E = m c squared?!

Small equation - big meaning

When I first started this blog I was planning on a post about Einsteins most famous equation. Then I got a bit side tracked. But I am back on track now, so lets see how we go.

E = m c² 

Probably the most famous equation in the world, certainly of the 20th century. This tiny equation that says such a lot. The equation showing mass-energy equivalence. While many people can quote, I suspect that a smaller number can tell you what each of the components of the equation are. So let's start there.

The equation has 3 components, E, m and c. We will tackle them in reverse order.

c - is the speed of light in a vacuum. By this we mean approximately 186,282 miles per second, or exactly 299,792,458 meters per second. We say "in a vacuum" to indicate that we are talking about the maximum speed of light. Light travels more slowly in materials such as water.

The closest we get to a vacuum is the space between galaxies, so strictly speaking we are saying the speed of light as it crosses these vast regions of inter galactic space. The speed of light in a vacuum is constant. This is the value of c that we are referring to in Einstein's E=mc² .

m - mass. Philosophically it can get a bit tricky here. Most of us get out mass from standing on scales. We weigh our selves. What we are doing here is getting our weight. Weight is our mass multiplied by the acceleration due to gravity. So what we are describing is a quantity of inertia! That said we know things have mass. Take a proton or the electron. From various experiments we have worked out that that have a well defined mass. So we'll go with that for now, I'll cover mass in detail in another post.

E - energy. Once again this is a bit of an oddity. We believe that energy can neither be created (produced) nor destroyed by itself. It can only be transformed from one state to another. Energy is an amount of something. If we have the same amount of something, then we have the same amount of energy.

So now we know what the 3 terms are that make up our wonder equation. The next thing to consider is how Einstein managed to show they were related. When you look at it, it is an amazingly simply equation, you would have thought that someone would have stumbled upon it way back when. The answer is that deriving the equation and understanding what it means takes a giant leap of the imagination. This is how Einstein made the leap...

Note: If your maths is a bit weak, take this slow and you'll get it, although it may take a couple of reads. Hang in there, its worth it.

Light, any electromagnetic radiation for that matter, has momentum. This can be measure and is found to be

P_photon = h / λ    --- (1)


P_photon - momentum
h - Planck's constant
λ - wave length of light

the shorter the wavelength the higher momentum. So gamma rays have the highest momentum. A photon also has energy and this is given by

E = P_photon c    ---(2) or P_photon = E / c

E - energy
P_photon - momentum of the photon
c - speed of light

Now image a train carriage that has a length of L (see diagram near the bottom). The carriage has a mass of M. The carriage is symmetrical in shape and mass. At the right hand end is a radioactive source. This source gives out a single photon. Hey, this is a thought experiment, so it can happen. The photon travels from the right end of the carriage to the opposite left end. It takes a finite time for the photon to reach the other end of the carriage. Now as soon as the photon takes off, the carriage recoils and takes off in the opposite direction.

The photon goes to the left, the carriage goes to the right.

When the photon reaches the other end of the carriage it is absorbed completely by the carriage wall and  the carriage stops. During the time of flight of the photon the the carriage has shifted, Δx.

Next bit is important - because the carriage has not been acted on by any external forces the center of mass of the carriage cannot have changed. But the carriage has actually moved to the side a distance, Δx. The only explanation is that the mass of the carriage has been redistributed slightly. The only thing that moved was a photon from one side to the other. The implication then is that the photon ray must have mass, m.

When the photon takes off with its momentum P_photon, the law of conservation of momentum tells us that the carriage goes in the opposite direction with momentum P_car ,and the two are the same, so

P_car = P_photon

now from standard mechanics we have

P_car = M v_car

P_car - momentum of the carriage
M - mass of the carriage
v_car - velocity of the carriage

P_car = M v_car  = P_photon = E / c   --- from equation 2 above, so

M v_car  = E / c    --- (3)  or  v_car  = E / (c M)


The next thing is to work out how long it takes the photon to travel from one end of the carriage. It travels at the speed of light, c, and has to travel the length of the carriage minus Δx the distance the carriage has traveled to meet the photon. Δx is small compared with L so we ignore it. So we have

t = L / c     --- (4)  , this is just the time it takes light to travel a distance L.

The velocity of the carriage is given by the distance travelled divided by the time it takes, so during the time of flight of the photon this is

v_car = Δx / t    ---(5) (for example, you go 60 miles in 2 hrs, v = 60/2 = 30 mph)

the t in (4) and (5) is the same so we can do some re-arranging to give

v_car = Δx c / L    ---(6), but take a look at (3) above, we have an equation for v_car so again re-arranging

E / (c M) = Δx c / L ---(7)

Now the center of the box is initially at a x=0, this is also the center of mass is x_m. After the photon has done its thing x_m is still the same, because the carriage has not been acted on by any external forces the center of mass of the carriage cannot have changed. The carriage has moved though by a distance Δx. This has happened because we have redistributed the mass of the carriage, a photon moved from one side to the other, taking a mass, m, with it.

This bit about center of mass is important, you have to get this part to crack it. You have to understand how we get from equation 8 to 9 below.

Look at the diagram below carefully. The top rectangle is the carriage before the photon is fired, the bottom of carriage after the photon is fired. The carriage has a mass of M and we imagine that it is distributed evenly at either end of the carriage. 

The carriage before and after the photon moves from right to left

It turns out that the center of mass is given by

Mass_left x Distance_left = Mass_right x Distance_right  . --- (8)

In the top carriage above (8) is just

M/2 x L/2 = M/2 x L/2   , which is the same on both sides and so is obviously true.

Now we have said that the center of mass is the same after the event, but after the photon has moved from right to left and the carriage has moved from left to right and we have

(M/2 + m) ( L/2 - Δx ) = (M/2 - m) ( L/2 + Δx )

expanding this gives a bit of a mess, take your time with this

(M/2)(L/2) + mL/2 - (M/2)Δx - mΔx = (M/2)(L/2) - mL/2 + (M/2)Δx - mΔx

The first term on each side is the same, so is the last, so we can simply remove them. Leaving

mL/2 -(M/2)Δx = - mL/2 + (M/2)Δx

doing a bit of swapping we get

m L = M Δx    --- (9) which becomes  m / M = Δx / L  --- (9a)

taking (7) we can replace Δx / L on the right with m / M, so we now have

E / (c M) = c m / M    and we are now on the home straight

E = c m ( c M ) / M = c m c  , we have just cancelled the Ms

and so with one last bit of re-arranging we get...

E = m c² 

Albert Einstein, that really is beautiful. Thank You.