4
$\begingroup$

I'm reading PBRT and am stuck in chapter 5.4 about radiometry. In particular:

We define the irradiance as the average density flux arriving at a surface with units $\frac{W}{m^2}$. So for a point light source, we have: $E = \frac{\Phi}{4 \pi r^2}$ since the area of a sphere is $4 \pi r^2$. Where $\Phi$ is the flux or power.

A (to me) similar concept is intensity which is the amount of power per angle. Again, for a sphere with a point light at the center, this is $I = \frac{\Phi}{4 \pi r^2}$ with the unit $[\frac{W}{sr}]$ (watt over steradian)

Now, the book defines radiance for a point $p$ as $L = \frac{d\Phi}{d\omega dA^\perp}$ in units $[\frac{W}{sr\cdot m^2}]$. Here, $\omega$ is the direction where the light comes from, $A^\perp$ is the projected are of $A$ as seen here:

enter image description here

This means that practically, when I implement a point light source with a given power that shines at a point $p$, I need to do the following to arrive at radiance:

  • Divide by $4 \pi r^2$ to convert power into $[\frac{W}{sr}]$, or in other words, intensity.
  • Given intensity, I need to divide it by $4 \pi r^2$ and multiply by $\cos \theta$ to arrive at $[\frac{W}{sr\cdot m^2}]$, the final radiance. The multiplication by $\cos \theta$ is to project $A$ to $A^\perp$ and is the dot product of the surface normal $n$ with the direction $w$ (as both are normalized).

For both calculations, $r$ is the distance between the light source and my point $p$.

However, when I look at the source, this is not what happens. The point light returns intensity divided by $r^2$ as seen here:

return I / DistanceSquared(pLight, ref.p);

and the integrator then multiplies it with the dot product (and the brdf) in the whitted integrator

L += f * Li * AbsDot(wi, n) / pdf;

So what is wrong in my derivation? Why do we "only" divide once by $4\pi r^2$ (to get the Intensity I) and not twice? Aren't we missing either the power per area or the power per steradian?

sources: http://www.pbr-book.org/3ed-2018/Color_and_Radiometry/Radiometry.html

$\endgroup$
3
$\begingroup$

Your definition for radiant intensity is wrong: it should be just $\Phi / 4\pi$. There are only $4\pi$ steradians in a sphere no matter how big it is, so $r$ doesn't come into it.

Also note that you can't calculate radiance for a point source—it would be infinite, due to the fact the point source emits a finite amount of flux compressed into zero size. It subtends zero solid angle, from the receiver's point of view, so the $d\omega$ factor in the denominator of radiance would be zero. The usable quantities are radiant intensity in a certain direction from the point source (which could vary with direction, for a non-omnidirectional light), and irradiance in a certain direction and distance. As you've seen, irradiance is obtained from radiant intensity by dividing by $r^2$—you could think of this as "area per steradian", as it's the conversion factor from $4\pi$ steradians to $4\pi r^2$ area of a sphere; then the units work out.

Where radiance would actually show up is when dealing with an area light rather than a point light. Then you would have the flux being distributed over a finite solid angle from the receiver's point of view, and you'd get the incident irradiance by integrating the light's radiance over that solid angle. The emitted radiance would be $\Phi/(2\pi A_\text{light})$, assuming it's emitted uniformly over the surface of the light and into all directions from each point. This is only $2\pi$ since it's only emitting into the outward-facing hemisphere. Also note that you don't do any division by $r^2$ for radiance—with area lights, the distance attenuation comes naturally as a result of the light subtending less solid angle from the receiver's point of view, when the receiver is farther away.

$\endgroup$
1
  • 1
    $\begingroup$ thank you for this comprehensive answer, this cleared a bunch of confusion for me! $\endgroup$
    – lyinch
    Oct 6 '20 at 13:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.