How does relativity affect the detection of atmospheric muons?

[L_ucifer]

TL;DR Summary: I got this question on a quiz for a Coursera course on special relativity, and I'm confused about the answer. I've detailed my thinking below any help on where I went wrong would be greatly appreciated.

Question.
Muons are unstable fundamental particles. In its own rest frame, the average lifetime of a muon is t0=2×10^−6s. Thus, if initially we have No muons, after time t, the number of remaining muons is N=No e^−t/t0.

Due to collision between cosmic rays and the earth's atmosphere, No=300,000 muons are created per square meter per minute. Suppose all the muons are created at a height of 10km from the earth surface, moving towards the center of the earth at v=0.98c and can travel through the atmosphere without interactions.

Let's now compare the predictions of Newtonian mechanics and relativity. In Newtonian mechanics, we would on average detect N_Newtonian muons per square meter per minute on the earth surface. In reality, relativity is important and we can detect N muons per square meter per minute.

a. N_Newtonian=1, N=0.01
b. N_Newtonian=1, N=10,000
c. N_Newtonian=0.01, N=10,000
d. N_Newtonian=1, N=1
e. N_Newtonian=0.01, N=1
f. N_Newtonian=0.01, N=0.01Answer.
N_Newtonian is fairly easy to find. We just find the time taken for the muons to reach the surface (distance / speed ) and then put this time in the decay equation. The answer we get is 0.01.

For N, the correct logic is you need to find the relative time of a muon's lifetime. So the equation is N=No e^−t/(t0 * gamma) and t is the same. This means the correct answer is c.
My logic for N is instead of finding the relative time of a muon's lifespan, we find the relative time for the muon to reach the surface. So t = distance/speed * gamma. I'm not sure why this is wrong and why we find the relative time for a muon's lifetime instead?
EDIT:
Here's the overall equation for the correct solution:
No = 300,000
t0 = 2 * 10^-6 s
t = distance / speed = 10,000 / 0.98c
N = No * e ^ -t/(t0 * gamma)

Here's the overall equation for my solution (which is wrong):
No = 300,000
t0 = 2 * 10^-6 s
t = distance / speed = 10,000 / 0.98c
N = No * e^-(t * gamma)/t0.

 

[PeroK]

My logic for N is instead of finding the relative time of a muon's lifespan, we find the relative time for the muon to reach the surface. So t = distance/speed * gamma. I'm not sure why this is wrong and why we find the relative time for a muon's lifetime instead?

Why didn't this give the same numerical answer?

 

[L_ucifer]

Why didn't this give the same numerical answer?

Edited the question and adding the same changes here:
Here's the overall equation for the correct solution:
No = 300,000
t0 = 2 * 10^-6 s
t = distance / speed = 10,000 / 0.98c
N = No * e ^ -t/(t0 * gamma)

Here's the overall equation for my solution (which is wrong):
No = 300,000
t0 = 2 * 10^-6 s
t = distance / speed = 10,000 / 0.98c
N = No * e^-(t * gamma)/t0.

 

[Ibix]

What's your reasoning for putting the $\gamma$ where you did in the last equation?

 

[PeroK]

Edited the question and adding the same changes here:
Here's the overall equation for the correct solution:
No = 300,000
t0 = 2 * 10^-6 s
t = distance / speed = 10,000 / 0.98c
N = No * e ^ -t/(t0 * gamma)

The logic here is that in the ground frame the muon's lifetime is dilated. If we use primes for measurements in the ground frame, then:
$$N = N_0 e^{-\frac{t'}{t'_0}} = N_0 e^{-\frac{t'}{\gamma t_0}} = N_0 e^{-\frac{d'}{v\gamma t_0}}$$

Here's the overall equation for my solution (which is wrong):
No = 300,000
t0 = 2 * 10^-6 s
t = distance / speed = 10,000 / 0.98c
N = No * e^-(t * gamma)/t0.

This makes no sense to me. If we use the muon frame, then we have:
$$N = N_0 e^{-\frac{t}{t_0}} = N_0 e^{-\frac{d}{v\gamma t_0}} = N_0 e^{-\frac{d'}{v\gamma t_0}}$$Since the distance to the ground in the muon frame is $d = \dfrac{d'}{\gamma}$, where $d'$ is the distance to the ground in the Earth frame.

 

[L_ucifer]

What's your reasoning for putting the $\gamma$ where you did in the last equation?

I put gamma there because I thought it'd take more time for the muons to reach the ground, instead of thinking of the muon's lifetime becoming dilated.

 

[PeroK]

I put gamma there because I thought it'd take more time for the muons to reach the ground, instead of thinking of the muon's lifetime becoming dilated.

You mean the faster the muon moves, the more time it takes to reach the ground?

 

[L_ucifer]

You mean the faster the muon moves, the more time it takes to reach the ground?

No I mean I thought that since time is slowed for the muons when looking at them from a ground frame, the muons would actually take longer to reach the Earth's surface. This is why in my solution the time for a muon to reach the Earth's surface is greater in the relative solution than in the Newtonian solution. In mathematical terms, my understanding was: t_Newtonian =10000/0.98c < t_relative = t_Newtonian * gamma

EDIT: grammar

 

[PeroK]

No I mean I thought that since time is slowed for the muons when looking at them from a ground frame, the muons would actually take longer to reach the Earth's surface.

Quite the opposite! Do the maths, as they say. In the ground frame, a muon travels a distance $d$ at speed $v$. How much time passes for the muon?

 

[kuruman]

No I mean I thought that since time is slowed for the muons when looking at them from a ground frame, the muons would actually take longer to reach the Earth's surface.

Longer than what? Don't forget that in the muon frame the distance to the Earth's surface is shorter than $d$. The muon may live longer in its own frame but it doesn't have to travel that far.

 

[PeroK]

Longer than what? Don't forget that in the muon frame the distance to the Earth's surface is shorter than $d$. The muon may live longer in its own frame but it doesn't have to travel that far.

I don't follow this at all. The muon's lifetime ($t_0$) is defined in its rest frame. In the ground frame, it has a longer lifetime than this - of $\gamma t_0$.

 

[kuruman]

I guess what I wanted to say didn't come out right. It's better with equations.
If one calculates $v$ in the muon's frame, the distance is $\frac{d}{\gamma}$ and the time is $t_0$ so $v=(\frac{d}{\gamma})\frac{1}{t_0}$.
If one calculates $v$ in the Earth's frame, the distance is $d$ and the time is $\gamma t_0$ so $v=d\frac{1}{\gamma t_0}$.
I fixed the offending statement in the earlier post.