Momentum of massless particle after decay

[nonequilibrium]

Homework Statement

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..."

Homework Equations

energy² = mass² c^4 + p² c²

momentum is conserved

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 $p = \frac{M^2 - m^2}{2M} c$

 

[George Jones]

Using conservation of energy we get $M^2 c^4 = m^2 c^4 + p_m^2 c^2 + p_\gamma^2 c^2$

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

 

[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.

 

[nonequilibrium]

Oh jeesh, thank you

 

[George Jones]

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