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

[vladimir69]

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?

 

[tiny-tim]

hi vladimir69!

(have an omega: ω and a delta: δ )

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?

try it without the relativity (and yes, ω = |k|)

 

[vladimir69]

I thought this question must use relativistic principles since it was asked in the special relativity chapter.
The solution does it as
$$\mathbf{P_{2}}=\mathbf{P_{1}}-\mathbf{K}$$
$$\mathbf{P_{2}}^2=\mathbf{P_{1}}^2-2\mathbf{KP_{1}}$$
$$(m_{0}-\delta)^2=m_{0}^2-2m_{0}\omega$$
Solving for omega reveals the correct result, just wondering why my method didn't work