After a nuclear decay is it a necessity that the total energy of the products is more than the energy of the original particle before decaying? (NB: by 'energy' I don't intend to include mass-energy as well, unless by definition the term 'total energy' happens to include that in the realm of particle physics)

I was told that this might depend on whether the total mass of the products is less than the mass of the decaying particle. I'm not sure of the veracity of this claim.

I've heard the Higgs Boson can decay to heavier particles than itself. But, I'm not sure about other decays, such as the one I described, which is not a Higgs Boson.

For example, is the statement "the mass energy of the original particle is equal to the total energy of the decay products" false?

  • $\begingroup$ Particle physicists do tend to include mass in "energy" most (but not quite all) of the time. But I can take it that you are interested in the relationship between the pre- and post-decay kinetic energies, no? $\endgroup$ – dmckee May 10 '14 at 17:49
  • $\begingroup$ @dmckee yes, particularly in whether the last statement I quoted is true/false $\endgroup$ – hb20007 May 10 '14 at 17:53
  • $\begingroup$ If you examine the situation in the rest frame of the progenitor, then it should be absolutely clear that the kinetic energies can't be the same. From that I would conclude that the question is asking you about all the energy contributions, including the masses. $\endgroup$ – dmckee May 10 '14 at 18:29
  • $\begingroup$ @dmckee oh, I think you're right then $\endgroup$ – hb20007 May 10 '14 at 19:46

Total energy, including rest mass, is conserved in all known decays. This was not always obvious: prior to Pauli's proposal of a massless, essentially noninteracting neutrino, some physicists were prepared to conclude that energy conservation is a macroscopic, statistical phenomenon, and that energy was not conserved in nuclear beta decays.

In a decay, you have an initial state with a single particle. According to special relativity there always exists a reference frame in which that particle is at rest. In that special reference frame, the decay products in the final state have some kinetic energy and move apart. In order for energy to be conserved, the rest masses of all the particles must add up to less than the rest mass of the initial state.

The only fundamental particle heavier than the Higgs boson is the top quark. I'm quite sure that there are no decays of an isolated Higgs with a top quark in the final state. It's possible there could be a reaction between a Higgs and another energetic particle that has a top quark in the final state, but what little I know about collider experiments makes me doubt that has been observed, either.

The statement "the mass energy of the original particle is equal to the total energy of the decay products" is true in every known decay reaction.

  • $\begingroup$ I just have one more question. You said the statement is true in all known decay reactions. Is this in the special reference frame where the particle is at rest? $\endgroup$ – hb20007 May 10 '14 at 20:22
  • $\begingroup$ I see it that way because the KE of the original particle doesn't seem to be accounted for. $\endgroup$ – hb20007 May 10 '14 at 20:24
  • 2
    $\begingroup$ @hb2007 Energy (and also momentum) is conserved in all references frames. The arithmetic is easier in the rest frame of the decaying particle, but observers in all inertial reference frames will agree that total energy is conserved. In fact, in special relativity, energy and momentum make a four-vector $(E,\vec p)$ that transforms just the same way as the more familiar combinations of time and position coordinates $(t, \vec x)$. $\endgroup$ – rob May 10 '14 at 20:37

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