# Rest frame angular distribution of meson decay into two photons

1. Dec 3, 2012

### slimjim

1. The problem statement, all variables and given/known data

consider a pion decaying into 2 photons.
In the rest frame of the pion, the two photons must emerge back-to-back photons are equally likely to emerge in any direction.

determine the rest frame angular distribution of the emerging photons.

2. Relevant equations

ΔP=f(θ)Δθ is the probability that a photon emerges with polar
angle θ within some infinitesimal interval Δθ in the angle,

and f(θ) is the rest frame angular distribution.

3. The attempt at a solution

in the pion frame, the photons emerge back to back, and since there is no preferred decay direction, the probabilities of a photon emerging in any direction are the same.

so f(theta) is a constant?

Last edited by a moderator: Dec 4, 2012
2. Dec 4, 2012

### vela

Staff Emeritus
I think the problem is asking you to calculate $f(\theta)$ for the lab frame. The pion is moving in this frame. If you were only looking at the decay in the the pion's rest frame, then yes, $f(\theta)$ would be a constant because the decay is isotropic.

3. Dec 6, 2012

### andrien

where did you find this problem.

4. Dec 7, 2012

### slimjim

I paraphrased it from a hw problem. The actual hw question asks to explain why f(θ)=sinθ is the rest frame angular distribution. Which is even less reassuring...

but the question doesn't specify the direction in which the pion is traveling in the lab frame, and θ is the usual polar angle, relative to some choice of z-axis.

Last edited: Dec 7, 2012
5. Dec 7, 2012

### slimjim

Say theta is taken from the direction of the pions momentum in the lab frame (call it +z). Then the pion decays into two photons, that when boosted to the lab frame, give the same total momentum as the pion initially.

So if the photons emerge along the z axis, could boosting to the lab frame violate cons. of momentum? I dont see how but this might give rise to the sin(theta) distribution.

6. Dec 7, 2012

### diazona

Think about this: equal probability of emerging in any direction means the probability distribution is uniform over a two-dimensional sphere. You could write this as ΔP=f(θ,φ)ΔθΔφ. What would f(θ,φ) be in that case? Remember to think the properties of spherical coordinates.

7. Dec 7, 2012

### slimjim

oh i see. picturing a sperical surface surrounding the pion (in its own frame), the probability of a photon emerging through some infinitesimally small surface element would be constant.

for a sphere, dA=sin(theta)d∅dθ; ΔP=sin(theta)ΔθΔ∅

phi and theta are orthogonal, so ignoring phi, ΔP=sin(theta)Δθ