Homework Helper

## Main Question or Discussion Point

Hi!

In my book Particle Physics by Martin & Shaw, eq 2.15:

$$\tau _l = \dfrac{B(l^- \rightarrow e^-\bar{\nu }_e\nu _l )}{\Gamma (l^-\rightarrow e^-\bar{\nu }_e\nu _l )}$$

Where B is branching ratio and Gamma the decay rate and l is a lepton.

Why is B included? :S I thought the lifetime just was the inverse of the decay rate...

Related High Energy, Nuclear, Particle Physics News on Phys.org
nrqed
Homework Helper
Gold Member
Hi!

In my book Particle Physics by Martin & Shaw, eq 2.15:

$$\tau _l = \dfrac{B(l^- \rightarrow e^-\bar{\nu }_e\nu _l )}{\Gamma (l^-\rightarrow e^-\bar{\nu }_e\nu _l )}$$

Where B is branching ratio and Gamma the decay rate and l is a lepton.

Why is B included? :S I thought the lifetime just was the inverse of the decay rate...
Because this decay rate is not the total decay rate, it's only the decay rat efor that particular mode. The lifetime is the inverse of the total decay rate.

Hi!

In my book Particle Physics by Martin & Shaw, eq 2.15:

$$\tau _l = \dfrac{B(l^- \rightarrow e^-\bar{\nu }_e\nu _l )}{\Gamma (l^-\rightarrow e^-\bar{\nu }_e\nu _l )}$$

Where B is branching ratio and Gamma the decay rate and l is a lepton.

Why is B included? :S I thought the lifetime just was the inverse of the decay rate...
Well, assuming that;

$$\Gamma (l^-\rightarrow e^-\bar{\nu }_e\nu _l ) < \Gamma (l^-\rightarrow anything)$$

you must compensate for the longer lifetime that would occur if you restricted the lepton to that one decay mode. Compensating will require you to consider the ratio of this decay mode versus any available decay mode, which is the branching ratio in the posted equation. Thus, dividing the branching ratio, which is less than or equal to one, by the partial width, which is less than or equal to the total width, is the logical solution.