# Widom-Larsen 3: Ponderomotive (quiver) energy

In Widom-Larsen theory, it is argued that there is a region at the surface of a metal-hydrogen (or -deuterium) system where electrons have an insanely high mass, as much as 10.5MeV/c², because of the electromagnetic environment they are in.

In the the previous post, I argued that you should understand that “insanely high mass” implies (or maybe is equivalent to) “insanely high energy”. In this post I will explain what exactly this energy is, according to Widom-Larsen theory. It’s simpler than you might think!

The electron’s energy is ponderomotive energy (also called “quiver energy“). What does that mean? There is an oscillating electric field that pushes the electron back and forth. (“Quivering” motion.) If the electric field pushes hard enough, and the frequency is low enough, then the electron winds up with a lot of kinetic energy (on average).

With enough kinetic energy, it can smash into a proton at high speed to make a neutron (“electron capture”). That’s what happens according to Widom-Larsen theory.

The Widom-Larsen paper seems to describe something much more complicated and subtle than this simple classical physics effect. Indeed, they calculate it not with freshman-level physics but with quantum field theory.

Therefore, you may ask: How do I know that the electromagnetic interaction in Widom-Larsen theory is simply a classical ponderomotive effect, and not something more complicated and subtle? Two ways:

(1) In this later paper by Srivastava & Widom & Larsen, they describe the effect in exactly this way. (They don’t use the word “ponderomotive” or “quiver”, but it’s obviously the same thing).

our first task is to find a mechanism within a condensed matter system which can supply MeV scale energies to accelerate an electron to overcome the threshold barrier. As electrons are accelerated by an electric field through the equation $\dot{p} = (-e)\mathbf{E}$, let us assume that on a metal surface a sinusoidal electric field exists: $\mathbf{E} = \mathbf{E}_0 \cos(\Omega t)$ of frequency Ω. The average change in the momentum (Δp) is then easily obtained through $(\Delta p)^2 = e^2 \bar{E}^2/\Omega^2$. The average (squared) total energy $\bar{K}^2$ for an electron of rest mass m with an original small momentum p is given by

$\bar{K}^2 = (mc^2)^2 + (c\mathbf{p})^2 + \frac{e^2c^2 \bar{E}^2}{\Omega^2} = (c\mathbf{p})^2 + (mc^2)^2 \left[ 1 + \frac{\bar{E}^2}{\mathcal{E}^2} \right] \; ,$

where the relevant scale of the required electric field for the neutron production is set by $\mathcal{E} = (mc/\hbar)(\hbar \Omega/e)$ [25].

(2) Peter Hagelstein has calculated the ponderomotive energy for an arbitrary (non-monochromatic) electric field, and got the exact same expression as the original Widom-Larsen paper derived using quantum field theory.

Equation (19) in Hagelstein’s simple classical ponderomotive calculation:

$\text{mass enhancement factor} = \sqrt{1 + \frac{q^2}{m^2c^2} \text{Tr} \int_{-\infty}^\infty \frac{S_{EE}(\omega)}{\omega^2} \frac{d\omega}{2\pi}}$

Equation (14) in Widom-Larsen’s quantum-field-theory calculation:

$\text{mass enhancement factor} = \sqrt{1 + (\frac{e}{M_e c})^2\int_{-\infty}^\infty S_{EE}(\omega)\frac{d\omega}{\omega^2}}$

OK well there’s a 2π discrepency and some different notation, but this is obviously the same thing.

So I don’t think there is any doubt that the electron energy in Widom-Larsen theory is classical ponderomotive energy.

Incidentally, in my last post I said there should be a force, $F=-(\nabla m)c^2$, pushing electrons away from the high-mass regions. What is that force? Now we know: It’s the ponderomotive force! (This was previously pointed out by Vysotskii.)

Could an electron really have 10MeV of ponderomotive energy? In some systems, like maybe using the world’s most powerful laser, it is surely possible. But in cold-fusion experiments, it would be awfully hard to believe. You can read Hagelstein’s analysis for a description of what the electron trajectory would really look like, and what consequences that would have. The electron would move many microns back and forth at nearly the speed of light (Many microns? But the field is caused by a surface monolayer! It would only extend over a nanometer or two!) It would lose a huge amount of energy to radiation and other losses, so that sustaining the motion would require many gigawatts per square centimeter, etc. etc.