I'm writing a ray tracer. Assumptions which led to this point:

  • Rays are in units of spectral radiance (watts per steradian per square metre) - this appears to be correct based on the subsequent maths which have to be used to convert the ray to a colour. (Actually I also do it spectrally, so there is a per-nanometre term as well, which I'm omitting to keep this question simpler.)
  • Point lights will be specified in units of luminous flux (watts)

So given the luminous flux, I have to compute the radiance. The equation given for this is,

$$ L_{e,\Omega} = {\partial^2 \Phi_e \over \partial \Omega \partial A\cos\theta} $$


  • $L_{e,\Omega}$ is the radiance
  • $\partial$ are partial derivatives
  • $\Phi_e$ is the luminous flux
  • $\Omega$ is the solid angle
  • $A$ is the area
  • resulting units: watts per steradian per square metre.

But point lights have no area, so it wasn't immediately clear how to deal with this equation. Also, sources differ in what they mean by the solid angle and the area.

In any case, after thinking about it for a while, my reasoning is as follows:

  1. Suppose I want the radiance at a given point, distance $R$ (in metres) from the light.
  2. I construct a sphere of radius $R$ around the point light.
  3. Since it's a sphere, the whole surface is normal to the source, so $\cos \theta$ is always 1.
  4. Also since it's a sphere, the radiance at all points must be the same, so I should get the same result for any area I choose.
  5. I choose to use the entire sphere. Therefore:
    • $\partial \Phi_e$ is just $\Phi_e$
    • $\partial \Omega$ for the entire sphere is just $4\pi$ steradians
    • $\partial A \cos \theta$ for the entire sphere is just $4\pi R^2$

So I get,

$$ L_e = {\Phi_e \over 4\pi \cdot 4\pi R^2} $$

It at least comes out showing that it depends inversely by the square of the distance. But what makes me less than confident about this is the appearance of a $\pi^2$ in the result.

So my question is, is this reasoning sound? How are the area and the solid angle meant to work - is it the area at the point you're observing the light from, and the solid angle covering that area? Is this effectively putting the solid angle into the equation twice?

I've talked to a few people independently about this stuff, and all of them seem confused as to how a "per square metre" term ends up in the units at all, because the "per steradian" term takes care of the inverse square law and then there's no perception that a second term should be needed. Myself, the whole time I was under the impression that the "per square metre" was taking care of the inverse square law, but that turns out not to be the case at all. (LOL)


3 Answers 3


As you remarked, radiance is only a measure for a illuminated area, the point source just has a certain luminous flux $\Phi$ in Watt, this luminous decreases with 1/ area of sphere so the luminous flux at distance r from the point source is $\frac{\Phi(r)}{4\pi r^2}$ with this $\Phi$ you can than calculate the radiance of your object.

  • $\begingroup$ So maybe what I should be doing is assuming the point light is actually a very small sphere and then observing that sphere at a distance and trying to calculate that. I can't tell intuitively whether the result would differ. I also found a reference in the PBR book which also says, what a point light has is radiant intensity $I$, and they then go on to use just $I/R^2$. $\endgroup$
    – Hakanai
    Sep 29, 2022 at 3:23
  • $\begingroup$ A ray tracer asks for the incoming intensity, and than calculates the radiance of the illuminated object. So I don't understand, why you need the radiance of your source of illumination? $\endgroup$
    – trula
    Sep 29, 2022 at 9:25
  • $\begingroup$ I guess because all other rays are in units of radiance, it made sense for the ray cast to the light to have the same units. $\endgroup$
    – Hakanai
    Sep 30, 2022 at 1:58

I have the same problem: We must consider some differential solid angle ($d\omega$) excitant from a differential area ($dA$) from the point light source. However, since a point light is infinitely small, it has no area.

I think that while the derivation of the inverse square law does take some $dA$ on the light source, the integral removes the term. Consequently, it can be applied to point lights. I think that the inverse square law is simply the definition of radiance from a point light at distance $r$.

I found this derivation on hyperphysics which shows this: $$E = \frac{\Phi}{A} = \frac{I\Omega}{A} = \frac{IA}{Ar^2} = \frac{I}{r^2}.$$

Side-note: From my understanding, "per square metre" is what "takes care" of the inverse square law. Since the surface area of a steradian scales with the radius, and the intensity of light is measured in watts per steradian, the intensity is independent of radius. The "$\text{m}^2$" term is the surface area of the sphere subtended by the solid angle at radius $r$, meaning it accounts for the increase in surface area of the steradian as the radius increases; hence, the inverse-square-law.

Apologies for the sub-par answer: I'm not an expert, but I hope this can help someone.

  • $\begingroup$ I ended up parking my own work until I come up with some better ideas, but I did manage to dig inside other ray tracers to see what they were doing, and at least in the case of the PBR book, they use a kind of sampling where objects which are larger or closer to the object being lit will simply get more samples taken, thus get more brightness. So the inverse square law kind of gets applied as a side-effect of how the sampling works, while the inverse square law on a point light is a hack because otherwise a point light would never get sampled at all as it has no area. $\endgroup$
    – Hakanai
    Jan 17, 2023 at 6:36

The way you calculate the radiance is actually to regard center position as a receiver with differential area $\mathrm{d}A$ absorbing light coming from any direction (all directions from two hemispheres), but that's not how you get radiance along a line (a specific direction). See my illustration below. 1

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