<p>First of all, <strong>irradiance</strong> at a certain point of a surface is the density of radiant flux (power) per unit of surface area, while <strong>radiance</strong> at a certain point of a surface in a certain direction is the density of radiant flux per unit of surface area and unit of solid angle. Vaguely speaking, <strong>irradiance</strong> is the amount of light incoming to a certain point from possibly all directions, while <strong>radiance</strong> is the amount of light incoming to a point from a single direction. Why on Earth do we talk about irradiance from a direction then?</p>
<p>The thing is that the BRDF definition assumes that there is radiance <span class="math-container">$L_i\left(\omega_i\right)$</span> incoming to point <span class="math-container">$x$</span> through an infinitesimal cone <span class="math-container">$\mathrm{d}\omega_{i}$</span> around the direction <span class="math-container">$\omega_i$</span> with solid angle size denoted <span class="math-container">$\mathrm{d}\sigma\left(\omega_{i}\right)$</span>, which then generates some irradiance on the surface. This &quot;partial&quot; irradiance is then denoted <span class="math-container">$\mathrm{d}E\left(\omega_{i}\right)$</span>. This incident light can be partially absorbed by the surface and partially scattered in all directions. The out-scattered contribution of <span class="math-container">$\mathrm{d}E\left(\omega_{i}\right)$</span> to radiance outgoing in direction <span class="math-container">$\omega_o$</span> is denoted as <span class="math-container">$\mathrm{d}L_{o}\left(\omega_{o}\right)$</span>.</p>
<p>It can be shown experimentally that the amount of outgoing radiance <span class="math-container">$\mathrm{d}L_{o}\left(\omega_{o}\right)$</span> is proportional to the incoming irradiance <span class="math-container">$\mathrm{d}E\left(\omega_{i}\right)$</span> and the BRDF is then defined simply as the ratio between the two:</p>
<p><span class="math-container">$$
f_{r}\left(\omega_{i}\rightarrow\omega_{o}\right) = \frac{\mathrm{d}L_{o}\left(\omega_{o}\right)}{\mathrm{d}E\left(\omega_{i}\right)}
$$</span></p>

CC BY-SA 3.0
2018-04-22T11:06:59.993Z
I think, I currently have understood this concept better. The way I think about this function is to add dw term so BRDF = ((dL / dw) / (dE / dw)). dE / dw is input radiance from this w direction. dL / dw is partial outgoing radiance that is contributed from input direction w. . Is there any error in the way I see this now?
kevinyu
CC BY-SA 4.0
2021-09-29T06:19:49.013Z
Or even more vague; irradiance is incoming light, radiance outgoing light.
jackw11111
CC BY-SA 4.0
2018-04-15T12:58:41.717Z
2021-02-22T12:10:53.687Z
2021-02-22T12:10:53.687Z
ivokabel
<p>It is helpful if you always look at the units that a certain physical quantity measures. Since you use Real-Time Rendering, I'll also quote from that (3rd edition). Also, for the sake of completeness, I'll go through all quantities and units related. I will however assume you understand solid angles. The time <span class="math-container">$t$</span> is measured in <em>seconds</em> <span class="math-container">$\left[s\right]$</span> and the solid angles <span class="math-container">$\omega$</span> are measured in <em>steradians</em> <span class="math-container">$\left[sr\right]$</span>.</p>
<ol>
<li><p><em>radiant energy <span class="math-container">$Q$</span></em> (in <em>joules</em>, <span class="math-container">$\left[J\right]$</span>) measures the energy, i.e. the energy of a photon times the number of photons.</p>
</li>
<li><p><em>radiant flux <span class="math-container">$\Phi$</span></em> (in <em>watts</em>, <span class="math-container">$\left[W\right]= \left[\frac{J}{s}\right])$</span> measures the energy per time, e.g. don't just count the number of photons but the number of photons per second. <span class="math-container">$\Phi = \frac{Q}{t}$</span></p>
</li>
<li></li>
</ol>
<p>a) <em>irradiance <span class="math-container">$E$</span></em> (in <em>watts per square meter</em>, <span class="math-container">$\left[\frac{W}{m^2}\right]$</span>) measures the energy per time and surface area, or the flux per surface area. <span class="math-container">$E = \frac{Q}{t A} = \frac{\Phi}{A}$</span></p>
<p>b) <em>radiosity <span class="math-container">$M$</span></em> (in some papers also <span class="math-container">$B$</span>) is the same as irradiance, only it's leaving a surface and not arriving at it</p>
<ol start="4">
<li><em>radiance <span class="math-container">$L$</span></em> (<em>watts per square meter per steradian</em>, <span class="math-container">$\left[\frac{W}{m^2 sr}\right]$</span>) is the radiant flux per area and solid angle, or the irradiance per solid angle. <span class="math-container">$L = \frac{\Phi}{A w}$</span></li>
</ol>
<p>Now there is one thing to consider: <span class="math-container">$E$</span> is measured with regards to a surface <span class="math-container">$A$</span> that is perpendicular to the light direction (in other words, the normal of the surface is parallel to the light direction). Therefore we project <span class="math-container">$A$</span> onto a plane that fulfills this requirement. If the angle between the surface normal and the light direction is <span class="math-container">$\theta$</span>, then our projected surface <span class="math-container">$A_{proj}$</span> is calculated thus: <span class="math-container">$$A_{proj} = \frac{A}{\cos\theta}$$</span></p>
<p><a href="https://i.sstatic.net/aJrFR.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/aJrFR.png" alt="Visualization of a surface being projected onto a plane, where the projected surface has a different size based on the angle of projection, from &quot;Parameterbasierte Texturgenerierung und echtzeitfähiges Rendering von nassen und trockenen Straßenoberflächen in kamerabasierten ADAS Tests&quot;, Tim Lobner" /></a></p>
<p>With this, we can of course also further our radiance calculation to <span class="math-container">$L = \frac{E}{\omega}$</span></p>
<p>But still, <span class="math-container">$E$</span> only considers the amount of energy per time, not from which direction it comes. Why is this important? Because of the way you usually look at lighting in computer graphics. You calculate how much light is being reflected from a surface to your viewer (/camera), which also means, that you want to know from which light source it originates (seeing as you would like to have the right amount of energy and the correct color). Additionally, you usually use point lights, meaning that you can view the lighting calculation as that of a ray from a single point (the light source) onto a single point on a surface (your pixel/fragment) and then to your viewer. These directions are written in the matter of solid angles, or to make the theory even easier, in differential solid angles. Another point is that your surface may reflect light differently depending on the where it comes from, which also makes important the direction part.</p>
<p>So to sum this up a bit:</p>
<blockquote>
<p>I can't visually understand what does irradiance of a certain direction means?</p>
</blockquote>
<p>It basically means photons from a specific light source, not from any place in space.</p>
<blockquote>
<p>And what is the difference with radiance of certain direction?</p>
</blockquote>
<p>I hope it is clear, that radiance is irradiance from a certain direction. If not, please try to specify which part still bothers you.</p>
<blockquote>
<p>They both represents the power of light</p>
</blockquote>
<p>Yes they do. In fact, <em>steradian</em> is a dimensionless unit, since it is <span class="math-container">$\left[\frac{m^2}{m^2}\right]$</span>, and therefore it doesn't really add anything. I see how this is confusing. I hope I could clear up why you do this.</p>
<blockquote>
<p>What does irradiance of a certain direction means? Doesn't we associate a direction with radiance?</p>
</blockquote>
<p>Careful. We don't associate direction (other than considering only surfaces perpendicular to the light direction) with irradiance. We do however with radiance.</p>
<blockquote>
<p>Isn't radiance defined as irradiance in a single direction?</p>
</blockquote>
<p>Yes, you can say that.</p>
<blockquote>
<p>Is E(v) &lt;= L(v) where v is direction?</p>
</blockquote>
<p>I'd say it is the other way round, <span class="math-container">$L(v) \leq E(v)$</span>, since <span class="math-container">$E(v)$</span> would consider any light source emitting light onto a surface, whereas <span class="math-container">$L(v)$</span> only considers light form <span class="math-container">$v$</span> (if you consider <span class="math-container">$v$</span> as the light source to surface direction. In my writing earlier, this is <span class="math-container">$\omega$</span> and in Real Time Rendering specifically, this is <span class="math-container">$\omega$</span> as well, or <span class="math-container">$l$</span> in chapter 5 as well as in the BRDF Theory chapter). Also, remember that these two physical quantities don't have the same units and should not really be compared this way.</p>

CC BY-SA 4.0
2018-04-14T16:49:34.570Z
2024-04-16T10:01:50.703Z
2024-04-16T10:01:50.703Z
Tare
<p>To be honest, terms like these are very confusing as they aren't clear cut and on one side of the border. They are more grayish.</p>
<p>I'm gonna tell you how I convinced myself, as I too had this confusion as soon as I read your question. But I managed to convince myself through this argument.</p>
<p>First of all we are gonna clear up 4 terms, Radiance, Irradiance, Differential radiance and Differential Irradiance.</p>
<p><strong>&quot;Radiance&quot; is what you say associated with a certain direction. To be more formal and according to wikipedia,</strong></p>
<blockquote>
<p>It's the amount of radiant flux emitted/transmitted/received per unit
projected area, per unit solid angle.</p>
</blockquote>
<p><strong>Next is differential radiance. We can think of it as an infinitesimal quantity of radiance emitted or recieved in a very small solid angle <span class="math-container">$d\omega$</span>.</strong></p>
<p><strong>Next is Irradiance. Irradiance isn't normally associated with a direction. According to Wikipedia it's</strong></p>
<blockquote>
<p>Radiant flux received by a surface per unit area</p>
</blockquote>
<p>But more commonly and what makes more sense to me, and to the answer of your question, think of irradiance as the integration of radiances over a set of directions.</p>
<p>So we can say</p>
<p><span class="math-container">$E = \displaystyle\int_{\Omega} L(\omega)\;  \omega.n \;d\omega\\\omega \in \Omega$</span></p>
<p>So if we integrate the radiances from every direction that leads us to the original definition of irradiance where direction isn't of concern. However usually we are concerned with only a subset of all the directions such as the Upper hemisphere or the lower hemisphere. This means for example,</p>
<p><span class="math-container">$\Omega = \{ \omega : \omega.n \geq 0 \}$</span></p>
<p>As we can see here, we have limited the irradiance to a set of directions, the upper hemisphere. This doesn't necessarily change it into radiance which is associated by direction. Instead what this means is, when calculating irradiance we are concerned with the light coming only from these directions, although we haven't incorporated the directional quantity into the formula like with radiance.</p>
<p>This is the difference between irradiance from a certain direction and radiance. Think of it like this. You are holding a paper and there are 2 light bulbs in front of you. You want to measure the irradiance. Normally it would just be the radiant flux received by both bulbs per unit area. But now let's say I limit the direction so I am only concerned with the first bulb. Note that I am still calculating the &quot;irradiance&quot;. If I move farther away the flux will decrease thus the irradiance even tho I am concerned with a specific direction. However this isn't the case with radiance where moving farther away won't change it since we divide by the solid angle too balancing the change.</p>
<p><strong>The last quantity is differential irradiance. I thought of it as a tiny amount of irradiance from a specific direction. (Again direction gets involved)</strong></p>
<p>If you think of irradiance as not assosciated with direction at all, even then when you try to think of differential irradiance you are gonna say it's a tiny amount of irradiance from a small range of direction or maybe a specific direction. That's the reason why it's small.</p>
<p>But if you think of irradiance as the sum of all the radiances over a specific set of directions. You'll see that it makes things clearer and you'll naturally arrive at the conclusion that differential irradiance will then refer to an irradiance from a specific direction.</p>
<p>So coming back to your question at last, I hope you might have gotten some intuition as to what &quot;irradiance in a direction means&quot;.</p>
<p>Mathematically proving it is no hard feat tho. The answer <a href="https://computergraphics.stackexchange.com/questions/3757/why-isnt-a-brdf-a-ratio-of-radiances">here</a> explains it quite well. I'm just gonna give a brief explanation. We know the rendering equation is given as</p>
<p><span class="math-container">$L_{outgoing} = L_{emission} + \displaystyle\int_{\Omega} L_{incoming} \;f_{BRDF}(\omega_i, \omega_o)\; \omega_i.n \; d\omega_i$</span></p>
<p>Assuming for the moment that emission part is zero. we end up with,</p>
<p><span class="math-container">$L_{outgoing} = \displaystyle\int_{\Omega} L_{incoming} \;f_{BRDF}(\omega_i, \omega_o)\; \omega_i.n \; d\omega_i$</span></p>
<p>Now as I wrote before, if you forget the BRDF for the time being, we are just integrating the radiances over a given set of direction which is the same as irradiance.</p>
<p>If we look at one instance of this summation/integration, it's gonna be</p>
<p><span class="math-container">$dL_{outgoing} =  L_{incoming} \; \omega_i.n \; d\omega_i \; f_{BRDF}(\omega_i, \omega_o)$</span></p>
<p><span class="math-container">$dL_{outgoing} =  dE \; f_{BRDF}(\omega_i, \omega_o)$</span></p>
<p>We put <span class="math-container">$d$</span> with the outgoind radiance and irradiance because it's a very small part (we are looking at just one instance of that summation/integration)</p>
<p><span class="math-container">$f_{BRDF}(\omega_i, \omega_o) = \displaystyle\frac{dL_{outgoing}}{dE} $</span></p>
<p>Which is the ratio of the outgoing radiance to the incoming irradiance.</p>
<p>Again this was just the way I convinced myself and might have some mistakes. Though this is the best I came up with.</p>

CC BY-SA 3.0
2018-04-22T11:14:39.463Z
I think, I currently have understood this concept better. The way I think about this function is to add dw term so BRDF = ((dL / dw) / (dE / dw)). dE / dw is input radiance from this w direction. dL / dw is infinitesmall partial outgoing radiance that is contributed from input direction w. . Is there any error in the way I see this now?
kevinyu
CC BY-SA 3.0
2018-04-23T10:37:07.503Z
I don't quite understand your concept. First of all $dL$ already has a $d\omega$ term in it. So $dL / d\omega$ isn't radiance. Dunno what to call it. Don't try to make it a ratio of radiances.
gallickgunner
CC BY-SA 3.0
2018-04-24T12:23:28.743Z
Well dL/dw is not radiance. But it is **infinitesimal** outgoing radiance that is contributed from radiance from w direction hitting the surface.The outgoing radiance is the integral of this infinitesimal outgoing radiance from every input direction.
kevinyu
CC BY-SA 3.0
2018-04-24T17:06:13.573Z
Nope I still don't understand. First of all, $L_o$ is the outgoing radiance, then that makes $dL_o$ the infinitesimal outgoing radiance not $dL_o/d\omega$. Your second part is somewhat right, the total outgoing radiance is the integral of the infinitesimal outgoing radiances. However that seems like a recursive definition so although you might or might not have this in mind, do keep note that you are actually integrating the differential irradiances from every direction and multiplying by the BRDF to get $L_o$
gallickgunner
CC BY-SA 3.0
2018-04-16T18:08:32.713Z
2018-04-16T18:08:32.713Z
2020-06-18T08:32:12.800Z
gallickgunner
<p>Currently I am reading the BRDF section from Real Time Rendering and I am having a hard time to visually understand the definition of this function. BRDF is the ratio between the radiance in outgoing direction and irradiance of incoming direction. I can't visually understand what does irradiance of a certain direction means? And what is the difference with radiance of certain direction? They both represents the power of light. What does irradiance of a certain direction means? Doesn't we associate a direction with radiance? Isn't radiance defined as irradiance in a single direction? I feel like i misunderstand something here. Is E(v) &lt;= L(v) where v is direction?</p>

<brdf>
What is the difference between radiance and irradiance in BRDF
kevinyu
# What is the difference between radiance and irradiance in BRDF

(Asked by: username_0 on 2018-04-13)

Currently I am reading the BRDF section from Real Time Rendering and I am having a hard time to visually understand the definition of this function. BRDF is the ratio between the radiance in outgoing direction and irradiance of incoming direction. I can't visually understand what does irradiance of a certain direction means? And what is the difference with radiance of certain direction? They both represents the power of light. What does irradiance of a certain direction means? Doesn't we associate a direction with radiance? Isn't radiance defined as irradiance in a single direction? I feel like i misunderstand something here. Is E(v) <= L(v) where v is direction?

**Tags:** brdf


---

## 3 answers

### Answer by username_1 on 2018-04-15. Score: 16

First of all, **irradiance** at a certain point of a surface is the density of radiant flux (power) per unit of surface area, while **radiance** at a certain point of a surface in a certain direction is the density of radiant flux per unit of surface area and unit of solid angle. Vaguely speaking, **irradiance** is the amount of light incoming to a certain point from possibly all directions, while **radiance** is the amount of light incoming to a point from a single direction. Why on Earth do we talk about irradiance from a direction then?

The thing is that the BRDF definition assumes that there is radiance $L\_i\left(\omega\_i\right)$ incoming to point $x$ through an infinitesimal cone $\mathrm{d}\omega\_{i}$ around the direction $\omega\_i$ with solid angle size denoted $\mathrm{d}\sigma\left(\omega\_{i}\right)$, which then generates some irradiance on the surface. This "partial" irradiance is then denoted $\mathrm{d}E\left(\omega\_{i}\right)$. This incident light can be partially absorbed by the surface and partially scattered in all directions. The out-scattered contribution of $\mathrm{d}E\left(\omega\_{i}\right)$ to radiance outgoing in direction $\omega\_o$ is denoted as $\mathrm{d}L\_{o}\left(\omega\_{o}\right)$.

It can be shown experimentally that the amount of outgoing radiance $\mathrm{d}L\_{o}\left(\omega\_{o}\right)$ is proportional to the incoming irradiance $\mathrm{d}E\left(\omega\_{i}\right)$ and the BRDF is then defined simply as the ratio between the two:

$$
f\_{r}\left(\omega\_{i}\rightarrow\omega\_{o}\right) = \frac{\mathrm{d}L\_{o}\left(\omega\_{o}\right)}{\mathrm{d}E\left(\omega\_{i}\right)}
$$

> **Comments:**
> **username_0**: I think, I currently have understood this concept better. The way I think about this function is to add dw term so BRDF = ((dL / dw) / (dE / dw)). dE / dw is input radiance from this w direction. dL / dw is partial outgoing radiance that is contributed from input direction w. . Is there any error in the way I see this now?  
> **username_2**: Or even more vague; irradiance is incoming light, radiance outgoing light.  

---

### Answer by username_3 on 2018-04-14. Score: 15

It is helpful if you always look at the units that a certain physical quantity measures. Since you use Real-Time Rendering, I'll also quote from that (3rd edition). Also, for the sake of completeness, I'll go through all quantities and units related. I will however assume you understand solid angles. The time $t$ is measured in *seconds* $\left[s\right]$ and the solid angles $\omega$ are measured in *steradians* $\left[sr\right]$.

1. *radiant energy $Q$* (in *joules*, $\left[J\right]$) measures the energy, i.e. the energy of a photon times the number of photons.
2. *radiant flux $\Phi$* (in *watts*, $\left[W\right]= \left[\frac{J}{s}\right])$ measures the energy per time, e.g. don't just count the number of photons but the number of photons per second. $\Phi = \frac{Q}{t}$

a) *irradiance $E$* (in *watts per square meter*, $\left[\frac{W}{m^2}\right]$) measures the energy per time and surface area, or the flux per surface area. $E = \frac{Q}{t A} = \frac{\Phi}{A}$

b) *radiosity $M$* (in some papers also $B$) is the same as irradiance, only it's leaving a surface and not arriving at it

4. *radiance $L$* (*watts per square meter per steradian*, $\left[\frac{W}{m^2 sr}\right]$) is the radiant flux per area and solid angle, or the irradiance per solid angle. $L = \frac{\Phi}{A w}$

Now there is one thing to consider: $E$ is measured with regards to a surface $A$ that is perpendicular to the light direction (in other words, the normal of the surface is parallel to the light direction). Therefore we project $A$ onto a plane that fulfills this requirement. If the angle between the surface normal and the light direction is $\theta$, then our projected surface $A\_{proj}$ is calculated thus: $$A\_{proj} = \frac{A}{\cos\theta}$$

[![Visualization of a surface being projected onto a plane, where the projected surface has a different size based on the angle of projection, from "Parameterbasierte Texturgenerierung und echtzeitfähiges Rendering von nassen und trockenen Straßenoberflächen in kamerabasierten ADAS Tests", Tim Lobner](https://i.sstatic.net/aJrFR.png)](https://i.sstatic.net/aJrFR.png)

With this, we can of course also further our radiance calculation to $L = \frac{E}{\omega}$

But still, $E$ only considers the amount of energy per time, not from which direction it comes. Why is this important? Because of the way you usually look at lighting in computer graphics. You calculate how much light is being reflected from a surface to your viewer (/camera), which also means, that you want to know from which light source it originates (seeing as you would like to have the right amount of energy and the correct color). Additionally, you usually use point lights, meaning that you can view the lighting calculation as that of a ray from a single point (the light source) onto a single point on a surface (your pixel/fragment) and then to your viewer. These directions are written in the matter of solid angles, or to make the theory even easier, in differential solid angles. Another point is that your surface may reflect light differently depending on the where it comes from, which also makes important the direction part.

So to sum this up a bit:

> I can't visually understand what does irradiance of a certain direction means?

It basically means photons from a specific light source, not from any place in space.

> And what is the difference with radiance of certain direction?

I hope it is clear, that radiance is irradiance from a certain direction. If not, please try to specify which part still bothers you.

> They both represents the power of light

Yes they do. In fact, *steradian* is a dimensionless unit, since it is $\left[\frac{m^2}{m^2}\right]$, and therefore it doesn't really add anything. I see how this is confusing. I hope I could clear up why you do this.

> What does irradiance of a certain direction means? Doesn't we associate a direction with radiance?

Careful. We don't associate direction (other than considering only surfaces perpendicular to the light direction) with irradiance. We do however with radiance.

> Isn't radiance defined as irradiance in a single direction?

Yes, you can say that.

> Is E(v) <= L(v) where v is direction?

I'd say it is the other way round, $L(v) \leq E(v)$, since $E(v)$ would consider any light source emitting light onto a surface, whereas $L(v)$ only considers light form $v$ (if you consider $v$ as the light source to surface direction. In my writing earlier, this is $\omega$ and in Real Time Rendering specifically, this is $\omega$ as well, or $l$ in chapter 5 as well as in the BRDF Theory chapter). Also, remember that these two physical quantities don't have the same units and should not really be compared this way.


---

### Answer by username_4 on 2018-04-16. Score: 4

To be honest, terms like these are very confusing as they aren't clear cut and on one side of the border. They are more grayish.

I'm gonna tell you how I convinced myself, as I too had this confusion as soon as I read your question. But I managed to convince myself through this argument.

First of all we are gonna clear up 4 terms, Radiance, Irradiance, Differential radiance and Differential Irradiance.

**"Radiance" is what you say associated with a certain direction. To be more formal and according to wikipedia,**

> It's the amount of radiant flux emitted/transmitted/received per unit
> projected area, per unit solid angle.

**Next is differential radiance. We can think of it as an infinitesimal quantity of radiance emitted or recieved in a very small solid angle $d\omega$.**

**Next is Irradiance. Irradiance isn't normally associated with a direction. According to Wikipedia it's**

> Radiant flux received by a surface per unit area

But more commonly and what makes more sense to me, and to the answer of your question, think of irradiance as the integration of radiances over a set of directions.

So we can say

$E = \displaystyle\int\_{\Omega} L(\omega)\; \omega.n \;d\omega\\\omega \in \Omega$

So if we integrate the radiances from every direction that leads us to the original definition of irradiance where direction isn't of concern. However usually we are concerned with only a subset of all the directions such as the Upper hemisphere or the lower hemisphere. This means for example,

$\Omega = \{ \omega : \omega.n \geq 0 \}$

As we can see here, we have limited the irradiance to a set of directions, the upper hemisphere. This doesn't necessarily change it into radiance which is associated by direction. Instead what this means is, when calculating irradiance we are concerned with the light coming only from these directions, although we haven't incorporated the directional quantity into the formula like with radiance.

This is the difference between irradiance from a certain direction and radiance. Think of it like this. You are holding a paper and there are 2 light bulbs in front of you. You want to measure the irradiance. Normally it would just be the radiant flux received by both bulbs per unit area. But now let's say I limit the direction so I am only concerned with the first bulb. Note that I am still calculating the "irradiance". If I move farther away the flux will decrease thus the irradiance even tho I am concerned with a specific direction. However this isn't the case with radiance where moving farther away won't change it since we divide by the solid angle too balancing the change.

**The last quantity is differential irradiance. I thought of it as a tiny amount of irradiance from a specific direction. (Again direction gets involved)**

If you think of irradiance as not assosciated with direction at all, even then when you try to think of differential irradiance you are gonna say it's a tiny amount of irradiance from a small range of direction or maybe a specific direction. That's the reason why it's small.

But if you think of irradiance as the sum of all the radiances over a specific set of directions. You'll see that it makes things clearer and you'll naturally arrive at the conclusion that differential irradiance will then refer to an irradiance from a specific direction.

So coming back to your question at last, I hope you might have gotten some intuition as to what "irradiance in a direction means".

Mathematically proving it is no hard feat tho. The answer [here](https://computergraphics.stackexchange.com/questions/3757/why-isnt-a-brdf-a-ratio-of-radiances) explains it quite well. I'm just gonna give a brief explanation. We know the rendering equation is given as

$L\_{outgoing} = L\_{emission} + \displaystyle\int\_{\Omega} L\_{incoming} \;f\_{BRDF}(\omega\_i, \omega\_o)\; \omega\_i.n \; d\omega\_i$

Assuming for the moment that emission part is zero. we end up with,

$L\_{outgoing} = \displaystyle\int\_{\Omega} L\_{incoming} \;f\_{BRDF}(\omega\_i, \omega\_o)\; \omega\_i.n \; d\omega\_i$

Now as I wrote before, if you forget the BRDF for the time being, we are just integrating the radiances over a given set of direction which is the same as irradiance.

If we look at one instance of this summation/integration, it's gonna be

$dL\_{outgoing} = L\_{incoming} \; \omega\_i.n \; d\omega\_i \; f\_{BRDF}(\omega\_i, \omega\_o)$

$dL\_{outgoing} = dE \; f\_{BRDF}(\omega\_i, \omega\_o)$

We put $d$ with the outgoind radiance and irradiance because it's a very small part (we are looking at just one instance of that summation/integration)

$f\_{BRDF}(\omega\_i, \omega\_o) = \displaystyle\frac{dL\_{outgoing}}{dE} $

Which is the ratio of the outgoing radiance to the incoming irradiance.

Again this was just the way I convinced myself and might have some mistakes. Though this is the best I came up with.

> **Comments:**
> **username_0**: I think, I currently have understood this concept better. The way I think about this function is to add dw term so BRDF = ((dL / dw) / (dE / dw)). dE / dw is input radiance from this w direction. dL / dw is infinitesmall partial outgoing radiance that is contributed from input direction w. . Is there any error in the way I see this now?  
> **username_4**: I don't quite understand your concept. First of all $dL$ already has a $d\omega$ term in it. So $dL / d\omega$ isn't radiance. Dunno what to call it. Don't try to make it a ratio of radiances.  
> **username_0**: Well dL/dw is not radiance. But it is **infinitesimal** outgoing radiance that is contributed from radiance from w direction hitting the surface.The outgoing radiance is the integral of this infinitesimal outgoing radiance from every input direction.  
> **username_4**: Nope I still don't understand. First of all, $L_o$ is the outgoing radiance, then that makes $dL_o$ the infinitesimal outgoing radiance not $dL_o/d\omega$. Your second part is somewhat right, the total outgoing radiance is the integral of the infinitesimal outgoing radiances. However that seems like a recursive definition so although you might or might not have this in mind, do keep note that you are actually integrating the differential irradiances from every direction and multiplying by the BRDF to get $L_o$  

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