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

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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