A muon is created 54.0 km above the surface of the earth (as measured in the ear

Question

A muon is created 54.0 km above the surface of the earth (as measured in the earth's frame). The average lifetime of a muon, measured in its own rest frame, is 2.20 us , and the muon we are considering has this lifetime. In the frame of the muon, the earth is moving toward the muon with a speed of 0.984 c.

In the muon's frame, what is its initial height above the surface of the earth?

In the muon's frame, how much closer does the earth get during the lifetime of the muon?

What fraction is this of the muon's original height, as measured in the muon's frame?

In the earth's frame, what is the lifetime of the muon?

In the earth's frame, how far does the muon travel during its lifetime?

What fraction is this of the muon's original height in the earth's frame

thank you!

Explanation / Answer

a) muon perceived height

= 54.0 km * (1 - v2/c2)1/2

= 54.0 km * (1 - 0.9842)1/2

= 9.62 km. [Shorter than   Earth frame measurement. ]

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b) distance closer (Earth frame) = v * ( 2.20  x10 - 6 seconds)

= 0.984 * 3 * 108 * ( 2.20  x10 - 6 seconds)

= 649 m.

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c) v * ( 2.20  x 10 - 6 seconds) / [54 km*(1 - v2/c2)1/2 ]

= 0.067 = 6.7 %

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d) earth time = (2.20  x10 - 6) / (1 -0.9892)1/2

=  1.23 * 10-5 s. [ Longer than muon frame measurement. ]

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e) earth measurement = v * (earth time)

= 0.984 * 3 * 108 * 1.23 * 10-5 s

= 3630 m

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f) v * (earth time)/54 km

= 0.984 * 3 * 108 * 1.23 * 10-5 s / 54 km

= 0.067 = 6.7 %

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