Proving a Uniform Energy Distribution of B in Decay A -> B + C

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Homework Help Overview

The discussion revolves around the decay process A -> B + C, specifically analyzing the energy distribution of particle B in the lab frame when A is not at rest. The original poster seeks to demonstrate that this energy distribution is uniform, assuming B has negligible mass.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to establish the 4-momenta of particles A and B in different frames and calculates the maximum and minimum energy of B. They express uncertainty about demonstrating the uniformity of the energy distribution.
  • Some participants suggest calculating the energy of B as a function of the angle θ and propose showing that the differential energy distribution dE/dΩ is constant.
  • Another participant points out the independence of the energy distribution on the azimuthal angle φ' and discusses integrating out φ' to simplify the analysis.

Discussion Status

Participants are actively exploring the relationship between energy and angular distribution, with some guidance provided on how to approach the isotropy of the decay process. There is a focus on using mathematical relationships to show the uniformity of the energy distribution, but no consensus has been reached yet.

Contextual Notes

The discussion involves assumptions about the mass of particle B and the isotropic nature of the decay process, which are critical to the analysis but remain under examination.

ryanwilk
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Homework Statement



Consider the decay A -> B + C (where A is not at rest). In the rest frame of A, B is emitted in a random direction (all directions have equal probability) and I need to show that in the lab frame, the energy distribution of B is uniform.

(We assume that B has negligible mass)


Homework Equations



909c36cf59b2e9e46302e4c4fa1062ee.png


(Let c=1)

The Attempt at a Solution



So I started by writing down the 4-momenta of A and B in the rest frame of A (choosing the momentum of B to be along the x axis):

PA' = (mA,0,0,0)
PB' = (EB',EB',0,0)

In the lab frame:

PA = (EA,(EA2-mA2)1/2,0,0)
PB = (EB,EB,0,0)

Lorentz boosting along the x-axis, I can determine the maximum and minimum energy that B can have:

<br /> E_B^{\mathrm{min,max}} = \frac{E_A}{2} \bigg(1-\frac{m_C^2}{m_A^2}\bigg) \bigg(1 \pm \sqrt{1-\frac{m_A^2}{E_A^2}}\bigg)<br />

using:

<br /> \gamma = \frac{E_A}{m_A}\&gt;,\gamma \beta = \frac{p_A}{m_A},\&gt;E_A&#039; = \frac{m_A^2-m_C^2}{2m_A}<br />

I could also Lorentz boost in all other directions to get expressions for the energy. But I have no idea how to show that each of these energies in equally probable?

Any help would be appreciated.

Thanks!
 
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Try calculating EA as a function of θ and show that the dE/dΩ is a constant.
 
vela said:
Try calculating EA as a function of θ and show that the dE/dΩ is a constant.

Ok, so using 3D spherical polars, I get E_B(\theta&#039;) = \frac{E_A}{2} \bigg(1-\frac{m_C^2}{m_A^2}\bigg) \bigg(1 + \sqrt{1-\frac{m_A^2}{E_A^2}}\&gt;\mathrm{cos}(\theta &#039;) \bigg)

(I defined it so that there is max. energy at θ'=0 and min. at θ'=π)

Then dΩ = sin(θ')dθ'dφ' and so since EB is independent of φ', dE/dΩ is a constant?
 
Last edited:
Not quite. Nothing depends on φ', so you can integrate that out and deal only with θ', so you have dΩ' = 2π sin θ' dθ' where θ' goes from 0 to π. It's convenient to change variables to cos θ' so you have dΩ' = d(cos θ') where cos θ' runs from -1 to 1. (The minus sign goes into switching the order of the limits.) When you say that a decay is isotropic, that means that the distribution N is flat as a function of cos θ'. What you want to show is that dN/dEB is constant, using the fact that\frac{dN}{dE_B} = \frac{dN}{d(\cos \theta&#039;)} \frac{d(\cos\theta&#039;)}{dE_B}(which is simply an application of the chain rule).
 

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