<p><span class="math-container">$cos (\theta_i)$</span>, or <span class="math-container">$\hat \omega_i ·\hat n$</span>, accounts for the effective scaling of differential surface area at oblique angles. BRDF is the change in reflected radiance (L, W/m^2/sr) divided by the change in irradiance (E, W/m^2). Irradiance is radiance integrated over solid angle,</p>
<p><span class="math-container">$E = \int_Ω L_i(\omega_i) (\hat \omega_i ·\hat n) d\omega_i $</span></p>
<p>where <span class="math-container">$\hat \omega_i$</span> is a unit vector pointing toward the light source with its tail anchored to the differential area of surface reflection. Imagine a hemispherical ceiling with its origin at the <span class="math-container">$\hat \omega_i$</span> tail. The head of <span class="math-container">$\hat \omega_i$</span> would be integrated over a small area of the ceiling. By definition, solid angle (<span class="math-container">$\Omega$</span>) is that small ceiling area divided by the squared radial distance. The <span class="math-container">$\hat \omega_i ·\hat n$</span> term converts the oblique differential surface area (as would be viewed along <span class="math-container">$-\omega_i$</span>) into normal differential surface area (as would be viewed along <span class="math-container">$-n$</span>).</p>
<p>By the definition of the dot product,</p>
<p><span class="math-container">$\hat \omega_i ·\hat n = |\hat \omega_i||\hat n| cos (\theta_i) = cos (\theta_i)$</span></p>
<p>The differential irradiance is therefore</p>
<p><span class="math-container">$dE = L_i(\omega_i)(\hat \omega_i ·\hat n) d\omega_i = L_i(\omega_i) cos (\theta_i) d\omega_i$</span></p>
<p>and the ratio of differential output radiance to differential irradiance is</p>
<p><span class="math-container">$BRDF(\omega_i,\omega_r) = \frac{dL_r(\omega_r)} {L_i(\omega_i) cos (\theta_i) d\omega_i}$</span></p>
<p>The same conversion of oblique differential surface area to normal differential surface area is required by the rendering equation,</p>
<p><span class="math-container">$L_o = L_e + \int_Ω BRDF(\omega_i,\omega_r)  L_i(\omega_i)(\hat \omega_i ·\hat n) d\omega_i$</span></p>

CC BY-SA 4.0
2023-02-08T19:20:43.673Z
2023-02-08T23:03:56.810Z
2023-02-08T23:03:56.810Z
stp
<p>I am somewhat confused with the following issue reading about the rendering equation and BRDFs. That is, the BRDF is usually defined as surface radiance <span class="math-container">$Lr(ω_r)$</span> over irradiance <span class="math-container">$L_i(ω_i)cosθ_i$</span> for an infinitesimal solid angle like so:</p>
<p><span class="math-container">$$
fr(ωi,ωr) = \frac{dL_r(ω_r)}{L_i(ω_i)cosθ_idω_i}
$$</span></p>
<p>And the integrand of the rendering equation is given by:</p>
<p><span class="math-container">$$f_r(ω_i, ω_o)L_i(x,ω_i)cos(ω_i\cdot\vec{n})dω_i$$</span></p>
<p>I conclude that <span class="math-container">$cosθ_i$</span> = <span class="math-container">$cos(ω_i\cdot\vec{n})$</span> which raises the question why bothering about the cosine term at all since it cancels out anyway? <strong>There are some posts on why the brdf is radiance over irradiance (instead of radiance of radiance). This makes sense on its own to get independence of the incoming radiance direction. However, in combination with the rendering equation it gets confusing again because it seems useless.</strong></p>

CC BY-SA 4.0
2021-11-30T13:17:28.500Z
The brdf is formally the ratio you have written above, but note that $L$ is unknown. Generally you have to define the scene geometry $\mathcal{M}, \, N_{\mathcal{M}}$ (implicitly it is in the ray-tracing operstor $r(x,\omega)$ which is implicitly in $L_i$), the brdf $f_r$ at every point of every surface, and the light sources $L_e$, for there to be a solution for $L$. Can you solve the opposite problem of finding the brdf when given $L$, $L_e$, and the scene geometry? Sometimes you can, and that's what inverse rendering is concerned with. So in your case $f$ is a given function, e.g. $C/\pi$.
lightxbulb
<rendering><brdf><physically-based>
Cosine in Rendering Equation and BRDF cancel out: Why cosine at all?
Sam
# Cosine in Rendering Equation and BRDF cancel out: Why cosine at all?

(Asked by: username_0 on 2021-11-30)

I am somewhat confused with the following issue reading about the rendering equation and BRDFs. That is, the BRDF is usually defined as surface radiance $Lr(ω\_r)$ over irradiance $L\_i(ω\_i)cosθ\_i$ for an infinitesimal solid angle like so:

$$
fr(ωi,ωr) = \frac{dL\_r(ω\_r)}{L\_i(ω\_i)cosθ\_idω\_i}
$$

And the integrand of the rendering equation is given by:

$$f\_r(ω\_i, ω\_o)L\_i(x,ω\_i)cos(ω\_i\cdot\vec{n})dω\_i$$

I conclude that $cosθ\_i$ = $cos(ω\_i\cdot\vec{n})$ which raises the question why bothering about the cosine term at all since it cancels out anyway? **There are some posts on why the brdf is radiance over irradiance (instead of radiance of radiance). This makes sense on its own to get independence of the incoming radiance direction. However, in combination with the rendering equation it gets confusing again because it seems useless.**

**Tags:** rendering, brdf, physically-based

> **Comments:**
> **username_1**: The brdf is formally the ratio you have written above, but note that $L$ is unknown. Generally you have to define the scene geometry $\mathcal{M}, \, N_{\mathcal{M}}$ (implicitly it is in the ray-tracing operstor $r(x,\omega)$ which is implicitly in $L_i$), the brdf $f_r$ at every point of every surface, and the light sources $L_e$, for there to be a solution for $L$. Can you solve the opposite problem of finding the brdf when given $L$, $L_e$, and the scene geometry? Sometimes you can, and that's what inverse rendering is concerned with. So in your case $f$ is a given function, e.g. $C/\pi$.  

---

## 1 answers

### Answer by username_2 on 2023-02-08. Score: 1 | Accepted answer

$cos (\theta\_i)$, or $\hat \omega\_i ·\hat n$, accounts for the effective scaling of differential surface area at oblique angles. BRDF is the change in reflected radiance (L, W/m^2/sr) divided by the change in irradiance (E, W/m^2). Irradiance is radiance integrated over solid angle,

$E = \int\_Ω L\_i(\omega\_i) (\hat \omega\_i ·\hat n) d\omega\_i $

where $\hat \omega\_i$ is a unit vector pointing toward the light source with its tail anchored to the differential area of surface reflection. Imagine a hemispherical ceiling with its origin at the $\hat \omega\_i$ tail. The head of $\hat \omega\_i$ would be integrated over a small area of the ceiling. By definition, solid angle ($\Omega$) is that small ceiling area divided by the squared radial distance. The $\hat \omega\_i ·\hat n$ term converts the oblique differential surface area (as would be viewed along $-\omega\_i$) into normal differential surface area (as would be viewed along $-n$).

By the definition of the dot product,

$\hat \omega\_i ·\hat n = |\hat \omega\_i||\hat n| cos (\theta\_i) = cos (\theta\_i)$

The differential irradiance is therefore

$dE = L\_i(\omega\_i)(\hat \omega\_i ·\hat n) d\omega\_i = L\_i(\omega\_i) cos (\theta\_i) d\omega\_i$

and the ratio of differential output radiance to differential irradiance is

$BRDF(\omega\_i,\omega\_r) = \frac{dL\_r(\omega\_r)} {L\_i(\omega\_i) cos (\theta\_i) d\omega\_i}$

The same conversion of oblique differential surface area to normal differential surface area is required by the rendering equation,

$L\_o = L\_e + \int\_Ω BRDF(\omega\_i,\omega\_r) L\_i(\omega\_i)(\hat \omega\_i ·\hat n) d\omega\_i$


---

