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Oxymoron
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A muon is created 3km above the Earth's surface heading downward at a speed of [itex]0.98c[/itex]. It is able to survive [itex]2.2\mu s[/itex] in its own frame before it decays.
(1) The muon travels a distance of 647m before it decays in its frame
[tex]d = vt = (0.98\times 3\times 10^8m/s)(2.2\times 10^{-6}s) = 647m[/tex]
(2) To an observer on Earth the muon's lifetime is longer. To the Earth observer, [itex]11.2\mu s[/itex] have passed in the time it took for the muon to decay.
[tex]\gamma_{v} = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{1}{\sqrt{1-\frac{(0.98)^2c^2}{c^2}}} = 5.03[/tex]
[tex]\Delta t = \gamma_{v}\Delta t_0 = 5.03 \times 2.2\times 10^{-6}s = 11.1\mu s[/tex]
(3) To the Earth observer the muon now travels a distance of 3.25km
[tex]d = vt = (0.98\times 3 \times 10^8m/s)(11.1\times 10^{-6}s) = 3.25km[/tex]
From these calculations, the muon actually hits the Earth in the observer's frame on Earth. But in the muon's frame it doesn't travel far enough. Who is right? If this happened in real life, would the muon hit the ground or not (to an observer on the ground or to a muon).
Lets take this a little further.
Does the muon actually percieve the distance to the ground as being 3.5km? To the observer on the ground the distance is well and truly 3.5km. But length contraction says that length contracts in the direction of motion. So the observer measures the length to be 3.5km in his frame, but that is simply because he is not moving relative to the distance.
The muon, however, IS moving relative to the distance. In fact the muon thinks that the distance to the ground is 597m, NOT 3.5km!
[tex]L = \frac{L_0}{\gamma_{v}} = \frac{3km}{5.03} = 597m[/tex]
So according to the muon, the distance to the ground is only 597m, which it will travel through in a mere [itex]2.03\mu s[/itex].
[tex]t = \frac{L}{u} = \frac{597m}{0.98\times 3\times 10^8m/s} = 2.03\mu s[/tex]
Therefore, according to the muon it hits the Earth.
If all my reasoning is correct here, my initial problem of discerning who was right (did the muon hit or not?) was incomplete. I hadn't adjusted for length contraction for the moving object. I this the reason why I had my dilemma?
Is it safe to say that the muon is at rest and the Earth is moving toward it at [itex]0.98c[/itex], in the muons frame? If so, I should get the same answer right?
Forgive my elementary-ness, I've only just begun studying this stuff!
(1) The muon travels a distance of 647m before it decays in its frame
[tex]d = vt = (0.98\times 3\times 10^8m/s)(2.2\times 10^{-6}s) = 647m[/tex]
(2) To an observer on Earth the muon's lifetime is longer. To the Earth observer, [itex]11.2\mu s[/itex] have passed in the time it took for the muon to decay.
[tex]\gamma_{v} = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{1}{\sqrt{1-\frac{(0.98)^2c^2}{c^2}}} = 5.03[/tex]
[tex]\Delta t = \gamma_{v}\Delta t_0 = 5.03 \times 2.2\times 10^{-6}s = 11.1\mu s[/tex]
(3) To the Earth observer the muon now travels a distance of 3.25km
[tex]d = vt = (0.98\times 3 \times 10^8m/s)(11.1\times 10^{-6}s) = 3.25km[/tex]
From these calculations, the muon actually hits the Earth in the observer's frame on Earth. But in the muon's frame it doesn't travel far enough. Who is right? If this happened in real life, would the muon hit the ground or not (to an observer on the ground or to a muon).
Lets take this a little further.
Does the muon actually percieve the distance to the ground as being 3.5km? To the observer on the ground the distance is well and truly 3.5km. But length contraction says that length contracts in the direction of motion. So the observer measures the length to be 3.5km in his frame, but that is simply because he is not moving relative to the distance.
The muon, however, IS moving relative to the distance. In fact the muon thinks that the distance to the ground is 597m, NOT 3.5km!
[tex]L = \frac{L_0}{\gamma_{v}} = \frac{3km}{5.03} = 597m[/tex]
So according to the muon, the distance to the ground is only 597m, which it will travel through in a mere [itex]2.03\mu s[/itex].
[tex]t = \frac{L}{u} = \frac{597m}{0.98\times 3\times 10^8m/s} = 2.03\mu s[/tex]
Therefore, according to the muon it hits the Earth.
If all my reasoning is correct here, my initial problem of discerning who was right (did the muon hit or not?) was incomplete. I hadn't adjusted for length contraction for the moving object. I this the reason why I had my dilemma?
Is it safe to say that the muon is at rest and the Earth is moving toward it at [itex]0.98c[/itex], in the muons frame? If so, I should get the same answer right?
Forgive my elementary-ness, I've only just begun studying this stuff!
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