Question

The proper lifetime of a certain particle is 161.0 ns (a) How long does it live in the laboratory if it moves at v- 0.945c? ns (b) How far does it travel in the laboratory during that time? (c) What is the distance traveled in the laboratory according to an observer moving with the particle?

Answer 1

(a) 492.2 ns

The lifetime of the particle in the frame of reference of the laboratory is given by

$T= \frac{T_0}{\sqrt{1-(\frac{v}{c})^2}}$

where

$T_0$ is the proper lifetime

v is the speed of the particle

c is the speed of the light

For this particle we have:

$T_0 = 161 ns$

$v=0.945c$

Substituting into the equation, we get:

$T= \frac{161}{\sqrt{1-(0.945)^2}}=492.2 ns$

(b) 139.5 m

In the reference frame of the laboratory, the distance travelled by the particle is given by

$L = v T$

where

v is the speed of the particle

T is the lifetime of the particle in the laboratory frame of reference (found in part a)

The speed of the particle is

$v=0.945c=0.945(3\cdot 10^8)=2.84\cdot 10^8 m/s$

The lifetime of the particle in the laboratory's frame of reference is

$T=492.2 ns = 492.2\cdot 10^{-9}s$

And substituting into the equation, we find:

$L=(2.84\cdot 10^8)(492.2\cdot 10^{-9})=139.5 m$

(c) 426.5 m

Here we have to calculate the distance travelled by the particle in its frame of reference. This can be calculated by using the equation

$L=L_0 \sqrt{1-(\frac{v}{c})^2}$

where

$L_0$ is the distance travelled measured by an observer moving with the particle

L is the distance travelled measured by an observer in the laboratory

We already know that

L = 139.5 m

So, solving the formula for $L_0$,

$L_0 = \frac{L}{\sqrt{1-(\frac{v}{c})^2}}=\frac{139.5}{\sqrt{1-(0.945)^2}}=426.5 m$