<p>You are right to be confused. What I think they should have written:
$$
L( x \leftarrow \Psi ) = L_{in} \delta(\Psi - \alpha)
$$
using $ \alpha $ instead of $ \Theta $, which is already used as a dummy variable in the integral.</p>

<p>You should look up the Dirac delta function to learn its meaning and properties. In this context, you can imagine the $ L$ above as representing a very concentrated beam (a laser) coming from the angle $ \alpha $. Practically, to do the integral over $ \Psi $ with $ \delta (\Psi - \alpha) $ present in the integrand, remove the integration and replace all occurrences of $ \Psi $ with $ \alpha $. Then it should be clear how they arrive at the next line, which should read, for all $ \alpha $: $$ \int f_r(x, \alpha \rightarrow \Theta) \; \cos(N_x, \Theta)\; d \omega_\Theta \leq 1 $$.</p>

<p>The fact that this condition is sufficient follows from two facts: 1. that any function (e.g. $ L $) can be approximated by a sum of many $ \delta $ functions, and 2. that everything is linear. In other words, if I write $ N(L) $ and $ D(L) $ for the numerator and denominator of the left hand side of (2.21), then you can see that if $ L = L_1 + L_2 $, then $ N(L_1+L_2) = N(L_1) + N(L_2) $ and $ D(L_1+L_2) = D(L_1) + D(L_2) $. So if I know $ N(L_1) \leq D(L_1) $ and $ N(L_2) \leq D(L_2) $, then I know $ N(L) \leq D(L) $ since $$ N(L) = N(L_1) + N(L_2) \leq D(L_1) + D(L_2) = D(L) $$. </p>

CC BY-SA 3.0
2017-02-25T17:31:52.827Z
2017-02-25T19:49:27.607Z
2017-02-25T19:49:27.607Z
StinkySkunk
<p>I'm reading <a href="http://www.advancedglobalillumination.com/" rel="noreferrer">Advanced Global Illumination</a>.</p>

<p>Here is the part confusing me:</p>

<p><a href="https://i.sstatic.net/CDfwS.png" rel="noreferrer"><img src="https://i.sstatic.net/CDfwS.png" alt="Advanced Global Illumination 2nd Edition, page 34"></a></p>

<ol>
<li><p>What do the second equation and $\delta$-function mean?</p></li>
<li><p>Why the third equation is a sufficient condition even though a reason is given?</p></li>
</ol>

CC BY-SA 3.0
2017-02-24T23:12:27.960Z
I'm working through the same book and coincidentally went through that chapter (again) in the last few days. Are you aware of the [UC Davis lectures on YouTube](https://www.youtube.com/playlist?list=PLslgisHe5tBPckSYyKoU3jEA4bqiFmNBJ) that use this book? Unfortunately the lecturer doesn't address your questions specifically in the [BRDF lecture](https://www.youtube.com/watch?v=ytRrjf9OPHg). I was formulating an answer for you but I'm not **fully** understanding it either, so I'll not reply as I don't want to misinform. I will try to write my thoughts about it when I have time though...
PeteUK
<rendering><brdf>
Energy conservation of BRDF
chaosink
# Energy conservation of BRDF

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

I'm reading [Advanced Global Illumination](http://www.advancedglobalillumination.com/).

Here is the part confusing me:

[![Advanced Global Illumination 2nd Edition, page 34](https://i.sstatic.net/CDfwS.png)](https://i.sstatic.net/CDfwS.png)

1. What do the second equation and $\delta$-function mean?
2. Why the third equation is a sufficient condition even though a reason is given?

**Tags:** rendering, brdf

> **Comments:**
> **username_1**: I'm working through the same book and coincidentally went through that chapter (again) in the last few days. Are you aware of the [UC Davis lectures on YouTube](https://www.youtube.com/playlist?list=PLslgisHe5tBPckSYyKoU3jEA4bqiFmNBJ) that use this book? Unfortunately the lecturer doesn't address your questions specifically in the [BRDF lecture](https://www.youtube.com/watch?v=ytRrjf9OPHg). I was formulating an answer for you but I'm not **fully** understanding it either, so I'll not reply as I don't want to misinform. I will try to write my thoughts about it when I have time though...  

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## 1 answers

### Answer by username_2 on 2017-02-25. Score: 4 | Accepted answer

You are right to be confused. What I think they should have written:
$$
L( x \leftarrow \Psi ) = L\_{in} \delta(\Psi - \alpha)
$$
using $ \alpha $ instead of $ \Theta $, which is already used as a dummy variable in the integral.

You should look up the Dirac delta function to learn its meaning and properties. In this context, you can imagine the $ L$ above as representing a very concentrated beam (a laser) coming from the angle $ \alpha $. Practically, to do the integral over $ \Psi $ with $ \delta (\Psi - \alpha) $ present in the integrand, remove the integration and replace all occurrences of $ \Psi $ with $ \alpha $. Then it should be clear how they arrive at the next line, which should read, for all $ \alpha $: $$ \int f\_r(x, \alpha \rightarrow \Theta) \; \cos(N\_x, \Theta)\; d \omega\_\Theta \leq 1 $$.

The fact that this condition is sufficient follows from two facts: 1. that any function (e.g. $ L $) can be approximated by a sum of many $ \delta $ functions, and 2. that everything is linear. In other words, if I write $ N(L) $ and $ D(L) $ for the numerator and denominator of the left hand side of (2.21), then you can see that if $ L = L\_1 + L\_2 $, then $ N(L\_1+L\_2) = N(L\_1) + N(L\_2) $ and $ D(L\_1+L\_2) = D(L\_1) + D(L\_2) $. So if I know $ N(L\_1) \leq D(L\_1) $ and $ N(L\_2) \leq D(L\_2) $, then I know $ N(L) \leq D(L) $ since $$ N(L) = N(L\_1) + N(L\_2) \leq D(L\_1) + D(L\_2) = D(L) $$.


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