Martin & Shaw: Problem 3.5 - Understanding \tau ≈ c

[Daaavde]

From "Martin & Shaw", regarding problem 3.5 (sometimes it uses natural units "c=1"):

"The particle has $\gamma = E/m \approx 10$, hence $\tau \approx c$ (?) and the average distance is $d \approx c \gamma \tau \approx 3 \times 10^{-14}m$ if we assume a lifetime for the particle at rest of $10^{-23}$."

I don't understand why $\tau \approx c$.
Since $v = c \sqrt{1 - \frac{1}{\gamma^2}}, v = 0.994 c$ and I would agree that $v \approx c$, not $\tau$.

 

[Dale]

Hmm, I am with you on that. In natural units where c=1 you get energy and mass in the same units, but not speed and time.

 

[jtbell]

It's probably a typographical error, and it was supposed to say $v \approx c$. That gives his result for the average distance: $d = vt = v \gamma \tau \approx c \gamma \tau$.