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DeutschFox
10-26-2000, 10:18 PM
I understand that it can be shown that kinetic energy at speeds v<<c can be derived from the relativistic formula E=mc^2 to show the Newtonian KE=(1/2)mv^2. Hmmmm?

scr4
10-26-2000, 10:29 PM
Well, yes, but you need another formula. The e=mc2 includes both kinetic energy and rest mass energy, because m is relativistic mass which gets bigger as the speed increases. The kinetic energy is ke=mc2 - m0c2 where m0 is the rest mass of the particle. Relativistic mass is m0/sqrt(1-v2/c2). Plug that into the ke formula and I think you can simplify it down to ke=1/2 m0 v2, if you take the limit as v/c << 1.

Opus1
10-26-2000, 11:51 PM
You need to use a Taylor's expansion to reduce it further. Sqrt[1-x]~= 1-(x/2) for very small x. In this particular case, x = v^2/c^2. Substitute 1-(2v^2/c^2) for Sqrt[1-v^2/c^2] and you'll reduce it down.