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

[ryanwilk]

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

(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):

P_A' = (m_A,0,0,0)
P_B' = (E_B',E_B',0,0)

In the lab frame:

P_A = (E_A,(E_A²-m_A²)^1/2,0,0)
P_B = (E_B,E_B,0,0)

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

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

using:

$$\gamma = \frac{E_A}{m_A}\>,\gamma \beta = \frac{p_A}{m_A},\>E_A' = \frac{m_A^2-m_C^2}{2m_A}$$

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!

 

[vela]

Try calculating E_A as a function of θ and show that the dE/dΩ is a constant.

 

[ryanwilk]

Try calculating E_A as a function of θ and show that the dE/dΩ is a constant.

Ok, so using 3D spherical polars, I get $$E_B(\theta') = \frac{E_A}{2} \bigg(1-\frac{m_C^2}{m_A^2}\bigg) \bigg(1 + \sqrt{1-\frac{m_A^2}{E_A^2}}\>\mathrm{cos}(\theta ') \bigg)$$

(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 E_B is independent of φ', dE/dΩ is a constant?

 

[vela]

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/dE_B is constant, using the fact that$$\frac{dN}{dE_B} = \frac{dN}{d(\cos \theta')} \frac{d(\cos\theta')}{dE_B}$$(which is simply an application of the chain rule).

 

[paranormal]

To show that the energy distribution of B is uniform in the lab frame, we can use the principle of relativity. This principle states that the laws of physics are the same in all inertial frames of reference. In this case, the lab frame and the rest frame of A are both inertial frames, and the energy distribution of B should be the same in both frames.

Therefore, if we can show that the energy distribution of B is uniform in the rest frame of A, then it must also be uniform in the lab frame. In the rest frame of A, B is emitted in a random direction with equal probability. This means that the energy of B can take on any value with equal probability, resulting in a uniform energy distribution.

When we Lorentz boost to the lab frame, we are essentially rotating the rest frame of A. However, this rotation does not affect the energy distribution of B. Since the energy distribution is uniform in the rest frame of A, it remains uniform in the lab frame. Therefore, the energy distribution of B is uniform in the lab frame, as desired.

In conclusion, the principle of relativity and the fact that B is emitted in a random direction with equal probability in the rest frame of A, guarantees a uniform energy distribution of B in the lab frame.