By Firk F.W.K.

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Although the lifetime of the muon is 2µs in its rest frame, in the rest frame of the Earth, it is moving very rapidly, and therefore the interval between its creation and decay is no longer ∆t but rather ∆tE = γ∆tµ = 2 x 10 − 6 sec / √ [1 − (V/c)2 ] where V is the speed of the muon relative to Earth, and ∆tµ is the lifetime of the muon in its rest frame. ∆tE is its lifetime in the Earth’s frame. We see that if γ ≥ 50, the muon will reach the Earth. (For then, ∆tE ≥ 10 −4 s , and therefore H E ≥ 30,000m).

It is worth noting that Newton considered light to consist of particles; he did not discuss the properties of his particles. In the early 1800’s, Soldner actually calculated the deflection of a beam of “light-particles” in the presence of a massive object! Einstein was not aware of this earlier work). Let us consider a photon of initial frequency fS, emitted by a massive star of mass MS, and radius R. The gravitational potential energy, V, of a mass m at the surface of the star, is given by a standard result of Newton’s Theory of Gravitation; it is V(surface) = − GMSm/R.

This cannot be! The paradox stems 29 from the fact that the twin, B, who leaves the Earth, must accelerate from the Earth, must slow down at the star, turn around, accelerate away from the star, and slow down on reaching Earth. In so doing, the B-twin has shifted out of inertial frames and into accelerating frames. Special Relativity does not hold throughout the entire journey of the twin in the spacecraft The A-twin is always at rest in the inertial frame of the Earth. There is a permanent asymmetry in the space-time behavior of the twins.

### Age of Einstein (intro to relativity) by Firk F.W.K.

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