The converse of this question is perhaps "why are metals shiny". From what I understand, metals are covered by a sea of free electrons that oscillate in response to incident light/EM wave, and the oscillation is in turn associated with another EM wave travelling in the opposite direction from the first one. Hence the reflected light back to our eyes. (correct me if I am wrong. Also I am not too clear on the precise mechanisms of EM waves. For example, how does the electric field in incoming wave cause the electrons to move if there are no charged particles in the wave? This may be a self-explanatory question in that electric fields by definition affect charges, but how? I know this is a really basic concept, but I've never been able to get it. How is the oscillation of electrons associated with the new wave, what determines its direction? Does this oscillation use up some sort of energy, if so, where from?)

So with non-metals, in which electrons are fixed in place, is the dull appearance because of the lesser extent to which the electrons can oscillate? How does this relate to the intensity of the reflected light (maybe because of the smaller amplitude of the reflected wave?) Or, does the reflection occur via a completely different mechanism?

The other aspect is absorption. Then the question becomes, what is the difference between metals and non-metals that make the rate of absorption different, if the dullness is due to the greater absorption of light by materials such as cloth etc. ? What is meant by absorb more light anyway, which quantity does 'more' correspond to apart from the bright/dim perceived by our eyes?

There are so many questions here but any help is greatly appreciated!


"Shininess" is basically a property that is based on the proportion of specular reflection to diffuse reflection. Specular reflection basically means the directions of parallel rays relative to each other are preserved, while diffuse means they are disrupted. This is because what "shine" really is is an image (a virtual image, more precisely) of the light source (perhaps distorted), and thus the rays coming out must "look like they came from a light source" meaning that their geometric relationships must be suitably preserved so as to present the same pattern to the eye as a light source of that shape and behind the object would present its rays to the eye. Since the only real light source is the one shining on the object, this means the reflection must suitably preserve the relative geometry of the rays as they reached it having come from the light source illuminating.

This is not due to the reflectivity of materials, rather it is due to the smoothness of their surface. If you rough up metal, it won't be shiny. The rougher it is, the less "shine" it will have as the rays will be more disturbed from their original configuration when impinging upon the surface. Smooth plastic is shiny as well. Smooth coal is shiny! Look at these pieces of lignite - lowest grade of coal - that has been smoothed into nice spheres:


Cloth will never be shiny because as a fibrous material, its surface is innately irregular due to the fibers. Matte paint forms a rough microstructure on its surface when it dries - see:


(sorry I don't have source better than WP at the moment to keep things quick - if you can find one though, please suggest it.)

You also asked another question about how impinging EM waves cause charges to move when they "don't contain any charges". The answer to this is that charges don't respond to other charges, they respond to EM fields. The force law is $F = qE$ (yes I'm ignoring magnetic fields for simplicity), force is charge times electric field, not charge times another charge. Coulomb's law is not the most general force expression, and you should think of it as not $F = k\frac{Qq}{r^2}$ but rather $F = q\left(k \frac{Q}{r^2}\right)$ where the parenthesized term is an electric field produced by the "master" charge $Q$ and the charge of interest $q$ (which is which is arbitrary) is experiencing a force due to $Q$'s field. It is thus by producing an EM field that charges "respond to other charges", but they only directly respond to the field itself (which is very important to understand if you want to understand EM waves, since the same principle also applies to changes in the field: a change in the field only affects the field immediately next to that, not immediately a distant charge, and thus a wave propagates.). Since an EM wave is composed of electric and magnetic fields, when it impinges upon charges they will react to it because, as said, they respond to fields.

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  • $\begingroup$ What still troubles me is how rough metal surfaces can still be shiny like this.shutterstock.com/image-photo/… $\endgroup$ – ABC Dec 20 '17 at 4:49
  • $\begingroup$ @ABC : How rough are you thinking? Just a few dings or rough (like mild sandpaper, or the surface of paper or wood) to the touch? My glasses have a metal surface on them that's rough-rough (as in rough to the touch meaning you can feel friction with it) though not as rough as paper or wood. Its reflections are more fuzzy. The smooth surfaces reflect shinily. And my computer keyboard has plastic surfaces that are smooth. They are shiny even though they are not metal. $\endgroup$ – The_Sympathizer Dec 20 '17 at 4:54
  • $\begingroup$ because in a electroplating experiment I did, all the copper deposited as black soot, even though there's no chemical reason to suggest that it may have formed an impure substance. The explanation I found was this :""When you have too much current flow, what also happens is the moment that an ion of copper gets to the cathode, it is reduced instantaneously with no opportunity for proper crystal growth, and it forms a powder of tiny, non adherent individual specs of metal which appear black " but I still don't get the reason behind the appearance in terms of the atomic structure $\endgroup$ – ABC Dec 20 '17 at 4:57
  • $\begingroup$ Ah I see you have a picture. That surface has unevenness at different levels or scales. It looks like the small-scale bumps are rather smoother than the large-scale ones. Thus they will be shinier, but the shine will be broken up by the larger-scale irregularities, producing the effect you see. But already you can see it's less shiny than a perfectly smooth piece would be. $\endgroup$ – The_Sympathizer Dec 20 '17 at 5:00
  • $\begingroup$ @ABC unless that's what metal dust is meant to look like. but still, why would copper start absorbing all the frequencies of light? $\endgroup$ – ABC Dec 20 '17 at 5:08

The most important aspect for the difference between matt & shiny surfaces is the roughness of the surface; if it is rough it scatters light in many directions looking rough, whereas if it is smooth light is reflected in the same manner and so appearing shiny.

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