How Does Photon Energy Relate to Mass Loss in Particle Decay?

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SUMMARY

The discussion focuses on the relationship between photon energy and mass loss during particle decay, specifically addressing the equation for photon energy as ω=δ(1-δ/2m₀) in the particle's rest frame. Participants analyze the conservation of momentum and energy, utilizing equations such as P₁ = P₂ + K and P₂² = P₁² - 2K·P₁. The initial attempt to derive the photon energy resulted in an incorrect conclusion of ω=δ, prompting further exploration of relativistic principles and the correct application of momentum conservation.

PREREQUISITES
  • Understanding of special relativity principles
  • Familiarity with conservation of momentum and energy
  • Knowledge of photon momentum represented as (ω, k)
  • Ability to manipulate algebraic equations involving mass and energy
NEXT STEPS
  • Study the derivation of photon energy in particle decay scenarios
  • Learn about the implications of mass-energy equivalence in special relativity
  • Explore advanced topics in relativistic momentum and energy conservation
  • Investigate the role of decay processes in particle physics
USEFUL FOR

Students and educators in physics, particularly those focusing on particle physics and special relativity, as well as researchers interested in the implications of mass loss during particle decay.

vladimir69
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Homework Statement


Set the speed of light c=1.
A particle of rest mass m_{0} decays at rest into a photon and loses rest mass \delta in the bargain. Show that the photon energy is \omega=\delta(1- \frac{\delta}{2m_{0}}) in the particle's rest frame before the decay.

Homework Equations


\mathbf{K}=(\omega,k) is the photon momentum
\mathbf{P_{1}}=(m_{0},0) is the initial momentum of the particle
\mathbf{P_{2}}=(\gamma(m_{0}-\delta),\gamma(m_{0}-\delta)v) is the final momentum of the particle

initial momentum = final momentum
"dot" product invariant
(a,b) \cdot (x,y) = ax-by

The Attempt at a Solution


Firstly conservation of momentum
\mathbf{P_{1}}=\mathbf{P_{2}}+\mathbf{K}
\mathbf{P_{1}}\cdot \mathbf{P_{1}} = \mathbf{P_{2}} \cdot \mathbf{P_{2}} +2\mathbf{K}\cdot\mathbf{P_{2}}
The first equation gives us
m_{0} = \gamma(m_{0}-\delta) +\omega
0= \gamma(m_{0}-\delta)v + k
The second equation gives us
m_{0}^2=(m_{0}-\delta)^2+2(\omega\gamma(m_{0}-\delta)-kv\gamma(m_{0}-\delta))
Stirring all these into the mixing pot and going on an algebraic safari out pops
\omega=\delta which of course is wrong

But just where did I go wrong?

Isn't it true that \omega-k=0 in a world of just 1 spatial dimension and 1 dimension of time?
 
Last edited:
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hi vladimir69! :smile:

(have an omega: ω and a delta: δ :wink:)
vladimir69 said:
Set the speed of light c=1.
A particle of rest mass m_{0} decays at rest into a photon and loses rest mass \delta in the bargain. Show that the photon energy is \omega=\delta(1- \frac{\delta}{2m_{0}}) in the particle's rest frame before the decay.

hmm … do you really think this particle is going to move at a relativistic speed? :wink:

try it without the relativity (and yes, ω = |k|) :smile:
 
I thought this question must use relativistic principles since it was asked in the special relativity chapter.
The solution does it as
<br /> \mathbf{P_{2}}=\mathbf{P_{1}}-\mathbf{K}<br />
<br /> \mathbf{P_{2}}^2=\mathbf{P_{1}}^2-2\mathbf{KP_{1}}<br />
(m_{0}-\delta)^2=m_{0}^2-2m_{0}\omega
Solving for omega reveals the correct result, just wondering why my method didn't work
 

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