There are 3 modes of decay via which a $$\Omega ^{-}$$ particle can decay

This is one of the decay:

$$\Omega ^{-} \rightarrow \Xi ^{0} + \pi^{-}$$

Baryon number is conserved. Strangeness number is not conserved so that would imply that the interaction involved is weak Apparently, the rest mass isn't conserved either.

$$\Omega ^{-} = 1672\frac{\rm MeV}{c^{2}}$$ $$\Xi ^{0} = 1315\frac{\rm MeV}{c^{2}}$$ $$\pi^{-} = -140\frac{\rm MeV}{c^{2}}$$

Really, what does this mean? How is this a possible decay mode if the rest mass isn't conserved?

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    $\begingroup$ The mass deficit goes into the kinetic energy of the decay products $\endgroup$ – John Rennie Jun 5 '15 at 10:10
  • $\begingroup$ @JohnRennie John, could you elaborate? $\endgroup$ – Physkid Jun 5 '15 at 10:14
  • $\begingroup$ possible duplicate of Is (rest) mass conserved in special relativity? $\endgroup$ – innisfree Jun 5 '15 at 10:15
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    $\begingroup$ @Physkid: in quantum field theory mass can change into (kinetic) energy and vice versa. That's how two 1GeV protons can produce a 125GeV Higgs boson in the LHC. In this case mass is being converted into kinetic energy. If you add up the total energy given by $E^2 = p^2c^2 + m^2c^4$ before and after you'll find it is constant. $\endgroup$ – John Rennie Jun 5 '15 at 10:38
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    $\begingroup$ No, in the standard model, all masses are positive. $\endgroup$ – Omry Jun 5 '15 at 11:02

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