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An ultra-relativistic particle with $pc>>Mc^2$ is supposed to decay into two massive particles of mass $m$. I've been told that this is impossible and have tried the following to reach that conclusion:

The four momentum of the ultra-relativistic particle in its rest frame is $p^{\mu} = (M,0)$ and the decay products will have identical energy and momenta so writing the four momentum of the second particle in terms of the first and the parent particle: $p_2^{\mu} = p^{\mu} -p_1^{\mu}$ we can take magnitudes and find: $-m^2+m^2 = -M^2 +2ME$ where $E$ is the energy of the decay products. So $E =M/2$ and the magnitude of the daughter's four momenta are: $|p| = \frac{\sqrt{M^4 -4M^2 m^2}}{2M}$. I don't see how to use the ultra-relativistic condition to reach a contradiction here.

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  • $\begingroup$ Compare to the alleged process $\gamma \to e^+ + e^-$ in free space (about which we have quite a few questions). $\endgroup$ – dmckee --- ex-moderator kitten Apr 15 '18 at 18:31
  • $\begingroup$ Is it massive as $m>>M$? $\endgroup$ – Oktay Doğangün Apr 15 '18 at 18:35
  • $\begingroup$ So because the energy of an ultra-relativistic particle is almost entirely momentum we can approximate its energy as just being momentum and then it's impossible to conserve energy and linear momentum at the same time as for photon decay? $\endgroup$ – John Doe Apr 15 '18 at 18:54
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The “ultra-relativistic” part is a statement about motion relative to an observer, e.g. you. The decaying particle simply doesn’t care about that: it’s decaying or not in its own rest frame.

So if it can decay, it eventually will.

The only thing “ultra-relativistic” brings to the discussion is how long you observe it to take.

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"I've been told that this is impossible..." Who told you that? The particle will decay, but it's lifetime will be extended by a factor of E/M. this means that it will probably leave your laboratory before it decays. Your not seeing it decay is different than "impossible".

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