Hello and thank you for your question.
The best explanation that I've seen on the Web of how the speed of
light gets into the formula is at
Mass Energy Equivalence
The algebra on that page gets a little confusing, but thankfully
there's no calculus! Because Einstein used photons as an example of a
body having a certain amount of mass and a certain amount of energy,
and given that photons travel at the speed of light, his formula ended
up with 'c' in it, and in fact c^2
The photon example turns out to yield a formula that's true for all
matter, even matter standing still, and the c^2 stays in that formula
Let me try to talk you through the calculation that's explained on that page.
Einstein knew (because he had figured it out but actual experiments
prove it to be so) that the the energy of a photon of light is a fixed
multiple of its frequency.
E = h * [greek letter 'mu']
The constant multiplier 'h' is called Planck's constant, and 'mu' is
the frequency of the light [it looks like the letter 'v' but it isn't
velocity, it's frequency]. The name of this principle is the
So if two photons of frequency v jump off in opposite directions
the total energy of the two of them is
E = hv + hv
However, because of the Doppler effect [like the change in frequency
of a train whistle as it moves toward you or away from you] the
frequency of one photon is increased, relatively speaking, (call it
v+) and the frequency of the other photon is decreased (call it v-).
In other words, one photon is emitted in the +x? direction and the
other in the -x? direction. The photons have frequencies v+ and v-
respectively. The velocity of the body remains unchanged. Due to
Doppler shift the photon moving in the +x direction is red shifted and
the photon moving in the -x direction is blue shifted.
And because of conservation of momentum, and the equal and opposite
direction of the two photons, the total momentum is the same as it was
before they jumped off. So if the starting condition had momentum
zero (that is, we see it standing still at the start), the total
momentum remains zero although both photons are now headed away from
This is where 'c' gets into the picture, because momentum is mass
times velocity, and the velocity of each photon is the speed of light,
The rest of the algebra on that page is pretty hard to follow, but
you'll notice that the doppler shift formula [the one that has greek
letter beta in it that I have to type here as 'B'] has a fraction
The ^1/2 in that formula is a square root, and the algebra requires
squaring that as part of the solution, and that's how 'c' ends up
being squared too.
When you get to the end of the algebra, you discover that "if a body
emits radiation having an energy E then there is a decrease in the
body?s inertial mass in the amount Dm = E/c2. This holds even when the
body is at rest. Therefore the mass of a body is related to its energy
content [and a] body at rest has a rest energy E0 = m0c^2."
Here's a scientific paper that does the actual calculation, not once but twice!
Two exact derivations of the mass/energy relationship, E=mc2
I don't know if you'll find this paper easier or harder to follow.
But here you'll see a relativistic formula that you might have learned
in high school physics, the one that has (1-v^2/c^2)^(1/2) which is a
measure of how an object's mass increases as its speed increases.
This too turns out to tie up with E=mc^2 so at least if you accept the
increase in mass formula (which like the photoelectic effect has been
experimentally proven) then that also counts as an explanation of how
the speed of light gets into the picture.
I'll finish with an item of historical interest.
The Past, Present and Future of the Mass Energy Equation E = mc^2
"Some scientists believe that Einstein should not be given credit as
he has published an unreferenced paper deliberately hiding the
previous discoveries and status of E=mc2. The mass-energy
inter-conversion holds good universally e.g. in chemical reactions,
nuclear reactions, reactions taking place in heavenly bodies and
Note also the last sentence above about the universality of the
mass-energy conversion. That was the topic of the Google Answers
question cited by welte-ga, and you'll note in the comments to the
other question my consternation that the posted Answer seemed to say
that matter doesn't really get changed to energy in a standard
chemical reaction. But it does!
Search terms used:
"why c squared"
photon shift momentum
Fadner, Am. J. Phys. Feb 1988
Thanks again for bringing us your question. If you think I can make
this any clearer (although I've really done my best here) you may
Google Answers Researcher