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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 \(Mc^2\).

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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.

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Most popular questions from this chapter

Prove that for any two matrices \(A\) and \(B\), where \(A\) has as many columns as \(B\) has rows, the transpose of \(A B\) satisfies \((A B)=\tilde{B} \tilde{A}\)

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The relativistic kinetic energy of a particle is \(T=(\gamma-1) m c^{2} .\) Use the binomial series to express \(T\) as a series in powers of \(\beta=v / c .\) (a) Verify that the first term is just the non relativistic kinetic energy, and show that to lowest order in \(\beta\) the difference between the relativistic and non relativistic kinetic energies is \(3 \beta^{4} m c^{2} / 8 .\) (b) Use this result to find the maximum speed at which the non relativistic value is within \(1 \%\) of the correct relativistic value.

Let \(\Lambda_{\mathrm{B}}(\theta)\) denote the 4 \(\times 4\) matrix that gives a pure boost in the direction that makes an angle \(\theta\) with the \(x_{1}\) axis in the \(x_{1} x_{2}\) plane. Explain why this can be found as \(\Lambda_{\mathrm{B}}(\theta)=\Lambda_{\mathrm{R}}(-\theta) \Lambda_{\mathrm{B}}(0) \Lambda_{\mathrm{R}}(\theta)\) where \(\Lambda_{\mathrm{R}}(\theta)\) denotes the matrix that rotates the \(x_{1} x_{2}\) plane through angle \(\theta\) and \(\Lambda_{\mathrm{B}}(0)\) is the standard boost along the \(x_{1}\) axis. Use this result to find \(\Lambda_{\mathrm{B}}(\theta)\) and check your result by finding the motion of the spatial origin of the frame \(\mathcal{S}\) as observed in \(\mathcal{S}^{\prime}\).

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