<p>There are  a couple of ways to answer this question: an algebraic way and a geometric way.</p>

<p>Algebraically, we can identify the units that the BRDF must have by looking at its place in the rendering equation. The classic rendering equation is:</p>

<p>$$L_\text{outgoing}(\omega) = L_\text{emitted}(\omega) + \int_\Omega L_\text{incoming}(\omega') \, f_\text{BRDF}(\omega, \omega') \, (n \cdot \omega') \, d\omega'$$</p>

<p>The output value on the left is a radiance, so the result of the integral must also be a radiance. The integrand contains a radiance multiplied by a solid angle $d\omega'$, so something else in the integrand has to cancel out that factor of solid angle. The $n \cdot \omega'$ factor is dimensionless, and the only other thing there is the BRDF&mdash;so to make the whole thing work out, the BRDF must have units of inverse solid angle. Equivalently, the BRDF can be seen as a ratio of radiance to irradiance, since they differ by a factor of solid angle in the denominator of radiance.</p>

<p>Another way to see it is that the BRDF plays a role similar to a probability density. If you look at how probability densities work, they have units inverse to the volume of their domain. For instance, a 1D probability density has units of inverse length (probability per unit length, but probability itself is dimensionless), a 2D one has units of inverse area, and so on. The BRDF acts much like a probability density defined on the hemisphere, giving a likelihood for a photon coming in from a given direction to be reflected into some other direction. So, like any other probability density on a spherical domain, it has units of inverse solid angle.</p>

<p>Geometrically, we can get right down to brass tacks and take apart what's going on in the integral in the rendering equation. Recall that an integral means to subdivide the domain into tiny pieces and sum the integrand over all the pieces (in the limit as the pieces get infinitesimally small). Let's look at one such piece. The integrand should result in an infinitesimal amount of radiance $dL$, since we're going to sum over many pieces to arrive at a finite outgoing radiance. So a single infinitesimal piece of the integral looks like:</p>

<p>$$dL = L_\text{incoming} \, f_\text{BRDF} \, (n \cdot \omega') \, d\omega'$$</p>

<p>If we regroup the factors a bit, the combination $L_\text{incoming} \, (n \cdot \omega') \, d\omega'$ calculates the irradiance on the surface due to the light coming from the infinitesimal solid angle $d\omega'$. Since it's arriving from an infinitesimal amount of solid angle, it produces an infinitesimal irradiance $dE$.</p>

<p>$$dL = f_\text{BRDF} \, dE$$</p>

<p>or</p>

<p>$$f_\text{BRDF} = \frac{dL}{dE}$$</p>

<p>So the BRDF acts as a proportionality constant between the infinitesimal irradiance arriving at the surface from an infinitesimal solid angle, and the infinitesimal outgoing radiance generated thereby. It couldn't be a ratio of radiances, because we have a <em>finite</em> incoming radiance, and we need an <em>infinitesimal</em> outgoing radiance if we want to sum up many pieces of the integral and get a finite result. To make that happen, the BRDF would have to be infinitesimal-valued, which...isn't a thing, in standard mathematics. :)</p>

<p>I hope some of this helps. There are a variety of equivalent ways to look at this problem, as with so many things in math and physics.</p>

CC BY-SA 4.0
2021-05-11T13:01:55.660Z
This answer kinda explains it as if BRDFs are derived from the rendering equation while the concept "BRDF" is actually older. Looking at BRDFs isolated: Why is it not the ratio between rediances?
akuzminykh
CC BY-SA 4.0
2021-05-11T16:56:03.867Z
@akuzminykh It's not really about the rendering equation specifically, but about how integrals work. The point of a BRDF is to integrate incoming radiance over the whole hemisphere and produce output radiance. The integral will introduce a unit of solid angle (or more generally, a unit of "volume" of whatever the domain of the integral is), so the integrand must be a density in that unit. As I said in the answer, it's the same as how probability densities work.
Nathan Reed
CC BY-SA 4.0
2021-05-11T19:23:51.147Z
@NathanReed But isn't the irradiance in context of BRDF only including radiant flux coming from a specific direction, not the whole hemishere? When I e.g. read [this](https://graphics.stanford.edu/courses/cs448-05-winter/papers/nicodemus-brdf-nist.pdf), I just cannot pinpoint where the decision based on reasoning is made to let it be irradiance at all. Maybe I just need to read it a few times more ..
akuzminykh
CC BY-SA 4.0
2021-05-11T21:38:09.943Z
@akuzminykh That the BRDF is defined in terms of irradiance on a surface is in part just a convention. You might compare it to the phase functions found in volume rendering, which play a similar role to the BRDF—deciding how much light from one direction scatters to another. In that case there is no surface and no irradiance (no $n \cdot \omega$ factor), but the phase function is still a density over solid angle. If you're still having trouble, again I would urge thinking about probability densities and how their units work. It is the same thing.
Nathan Reed
CC BY-SA 4.0
2021-05-21T02:27:38.907Z
@NathanReed Thank you for the additional explanations!
akuzminykh
CC BY-SA 3.0
2017-03-11T11:50:05.027Z
I like your explanation very much. I get the arguments that there must be the inverse solid angle factor in BRDF, but what about the cosine factor? If we could drop the cosine term from BRDF, then we could drop if from the integral in the rendering equation, couldn't we? The only reason I can see is in the correct/current formulation the denominator can be seen as irradiance...
ciechowoj
CC BY-SA 3.0
2016-07-22T19:42:31.390Z
2016-07-22T19:42:31.390Z
Nathan Reed
<p>I'm Learning about BRDFs and wondering why the BRDF is defined as the ratio of outgoing radiance to a given direction and incoming irradiance from another direction. Why isn't the BRDF defined as ratio of radiances?</p>

CC BY-SA 3.0
2016-07-15T21:06:23.883Z
I would write an answer if I had time, but in a concise way: because of definition. Roughly speaking radiance measure OUTGOING light in a certain direction (or better: radiant flux per solid angle). Irradiance is INCOMING light from a certain direction (or better radiant flux received per unit area. BRDF is describing the ratio of *outgoing* light to *incoming* light
cifz
CC BY-SA 3.0
2016-07-22T21:13:54.303Z
The short answer is: _"because then it wouldn't be bidirectional"_. It's been a while, but I believe my [alternate formulation](http://geometrian.com/programming/tutorials/clearer_re/index.php) of the rendering equation works out to be using a 1:1 reflectance function.
geometrian
CC BY-SA 4.0
2021-05-11T12:52:28.347Z
@cifz Irradiance is $W/m^{2}$, it's not dependent on directions.
akuzminykh
CC BY-SA 4.0
2021-11-28T18:44:06.377Z
@cifz Saying that irradiance is incoming and radiance is outgoing light can be very misleading for beginners as it begs you to draw the conclusion that this is the actual difference between those two terms. In fact, radiance can be incoming and outgoing and irradiance has nothing to do with direction. Its light from any direction hitting a surface (you made that clear in the braces, but it should be emphasized more)
Sam
<brdf>
Why isn't a BRDF a ratio of radiances?
PeteUK
# Why isn't a BRDF a ratio of radiances?

(Asked by: username_0 on 2016-07-15)

I'm Learning about BRDFs and wondering why the BRDF is defined as the ratio of outgoing radiance to a given direction and incoming irradiance from another direction. Why isn't the BRDF defined as ratio of radiances?

**Tags:** brdf

> **Comments:**
> **username_1**: I would write an answer if I had time, but in a concise way: because of definition. Roughly speaking radiance measure OUTGOING light in a certain direction (or better: radiant flux per solid angle). Irradiance is INCOMING light from a certain direction (or better radiant flux received per unit area. BRDF is describing the ratio of *outgoing* light to *incoming* light  
> **username_2**: The short answer is: _"because then it wouldn't be bidirectional"_. It's been a while, but I believe my [alternate formulation](http://username_2.com/programming/tutorials/clearer_re/index.php) of the rendering equation works out to be using a 1:1 reflectance function.  
> **username_3**: @username_1 Irradiance is $W/m^{2}$, it's not dependent on directions.  
> **username_4**: @username_1 Saying that irradiance is incoming and radiance is outgoing light can be very misleading for beginners as it begs you to draw the conclusion that this is the actual difference between those two terms. In fact, radiance can be incoming and outgoing and irradiance has nothing to do with direction. Its light from any direction hitting a surface (you made that clear in the braces, but it should be emphasized more)  

---

## 1 answers

### Answer by username_5 on 2016-07-22. Score: 15 | Accepted answer

There are a couple of ways to answer this question: an algebraic way and a geometric way.

Algebraically, we can identify the units that the BRDF must have by looking at its place in the rendering equation. The classic rendering equation is:

$$L\_\text{outgoing}(\omega) = L\_\text{emitted}(\omega) + \int\_\Omega L\_\text{incoming}(\omega') \, f\_\text{BRDF}(\omega, \omega') \, (n \cdot \omega') \, d\omega'$$

The output value on the left is a radiance, so the result of the integral must also be a radiance. The integrand contains a radiance multiplied by a solid angle $d\omega'$, so something else in the integrand has to cancel out that factor of solid angle. The $n \cdot \omega'$ factor is dimensionless, and the only other thing there is the BRDF—so to make the whole thing work out, the BRDF must have units of inverse solid angle. Equivalently, the BRDF can be seen as a ratio of radiance to irradiance, since they differ by a factor of solid angle in the denominator of radiance.

Another way to see it is that the BRDF plays a role similar to a probability density. If you look at how probability densities work, they have units inverse to the volume of their domain. For instance, a 1D probability density has units of inverse length (probability per unit length, but probability itself is dimensionless), a 2D one has units of inverse area, and so on. The BRDF acts much like a probability density defined on the hemisphere, giving a likelihood for a photon coming in from a given direction to be reflected into some other direction. So, like any other probability density on a spherical domain, it has units of inverse solid angle.

Geometrically, we can get right down to brass tacks and take apart what's going on in the integral in the rendering equation. Recall that an integral means to subdivide the domain into tiny pieces and sum the integrand over all the pieces (in the limit as the pieces get infinitesimally small). Let's look at one such piece. The integrand should result in an infinitesimal amount of radiance $dL$, since we're going to sum over many pieces to arrive at a finite outgoing radiance. So a single infinitesimal piece of the integral looks like:

$$dL = L\_\text{incoming} \, f\_\text{BRDF} \, (n \cdot \omega') \, d\omega'$$

If we regroup the factors a bit, the combination $L\_\text{incoming} \, (n \cdot \omega') \, d\omega'$ calculates the irradiance on the surface due to the light coming from the infinitesimal solid angle $d\omega'$. Since it's arriving from an infinitesimal amount of solid angle, it produces an infinitesimal irradiance $dE$.

$$dL = f\_\text{BRDF} \, dE$$

or

$$f\_\text{BRDF} = \frac{dL}{dE}$$

So the BRDF acts as a proportionality constant between the infinitesimal irradiance arriving at the surface from an infinitesimal solid angle, and the infinitesimal outgoing radiance generated thereby. It couldn't be a ratio of radiances, because we have a *finite* incoming radiance, and we need an *infinitesimal* outgoing radiance if we want to sum up many pieces of the integral and get a finite result. To make that happen, the BRDF would have to be infinitesimal-valued, which...isn't a thing, in standard mathematics. :)

I hope some of this helps. There are a variety of equivalent ways to look at this problem, as with so many things in math and physics.

> **Comments:**
> **username_6**: I like your explanation very much. I get the arguments that there must be the inverse solid angle factor in BRDF, but what about the cosine factor? If we could drop the cosine term from BRDF, then we could drop if from the integral in the rendering equation, couldn't we? The only reason I can see is in the correct/current formulation the denominator can be seen as irradiance...  
> **username_3**: This answer kinda explains it as if BRDFs are derived from the rendering equation while the concept "BRDF" is actually older. Looking at BRDFs isolated: Why is it not the ratio between rediances?  
> **username_5**: @username_3 It's not really about the rendering equation specifically, but about how integrals work. The point of a BRDF is to integrate incoming radiance over the whole hemisphere and produce output radiance. The integral will introduce a unit of solid angle (or more generally, a unit of "volume" of whatever the domain of the integral is), so the integrand must be a density in that unit. As I said in the answer, it's the same as how probability densities work.  
> **username_3**: @username_5 But isn't the irradiance in context of BRDF only including radiant flux coming from a specific direction, not the whole hemishere? When I e.g. read [this](https://graphics.stanford.edu/courses/cs448-05-winter/papers/nicodemus-brdf-nist.pdf), I just cannot pinpoint where the decision based on reasoning is made to let it be irradiance at all. Maybe I just need to read it a few times more ..  
> **username_5**: @username_3 That the BRDF is defined in terms of irradiance on a surface is in part just a convention. You might compare it to the phase functions found in volume rendering, which play a similar role to the BRDF—deciding how much light from one direction scatters to another. In that case there is no surface and no irradiance (no $n \cdot \omega$ factor), but the phase function is still a density over solid angle. If you're still having trouble, again I would urge thinking about probability densities and how their units work. It is the same thing.  
> **username_3**: @username_5 Thank you for the additional explanations!  

---

