# Two-body decay: Heavier particles live longer than light particles

From Fermi's Golden Rule one can derive that the decay rate for a two-particle decay ($$A\to B+C$$) is given by

$$\Gamma = \frac{p^*}{32\pi^2m_A^2} \int |{\cal M}|^2 d\Omega,$$

where the absolute value of the momenta of the outgoing particles is given by

$$p^* = \frac{1}{2m_A} \sqrt{\left[ m_A^2 - (m_B + m_C)^2 \right] \left[ m_A^2 - (m_B - m_C)^2 \right] }.$$

$$\cal M$$ is the matrix element, and $$m_{A,B,C}$$ are the masses of the particles involved. (This is textbook knowledge, cf. Griffiths, Thomson, or Wikipedia.)

Now the lifetime of a particle is given by $$\tau = 1/\Gamma$$, and from the above equation we should be able to tell how the lifetime goes with the mass $$m_A$$ (assuming $$m_{B,C}$$ stay constant).

For $$m_A \gg m_{B,C}$$, we get $$p^* \sim \frac{\sqrt{m_A^4}} {m_A} = m_A$$, thus $$\Gamma \sim \frac1{m_A}$$ and $$\tau \sim m_A$$, i.e. heavier particles live longer than light particles.

Where have I gone wrong?

You know $$|{\cal M}|^2$$ must have dimensions of [mass]$$^2$$, if Γ has to have dimensions of [mass]. In a strong, "normal", interaction, the mass scale in your limit is then set by $$m_A^2$$, and so, dividing by your phase-space $$m_A$$, you see that $$\Gamma \sim m_A$$, as you appear to appreciate. The heavier the particle, the shorter it lives.
This is not all: for some freak cases involving chirality-suppressing weak decays, like those of charged leptons, you have $$|{\cal M}|^2\propto m_A^6/M_W^4$$, so that $$\Gamma \sim m_A^5/M_W^4$$, which reminds you why the tau lepton is so much shorter-lived than the muon.