From my understanding, the specular color usually refers to the amount of light that is reflected when the surface is lit at normal incidence, and is noted $F_0$ or $R_0$. Moreover, for non metal materials, this value is calculated from the index of refraction of the material $n$ with the formula deduced from the Fresnel equations (in which 1 is the index of refraction of air or void): $$F_0 = \frac{(n - 1)^2}{(n + 1)^2} $$

According to this list of refractive indices on Wikipedia:

  • Solid materials typically have $n$ between 1.46 (fused quartz) and 2.69 (Moissanite). That would mean an $F_0$ between 0.03 and 0.21.
  • Liquids typically have $n$ between 1.33 (water) and 1.63 (carbon disulfide). That would mean an $F_0$ between 0.02 and 0.057, if I am not mistaken.
  • Gases typically have $n \approx 1$, so I guess we can safely assume an $F_0$ of 0.

All these values are very low; even crystals with high refractive indices like diamond ($F_0=0.17$) and moissanite ($F_0=0.21$) hardly exceed 20%. Yet most metals have $F_0$ values above 50%. Moreover, I have read multiple times that the formula mentioned above doesn't apply for metals (which can be easily confirmed by trying to use it and see completely wrong results), but I haven't found any further explanation.

What phenomenon explains this difference? How can I calculate $F_0$ for a metal (in particular if the medium it is in contact with has an IoR different than 1, like water)?

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    $\begingroup$ Doesn't this belong on Physics.SE? $\endgroup$ – Kyle Strand Jan 4 '17 at 18:31
  • $\begingroup$ Although many computer graphics questions involve physics, this is clearly a question looking for answers from computer graphics experts, and would not be a good fit on physics.SE. $\endgroup$ – trichoplax Jan 5 '17 at 10:40

Warning: I am not a physicist.

As Dan Hulme already explained, light can't travel through metals, so dealing with IOR is a lot more... complex. I will answer why that happens and how to calculate the reflection coefficient.

Explanation: Metals are filled with free electrons. Those electrons react to external fields and reposition until electrostatic equilibrium is met (the electric field is zero inside a conductor in electrostatic equilibrium). When electromagnetic waves hit a metallic surface, the free electrons move until the field that they create cancels the field of the incoming wave. Those electrons grouped together radiate a wave going out nearly the same as the one that hit the surface (i.e. with very low attenuation). How much is attenuated depends on the material properties.

From this explanation it is clear that conductivity is a key part of the high reflection coefficient on metals.

Math-wise, what you are missing is the complex index of refraction. On good conductors, such as metals, the complex term of the IOR is relevant and key for explaining this phenomena.

Practically, in rendering, achieving good metal parameters is more visual based. Artists adjust to their preference until it looks believable. Often you see a metalness parameter with specific handling for materials marked as metal.

Involved answer:

The complex index of refraction can be seen if we use Ohm's Law $J = \sigma \vec{E}$, which holds for conductors, on the Ampère-Maxwell equation using sinusoidal waves $\vec{E} = e^{i\omega t}$:

$$ \vec{\nabla} \times \vec{H} = \sigma\vec{E} + \frac{\partial \vec{D}}{\partial t} = \sigma \vec{E} + i\omega \epsilon \vec{E} $$ $$ = i\omega \left( \epsilon - i \frac{\sigma}{\omega} \right)\vec{E} = i \omega \epsilon_m\vec{E} $$

Note how we can interpret that whole term as a complex permittitivity $\epsilon_m$ and that $\sigma$ is the conductivity of the material.

This affects the IOR, as its definition is given by:

$$ n' = \sqrt{\frac{\epsilon_m}{\epsilon_0}} = \sqrt{\frac{\left(\epsilon - i \sigma / \omega\right)}{\epsilon_0}} = n_{\text{real}} + in_{\text{img}} $$

This shows how $n'$ can be complex. Also, note how very good conductors have a relevant complex term, as $\sigma \gg \epsilon_0 \omega$. Since it would take a lot, I will skip some steps with a reference, page 27: it can be shown that, since $\sigma \gg \epsilon_0\omega$, (we are dealing with $\omega$ of the visible spectrum): $$ n_{\text{real}} \approx n_{\text{img}} $$

and reflection from metals with normal incidence, from a medium with IOR $n$, given that $n' \gg n$:

$$ R = \frac{(n_{\text{real}} - n)^2 + n_{\text{img}}^2}{(n_{\text{real}} + n)^2 + n_{\text{img}}^2} \approx 1 $$

Agreeing that a good conductor is, in general, a good reflector.

The famous Introduction to Electrodynamics from Griffiths, pages 392-398, explains this and a lot more in a similar fashion.

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  • $\begingroup$ This is exactly the kind of detail I was hoping for when posting the question; thanks a lot! I've tried running the numbers again with the complex values, and I get results a lot more close to expected. So what you're describing about electrostatic equilibrium is basically $\nabla B=0$? $\endgroup$ – Julien Guertault Jan 6 '17 at 0:23

Look at the refractive index of several metals. They are all complex numbers and the math does work out when you put this into the fresnel equation: you get the expected high reflectivity at all angles.

There are also subtle color shifts because the index depends on wavelength. This is actually used in rendering but it is not common. The function is sometimes named "conductor fresnel" but it's really the same fresnel equation with complex numbers.

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The refractive index is related to the speed at which light travels through the medium, and only applies to materials which are at least partially transparent. Metals are electrically conductive, so they are opaque, so light can't travel through them at any speed, so they don't have a refractive index.

This is why Fresnel's law doesn't apply: it's for predicting what fraction of the incoming light is reflected vs. transmitted. No light is transmitted through the material: everything that isn't absorbed is reflected, either as a specular reflection (if the surface is smooth) or as diffuse scatter (if the surface is rough).

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    $\begingroup$ Strictly speaking, light does travel through metals but gets attenuated very rapidly, so that it doesn't penetrate more than a few microns below the surface. (Very thin layers of metal are partially transparent—the gold film on spacesuit helmets, for instance.) That's what the imaginary component of the IOR measures: the rate of attenuation. And Fresnel's law applies just as much to metals as to anything else, as seen in the other answers. $\endgroup$ – Nathan Reed Jan 4 '17 at 22:03

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