Based on the observed behaviour of muons, assuming one comes into being at 10km and is travelling at 0.9997% lightspeed toward an observer on earth. Gamma for this speed is about 41. (I have seen various variations on the calculations but this doesn't really effect the question.)
In 0.75 microseconds the muon travels about 225 metres (300 metres per microsecond). With time dilation at this speed the observer will see the muon exist for 225*41=9225 metres (ok, not exactly 10km but near enough for the question)
My question is, from the muon's frame, and under the relativistic effects of 41 gamma, does it see time flow faster outside it's frame?
Baring in mind it hits earth's surface in 0.75 microseconds from it's frame, the earth must rotate to the same physical point that both it and the observer on earth see's it hit (an imaginary X on earth's surface)
If time flows more slowly outside the muon's frame, then in 0.75 microseconds it will miss the X and hit a different spot than the observer see's it hit, obviously this can't be right.
It makes sense to think time flows faster, and the muon would see the earth rotating 41 times faster, then it would hit the right spot
But I read that in relativity from the muon's frame a clock on earth would appear slower not faster..
So how does it manage to hit the right spot considering earth's rotation and difference in measured descent time between the two reference frames?