<p>Your definition for radiant intensity is wrong: it should be just <span class="math-container">$\Phi / 4\pi$</span>. There are only <span class="math-container">$4\pi$</span> steradians in a sphere no matter how big it is, so <span class="math-container">$r$</span> doesn't come into it.</p>
<p>Also note that you can't calculate <em>radiance</em> 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 <span class="math-container">$d\omega$</span> 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 <span class="math-container">$r^2$</span>—you could think of this as &quot;area per steradian&quot;, as it's the conversion factor from <span class="math-container">$4\pi$</span> steradians to <span class="math-container">$4\pi r^2$</span> area of a sphere; then the units work out.</p>
<p>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 <span class="math-container">$\Phi/(2\pi A_\text{light})$</span>, assuming it's emitted uniformly over the surface of the light and into all directions from each point. This is only <span class="math-container">$2\pi$</span> since it's only emitting into the outward-facing hemisphere. Also note that you don't do any division by <span class="math-container">$r^2$</span> 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.</p>

CC BY-SA 4.0
2020-10-06T13:47:29.293Z
thank you for this comprehensive answer, this cleared a bunch of confusion for me!
lyinch
CC BY-SA 4.0
2022-08-02T12:09:23.393Z
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 = $\frac{ \Phi }{ (4\pi)^2 r^2 }$.
Gab
CC BY-SA 4.0
2022-08-03T19:57:53.420Z
@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.
Nathan Reed
CC BY-SA 4.0
2022-08-06T14:04:49.193Z
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?
Gab
CC BY-SA 4.0
2022-08-08T14:42:40.943Z
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).
Nathan Reed
CC BY-SA 4.0
2020-10-03T18:18:31.463Z
2020-10-03T18:43:43.177Z
2020-10-03T18:43:43.177Z
Nathan Reed
<p>I'm reading PBRT and am stuck in chapter 5.4 about radiometry. In particular:</p>
<p>We define the <em>irradiance</em> as the average density flux arriving at a surface with units <span class="math-container">$\frac{W}{m^2}$</span>. So for a point light source, we have: <span class="math-container">$E = \frac{\Phi}{4 \pi r^2}$</span> since the area of a sphere is <span class="math-container">$4 \pi r^2$</span>. Where <span class="math-container">$\Phi$</span> is the flux or power.</p>
<p>A (to me) similar concept is <em>intensity</em> which is the amount of power per angle. Again, for a sphere with a point light at the center, this is <span class="math-container">$I = \frac{\Phi}{4 \pi r^2}$</span> with the unit <span class="math-container">$[\frac{W}{sr}]$</span> (watt over steradian)</p>
<p>Now, the book defines <em>radiance</em> for a point <span class="math-container">$p$</span> as <span class="math-container">$L = \frac{d\Phi}{d\omega dA^\perp}$</span> in units <span class="math-container">$[\frac{W}{sr\cdot m^2}]$</span>. Here, <span class="math-container">$\omega$</span> is the direction where the light comes from, <span class="math-container">$A^\perp$</span> is the projected are of <span class="math-container">$A$</span> as seen here:</p>
<p><a href="https://i.sstatic.net/co4VV.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/co4VV.png" alt="enter image description here" /></a></p>
<p>This means that practically, when I implement a point light source with a given power that shines at a point <span class="math-container">$p$</span>, I need to do the following to arrive at radiance:</p>
<ul>
<li>Divide by <span class="math-container">$4 \pi r^2$</span> to convert power into <span class="math-container">$[\frac{W}{sr}]$</span>, or in other words, intensity.</li>
<li>Given intensity, I need to divide it by <span class="math-container">$4 \pi r^2$</span> and multiply by <span class="math-container">$\cos \theta$</span> to arrive at <span class="math-container">$[\frac{W}{sr\cdot m^2}]$</span>, the final radiance. The multiplication by <span class="math-container">$\cos \theta$</span> is to project <span class="math-container">$A$</span> to <span class="math-container">$A^\perp$</span> and is the dot product of the surface normal <span class="math-container">$n$</span> with the direction <span class="math-container">$w$</span> (as both are normalized).</li>
</ul>
<p>For both calculations, <span class="math-container">$r$</span> is the distance between the light source and my point <span class="math-container">$p$</span>.</p>
<p>However, when I look at the source, this is not what happens. The point light returns intensity divided by <span class="math-container">$r^2$</span> as seen <a href="https://github.com/mmp/pbrt-v3/blob/master/src/lights/point.cpp#L44" rel="nofollow noreferrer">here</a>:</p>
<pre><code>return I / DistanceSquared(pLight, ref.p);
</code></pre>
<p>and the integrator then multiplies it with the dot product (and the brdf) <a href="https://github.com/mmp/pbrt-v3/blob/master/src/integrators/whitted.cpp#L79" rel="nofollow noreferrer">in the whitted integrator</a></p>
<pre><code>L += f * Li * AbsDot(wi, n) / pdf;
</code></pre>
<p>So what is wrong in my derivation? Why do we &quot;only&quot; divide once by <span class="math-container">$4\pi r^2$</span> (to get the Intensity I) and not twice? Aren't we missing either the power per area or the power per steradian?</p>
<p>sources:
<a href="http://www.pbr-book.org/3ed-2018/Color_and_Radiometry/Radiometry.html" rel="nofollow noreferrer">http://www.pbr-book.org/3ed-2018/Color_and_Radiometry/Radiometry.html</a></p>

<raytracing><pathtracing>
Deriving radiance from irradiance and intensity
lyinch
# Deriving radiance from irradiance and intensity

(Asked by: username_0 on 2020-10-03)

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](https://i.sstatic.net/co4VV.png)](https://i.sstatic.net/co4VV.png)

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](https://github.com/mmp/pbrt-v3/blob/master/src/lights/point.cpp#L44):

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

```

and the integrator then multiplies it with the dot product (and the brdf) [in the whitted integrator](https://github.com/mmp/pbrt-v3/blob/master/src/integrators/whitted.cpp#L79)

```
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>

**Tags:** raytracing, pathtracing


---

## 1 answers

### Answer by username_1 on 2020-10-03. Score: 5 | Accepted answer

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.

> **Comments:**
> **username_0**: thank you for this comprehensive answer, this cleared a bunch of confusion for me!  
> **username_2**: 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 = $\frac{ \Phi }{ (4\pi)^2 r^2 }$.  
> **username_1**: @username_2 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.  
> **username_2**: 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?  
> **username_1**: 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).  

---

