1. Apr 7, 2008

### malawi_glenn

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...

2. Apr 7, 2008

### nrqed

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.

3. Apr 7, 2008

### mormonator_rm

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.

4. Apr 8, 2008

### malawi_glenn

Ok i think I understand now, perhaps I have not understand the concept of lifetime proper yet. Thanx!