Is Lepton Flavour Universality an accidental symmetry of the Standatd Model? If it is, why? How does it emerge from the Standard Model?
Particle Physics – Is Lepton Flavour Universality an Accidental Symmetry of the Standard Model?
leptonsparticle-physicsstandard-model
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Let me address your questions one by one.
Why is it said that lepton number is conserved in Standard Model (SM)? How do I know that lepton number is an Abelian charge?
The SM Lagrangian is invariant under the fermion transformations, $$ \psi \to e^{iL\theta}\psi $$ where $L$ is assigned such that $e^-$, $\mu^-$ and $\tau^-$ leptons and lepton-neutrinos have $L=1$, whilst their antiparticles have $L=-1$, and everything else has $L=0$. This global $U(1)$ symmetry corresponds to lepton number conservation - $L$ is what we call lepton number. Lepton number is by construction the Abelian charge corresponding to that $U(1)$ global symmetry.
Why is this conservation not as sacred as electric charge conservation?
The lepton $U(1)$ global symmetry is accidental. We simply cannot write a gauge invariant, renormalizable operator in our Lagrangian that breaks conservation of lepton number. There is no reason to expect that physics beyond the SM respects lepton number conservation.
The electric charge $U(1)_{em}$ local symmetry was a principle on which the SM was built. The SM would not be renormalizable if $U(1)_{em}$ was explicitly or anomalously broken. It would be catastrophic if $U(1)_{em}$ were broken.
How does one mathematically distinguish between lepton number and electric charge?
They are the conserved charges associated with different $U(1)$ symmetries. In general, when you have multiple $U(1)$ symmetries, charge assignment is somewhat arbitrary, since one can pick different linear combinations of the original $U(1)$ generators as the symmetries. In this case, however, the $U(1)_{em}$ is local, so there is no mixing with the global lepton number $U(1)$.
When energy in a given reaction is much larger than the masses of all leptons, these masses can be neglected. For example if you consider $W$ bosons decays, then $$M_W \gg m_\tau.$$ Tau is the heaviest among the charged leptons, so if you can neglect mass of the tau, you also can neglect masses of other leptons. But now you neglected the only quantity that distinguish between leptons. Except mass, all other characteristics (spin, charge, etc) are the same for $\tau$, $\mu$ and $e$. Since there is nothing that differentiate between leptons in the massless limit, all the decay rates and cross-sections have to be equal. For example $$\Gamma(W^+\rightarrow e^+ \nu_e )=\Gamma(W^+\rightarrow \mu^+ \nu_\mu ).$$ And this is what we call the lepton universality.
Best Answer
It is not accidental, at least in the sense that this term is used.
The accidental symmetry is the symmetry that is forced on you by the renormalizability - i.e. for a given field set you can't write the interaction that would violate such symmetry. On the other hand nonrenormalizable interactions or renormalizable interactions with extra fields (these two options are related) may violate symmetry. The example is the baryonic number conservation which is easily violated in many extensions of the standard model.
On the other hand the lepton universality is the property of the non-abelian gauge invariance. In qft the non-abelian gauge interaction is determined solely by the representation of the field and the coupling constant, common for all the fields. I.e. you can't have $SU(2)$ doublet interacting with $W$ boson 1% stronger than another doublet.
Of course this constraint is weaker for $U(1)$ interaction, such as hypercharge field. But if you change somewhat $Y$ for some lepton, this may ruin the gauge invariance of the mass terms and the triangle anomaly cancellation.