Muons have a "proper lifetime" = 2.2 micro seconds ("proper lifetime" = lifetime of a muon which is at rest). These Muons are created in the upper atmosphere at a height of 4700m above the earth's surface. So, a muon travelling at a speed of 0.99c can travel only a distance of 653.4 m if you do distance = speed * time (the non-relativistic calculation). So you would not expect to find any muons to survive and reach the surface of the earth. But the muons manage to travel almost 4700m and are detected near the surface of the earth. So now we need to explain it using Special Theory of Relativity. Let me try Time Dilation. According to me( I am an observer in the frame of rest on earth), the muon is travelling is travelling at a speed v=0.99c. So it'll will suffer length contraction, but that'll not help me in my calculation. The muon is supposed to decay after 2.2 microseconds (that's the characteristic feature of a muon). But i know that the time interval 't' as observed by me for the muon, becomes larger given by t= gamma* to where to= proper lifetime and t= observed lifetime. So if to=2.2microsecond, t=15.6 microseconds. I think i've understood it correctly so far. So a fast-moving muon appears to live for a longer time. WHAT I DON'T UNDERSTAND IS WHEN THEY SAY THAT: LENGTH CONTRACTION: Muon sees shorter length (by gamma = 7.1). Does that mean that we're now considering the Muon to be at rest and the surrounding space to be moving towards it (in a direction perpendicular to the surface of the earth) at a speed of 0.99c? So the space gets contracted from Lo to L Where L= 653.4m. So the Muon thinks that it has travelled ONLY 653.4m, but we will see that the muon has actually travelled a distance of Lo= 4631.8 m and thus has managed to reach the surface of the earth. Oh! Now i seem to have understood it after all!:P Am i correct?