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Why E=mc^2 is different from E=1/2 mv^2?
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[QUOTE="RPinPA, post: 6066996, member: 651116"] Here's the proof. ##\gamma = 1/\sqrt{1-(v/c)^2} = [1-(v/c)^2]^{-(1/2)}##. It can be shown with calculus that when ##\epsilon## is a small number, which we write as ##\epsilon \ll 1## or "##\epsilon## is much less than 1" then a good approximation to ##(1 - \epsilon)^n## is ##1 - n \epsilon##. We have ##\epsilon = (v/c)## and ##n = -0.5##. So when ##(v/c) \ll 1##, ##\gamma## is approximately ##1 - (-0.5)(v/c)^2## or ##1 + 0.5(v/c)^2##, to a very good approximation. How good? Let's say you're moving at 1/10 of the speed of light, ##(v/c)^2 = 0.01##. Then the exact value of ##\gamma## is 1.0050387... while the approximation gives ##1 + 0.5*0.01 = 1.005##. And the smaller v is, the better the approximation. So ##(\gamma - 1)## is approximately ##1 + 0.5(v/c)^2 - 1 = 0.5v^2/c^2## and ##KE = (\gamma - 1)mc^2 = (0.5v^2/c^2)* mc^2 = 0.5mv^2##. To "very good approximation", for low velocities. [/QUOTE]
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Why E=mc^2 is different from E=1/2 mv^2?
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