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

1. May 14, 2018

### 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.

2. May 14, 2018

### Staff: Mentor

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.

3. May 15, 2018

### DrStupid

Yes, they do. If you realise 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.

4. May 18, 2018

### vanhees71

You can get it for the contribution of electrmagnetic fields to the mass. The most simple example is a charged capacitor. The mass increase compared to the uncharged capacitor indeed turns out to be
$$\Delta m c^2 = \frac{C}{2} U^2,$$
where $C$ is the capacitance and $U$ the voltage on the capacitor. Note that you have to calculate this in the rest frame of the capacitor.

Of course there's only one sensible definition of mass in relativty, and that's the invariant mass of an object, but that should be clear within this forum by now!

5. May 18, 2018

### snoopies622

I’m fascinated by this idea that transverse waves – including waves with no mass at all in the classical sense – carry momentum. I once saw a derivation here using continuous Lagrangian mechanics that transverse mechanical waves carry momentum p=E/c, which (coincidentally?) is the same value as the momentum of a photon.