0
$\begingroup$

Antimatter could be produced by inverse annihilation from kinetic energy of particles. This is, however, not useful to produce any energy.

Is there any pathway (like set of reactions between sub atomic particles) to produce some antiparticles from normal matter using its rest mass.

I think I read in some popularization book that such process are possible governed by weak nuclear force (rather than strong), or maybe by some super-symmetry, but that they are not very efficient. But I don't remember it exactly. (EDIT - I think baryogenesis and symetry violation by weak force are examples ... but I guess this is not usefull for any artificial antimatter production by houmans? is it?)

I understand that you have to fulfil conservation laws (conservation of electric charge, spin, isospin, lepton / baryon number ... ). But, maybe you can use some heavy particles of normal matter and some light particles of anti matter and gain some heavy particles of antimatter while net change of all conserved properties would be still zero.

e.g. production of electron-positron pairs or neutrino anti-neutrino pairs by inverse annihilation does not cost lot of energy.

If you can use this positrons and anti-neutrinos, do some reactions with some heavy particles of normal matter (like proton, neutron ... ) and produce some heavy antiparticles (like antineutrons, antiprotons, antitaouons, pions ... ) and then annihilate this heavy antiparticles with corresponding normal matter particle ... you would possibly gain some energy.

the "charge" and other conserved properties would be basically transferred to the lighter particles.

$\endgroup$
  • $\begingroup$ $\beta$-decay will net you some antiparticles. $\endgroup$ – Ryan Unger Feb 14 '15 at 21:38
  • $\begingroup$ ah, yes, but just 0.5 MeV :-( less than many nuclear reactions. I was thinking about some process which if combined with anihilation will convert to energy considerable share of rest mass of the source of normal matter. $\endgroup$ – Prokop Hapala Feb 14 '15 at 22:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.