# Can one get E=mc^2 using classical EM theory?

snoopies622
I'm wondering if one can arrive at $E=mc^2$ using only the physics of the late 19th century, in the following way:

As light waves pass over an electrically charged particle, they push it in the direction of the wave motion, transferring both (kinetic) energy and momentum to the particle. Let's call these quantities $\Delta KE$ and $\Delta p$. So after the waves have passed over the particle, they've lost $\Delta$KE of energy and $\Delta$p of momentum.

As far as I know, the physics of the late 19th century regarded light waves as massless. But what if we plug in the classical Newtonian definition of momentum p=mv and — since the light waves won't change speed — say that the waves have "lost mass" according to $\Delta m = \Delta p / v$ where v is the speed of light $c=1/ \sqrt{\epsilon_o \mu_o}$.

Do the quantities $\Delta KE$ and $\Delta m$ work out such that $\Delta KE = c^2 \Delta m$? I'm guessing yes but frankly don't have the experience working out the mathematics of electromagnetism to find out for myself.

Thanks.

Mentor
Do the quantities $\Delta KE$ and $\Delta m$ work out such that $\Delta KE = c^2 \Delta m$? I'm guessing yes but frankly don't have the experience working out the mathematics of electromagnetism to find out for myself.
It won't work and you don't need any E&M to see the problem. Just try deriving ##\Delta{m}## from the classical expression for kinetic energy ##E_k=mv^2/2## and comparing with the value you get when you start, as you did, with the expression for momentum; you'll find that your model of light retaining its speed while transferring energy and momentum according to the classical formulas is internally inconsistent.

This inconsistency is just a symptom of the deeper problem: the classical expressions for energy and momentum are approximations based on the assumption that the relativistic effects that lead to ##E=mc^2## are negligible. They don't apply when, as in your example, that assumption is invalid.

DrStupid
Do the quantities $\Delta KE$ and $\Delta m$ work out such that $\Delta KE = c^2 \Delta m$?

Yes, they do. If you realize that a change of E results in a proportional change of m, than the work-energy-theorem leads to

$k \cdot dm: = dE = F \cdot ds = \dot p \cdot ds = m \cdot \dot v \cdot ds + v \cdot \dot m \cdot ds = m \cdot v \cdot dv + v^2 \cdot dm$

and the solution of the resulting differential equation

$\frac{{dm}}{m} = \frac{{v \cdot dv}}{{k - v^2 }}$

is

$m = \frac{{m_0 }}{{\sqrt {1 - \frac{{v^2 }}{k}} }}$

which has real solutions for

$v^2 < k$

only. That means that the proportionality factor k is the square of a maximum speed that can't be reached or even exceeded by the particle. As we know today, this speed limit is the speed c of plain light waves in vacuum.

But there are two problems:

1. The proportionality dE=c²·dm is not Einstein's mass-energy equivalence. The integration of

$dE = c^2 \cdot dm$

just results in the kinetic energy

$E - E_0 = c^2 \cdot m - c^2 \cdot m_0$

The equivalence

$E_0 = c^2 \cdot m_0$

between rest energy and rest mass remains unknown and can't be derived the way above.

2. The derivation works in a special frame of reference only. It fails as soon as you change to a moving frame of reference.

snoopies622
$$\Delta m c^2 = \frac{C}{2} U^2,$$