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I'm a 3D artist trying to learn the basic (or perhaps even intermediate) level of physics of photorealistic rendering. But most artist and tutorials on the internet have little to no clue of the physics behind all the settings they tweaking to get a good rendering. I want to understand what it is I'm actually doing :)

So my question is: Why don't dielectric materials have coloured reflections like conductors? Metals like copper tint the reflection with a red-ish color, but a red rubber ball does not, why is that?

In this image you can see the reflection of the white and blue pencil is being tinted slightly yellow in the brass material.

enter image description here

I was once told that the white reflection was caused by the reflection of the coating material, and the coloured diffuse reflection was the reflection of the pigments inside. That makes sense for like oil based paint were you actually mix pigments in oil, and the oil becomes the white reflective coating layer, and the same thing goes for plastics I guess. The plastic is actually white or transparent but they add all the colour pigments in it.

In this image you can see the white reflections in the blue plastic, the mirror image in the plastic is NOT being tinted blue.

enter image description here

But what about, let's say an red apple? Is the apple peel some sort of oily coating layer filled with red pigments? Or is it just made out of "apple peel molecules" and it is simply the physical/chemical properties of these atomic/molecular bonds that effect the light in that way?

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  • $\begingroup$ A red rubber ball does colour the reflected light. It only reflects less light than a shiny copper surface and the reflection is diffuse so there will be no distinct spot of reflected light. Nevertheless the effect can easily be observed. $\endgroup$
    – Sebastian Riese
    Commented Oct 19, 2015 at 23:44
  • $\begingroup$ But if you look at a shiny plastic surface you see a mirror image. But if you look at a shiny coloured metallic surface the mirror image is tinted in the colour of the metal. Why does this happen for metals but not for plastics? $\endgroup$
    – Kristoffer Helander
    Commented Oct 20, 2015 at 6:56
  • $\begingroup$ Oh, now I get it. I'll try to write an answer. $\endgroup$
    – Sebastian Riese
    Commented Oct 20, 2015 at 7:31
  • $\begingroup$ I've updated the original post with some more details. $\endgroup$
    – Kristoffer Helander
    Commented Oct 20, 2015 at 7:32

4 Answers 4

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The short answer is: Because metals are really absorptive (which comes from the fact that the nearly free electrons in the metal follow the oscillations of the radiation thereby depleting its energy), but some only in part of the visible range.

The reflectivity of a material is given by the Fresnel equations in terms of the index of refraction. They describe the angle dependency and further tell you that the higher the difference in the index of refraction the more light will be reflected at the interface.

It is important to understand, that the index of refraction in general can be a complex number. The imaginary part of the index of refraction describes the absorption of the material, while the (well known) real part describes the usual "optical density" causing refraction. So there are two possibilities for a material to reflect strongly: Either because it has a large real part of the index of refraction (like diamond) or because it absorbs light strongly (like metals). The latter effect can also be seen with lines written using a dark overhead transparency marker: they reflect in the colour range that does not pass through.

So, the reflection on the surface of metals is mainly due to the imaginary part of the index of refraction (that is, the absorptivity). For coloured metals like copper or gold the so called "plasma frequency" of the metal above which the metal begins to loose its strong absorptivity is in the visible range (or in the near UV). Therefore such metals only reflect a portion of the spectrum, well you get a tinted reflection.

The other materials (plastic, glass, apples) have one thing in common: they have a relatively low absorptivity (while for metals the wave only enters a few nanometers, the other materials range from transparent to waves entering at least several micrometers; the absorption caused by pigments in the material is typically much weaker than the one in metals). This means that the reflection is caused by the change of the real part of the index of refraction. As most materials are only slightly dispersive in the optical range, this means that all frequencies are reflected more or less equally, therefore the reflection is not tinted.

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  • $\begingroup$ Wow! That's an amazing reply - thanks! It did however raise more questions then answers because I only have limited understanding of physics beyond high school level. But you gave me a lot of things to search for, so I will read up on this and get back to you to see if I have understood it correctly. Again thanks! $\endgroup$
    – Kristoffer Helander
    Commented Oct 20, 2015 at 15:01
  • $\begingroup$ One very confusing part of all this is that you say that metals are really absorptive, that sounds backwards, if the metall absorbs light very well, then how come there is a reflection at all - shouldn't it just be a very diffuse and dark surface? Like a matte black object, kind of like a piece of charcoal? $\endgroup$
    – Kristoffer Helander
    Commented Oct 20, 2015 at 15:22
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    $\begingroup$ While this seems paradoxical at first, a perfectly absorbing surface will reflect all light. Perhaps absorptive is not the perfect word. The point is, that metals attenuate the electromagnetic wave within a few nanometers. Not much of the energy of the incident wave actually reaches deep into to metal. This is similar to a surface with the boundary condition that the field vanish there (for which we know there is total reflection). Materials that are black attenuate the light more slowly or have a rough surface, so that the reflection is suppressed by multiple reflections. $\endgroup$
    – Sebastian Riese
    Commented Oct 20, 2015 at 15:40
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Sebastian Riese's answer is very insightful and detailed, but it gets one thing wrong, and for sake of correctness and completeness (and since your questions specifically refers to copper), I think the explanation is better suited as an answer than a comment.

Specifically, Sebastians's answer reads

For coloured metals like copper or gold the so called "plasma frequency" of the metal above which the metal begins to loose its strong absorptivity is in the visible range (or in the near UV). Therefore such metals only reflect a portion of the spectrum, well you get a tinted reflection.

This is not correct, since the plasma frequencies of coloured metals such as copper ($\nu = 2.61$ THz) and gold ($\nu = 2.18$ THz) lie entirely in the UV, and one would therefore expect them to be completely reflective at the visible frequencies.

The reason for their coloured reflection is due to Interband Absorption. As Sebastian said, when metals are hit by EM radiation at frequencies below the plasma frequency, incoming radiation is almost completely reflected, but penetrates slightly into the metal (see Skin Effect). In the case of most metals (those that appear as having the typical "metallic" color) in this small length absorption by the free electrons of the metal occur, which eliminates the small fraction of the incoming wave that wasn't immediately reflected, so that it doesn't penetrate into the medium.

But for some metals such as gold and copper, some visible frequencies can get absorbed strongly in that little penetration length, because they have the correct energy to excite electrons in the metal from one band to another. This causes some visible frequencies to be missing in the reflected light, which therefore appears coloured.

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  • $\begingroup$ What's the difference between "absorbed by free electrons" and "absorbed and excited"? What happens to the light? Does the "absorption by free electrons" cause the almost complete reflection, minus the "absorbed and excited wavelengths"? Or is the "absorbed by free electrons" totally lost, plus reflected "absorbed and excited wavelengths"? $\endgroup$
    – user19087
    Commented Aug 28, 2021 at 14:34
  • $\begingroup$ Your insight is very interesting, thanks for the correction – I've definitely seen the plasma frequency explanation for the colour of gold and copper in a lot of books and other didactic material. Can you point me to a source where there's numbers for the plasma frequencies and where the intraband contributions are quantified (googling for this was surprisingly ineffective)? And I think there's a typo in your plasma frequency numbers, 2 THz corresponds to a wavelength of about 150 µm. $\endgroup$
    – Sebastian Riese
    Commented Jan 23, 2022 at 15:37
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Plastics and biologic materials are very similar to the paint example you gave: they are basically transparent (well, translucent), and their color come from pigments inside. So the first air/material dioptre the light encounter is (mostly) color-neutral (at least for smooth surfaces), and specular is caused by the Fresnel term on this surface. Diffuse color really comes from the light entering the medium and diffusing once or more on pigments (+ligand, possibly).

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It seems like the answers here are a bit "too clever" - one can simply look into the Photoelectric Effect in order to understand why Metals are Much more reflective than any other dielectric material.

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    $\begingroup$ Could you expand on why you think the photoelectric effect helps? $\endgroup$
    – Kyle Kanos
    Commented Feb 13, 2019 at 10:56

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