# Formulas for kinetic energy

I was reading ABC of relativity from Bertrand Russell and some formulas about kinetic energy caused me some problems. Here is the extract :

The kinetic energy is, in the usual form $\frac{m}{\sqrt{-1-v^2}}$. As we've seen before, energy can be gained or lost, so if we want to, we can add an arbitrary quantity. Therefore, we can take the energy using $\frac{1}{2}mv^2$. (...) But it is better to use the formula as $\frac{m}{\sqrt{1-v^2}}$, because it is valid for the conservation law.

First of all I have to admit that I am just a high school student and I've only seen the 2nd formula to calculate the kinetic energy. Supposing that the other forumulas are equal to the second (even if only in special cases) seems to be impossible, just because of the square-root wich value will always be negative in the first one and negative for $v>1$ for the third one, wich of course will give a complex number as an energy. Don't know about high-level physics but this has no meaning to me.

Can soneone explain me what are these formulas and where they come from?

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The first form, with the -1, is a typo. It should be $m/\sqrt{1-v^2}$. –  Michael Brown May 28 '13 at 15:48
Then what about if $v>1$ or just $v=1$? –  moray95 May 28 '13 at 15:52
It never is. This formula is written in units where the speed of light $c=1$. So when you say $v>1$ you mean "what happens when something moves faster than light?" - which is impossible for massive bodies for various reasons you'll find by searching this site. If you want to restore ordinary units you get $E = mc^2 / \sqrt{1-v^2/c^2}$. This is the generalisation of $E=mc^2$ (which only applies to bodies at rest) to moving bodies. For massless particles you have $v=c$, but the formula for energy is different as well: $E = pc$, where $p$ is the momentum. –  Michael Brown May 28 '13 at 15:55
Okay, thanks a lot didn't know about the fact $c=1$. Get it now :) –  moray95 May 28 '13 at 15:58

The formula $\frac{m}{\sqrt{1-v^2}}$ is actually for the total energy, not the kinetic energy. The formula for the kinetic energy is actually $(\frac{1}{\sqrt{1-v^2}} - 1)m$. And if you do a Taylor expansion of this formula to second order, you'll find that you actually do recover $KE = \frac{1}{2}mv^2$. You get this from the aforementioned total energy equation $E = \frac{m}{\sqrt{1-v^2}}$, and then you subtract the potential (rest) energy $E_{rest} = m$.