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- Why is time dilation calculated as $$t = \gamma T$$ and not $$t = \gamma(T - x^{'} v/c^2)$$ ?
From the Lorentz transformation equations we know that $$t = \gamma(t^{'} - x^{'} v/c^2)$$
but for the Muon decay example where the setup is as follows :
"Assume for simplicity that a certain muon is created at a height of 50 km, moves straight downward, has a speed v = .99998 c, decays in exactly T = 2 · 10−6 seconds, and doesn’t collide with anything on the way down.11 Will the muon reach the earth before it decays? "
This particular example is from Morin's book. The solution is that $$t = \gamma T$$ but why is it not $$t = \gamma(T - x^{'} v/c^2)$$ ?
Is the idea that x' (where the muon decays in its reference frame) is 0 ? because consider a reference that is offset from the muon by a distance d perpendicular in direction of velocity (so it is beside it) and is moving at the same velocity as the muon. Then x' = d and v is the speed of the muon. but then it is a different answer than that in Morin's book. Why is this the case?
but for the Muon decay example where the setup is as follows :
"Assume for simplicity that a certain muon is created at a height of 50 km, moves straight downward, has a speed v = .99998 c, decays in exactly T = 2 · 10−6 seconds, and doesn’t collide with anything on the way down.11 Will the muon reach the earth before it decays? "
This particular example is from Morin's book. The solution is that $$t = \gamma T$$ but why is it not $$t = \gamma(T - x^{'} v/c^2)$$ ?
Is the idea that x' (where the muon decays in its reference frame) is 0 ? because consider a reference that is offset from the muon by a distance d perpendicular in direction of velocity (so it is beside it) and is moving at the same velocity as the muon. Then x' = d and v is the speed of the muon. but then it is a different answer than that in Morin's book. Why is this the case?