Relativity of Length: Understanding Length Contraction & Time Dilation

[Fernando Rios]

Homework Statement

An unstable particle is created in the upper atmosphere from
a cosmic ray and travels straight down toward the surface of the
earth with a speed of relative to the earth. A scientist at
rest on the earth’s surface measures that the particle is created at an
altitude of a) As measured by the scientist, how much
time does it take the particle to travel the to the surface of
the earth? b) Use the length contraction formula to calculate the distance
from where the particle is created to the surface of the earth as
measured in the particle’s frame. c) In the particle’s frame, how
much time does it take the particle to travel from where it is created
to the surface of the earth? Calculate this time both by the time dilation
formula and also from the distance calculated in part (b). Do
the two results agree?

Relevant Equations

delta_t = delta_t_0/sqrt(1-v^2/c^2)

delta_L = L_0*sqrt(1-v^2/c^2)

a) We use the definition of speed:
v = delta_L/delta_t

delta_t = delta_L/v = 45000 m/(0.99540*3*10^8 m/s) = 1.55*10^-4 s

b) We use the length contraction equation:
delta_L = L_0*sqrt(1-v^2/c^2)

L_0 = delta_L/sqrt(1-v^2/c^2) = 45000 m/sqrt(1-0.99540^2) = 469698 m

However, the solution shows the following:
L_0 = delta_L*sqrt(1-v^2/c^2) = 45000 m*sqrt(1-0.99540^2) = 4310 m

Shouldn't the length in the particle's frame be greater?

c) We use the time dilation equation:
delta_t = delta_t_0/sqrt(1-v^2/c^2)

delta_t_0 = delta_t*sqrt(1-v^2/c^2) = (1.51*10^-4 s)*sqrt(1-0.99540^2) = 1.44*10^-5 s

We use the definition of velocity:
v = L_0/delta_t_0

delta_t_0 = L_0/v = 469698 m/(0.99540*3*10^8 m/s) = 1.51*10^-4 s

However, the solution shows the following:
delta_t_0 = L_0/v = 4310 m/(0.99540*3*10^8 m/s) = 1.44*10^-4 s

What am I doing wrong?

 

[anuttarasammyak]

a) We use the definition of speed:
v = delta_L/delta_t

$$t= L_0 / v$$
because distance is still at the Earth. I read $L_0=45,000 \ m$

b) We use the length contraction equation:
delta_L = L_0*sqrt(1-v^2/c^2)

L_0 = delta_L/sqrt(1-v^2/c^2) = 45000 m/sqrt(1-0.99540^2) = 469698 m

L is distance measured in particle frame of reference
$$L=L_0 \sqrt{1-\frac{v^2}{c^2}}=\frac{L_0}{ \gamma}$$
I read $v=0.9954c,\ \gamma=10.44$

I will leave the followings to you.

 

[Orodruin]

Homework Statement:: An unstable particle is created in the upper atmosphere from
a cosmic ray and travels straight down toward the surface of the
earth with a speed of relative to the earth. A scientist at
rest on the earth’s surface measures that the particle is created at an
altitude of a) As measured by the scientist, how much
time does it take the particle to travel the to the surface of
the earth? b) Use the length contraction formula to calculate the distance
from where the particle is created to the surface of the Earth as
measured in the particle’s frame. c) In the particle’s frame, how
much time does it take the particle to travel from where it is created
to the surface of the earth? Calculate this time both by the time dilation
formula and also from the distance calculated in part (b). Do
the two results agree?
Relevant Equations:: delta_t = delta_t_0/sqrt(1-v^2/c^2)

delta_L = L_0*sqrt(1-v^2/c^2)

Shouldn't the length in the particle's frame be greater?

No. The rest frame of the distance is the Earth’s rest frame. It is therefore contracted in the particle’s rest frame.

Also note that part (c) of the statement is badly formulated.
”In the particle’s frame, how much time does it take the particle to travel”
The particle is at rest in its rest frame and therefore does not travel in that frame by definition. In that frame, it is the Earth’s surface that travels.

 

[Fernando Rios]

Thank you for your answer. It really helped me.