<p>To start, I highly suggest reading Naty Hoffman's Siggraph <a href="http://blog.selfshadow.com/publications/s2015-shading-course/hoffman/s2015_pbs_physics_math_slides.pdf" rel="noreferrer">presentation</a> covering the physics of rendering. That said, I will try to answer your specific questions, borrowing images from his presentation.</p>

<p>Looking at a single light particle hitting a point on the surface of a material, it can do 2 things: reflect, or refract. Reflected light will bounce away from the surface, similar to a mirror. Refracted light bounces around inside the material, and may exit the material some distance away from where it entered. Finally, every time the light interacts with the molecules of the material, it loses some energy. If it loses enough of its energy, we consider it to be fully absorbed.</p>

<p>To quote Naty, "Light is composed of electromagnetic waves. So the optical properties of a substance are closely linked to its electric properties." This is why we group materials as metals or non-metals.</p>

<p>Non metals will exhibit both reflection and refraction.
<a href="https://i.sstatic.net/ECwfp.png" rel="noreferrer"><img src="https://i.sstatic.net/ECwfp.png" alt="Non-Metals"></a></p>

<p>Metallic materials only have reflection. All refracted light is absorbed.
<a href="https://i.sstatic.net/JyBy3.png" rel="noreferrer"><img src="https://i.sstatic.net/JyBy3.png" alt="Metals"></a></p>

<p>It would be prohibitively expensive to try to model the light particle's interaction with the molecules of the material. We instead, make some assumptions and simplifications.</p>

<p><a href="https://i.sstatic.net/PXEQW.png" rel="noreferrer"><img src="https://i.sstatic.net/PXEQW.png" alt="Simplifying refraction"></a></p>

<p>If the pixel size or shading area is large compared to the entry-exit distances, we can make the assumption that the distances are effectively zero.
For convenience, we split the light interactions into two different terms. We call the surface reflection term "specular" and the term resulting from refraction, absorption, scattering, and re-refraction we call "diffuse".
<a href="https://i.sstatic.net/85Pmg.png" rel="noreferrer"><img src="https://i.sstatic.net/85Pmg.png" alt="Splitting into diffuse and specular"></a></p>

<p>However, this is a pretty large assumption. For most opaque materials, this assumption is ok and doesn't differ too much from real-life. However, for materials with any kind of transparency, the assumption fails. For example, milk, skin, soap, etc. </p>

<p>A material's observed color is the light that is not absorbed. This is a combination of both the reflected light, as well as any refracted light that exits the material. For example, a pure green material will absorb all light that is not green, so the only light to reach our eyes is the green light.</p>

<p>Therefore an artist models the color of a material by giving us the attenuation function for the material, ie how the light will be absorbed by the material. In our simplified diffuse/specular model, this can be represented by two colors, the diffuse color, and the specular color. Back before physically-based materials were used, the artist would arbitrarily choose each of these colors. However, it should seem obvious that these two colors should be related. This is where the albedo color comes in. For example, in UE4, they calculate diffuse and specular color as follows:</p>

<pre><code>DiffuseColor = AlbedoColor - AlbedoColor * Metallic;
SpecColor = lerp(0.08 * Specular.xxx, AlbedoColor, Metallic)
</code></pre>

<p>where Metallic is 0 for non-metals and 1 for metals. The 'Specular' parameter controls the specularity of an object (but it's usually a constant 0.5 for 99% of materials)</p>

CC BY-SA 3.0
2015-09-18T06:59:34.400Z
Why does the incident ray is split after reflection/refraction? If the light is a particle does that means that that this particle recursively split? And if the light is a wave does that means that is splits by frequency (but in this case why it splits after second/third/etc hit)?
nikitablack
CC BY-SA 3.0
2015-09-18T07:03:06.197Z
The particle does not split. Rather, the images show the potential paths it could take.
RichieSams
CC BY-SA 3.0
2015-09-18T09:09:51.863Z
Many particles will hit the (nearly) same spot from the (nearly) same angle. For every particle going out there is (usually) a particle that went in. That means that averaged out the *beam* of particles from a certain angle on a certain spot gets split up in several (a lot) reflections.
ratchet freak
CC BY-SA 3.0
2015-09-18T10:58:27.123Z
Great answer shedding light on most of my questions. Why is the specular part of non-metals colorless and not affected by the albedo? How and where does [polarization](http://filmicgames.com/archives/547) come into play?
David Kuri
CC BY-SA 3.0
2015-09-18T11:01:51.843Z
_"A material's observed color is the light that is not absorbed."_ At this point it might be worth referencing the [Are there common materials that aren't represented well by RGB?](https://computergraphics.stackexchange.com/questions/203/are-there-common-materials-that-arent-represented-well-by-rgb) discussion, as fluorescent materials spring to mind.
Simon F
CC BY-SA 3.0
2015-09-18T14:55:06.097Z
@DavidKuri I don't know the specific physical reason to *why* metals absorb all diffracted light and non-metals highly attenuate reflected light (IE. black diffuse for metals, and very small, monochrome specular for non-metals). Rather, it's just an observed phenomenon in nature. If anyone has a link or explanation, I would love to see it.
RichieSams
CC BY-SA 3.0
2015-09-18T15:01:41.967Z
@DavidKuri For polarization, you have to look into the other way of representing light, ie. waves. Polarization filters, for example in cameras, use long polymer strands  to block out certain wave orientations. A similar, but much less pronounced, process happens when the light wave interacts with all materials. This is one of the ways that light 'loses' energy.
RichieSams
CC BY-SA 3.0
2015-09-25T03:02:51.627Z
@RichieSams : the explanation is held in Maxwell 4 equations of electromagnetism. Because of conduction there can be no wave intensity at the surface of a metal. This is the reason why grids protects micro-wave doors.
v.oddou
CC BY-SA 3.0
2015-09-17T17:02:42.913Z
2015-09-17T21:52:08.073Z
2015-09-17T21:52:08.073Z
RichieSams
<p>I was actually wondering about exactly this a few days ago.  Not finding any resources within the graphics community, I actually walked over to the Physics department at my university and <em>asked</em>.</p>

<p>It turns out that there are a lot of lies we graphics people believe.</p>

<hr>

<p>First, when light hits a surface, the Fresnel equations apply.  The proportions of reflected/refracted light depend on them.  You probably knew this.</p>

<p><strong>There's no such thing as a "specular color"</strong></p>

<p>What you might not have known is that the Fresnel equations vary based on wavelength, because the <em>refractive index</em> varies based on wavelength.  The variation is relatively small for dielectrics (dispersion, anyone?), but can be enormous for metals (I presume this has to do with these materials' differing electric structures).</p>

<p>Therefore, the Fresnel reflection term varies by wavelength, and therefore <em>different wavelengths are reflected preferentially</em>.  Seen under broad-spectrum illumination, this is what leads to specular color.  But in particular, there is <em>no absorption</em> that magically happens at the surface (the other colors are just refracted).</p>

<p><strong>There's no such thing as "diffuse reflection"</strong></p>

<p>As Naty Hoffman says in the talk linked in the other answer, this is really an approximation to outscattered subsurface scattering.</p>

<p><strong>Metals DO transmit light</strong></p>

<p>Naty Hoffman is wrong (more precisely, simplifying).  Light does <em>not</em> get absorbed immediately by metals.  In fact, it will pass quite handily through materials several nanometers thick.  (For example, for gold, it takes 11.6633nm to attenuate 587.6nm light (yellow) by half.)</p>

<p>Absorption, as in dielectrics, is due to the Beer-Lambert Law.  For metals, the absorption coefficient is just much larger (α=4πκ/λ, where κ is the imaginary component of the refractive index (for metals ~0.5 and up), and λ is given in <em>meters</em>).</p>

<p>This transmission (or more accurately the <a href="https://en.wikipedia.org/wiki/Subsurface_scattering">SSS</a> it produces) is actually responsible for a significant portion of metals' colors (although it is true that metals' appearances are dominated by their specular).</p>

CC BY-SA 3.0
2015-09-21T06:07:42.403Z
Thank you! I only knew the simplifications. These extra details are awesome
RichieSams
CC BY-SA 3.0
2015-09-21T20:28:47.780Z
This is a fascinating answer. Could you clarify/link the acronym SSS please?
trichoplax is on Codidact now
CC BY-SA 3.0
2015-09-22T09:48:26.563Z
Thanks :) If you clarify it in the question, it will survive deletion of the comments (which are not guaranteed to be long lived). I've edited in a link and hover text which hopefully leaves your intended presentation intact.
trichoplax is on Codidact now
CC BY-SA 3.0
2015-09-22T13:38:11.347Z
While i do appreciate the pedantry of this answer. Sub surface scattering is considered a mm scale effect while its true that at molecular ranges everything passes the surface to some degree. But the base constraint is that we are generally counting mm scale effects and trying to abstract lower levels as statistical models. Hence micrometer is equal to immediately as most pixels see much greater area than this. Same applies to color which does not meaningfully exist in physics the same way as our eyes and brain precieve it
joojaa
CC BY-SA 3.0
2016-04-08T08:03:34.527Z
Do you have sources to recommend on the fact that metals do transmit light? What is the order of magnitude of the contribution to the perceived color?
Julien Guertault
CC BY-SA 3.0
2016-04-08T16:25:32.930Z
@JulienGuertault It's fairly standard knowledge, and Google returns useful results. E.g. [this](https://www.reddit.com/r/askscience/comments/37ktye/would_it_be_possible_to_make_a_thin_enough_sheet/crnsmgw) "Below 50 nm thickness, the light transmitted through gold is green (imagine all the colours that aren't reflected are coming through . . .)." Probably appearance depends. Metals reflect a lot, even with thin coatings. The amount refracted depends on the extinction coefficient and thickness.
geometrian
CC BY-SA 3.0
2015-09-21T04:47:42.927Z
2015-09-22T09:51:46.387Z
2015-09-22T09:51:46.387Z
geometrian
<p>The classical way of shading surfaces in real-time computer graphics is a combination of a (Lambertian) diffuse term and a specular term, most likely Phong or Blinn-Phong.</p>

<p><a href="https://i.sstatic.net/VOfxv.png" rel="noreferrer"><img src="https://i.sstatic.net/VOfxv.png" alt="Image from Wikipedia"></a></p>

<p>Now with the trend going towards physically-based rendering and thus material models in engines such as <a href="http://www.frostbite.com/wp-content/uploads/2014/11/course_notes_moving_frostbite_to_pbr.pdf" rel="noreferrer">Frostbite</a>, <a href="http://blog.selfshadow.com/publications/s2013-shading-course/karis/s2013_pbs_epic_slides.pdf" rel="noreferrer">Unreal Engine</a> or <a href="http://blog.selfshadow.com/publications/s2013-shading-course/karis/s2013_pbs_epic_slides.pdf" rel="noreferrer">Unity 3D</a> these BRDFs have changed. For example (a pretty universal one at that), the latest Unreal Engine still uses Lambertian diffuse, but in combination with the Cook-Torrance microfacet model for specular reflection (specifically using GGX/Trowbridge-Reitz and a modified Slick approximation for the Fresnel term). Furthermore, a 'Metalness' value is being used to distinguish between conductor and dielectric.</p>

<p>For dielectrics, diffuse is colored using the albedo of the material, while specular is always colorless. For metals, diffuse is not used and the specular term is multiplied with the albedo of the material.</p>

<p>Regarding real-world physical materials, does the strict separation between diffuse and specular exist and if so, where does it come from? Why is one colored while the other is not? Why do conductors behave differently?</p>

<physically-based><physics><brdf>
How physically-based is the diffuse and specular distinction?
David Kuri
# How physically-based is the diffuse and specular distinction?

(Asked by: username_0 on 2015-09-17)

The classical way of shading surfaces in real-time computer graphics is a combination of a (Lambertian) diffuse term and a specular term, most likely Phong or Blinn-Phong.

[![Image from Wikipedia](https://i.sstatic.net/VOfxv.png)](https://i.sstatic.net/VOfxv.png)

Now with the trend going towards physically-based rendering and thus material models in engines such as [Frostbite](http://www.frostbite.com/wp-content/uploads/2014/11/course_notes_moving_frostbite_to_pbr.pdf), [Unreal Engine](http://blog.selfshadow.com/publications/s2013-shading-course/karis/s2013_pbs_epic_slides.pdf) or [Unity 3D](http://blog.selfshadow.com/publications/s2013-shading-course/karis/s2013_pbs_epic_slides.pdf) these BRDFs have changed. For example (a pretty universal one at that), the latest Unreal Engine still uses Lambertian diffuse, but in combination with the Cook-Torrance microfacet model for specular reflection (specifically using GGX/Trowbridge-Reitz and a modified Slick approximation for the Fresnel term). Furthermore, a 'Metalness' value is being used to distinguish between conductor and dielectric.

For dielectrics, diffuse is colored using the albedo of the material, while specular is always colorless. For metals, diffuse is not used and the specular term is multiplied with the albedo of the material.

Regarding real-world physical materials, does the strict separation between diffuse and specular exist and if so, where does it come from? Why is one colored while the other is not? Why do conductors behave differently?

**Tags:** physically-based, physics, brdf


---

## 2 answers

### Answer by username_1 on 2015-09-17. Score: 30 | Accepted answer

To start, I highly suggest reading Naty Hoffman's Siggraph [presentation](http://blog.selfshadow.com/publications/s2015-shading-course/hoffman/s2015_pbs_physics_math_slides.pdf) covering the physics of rendering. That said, I will try to answer your specific questions, borrowing images from his presentation.

Looking at a single light particle hitting a point on the surface of a material, it can do 2 things: reflect, or refract. Reflected light will bounce away from the surface, similar to a mirror. Refracted light bounces around inside the material, and may exit the material some distance away from where it entered. Finally, every time the light interacts with the molecules of the material, it loses some energy. If it loses enough of its energy, we consider it to be fully absorbed.

To quote Naty, "Light is composed of electromagnetic waves. So the optical properties of a substance are closely linked to its electric properties." This is why we group materials as metals or non-metals.

Non metals will exhibit both reflection and refraction.
[![Non-Metals](https://i.sstatic.net/ECwfp.png)](https://i.sstatic.net/ECwfp.png)

Metallic materials only have reflection. All refracted light is absorbed.
[![Metals](https://i.sstatic.net/JyBy3.png)](https://i.sstatic.net/JyBy3.png)

It would be prohibitively expensive to try to model the light particle's interaction with the molecules of the material. We instead, make some assumptions and simplifications.

[![Simplifying refraction](https://i.sstatic.net/PXEQW.png)](https://i.sstatic.net/PXEQW.png)

If the pixel size or shading area is large compared to the entry-exit distances, we can make the assumption that the distances are effectively zero.
For convenience, we split the light interactions into two different terms. We call the surface reflection term "specular" and the term resulting from refraction, absorption, scattering, and re-refraction we call "diffuse".
[![Splitting into diffuse and specular](https://i.sstatic.net/85Pmg.png)](https://i.sstatic.net/85Pmg.png)

However, this is a pretty large assumption. For most opaque materials, this assumption is ok and doesn't differ too much from real-life. However, for materials with any kind of transparency, the assumption fails. For example, milk, skin, soap, etc.

A material's observed color is the light that is not absorbed. This is a combination of both the reflected light, as well as any refracted light that exits the material. For example, a pure green material will absorb all light that is not green, so the only light to reach our eyes is the green light.

Therefore an artist models the color of a material by giving us the attenuation function for the material, ie how the light will be absorbed by the material. In our simplified diffuse/specular model, this can be represented by two colors, the diffuse color, and the specular color. Back before physically-based materials were used, the artist would arbitrarily choose each of these colors. However, it should seem obvious that these two colors should be related. This is where the albedo color comes in. For example, in UE4, they calculate diffuse and specular color as follows:

```
DiffuseColor = AlbedoColor - AlbedoColor * Metallic;
SpecColor = lerp(0.08 * Specular.xxx, AlbedoColor, Metallic)

```

where Metallic is 0 for non-metals and 1 for metals. The 'Specular' parameter controls the specularity of an object (but it's usually a constant 0.5 for 99% of materials)

> **Comments:**
> **username_2**: Why does the incident ray is split after reflection/refraction? If the light is a particle does that means that that this particle recursively split? And if the light is a wave does that means that is splits by frequency (but in this case why it splits after second/third/etc hit)?  
> **username_1**: The particle does not split. Rather, the images show the potential paths it could take.  
> **username_3**: Many particles will hit the (nearly) same spot from the (nearly) same angle. For every particle going out there is (usually) a particle that went in. That means that averaged out the *beam* of particles from a certain angle on a certain spot gets split up in several (a lot) reflections.  
> **username_0**: Great answer shedding light on most of my questions. Why is the specular part of non-metals colorless and not affected by the albedo? How and where does [polarization](http://filmicgames.com/archives/547) come into play?  
> **username_4**: _"A material's observed color is the light that is not absorbed."_ At this point it might be worth referencing the [Are there common materials that aren't represented well by RGB?](https://computergraphics.stackexchange.com/questions/203/are-there-common-materials-that-arent-represented-well-by-rgb) discussion, as fluorescent materials spring to mind.  
> **username_1**: @username_0 I don't know the specific physical reason to *why* metals absorb all diffracted light and non-metals highly attenuate reflected light (IE. black diffuse for metals, and very small, monochrome specular for non-metals). Rather, it's just an observed phenomenon in nature. If anyone has a link or explanation, I would love to see it.  
> **username_1**: @username_0 For polarization, you have to look into the other way of representing light, ie. waves. Polarization filters, for example in cameras, use long polymer strands  to block out certain wave orientations. A similar, but much less pronounced, process happens when the light wave interacts with all materials. This is one of the ways that light 'loses' energy.  
> **username_5**: @username_1 : the explanation is held in Maxwell 4 equations of electromagnetism. Because of conduction there can be no wave intensity at the surface of a metal. This is the reason why grids protects micro-wave doors.  

---

### Answer by username_6 on 2015-09-21. Score: 23

I was actually wondering about exactly this a few days ago. Not finding any resources within the graphics community, I actually walked over to the Physics department at my university and *asked*.

It turns out that there are a lot of lies we graphics people believe.

---

First, when light hits a surface, the Fresnel equations apply. The proportions of reflected/refracted light depend on them. You probably knew this.

**There's no such thing as a "specular color"**

What you might not have known is that the Fresnel equations vary based on wavelength, because the *refractive index* varies based on wavelength. The variation is relatively small for dielectrics (dispersion, anyone?), but can be enormous for metals (I presume this has to do with these materials' differing electric structures).

Therefore, the Fresnel reflection term varies by wavelength, and therefore *different wavelengths are reflected preferentially*. Seen under broad-spectrum illumination, this is what leads to specular color. But in particular, there is *no absorption* that magically happens at the surface (the other colors are just refracted).

**There's no such thing as "diffuse reflection"**

As Naty Hoffman says in the talk linked in the other answer, this is really an approximation to outscattered subsurface scattering.

**Metals DO transmit light**

Naty Hoffman is wrong (more precisely, simplifying). Light does *not* get absorbed immediately by metals. In fact, it will pass quite handily through materials several nanometers thick. (For example, for gold, it takes 11.6633nm to attenuate 587.6nm light (yellow) by half.)

Absorption, as in dielectrics, is due to the Beer-Lambert Law. For metals, the absorption coefficient is just much larger (α=4πκ/λ, where κ is the imaginary component of the refractive index (for metals ~0.5 and up), and λ is given in *meters*).

This transmission (or more accurately the [SSS](https://en.wikipedia.org/wiki/Subsurface_scattering) it produces) is actually responsible for a significant portion of metals' colors (although it is true that metals' appearances are dominated by their specular).

> **Comments:**
> **username_1**: Thank you! I only knew the simplifications. These extra details are awesome  
> **username_7**: This is a fascinating answer. Could you clarify/link the acronym SSS please?  
> **username_7**: Thanks :) If you clarify it in the question, it will survive deletion of the comments (which are not guaranteed to be long lived). I've edited in a link and hover text which hopefully leaves your intended presentation intact.  
> **username_8**: While i do appreciate the pedantry of this answer. Sub surface scattering is considered a mm scale effect while its true that at molecular ranges everything passes the surface to some degree. But the base constraint is that we are generally counting mm scale effects and trying to abstract lower levels as statistical models. Hence micrometer is equal to immediately as most pixels see much greater area than this. Same applies to color which does not meaningfully exist in physics the same way as our eyes and brain precieve it  
> **username_9**: Do you have sources to recommend on the fact that metals do transmit light? What is the order of magnitude of the contribution to the perceived color?  
> **username_6**: @username_9 It's fairly standard knowledge, and Google returns useful results. E.g. [this](https://www.reddit.com/r/askscience/comments/37ktye/would_it_be_possible_to_make_a_thin_enough_sheet/crnsmgw) "Below 50 nm thickness, the light transmitted through gold is green (imagine all the colours that aren't reflected are coming through . . .)." Probably appearance depends. Metals reflect a lot, even with thin coatings. The amount refracted depends on the extinction coefficient and thickness.  

---

