<p>This is an interesting question. BTW, I am no expert in  the fundamental theory of ray tracing, so this answer can be flawed.</p>
<p>As far as I understand, I believe Eric Veach discusses the thermodynamic equilibrium only for BSDF local transport purpose, and thermodynamic equilibrium can not be reached in the scene to be rendered: we have a <strong>source</strong> in the scene, and if the scene is a thermal enclosure, the temperature should go up continuously, since the heat can not be transported elsewhere and the energy won't disappear itself.</p>
<p>In this sense, we can never have the thermodynamic equilibrium since the source will eventually drive the energy within the enclosure to... you know, nuclear fusion of some sort. The thermodynamic equilibrium can be achieved, if the system relies solely on the emission and absorption of thermal radiation between objects and the media within. In rendering, the source continuously injects <em>new energy</em> into the system, constantly breaking the equilibrium, but yes, there should be a way for the system to <em>vent the heat within</em>, which is actually simple:</p>
<ul>
<li>We ignore the thermal radiation (most of the time, one exception is the blackbody related emission) and the temperature influences. So when the visible radiance is absorbed (for example, albedo (1, 0, 1) absorbs all the green light), the energy is just considered to have vanished (since it won't impact the rendering result).</li>
<li>For non-closed scene, radiation can simply escape from the system.</li>
</ul>
<p>So what kind of steady state can we achieve? First off, steady state does not mean the thermodynamic equilibrium. The radiance distribution is steady, when the energy gain (from the continuous emitter) and the energy loss (radiance absorption, escaped radiance) reach a steady state. A toy example: you have a water pipe that pumps 3L of water to you <strong>per second</strong>, the water is spilled out elsewhere 1L <strong>per second</strong>, so you will have a steady state of gaining water (2L <strong>per second</strong>). I use bold-face to emphasize that we only accumulate radiance with the time range of the shutter, so please don't misunderstand that hey, we have 2L per second so the water is getting more and more and therefore there is nothing steady --- this is not correct.</p>
<p>For rendering, there is indeed <a href="https://cs.dartmouth.edu/%7Ewjarosz/publications/jarabo14framework.html" rel="nofollow noreferrer">transient state rendering</a> though: when the shutter is just too fast that we can capture the photon from <strong>non-continuous sources</strong>. If you have more question on transient state rendering, we can have a discussion room for this since it is my current research interest. Hope this clarifies.</p>

CC BY-SA 4.0
2024-09-19T15:25:02.287Z
After pondering for a while on how to ensure that the absorbed energy equals the emitted energy, I understand what you mean. Because the light will continue to bounce, and each time it bounces, a portion of the energy will be absorbed. Eventually, the entire light will be absorbed, while the light source continues to emit new light, which achieves a dynamic balance!
ckf104
CC BY-SA 4.0
2024-09-18T05:00:57.780Z
2024-09-18T05:00:57.780Z
Enigmatisms
<p>in chapter 6 of <a href="https://graphics.stanford.edu/papers/veach_thesis/thesis.pdf" rel="nofollow noreferrer">Eric Veach's paper</a>, Veach has proved general reciprocity of BSDF by <em>Kirchhoff’s equilibrium radiance law</em> and <em>detailed balance</em>
<span class="math-container">$$
\frac{f_s(w_i \rightarrow w_o)}{\eta_o^2} = \frac{f_s(w_o \rightarrow w_i)}{\eta_i^2}
$$</span>
And <em>Kirchhoff’s equilibrium radiance law</em> states that expression
<span class="math-container">$$
\frac{L_v(x,w,v)}{\eta^2(x,w,v)}
$$</span>
should be constant within a scene with thermodynamic equilibrium.</p>
<p>I feel confused about this because there is no need to do any rendering if <span class="math-container">$L$</span> almost consant in thermodynamic equilibrium. The rendered image always has only one single color despite camera settings.</p>
<p>This contradiction may show that we actually render a scene which doesn't reach thermodynamic equilibrium. But it seems to me that we always the scene has reached a steady state before rendering, where distribution of radiance doesn't change over time. For example, in <a href="https://www.pbr-book.org/4ed/Radiometry,_Spectra,_and_Color/Radiometry" rel="nofollow noreferrer">section 4.1 of pbrt4</a>, it states</p>
<blockquote>
<p>Steady state: Light in the environment is assumed to have reached equilibrium, so its radiance distribution is not changing over time. This happens nearly instantaneously with light in realistic scenes, so it is not a limitation in practice. Note that phosphorescence also violates the steady-state assumption.</p>
</blockquote>
<p>In summary, I have two questions</p>
<ul>
<li>Will rendered image always have only one single color in thermodynamic equilibrium?</li>
<li>What does <em>steady state</em> mean? If it doesn't mean thermodynamic equilibrium, why doesn't radiance distribution change over time?</li>
</ul>

<brdf><pbr><pbrt><bsdf>
The reciprocity of BSDF and equilibrium of rendered scene
ckf104
# The reciprocity of BSDF and equilibrium of rendered scene

(Asked by: username_0 on 2024-09-17)

in chapter 6 of [Eric Veach's paper](https://graphics.stanford.edu/papers/veach_thesis/thesis.pdf), Veach has proved general reciprocity of BSDF by *Kirchhoff’s equilibrium radiance law* and *detailed balance*
$$
\frac{f\_s(w\_i \rightarrow w\_o)}{\eta\_o^2} = \frac{f\_s(w\_o \rightarrow w\_i)}{\eta\_i^2}
$$
And *Kirchhoff’s equilibrium radiance law* states that expression
$$
\frac{L\_v(x,w,v)}{\eta^2(x,w,v)}
$$
should be constant within a scene with thermodynamic equilibrium.

I feel confused about this because there is no need to do any rendering if $L$ almost consant in thermodynamic equilibrium. The rendered image always has only one single color despite camera settings.

This contradiction may show that we actually render a scene which doesn't reach thermodynamic equilibrium. But it seems to me that we always the scene has reached a steady state before rendering, where distribution of radiance doesn't change over time. For example, in [section 4.1 of pbrt4](https://www.pbr-book.org/4ed/Radiometry,_Spectra,_and_Color/Radiometry), it states

> Steady state: Light in the environment is assumed to have reached equilibrium, so its radiance distribution is not changing over time. This happens nearly instantaneously with light in realistic scenes, so it is not a limitation in practice. Note that phosphorescence also violates the steady-state assumption.

In summary, I have two questions

* Will rendered image always have only one single color in thermodynamic equilibrium?
* What does *steady state* mean? If it doesn't mean thermodynamic equilibrium, why doesn't radiance distribution change over time?

**Tags:** brdf, pbr, pbrt, bsdf


---

## 1 answers

### Answer by username_1 on 2024-09-18. Score: 2 | Accepted answer

This is an interesting question. BTW, I am no expert in the fundamental theory of ray tracing, so this answer can be flawed.

As far as I understand, I believe Eric Veach discusses the thermodynamic equilibrium only for BSDF local transport purpose, and thermodynamic equilibrium can not be reached in the scene to be rendered: we have a **source** in the scene, and if the scene is a thermal enclosure, the temperature should go up continuously, since the heat can not be transported elsewhere and the energy won't disappear itself.

In this sense, we can never have the thermodynamic equilibrium since the source will eventually drive the energy within the enclosure to... you know, nuclear fusion of some sort. The thermodynamic equilibrium can be achieved, if the system relies solely on the emission and absorption of thermal radiation between objects and the media within. In rendering, the source continuously injects *new energy* into the system, constantly breaking the equilibrium, but yes, there should be a way for the system to *vent the heat within*, which is actually simple:

* We ignore the thermal radiation (most of the time, one exception is the blackbody related emission) and the temperature influences. So when the visible radiance is absorbed (for example, albedo (1, 0, 1) absorbs all the green light), the energy is just considered to have vanished (since it won't impact the rendering result).
* For non-closed scene, radiation can simply escape from the system.

So what kind of steady state can we achieve? First off, steady state does not mean the thermodynamic equilibrium. The radiance distribution is steady, when the energy gain (from the continuous emitter) and the energy loss (radiance absorption, escaped radiance) reach a steady state. A toy example: you have a water pipe that pumps 3L of water to you **per second**, the water is spilled out elsewhere 1L **per second**, so you will have a steady state of gaining water (2L **per second**). I use bold-face to emphasize that we only accumulate radiance with the time range of the shutter, so please don't misunderstand that hey, we have 2L per second so the water is getting more and more and therefore there is nothing steady --- this is not correct.

For rendering, there is indeed [transient state rendering](https://cs.dartmouth.edu/%7Ewjarosz/publications/jarabo14framework.html) though: when the shutter is just too fast that we can capture the photon from **non-continuous sources**. If you have more question on transient state rendering, we can have a discussion room for this since it is my current research interest. Hope this clarifies.

> **Comments:**
> **username_0**: After pondering for a while on how to ensure that the absorbed energy equals the emitted energy, I understand what you mean. Because the light will continue to bounce, and each time it bounces, a portion of the energy will be absorbed. Eventually, the entire light will be absorbed, while the light source continues to emit new light, which achieves a dynamic balance!  

---

