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I am currently a year 12 Australian student and we are just starting on the special relativity topic, and I got confused on length contraction.

Muons are created in the upper atomosphere with speeds of $0.99c$ or more. Their average life-time is $2.2\mu$ seconds measured at low speeds in the lab. A simple calculation shows that most should only travel about $660$ metres before decaying. Thus, very few muons should ever reach sea level. Using relativistic mechanics, calculate how far a muon can travel according to an observer on Earth.

Firstly, I used time dilation and that worked perfectly fine, got the answer about 4600 m. But when I was trying to solve it by length contraction, I got confused.

$$ L=L_0\frac{1}\gamma $$ with $\gamma$ the Lorentz factor

$L_0$ is the proper length, which is the length measured by a person on the muon (If we can!). Then I substituted $660$ metres (The distance muon travels from a person on muon) into it. And I got an answer less than 100! I must have done something silly and not understood some part of special relativity, can anyone point it out for me please?

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closed as off-topic by Kyle Kanos, Jon Custer, ACuriousMind Aug 23 '17 at 12:21

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  • $\begingroup$ Can you be more specific what numbers you plug in? $\endgroup$ – pfnuesel Aug 23 '17 at 10:14
  • $\begingroup$ I substituted 660 metres which is a approximate value suggested by the question itself. (distance muon travels from its own perspective) $\endgroup$ – L.H.Yu Aug 23 '17 at 11:29
  • $\begingroup$ Please note that homework-like questions and check-my-work questions are generally considered off-topic here. We intend our questions to be potentially useful to a broader set of users than just the one asking, and prefer conceptual questions over those just asking for a specific computation. $\endgroup$ – ACuriousMind Aug 23 '17 at 12:21
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I think you mean $4600~\text{m}$, not $\text{km}$.

This is slightly less natural to do it with length contraction, but here we go anyway! From the muon's frame of reference, it is the earth's atmosphere that is moving towards them. After approximately $660~\text{m}$, the muon will on average decay. But that $660~\text{m}$ is contracted, since the atmosphere is moving relative to the muon at $0.99c$. The length contraction formula is: $$L = \frac{L_0}{\gamma},$$ Here $L$ is what the muon sees, $660~\text{m}$, and we're interested in the true length, $L_0$. That gives us $L_0 = \gamma L$, which using your numbers gives $L_0 \approx 4600~\text{m}$.

Note that your formula for length contraction is incorrect. Note that $\gamma\geq1$ always, and the length should be getting smaller. Length contracts, and time dilates, which is what allows the speed of light to remain constant in all frames.

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  • $\begingroup$ Is it because, since we are looking for how an observer on earth will observe, we are actually looking for L0 and L is 660m rather than the other-war around? $\endgroup$ – L.H.Yu Aug 23 '17 at 11:51
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Well, in some way you are right, but I think you haven't fully understood it. 660 meters in the eyes of an observer on earth will be shorter in the eyes of an observer on the muon, as we take the muon as a static reference frame, and the earth will be then moving at the speed of $v$, $$ L=L_0\sqrt{1-\frac{v^2}{c^2}} $$ here $L$ is what the observer on the muon sees,and $L_0$ is the distance (660 meters) on earth (now the earth is moving), so the time the muon needs to get to the sea level is only $L/v$.

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