1
$\begingroup$

In chemical reactions, there are usually energy barriers that must be overcome before a reaction can take place. It seems that such a phenomena also exists at the atomic level as well. For instance, a certain amount of energy must be provided to two nuclei to bring them close enough together to fuse and release more energy than was input. In general, I read things like "energetically favorable" or "a lower energy state is available".

However, in examining the behavior of anti-matter, it seems almost as if there is no such barrier between matter and anti-matter: they annihilate immediately on contact, converting most or all of their matter into energy. It seems like such a sharp divide between total annihilation with virtually no extra energy added to the equation, versus almost total stability for either matter or anti-matter by itself. Why such the big difference? Why does anti-matter seem to have no constraints to "decaying" into energy in the presence of matter (and vice versa) like other processes do? I know there are certain quantities like charge which must be conserved but it seems like a pretty abstract thing to say we don't all spontaneously decay into photons simply because charge must be conserved. Why isn't there some kind of energy barrier between matter and anti-matter which must be overcome before they can react? If anti-matter had negative mass, then it would have negative energy, in which the lack of such a barrier would make more sense. However, I don't know of any evidence that suggests anti-matter has negative mass.

My naive guess would be that since a particle and anti-particle have opposite spin, they sort of "slam into" each other and cancel out, causing the particle to "break apart" into a free wave. However, since my understanding of spin is that it isn't really the same kind of spin we would observe at the macroscopic scale, i.e. actual physical rotation, this seems like an unreal explanation to me.

$\endgroup$
  • $\begingroup$ Activation energies are generally a feature of the system already being settled into a local energy minimum (think about a ball in a shallow depression at the top of a tall hill). That isn't the case when you have loose particle-antiparticle pairs flying around so there is no expectation that you would see such a thing. Nor is there any need to guess about these things: they are well explained with even basic facts about particles, atoms and molecules (just charge, and what particles are stable mostly). $\endgroup$ – dmckee Mar 22 '14 at 17:28

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.