All the decays I read about involve electron - positron annihilation.
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4 Answers
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No. There are loads of conserved quantities in decay processes like the one you are talking about. Lepton number conservation and charge conservation are the most notable ones for the case of a positron. Also, the sum of the masses of decay products should always be lesser than the mass of the initial particle. (This naturally follows from energy conservation.)
A positron can't decay because there is no lighter particle it can decay into, while conserving charge and lepton number. (This, however, is possible in electron-positron annihilation, because the electron and positron cancel each other's charge and lepton number perfectly.)
answered Sep 6, 2015 at 7:46
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$\begingroup$ Would a tau and a positron bounce off each other? $\endgroup$– JitterSep 6, 2015 at 13:57
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$\begingroup$ If we assume the tau to be stable, I think they'd behave how positive and negative charges are expected to behave when in close proximity! (or so I think. I'm not sure. Interesting.) $\endgroup$ Sep 6, 2015 at 14:01
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2$\begingroup$ a tau and a positron would scatter off each other( muon+ and electrons do routinely ) if enough energy existed more particles could be produced and transmutations . Probability would be small as the tau is unstable and decays fast ~10^-13seconds $\endgroup$– anna vSep 6, 2015 at 14:32
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$\begingroup$ lesser than the mass + KE $\endgroup$– user46925Jan 15, 2016 at 15:50
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If electric charge is conserved, then electrons and positrons can't decay. We don't know if electric charge is conserved. The most conservative limits for the stability of the electron we seem to have seem to suggest a lifetime of approx. 4.6e26 years from a lab measurement (at BOREXINO, I believe) and >1e39 years from cosmological arguments. Both are far short of typical lifetimes of the universe that are being discussed by cosmologists (1e100 years and more). See K.A. Olive et al. (Particle Data Group), Chin. Phys. C38, 090001 (2014) (URL: http://pdg.lbl.gov).
From an experimental perspective it is always better to use established limits on lifetime/symmetry breaking than to say "x is conserved therefor y doesn't happen". The limits are the state of our current knowledge, the latter statement is an extrapolation to a fundamentally untestable hypothesis.
answered Sep 6, 2015 at 10:33
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1$\begingroup$ new measures of the lower bound published A test of electric charge conservation with Borexino : $6.6*10^{28}$ years $\endgroup$– user46925Jan 15, 2016 at 15:48
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$\begingroup$ @igael: Cool! Thanks for the link. Yeah... that's a very long time, for sure. $\endgroup$ Jan 15, 2016 at 16:50
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Yes a positron can interact without encountering an electron. Strictly speaking, it is not a decay since it must encounter another particle because as it is said in another answer, the positron is a stable particle (in the vacuum), so it cannot decay on its own. An example of interaction not involving an electron: $$e^+ + \mu^- \to \bar{\nu_e} +\nu_{\mu}$$ It proceeds via the weak interaction (a $W$ boson being exchange in this case). Now in the nature, it is easier to encounter an electron (part of any atom) than a muon! So the most frequent reaction does involve an electron: $$e^+ + e^- \to \gamma + \gamma$$ which is an electromagnetic decay (much more likely than a weak interaction decay).
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1$\begingroup$ These are interactions, not decays. It is wrong to mix up vocabularies. annihilation is an alternate term for the disappearance of a positron. not decay $\endgroup$– anna vSep 11, 2015 at 13:12
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$\begingroup$ @annav: Strictly speaking, I agree with you but the formulation of the question mentions the electron-positron annihilation as a decay. So it was not clear whether MAXIMILLION wanted to know if a positron can disappear by interacting with something else than en electron or if he just asked why the positron was a stable particle. BTW, I used the word decay with double quotes suggesting an abuse of language. $\endgroup$– PaganiniSep 11, 2015 at 14:38
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$\begingroup$ As long as this post is using the word "decay" to describe these reaction it is contributing to the misunderstanding of the OP and other reader that come along. Don't encourage this kind of misunderstanding, correct it. $\endgroup$ Feb 22, 2016 at 14:43
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$\begingroup$ @dmckee: done, I changed decay for interaction. $\endgroup$– PaganiniFeb 26, 2016 at 19:37
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As several answers have stated already: A positron by itself is not known to decay at all. But if you are considering "encounters" in the course of which a given positron ceases to exist, then how about the "absorption" of a positron by a neutron, leaving a proton, accompanied by (emission of) an anti-electron-neutrino: $$\mathbf n + \mathbf e^+ \rightarrow \mathbf p + \overline \nu_e,$$ i.e. in terms of the relevant quarks: $$\mathbf d + \mathbf e^+ \rightarrow \mathbf u + \overline \nu_e.$$
