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When beta-decay occurs an electron or a positron is emitted along a neutrino or an antineutrino. The energy spectrum of the beta particles is continuous because, as I read, the energy is shared between the two emitted bodies.

My question is: what determines the distribution of energy here? That is, what determines with how much energy each particle is emitted? And for that matter: why is this sharing of the energy continuous?

carllacan
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  • You require calculations regarding the weak force for this. Since you are trying to find out the energy given to each particle, the force acted upon each particle is required. For this, you need to consider the speed of the particles involved in the beta decay. –  Jan 18 '14 at 21:00

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The process must conserve all of energy, momentum and angular momentum. Within the possible final states that obey those limits it appears to be entirely random (or to be controlled by some inaccessible, non-local hidden variable if you prefer those kinds of QM interpretations).

In any case, the energy distribution of any particular decay can not be predicted, though the distributions of a large number is both predictable (on the basis of available phase space) and well measured.

dmckee --- ex-moderator kitten
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  • dmckee: "The process must conserve all of energy, momentum and angular momentum. [...] the distributions of a large number is both predictable (on the basis of available phase space) and well measured." -- Would you please elaborate how "available phase space" of the indicated decay process(es) is to be compared, e.g. comparing for any two specific (allowed) "energy" values of the generated "beta particles"? (I had voted up the OP question a while ago precisely in order to learn more about how to compare "available phase space".) – user12262 Jan 19 '14 at 09:29
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    @user12262 I talk about phase space in my answer to "Why is the (free) neutron lifetime so long?", and about how one applies the calculation to writing a Monte Carlo generator for a three particle final state in another answer (assuming that you will use weighted events). The bit where I say "parameterize the physics" is where you enforce the conservation rules. – dmckee --- ex-moderator kitten Jan 19 '14 at 15:23