When are decays into quarks kinematically possible?

Question

Consider the decay of a particle, $X$, into two quarks $q_1$ and $q_2$. Naively one would expect such a decay to be kinematically possible if: \[ m_Xc^2 \ge m_{q_1}c^2+m_{q_2}c^2\] however, I think that such a relation is wrong and the correct statment is that we require: \[ m_Xc^2 \ge m_{H_1}c^2+m_{H_2}c^2\] where $m_{H_i}$ represents the mass of the lightest hadron containing the quark $q_i$. Is this correct and please can you explain either way. Also what happens if one of the quarks is a top quark?

Example

For example I would say that the $\tau$ lepton can't decay into $\nu_\tau \bar c d$ since the lightest meson with a $\bar c$ quark is heavier then the $\tau$ particle, although $m_c\lt m_\tau$.

Answer 1

All interactions have to obey conservation of quantum numbers that identify a particle in the table of particles in the standard model. Depending on the exchange forces, i.e. strong, or weak or electromagnetic, rules have to be followed.

The necessity of using the second of your two inequalities is due to the fact that particle physics works with special relativity mathematics. In contrast to the leptons , quarks cannot appear by themselves and are bound within hadrons. The mass of the hadrons is not just an addition of the quark masses, but due to the strong interaction there exists within the hadron a sea of quarks, antiquarks and gluons whose four vectors add up to give the invariant mass of the specific hadron. See a relevant question and answers here. The decays have to happen into quarks that are bound within a hadron and total energy has to be conserved.

In this sense your example is not energetically possible because the valence c quark mass contributes too much in the lightest massive charmed meson whose mass is 100 MeV larger than the tau mass.