By P. W Bridgman

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**Extra info for A sophisticate's primer of relativity**

**Example text**

Indeed, let us consider again the inertial frames S and S , and suppose that A and B are two events which are simultaneous in S, namely t A = t B ( t = 0). When observed by S the two events will be separated by a time interval t = γ (V ) V c2 x = 0. 69) This implies that two events which are simultaneous in S, but occur at different points, are not simultaneous with respect to a frame S in motion with respect to S. 8 Note that the time dilation is a relative effect, that is if we have a clock at rest in S, from Eq.

Suppose an observer in S is measuring the time lapse t = t B − t A between two events A and B which occur in S at the same place but at different times, so that 24 1 Special Relativity Fig. 9 Event A causing event B t B > t A (for instance two successive positions of the second hand of a clock at rest in S ). If the events occur in S along the x -axis, we then suppose x = x B − x A = 0. 57) it then follows that x = V t. 60) we find t = γ (V ) t− V c2 x = tγ (V ) 1 − V2 c2 = t . 67) We conclude that the time lapse measured in S is t = γ (V ) t > t.

These muons are actually detected in our laboratories, so that their velocity should be high enough as to reach our detectors before they decay. If we were to perform a computation using classical Newtonian mechanics, which is based on the assumption of absolute time, the mean lifetime of a muon is the same in every inertial frame. Therefore the minimum velocity v for a μ particle to reach the surface of the earth would be approximately given by the height h of the atmosphere divided its mean lifetime.