Rest frame angular distribution of meson decay into two photons

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Homework Help Overview

The discussion centers around a problem involving the decay of a pion into two photons, specifically examining the angular distribution of the photons in the pion's rest frame. The participants explore the implications of isotropic decay and how this relates to the distribution function f(θ).

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss whether the angular distribution f(θ) is constant in the rest frame due to isotropy or if it changes when considering a lab frame where the pion is moving. There is also consideration of how the distribution might relate to conservation laws and spherical coordinates.

Discussion Status

The conversation is ongoing, with participants raising questions about the assumptions of the problem and exploring different interpretations of the angular distribution. Some guidance has been offered regarding the uniformity of the distribution over a spherical surface, but no consensus has been reached.

Contextual Notes

There is a lack of specification regarding the direction of the pion's momentum in the lab frame, which is noted as a potential source of confusion. The original problem also asks for an explanation of why f(θ) equals sin(θ) in the rest frame, adding complexity to the discussion.

slimjim
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Homework Statement



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.


Homework 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.

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?
 
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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.
 
where did you find this problem.
 
I paraphrased it from a homework problem. The actual homework 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:
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 don't see how but this might give rise to the sin(theta) distribution.
 
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.
 
diazona said:
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.

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
 

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