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


1 Answer 1


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.

  • 1
    $\begingroup$ thank you for this comprehensive answer, this cleared a bunch of confusion for me! $\endgroup$
    – lyinch
    Oct 6, 2020 at 13:47
  • $\begingroup$ I don't get this part: "irradiance is obtained from radiant intensity by dividing by $r^2$". Isn't the intensity distributed over a sphere of area $4\pi r^2$ ? Overall, that would amount to: Pseudo-L = $\frac{ \Phi }{ (4\pi)^2 r^2 }$. $\endgroup$
    – Gab
    Aug 2, 2022 at 12:09
  • $\begingroup$ @Gab The radiant intensity already has the 1/4π embedded in it (it's $\Phi/4\pi$). Radiant intensity is already per-steradian, so you don't need to put in another factor of 1/4π to get the irradiance. $\endgroup$ Aug 3, 2022 at 19:57
  • $\begingroup$ Yes, but then what did you mean by the term area when you said "think of this as 'area per steradian'" ? Which area did you mean? $\endgroup$
    – Gab
    Aug 6, 2022 at 14:04
  • $\begingroup$ I was describing $r^2$ as a conversion factor from steradians to area on the sphere. $4\pi$ steradians -> $4\pi r^2$ area. So if you turn it around a bit, then $r^2$ is the area corresponding to 1 steradian. It converts between quantities that are per-steradian (radiant intensity) and those that are per area (irradiance). $\endgroup$ Aug 8, 2022 at 14:42

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