# I Why E=mc^2 is different from E=1/2 mv^2?

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1. Oct 4, 2018

### sush

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

2. Oct 4, 2018

### RPinPA

First of all, Einstein's equation is for the total energy, while $(1/2)mv^2$ is the part of the energy just due to motion. Even when something is at rest, its mass is equivalent to energy, and in fact could be converted completely to energy if it came into contact with an equal amount of antimatter.

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.

3. Oct 4, 2018

### RPinPA

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.

4. Oct 4, 2018

### Mister T

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.