<p>The dot product basically is the projection of one vector onto the other. Using it here allows to take the incident angle into account. E.g. when the Sun is directly at the horizon, the light rays are all parallel the the surface, and nothing will hit the surface, thus intensity is 0 (<span class="math-container">$(1,0,0) \cdot (0,0,1) = 0$</span>). Vice versa, when the incident angle of the incoming light is 90° to the surface, then the light is not streched over a larger surface, but the maximum intensity of the sunlight reaches the surface (<span class="math-container">$(0,0,1) \cdot (0,0,1) = 1$</span>).</p>

CC BY-SA 4.0
2024-12-08T05:18:39.783Z
Hi planetmaker, thx for the reply! Your answer inspired me to further specify my question, so I re-edit it. Would you take a look?
Splinter_X
CC BY-SA 4.0
2024-12-08T10:19:07.327Z
I'm not sure I understand or can follow the argument you make in the edit. A definition is a definition - and often chosen for convenience. I sure can re-define the meter to be a foot long - but that's not useful.
planetmaker
CC BY-SA 4.0
2024-12-07T18:54:01.960Z
2024-12-07T18:54:01.960Z
planetmaker
<p>Oookay I get the picture now! The crux here is that <em>specific intensity is also known as surface brightness</em>.</p>
<p>We start with imagining holding a sheet of paper up to the light. We now have a nice approximation of a Lambert surface, which means it's a surface element radiating isotropically. In fact, to better visualize specific intensity/surface brightness let's make it a red (any color) piece of paper. Like stained glass in churches, the paper would appear almost white if the light bulb that shines through it is extremely bright, and redder if the light bulb dims. We set an extremely bright bulb for the experiment.</p>
<p>Now, if you stare right into the paper (in its normal direction), and tilt your head sidewise to gradually look into its edge, increasing <span class="math-container">$&lt;\hat n, \hat k&gt;$</span> all the way from <span class="math-container">$0^\circ$</span> to <span class="math-container">$90^\circ$</span>, your eyes sting less because there is less energy going into your eye, but that's only because the paper is taking up less a solid angle as it gradually thins to a line. At the same time, the paper's apparent color does not change, an indicator that it stays as bright, or remains constant in energy per unit solid angle.</p>
<p>Consider stopping midway at <span class="math-container">$&lt;\hat n, \hat k&gt; = 60^\circ$</span>, and re-examine our definition for specific flux. If there's no <span class="math-container">$\hat n \cdot \hat k $</span>, we would now have a specific intensity that's only half of that in the normal direction to begin with, as the paper's apparent area (which is an equivalent of its solid angle) has been squeezed to half because of our tilted viewpoint, making its energy halved while the denominator is fixed. The dot product is placed there to account for the changing solid angle, so that we can describe the paper's innate property that its brightness (color) stays the same in any direction.</p>

CC BY-SA 4.0
2024-12-11T11:30:23.993Z
2024-12-17T08:21:23.920Z
2024-12-17T08:21:23.920Z
Splinter_X
<p>I'm reading section 2.3 in Sara Seager's <em>Exoplanet Atmospheres</em>, and have encountered the definition for specific intensity as such:</p>
<p>“The intensity <span class="math-container">$I$</span> is formally defined as the amount of energy passing through a surface area <span class="math-container">$dA$</span>, within a differential solid angle <span class="math-container">$d\Omega$</span> centered about <span class="math-container">$\hat{n}$</span>, per frequency interval, per unit time”</p>
<p><a href="https://i.sstatic.net/pvtK4Ofg.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/pvtK4Ofg.png" alt="Figure 2.1 Definition of the specific intensity I(x, , v, t)" /></a></p>
<p>which gives us the definition:
<a href="https://i.sstatic.net/6Ht5HbWB.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/6Ht5HbWB.png" alt="enter image description here" /></a></p>
<p>My question is, why bother including the dot product, instead of just using <span class="math-container">$I(\vec x, \hat{n }, \nu, t)= \frac{dE}{dAd\Omega d\nu dt}$</span>? I mean, <span class="math-container">$\hat n$</span> is already included among the variables for <span class="math-container">$I$</span> (as we have <span class="math-container">$I(\vec x, \hat{n}, \nu, t)$</span>), and <span class="math-container">$\hat k$</span> is just the normal vector of dA, which should be possible to infer from <span class="math-container">$\vec x$</span>, given the geometry of the object. So technically you can shove any difference made by the absence of <span class="math-container">$\hat{n}\cdot \hat{k}$</span> into a different-valued <span class="math-container">$I(\vec x, \hat{n }, \nu, t)$</span>.</p>
<p>To dicuss with something specific, let's take a random surface element radiating energy in all directions and try to describe it. Let's then look at a specific direction of <span class="math-container">$\hat{n_0}$</span> where the angle between <span class="math-container">$\hat{n_0}$</span> and <span class="math-container">$\hat{k}$</span> is <span class="math-container">$60^{\circ}$</span>, which means <span class="math-container">$\hat{n_0}\cdot\hat{k}=\frac 1 2$</span>. Now, since we're dealing with the same radiating surface element, <span class="math-container">$\frac{dE}{dAdtd\nu d\Omega}$</span> is just the same for both definitions, and we have <span class="math-container">$I(\vec x, \hat{n_0}, \nu, t)_{definition with the dot product}=2\cdot I(\vec x, \hat{n_0}, \nu, t)_{no dot product}$</span>.</p>
<p>With the example above, my question then equates to this: as for describing the innate property of our radiating element, why is it more intuitional or fundamental to opt for the dot product and have the two-fold <span class="math-container">$I(\vec x, \hat{n_0}, \nu, t)$</span> at <span class="math-container">$&lt;\hat{n_0}, \hat{k}&gt;=60^\circ$</span>?</p>
<p>I'm dimly aware that the dot product is there to eliminate the influence of the <span class="math-container">$I(\vec x, \hat{n}, \nu, t)$</span> definition being geometry or directional dependent or something, though I would really appreciate if someone may provide a more lucid argument, and preferably with the example I had provided above.</p>
<p>Any help is appreciated!</p>

<observational-astronomy>
Why is there a vector dot product term in the denominator in the definition for specific intensity?
Splinter_X
# Why is there a vector dot product term in the denominator in the definition for specific intensity?

(Asked by: username_0 on 2024-12-07)

I'm reading section 2.3 in Sara Seager's *Exoplanet Atmospheres*, and have encountered the definition for specific intensity as such:

“The intensity $I$ is formally defined as the amount of energy passing through a surface area $dA$, within a differential solid angle $d\Omega$ centered about $\hat{n}$, per frequency interval, per unit time”

[![Figure 2.1 Definition of the specific intensity I(x, , v, t)](https://i.sstatic.net/pvtK4Ofg.png)](https://i.sstatic.net/pvtK4Ofg.png)

which gives us the definition:
[![enter image description here](https://i.sstatic.net/6Ht5HbWB.png)](https://i.sstatic.net/6Ht5HbWB.png)

My question is, why bother including the dot product, instead of just using $I(\vec x, \hat{n }, \nu, t)= \frac{dE}{dAd\Omega d\nu dt}$? I mean, $\hat n$ is already included among the variables for $I$ (as we have $I(\vec x, \hat{n}, \nu, t)$), and $\hat k$ is just the normal vector of dA, which should be possible to infer from $\vec x$, given the geometry of the object. So technically you can shove any difference made by the absence of $\hat{n}\cdot \hat{k}$ into a different-valued $I(\vec x, \hat{n }, \nu, t)$.

To dicuss with something specific, let's take a random surface element radiating energy in all directions and try to describe it. Let's then look at a specific direction of $\hat{n\_0}$ where the angle between $\hat{n\_0}$ and $\hat{k}$ is $60^{\circ}$, which means $\hat{n\_0}\cdot\hat{k}=\frac 1 2$. Now, since we're dealing with the same radiating surface element, $\frac{dE}{dAdtd\nu d\Omega}$ is just the same for both definitions, and we have $I(\vec x, \hat{n\_0}, \nu, t)\_{definition with the dot product}=2\cdot I(\vec x, \hat{n\_0}, \nu, t)\_{no dot product}$.

With the example above, my question then equates to this: as for describing the innate property of our radiating element, why is it more intuitional or fundamental to opt for the dot product and have the two-fold $I(\vec x, \hat{n\_0}, \nu, t)$ at $<\hat{n\_0}, \hat{k}>=60^\circ$?

I'm dimly aware that the dot product is there to eliminate the influence of the $I(\vec x, \hat{n}, \nu, t)$ definition being geometry or directional dependent or something, though I would really appreciate if someone may provide a more lucid argument, and preferably with the example I had provided above.

Any help is appreciated!

**Tags:** observational-astronomy

---

## 2 answers

### Answer by username_1 on 2024-12-07. Score: 2

The dot product basically is the projection of one vector onto the other. Using it here allows to take the incident angle into account. E.g. when the Sun is directly at the horizon, the light rays are all parallel the the surface, and nothing will hit the surface, thus intensity is 0 ($(1,0,0) \cdot (0,0,1) = 0$). Vice versa, when the incident angle of the incoming light is 90° to the surface, then the light is not streched over a larger surface, but the maximum intensity of the sunlight reaches the surface ($(0,0,1) \cdot (0,0,1) = 1$).

> **Comments:**
> **username_0**: Hi planetmaker, thx for the reply! Your answer inspired me to further specify my question, so I re-edit it. Would you take a look?  
> **username_1**: I'm not sure I understand or can follow the argument you make in the edit. A definition is a definition - and often chosen for convenience. I sure can re-define the meter to be a foot long - but that's not useful.  

---

### Answer by username_0 on 2024-12-11. Score: 1

Oookay I get the picture now! The crux here is that *specific intensity is also known as surface brightness*.

We start with imagining holding a sheet of paper up to the light. We now have a nice approximation of a Lambert surface, which means it's a surface element radiating isotropically. In fact, to better visualize specific intensity/surface brightness let's make it a red (any color) piece of paper. Like stained glass in churches, the paper would appear almost white if the light bulb that shines through it is extremely bright, and redder if the light bulb dims. We set an extremely bright bulb for the experiment.

Now, if you stare right into the paper (in its normal direction), and tilt your head sidewise to gradually look into its edge, increasing $<\hat n, \hat k>$ all the way from $0^\circ$ to $90^\circ$, your eyes sting less because there is less energy going into your eye, but that's only because the paper is taking up less a solid angle as it gradually thins to a line. At the same time, the paper's apparent color does not change, an indicator that it stays as bright, or remains constant in energy per unit solid angle.

Consider stopping midway at $<\hat n, \hat k> = 60^\circ$, and re-examine our definition for specific flux. If there's no $\hat n \cdot \hat k $, we would now have a specific intensity that's only half of that in the normal direction to begin with, as the paper's apparent area (which is an equivalent of its solid angle) has been squeezed to half because of our tilted viewpoint, making its energy halved while the denominator is fixed. The dot product is placed there to account for the changing solid angle, so that we can describe the paper's innate property that its brightness (color) stays the same in any direction.


---

