<p>A perfect Lambert reflector actually reflects light in a cosine distribution - that is, the amount of light per unit area reflected in any given direction <span class="math-container">$R$</span> is proportional to <span class="math-container">$N \cdot R$</span>. The reason the radiance appears constant for all angles is that as the view direction moves away from the normal, the reflected light per unit area decreases, but the surface area per projected beam area (and hence per pixel) increases by the same amount - the cosine factors cancel out.</p>
<p>Given this distribution, to be energy conserving, the probability density for the reflected light in all possible directions cannot sum to more than 1. If you integrate the cosine function over the whole hemisphere around the normal, you get <span class="math-container">$\pi$</span>, so this is the normalization factor you need.</p>

CC BY-SA 3.0
2018-01-25T07:15:10.197Z
Would it be reasonable to say that the cosine distribution integrates to $\pi$ because $cos$ is a one-dimensional function? So integrating it over a hemispherical domain gives you the solid angle consumed by the disk passing through the equator?
Paul Ferris
CC BY-SA 3.0
2018-01-25T10:02:18.347Z
Yeah that's an interesting way of looking at it. Intuitively, any infinitesimal patch of the hemisphere's surface will have its projected area onto the equatorial disk scaled by N.L, so the sum of the projected areas is just the area of the disk. Spherical calculus is not really my strong suit though ;)
russ
CC BY-SA 3.0
2018-01-25T19:39:15.643Z
I guess that's understandable :P. Thanks for the answers.
Paul Ferris
CC BY-SA 4.0
2018-01-25T06:41:28.923Z
2021-08-03T12:44:29.750Z
2021-08-03T12:44:29.750Z
russ
<p>The Lambertian BRDF is defined as $R / \pi$, where the division by $\pi$ is there to maintain conservation of energy. The derivation is provided <a href="http://rorydriscoll.com/2009/01/25/energy-conservation-in-games/" rel="noreferrer">here</a>.</p>

<p>The maths makes sense, and I can understand where the BRDF comes from in a mechanical sense. But I'm struggling to understand what it <em>represents</em>; how does division by $\pi$ relate to reflectance within the unit hemisphere?</p>

<brdf><diffuse>
Intuitively understanding conservation of energy for the Lambertian BRDF
Paul Ferris
# Intuitively understanding conservation of energy for the Lambertian BRDF

(Asked by: username_0 on 2018-01-25)

The Lambertian BRDF is defined as $R / \pi$, where the division by $\pi$ is there to maintain conservation of energy. The derivation is provided [here](http://rorydriscoll.com/2009/01/25/energy-conservation-in-games/).

The maths makes sense, and I can understand where the BRDF comes from in a mechanical sense. But I'm struggling to understand what it *represents*; how does division by $\pi$ relate to reflectance within the unit hemisphere?

**Tags:** brdf, diffuse


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## 1 answers

### Answer by username_1 on 2018-01-25. Score: 5 | Accepted answer

A perfect Lambert reflector actually reflects light in a cosine distribution - that is, the amount of light per unit area reflected in any given direction $R$ is proportional to $N \cdot R$. The reason the radiance appears constant for all angles is that as the view direction moves away from the normal, the reflected light per unit area decreases, but the surface area per projected beam area (and hence per pixel) increases by the same amount - the cosine factors cancel out.

Given this distribution, to be energy conserving, the probability density for the reflected light in all possible directions cannot sum to more than 1. If you integrate the cosine function over the whole hemisphere around the normal, you get $\pi$, so this is the normalization factor you need.

> **Comments:**
> **username_0**: Would it be reasonable to say that the cosine distribution integrates to $\pi$ because $cos$ is a one-dimensional function? So integrating it over a hemispherical domain gives you the solid angle consumed by the disk passing through the equator?  
> **username_1**: Yeah that's an interesting way of looking at it. Intuitively, any infinitesimal patch of the hemisphere's surface will have its projected area onto the equatorial disk scaled by N.L, so the sum of the projected areas is just the area of the disk. Spherical calculus is not really my strong suit though ;)  
> **username_0**: I guess that's understandable :P. Thanks for the answers.  

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