How Does Conservation of Momentum Apply to Particle Decay?

[Wledig]

Homework Statement

Consider the decay of a particle of mass M, at rest, into two particles with masses $m_1$ and $m_2$, both nonzero. With an appropriate choice of axes, the momentum vectors of the final particle can be written: $$p_1 = (E_1,0,0,k)$$ $$p_2 = (E_2,0,0,-k)$$ with $E_1^2 = k^2 + m_1^2, E_2^2 = k^2 + m_2^2$.


a) Show that $k = \dfrac{\sqrt{(M^4 -2M^2(m_1^2+m_2^2)+(m_1^2-m_2^2)^2}}{2M}$



b) Take the limit $m_2 \rightarrow 0$ and show that this reproduces the result for the decay into one massive and one massless particle.

c) Find formulae for $E_1$ and $E_2$ in terms of M, m1, m2.

Relevant Equations

Energy momentum relation: $E^2 = p^2 + m^2$

Attempt at solution:

By conservation of momentum: $$P = (M,0,0,0) = p_1 + p_2 = (E_1 + E_2, 0, 0,0)$$ thus
$$ M = E_1 + E_2 = 2k^2 + m_1^2 + m_2^2$$
Now $$E_1^2 - E_2^2 = m_1^2 - m_2^2 = (m_1 + m_2)(m_1-m_2)$$
$$ = M(m_1-m_2) = (2k^2+m_1^2+m_2^2)(m_1-m_2)$$
Isolating k: $$ k = \sqrt{\dfrac{M-m_1^2-m_2^2}{2}}$$

 

[vela]

$M \ne m_1+m_2$. Also, $M$ can't be equal to $m_1^2+m_2^2+2k^2$. The units don't work out.

 

[Wledig]

How come?

 

[vela]

Mass isn't conserved in special relativity.

 

[Wledig]

You're right. Is it fair to say though that $P = p_1 + p_2$ and $p_1^2 = m_1^2$, $p_2^2 = m_2^2$?

 

[vela]

You're right. Is it fair to say though that $P = p_1 + p_2$ and $p_1^2 = m_1^2$, $p_2^2 = m_2^2$?

Yes. Try squaring $p_1 = P - p_2$.

 

[Wledig]

Ok, I think I got it. Squaring this term like you suggested gives:

$$ m_1^2 = P^2 - 2P\cdot p_2 + p_2^2$$
$$m_1^2 = M^2 - 2ME_2 + m_2^2$$

Isolating $E_2$:
$$ E_2 = \dfrac{(M^2-m_1^2-m_2^2)}{2M}$$
Which if we plug it into:
$$ \vec{p_2}^2 = k^2 = E_2^2 - m_2^2 $$

Returns the relation asked, if I didn't mess up the calculation.

https://www.physicsforums.com/threads/equivalent-representations-for-dirac-algebra.973347/Can you help me with this one?