🗒️ Note: remember we are still working in natural units

📖 Definitions

• Decay rate: $\Gamma$ with units $E$ (It can also be called width since it related to the width of a particle’s resonance, following the figure)

  

  This comes from the time energy uncertainty $\Gamma=2\Delta E=\frac{\hbar}{\tau}$ (from $\Delta E\,\tau\gtrsim \frac{\hbar}{2}$)

• Lifetime: $\tau=1/\Gamma$ with units $E^{-1}$

💃 Example:

• Full width $\Gamma$ is the main decay rate

• We have partial decay rates in red

  $$ {\Gamma(x\to Y) = \Gamma_\text{tot} (x) \cdot B(X\to Y)} $$

• Blue is a showcase of lepton universality

Phase space

The picture above is incomplete because the results are for a “flat” infinite phase space, in practice the phase space will impact the decay rate $\Gamma$

• We define the matrix element $\mathcal M$

  $$ \mathcal M(q^2)=\frac{g^2}{q^2-M^2_Z} $$

  where $g\propto \alpha_\text{w}$ is coupling strength of the weak force, $q$ is the four momentum transfer between the interacting particles and $M_Z$ is the mass of the $Z$ exchange boson. The units of $\mathcal M$ are $E^{-2}$

• We can then define our decay rate using it as

  $$ \Gamma = K|\mathcal M |^2 $$

  • where $K$ has units of $E^5$ to compensate for the $|\mathcal M|^2$

  If we now look at the only value remaining in the decay its the lepton $\ell$ mass thus we can say

  $$ K\propto m_\ell ^5 $$

💼 Case: lets now look the observable ${\tau_\tau}/{\tau_\mu}$ for a tau decay against a mu decay

• Looking first at $\tau_\tau$ we write

  $$ \tau_{\tau}=\frac{1}{\Gamma_\text{tot}(\tau)}=\frac{B(\tau ^-\to e^- \overline \nu_e \nu_\tau) }{\Gamma(\tau ^- \to e^- \overline \nu_e \nu_\tau)}\propto \frac{B(\tau ^- \to e^- \overline \nu e \nu\tau)}{m^5_\tau} $$

  where we used $\Gamma x= \Gamma\text{tot}/B_x$ and $B$ is the branching fraction

• Then at $\tau_\mu$ in a similar manner we write

  $$ \tau_\mu = \frac{1}{\Gamma \text{tot} (\mu)} =\frac{B(\mu^- \to e^- \overline \nu_e \nu\mu)}{\Gamma(\mu^- \to e^- \overline \nu_e \nu_\mu)}\propto \frac{B(\mu^- \to e^- \overline \nu_e \nu_\mu)}{m^5_\mu} $$

• Now we put it together to get

  $$ \frac{\tau_\tau}{\tau\mu} =\frac{B(\tau^- \to e^- \overline \nu_e \nu_\tau)}{B(\mu^- \to e^- \overline \nu_e \nu_\mu)}\cdot \frac{m_\mu^5}{m_\tau^5} $$

which we can calculate to give about $1.328\cdot 10^{-7}$ which is close to the experimental results