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

<p>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.</p>

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

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

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

<p>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 $L(x\rightarrow\Theta)$.</p>

<p>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.</p>

CC BY-SA 3.0
2018-02-09T13:38:35.383Z
You forgot the foreshortening term in the equation.
user8469759
CC BY-SA 3.0
2018-02-09T14:11:37.297Z
Yes but it seems like you managed to understand it so it's not a big deal
Sebastián Mestre
CC BY-SA 3.0
2018-02-09T14:26:48.360Z
I like your reply XD. I see what you mean anyway.
user8469759
CC BY-SA 3.0
2018-02-15T05:26:21.993Z
If this answer helped you, it would be great if you upvoted it or selected it as correct.
Sebastián Mestre
CC BY-SA 3.0
2018-02-09T03:07:51.730Z
2018-02-09T03:07:51.730Z
Sebastián Mestre
<p>In relation to <a href="https://computergraphics.stackexchange.com/questions/6221/why-is-the-approximation-valid-in-the-formula-provided-by-brian-karis?noredirect=1#comment9519_6221">this question</a>, in the accepted answer I have a trivial question probably.</p>

<p>W.r.t. the following integral</p>

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

<p>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 <a href="https://books.google.co.uk/books?id=TB1jDAAAQBAJ&amp;printsec=frontcover&amp;dq=advanced%20global%20illumination&amp;hl=en&amp;sa=X&amp;ved=0ahUKEwighpjR5JPZAhWJLMAKHbOZC4oQ6AEIKTAA#v=onepage&amp;q=advanced%20global%20illumination&amp;f=false" rel="nofollow noreferrer">this book</a>. In such book the brdf $f_r(l,v)$ is <strong><em>defined</em></strong> as</p>

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

<p>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?</p>

<p>A note about the notation used.</p>

<p>$L(x\rightarrow\Theta)$ : 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 $x$</p>

CC BY-SA 3.0
2018-02-07T13:33:42.107Z
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.
Tare
CC BY-SA 3.0
2018-02-07T16:06:45.253Z
@Tare I hope is clearer now
user8469759
<mathematics><physically-based><physics>
Why are BRDF and the lighting uncorrelated?
user8469759
# Why are BRDF and the lighting uncorrelated?

(Asked by: username_0 on 2018-02-07)

In relation to [this question](https://computergraphics.stackexchange.com/questions/6221/why-is-the-approximation-valid-in-the-formula-provided-by-brian-karis?noredirect=1#comment9519_6221), 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](https://books.google.co.uk/books?id=TB1jDAAAQBAJ&printsec=frontcover&dq=advanced%20global%20illumination&hl=en&sa=X&ved=0ahUKEwighpjR5JPZAhWJLMAKHbOZC4oQ6AEIKTAA#v=onepage&q=advanced%20global%20illumination&f=false). 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 $x$

**Tags:** mathematics, physically-based, physics

> **Comments:**
> **username_1**: 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.  
> **username_0**: @username_1 I hope is clearer now  

---

## 1 answers

### Answer by username_2 on 2018-02-09. Score: 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.

> **Comments:**
> **username_0**: You forgot the foreshortening term in the equation.  
> **username_2**: Yes but it seems like you managed to understand it so it's not a big deal  
> **username_0**: I like your reply XD. I see what you mean anyway.  
> **username_2**: If this answer helped you, it would be great if you upvoted it or selected it as correct.  

---

