In the context of global illumination and the scattering equation, do we need to understand "light sources" as surfaces with a "purely emissive" material on which no scattering is present?

In contrast, any surface in the scene geometry could be equipped with a self-emissive material and hence being somehow a "light source" (it's kind of a philosophical question, I guess).

However, since we often talk about light-carrying paths of "length" $k$, where one endpoint is located on the camera and the other on a light source, I guess that's how we need to see it. Am I correct or is there something I'm missing?

  • $\begingroup$ It can have a brdf, there's no "no scattering is present" requirement. $\endgroup$ – lightxbulb Dec 26 '19 at 7:20
  • $\begingroup$ @lightxbulb But isn't it the case that we usually stop tracing after we hit a light source (when we start a random walk on the camera)? $\endgroup$ – 0xbadf00d Dec 26 '19 at 7:35
  • $\begingroup$ No, we usually do not, unless the brdf is perfectly absorbing. If you did this for light sources with a brdf that is not perfectly absorbing you will get a wrong solution of the rendering equation. Whether you decide to model all your light sources as perfectly absorbing or not in practice is a separate question that does not affect the theory. $\endgroup$ – lightxbulb Dec 26 '19 at 10:36
  • $\begingroup$ @lightxbulb Please take note of my comment below httpdigest's answer. $\endgroup$ – 0xbadf00d Dec 27 '19 at 13:59

Like @lightxbulb said, a light source (i.e. an emissive surface whose L_e term in the rendering equation is greater than zero) can also have a BRDF. People usually do not model light sources really accurately down to the Tungsten filament geometry inside of a glass-enclosed gas chamber for incandescent light bulbs. Such light sources usually are modelled as simple geometries, like spheres, capsules, lines, rectangles or more generic polygons, etc. So, most often, light sources are usually not exactly physically accurate anyways, because it would be too computationally expensive (imagine computing the intersection of a ray with the Tungsten filament in addition to account for a very little bit of volumetric scattering through the gas). Therefore, in the simplest cases, such an incandescent light source is down to a simple sphere anyways. In that case, it would make most sense to model this light source with a translucent BTDF (Bidirectional Transmittance Distribution Function), because most of the rays will probably not hit the filament but would go straight through the glass. What we would have to consider, however, is Fresnel and retro-reflection on the glass surface of the light bulb. But it all depends on how accurately you want to model your light sources. If, on the other hand, your light source is a milk glass light bulb, then you would probably need a BRDF + BTDF (or more generally a BSDF - bidirectional scattering distribution function) to account for the light that is reflected off the glass surface and the amount of light that is trasmitting through the glass.

In any case, stopping a ray at the light source (as you pointed out) would only be actually physically correct, if the surface of the light source is a black body not reflecting any light but only emitting light due to its temperature.

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  • $\begingroup$ Thank you for your answer. Well, I guess what's confusing me is that in the context of the path integral formulation of the light transport equation it seems like that we only consider the emitted radiance from the first vertex. To be precise, if we consider a path $(x_0,\ldots,x_k)$ of length $k$, where $x_k$ is on the camera, there's only the emitted radiance $L_e(x_k\to x_{k-1})$ occurring in the expansion of the recursive area form of the LTE. Or am I missing something? $\endgroup$ – 0xbadf00d Dec 27 '19 at 13:58
  • $\begingroup$ If you are referring to the path integral formulation as done in Veach's thesis on chapter 8, then a path starting from the eye and ending at a light source is not meant to say that we always stop there. If the light source is a surface reflecting or transmitting light, there will still be other paths along the same vertices as the first path from the eye to the light source but we do not consider the surface a light source then, but keep on going further along more path vertices until we hit another light source. And the final result is the accumulations of all such paths. $\endgroup$ – httpdigest Dec 27 '19 at 15:14
  • $\begingroup$ Yes, I know. I was actually talking about the measurement contribution function as defined on page 223 of Veach's thesis. $\endgroup$ – 0xbadf00d Dec 27 '19 at 15:34
  • $\begingroup$ I am not sure I still understand what your question is now. Want exactly are you asking? $\endgroup$ – httpdigest Dec 27 '19 at 15:47
  • $\begingroup$ @0xbadf00d The earlier $L_e$ are simply absorbed by the earlier terms of the Neumann expansion. When you do recursive bounces, and are not actually restarting the path for each path length you are really estimating the first $M$ terms of the Neumann expansion, where $M$ is the number of bounces you do. That is you actually evaluate $M$ estimators. Reusing the throughput you have computed on the path is just an optimization, one could technically regenerate paths from scratch for each path length. $\endgroup$ – lightxbulb Dec 27 '19 at 16:46

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