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This is chapter 37 problem number 48.
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A muon is created 55 kilometers above the surface, a measure from the earth.
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So our proper length is 55 kilometers.
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Average lifetime over muon is also given to us.
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Average lifetime over muon in its own frame is given to us.
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So for the muon, it's going to be the proper time, right? 2 .2 microseconds.
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So in the frame of the meon, earth is moving towards the muleon with the speed of .986c.
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Now in part a, we are asked to calculate in mewon's frame.
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What is its initial height? what is the initial height? remember, initial height of the mion is given to us as 55 kilometers in the earth's frame.
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And now we are to calculate it in the mion's frame.
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So the equation that we're going to use is l equals l0 over gamma, right? in this case, we know what l0 is.
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That's the proper length.
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It's 55 kilometers, so let's convert kilometers into meters times 1 over gamma factor.
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1 over gamma factor is square root of 1 minus v squared, right? v is 0 .986c square over c squared.
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So when we do with the algebra, what we find is 9 .17 times center of 3 meters.
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So again, let me explain you the gamma factor here.
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The gamma factor is 1 over square of 1 minus v squared over c squared.
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That's the formula for it.
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Since we have 1 over gamma, then if we take the inverse of it, that equals to 1 over v square over c squared.
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And that's where this term comes from that you see here.
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So in the mion's frame, the initial height is 9 .17 times 10 to 0 .3 meters...