Help me understand Lepton universality

I understand that we have three pairs of Leptons (generations).

$(\nu_e , e^-), (\nu_{\mu}, \mu), (\nu_{\tau}, \tau^-)$

But what is this principle of Lepton universality?

I have obviously tried googling it before asking in here, but all of the sites tend to just quickly run over it and I am not quite sure I understand exactly what it covers. So can someone please help me understand it or maybe give me a good source.

• Lepton universality means all leptons behave the same, except when they don't. – ACuriousMind Jan 14 '16 at 16:36
• A few more details perhabs? – Nillo Jan 14 '16 at 16:38
• from wikipedia: "The coupling of the leptons to gauge bosons are flavour-independent (i.e., the interactions between leptons and gauge bosons are the same for all leptons)". In other words: the all leptons couple to the electroweak field in the same manner. – AccidentalFourierTransform Jan 14 '16 at 16:43

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.
• for relative branching fractions such as $\frac{\Gamma[ \, \pi^- \rightarrow \, e^- \overline\nu \, ]}{\Gamma[ \, \pi^- \rightarrow \, \mu^- \overline\nu \, ]}$ and $\frac{\Gamma[ \, \tau^- \rightarrow \, \pi^- \nu \, ]}{\Gamma[ \, \pi^- \rightarrow \, \mu^- \overline\nu \, ]}$, or