 You need to have an idea of the following concepts if this page is to make sense to you:

• Conservation of energy
• Conservation of momentum

It would be beneficial for you to have an insight into:

• Special relativity
• The doppler effect

Have you ever wondered where the equation $E = mc^{2}$ came from? It is such a simple, yet elegant equation that links mass and energy together, but how did Einstein come up with it? The idea that energy and mass are interlinked using… the speed of light, even though anything with a mass can’t actually travel at the speed of light? It seems a bit mad.

Firstly, the concept of special relativity (a theory developed by a number of Physicists including Albert A. Michelson, Hendrik Lorentz, Henri Poincaré and Albert Einstein) was used. The main and crucial step requires special relativity. Imagine a point in space that suddenly emits a flash of light. We can say that this flash of light has energy $E$. This means that the source of light has lost an amount of energy that we can say is equal to $-E$. If we say that this energy is emitted in all directions (i.e. the light sends out a flash that would travel outwards in all directions, all at the same speed), we can safely assume that the velocity of this point source will not change, $v_{before} = v_{after}$. Whereas, if the energy was sent out all in one direction (and therefore the momentum of the energy was in that same direction, we could assume that the point source would end up having momentum in the opposite direction.

If you were observing this flash in space but you were moving past it at some velocity, you would consider yourself to be stationary (because every other star and galaxy is so far away that they appear stationary) and the source of light to have some velocity. Relative to you this flash has the same velocity as you but in the opposite direction (to your point of view).

If you could measure it, you would find that the point source would have a kinetic energy, $E_{k_{1}}$. After the flash of light was emitted you would measure the point source as having $E_{k_{1}} - E$.

At this point, because the point source is moving, special relativistic effects have to be taken into account. As such the point source is effected by the relativist doppler effect. This means that the amount of energy that the point source decreases by as it emits the light is $E(1 - \frac{v^{2}}{2c^{2}})$

where $E$ is the energy emitted by the point source during the flash of light, therefore is the decrease (or change in energy) and should be written as $\Delta E$ $v$ is the velocity of the moving object $c$ is the speed of light, equal to $299,792,458 ms^{-1}$ So now we can write that as an observed you would measure the sources energy after the emission of light as: $E_{k_{1}} - \Delta E(1 - \frac{v^{2}}{2c^{2}})$

What if you were observing the flash and were stationary?

If you were stationary instead of moving past the point source as it flashed, there would be no relativistic effect. You would measure the energy fo the light emitted as $-E_{1}$. Lets say you then speed up so that you were moving past the point source now (so as to compare the scenario to the previous one).

The point source would now be measure to have a kinetic energy of $E_{k_{2}}$.

The total energy in this scenario is therefore $-E + E_{k_{2}}$

It should not matter whether an external observer, observing the flash, is moving or not – otherwise the whole world of Physics would depend on an observer, or a number of observers etc… and this just does not happen. Therefore the energy involved in the first scenario where you are moving past the point source of light will equal the second scenario where relativistic effects are not involved: $-E + E_{k_{2}}$ is equal to $E_{k_{1}} - \Delta E(1 - \frac{v^{2}}{2c^{2}})$ $-E + E_{k_{2}} = E_{k_{1}} - \Delta E(1 - \frac{v^{2}}{2c^{2}})$  $-E + E_{k_{2}} = +E_{k_{1}} - \Delta E(1 + \frac{v^{2}}{2c^{2}})$

This can simplify to give: $-E + E_{k_{2}} = E_{k_{1}} - E - \frac{\Delta Ev^{2}}{2c^{2}}$

The $-E$‘s can cancel to give: $E_{k_{2}} = E_{k_{1}} - \frac{\Delta Ev^{2}}{2c^{2}}$

Rearranging this to compare the kinetic energy before and after the flash gives: $E_{k_{1}} = E_{k_{2}} + \frac{\Delta Ev^{2}}{2c^{2}}$

where; $E_{k_{1}}$ is the kinetic energy before the flash $E_{k_{2}}$ is the kinetic energy after the flash $\frac{\Delta Ev^{2}}{2c^{2}}$ is the difference in the two kinetic energies

Since $E_{k} = \frac{1}{2}mv^{2}$: $\frac{1}{2}mv_{before}^{2} = \frac{1}{2}mv_{after}^{2} + \frac{\Delta Ev^{2}}{2c^{2}}$

As discussed previously, $v_{before} = v_{after}$, additionally since the object does not speed up or slow down we can write that $v_{before} = v_{after} = v$, so we can simplify to: $\frac{1}{2}mv^{2} = \frac{1}{2}mv^{2} + \frac{\Delta Ev^{2}}{2c^{2}}$

When an object emits energy something needs to change, since energy has to conserved ad therefore cannot be created! Therefore, if the velocity before and after doesn’t change, this means that the mass must. So we should re-write the formula above as: $\frac{1}{2}m_{1}v^{2} = \frac{1}{2}m_{2}v^{2} + \frac{\Delta Ev^{2}}{2c^{2}}$

Dividing through $\frac{1}{2}$ and $v^{2}$ gives: $m_{1} = m_{2} + \frac{\Delta E}{c^{2}}$

Rearranging for $E$ gives: $\Delta E = (m_{1} - m_{2})c^{2}$

where; $m_{1} - m_{2}$ is the change in mass, written as $\Delta m$ $\Delta E$ is the difference in energy due to the emission of light $\Delta E = \Delta mc^{2}$

So there you have it… quite elegant really!

Minutephysics have a nice video which puts this together, in only 2 minutes!