<p>I think I got it! </p>

<p>Because $cos(\theta)$ integrates to $\pi$ over the hemisphere (and not $2\pi$). And the incoming light is multiplied by $cos(\theta)$ (and the BRDF).</p>

CC BY-SA 3.0
2016-11-28T06:13:26.820Z
2016-11-28T06:13:26.820Z
Supernormal
<p>The apparent brightness of a Lambertian surface is the same regardless of viewing angle. This is accomplished mathematically by the Lambertian BRDF cosine function, which counteracts the 1/cosine change in differential surface area at oblique angles. Since BRDF is reflected radiance (W/m^2/sr) over irradiance (W/m^2), the value for a hemisphere is the albedo divided by the projected solid angle of a hemisphere.</p>
<p>The area projected by a hemisphere is that of a circle, <span class="math-container">$\pi r^2$</span>, and the definition of solid angle is <span class="math-container">$\Omega = A/r^2$</span>. It follows that <span class="math-container">$\Omega_{h} = \pi r^2/r^2 = \pi$</span>. Converting from irradiance to reflected radiance, i.e., <em>applying the BRDF</em>, is as simple as multiplying by <span class="math-container">$albedo/\Omega_{h}=albedo/\pi$</span>, where <span class="math-container">$albedo=1$</span> for a non-emissive surface.</p>

CC BY-SA 4.0
2023-02-08T23:26:34.807Z
2023-02-09T19:19:42.497Z
2023-02-09T19:19:42.497Z
stp
<p>Why is a Lambertian BRDF normalised by dividing by $\pi$? Since the area of a unit sphere is $4 \pi$, and the area of the half sphere above the surface is $2 \pi$, shouldn't it rather be $1/(2\pi)$?</p>

<brdf><pbr>
Why normalise Lambertian BRDF by 1/pi?
Supernormal
# Why normalise Lambertian BRDF by 1/pi?

(Asked by: username_0 on 2016-11-28)

Why is a Lambertian BRDF normalised by dividing by $\pi$? Since the area of a unit sphere is $4 \pi$, and the area of the half sphere above the surface is $2 \pi$, shouldn't it rather be $1/(2\pi)$?

**Tags:** brdf, pbr


---

## 2 answers

### Answer by username_0 on 2016-11-28. Score: 13 | Accepted answer

I think I got it!

Because $cos(\theta)$ integrates to $\pi$ over the hemisphere (and not $2\pi$). And the incoming light is multiplied by $cos(\theta)$ (and the BRDF).


---

### Answer by username_1 on 2023-02-08. Score: 1

The apparent brightness of a Lambertian surface is the same regardless of viewing angle. This is accomplished mathematically by the Lambertian BRDF cosine function, which counteracts the 1/cosine change in differential surface area at oblique angles. Since BRDF is reflected radiance (W/m^2/sr) over irradiance (W/m^2), the value for a hemisphere is the albedo divided by the projected solid angle of a hemisphere.

The area projected by a hemisphere is that of a circle, $\pi r^2$, and the definition of solid angle is $\Omega = A/r^2$. It follows that $\Omega\_{h} = \pi r^2/r^2 = \pi$. Converting from irradiance to reflected radiance, i.e., *applying the BRDF*, is as simple as multiplying by $albedo/\Omega\_{h}=albedo/\pi$, where $albedo=1$ for a non-emissive surface.


---

