Special Relativity and muon decay

In summary, the conversation is discussing the likelihood of a muon, created in Earth's upper atmosphere, surviving its trip to the Earth's surface before decaying. This is affected by time dilation, which is represented by the formula T=\frac{T_{0}}{\sqrt{1-(v^{2}/c^{2})}} and the probability equation P = 1 - e^{-Δt/τ}. The conversation also mentions the four different times involved in the problem and how they can be calculated using the time-dilation formula.
  • #1
astr0
17
0
The proper mean lifetime of a muon is 2.20 µs, which is denoted as τ. Consider a muon, created in Earth's upper atmosphere, speeding toward the surface 8.00 km below, at a speed of 0.980c. What is the likelihood that the muon will survive its trip to Earth's surface before decaying? The probability of a muon decaying is given by P = 1 - [tex]e^{-Δt/τ}[/tex], where Δt is the time interval as measured in the reference frame in question. Also, calculate the probability from the point of view of an observer moving with the muon.

I figured that this is dealing with time dilation, so I used the formula T=[tex]\frac{T_{0}}{\sqrt{1-(v^{2}/c^{2})}}[/tex]

I know that v = 0.980c
And that [tex]T_{0}[/tex] = 2.2x[tex]10^{-6}[/tex] s

But doing this and solving for T, then plugging T into the probability equation does not give me the correct answer. What am I missing? Do I need to somehow account for the height?
 
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  • #2
What does the T you calculated represent physically?
 
  • #3
The time in the frame of reference that is outside of the muon.
 
  • #4
What time specifically? If you understand this, you should be able to solve the problem.
 
  • #5
Then I'm not sure I understand it.
 
  • #6
Well, you have at least four different times in this problem:

  1. Δt in the Earth's frame,
  2. τ in the Earth's frame,
  3. Δt in the muon's frame, and
  4. τ in the muon's frame.
Which one is equal to T0=2.2x10-6 s and which one is equal to the time you calculated using the time-dilation formula? How can you calculate the others?
 

FAQ: Special Relativity and muon decay

1. What is special relativity?

Special relativity is a theory of physics that describes the relationship between space and time. It states that the laws of physics are the same for all observers in uniform motion, and that the speed of light in a vacuum is constant regardless of the observer's frame of reference.

2. How does special relativity relate to muon decay?

Special relativity plays a crucial role in understanding the phenomenon of muon decay. According to the theory, as a muon travels at high speeds, its internal clock slows down relative to a stationary observer. This time dilation allows the muon to travel further before decaying, as measured by the stationary observer.

3. What is muon decay?

Muon decay is a process in which a muon particle transforms into other particles, such as an electron and two neutrinos. This decay occurs due to the instability of the muon, which has a half-life of about 2.2 microseconds.

4. What are the implications of muon decay for special relativity?

The discovery of muon decay provided strong evidence for the validity of special relativity. The observed time dilation of the muon's decay process is consistent with the predictions of the theory, demonstrating its accuracy in describing the behavior of particles at high speeds.

5. How is muon decay used in experiments?

Muon decay is used in various experiments to study the properties of particles and to test the predictions of special relativity. For example, the time dilation effect can be observed by measuring the difference in the decay rates of muons traveling at different speeds in a particle accelerator. This has led to important discoveries in particle physics and has further confirmed the validity of special relativity.

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