Relativity : Muons traveling Qns

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SUMMARY

The discussion centers on the behavior of muons moving at a speed of 0.9994c in a circular storage ring with a radius of 500 meters. Calculations reveal that muons are expected to complete approximately 6.06 trips around the ring before decaying, while in their own frame, they perceive only about 0.21 trips due to length contraction. The analysis confirms that there are no contradictions in the results, as time dilation allows muons to travel further than they perceive. The key equations utilized include Lorentz contraction (L = L_0/γ) and time dilation (ΔT = γΔT_0).

PREREQUISITES
  • Understanding of special relativity concepts, including time dilation and length contraction.
  • Familiarity with Lorentz transformations and the Lorentz factor (γ).
  • Basic knowledge of circular motion and its equations.
  • Ability to perform calculations involving relativistic speeds (e.g., 0.9994c).
NEXT STEPS
  • Study the derivation and applications of the Lorentz factor (γ) in special relativity.
  • Explore the implications of time dilation and length contraction in high-speed particle physics.
  • Learn about experimental evidence supporting the predictions of special relativity, such as muon decay rates.
  • Investigate the mathematical modeling of circular motion in relativistic contexts.
USEFUL FOR

Students of physics, particularly those studying special relativity, as well as educators and anyone interested in the behavior of high-speed particles like muons in circular motion.

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Homework Statement


Muons move in circular orbits at a speed of 0.9994c in a storage ring which has a radius of 500m. How many trips around the storage ring do we expect the muons to make before they decay? How many trips around the storage ring do the muons think they have made? Are there any contradictions?
[ The question didn't give the by what time will muons decay, so i did a little search and took 2.2*10^-6 sec as the time before the decay.)

Homework Equations


[tex]L = L_0/\gamma[/tex]
[tex]\Delta T = \gamma \Delta T_0[/tex]

The Attempt at a Solution



(i) v = 0.9994c, r=500m, C=2*pi*500=1000*pi, [tex]\Delta T_0 = 2.2*10^-6[/tex]

Therefore using [tex]\Delta T = \gamma \Delta T_0[/tex],
[tex]\Delta T = 6.35*10^-5[/tex]
Distance moved = (0.9994)(3.0*10^8)(6.35*10^-5)=19,038.57m
Trips around storage ring = 19,038.57/1000pi = 6.06 times.

(ii)
Distance moved in the frame of muons = 0.9994c*2.2*10^-6 = 659.604
Trips around the ring as in frame of muons = 659.604/1000pi = 0.20996

(iii)
There are not contradiction since in the frame of muon, the ring or the distance traveled by it has been length contracted. Therefore, the muon actually traveled more than it seem to have traveled.

In the frame of an observer however the muon although should decay in 2.2micro seconds didn't due to time dilation. Thus with this apparent increase in time, the muon can cover more distance and hence have more trips around the rings.
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Can anyone check my answer for part i ii and iii. I believe i got i and ii correct. But most importantly is my explanation in iii in anyway wrong or unclear or lacking?

Any help will be greatly appreciated
 
Last edited:
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Ermm...any help?
 

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