<p>Flux, in terms of power per unit area is in general direction-dependent, unless the radiation is isotropic.</p>
<p>For example, the power emitted by a dipole is zero along its axis and at a maximum for a polar angle of 90 degrees. We can therefore discuss the flux emitted in particular directions.</p>
<p>The second question relates to the definition of specific intensity, which involves emission from a surface element to which <span class="math-container">${\bf n}$</span> is a normal. To express flux per unit area for an arbitrarily oriented area requires us to calculate the projection of that area, which is <span class="math-container">$dA\  {\bf n}\cdot {\bf k}$</span>.</p>

CC BY-SA 4.0
2024-12-10T13:54:23.733Z
Thanks for the dipole analogy for the first question! That did make things feel more natural.
Splinter_X
CC BY-SA 4.0
2024-12-10T14:03:35.543Z
For the second question, $\hat n$ here is not the normal to a surface element, but rather a changing vector that's meant to indicate different radiation directions. You can notice in the $F_k(\vec x, \nu, t)$ expression that the integration happens over dΩ, which corresponds to different points of $\hat{n}$.  Also, $\hat k$ is the direction we're interested in. So it's not surface normal $\cdot$ emission direction in this case.
Splinter_X
CC BY-SA 4.0
2024-12-10T14:45:03.910Z
@Splinter_X dA is normal to to n in the definition of specific intensity. That it isn't normal to the surface you define is the reason for the scalar product.
ProfRob
CC BY-SA 4.0
2024-12-09T09:03:29.340Z
2024-12-10T07:27:16.880Z
2024-12-10T07:27:16.880Z
ProfRob
<p>I'm working with sections 2.3-2.4 in Sara Seager's <em>Exoplanet Atmospheres</em>. The book has these following definitions:
<a href="https://i.sstatic.net/eANHtqDv.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/eANHtqDv.png" alt="enter image description here" /></a>
Specific Intensity <span class="math-container">$I(\vec x, \hat n, \nu, t)$</span>: <span class="math-container">$dE(\nu, t) = I(\vec x, \hat n, \nu, t)\hat n\cdot \hat k d\Omega dA d\nu dt$</span></p>
<p>Flux: <span class="math-container">$\vec{F}(\vec x, \nu, t) = \int_\Omega I(\vec x, \hat n, \nu, t)\hat n d\Omega$</span></p>
<p>Flux in a specific direction: <span class="math-container">$F_k(\vec x, \nu, t)=\int_\Omega I(\vec x, \hat n, \nu, t)\hat n\cdot \hat k d\Omega$</span>. The book says that with this <span class="math-container">$F_k$</span> 'we are essentially describing the energy flow in one direction'.</p>
<p>My question lies with the last definition. Saying 'flux in a specific direction' is a bit like saying 'mass at a specific point,'. It doesn't make sense to me. Flux is an integration of specific intensity over a solid angle and thus cannot be ascribed to a single 'direction' (just as it's only possible to ascribe a density but not mass value to a point.)</p>
<p>And I don't get the integration either. Say we have an observer looking at a specific surface element <span class="math-container">$dA$</span> on a distant planet, and the direction connecting <span class="math-container">$dA$</span> with the observer is <span class="math-container">$\vec k$</span>. Then any beam of radiation with pointing <span class="math-container">$\hat n \ne \hat k$</span> would simply miss our observer and should not impact the flux in the observer's direction. Yet the textbook definition <span class="math-container">$F_k(\vec x, \nu, t)=\int_\Omega I(\vec x, \hat n, \nu, t)\hat n\cdot \hat k d\Omega$</span> indicates these mispointing beams are also contributing as well, except with a <span class="math-container">$\hat n\cdot \hat k$</span> projection factor. Say there's a beam pointing 60<span class="math-container">$^\circ$</span> away from the observer, and somehow the observer still get to see half of that energy...???
If you ask me I'd say flux in a specific direction is <span class="math-container">$F_k(\vec x, \nu, t)=\int_{\Omega Telescope} I(\vec x, \hat n, \nu, t)d\Omega$</span>.</p>
<p><a href="https://i.sstatic.net/f1bzZ16t.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/f1bzZ16t.jpg" alt="enter image description here" /></a></p>
<p>What am I getting wrong at the meaning of <span class="math-container">$F_k$</span>? The textbook's context is included here for sake of completeness. <a href="https://i.sstatic.net/AJBhIPx8.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/AJBhIPx8.png" alt="enter image description here" /></a></p>

<observational-astronomy><exoplanet><photometry>
Surface flux in a specific direction: why is specific intensity in other directions allowed to contribute?
Splinter_X
# Surface flux in a specific direction: why is specific intensity in other directions allowed to contribute?

(Asked by: username_0 on 2024-12-08)

I'm working with sections 2.3-2.4 in Sara Seager's *Exoplanet Atmospheres*. The book has these following definitions:
[![enter image description here](https://i.sstatic.net/eANHtqDv.png)](https://i.sstatic.net/eANHtqDv.png)
Specific Intensity $I(\vec x, \hat n, \nu, t)$: $dE(\nu, t) = I(\vec x, \hat n, \nu, t)\hat n\cdot \hat k d\Omega dA d\nu dt$

Flux: $\vec{F}(\vec x, \nu, t) = \int\_\Omega I(\vec x, \hat n, \nu, t)\hat n d\Omega$

Flux in a specific direction: $F\_k(\vec x, \nu, t)=\int\_\Omega I(\vec x, \hat n, \nu, t)\hat n\cdot \hat k d\Omega$. The book says that with this $F\_k$ 'we are essentially describing the energy flow in one direction'.

My question lies with the last definition. Saying 'flux in a specific direction' is a bit like saying 'mass at a specific point,'. It doesn't make sense to me. Flux is an integration of specific intensity over a solid angle and thus cannot be ascribed to a single 'direction' (just as it's only possible to ascribe a density but not mass value to a point.)

And I don't get the integration either. Say we have an observer looking at a specific surface element $dA$ on a distant planet, and the direction connecting $dA$ with the observer is $\vec k$. Then any beam of radiation with pointing $\hat n \ne \hat k$ would simply miss our observer and should not impact the flux in the observer's direction. Yet the textbook definition $F\_k(\vec x, \nu, t)=\int\_\Omega I(\vec x, \hat n, \nu, t)\hat n\cdot \hat k d\Omega$ indicates these mispointing beams are also contributing as well, except with a $\hat n\cdot \hat k$ projection factor. Say there's a beam pointing 60$^\circ$ away from the observer, and somehow the observer still get to see half of that energy...???
If you ask me I'd say flux in a specific direction is $F\_k(\vec x, \nu, t)=\int\_{\Omega Telescope} I(\vec x, \hat n, \nu, t)d\Omega$.

[![enter image description here](https://i.sstatic.net/f1bzZ16t.jpg)](https://i.sstatic.net/f1bzZ16t.jpg)

What am I getting wrong at the meaning of $F\_k$? The textbook's context is included here for sake of completeness. [![enter image description here](https://i.sstatic.net/AJBhIPx8.png)](https://i.sstatic.net/AJBhIPx8.png)

**Tags:** observational-astronomy, exoplanet, photometry


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## 1 answers

### Answer by username_1 on 2024-12-09. Score: 1

Flux, in terms of power per unit area is in general direction-dependent, unless the radiation is isotropic.

For example, the power emitted by a dipole is zero along its axis and at a maximum for a polar angle of 90 degrees. We can therefore discuss the flux emitted in particular directions.

The second question relates to the definition of specific intensity, which involves emission from a surface element to which ${\bf n}$ is a normal. To express flux per unit area for an arbitrarily oriented area requires us to calculate the projection of that area, which is $dA\ {\bf n}\cdot {\bf k}$.

> **Comments:**
> **username_0**: Thanks for the dipole analogy for the first question! That did make things feel more natural.  
> **username_0**: For the second question, $\hat n$ here is not the normal to a surface element, but rather a changing vector that's meant to indicate different radiation directions. You can notice in the $F_k(\vec x, \nu, t)$ expression that the integration happens over dΩ, which corresponds to different points of $\hat{n}$.  Also, $\hat k$ is the direction we're interested in. So it's not surface normal $\cdot$ emission direction in this case.  
> **username_1**: @username_0 dA is normal to to n in the definition of specific intensity. That it isn't normal to the surface you define is the reason for the scalar product.  

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