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I started thinking about this in a discussion in comments.

One can start by thinking of the reflection of visible light by most metals as similar to the reflection of radio waves in that it's an interaction between an electromagnetic wave and an infinite half-plane of dense electron plasma. So long as the frequency is somewhat below the plasma frequency, this is a good starting point. Once in a while there are strong, obvious atomic effects such as gold metal being gold in color but many/most crystalline, amorphous and molten metals have high specular reflectivity and little obvious spectral variation, i.e. most of them look more or less "silvery".

Are there any crystallographic effects in the specular reflection from smooth metal surfaces that are visibly detectable either by looking at a reflection, or doing a simple experiment without an ellipsometer or other special equipment? Perhaps a simple polarizer film or plastic diffraction grating or something around a school science lab or home?

A visible "crystallographic effects" might be for example a difference in appearance or easy-to-observe optical behavior between

  1. two polished faces of a metal crystal having different crystallographic orientations,
  2. two polished faces of two metal samples of different crystallographic grain size, or one being totally amorphous
  3. a polished face of a metal crystal and the surface of the metal liquid at nearly the same temperature.

or it could be something visible that "happens" at a certain angle or under a certain condition or color of illumination for one sample but not another.

uhoh
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  • (Gold being gold color really has nothing to do with 'relativistic electrons' but that is another question - https://chemistry.stackexchange.com/questions/16633/why-is-gold-golden). A metal with an asymmetric crystal structure might have a shot. Certainly grain size has an effect. – Jon Custer May 28 '19 at 14:40
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    @JonCuster it's better to put a comment about 'relativistic electrons' on a post that mentions 'relativistic electrons'. Let's not start an unrelated conversation here, thanks! – uhoh May 28 '19 at 14:42
  • @uhoh - The "gold metal being gold in color link" mentions it. – mmesser314 May 28 '19 at 14:48
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    @mmesser314 so the comment belongs on the linked post, not here. That way people reading the post will also see the comment. That way the comment doesn't have to excuse itself by saying "but that is another question" – uhoh May 28 '19 at 14:49
  • Well, that comment was addressed at points made in your question that are (i) incorrect, and (ii) have nothing to do with the core of your question. Your move... – Jon Custer May 28 '19 at 14:55
  • @JonCuster that answer contains only one sentence with the word relativistic, and it is in a block quote of another answer in Physics SE, and a tiny part of the linked answer at that. Your comment here is two links away from where it belongs. The only thing I've said here is that it is an atomic effect which is correct according to your answer as well. – uhoh May 28 '19 at 15:02
  • The OP is very interested in contrasting the reflection of radio waves with the reflection of visible light from polished metals with typical grain size – Paul Young May 28 '19 at 16:33
  • @PaulYoung this question is only about visible light, it's the discussion that I link to in the first sentence that compares visible light with radio. – uhoh May 28 '19 at 16:51
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    @uhoh - I stand corrected ... I guess I was thinking about what was on my mind LOL – Paul Young May 28 '19 at 17:15

2 Answers2

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Maybe it partially depends on what is meant by "smooth". A coin, viewed in nondiffuse light, often displays a speckle pattern that shifts as the coin is tilted or as the point of view is shifted. Perhaps that pattern relates to the microcrystalline structure of the coin's metal. This paper relates to the topic.

S. McGrew
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  • Each of my examples mentioned polished surfaces, so polished is the kind of smooth that I am thinking of. It is true that chemical processes can reveal crystalline texture by producing surface topography, but that would no longer be smooth. – uhoh May 28 '19 at 14:45
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Sticking with looking for metals with lower symmetry than usual (i.e. not fcc or bcc), one comes to metals such as Antimony and Bismuth. Pulling up an early paper (Optical Properties of Antimony and Bismuth Crystals, A.P. Lenham et al., J. Optical Soc. Americal 55(9) 1072-1074 (1965)), they measured the optical properties parallel to and perpendicular to the basal planes.

In the figures, the difference in (n,k) in the visible looks pretty small, compared with the infrared. However, early on in the paper they note 'When viewed under a polarizing microscope, the anisotropy was pronounced.'

A question to ask, of course, is about the surface quality and the possibility of, say, oxidation. As noted in the paper,

In studies of the optical parameters of metals, great attention must be given to surface preparation. The absolute values of the optical constants are often found to be sensitive to the method of surface preparation, but the shapes of the dispersion curves and especially the energies of the absorption maxima and minima, and the sense of the anisotropy, are preserved for various sensible surface preparations.

Which, of course, gives no definitive description of what 'sensible' means. But, the bismuth was both hand polished (diamond grit) and electropolished, while the antimony was only hand polished. For thin oxides on the surface, there should be little influence on the relative optical properties, as noted above, because they will be optically thin.

Cadmium also shows reasonable anisotropy.

Continuing on, the authors have a variety of other papers on other materials. One you may or may not consider a metal, graphite, is covered in The Observatory (1966), showing a larger anisotropy there than for the more traditional metals.

Again, the key is to look for non-cubic metals. In addition the 'poorer' a metal likely the better. As a counterpoint, the noble metals (Cu, Ag, Au) are all fcc and all have nearly-free-electron like Fermi surfaces, so exhibit a very high degree of symmetry.

Jon Custer
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  • So how would I be able to see these effects? How does the data in these 1960's papers that I can't see yet indicate that there is a visible effect. In what way would it become visible? – uhoh May 28 '19 at 23:55