E = mc² (High School)

The derivation  of Einstein's famous formula is not that hard if you know calculus, use the familiar physics definitions for force and energy and the Lorentz Transformation.  However, even the PBS series on the subject does not go through the "nuts and bolts" of the derivation of E=mc², which was a starred exercise in this high school physics book in 1972.  We show what parts of the relationship would have been understood by the famous physicists: Newton, Faraday, Lorentz and Einstein.

Newton

The first formula is the definition of Force, which comes from Newton:

F = dp/dt

or force is the change in momentum over time.  Since momentum is defined as

p = m * v

or mass times velocity, then we can follow the rules of differentiating a product to get:

F= m * dv/dt + v * dm/dt

The first term is the familiar force = mass times acceleration, since dv/dt is the definition of acceleration.

Comment: Since Newton invented calculus, and in particular invented calculus to explain physics, he probably wrote the above differential at some point. If he did, he probably discarded the second term, assuming mass was a constant and so dm/dt = 0.  This mistake by one of the greatest minds on the planet was due to the fact that he didn't know about Energy or the Michelson-Morley experiment.

Faraday

The second formula is the definition of Energy which comes from Faraday's time.

E = _s0∫^s1 F * ds

or energy is the integral of force over distance (from s₀ to s₁). Plugging in the above formula for force, we get:

E = _s0∫^s1 m * dv * ds/dt + v * dm * ds/dt

but v = ds/dt (velocity is the change in distance over time), so

E = _s0∫^s1 m * v * dv + v² * dm

Comment: Newton was not familiar with the concept of energy. In particular, if he wanted to figure out the velocity of a roller coaster at various heights, he would have had to integrate momentum all along its path.  Nowadays, the more familiar approach is to use formulas for Kinetic and Potential energy and to calculate a roller coaster's final velocity in terms of its initial velocity and change in height.  Faraday knew about energy, but he was not familiar with the Michelson-Morley experiment, which Lorentz codified below.

Lorentz

The third formula come from Lorentz.  It predates Einstein's work on relativity and is called the Lorentz Transformation.

m = m₀/(1 - (v/c)²)^1/2

or the mass of a body in motion is a function of its rest mass (m₀), and how fast it travels (v).
"c" is the speed of light.  Note that this formula shows that if velocity ever hits the speed of light, mass becomes infinite.  Note also that if a particle's  velocity becomes greater than the speed of light, it's mass becomes imaginary!

We can see how mass changes with velocity by differentiating mass by velocity.
I have included each term as I differentiate "down" into the center term.

dm/dv = m₀ * (1 - (v/c)²)^-3/2  *  -1/2  *  -2 * v/c²

"simplifying"

dm/dv = (m₀ * v/c²) * (1 - (v/c)²)^-3/2

But we can solve the Lorentz transformation for m₀ and plug it in to the above equation giving

dm/dv = m * v/(c² * (1 - (v/c)²))

rearranging...

m * v * dv = c² * (1 - (v/c)²) * dm

Simplifying...


m * v * dv = c² * dm - v² * dm

Comment: Lorentz had all the pieces.  Einstein put them together.  Lorentz is known to physicists.  Einstein is known to everybody.  It couldn't have made Lorentz very happy that he was one high school physics problem away from figuring this out.

Einstein

We can plug the above m*v*dv term into Faraday's Energy formula above, giving:

E = _s0∫^s1 c²*dm - v² * dm + v² * dm

or

E = _s0∫^s1 c² * dm

This brings us to our final step, requiring us to assume we can change the limits of the definite integral.

E = _m0∫^m1 c² * dm

solving...

E = (m₁ - m₀) * c²

If we convert all of "m", we can plug in the Lorentz transformation to get the formula on PBS' website.

Acknowledgements:

This problem was submitted by our father.