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

AI Thread Summary
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.
vladimir69
Messages
124
Reaction score
0

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:
Physics news on Phys.org
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
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top