Einsteins theory of relativity and formula of kinetic energy [duplicate]

Question

After seeing both Einstein's theory of relativity ($E=mc^2$) and the formula of kinetic energy (K.E=$\frac{mv^2}{2}$), I noticed that that they are mostly the same, with the only difference that Einstein's theory uses a specific speed, namely of light, while the kinetic formula uses the object's speed. Is the similarity a coincidence, like with density [mass/volume] and concentration [mass of solute/volume of solution]? Was one derived from the other?

Answer 1

The two are indeed related. The relativistic expression for the total energy is:

$$ E^2 = p^2c^2 + m^2c^4 \tag{1} $$

where $p$ is the relativistic momentum:

$$ p = \frac{mv}{\sqrt{1 - v^2/c^2}} $$

and $m$ is the rest mass. If the object isn't moving then $p=0$ and equation (1) becomes:

$$ E = mc^2 $$

which is of course the well known expression for the energy equivalent to a mass $m$. To recover the low energy approximation for the kinetic energy is a bit more involved. We start by writing equation (1) as:

$$ E^2 = \frac{m^2v^2c^2}{1 - v^2/c^2} + m^2c^4 $$

and rearrange this to:

$$ E = mc^2\left(1 + \frac{v^2}{c^2 - v^2}\right)^{1/2} $$

Then if $v \ll c$ we can use a binomial expansion to approximate the right hand side:

$$ E \approx mc^2\left(1 + \tfrac{1}{2}\frac{v^2}{c^2 - v^2}\right) $$

and multiplying out gives:

$$ E \approx mc^2 + \tfrac{1}{2}mv^2\frac{c^2}{c^2 - v^2} $$

and since $v \ll c$ that means $c^2 - v^2 \approx c^2$ and the equation simplifies further to:

$$ E \approx mc^2 + \tfrac{1}{2}mv^2 $$

This is the total energy so we subtract off the rest mass energy $mc^2$ to get the kinetic energy:

$$ T = E - mc^2 = \tfrac{1}{2}mv^2 $$

and this is just the usual low energy expression for the kinetic energy.

So it isn't the case that the rest energy and kinetic energy equations are similar because one is derived from the other, but rather that they are both derived from the same equation for the total energy.