<p>You are not integrating the flux in all directions as indicated in your sketch. The integration is summing up the contributions over all (visible) latitudes and longitudes to the flux in a particular direction; in this case, along the x-axis.</p>

CC BY-SA 4.0
2024-12-10T13:42:08.210Z
Oh yeah, I did put the wrong coordinate diagram before the integration equation. Sorry for the mistake. The problem is now fixed, but I'm still puzzled over why the 'visible' latitudes and longitudes, like you said, is exactly all that of half a hemisphere for every random surface elements? I'm trying to show in the sketch that the red pointings should not be taken into the limit because they are pointing inward the planet
Splinter_X
CC BY-SA 4.0
2024-12-10T14:41:31.740Z
@Splinter_X as I said, it's a double integral over all latitudes and longitudes, not a quadruple integral that includes all directions as well.
ProfRob
CC BY-SA 4.0
2024-12-10T07:04:34.937Z
2024-12-10T19:37:50.123Z
2024-12-10T19:37:50.123Z
ProfRob
<p>I have this integral limit question with how Sara Seager calculates the surface flux in her <em>Exoplanet Atmospheres</em>. As I understand it, the surface flux is the outgoing flux at a particular point on a planet's surface, and it's calculated by integrating the specific intensity at this point across a solid angle range of radiation pointings, which is <span class="math-container">$F_k(\vec x)=\int_{\Omega}I(\vec x, \hat n, \nu, t)\hat n \cdot \hat k d\Omega$</span></p>
<p>The book works with a planetary latitude and longitude coordinate system, like this
<a href="https://i.sstatic.net/wC4kgVY8.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/wC4kgVY8.png" alt="enter image description here" /></a>
And, after some work of expressing the <span class="math-container">$\hat n \cdot \hat k$</span> and <span class="math-container">$d\Omega$</span>, the book goes on to write<a href="https://i.sstatic.net/xIr5OsiI.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/xIr5OsiI.png" alt="enter image description here" /></a>
And <span class="math-container">$-\pi/2 &lt; \theta_n &lt; \pi/2$</span>, <span class="math-container">$-\pi/2 &lt; \phi_n &lt; \pi/2$</span> is the half sphere in front of the yoz plane in the +x direction. I get that we only see radiation that's not pointing away from us. However, for any points apart from the center, part of the radiation pointing in this range would go inside the sphere. For example, look down from the north pole into the xoy equatorial plane and examine a surface element at the planet's equator. Here red arrows indicate radiation going inside the surface, blue leaving. <a href="https://i.sstatic.net/6HPdbPhB.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/6HPdbPhB.jpg" alt="" /></a>
So isn't it more accurate to use a solid angle range intersected by both yoz plane (in the +x direction) and the tangential plane (in the outward direction)? Which means keeping only the blue pointings? Although dealing with the tangential plane for every surface element would be bit of a headache.</p>
<p>Original text is included for verification.
<a href="https://i.sstatic.net/19o7le73.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/19o7le73.png" alt="enter image description here" /></a>
<a href="https://i.sstatic.net/82CcfvOT.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/82CcfvOT.png" alt="enter image description here" /></a></p>

<observational-astronomy><exoplanet><photometry>
Why would surface flux include radiation pointing inward the target?
Splinter_X
# Why would surface flux include radiation pointing inward the target?

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

I have this integral limit question with how Sara Seager calculates the surface flux in her *Exoplanet Atmospheres*. As I understand it, the surface flux is the outgoing flux at a particular point on a planet's surface, and it's calculated by integrating the specific intensity at this point across a solid angle range of radiation pointings, which is $F\_k(\vec x)=\int\_{\Omega}I(\vec x, \hat n, \nu, t)\hat n \cdot \hat k d\Omega$

The book works with a planetary latitude and longitude coordinate system, like this
[![enter image description here](https://i.sstatic.net/wC4kgVY8.png)](https://i.sstatic.net/wC4kgVY8.png)
And, after some work of expressing the $\hat n \cdot \hat k$ and $d\Omega$, the book goes on to write[![enter image description here](https://i.sstatic.net/xIr5OsiI.png)](https://i.sstatic.net/xIr5OsiI.png)
And $-\pi/2 < \theta\_n < \pi/2$, $-\pi/2 < \phi\_n < \pi/2$ is the half sphere in front of the yoz plane in the +x direction. I get that we only see radiation that's not pointing away from us. However, for any points apart from the center, part of the radiation pointing in this range would go inside the sphere. For example, look down from the north pole into the xoy equatorial plane and examine a surface element at the planet's equator. Here red arrows indicate radiation going inside the surface, blue leaving. [![](https://i.sstatic.net/6HPdbPhB.jpg)](https://i.sstatic.net/6HPdbPhB.jpg)
So isn't it more accurate to use a solid angle range intersected by both yoz plane (in the +x direction) and the tangential plane (in the outward direction)? Which means keeping only the blue pointings? Although dealing with the tangential plane for every surface element would be bit of a headache.

Original text is included for verification.
[![enter image description here](https://i.sstatic.net/19o7le73.png)](https://i.sstatic.net/19o7le73.png)
[![enter image description here](https://i.sstatic.net/82CcfvOT.png)](https://i.sstatic.net/82CcfvOT.png)

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

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## 1 answers

### Answer by username_1 on 2024-12-10. Score: 1

You are not integrating the flux in all directions as indicated in your sketch. The integration is summing up the contributions over all (visible) latitudes and longitudes to the flux in a particular direction; in this case, along the x-axis.

> **Comments:**
> **username_0**: Oh yeah, I did put the wrong coordinate diagram before the integration equation. Sorry for the mistake. The problem is now fixed, but I'm still puzzled over why the 'visible' latitudes and longitudes, like you said, is exactly all that of half a hemisphere for every random surface elements? I'm trying to show in the sketch that the red pointings should not be taken into the limit because they are pointing inward the planet  
> **username_1**: @Splinter_X as I said, it's a double integral over all latitudes and longitudes, not a quadruple integral that includes all directions as well.  

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