feynman-diagramsradiationweak-interaction
From https://en.wikipedia.org/wiki/Beta_decay
In $β^+$ decay, or positron emission, the weak interaction converts an atomic nucleus into a nucleus with atomic number decreased by one, while emitting a positron ($e^+$) and an electron neutrino ($ν_e$).
If the positron is emitted, why is its arrow directed towards the $W^+$?
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:
Best Answer
The motivation of this convention in Feynman diagrams is that the positron is antimatter, whereas the electron neutrino is matter. By CPT symmetry, an emitted positron is equivalent to an absorbed electron travelling backwards in time. Under time reversal, the rightmost interaction vertex is equivalent to $e^-\to\nu_e+W^-$, and the arrow directions on the "electron" and neutrino are less surprising.