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?

⁺ or rather, its electromagnetic spectral resonances


I think to first order you are looking for 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 the absorption as a function of wavelength, you're really describing the Raman spectrum of an electron, albeit at high energies.


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, but near the electron mass you have Compton scattering which is inelastic.

At much higher energies you get a roughly wavelength independent response once again, although I would guess other contributions can play a role as well depending on the exact energy scale (e.g. Schwinger limit for non-linear electrodynamics).

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's 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.

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

Which shows that the "color" of the electron depends on the exact wavelength of the illumination light and on the angle the observer is 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.

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  • $\begingroup$ Great, thanks! Yes, the Klein-Nishina cross-section looks like what I sought. I'll try and find some time to simulate the (angle-dependent) colour an electron would therefore "look" when illuminated slowly by a "single photon at a time" sun 🤔 $\endgroup$ – Tobias Kienzler Feb 25 '19 at 7:27

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.

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  • $\begingroup$ Isn't Thomson scattering only valid for low photon energies? $\endgroup$ – Tobias Kienzler Apr 19 '17 at 13:21
  • $\begingroup$ @TobiasKienzler - You are right. But he is asking about the low energy limit. $\endgroup$ – mmesser314 Apr 19 '17 at 13:22
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    $\begingroup$ He is me ;) And I'm also asking about the not-so-low-energy QED effects, though I'm not so sure Compton scattering is then already enough $\endgroup$ – Tobias Kienzler Apr 19 '17 at 13:24
  • $\begingroup$ Ok, let me reformulate the "colour" part as "lowest energy approximation which is non-trivial" $\endgroup$ – Tobias Kienzler Apr 19 '17 at 13:29
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    $\begingroup$ Sorry, I missed that. You are indeed you. I couldn't say anything about non-linear QED effects. Except that I wouldn't expose my eyes to light at high energies. $\endgroup$ – mmesser314 Apr 19 '17 at 13:31

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