The pion in flight

Problem

Summary: pion in flight. A charged pion has proper mean lifetime $\tau_0=26$ ns and travels at $v=0{,}98\,c$ in the laboratory. What average distance does it cover before decaying? Compare with the “Galilean” distance $v\tau_0$ and comment on the role of $\gamma$.

Solution

Lorentz factor. With $\beta = 0{,}98$:

$\gamma = \frac{1}{\sqrt{1-0{,}98^2}} = \frac{1}{\sqrt{1-0{,}9604}} = \frac{1}{\sqrt{0{,}0396}} \approx 5{,}03.$

“Galilean” distance (without relativity). If the pion only lived $\tau_0$ in the laboratory:

$d_{\text{cl}} = v\,\tau_0 = (0{,}98)(3\cdot 10^8)(26\cdot 10^{-9}) \approx 7{,}6\ \text{m}.$

Relativistic distance. In the laboratory the mean lifetime appears dilated by $\gamma$, so the pion travels

$d = v\,(\gamma\tau_0) = \gamma\,d_{\text{cl}} \approx 5{,}03 \times 7{,}6\ \text{m} \approx 38{,}5\ \text{m}.$

$\ev{d \approx 38{,}5\ \text{m} \quad (\text{versus } v\tau_0 \approx 7{,}6\ \text{m})}$

Role of $\gamma$. The distance travelled is amplified by exactly the factor $\gamma \approx 5$: time dilation stretches the pion’s lifetime in the laboratory fivefold, letting it travel much further than classical mechanics would predict. This is the same mechanism that explains cosmic-ray muons.

Links

Topics: Special relativity Concepts: Time dilation Skills: Changing reference frame