I'm sure you've seen this equation before, $E = mc^2$ , it's probably the most famous equation in all of Physics. While you may have heard of it you probably don't know what it means or how it came about. Einstein didn't just decide one day that E must equal mc^2 and it magically fitted into place with all of physics so far; it was actually a by-product of special relativity that just happens to be exceptionally useful!
So what does it mean? Well $E = mc^2$ means that mass and energy are equivalent. Basically that means that you can convert them into each other, which you can express mathematically as $E \propto m $. It just so happens that $c^2$ is the conversion factor; for those of you unfamiliar with physical constants c is the speed of light in a vacuum. So you will appreciate that $c^2$ is a phenomenally massive number, in fact it is $c^2 = 9 \times 10^{16}$, or 9 followed by 16 zero's!
The best way to see mass energy equivalence is through an example:
If I had an average cat, it probably has a mass of about 5kg, so if I put that back into our expression we'll have: \[E = 5 \times 9\times 10^{16} = 45 \times 10^{16} Joules\] so you can appreciate that written down it's:
450,000,000,000,000,000 Joules of energy, that's quite a lot!
In fact "Fat Man", the bomb dropped on Nagasaki had a yield of only 88,000,000,000 Joules. That means all the energy contained in your pet cat is about 5 millions times greater!
*(note: the total energy contained in the bomb was much larger than the cat as it is of greater mass; however its yield energy was not as high as its intrinsic energy)
So that's what $E = mc^2$ means, but how did Einstein arrive at it?
I'm going to have to set some axioms for this, else this blog post could span on a lot and get rather boring, so you'll have to take my word on a few things.
Firstly:
$\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2} }}$
Secondly:
$m = \gamma m_o$
All that means is that as you move faster your mass increases, even at slow speeds. This is due to the fact that the speed of light is a constant, but that is another story entirely.
So to derive $E = mc^2$ we just have to write the second expression in an alternate form. \[m = m_0(1 - \frac{v^2}{c^2})^{-\frac{1}{2}} \]And those of you familiar with A level maths should recognise this as a binomial expansion. It expands to: \[m = m_0(1 + \frac{v^2}{2c^2} + \frac{3v^4}{8c^4} + \cdots) \]This sequence converges very rapidly at low velocities as c is so high so: \[m \simeq m_0 + \frac{1}{2}m_0 v^2(\frac{1}{c^2}) \]If we multiply throughout by $c^2$ the sequence simplifies to: \[mc^2 \simeq m_0 c^2 + \frac{1}{2}m_0v^2 + \cdots \]If we assume an object is not moving then its velocity is 0 so: \[mc^2 = m_0c^2\]As this is an energy term so we can then say: \[E = mc^2 = m_0c^2\] as long as the object in question is not moving.
In actual fact we can use this binomially expanded equation while v is much less than c, even if the object is in motion. However in its usual form $E = mc^2$ is referring to the intrinsic energy of the object, i.e. its energy when it is not moving.
And there you have it, with nothing more than an understanding of binomial expansion it is possible to show that $E = mc^2$.
