## $E=mc^2$, a derivation

$E=mc^2$ is perhaps the most well known equation in the world. First proposed by a patent clerk working in Switzerland in his now world famous (but absolutely unreadable) paper ‘On the Electrodynamics of Moving Bodies’, compiled later by Minkowski, Einstein’s novel idea about how mass and energy are intricately linked and that the classical concept of time was incorrect shook the world. Now just about every single high school physics student enrolled in a respectable physics class will learn about the equation $E = mc^2$, but few knows its origin. In this article, I aim to use some simple physics and some maths to derive said equation.

We begin. with of course,

$$F=ma=\dv{\vec{p}}{t}$$

and the classical Newtonian kinetic energy:

$$T = \frac{1}{2}mv^2$$

We know from Lorentz transform that the following is true:

$$p=\gamma{}mv$$

Where $\gamma = \frac{1}{\sqrt{1-v^2}}$.

By combining eqns. 1 and 2, we can rewrite an expression for force relativistic force in the following form:

$$F = \dv{}{t}(\gamma{}mv)$$

By operating the derivative operator through and apply the chain rule and product rule where appropriate, we get this:

\begin{align*}F &= \dv{}{t}(\gamma mv) \\ &= \gamma m\dv{v}{t} + vm\dv{\gamma}{v}\dv{v}{t}\\ &= \gamma ma + vm (-\frac{1}{2})\frac{1}{\sqrt{1-v^2}^3}\cdot(-2v)\cdot a\\ &= \gamma ma + mv^2a\gamma^3 \\ &= \gamma ma(1+v^2\gamma^2) \\ &= \gamma ma(1+\frac{v^2}{1-v^2}) \\ &= \gamma ma (\frac{1-v^2+v^2}{1-v^2}) \\ F &= \gamma^3 ma\end{align*}

The end result should be expected, but let’s do a quick sanity check anyways. When $v \lll 1$, $\gamma$ does indeed vanish. Hopefully, you can also see that this is the reason why accelerating a body to near the speed of light is incredibly difficult. As we approach relativistic speeds and $\gamma$ is no longer negligible, acceleration becomes extremely inefficient. It is also important to note that the derivation we show here are only valid for one-dimensional systems. Generalizing our finding here to higher dimensions will surely make a fiendish exercise indeed!

Next, we consider the energy. Energy can be considered as the work done by the force, or more precisely:

$dT = dW = F\cdot dx$

A small change in energy is equal to a small change in work done, which in turn equals the force times the displacement. We may express this finding in integrals if we are looking for the total amount of energy along the path of travel (don’t worry, we will add the limits soon!)

$T = \int F\cdot dx$

Which we can conveniently express in a derivative w.r.t $t$:

$T = \int F \dv{x}{t} dt = \int \gamma ^3 m\dv{v}{t}\cdot v \cdot dt$

Since we are now integrating w.r.t $v$, the limits of integration are rather obvious: from $0$ to $v*$, assuming the object travels from rest, and reached final velocity $v*$.

$\int_{0}^{v*}\gamma{}^3mv \cdot dv$

Which, after subbing in $\gamma$,

$\int_{0}^{v*} \frac{mv}{\sqrt{1-v^2}^3}dv$

Please feel free to stop reading and perhaps try the above integral yourself.

The key is to use the substitution $z = 1-v^2$

$dz = 2vdv$

$m\int_1^{1-v*^2} -\frac{1}{2}m\frac{1}{z^{\frac{2}{3}}}dz$

Which is just

$-\frac{m}{2}\left[ z^{-\frac{1}{2}} \right]^{1-v*^2}_{1}$

$=\frac{m}{\sqrt{1-v*^2}}-m = m(\gamma -1)$

And lo and behold:

$T(v) = m(\gamma(v) -1)$

You don’t see it yet? During the derivation, we have used relativistic unit, where $c=1$, so in fact, in SI units, the above equation will look like this:

$T = mc^2(\gamma(v) – 1)$

It may seem suspicious to you that $c^2$ have just popped out of no where, but if you went back to the start and used the non-relativistic form of Lorentz transform, you will see that $c^2$ permeates our calculation.

But you may ask, why is the left hand side equal to $T$ instead of $E$? There are two types of mass, the first is what we are familiar with, gravitational mass. This mass characterises how an object interact with another object with gravitational forces. The other one is something called the inertial mass. The inertial mass characterises how difficult it is to accelerate something through space. It is important to note that the two concepts of mass are completely unrelated. It is important for us to embrace the idea the mass is just energy frozen in time. Thus, we can write an equation in the form:

$$E_{\text{total}}=T + E_{\text{mass}}$$

The total energy $E$ is the sum of the kinetic energy and its energy that manifests as what we call mass, subbing in what we know gives:

$$E_{\text{total}}=mc^2(\gamma(v) -1) + mc^2$$

$$E= \gamma mc^2$$

Or, in its full glory:

$$E = \frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}$$

The $\gamma$ factor is frequently ignored as it makes the equation bulky and less catchy, but nevertheless, it is an essential part of the equation.