The book defines operator $\mathcal{T}$ for the propagation of an exitant function as:

enter image description here

if I'm assuming that the position of the point $x$ is on the top-left surface, and that $\Psi$ is considered as the direction originating from point $y = r(x, \Theta)$ ($r$ being the raycast function) then what's the interpretation of $cos(N_{x}, \Psi)$ ? Why would we use the normal at point $x$ (on another surface!) instead of $cos(N_{r(x, \Theta)}, \Psi)$ ?

on the other hand, if I'm assuming that $x$ is the point on the surface at bottom-right, then where's the point $y = r(x, \Theta)$ ? And what's the physical meaning of $L(y \rightarrow -\Psi)$ ?

  • $\begingroup$ Their picture is misleading/wrong. See this instead: viclw17.github.io/2018/06/30/raytracing-rendering-equation $\endgroup$
    – lightxbulb
    Jun 11, 2022 at 6:27
  • $\begingroup$ @lightxbulb thank you for the resource, I fully understand the principles behind the standard rendering equation, however in the book they're doing a somewhat long derivation to show 4 different ways of calculating the flux of a set (with exitant/incident radiance/importance) part of this derivation involves the calculation of a function for exitant radiance, which unfortunately as you mentioned seems to be misleading $\endgroup$
    – Domenico
    Jun 11, 2022 at 15:53
  • $\begingroup$ Their image is misleading/wrong, but the math is ok. Here's the translation of the usual notation: $L(x, \omega_o) = L_e(x, \omega_o) + \int_{\Omega_x}f(x,\omega_o,\omega_i) L(r(x,\omega_i), -\omega_i) (N_x \cdot \omega_i) d\omega_i$. Set $\omega_o = \Theta$, $\omega_i = \Psi$ and you have their notation. Note that $(N_x \cdot \omega_i) = \cos\angle(N_x, \omega_i)$ and $L_i(x, \omega_i) = L(r(x,\omega_i), -\omega_i)$. The point $x$ is a point on some surface, and $\omega_o$ is the direction you want to compute the radiance exiting $x$ in: $L(x,\omega_o)$. $\endgroup$
    – lightxbulb
    Jun 11, 2022 at 16:12
  • $\begingroup$ @lightxbulb just to make sure I understood, you mentioned that the formula itself is ok, but in the comment you just wrote $L(r(x, \omega_{i}), - \omega_{i})$ while the formula in the book is using: $L(r(x, \omega_{o}), - \omega_{i})$ Am I missing something? $\endgroup$
    – Domenico
    Jun 11, 2022 at 20:48
  • $\begingroup$ I hadn't noticed that, $L(r(x, \Theta), -\Psi)$ ought to be $L(r(x,\Psi), -\Psi)$ instead. $\endgroup$
    – lightxbulb
    Jun 11, 2022 at 21:04


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