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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?

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    $\begingroup$ 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 $\endgroup$ – cifz Jul 15 '16 at 21:06
  • $\begingroup$ The short answer is: "because then it wouldn't be bidirectional". It's been a while, but I believe my alternate formulation of the rendering equation works out to be using a 1:1 reflectance function. $\endgroup$ – imallett Jul 22 '16 at 21:13
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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.

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  • $\begingroup$ 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... $\endgroup$ – ciechowoj Mar 11 '17 at 11:50

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