<p>Recently I've been digging deeper into the volumetric rendering (volumetric path tracing, to be specific). I already know that for surface rendering, there are <strong>two commonly used measures</strong> <a href="https://graphics.stanford.edu/courses/cs348b-03/papers/veach-chapter8.pdf" rel="nofollow noreferrer">(reference here)</a>: area product measure and solid angle measure, which can be related via this equation:</p>
<p><span class="math-container">$$
\frac{d \sigma(w_o)}{dA(\mathbf{x}')} = \frac{|\cos(\theta_i')|}{\Vert \mathbf{x} - \mathbf{x}'\Vert}
$$</span></p>
<p>which is physically intuitive: the geometric term (without visibility) here accounts for a change of measure --- how do we map an arbitrarily located differential area to a unit ball, as shown in the following figure. This is <strong>quite useful</strong> when we draw samples from another space other than the space we integrate.</p>
<p><a href="https://i.sstatic.net/y5WDb.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/y5WDb.jpg" alt="enter image description here" /></a></p>
<p>My questions are:</p>
<ul>
<li><p>What is the measure for integration in the volumetric medium? Is there any intuitive explanation resembling to the figure above? I personally find it difficult to get everything straight.</p>
</li>
<li><p>What is the relation for this &quot;volumetric measure&quot; with other measures, mathematically? I don't think I've seen much difference in the renderer (a missing cosine term, tops).</p>
</li>
<li><p>Say, during distance sampling (mean-free-path sampling), I devised a new approach but it does not sample in the original space. Is there any extra stuff I need to put into the rendering equation? Like a Jacobian? For example, I get a mean-free-path path sample but actually the sample is originated from another space, then can I directly use NEE? I don't feel right about <span class="math-container">$I / d^2$</span> term for sampling a light source (diffusive area or point), which should be correct if I use samples from the original space.</p>
</li>
</ul>

CC BY-SA 4.0
2023-07-22T05:45:24.300Z
P.S [reference PBR-book 16.3 Probability Densities](https://www.pbr-book.org/3ed-2018/Light_Transport_III_Bidirectional_Methods/Bidirectional_Path_Tracing#x1-ProbabilityDensities): Recall from Section 5.5 that the Jacobian of the mapping from solid angles to surface area involves the inverse squared distance and the cosine of angle between the geometric normal at next and wn (assuming next is a surface vertex—**if it is a point in a participating medium**, there is no cosine term). So why the measure looks so similar to area product measure? Surface measure is only defined on 2D space...?
Enigmatisms
CC BY-SA 4.0
2023-07-22T12:14:16.263Z
You can find the geometric term here: https://www.pbr-book.org/3ed-2018/Light_Transport_II_Volume_Rendering/The_Equation_of_Transfer
The change of variables yields only a $1/r^2$ in the purely volumetric case which is due to going from spherical coordinates to cartesian coordiates.
lightxbulb
CC BY-SA 4.0
2023-07-24T17:20:27.903Z
Thanks for this reference. I've read this before, but still I don't find volumetric measure so intuitive. Since I can easily relate the area product measure and the solid angle measure via the mathematical relationship between a arbitary differential area and one small piece of area on a unit sphere, for volumetric measure... I can't find a simple way to visualize the conversion and the relationship.
Enigmatisms
CC BY-SA 4.0
2023-07-25T13:57:00.483Z
The hemisphere to scene surface is less intuitive mathematically tbh. There you map from a differential element on the hemisphere around the normal at point $x$ to its projection on the scene surfaces around point $y=r(x,\omega)$. For volumetric you just take a step along direction $\omega$, so $x$ and $y$ are both volume elements in space. I think you can even interpret the $1/r^2$ as an inverse square law here. Imagine a "cone" (with curved base) cutout of the unit ball, then grow this cone outwards, it grows as $r^3$. Taking the differential yields $3r^2$. Does this motivate it better?
lightxbulb
CC BY-SA 4.0
2023-07-25T16:03:50.483Z
Yeah, that's better. Maybe the reason I think volumetric measure is less intuitive is that I found it difficult to visualize. Thanks a lot.
Enigmatisms
<mathematics><importance-sampling><volume-rendering><integral>
Measure for volumetric rendering equation
Enigmatisms
# Measure for volumetric rendering equation

(Asked by: username_0 on 2023-07-22)

Recently I've been digging deeper into the volumetric rendering (volumetric path tracing, to be specific). I already know that for surface rendering, there are **two commonly used measures** [(reference here)](https://graphics.stanford.edu/courses/cs348b-03/papers/veach-chapter8.pdf): area product measure and solid angle measure, which can be related via this equation:

$$
\frac{d \sigma(w\_o)}{dA(\mathbf{x}')} = \frac{|\cos(\theta\_i')|}{\Vert \mathbf{x} - \mathbf{x}'\Vert}
$$

which is physically intuitive: the geometric term (without visibility) here accounts for a change of measure --- how do we map an arbitrarily located differential area to a unit ball, as shown in the following figure. This is **quite useful** when we draw samples from another space other than the space we integrate.

[![enter image description here](https://i.sstatic.net/y5WDb.jpg)](https://i.sstatic.net/y5WDb.jpg)

My questions are:

* What is the measure for integration in the volumetric medium? Is there any intuitive explanation resembling to the figure above? I personally find it difficult to get everything straight.
* What is the relation for this "volumetric measure" with other measures, mathematically? I don't think I've seen much difference in the renderer (a missing cosine term, tops).
* Say, during distance sampling (mean-free-path sampling), I devised a new approach but it does not sample in the original space. Is there any extra stuff I need to put into the rendering equation? Like a Jacobian? For example, I get a mean-free-path path sample but actually the sample is originated from another space, then can I directly use NEE? I don't feel right about $I / d^2$ term for sampling a light source (diffusive area or point), which should be correct if I use samples from the original space.

**Tags:** mathematics, importance-sampling, volume-rendering, integral

> **Comments:**
> **username_0**: P.S [reference PBR-book 16.3 Probability Densities](https://www.pbr-book.org/3ed-2018/Light_Transport_III_Bidirectional_Methods/Bidirectional_Path_Tracing#x1-ProbabilityDensities): Recall from Section 5.5 that the Jacobian of the mapping from solid angles to surface area involves the inverse squared distance and the cosine of angle between the geometric normal at next and wn (assuming next is a surface vertex—**if it is a point in a participating medium**, there is no cosine term). So why the measure looks so similar to area product measure? Surface measure is only defined on 2D space...?  
> **username_1**: You can find the geometric term here: https://www.pbr-book.org/3ed-2018/Light_Transport_II_Volume_Rendering/The_Equation_of_Transfer The change of variables yields only a $1/r^2$ in the purely volumetric case which is due to going from spherical coordinates to cartesian coordiates.  
> **username_0**: Thanks for this reference. I've read this before, but still I don't find volumetric measure so intuitive. Since I can easily relate the area product measure and the solid angle measure via the mathematical relationship between a arbitary differential area and one small piece of area on a unit sphere, for volumetric measure... I can't find a simple way to visualize the conversion and the relationship.  
> **username_1**: The hemisphere to scene surface is less intuitive mathematically tbh. There you map from a differential element on the hemisphere around the normal at point $x$ to its projection on the scene surfaces around point $y=r(x,\omega)$. For volumetric you just take a step along direction $\omega$, so $x$ and $y$ are both volume elements in space. I think you can even interpret the $1/r^2$ as an inverse square law here. Imagine a "cone" (with curved base) cutout of the unit ball, then grow this cone outwards, it grows as $r^3$. Taking the differential yields $3r^2$. Does this motivate it better?  
> **username_0**: Yeah, that's better. Maybe the reason I think volumetric measure is less intuitive is that I found it difficult to visualize. Thanks a lot.  

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## 0 answers

