How Does a Muon's Travel Illustrate Relativity Principles?

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SUMMARY

The discussion centers on the travel of a muon and its implications for the principles of relativity. When moving at 0.99c, a muon has a lifetime of approximately 0.000015602 seconds as observed from Earth's frame, allowing it to travel about 463.32 meters. In contrast, using Newtonian mechanics, it would travel 653.4 meters before decaying. This discrepancy between the distances illustrates the validity of relativity, as the observed lifetime and distance traveled differ significantly from classical predictions.

PREREQUISITES
  • Understanding of special relativity principles
  • Familiarity with muon decay and lifetime measurements
  • Knowledge of the formula tm = ts / sqrt(1 - v^2/c^2)
  • Basic proficiency in scientific notation and unit conversions
NEXT STEPS
  • Study the implications of time dilation in special relativity
  • Learn about muon detection methods in particle physics
  • Explore the concept of relativistic velocity addition
  • Investigate experimental evidence supporting Einstein's theory of relativity
USEFUL FOR

Students of physics, educators teaching relativity concepts, and researchers in particle physics will benefit from this discussion on muon travel and its relation to relativistic effects.

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Homework Statement


A muon has a lifetime of 2.0x10^-6 seconds when at rest, after which time it decays into other particles
a) Ignore any effects of relativity discussed in this section. If the muon was moving at 0.99c, how far would it travel before decaying into other particles, according to Newtonian mechanics?
b) How long would the muon last, according to an observer in Earth's frame of reference?
c)How far would the muon actually travel, when viewed moving at 0.99c?
d) Compare the two distances travelled. Explain why this type of evidence is excellent support for the theory of relativity.

Homework Equations



tm = ts/ sqrt(1-v^2/c^2)

The Attempt at a Solution



a) (0.99)(3x10^8) = v
v = 2.97x10^8

v=dt
d = (2.97x10^8)(2.2x19^-6)
d = 653.4 m

b) tm = ts/ (1-v^2/c^2)
tm = 2.2x10^-6/0.141
tm = 0.000015602 s

c) d = vt
= (2.97x19^-6)(1.56x10^-6)
= 463.32 m

d) ?
 
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What are you calculating in (b) ? The expression is not your relevant formula and the calculation is way off !
Fix that and then c and d will be a lot more doable.
 
EgpYo said:
b) tm = ts/ (1-v^2/c^2)
tm = 2.2x10^-6/0.141
tm = 0.000015602 s

Ok, couple typos there. You forgot the square-root sign. And it might be helpful if you converted your tm into scientific notation.

EgpYo said:
c) d = vt
= (2.97x19^-6)(1.56x10^-6)
= 463.32 m

The (2.97x19^-6) looks like you intended to write (2.97x10^8) but your fingers got away from you.
Check the (1.56x10^-6) .
 

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