What is the significance of E=MC2?

Ronald C. Lasky, director of the Cook Engineering Design Center at Dartmouth College, explains the significance behind this hallowed equation:

It is the most famous equation in the world. Many can recite it—and attribute it to Albert Einstein—but few know its significance.

It tells us that mass and energy are related, and, in those rare instances where mass is converted totally into energy, how much energy that will be. The elegance with which it ties together three disparate parts of nature—energy, the speed of light and mass—is profound.

Here is where the equation of all equations comes from:

It was known for some time before Einstein’s insights that electromagnetic radiation (light, for example) possessed momentum. This quality of radiation is small in magnitude—after all, you needn’t worry about being knocked over by sunlight—but easily measurable. Applying an understanding of light’s momentum within a little thought experiment, it is possible to see how E = mc2 comes about.

Consider a cubic hollow box at rest in space with sides of length D and a mass of M. This box is also symmetrical in its mass distribution. One of the faces inside the box is coated with a fluorescing material, and, at a given moment, a photon (i.e., a particle of light) is emitted from that material, perpendicular to its surface. The momentum of this photon causes the box to move in the opposite direction as the photon, and it continues to move until the photon hits the opposite wall. During this time the box moves a very small distance, Δx.


Newton’s laws of mechanics tell us that the center of mass cannot move, because the box has not been acted upon by an outside force. However, in order to keep the center of mass constant, since the box has moved, some mass must have been transferred from the fluorescing side of the box to the absorbing side in the process of generating the photon and its striking the opposite side. Therefore the photon must have a mass, m.

So the photon, which also possesses energy E, is emitted from the fluorescing side of the box. Its momentum, Pphoton, is equal to its energy divided by the speed of light: Pphoton = E / c. The photon will impart this momentum to the box, causing the box to move a small distance, Δx, during the time, t, in which the photon travels to the opposite side of the box. The momentum of the box, Pbox, is also equal to its mass, M, times the velocity, vbox, at which it moves before the photon strikes its target. (Note: The box loses the photon’s mass, m, during this process, but this slight loss can be neglected here.) Hence:

Pphoton = Pbox = E / c = Mvbox

Then vbox = E / cM(1)

We can also determine the time it takes for the photon to travel across the box: it is equal to the length, between parallel faces, of the box (which is D), minus the amount the box moved in the opposite direction (Δx), divided by the speed of light, c. (The target will essentially have moved a slight distance closer, meaning the photon did not have to travel the full distance D.):

t = (D – Δx) / c

But, since Δx is a minute fraction of D, we essentially get:

t = D / c (2)

Now, since vbox = Δx / t , using equation 2, vbox can be restated as:

vbox = Δxc / D(3)

Substituting equation 3 for the term vbox in equation 1:

Δxc / D = E / cM

Next, we rearrange the terms to get:

ΔxM = ED / c2 (4)



Assuming the center of the box is initially at x = 0, this position is also the center of mass, xm. After the photon event, the box moves Δx to the left, as shown in the figure below, and the equivalent mass of the photon, m, is deposited on the opposite side. As mentioned above, we recall from Newton’s Laws that the center of mass must not change, because the box is not acted upon by an outside force. This concept is expressed in the center of mass equation below. The center of mass is at x = 0 in the left half of the equation and it is still at 0 after the photon strikes the opposite wall as described in the right half of the equation.

xm, initial = xm, final

0 = (–MΔx – mΔx + mD) / M

Grouping like terms:

0 = (mD – (M + m)Δx) / M

Solving for Δx:

Δx = mD / (M + m)

Since m is extremely small:

Δx = mD / M (5)

Substituting equation 5 into equation 4:

(mDM) / M = ED / c2

The mass of the box and D cancel out, leaving:

m = E / c2

Which rearranges to:

E = mc2