Observed lifetime of muons (relativity)

AI Thread Summary
The discussion revolves around calculating the observed lifetime of muons and the number remaining after traveling a distance of 3.0 km. The mean lifetime of muons is 2.2 microseconds, and when moving at 0.95c, the observed lifetime is calculated to be approximately 7.04 microseconds. The user initially miscalculated the remaining number of muons due to a mistake in entering the lifetime value into their calculator. After correcting this error, they found the answer aligned with the expected result of about 1.13 X 10^4 muons remaining. The problem was ultimately resolved, confirming the calculations for both the observed lifetime and remaining muons.
Ravenatic20
Messages
30
Reaction score
0

Homework Statement


The muon is an unstable particle that spontaneously decays into an electron and two neutrinos. If the number of muons at t = 0 is N_{o}, the number at time t is given by N = N_{0}e^{-t/\tau}, where \tau is the mean lifetime, equal to 2.2 \mu s. Suppose the muons move at a speed of 0.95c and there are 5.0 X 10^{4} muons at t = 0. (a) What is the observed lifetime of the muons? (b) How many muons remain after traveling a distance of 3.0 km?

Homework Equations


N = N_{0}e^{-t/\tau}
t = d/v
\tau ' = \gamma \tau
\gamma = 1/ \sqrt{1 - v^{2}/c^{2}}

The Attempt at a Solution


t = d/v => 3 km/0.95c = 1.05 X 10^{-5} where c is 3 X 10^{8}

In the Muon frame: \tau = 2.2\mu
In the Earth frame: \tau ' = \gamma \tau
\gamma = 1/ \sqrt{1 - v^{2}/c^{2}} => 1/ \sqrt{1 - (0.95c/c)^{2}}
\gamma = 3.2 \mu s
Plug this back into the equation for \tau '...
\tau ' = 7.04 \mu s
In the back of my book, the solution for part (a) is 7.1 \mu s. This answer (\tau ' = 7.04 \mu s) doesn't have anything to do with the answer to (a), does it?

Next, I need to plug in the numbers I have into N, which is N = N_{0}e^{-t/\tau}
If I read the problem right, N_{o} = 5.0 X 10^{4}. We know t, and we know the new value of tau. So it's plug and chug, right?
If so, this is what I got:
N = N_{0}e^{-t/\tau}
N = 5.0 X 10^{4}e^{-1.05 X 10^{-5}/7.04 \mu s}
N = 4.9 X 10^{-4}
This is not the same answer in the back of the book, nor does it make sense. The answer in the back of the book is 1.1 X 10^{-4}

Where am I going wrong? Please put me on track, thanks!
 
Physics news on Phys.org
I apologize for the shameless 'thread bumping', but I'm really stuck on this problem and could use some help. It's due in about 12 hours so that's why I'm rushing now. I don't like turning in HW I know is wrong.
 
I think your book answers have some problems. Sure, the gamma factor is about 3.2 (no units, it's dimensionless). So the observed lifetime is about 7.04 microseconds. Why the book says 7.1, I don't know. Now traveling for 3km at 0.95c takes about 10.5 microseconds. There should be a decrease in the number muons by a modest factor. Certainly not from 5*10^4 to 1.1*10^(-4). No, it doesn't make much sense.
 
Dick, thanks for catching that gamma is dimensionless. If anyone has any last minute ideas please post them. Thank you.
 
I found out what I was doing wrong. It's really stupid. When solving for N, my value for tau(7.04 microseconds), I didn't punch it into my calculator right. Was entering 7.04 instead of 7.04X10^-6. The answer I got then was right, about 1.13X10^4.

Mark this problem as solved. Thanks anyways guys.
 
Thread 'Help with Time-Independent Perturbation Theory "Good" States Proof'
(Disclaimer: this is not a HW question. I am self-studying, and this felt like the type of question I've seen in this forum. If there is somewhere better for me to share this doubt, please let me know and I'll transfer it right away.) I am currently reviewing Chapter 7 of Introduction to QM by Griffiths. I have been stuck for an hour or so trying to understand the last paragraph of this proof (pls check the attached file). It claims that we can express Ψ_{γ}(0) as a linear combination of...
Back
Top