during decay, the decay products must also have a net momentum of zero.
In the scenario described, the conservation of momentum requires the two emerging particles to move in opposite directions with equal but opposite momentum. This ensures that the sum of their momenta equals the initial zero momentum before decay. If these conditions cannot be met, such as in our case where the decay mass condition is violated, the process is not possible.
Particle Decay
Particle decay refers to the transformation of a particle into two or more different particles. In physics, understanding this process involves key conservation laws.
  • Energy conservation dictates that the total energy before and after the decay must be equal.
  • Momentum conservation ensures that momentum does not change before and after the event.
In the described problem, the particle of mass \( M \) is said to decay into two particles of mass \( m \). However, with \( M < 2m \), energy conservation cannot be achieved in any frame, including the rest frame. Moreover, if energy were somehow sufficient in a moving frame, momentum conservation would still fail, since in such a case, the decay products must each possess momentum resulting in net zero momentum—a contradiction under the conditions provided. Hence, the claim of decay is inherently flawed, violating these fundamental principles.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A particle of mass \(m_{a}\) decays at rest into two identical particles each of mass \(m_{b} .\) Use conservation of momentum and energy to find the speed of the outgoing particles.

If one defines a variable mass \(m_{\text {var }}=\gamma m\), then the relativistic momentum \(\mathbf{p}=\gamma m \mathbf{v}\) becomes \(m_{\text {var }} \mathbf{v}\) which looks more like the classical definition. Show, however, that the relativistic kinetic energy is not equal to \(\frac{1}{2} m_{\mathrm{var}} v^{2}\)

Newton's first law can be stated: If an object is isolated (subject to no forces), then it moves with constant velocity. We know that this is invariant under the Galilean transformation. Prove that it is also invariant under the Lorentz transformation. [Assume that it is true in an inertial frame \(\mathcal{S}\), and use the relativistic velocity-addition formula to show that it is also true in any other \(\mathcal{S}^{\prime} .\) ]

What is the factor \(\gamma\) for a speed of \(0.99 c\) ? As observed from the ground, by how much would a clock traveling at this speed differ from a ground-based clock after one hour (one hour as measured by the latter, that is)?

Consider the tale of the physicist who is ticketed for running a red light and argues that because he was approaching the intersection, the red light was Doppler shifted and appeared green. How fast would he have to have been going? \(\left(\lambda_{\text {red }} \approx 650 \mathrm{nm} \text { and } \lambda_{\text {green }} \approx 530 \mathrm{nm} .\right)\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free