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

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The discussion focuses on the relationship between photon energy and mass loss in particle decay, specifically deriving the photon energy formula in a particle's rest frame. The initial approach involved applying conservation of momentum and energy, but the user encountered an incorrect result of ω = δ. Another participant suggested using a different method that correctly applied relativistic principles, leading to the formula ω = δ(1 - δ/2m₀). The conversation highlights the importance of correctly applying relativistic equations in particle decay problems. Ultimately, the user seeks clarification on the failure of their initial method.
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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?
 
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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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