# Momentum of massless particle after decay

1. Jul 10, 2011

### nonequilibrium

1. The problem statement, all variables and given/known data
I must be overlooking something very simple:
"A particle of mass M decays from rest into two particles. One particle has mass m and the other particle is massless. The momentum of the massless particle is..."

2. Relevant equations
energy² = mass² c^4 + p² c²

momentum is conserved

3. The attempt at a solution
a quick drawing makes clear that $\vec p_{m}$ and $\vec p_{\gamma}$ (with gamma denoting the massless particle) are on the same line, and since the initial momentum was zero, we get $\vec p_{m} = - \vec p_{\gamma}$ and thus $p_m^2 = p_\gamma^2$.

Using conservation of energy we get $M^2 c^4 = m^2 c^4 + p_m^2 c^2 + p_\gamma^2 c^2 = m^2 c^4 + 2 p^2 c^2$ with p the sought for momentum.

However, the answer should be $p = \frac{M^2 - m^2}{2M} c$

2. Jul 10, 2011

### George Jones

Staff Emeritus
I don't think that this is what conservation of energy gives.

3. Jul 11, 2011

### Matterwave

You have put the squares of the energies added together. Conservation of energy is the energies themselves added together. You will then see that when you square it, you get cross terms. In other words, what is correct is:

$$E_i=E_{1,f}+E_{2,f}$$

What you have written there is:
$$E_i^2=E_{1,f}^2+E_{2,f}^2$$

You are missing the cross term.

4. Jul 11, 2011

### nonequilibrium

Oh jeesh, thank you

5. Jul 11, 2011

### George Jones

Staff Emeritus
Things are simpler if the the conservation of energy equation is rearranged before squaring.