Why is volumetric emission proportional to absorption coefficient?

 Posts: 75
 Joined: Sun Aug 19, 2012 3:24 pm
 Contact:
Why is volumetric emission proportional to absorption coefficient?
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 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) , 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
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 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) , 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?
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 outscattering 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, precomputed and optionally preintegrated transfer function texture.
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 outscattering 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, precomputed and optionally preintegrated transfer function texture.

 Posts: 75
 Joined: Sun Aug 19, 2012 3:24 pm
 Contact:
Re: Why is volumetric emission proportional to absorption coefficient?
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:
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:
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).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:
Re: Why is volumetric emission proportional to absorption coefficient?
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.I think it is required that emitting particles should not 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 ("inscattering"), and conversely: how much light is absorbed due to local absorption phenomena at x, and how much light is scattered away from x ("outscattering", distributed w.r.t. the phase function).
See e.g. Hadwiger et al. "Realtime Volume Graphics", p. 6:
(https://doc.lagout.org/science/0_Comput ... aphics.pdf)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
Outscattering + heat dissipation etc. ==> total absorption at point x contributed to a viewing ray in direction w
Inscattering + 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 outscattering 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+inscattering term: http://www.vis.unistuttgart.de/~amentm ... eprint.pdf (cf. Eq. 3 on page 3).
Hope I'm not misreading your question?

 Posts: 75
 Joined: Sun Aug 19, 2012 3:24 pm
 Contact:
Re: Why is volumetric emission proportional to absorption coefficient?
My current thinking process when reading the paper you lastly mentioned is like following:
0.
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.
0.
 Yes, I know.However, the mental model behind radiance transfer is not one that considers the interaction of individual particles.
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?
I found an interesting lecture script. http://www.ita.uniheidelberg.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 densityrelated 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.
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 densityrelated 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.
Re: Why is volumetric emission proportional to absorption coefficient?
That lecture script says:
"This is Kirchhoff’s law.It says that a medium in thermal equilibrium can have any emissivity jν and extinction αν, as long as their ratio is the Planck function."
Which sounds like they really CAN'T have any emissivity and extinction, but have to have them in a specific ratio.
For example, for green light of wavelength 570 nm and 2000 degrees K, that ratio is (from the Planck function) 6537. So the extinction is relatively small in comparison.
Later it says "If the temperature is constant along the ray, then the intensity will indeed exponentially approach [the Planck function]".
Anyway, the real reason I am replying is so I can share this video of a "black" flame.
The flame emits light but has no shadow (it seems fires don't have shadows), but can be made to have one and even appear black under singlefrequency lighting:
https://www.youtube.com/watch?v=5ZNNDA2WUSU
This seems to contradict the notion that media has to absorb light in order to emit it.... unless the amount absorbed is very tiny, as suggested by the lecture.
"This is Kirchhoff’s law.It says that a medium in thermal equilibrium can have any emissivity jν and extinction αν, as long as their ratio is the Planck function."
Which sounds like they really CAN'T have any emissivity and extinction, but have to have them in a specific ratio.
For example, for green light of wavelength 570 nm and 2000 degrees K, that ratio is (from the Planck function) 6537. So the extinction is relatively small in comparison.
Later it says "If the temperature is constant along the ray, then the intensity will indeed exponentially approach [the Planck function]".
Anyway, the real reason I am replying is so I can share this video of a "black" flame.
The flame emits light but has no shadow (it seems fires don't have shadows), but can be made to have one and even appear black under singlefrequency lighting:
https://www.youtube.com/watch?v=5ZNNDA2WUSU
This seems to contradict the notion that media has to absorb light in order to emit it.... unless the amount absorbed is very tiny, as suggested by the lecture.
Re: Why is volumetric emission proportional to absorption coefficient?
Ha! Now this comes a bit late, but I appreciate you posting this experiment. It is very cool indeed.
I think, in contrast to the assumptions in that part of the lecture, the lamp is not a black body. At least, obviously, its emission spectrum is does not follow Planck's law. Please don't ask when in reality the idealization as black body is justified ...
Somewhere I read that good emitters are generally also good absorbers, in the sense of material properties. The experiment displays this very well since the Sodium absorbs a lot of the light, whereas normal air and normal flame do not.
I think, in contrast to the assumptions in that part of the lecture, the lamp is not a black body. At least, obviously, its emission spectrum is does not follow Planck's law. Please don't ask when in reality the idealization as black body is justified ...
Somewhere I read that good emitters are generally also good absorbers, in the sense of material properties. The experiment displays this very well since the Sodium absorbs a lot of the light, whereas normal air and normal flame do not.