What colour is an electron?

Question

In order to explain the colour of an atom or molecule⁺, one considers the orbitals of the electrons surrounding it and the respective energy level differences. A single free electron does however not possess any "self-orbital", and thus no colour in this sense. But let's consider higher incoming photon energies - starting with 511 keV, an electron-positron pair can be temporarily created, so there is some interaction with electromagnetic radiation. Of course, 511 keV is far beyond what the human eye can perceive, so "colour" must be generalized into something like "spectral scattering cross section", and thus the title's question more correctly is:

Considering (non-linear) QED effects, what is the scattering spectrum of a single isolated electron in its rest-frame?

Of course, coming back to colour, one might then consider the question:

Given that spectrum, what is its lowest-but-non-trivial energy limit, and what (angle-dependent) colour does it provide when convoluted with sunlight (i.e. black body radiation at ~6000K) and the CIE standard observer, neglecting the under-saturation?

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⁺ or rather, its electromagnetic spectral resonances

Answer 1

I think to first order what you are looking for is the Klein Nishina cross-section. What is important here is that light can inelastically scatter from electrons, but can never be absorbed by a free electron. So instead of describing the color an electron by its absorption as a function of wavelength, you're really using its Raman spectrum, albeit at very high energies (x-rays and gamma rays instead of optical light).

https://en.m.wikipedia.org/wiki/Klein%E2%80%93Nishina_formula

In any case, the key quantity to consider is the ratio of the photon energy and electron mass of 512 keV. Well below this energy you have wavelength independent Thomson scattering (i.e. the electron is white and scatters all colors), but near the electron mass you have Compton scattering which is inelastic and angle-dependent.

At much higher energies compared to the electron rest mass, you get a roughly wavelength independent response once again. However, I would guess that other contributions will dominate the behavior depending on the exact energy scale (e.g. Schwinger limit for non-linear electrodynamics in the extreme limit).

At a more philosophical level, I think you can say the electron has a color only in the same sense that thin films or nanostructures have "color". In all these cases, it isn't the absorption of light giving the color, but instead it is the wave mechanics of the incoming and outgoing light.

Since the electron scattering cross section is purely given by the direction of the observer relative to the light source, it is very similar to something like the color of a wing of a butterfly, which also is geometry dependent.

Edit: For completeness, I will just repeat the exact formula of the scattered light energy vs incoming energy and angle below.

$$E_i - E_f = \Delta E = E_i\left(1-\frac{1}{1+\frac{E_i}{m_e c^2}\left(1-\mathrm{cos}(\theta)\right)}\right)$$

The above equation shows that the "color" of the electron depends on the exact wavelength of the illumination light and on the angle of the observer with the respect to the source. Moreover, the intensity of the inelastic light (i.e. light with different color than the illumination) is quite weak, and goes to zero away from the electron mass of 512 keV where you instead get wavelength-independent scattering again.

Answer 2

Mirrors are reflective because they are coated with a highly conductive metal. A perfect conductor is a perfect mirror. It reflects all wavelengths. It doesn't absorb or transmit any. A metal like copper is reddish because it isn't a good conductor at frequencies of bluer light.

Metals also constrain the position of electrons, and provide nuclei to keep the collection of electrons from repelling each other. The flat polished shape is needed to make incoming rays all reflect in the same direction, forming a good image. But a rough surface doesn't affect the spectrum.

So you might expect the same thing from a free electron. A free electron scatters light elastically. This is Thomson scattering. There is an angle dependence. But as this article shows, it is independent of wavelength.