# Particle decay formulae

#### Wledig

Problem 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: $$k = \sqrt{\dfrac{M-m_1^2-m_2^2}{2}}$$

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#### vela

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

How come?

#### vela

Staff Emeritus
Homework Helper
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

Staff Emeritus
Homework Helper
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.

Can you help me with this one?

"Particle decay formulae"

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