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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?

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1 Answer 1

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Where have I gone wrong?

Dimensional analysis.

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.

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  • $\begingroup$ Very clear answer, thanks! $\endgroup$ Commented Jun 5, 2020 at 6:55

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