Understanding Relativistic Muon Decay and Time Dilation

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SUMMARY

This discussion focuses on the relativistic muon decay and the implications of time dilation as described by the equations of special relativity. The lifetime of a muon is 2.197 microseconds, with a rest mass of 105.65 MeV and a total energy of 10 GeV. The user seeks clarity on the relationship between the moving time (deltaT') and the stationary time (deltaT), particularly how to calculate the distance traveled by the muon in the Earth's reference frame. The key equations involved include beta, gamma, and the transformation of time between reference frames.

PREREQUISITES
  • Understanding of special relativity concepts, including time dilation and length contraction.
  • Familiarity with the equations beta=v/c and gamma=1/sqrt(1-beta^2).
  • Knowledge of particle physics, specifically muon properties and decay processes.
  • Basic mathematical skills for manipulating equations and performing calculations.
NEXT STEPS
  • Study the derivation and implications of the Lorentz transformation equations.
  • Learn how to apply the concept of gamma in various relativistic scenarios.
  • Explore the relationship between energy, mass, and velocity in particle physics.
  • Investigate experimental evidence of time dilation effects in high-energy particle physics.
USEFUL FOR

This discussion is beneficial for students in nuclear physics, physicists interested in particle behavior at relativistic speeds, and educators teaching concepts of special relativity and time dilation.

Spurious J
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Given the lifetime of a muon as 2.197 microseconds, and the rest mass of 105.65MeV, and a total particle energy of 10GeV, I need to calculate how far, in the rest frame, the particle will travel before decay.

Homework Equations


Beta=v/c
Gamma=1/sqrt(1-beta^2)
deltaT'=gamma*deltaT
L'=L/gammaAttempts/Understanding Thus Far

The equations aren't complicated but I can't quite make them mesh with my understanding. If deltaT' is the moving time, dividing it by gamma to get the stationary period will always yield a shorter time amount of time passing for the stationary period; isn't this the opposite of what I want?

And, in order to figure out the total distance traveled, as measured in the rest frame, is it enough to multiply v*t'=L', then multiply by gamma to get the rest frame distance? Do I need to convert the time as well? Or would this lead to too many factors of gamma?
 
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deltaT is the proper time, the time that an event takes in the same reference frame where the event takes place. Here, deltaT would be 2.197 microseconds, since that's how long the muon thinks it lives.

After getting deltaT' and figuring out v, distance is just v*deltaT'. (In fact, distance is always v*deltaT' in any reference frame as long as v and deltaT' are both measured in said frame.)
 
But if deltaT' is the time frame of the muon, I must divide it by gamma to get the amount of time it'll take in the Earth's reference frame, right?

But that'll yield a smaller number. So, the particle would seem to decay quicker, and thus it's clock would be running fast according to the earth, right? I haven't taken Relativity yet, this is for a nuclear physics class, so I'm just running off of my own research over the years. It seems like the equation makes the wrong clock slow down, and I can't see (though I'm sure I am wrong) where I'm wrong in my understanding.

If deltaT' is 2.19 microseconds, then deltaT is .003 microseconds (gamma=70.66); but shouldn't it be if deltaT' passes 2.19 microseconds at .9999c, then the other observer measures the same event as lasting 154 microseconds?
 

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