Let's say I use a GGX distribution to model the NDF part of Cook-Torrance, such as here: https://de45xmedrsdbp.cloudfront.net/Resources/files/2013SiggraphPresentationsNotes-26915738.pdf​​

When the roughness is low, value of NDF can go up to well above 1, something in the range of hundreds of thousands. Therefore the whole BRDF would evaluate to a big number.

Now, say we have a single directional light source, a smooth surface (low roughness), and a camera. The half vector is equal to the surface normal, and we try to evaluate the amount of light reflected from surface to camera:

Lo(x, w) = brdf(x, h, v, n, l) * Li(x, l) * dot(n, l), where h=n

Then apparently, the outgoing radiance is stronger than the incoming radiance at the surface by hundreds of thousands?

That's my first question. If we convert incoming radiance to camera to our final pixel value, how does it make sense that reflected radiance can be stronger than the light source?

Now's my second question. I understand radiance's unit is a watts per area per steradian. To evaluate the actual light landed on the imaging sensor, I have to compute the rendering equation (integrate the same brdf*Li for all light incoming from every direction). After this integral, I'm sure the integrated radiance will have a value less than the original light source, since BRDFs are designed to integrate to 1 over hemisphere. However, doesn't this contradict the approximation we are using to compute pixel values in real-time graphics (using radiance directly to compute pixel value)?

Any help is appreciated, thanks.

I just took a shower and I might have this figured out. Here are what I think the errors in my mental model are.

correction #1:

To compute outgoing radiance, BRDF should be applied to the surface irradiance, not incoming radiance. My hypothesis is, it is not realistic to expect the surface to receive a good amount of light from a single light ray carrying some radiance. If anything, the irradiance should approach zero, as the only incoming radiance has a solid angle that is infinitesimal.

(edit: I guess what I'm trying to say is, it's nonsense to calculate surface irradiance from some radiance along a single incoming direction. They don't have the same units)

correction #2:

It's actually completely okay for the outgoing radiance along a single direction to approach infinity, even if it's reflected from a surface with a finite amount of irradiance. In my question, I was trying to draw equal signs between radiance and irradiance. This is wrong, as the outgoing radiance has an infinitesimal solid angle (also they use completely different units). Given a surface that reflects light around its hemisphere, it's completely reasonable to have radiance approach infinity along some set of directions. Since the integral of the outgoing radiance around the hemisphere is still finite or even small.

These are my intuitions so far. I don't think I chose the best wordings as it's still a little confusing in my head. Please add or correct me ?

@MJP Thanks MJP! That clears it up for me, I can always count on you ?