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Why is volumetric emission proportional to absorption coefficient?

Posted: Tue Nov 28, 2017 6:19 pm
by shocker_0x15
Hi,

I have a simple question as in the title.
Why is volumetric emission proportional to absorption coefficient?

I often see the volumetric emission term is written as
volumetric_emission_term_1.png
volumetric_emission_term_1.png (6.27 KiB) Viewed 362 times


I can also see another representation, volumetric emittance function (e.g. in Mark Pauly's thesis: Robust Monte Carlo Methods for Photorealistic Rendering of Volumetric Effects)
volumetric_emission_term_2.png
volumetric_emission_term_2.png (5.59 KiB) Viewed 362 times

, which has the unit of radiance divided by metre (that is W sr^-1 m^-3).

Do particles that emit light do not scatter light at all?

Thanks

Re: Why is volumetric emission proportional to absorption coefficient?

Posted: Tue Nov 28, 2017 9:17 pm
by szellmann
I don't quite understand the question: doesn't the first term refer to a particle density at an integration position x that emits radiance L_e in viewing direction w, and that the particle density absorbs radiance by w.r.t. sigma_a(x)? The emitted radiance is not proportional to the absorption, but is scaled by sigma_a(x). Let sigma_a := 1.0 be a constant for all x with respect to the density field, then your model only accounts for emission.

With L_b(xb,w)+\int_xb^x L_e(x,w) sigma_a(x), b being the position of a constantly radiating background light and \int_xb^x meaning integration over the viewing ray from the backlight to the integration position, you get the classical emission+absorption model that is e.g. used for interactive DVR in SciVis. In- and out-scattering can be incorporated in the equation. See Nelson Max's '95 paper on optical models for DVR for the specifics: https://www.cs.duke.edu/courses/cps296. ... dering.pdf

Also note that those models usually don't consider individual particles, but rather particle densities, and then derive coefficients e.g. by considering the projected area of all particles inside an infinitesimally flat cylinder projected on the cylinder cap.

Emission and absorption are sometimes expressed with a single coefficient in code for practical reasons, e.g. so that a single coefficient in [0..1] can be used to look up an RGBA tuple in a single, pre-computed and optionally pre-integrated transfer function texture.

Re: Why is volumetric emission proportional to absorption coefficient?

Posted: Mon Dec 04, 2017 2:31 pm
by shocker_0x15
Thanks for reply.

I can find the volumetric emission term which proportional to absorption coefficient for example in,
Jensen's Photon Mapping book, Spectral and Decomposition Tracking paper or Wojciech Jarosz's thesis.
In the Jarosz's thesis reference, there is the following sentence by the equation (4.12) in the page 60:
Media, such as fire, may also emit radiance, Le , by spontaneously converting other forms
of energy into visible light. This emission leads to a gain in radiance expressed as:
Jarosz_4.12.png
Jarosz_4.12.png (8.47 KiB) Viewed 247 times



I think it is required that emitting particles should not scatter light in order L_e^V to be represented as a decomposed form \sigma_a(x) L_e(x, w).

Re: Why is volumetric emission proportional to absorption coefficient?

Posted: Mon Dec 04, 2017 10:54 pm
by szellmann
I think it is required that emitting particles should not scatter light


I'm not quite sure if I understand how you come to this assumption and if I totally understand your question, but I don't see why particles that emit light shouldn't also scatter light.

However, the mental model behind radiance transfer is not one that considers the interaction of individual particles. The model rather derives the radiance in a density field due to emission, absorption, and scattering phenomena at certain sampling positions and in certain directions. So the question is: for position x, how much light is emitted by particles at or near x, how much light arrives there due to other particles scattering light into direction x ("in-scattering"), and conversely: how much light is absorbed due to local absorption phenomena at x, and how much light is scattered away from x ("out-scattering", distributed w.r.t. the phase function).

See e.g. Hadwiger et al. "Real-time Volume Graphics", p. 6:

Analogously, the total emission coefficient can be split into a source term q, which represents emission (e.g., from thermal excitation), and a scattering term


(https://doc.lagout.org/science/0_Comput ... aphics.pdf)

Out-scattering + heat dissipation etc. ==> total absorption at point x contributed to a viewing ray in direction w
In-scattering + emission ==> added radiance at point x along the viewing direction w

It is not about individual particles. The scattering equation is about the four effects contributing to the total radiance at a point x in direction w. There are no individual particles associated with the position x, you consider particle distributions and how they affect the radiance at x. The radiance increases if particles scatter light towards x, or if particles at (or near) x emit light. The radiance goes down due to absorption and out-scattering from the particle density at x. The point x is usually the sampling position that is encountered when marching a ray through the density field, and is not associated with individual particle positions.

I didn't find a more general source and am working with this paper anyway - the paper also shows the scattering equation and states that it has a combined emission+in-scattering term: http://www.vis.uni-stuttgart.de/~amentm ... eprint.pdf (cf. Eq. 3 on page 3).

Hope I'm not misreading your question?

Re: Why is volumetric emission proportional to absorption coefficient?

Posted: Fri Dec 08, 2017 6:34 pm
by shocker_0x15
My current thinking process when reading the paper you lastly mentioned is like following:
0.
However, the mental model behind radiance transfer is not one that considers the interaction of individual particles.

- Yes, I know.
1. eq. (1) says that contribution from source radiance Lm(x', w) is proportional to the extinction coefficient sigma_t(x').
- I can understand RTE of this form. Probability density with which light interact (one of scattering/absorption/emission) with medium at x' is proportional to particle density, that is sigma_t(x').
2. eq (3) says that once interaction happens, it is emission with probability (1 - \Lambda) and scattering with probability \Lambda.
- I can understand the latter because scattering albedo \Lambda is the probability that scattering happens out of some interaction. This is straightforward.
However I can't understand the former. The original question: Why is volumetric emission proportional to absorption coefficient?
I can understand absorption happens with the probability (1 - \Lambda), but cannot understand emission also happens with the probability (1 - \Lambda) :(

Now my question can be paraphrased as follows:
Shouldn't the probability emission happens be independent of absorption coefficient?


I'm sorry in case that the above explanation confuse you more and thank you for your kindness for detailed replying.

Re: Why is volumetric emission proportional to absorption coefficient?

Posted: Sat Dec 09, 2017 9:30 am
by dawelter
I found an interesting lecture script. http://www.ita.uni-heidelberg.de/~dulle ... pter_3.pdf
Section 3.3 Eq 3.9.

As I understand, ultimately it is a matter of definition motivated by thermodynamics of a special case.

I imagine that the same particles that block light along some beam also emit light of their own. So it makes sense that emission and absorption strength have a common density-related prefactor.

The reverse view from the point of importance being emitted into the scene seems more intuitive to me: Importance particle have a chance to interact with particles of the medium in proportion to their cross section. If they interact, the medium transfers energy to the imaging sensor.