In relation to this question, in the accepted answer I have a trivial question probably.

W.r.t. the following integral

$$ L_o(x\rightarrow \Theta) = \int_{\Omega_x} f_r(x,\Psi\rightarrow\Theta)L_i(x\leftarrow \Psi) cos(N_x,\Psi)d\omega_{\Psi} $$

in the accepted answer, one of the comments it is stated that $f_r(l,v) = f_r(x,\Psi\rightarrow\Theta)$ and $L_i(l) = L(x\leftarrow \Psi)$ are not correlated. As usual... I wanna understand why... My reference on this subject is this book. In such book the brdf $f_r(l,v)$ is defined as

$$ f_r(x,\Psi \rightarrow \Theta) = \frac{dL(x \rightarrow \Theta)}{dE(x \leftarrow \Psi)} = \frac{dL(x \rightarrow \Theta)}{L(x\leftarrow\Psi)cos(N_x,\Psi) d\omega_\Psi}, $$

From the last equation I can see that the two are actually related, unless there's something I'm missing. Why are they not correlated?

A note about the notation used.

$L(x\rightarrow\Theta)$ : this is the radiance from $x$ toward direction $\Theta$ $f_r(x,\Psi\rightarrow\Theta)$ is the BRDF coming from direction $\Psi$ in $x$ and exiting in direction $\Theta$ $L(x\leftarrow\Psi)$ : radiance from direction $\Psi$ toward $x$

  • $\begingroup$ can you specify, what the individual symbols ($\Theta, \Psi, ...$) represent? the symbols in BRDFs differ from book to book and I for one have learned with RealTimeRendering, Physically Based Rendering and Digital Modeling of Material Appearance, not your source. And even now, I can't recall how they did it individually from the top of my head, so I have a hard time grasping what is done in your equations. $\endgroup$
    – Tare
    Feb 7, 2018 at 13:33
  • $\begingroup$ @Tare I hope is clearer now $\endgroup$ Feb 7, 2018 at 16:06

1 Answer 1


When rendering an image you are trying to find the outgoing light from a point into the camera $L(x \rightarrow \Theta$).

To do this you solve "The rendering equation", which means you integrate the product of the brdf and the incoming light for every incoming direction.

$$L(x\rightarrow\Theta )=\int_\Omega f_r(x, \omega\rightarrow\Theta)\cdot L(x\leftarrow\omega)d\omega$$

Since you don't have the outgoing light you cannot use the definition of the brdf to calculate it.

$$f_r(x,\omega\rightarrow\Theta )=\frac{L(x\rightarrow\Theta)}{E(x\leftarrow\omega)}$$

Instead, you use a formula to compute $f_r(x,\omega\rightarrow\Theta )$ and some operation to find the incoming light $L(x\leftarrow\omega)$ (a raytrace, a env map sample, etc.), and use those to compute the outgoing light $L(x\rightarrow\Theta)$.

These two calculations are usually done in a completely separate way. So there is no way the incoming light can effect the brdf in any way.

  • $\begingroup$ You forgot the foreshortening term in the equation. $\endgroup$ Feb 9, 2018 at 13:38
  • $\begingroup$ Yes but it seems like you managed to understand it so it's not a big deal $\endgroup$ Feb 9, 2018 at 14:11
  • $\begingroup$ I like your reply XD. I see what you mean anyway. $\endgroup$ Feb 9, 2018 at 14:26
  • $\begingroup$ If this answer helped you, it would be great if you upvoted it or selected it as correct. $\endgroup$ Feb 15, 2018 at 5:26

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