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-((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= (161)/(√(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-((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-((v)/(c))^2}}=(139.5)/(√(1-(0.945)^2))=426.5 m`