Riding With a Particle: Measuring Its Lifetime

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SUMMARY

The discussion focuses on calculating the lifetime of a particle moving at 0.996c, as measured by a stationary observer and a hypothetical observer riding with the particle. The stationary observer measures the particle's lifetime as 3.00 x 10-8 seconds. The incorrect attempt at calculating the proper lifetime using the formula t = t(0)/(1 - v2/c2)1/2 resulted in an answer of 3 x 10-9 seconds. The suggestion made was to carry more significant figures in the calculations to achieve a more accurate result.

PREREQUISITES
  • Understanding of special relativity concepts, particularly time dilation
  • Familiarity with the formula for time dilation: t = t(0)/(1 - v2/c2)1/2
  • Knowledge of the speed of light (c) and its significance in relativistic physics
  • Ability to perform calculations involving significant figures
NEXT STEPS
  • Review the principles of time dilation in special relativity
  • Practice calculations using the time dilation formula with varying speeds
  • Explore the concept of proper time and how it differs from coordinate time
  • Learn about significant figures and their importance in scientific calculations
USEFUL FOR

Students and educators in physics, particularly those studying special relativity, as well as anyone interested in understanding particle physics and time measurement in high-speed scenarios.

cshelbythec
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A particle lives for a short time before breaking apart into other particles. Suppose it is moving at a speed of 0.996c, and an observer who is stationary in a laboratory measures the particle's lifetime to be 3.00 10-8 s.
What is the lifetime according to a hypothetical person who is riding along with the particle?




The Attempt at a Solution


I tried
t=t(0)/(1-v2/c2)^1/2
where t=3.00 10-8s and I'm finding t(0) proper.
my answer is 3 10-9 which is wrong. what am I doing wrong?
 
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cshelbythec said:
A particle lives for a short time before breaking apart into other particles. Suppose it is moving at a speed of 0.996c, and an observer who is stationary in a laboratory measures the particle's lifetime to be 3.00 10-8 s.
What is the lifetime according to a hypothetical person who is riding along with the particle?




The Attempt at a Solution


I tried
t=t(0)/(1-v2/c2)^1/2
where t=3.00 10-8s and I'm finding t(0) proper.
my answer is 3 10-9 which is wrong. what am I doing wrong?
i have forgotten this stuff, but try multiplying it if dividing doesn't work
 
cshelbythec said:
my answer is 3 10-9 which is wrong. what am I doing wrong?
I suggest carrying more significant figures in your answer.
 

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