# Mathematical Foundations of Radiative Transfer

I'm new to learning about ray tracing, but have been fairly confused by the mathematical foundations of it all. Specifically of radiometry/radiative transfer. I've internet searched a lot and skimmed a ton of references, but still very confused.

Here is a list of some of the things I have had to struggle a lot to understand:

1. Irradiance is a function of 3 variables $$E(p,n, \mathrm{in}/\mathrm{out})$$ and it's not really the derivative of anything with respect to anything else (in the function sense)

2. If $$p$$ is a point on an actual surface in a scene, irradiance doesn't make sense in directions that aren't the surface normal, except perhaps as limit.

3. Radiance is not meaningfully the derivative of irradiance (Wikipedia says it's the derivative w.r.t. étendue, which is a great way to get more confused).

4. Radiance can be defined without ever using the words "projected area", "solid angle" or "subtend".

5. You don't need to draw pictures with infinitesimals to think about radiance. I see infinitesimals in pictures a lot, e.g. Figures 1 and 2 Wikipedia's Lambert's cosine law article.

6. Radiance is preserved along paths (in vacuum). Is this an axiom of the theory or a theorem?

I've found the following approach more insightful than the usual presentations:

1. If you have an oriented surface, you can define the flux entering it $$\Phi(S)$$.

2. Irradiance is something that's integrated into flux: $$\Phi(S) = \int_SE(p, n_p, \mathrm{in}) \mathrm{d}S(p)$$

3. For two surfaces, we can define direct flux transport between them: the power that exist $$S_1$$ and directly enters $$S_2$$, $$\Phi(S_1, S_2)$$.

4. "Flux transport density" (not sure if correct term) is something that's integrated into flux transport. It is the function of 4 variables $$K(p_\text{source}, n_\text{source}, p_\text{target}, n_\text{target})$$ $$\Phi(S_1,S_2) = \int_{S_2} \int_{S_1} K(p, n_p, q, n_q) \mathrm{d}S_1(p) \mathrm{d}S_2(q)$$

5. "Light hull principle" consider two oriented disks $$D_p = D(p, n_p, r_p)$$ and $$D_q = D(q, n_q, r_q)$$ that are sufficiently small.

1. If $$(q-p) \cdot n_p \le 0$$ then $$\Phi(D_p, D_q) = 0$$

2. Taking the limit $$r_p \to 0$$ shows that $$(q-p) \cdot n_p$$ implies $$K(p, n_p, q, n_q) = 0$$

3. If $$(p-q) \cdot n_q \le 0$$ then $$\Phi(D_p, D_q) = 0$$

4. Taking the limit $$r_q \to 0$$ shows that $$(p-q) \cdot n_q$$ implies $$K(p, n_p, q, n_q) = 0$$

5. Let's assume $$(q-p) \cdot n_p > 0$$ and $$(p-q) \cdot n_q > 0$$ going forward. Maybe it's simplier to write $$|\cos(\theta_p)| < \pi/2$$, $$|\cos(\theta_q)| < \pi/2$$

6. Let's form the "light hull" of $$D_p$$ and $$D_q$$. This is a funny truncated cone like shape defined by $$H(D_p, D_q) = \{ t p' + (1-t) q' | p' \in D_p, q' \in D_q, t \in [0, 1] \}$$ It may later make sense to allow $$t$$ to go negative/greater than 1, extending the hull a bit.

7. Let $$a$$ be a point on the segment $$pq$$. Intersecting the light hull with a plane through $$a$$ gives as circular shaped section: $$H(a, n_a, D_p, D_q) = \{ b | (b - a) \cdot n_a = 0 \} \cap H(D_p, D_q)$$

8. In free space we have $$\Phi(D_p, D_q) = \Phi(D_p, H(a, n_a, D_p, D_q)) = \Phi(H(a, n_a, D_p, D_q), D_q)$$ I call this the light hull principle and is basically just geometry saying "if I put a transparent screen in the light's way, all the light will go through it".

9. Using the light hull principle, we see that we can replace $$D_q$$ with a similar disk-ish shape that's orthogonal to $$pq$$: $$\Phi(D_p, D_q) = \Phi(D_p, H(q, p-q, D_p, D_q))$$ Let's call this shape $$D_q'$$. For small $$r_p^2$$, $$r_q^2$$, the left hand side is $$\Phi(D_p, D_q) \approx K(p, n_p, q, n_q) \pi^2 r_p^2 r_q^2$$ Or more formally $$\frac{\partial^4 \Phi(D_p, D_q) }{(\partial r_p)^2 (\partial r_q)^2} \bigg|_{r_p = r_q = 0} = 4 \pi^2 K(p, n_p, q, n_q)$$ I don't want to go into too much detail here, but it's possible to prove for the right hand side that $$\frac{\partial^4 \Phi(D_p, H(q, p-q, D_p, D_q)) }{(\partial r_p)^2 (\partial r_q)^2} \bigg|_{r_p = r_q = 0} = 4 \pi^2 \cos(\theta_q) K\left(p, n_p, q, \frac{p-q}{||p-q||} \right)$$ It follows that $$K(p, n_p, q, n_q) = \cos(\theta_q) K\left(p, n_p, q, \frac{p-q}{||p-q||} \right)$$

10. Repeating the same argument at $$p$$ instead of $$q$$, we get $$K(p, n_p, q, n_q) = \cos(\theta_p) \cos(\theta_q) K\left(p, \frac{q-p}{||p-q||}, q, \frac{p-q}{||p-q||} \right)$$

11. Assuming now that the two disks are directly facing each other, i.e. $$n_p = \frac{q-p}{||p-q||} = -n_q$$, we can apply the light hull principle and take a limit/derivative as $$r_p, r_q \to 0$$ to show the inverse square law $$K(p, n_p, p+r n_p, -n_p) = \frac{1}{r^2} K(p, n_p, p+n_p, - n_p)$$

6. Let $$r = ||p-q||$$ and $$n = \frac{q-p}{r}$$. We now know that \begin{aligned} K(p, n_p, q, n_q) &= K(p, n, p+n, -n) \frac{\cos(\theta_p) \cos(\theta_q)}{r^2} \\ &=: L(p,n,\mathrm{out}) \frac{\cos(\theta_p) \cos(\theta_q)}{r^2} \\ &=: L(q,-n,\mathrm{in}) \frac{\cos(\theta_p) \cos(\theta_q)}{r^2} \end{aligned} Where $$L$$ is the defined by the equation above. We call $$L$$ radiance.

7. Taking the limit $$p \to q$$ shows that $$L(q, n, \mathrm{out}) = L(q, -n, \mathrm{in}) = L(p, n, \mathrm{out})$$. Therefore radiance is constant along rays (and $$K$$ is translation invariant).

The significance of this is that we have defined radiance and proved it invariant by using a simple geometric principle on finite (non-infinitesimal) surfaces and taking appropriate limits. It's a lot more work than saying irradiance over projected surface area times solid angle, but to me much more intuitive.

The light hull principle is an axiom/key "physical" fact, and everything else is a theorem.

My question is; is there a reference that deals with radiative transfer in general in a similar way as I've just dealt with radiance?

The closest thing I could find is Veach's PhD thesis where he defines some interesting concepts using measure theory in Chapter 3 and its appendix. The idea is that a scene is a system of $$N$$ non-interacting photons moving around. Radiometric quantities can then be defined by a snapshot of this system in time. I think it's a really cool idea and would love to see more of this. To me these are the true foundations. Sadly, I don't think Veach develops them in enough detail. (And I also don't see how a finite number of photons as Veach models them could create equilibrium in the sense that irradiance is independent of time)

Also, in Chapter 6, he presents a novel argument for the symmetry of the BRDF that's much more satisfying than "physical objects have symmetric BRDFs, now moving on..." that I've seen everywhere else.

I also found this paper by Lessig and Castro that says:

Radiative transfer describes the transport of electromagnetic energy in macroscopicenvironments, classically when polarization effects are neglected [37]. The theory originates in work by Bouguer [6, 7] and Lambert [22] in the 18thcentury where light intensity and its measurement were first studied systematically, cf. Fig. 1. In the 19thand early 20thcentury the theory was then extended to include transport and scattering effects [26, 9, 42, 43]. To this day, however, radiative transfer is a phenomenological theory with a mathematical formulation that still employs the concepts introduced by Lambert in the 18thcentury—and this despite the importance of the theory in a multitude of fields, such as medical imaging, remote sensing,computer graphics, atmospheric science, climate modelling, and astrophysics.

I don't know any physics so the paper is way over my head, but maybe it's the holy grail?

• "Sadly, I don't think Veach develops them in enough detail." - what details are you missing? You can define every quantity $\Phi$, $E$, $B$, $L$ as a density of your energy measurement function $Q$. If you want something even more involved: arxiv.org/pdf/1205.4447.pdf Jan 6 at 16:45
• @lightxbulb thanks very much, this seems like a great reference! Re: Veach, he defines these cool measures and goes on to not use them? He doesn't use them do define radiance, for instance. And his definition of irradiance is informal, almost "in passing". Lack of equilibrium is also a problem... Jan 6 at 17:06
• He gives an example of spectral radiance $L_{\lambda}$ as the Radon-Nikodym derivative of $Q$ with respect to the measure $\rho$. In a similar way you can define radiance $L$ as the Radon-Nikodym derivative of $Q$ w.r.t. a different measure (one that does not include $\lambda$). The definition for $\Phi, E, B, I$ is analogous, simply as Radon-Nikodym derivatives w.r.t. different measures. I do not know what equilibrium you are talking about. Jan 6 at 17:11
• @lightxbulb I need to read it again more carefully, will get back to you Jan 6 at 18:45
• @lightxbulb Re: equilibrium, I'm reading pbrt, which contains this paragraph: "Steady state: Light in the environment is assumed to have reached equilibrium, so its radiance distribution isn’t changing over time. This happens nearly instantaneously with light in realistic scenes, so it is not a limitation in practice." Jan 6 at 18:46

This is almost verbatim from Veach. Let the trajectory space be $$\mathbb{P} = \mathbb{R} \times \mathcal{M} \times \mathcal{S}^2 \times \mathbb{R}^+$$, corresponding respectively to: time, set of points of all scene surfaces, unit sphere (directions), wavelength. Let $$\mathcal{P}$$ be the set of measurable subsets of $$\mathbb{P}$$ and we have a measure $$\rho$$. Let $$\rho = l \times A \times \sigma^{\perp} \times l^+$$. Where $$l, l^+$$ are the Lebesgue measures on $$\mathbb{R}$$ and $$\mathbb{R}^+$$, $$A$$ is the Lebesgue measure on $$\mathcal{M}$$, and $$\sigma^{\perp}$$ is the projected solid angle measure $$(d\sigma^{\perp}(\omega) = |\cos\theta|\sin\theta\,d\theta\,d\phi$$). Then given the radiant energy measure $$Q(\mathcal{T}, \mathcal{D}, \Omega, \Lambda) \in [0,\infty]$$ ($$\mathcal{T} \times \mathcal{D} \times \Omega \times \Lambda \in \mathcal{P}$$), we can define the density $$L_{\lambda}$$ of $$Q$$ with respect to the measure $$\rho$$ as:

$$$$Q(\mathcal{T}, \mathcal{D}, \Omega, \mathcal{\Lambda}) = \int_{\Lambda}\int_{\Omega}\int_{\mathcal{D}}\int_{\mathcal{T}}L_{\lambda}(t, x, \omega,\lambda)d\rho(t,x,\omega,\lambda).$$$$

Then the density $$L_{\lambda}$$ is the Radon-Nikodym derivative of $$Q$$ w.r.t. $$\rho$$:

$$$$\frac{dQ}{d\rho}(t, x, \omega,\lambda) := L_{\lambda}(t, x, \omega,\lambda).$$$$

We can do a similar thing with respect to the appropriate measures:

Radiant flux: $$\begin{gather} \rho_1 = l \\ Q(\mathcal{T}, \mathcal{D}, \Omega, \mathcal{\Lambda}) = \int_{\mathcal{T}}\Phi(t, \mathcal{D}, \Omega, \mathcal{\Lambda})\,d\rho_1(t) \\ \frac{dQ}{d\rho_1}(t, \mathcal{D}, \Omega, \mathcal{\Lambda}) := \Phi(t, \mathcal{D}, \mathcal{\Omega}, \mathcal{\Lambda}). \end{gather}$$

Irradiance: $$\begin{gather} \rho_2 = l\times A \\ Q(\mathcal{T}, \mathcal{D}, \Omega, \mathcal{\Lambda}) = \int_{\mathcal{D}}\int_{\mathcal{T}}E(t, x, \Omega, \mathcal{\Lambda})\,d\rho_2(t,x) \\ \frac{dQ}{d\rho_2}(t, x, \Omega, \mathcal{\Lambda}) := E(t, x, \Omega, \mathcal{\Lambda}). \end{gather}$$

Intensity: $$\begin{gather} \rho_3 = l\times \sigma \\ Q(\mathcal{T}, \mathcal{D}, \Omega, \mathcal{\Lambda}) = \int_{\Omega}\int_{\mathcal{T}}I(t, \mathcal{D}, \omega, \mathcal{\Lambda})\,d\rho_3(t,\omega) \\ \frac{dQ}{d\rho_3}(t, \mathcal{D}, \omega, \mathcal{\Lambda}) := I(t, \mathcal{D}, \omega, \mathcal{\Lambda}). \end{gather}$$

Radiance: $$\begin{gather} \rho_4 = l\times A \times \sigma^{\perp} \\ Q(\mathcal{T}, \mathcal{D}, \Omega, \mathcal{\Lambda}) = \int_{\Omega}\int_{\mathcal{D}}\int_{\mathcal{T}}L(t, x, \omega, \mathcal{\Lambda})\,d\rho_4(t,x,\omega) \\ \frac{dQ}{d\rho_4}(t, x, \omega, \mathcal{\Lambda}) := L(t, x, \omega, \mathcal{\Lambda}). \end{gather}$$

• Just a comment: $\rho_4$ isn't really a product of measures since the $\sigma^\perp$ measure depends on the surface orientation. It's some kind of twisted product, see en.wikipedia.org/wiki/Markov_kernel#Semidirect_product or en.wikipedia.org/wiki/Disintegration_theorem Jan 21 at 11:10
• @user357269 The measure $\sigma^{\perp}_x(\Omega)$ is defined on page 77, eq (3.1) in Veach's thesis as $\sigma^{\perp}_x(\Omega) = \int_{\Omega}|\omega \cdot N_x|\,d\sigma(\omega)$. So yes, $\sigma^{\perp}$ depends on $x$, but that isn't really an issue. Jan 21 at 12:29