Question

An elementary particle produced in a laboratory experiment travels $\mathbf{0}\mathbf{.}\mathbf{230}\mathbf{}\mathit{m}\mathit{m}$through the lab at a relative speed of $\mathbf{0}\mathbf{.}\mathbf{960}\mathit{c}$before it decays (becomes another particle). (a) What is the proper lifetime of the particle? (b) What is the distance the particle travels as measured from its rest frame?

Short answer

1. The proper lifetime is the time in a particle’s rest frame is $0.22ps$.
2. The distance traveled measured from its rest frame is $0.820mm$.

Step-by-step solution

Time Dilation and Length contraction:

Suppose two consecutive events occur at the same place; the time interval measured in the same inertial frame of reference is called proper time. And the time interval measured in any other reference frame relative to that frame will be longer than the correct time.

The length of an object measured in the object's inertial frame of reference is called the proper length. When the length of this object is measured in any other inertial frame, the moving relative velocity is always shorter than the correct length. This is known as length contraction.

(a) Define the proper lifetime of the particle:

The formula given below is used to determine the time interval in another frame.

$∆t=\frac{∆t_{\circ }}{\sqrt{1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}}$

Here, $\Delta t_o$is the proper time, and $\Delta t$is the interval measured by an observer moving with a relative speed $u$.

Here, in the lab frame, the time taken until decay is

$$\begin{gathered} ∆t=\frac{0.230\times {10}^{-3}m}{0.960\times 3\times {10}^{8}m/s} \\ =0.08\times {10}^{-11}s \\ =0.8ps \end{gathered}$$

Inserting this value in time dilation equation;

$$\begin{gathered} ∆t=\frac{∆t_{\circ }}{\sqrt{1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}} \\ ∆t_{\circ }=∆t\left(\sqrt{1-\raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}\right) \\ =\left(0.8\times {10}^{-12}s\right)\left(\sqrt{1-{\left(0.96\right)}^2}\right) \\ =0.22ps \end{gathered}$$

(b) Determine the distance the particle travels as measured from its rest frame:

Write the equation for the length contraction as below.

$L=L_o\sqrt{1-{\beta }^2}$

$=\frac{L_o}{\gamma }$

The distance traveled measured from its rest frame is the proper length, therefore it can be calculated as

$L_o=\gamma L$

localid="1663057953120" $$\begin{gathered} =\frac{0.230\times {10}^{-3}m}{\sqrt{1-{\left(0.96\right)}^2}} \\ =0.820mm \end{gathered}$$