Special Relativity Muon Distance Travelled Question

AI Thread Summary
Muons, unstable particles with a mean lifetime of 2.2 microseconds, are produced in the upper atmosphere and travel at speeds close to light. Despite their short lifespan, they reach the Earth's surface from a height of 12.3 km due to relativistic effects. The discussion highlights the importance of using length contraction rather than assuming the muon is at rest while the Earth moves towards it. The calculation presented incorrectly assumes the muon's lifetime is sufficient to cover the entire distance without accounting for the effects of time dilation. Correctly applying length contraction will yield a different measurement of the atmospheric thickness as perceived by the muon.
ecneicS
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Homework Statement


uons are unstable subatomic particles that decay to electrons with a mean lifetime of 2.2 microseconds. They are produced when cosmic rays bombard the upper atmosphere about 12.3km above the Earth's surface, and they travel very close to the speed of light. It is known that these muons reach the ground. From the point of view of the muon, it still lives for only 2.2 microseconds, so how does it make it to the ground? What is the thickness of the 12.3 km of atmosphere through which the muon must travel, as measured by the muon? Assume speed of muon is .999c


Homework Equations


time dilation, length contraction


The Attempt at a Solution


I am not so much asking for a solution as I am asking why what I have done is wrong.

The scenario is that the muon is flying towards Earth at a really high speed, and we are given its lifetime in respect to the muon's frame of reference.

So to me it makes sense to think of the following: That the muon is at rest and that the atmosphere and the Earth are moving with the same speed towards the muon that the muon was moving towards the Earth in the previous frame of reference.

By doing this and assuming that the muon does actually hit the earth, it follows logically (in my head) that because we have the velocity of the Earth in respect to the muon's reference frame, and that we have the lifetime of the muon in respect to the muon's reference frame, that we can solve for the height of the atmosphere in respect to the muon's reference frame.

V=d/t
d=Vt
d=(0.999c)*(2.2*10^-6)
d=660m = 0.66km

Please, if you can, point out the error in what I've done or tell me if you agree with my result.
 
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ecneicS said:
So to me it makes sense to think of the following: That the muon is at rest and that the atmosphere and the Earth are moving with the same speed towards the muon that the muon was moving towards the Earth in the previous frame of reference.
Good.

By doing this and assuming that the muon does actually hit the earth, it follows logically (in my head) that because we have the velocity of the Earth in respect to the muon's reference frame, and that we have the lifetime of the muon in respect to the muon's reference frame, that we can solve for the height of the atmosphere in respect to the muon's reference frame.
Don't assume that the muon just barely makes it to the earth. The travel time to pass through the atmosphere will be less than the muon's lifetime.

Instead, use length contraction.
 
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