Understanding Special Relativity: Solving for Proper Distance and Lifetime

[jdub99]

Homework Statement

An unstable high-energy particle is created in the laboratory, and it moves at a speed of 0.978c. Relative to a stationary reference frame fixed to the laboratory, the particle travels a distance of 1.38x10^-3 m before disintegrating.

(a) What is the proper distance traveled?
(b) What is the distance measured by a hypothetical person traveling with the particle?
(c) What is the proper lifetime?
(d) What is the dilated lifetime?

Homework Equations

L₀/(gamma)=L

(gamma)=1/sqrt(1-(v/c)²)

The Attempt at a Solution

Stuck on part A:
V=.978c
L= 1.38x10^-3m
L₀=?

(gamma)=1/sqrt(1-(v/c)²)= 1/sqrt(1-(.978)²)= 4.79


L₀=L x (gamma)= (1.38x10^-3m)(4.79)= .00662m

Did something wrong and cannot figure it out. Please help!

 

[Doc Al]

L₀=L x (gamma)= (1.38x10^-3m)(4.79)= .00662m

Did something wrong and cannot figure it out. Please help!

According to the particle frame, is the distance traveled shorter or longer than seen in the lab frame?

 

[jdub99]

Distance should be greater for L₀ because (gamma) is going to always be greater than 1. Right? It should be shorter in the lab frame compared to particle frame.

 

[Doc Al]

Distance should be greater for L₀ because (gamma) is going to always be greater than 1. Right? It should be shorter in the lab frame compared to particle frame.

Realize that the distance is measured at rest in the lab frame, so I suppose that's the distance they want as the "proper distance". From the view of the particle frame, that distance is moving. What happens to moving lengths? (That's really part b, not part a. Oops!)

 

[jdub99]

Lengths contract when moving. so the particle should be the shorter distance. So part a is the given part of the question. Ok parts a and b make