- #1
Daaavde
- 30
- 0
From "Martin & Shaw", regarding problem 3.5 (sometimes it uses natural units "c=1"):
"The particle has [itex]\gamma = E/m \approx 10[/itex], hence [itex]\tau \approx c[/itex] (?) and the average distance is [itex]d \approx c \gamma \tau \approx 3 \times 10^{-14}m[/itex] if we assume a lifetime for the particle at rest of [itex]10^{-23}[/itex]."
I don't understand why [itex]\tau \approx c[/itex].
Since [itex]v = c \sqrt{1 - \frac{1}{\gamma^2}}, v = 0.994 c[/itex] and I would agree that [itex]v \approx c[/itex], not [itex]\tau[/itex].
"The particle has [itex]\gamma = E/m \approx 10[/itex], hence [itex]\tau \approx c[/itex] (?) and the average distance is [itex]d \approx c \gamma \tau \approx 3 \times 10^{-14}m[/itex] if we assume a lifetime for the particle at rest of [itex]10^{-23}[/itex]."
I don't understand why [itex]\tau \approx c[/itex].
Since [itex]v = c \sqrt{1 - \frac{1}{\gamma^2}}, v = 0.994 c[/itex] and I would agree that [itex]v \approx c[/itex], not [itex]\tau[/itex].