Scientists from various institutions have recently discovered that there might be a break in the application of the Standard Model, particularly with a fundamental principle called the lepton universality (lepton universality asserts that the interactions of leptons are the same, regardless of differences in masses and decay rates)

Their discovery comes from reviewing the data from three separate experiments — BaBar, LHCb and Belle — conducted in the United States, France/Switzerland, and Japan. Flavour-changing neutral currents making and breaking the standard model

All three experiments revealed that the ratio of B meson $B^0\to D^{(*)-}\tau^+\nu_\tau$ decays to $B^0\to D^{(*)-}\mu^+\nu_\mu$ decays may be larger than the standard model predicts

  • How to make sure that they’re true?

  • How this would play out?

  • What confirmation of these results will mean (especially for long term)?

  • What might cause this disagreement with the standard model?

  • Why we’re seeing that lepton universality is not satisfied as the Standard Model claims?

  • Do we need an entirely different model of elementary particle physics (to explain the peculiar behavior of the tau particle)?

  • Could a change in one lepton affect the others (as these principles often correlate with one another)?

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    $\begingroup$ If you're talking about the R(D) and R(D*) anomalies, then you should know that it's not tau decays which disagree with the SM. It's the ratio of decays of B mesons to final states containing either a tau or a muon. Otherwise I don't know of any tau decay anomaly from LHCb/BaBar/Belle (I assume those are the experiments you mean, although LHCb is just on the French side of the border). Can you link to a news article or journal paper to confirm what you are asking about? $\endgroup$ – dukwon Jun 23 '17 at 8:22
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    $\begingroup$ @dukwon 2 Accelerators Find Particles That May Break Known Laws of Physics $\endgroup$ – user159858 Jun 23 '17 at 8:37
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    $\begingroup$ Yeah, this is about R(D) and R(D*): the ratio of $B^0\to D^{(*)-}\tau^+\nu$ to $B^0\to D^{(*)-}\mu^+\nu$ $\endgroup$ – dukwon Jun 23 '17 at 8:40

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