Given the rendering equation:

$$ L_o(x, \theta_o) = L_e(x, \theta_o) + \int_{\Omega} \! f(x,\theta_i,\theta_o) L_i(x,\theta_i) \cos(\theta_i, \theta_o) \, \mathrm{d} \theta_i. $$

Rewriting it using linear operator $F$, as:

$$ L_o = L_e + F \circ L_i. $$

Consider the final state that $L = L_o = L_i$, then there are two ways to rewrite and interpret:

(1) (light tracing?) $$ L_e = (I - F) \circ L. $$

(2) (path tracing?) $$ L = (I + F + F^2 + F^3 + \cdots ) \circ L_e. $$


  • Are the two classifications the correct intrepretation?

  • Thinking about (2) as path tracing is easy, in that rays starting from $L$ go through multiple bounces and finally reaching $L_e$; while thinking of (1) as light tracing is difficult for me. Are there any intuitive explanations for both?

  • $\begingroup$ The first is not light tracing, nor is the second path tracing starting from the camera. in fact the second follows from the first only if a Neumann expansion is valid: $$(I-F)L = L_e$$ $$L= (I-F)^{-1}L_e = \sum_{k=0}^{\infty}F^kL_e$$ The equations just represent the steady state, they are unrelated to where you start from (you have no sensor sensitivity function anyways in this formulation). $\endgroup$
    – lightxbulb
    Commented Dec 30, 2019 at 7:40
  • $\begingroup$ That is to say, $L$ gives you the radiance at every point and in every direction of the scene, not just at the camera or light. $\endgroup$
    – lightxbulb
    Commented Dec 30, 2019 at 7:54
  • $\begingroup$ @lightxbulb Thanks for the explainations, would you mind expanding the comments into an answer? $\endgroup$
    – WDC
    Commented Dec 30, 2019 at 12:18

1 Answer 1


Having the rendering equation one can rewrite it in operator form:

$$L = L_e + TL$$ $$(I-T)L = L_e$$ $$L = (I-T)^{-1}L_e$$ $$L = \sum_{k=0}^{\infty}T^kL_e$$

The last equality holds if $T$ is a contraction ($\|T\| < 1$), and under very specific conditions if $T$ is not a contraction. An example of when the 4th equality does not hold would be picking an enclosed volume bounded by some surface, that is made of a non-energy conserving material, and putting a light source inside. Then the radiance will keep increasing with every bounce so at some point $L$ will be infinity everywhere on the inside of the volume. Meaning that the the series does not converge.

The 4th equality is really the Liouville-Neumann expansion. So you get an infinite sum of increasingly-dimensional integrals, each subsequent term being a bounce away from the previous. Each term gives you the contribution due to a light path of length equal to the index of the term.

More precisely, this solution gives you the steady state $L$ if it exists (we assume that light travels instantaneously, that it is incoherent, linear etc). Note that $L$ is simply the radiance function, so this formalism really gives you the solution at every point in the scene.

In practice you usually need only the radiance measured at the surface of a virtual camera film. So a straightforward optimization is to trace rays starting from the surface of the camera film. This does not guarantee that you get the solution for $L$ at every point of the scene, but is a lot more efficient. You can also trace rays starting at the light source, but you are usually also interested in their contribution to the film, so the measurement function (see Veach's thesis) has to also make it in there, and only a subset of lights may be of interest. Finally you can do photon mapping, or light tracing where you are interested of the radiance over the whole scene, and that would be the closest thing to the original formulation.


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