<p>In the BRDF curve, the total area under the curve is the albedo: what fraction of incident light is reflected in total (as opposed to being absorbed or transmitted). For a perfect reflector, the area under the curve sums to 1, because it reflects all of the incident light. This area limits how high the curve can be. For example, a <em>perfect</em> diffuse reflector has a BRDF that's a flat line at $\frac{1}{2\pi}$, because it reflects the light evenly across the whole hemisphere, the solid angle of a hemisphere is $2\pi$, and the area has to be 1.</p>

<p>In a <em>perfect</em> specular reflector, the area under the curve is still 1: it still reflects all of the incident light. But this time, it's reflecting all of that light in one direction. The BRDF curve is a single spike. If you draw a graph with an infinitely thin spike, and the area under the graph has to be 1, then the spike must also be infinitely high. It's the only way to make the area correct.</p>

<p>Obviously a real material can't be a <em>perfect</em> specular reflector, so it will have two differences. First, the area under the curve will be less than 1,  because some light is absorbed. This doesn't matter much to us. More importantly, the reflection peak can't be infinitely thin: the reflection will be blurred ever so slightly. This means the peak won't be infinitely high: the wider the peak, the less high it has to be to still have an area of 1.</p>

<hr>

<p>The remark about "$\delta$-functions" is because of a function (properly speaking, it's not a function but a <em>distribution</em>) called the Dirac $\delta$ (delta). This distribution is an infinitely thin, infinitely tall spike at $x = 0$, whose area is 1, just like our perfect specular BRDF.</p>

CC BY-SA 3.0
2017-02-24T16:37:34.243Z
Thanks for your time and patience! I've understand what you mean. However, I doubt that the total area under the BRDF curve(you mean the surface in the hemisphere coordinate system?) is the albedo. In my another question [Energy conservation of BRDF](http://computergraphics.stackexchange.com/questions/4768/energy-conservation-of-brdf), the last equation $\forall\Phi:\int_{\Omega_x}f_r(x,\Phi\to\Theta)cos(N_x,\Theta)d\omega_\Theta\le1$ is the necessary and sufficient condition for energy conservation.(To be continued)
chaosink
CC BY-SA 3.0
2017-02-24T16:37:41.743Z
So I think albedo maybe equal $\int_{\Omega_x}f_r(x,\Phi\to\Theta)cos(N_x,\Theta)d\omega_\Theta$ (I'm not sure since I haven't found any relative document for the time). An additional factor $cos(N_x,\Theta)$ is in the formula. From this formula, if we consider the situation that it equals 1 and the BRDF is constant, we can get the BRDF equals $\frac{1}{\pi}$ but not $\frac{1}{2\pi}$ since $\int_{\Omega_x}cos(N_x,\Theta)d\omega_\Theta=\pi$.
chaosink
CC BY-SA 3.0
2017-02-24T14:02:14.297Z
2017-02-24T14:02:14.297Z
Dan Hulme
<p>I know the BRDF of specular reflection is nonzero only in the reflection direction.</p>

<p>But why it is infinite?</p>

<p>A paragraph on page 36 of <a href="http://www.advancedglobalillumination.com/" rel="noreferrer">Advanced Global Illumination</a>:</p>

<p><a href="https://i.sstatic.net/GQgcj.png" rel="noreferrer"><img src="https://i.sstatic.net/GQgcj.png" alt="enter image description here"></a></p>

CC BY-SA 3.0
2017-02-24T12:58:23.320Z
Where are you getting this infinity from? Is this a formula you're using, or something you've read, or something about a particular shader?
Dan Hulme
CC BY-SA 3.0
2017-02-24T13:39:03.750Z
@DanHulme Question updated.
chaosink
CC BY-SA 3.0
2017-02-24T13:40:04.643Z
What's your maths background? Are you familiar with the Dirac delta distribution?
Dan Hulme
CC BY-SA 3.0
2017-02-24T13:44:02.973Z
@DanHulme Haven't heard of that... Learning CG always makes me worry about my math.
chaosink
<rendering><brdf>
Why the BRDF of specular reflection is infinite in the reflection direction?
chaosink
# Why the BRDF of specular reflection is infinite in the reflection direction?

(Asked by: username_0 on 2017-02-24)

I know the BRDF of specular reflection is nonzero only in the reflection direction.

But why it is infinite?

A paragraph on page 36 of [Advanced Global Illumination](http://www.advancedglobalillumination.com/):

[![enter image description here](https://i.sstatic.net/GQgcj.png)](https://i.sstatic.net/GQgcj.png)

**Tags:** rendering, brdf

> **Comments:**
> **username_1**: Where are you getting this infinity from? Is this a formula you're using, or something you've read, or something about a particular shader?  
> **username_0**: @username_1 Question updated.  
> **username_1**: What's your maths background? Are you familiar with the Dirac delta distribution?  
> **username_0**: @username_1 Haven't heard of that... Learning CG always makes me worry about my math.  

---

## 1 answers

### Answer by username_1 on 2017-02-24. Score: 6 | Accepted answer

In the BRDF curve, the total area under the curve is the albedo: what fraction of incident light is reflected in total (as opposed to being absorbed or transmitted). For a perfect reflector, the area under the curve sums to 1, because it reflects all of the incident light. This area limits how high the curve can be. For example, a *perfect* diffuse reflector has a BRDF that's a flat line at $\frac{1}{2\pi}$, because it reflects the light evenly across the whole hemisphere, the solid angle of a hemisphere is $2\pi$, and the area has to be 1.

In a *perfect* specular reflector, the area under the curve is still 1: it still reflects all of the incident light. But this time, it's reflecting all of that light in one direction. The BRDF curve is a single spike. If you draw a graph with an infinitely thin spike, and the area under the graph has to be 1, then the spike must also be infinitely high. It's the only way to make the area correct.

Obviously a real material can't be a *perfect* specular reflector, so it will have two differences. First, the area under the curve will be less than 1, because some light is absorbed. This doesn't matter much to us. More importantly, the reflection peak can't be infinitely thin: the reflection will be blurred ever so slightly. This means the peak won't be infinitely high: the wider the peak, the less high it has to be to still have an area of 1.

---

The remark about "$\delta$-functions" is because of a function (properly speaking, it's not a function but a *distribution*) called the Dirac $\delta$ (delta). This distribution is an infinitely thin, infinitely tall spike at $x = 0$, whose area is 1, just like our perfect specular BRDF.

> **Comments:**
> **username_0**: Thanks for your time and patience! I've understand what you mean. However, I doubt that the total area under the BRDF curve(you mean the surface in the hemisphere coordinate system?) is the albedo. In my another question [Energy conservation of BRDF](http://computergraphics.stackexchange.com/questions/4768/energy-conservation-of-brdf), the last equation $\forall\Phi:\int_{\Omega_x}f_r(x,\Phi\to\Theta)cos(N_x,\Theta)d\omega_\Theta\le1$ is the necessary and sufficient condition for energy conservation.(To be continued)  
> **username_0**: So I think albedo maybe equal $\int_{\Omega_x}f_r(x,\Phi\to\Theta)cos(N_x,\Theta)d\omega_\Theta$ (I'm not sure since I haven't found any relative document for the time). An additional factor $cos(N_x,\Theta)$ is in the formula. From this formula, if we consider the situation that it equals 1 and the BRDF is constant, we can get the BRDF equals $\frac{1}{\pi}$ but not $\frac{1}{2\pi}$ since $\int_{\Omega_x}cos(N_x,\Theta)d\omega_\Theta=\pi$.  

---

