# 為什麼BRDF和照明不相關？

W.r.t。以下積分

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

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

$L（x \ rightarrow \ Theta）$：這是從$x$向$\ Theta$方向的輻射$f_r（x，\ Psi \ rightarrow \ Theta）$是BRDF從$x$中的方向$\ Psi$進入並向$\ Theta$方向離開$L（x \ leftarrow \ Psi）$：方向$\ Psi$向$x$的輻射度

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.