Is a beta decay possible with out the emission of an electron in the process?

Is a beta decay possible with out the emission of an electron in the process ?

Beta decay involves the formation of a proton and an electron from a neutron.

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Since the definition of beta decay contains the emission of a $\beta^-$ particle (another word for an electron), the question is meaningless.

However, if you define beta decay as "neutron becomes a proton", then yes, it is possible.

This is "normal" beta decay:

A down quark becomes an up quark, in the process releasing a $W^-$ boson. The boson then decays into a $e,\bar \nu_e$ pair

However, the $W^-$ boson can also decay into a:

• $\tau,\bar\nu_\tau$ pair
• $\mu,\bar \nu_\mu$ pair
• A quark pair consisting of one of $\{d,s,b\}$ and one of $\{\bar u, \bar c, \bar t\}$. ($\pi^-,K^-$, etc)

in which case you will have a phenomenon similar to beta decay with different particle(s) coming out.

However, these are kinematically impossible. They may be possible with some photon capture/emission, but I doubt it.

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Worth mentioning that all these other processes are kinematically forbidden. :) –  Michael Brown Apr 3 '13 at 14:13
@MichaelBrown: Oh, right. Damn. Might be possible with some photon capture/emission, though (doubt it) –  Manishearth Apr 3 '13 at 14:15