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## Main Question or Discussion Point

Why kinetic energy is ½ m v2? Why it is different from Einstein’s equation for Energy E= m c2?

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Why kinetic energy is ½ m v2? Why it is different from Einstein’s equation for Energy E= m c2?

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RPinPA

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I prefer the notation and the language that "mass" refers to the invariant "rest mass". So when something is moving, its total energy is written as ##E = \gamma mc^2## where ##\gamma## is the relativistic factor ##1/\sqrt{1-(v/c)^2}## which is 1 when ##v=0## and approaches infinity as v approaches the speed of light.

That means that according to Einstein, the kinetic energy is the difference between total energy when moving and total energy when not moving.

$$KE = \gamma mc^2 - mc^2 = (\gamma - 1)mc^2$$.

It turns out that when v is small, then ##(1/2)mv^2## is a very good approximation to that. I'll post a followup with that proof.

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RPinPA

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##\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.

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Mister T

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It's not! ##\frac{1}{2}mv^2## is approximately equal to the kinetic energy, but that approximation is valid only for speeds that are a small fraction of the speed of light.Why kinetic energy is ½ m v2?

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