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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 $1/(2\pi)$?

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2 Answers 2

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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).

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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.

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