Special Relativity and muon decay

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astr0
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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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What does the T you calculated represent physically?
 
The time in the frame of reference that is outside of the muon.
 
What time specifically? If you understand this, you should be able to solve the problem.
 
Then I'm not sure I understand it.
 
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?