# Change of flavour, Weak interaction

I couldn't find a straight and clear answer to this question on the internet: Why is the weak interaction (charged), the only interaction which can change the flavours of the quarks?

• Charge currents also change the flavour of leptons as well. – AMS Jun 8 '16 at 17:23
• the model of weak, strong,electromagnetic interactions (SU(3)xSU(2)xU(1) )came from experimental observations: the standard model fits the data. – anna v Jun 8 '16 at 18:34

Lets look at the weak isospin doublet of for the first generation of quraks \begin{equation} \begin{pmatrix} u\\d' \end{pmatrix}_{L} \end{equation} Where $d'$ is represents the quantum mixing of $d$ and $s$ quark. From gauge theory, we know that, the role of gauge bosons is to transform a particle to another one, which living in the same mutiplet. So, if a $u$ quark turns into $d$ quark, that can be possible if a positively charge gauge boson can mediate such a transition.
Because interactions conserve the third component of weak isospin, $T_3$. So the incoming $T_3$ in a vertex must be equal to the outgoing $T_3$. For example: • Weak isospin is another flavor quantum number, associated to the weak interaction. With isospin you had a doublet with a quark up and a quark down. So you basically treated the quarks up and down as two different states of the same particle, assigning as the third component of isospin $I_3$=1/2 to the quark up and $I_3$=-1/2 to the quark down, creating a "doublet". With weak isospin it's the same, but your doublets are only for left-handed particles. So you have the doublet $(e^{-}, \nu_e)_L$, $(\mu^{-}, \nu_{\mu})_L$ and so on. – Luthien Jun 8 '16 at 20:05