Though beta positive decay's have been observed, in which a proton decays into a neutron, positron and an electron- neutrino; the why is it not the same as the exotic proton decay which is hypothetical and hasn't been observed?
[Physics] the difference between beta positive decay and proton decay
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Related Solutions
At first, consider two particles decay:
$A\rightarrow B + e^-$ Where A is initially rest. So $\vec{p_B}+\vec{p_{e^-}}=0$
now \begin{align} \frac{p_B^2}{2m_B}+\sqrt{p_{e^-}^2+m_{e^-}^2}&=E_{released} \\ \frac{p_{e^-}^2}{2m_B}+\sqrt{p_{e^-}^2+m_{e^-}^2}&=E_{released} \tag{1} \end{align}
see here you have uncoupled equation (equ.1) for $p_{e^-}$ .. So, solving above (equ.1) you will get a fixed $p_{e^-}$. Hence the energy($\sqrt{p_{e^-}^2+m_{e^-}^2}$) of the $\beta$ particle is always fixed in the two body $\beta$ decay. (you can find $p_{B}$ too using the momentum conservation formula)
At first, consider three particles decay:
$A\rightarrow B + e^-+\nu_e$ Where A is initially rest. So $\vec{p_B}+\vec{p_{e^-}}+\vec{\nu_e}=0$
so \begin{align} \frac{p_B^2}{2m_B}+\sqrt{p_{e^-}^2+m_{e^-}^2}+p_{\nu_e}&=E_{released} \\ \frac{(\vec{p_{e^-}}+\vec {p_{\nu_e}})^2}{2m_B}+\sqrt{p_{e^-}^2+m_{e^-}^2}+p_{\nu_e}&=E_{released} \tag{2} \end{align}
Now see in contrast to the two particles decay equation, here we have five unknowns $\big{(}p_{e^-},p_{\nu},p_{B},\theta\ (\rm the\ angle\ between\ \vec{p_{e^-}},\vec{p_{\nu}}),\phi\ (\rm the\ angle\ between\ \vec{p_{e^-}},\vec{p_{B}})\big)$ but four coupled equations *. so we can't solve them uniquely. That's also what happens physically. You will get different values of $p_{e^-},p_{\nu},p_{B},\theta,\phi$ satisfying the four coupled equations. Hence different $\beta$ particles will have different energy($\sqrt{p_{e^-}^2+m_{e^-}^2}$) maintaining the statistics of decay process. Hence the continuous spectra.
*The four coupled equations are equ.2 and three equations which we can get by taking Dot products of $\vec{p_{e^-}},\vec{p_{\nu}}\rm\ and\ \vec{p_{B}}$ with the momentum conservation($\vec{p_B}+\vec{p_{e^-}}+\vec{\nu_e}=0$ ) and remember they lie in a plane so two angles ($\theta\rm\ and\ \phi$) are sufficient.
At a first glance, W bosons are not needed at all in those scenarios: you could model the interaction by supposing a direct coupling exists between the 4 particles (neutron, proton, electron, neutrino) -- which is what was initially proposed by Fermi (see Fermi's 4-point Interaction).
Thus, both the $n+\nu \rightarrow p + e$ interaction and the decay $n\rightarrow p + e + \overline{\nu}$ can be diagrammatically represented by:
where time flows from left to right.
Problems, however, arise for this ansatz: as the center-of-mass energy (usually denoted $\sqrt{s}$) of the interacting neutron and neutrino in the left diagram rises, so does the cross-section $\sigma(s)$ computed in this context, in a nonsensical way (probabilities exceed 1).
One solution is to 'UV complete' the theory -- i.e. complete the theory by specifying behaviour in the ultraviolet/at high energies -- by abandoning the naïve contact interaction altogether and replacing it with the exchange of W bosons (this cannot happen just for the left diagram; one has changed the model). So, indeed, in this completed theory, even neutron decay is explained by exchange of a W boson.
Final comments:
- For low enough energies, the 4-point interaction description is accurate and still useful;
- In the Standard Model, the W boson couples to the quarks out of which the proton and neutron are made of.
Best Answer
Protons and neutrons are very similar particles. Although they have different charges, as far as the strong force is concerned they are almost identical. So changing a proton into a neutron and back isn't considered decay because you aren't changing the number of nucleons. More precisely the baryon number remains constant.
If we describe the proton and neutron as a bound state of three quarks$^1$ then a proton is two up and one down quarks while a neutron is two down and one up quarks. Interchanging them requires changing an up quark to a down quark and vice versa, which happens by emission/absorption of an electron or positron. The number of quarks doesn't change and the process is reversible.
The process we normally describe as proton decay is altogether more radical. There are actually several different possible mechanisms for proton decay, but they all involve the proton disappearing completely leaving behind just a positron and two photons. This has to involve the creation of a hypothetical and so far unobserved particle called an X boson. The Standard Model does not include the X boson so as far as the Standard model is concerned the proton cannot decay.
$^1$ Caution! Not literally true!