Comment by creamyhorror | Hacker News Reader

Comment by creamyhorror | Hacker News Reader

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> This electrical 'fluid' is an excellent reflector of light waves

Thanks, though I'm actually wondering about a deeper question: why is this so? Why do loose plasma-like electrons make metals reflective (or, roughly equivalently, how does adding extra electrons make a non-metal like water more reflective)?

To continue on this topic, the 'bands' I referred to are clusters of energy levels corresponding to the orbitals of the atoms. (I found this neat explanation: https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/... ). Apparently, metallic reflectivity comes down to the fact that metals have overlapping "valence" and "conduction" bands. This allows electrons in the lower "valence" band to be excited to the higher "conduction" band and then fall back down and emit photons - which is reflection.

Therefore, in the case of water with extra electrons, the additional electrons may be causing the valence and conduction bands to come closer together - maybe by filling up the valence band and bringing the top of the band closer to the conduction band?

Just wanted to speculate on the chemistry of this discovery a bit, I've always found the science interesting.

Because the electrons in the shared "metallic" electron cloud are in a myriad of quantum states, thus they are able to interact with myriad of "kind" of photons. (Of course the important thing is the wavelength = frequency = energy of the photon.) In particular for metals (like mirrors) where there's a lot of "free" electrons this interaction is through the EM field of all those electrons, not through "isolated" photon-atom pairs. The incoming photons are themselves traveling excitations of the EM field. The electrons - as charged particles - are nudged by the EM field. The moving charges then cause new EM waves, the outgoing photons.

In transparent stuff the electrons are bound to nuclei, and almost all of the photons in the human visible energy range just go through without interacting with the electrons. ( https://www.lehigh.edu/imi/teched/GlassProp/Slides/GlassProp... )

In opaque stuff, let's say a black cloth, the bound electrons are able to absorb almost all of visible light and then emit it as lower frequency (infrared) photons.

In mirrors and shiny stuff, the electrons are able to scatter most of the photons. (So the conversion to infrared photons - heat - is usually negligible.)

See this for more details: https://www.reddit.com/r/askscience/comments/1zi5hg/how_do_a...

Quite interesting - the details get ever more complex as you dig deeper. So the incoming photons interact with the oscillations of the free electrons (i.e. "plasmons"), causing new EM waves to be emitted. I take it that this interaction between photons and plasmons is equivalent to the changing of energy states of electrons and the release of photons as electrons fall to lower states.

So in non-metals, does the band gap between the valence and conduction bands get crossed by electrons excited by incoming photons? You said a black cloth is able to absorb and emit virtually all visible light, but as infrared, but how do band gaps work there?

I guess this is too advanced for a casual conversation with a layman like me, so thanks for the explanations. In the end, maybe it just boils down to "free electrons reflect light because that is what they do".

>So in non-metals, does the band gap between the valence and conduction bands get crossed by electrons excited by incoming photons?

It is an active area of research and they started exploring 'metallic ammonia' last year. recent paper looking at metallic ammonia using synchrotron x-ray PES in a refrigerated liquid microjet setup: https://science.sciencemag.org/content/368/6495/1086

They are currently known as 'metastable' nanostructures exhibiting plasmonic effects similar to noble nanoparticles.

Hmm, I was referring to typical non-metals rather than ones with excess electrons added, but thanks for the glimpse into current research!

I'm far from an expert either!

The important thing is that the sea of electrons in metals are able to interact with EM waves coherently, so phase is conserved, so the classic ray optics happens, an that's why we can see things in a mirror, while in case of a white non-mirror surface we have the absorption-emission, and in that case the emitted photons are going everywhere, so we just see a big homogeneous surface.

> I take it that this interaction between photons and plasmons is equivalent to the changing of energy states of electrons and the release of photons as electrons fall to lower states.

Yes, except "better", because it keeps the phase.

> band gap

So there's the optical band gap and the electric. (The electric is required for charge transport, so it's sometimes also called transport gap.)

So a cloth is an insulator, in an insulator the band gap is pretty big. The incoming photon does not lead to transport, but it leads to optical excitation. And then an emission of a photon with a lower energy.

In some instances the relatively high energy of the incoming photon gets converted through various radiationless processes into other forms of energy. For example molecular rotation and vibration (usually photons in the infrared and microwave range do this to matter). These also lead to at least one quantum number change [as every interaction in a quantum theory must], but in case of UV or visible light the electron getting excited and jumping to a higher atomic orbital shell is in an atom that's bound in a molecule. The incoming EM field wave is not high energy enough to "penetrate into" the atom, it interacts with the whole sum of the EM field of the molecule. (And the energy basically gets transferred into the molecular bond: https://en.wikipedia.org/wiki/Ultraviolet%E2%80%93visible_sp... )

See also: https://en.wikipedia.org/wiki/Hot_band

Thank you for the reply and for highlighting the importance of phase being preserved!

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