<p>If the specific intensity is isotropic, then indeed the net flux through a surface is zero.</p>
<p>I am not sure what you mean by a perfectly radiating source, but the radiation field say at the photosphere of a star is not isotropic and there is a net flux outwards.</p>

CC BY-SA 4.0
2024-12-10T13:46:38.693Z
Oh so surface elements at photospheres actually pour more energy outward than inward? This makes more sense. Thanks!
Splinter_X
CC BY-SA 4.0
2024-12-09T21:29:56.347Z
2024-12-10T06:43:57.740Z
2024-12-10T06:43:57.740Z
ProfRob
<p>I'm working on Section 2.3 from Sara Seager's <em>Exoplanets Atmosphere</em>, and went across this definition for radiation flux:
<span class="math-container">$\mathbf{F}(\mathbf x, \nu,t) = \int_{\Omega}I(\mathbf{x},\mathbf{\hat{n}},\nu,t)\mathbf{\hat{n}} d\Omega$</span>. Notice how here the specific intensity <span class="math-container">$I(\mathbf{x},\mathbf{\hat{n}},\nu,t)$</span> is a scalar and that the 'flux' <span class="math-container">$\mathbf{F}(\mathbf x, \nu,t)$</span> is a vector.</p>
<p>This differs from the 'flux' I was used to in electromagnetism and vector analysis, which is defined by a surface integral like <span class="math-container">$\Phi = \oint \mathbf{E}d\mathbf{a}$</span>. And the RHS would be a point multiplication which produces a scalar flux <span class="math-container">$\Phi$</span>, instead of a scalar.</p>
<p>Now this difference in definition produces an interesting result: <em>the net flux for an isotropic radiation surface element is zero</em>. I was very surprised to find it at first, but this conclusion is confirmed by various sources, including exercise problem 2 in the same chapter in <em>Exoplanets Atmosphere</em>, section 5 in this Max Planck Institute lecture note <a href="https://www.mpi-hd.mpg.de/personalhomes/frieger/HEA3.pdf" rel="nofollow noreferrer">https://www.mpi-hd.mpg.de/personalhomes/frieger/HEA3.pdf</a>, and page 4 in this book <a href="https://crossfield.ku.edu/A391_2020A/Rybicki_Ch1-firsthalf_Radiation.pdf" rel="nofollow noreferrer">https://crossfield.ku.edu/A391_2020A/Rybicki_Ch1-firsthalf_Radiation.pdf</a>. These sources often interpret the result as 'there is as much energy crossing surface element <span class="math-container">$dA$</span> in its normal direction <span class="math-container">$\mathbf{\hat n}$</span>  as in the <span class="math-container">$-\mathbf{\hat n}$</span> direction'.</p>
<p>It's hard to comprehend the conclusion if you borrow intuition from the electric field. <em>The electric flux surrounding an isolated charge that produces isotropic line fields is definitely not zero</em>, and a non-zero flux means that something is giving off electric field in all directions. That's what the Gaussian theorem is all about. Similarly, for an isotropic radiation source, I can't stop feeling that 'something is giving off energy in this surface element'. Yet the net radiation flux is zero, and the sources' interpretation suggests that energy is neither lost nor gained.
<a href="https://i.sstatic.net/pZtTmFfg.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/pZtTmFfg.jpg" alt="enter image description here" /></a>
So the analogy between the radiation field and the electric field is broken... but why? Mathematically speaking, how are these two fields different? (Why is the Gaussian theorem not applicable to radiation?) And physically speaking, how do you understand the statement that an isotropic radiating source is not losing energy? Is there a better intuition than the electric field to relate the radiation field to?</p>
<p>For the physical side, this discussion <a href="https://www.physicsforums.com/threads/is-the-net-flux-from-an-isotropic-source-zero.483418/" rel="nofollow noreferrer">https://www.physicsforums.com/threads/is-the-net-flux-from-an-isotropic-source-zero.483418/</a> differentiates an 'isotropic emitter' from 'a field point in isotropic radiation'. Nevertheless, it's still hard to convince myself that even 'a field point in isotropic radiation' is not losing energy.</p>

CC BY-SA 4.0
2024-12-08T10:23:33.367Z
Your conclusion that the surface integral around a homogenous source equals to 0 is mathematically wrong; it yields the total flux of the sources enclosed. (Your ignorance is not physics, but math). I suggest to study calculus: https://en.wikipedia.org/wiki/Surface_integral
planetmaker
CC BY-SA 4.0
2024-12-10T13:29:34.477Z
You have mistaken the definition of flux here for that in a standard calculus course.  Eq 2.3 integrates a scalar (specific intenisty) multiplied with a normal vector, which is why we end up with F as a vecotr, while the more common flux for a vector field integrates the dot product between a field vector and a vectored surface.
Splinter_X
CC BY-SA 4.0
2024-12-10T13:49:57.720Z
I think you're trying to indicate that the flux defined here should not be understood as the more common scalar energy, which would be a valid point.
Splinter_X
CC BY-SA 4.0
2024-12-18T18:24:22.117Z
I think there are two subtleties here that you missed: 1. The presence of charges is defined by the $div E$ - from which you have to integrate over the projected contributions onto a cube. That situation is fundamentally different from 2. the specific intensity, which can be multi-valued at the same location $\vec x$, due to it's dependence on the (additional dimension) direction $\vec n$. It doesn't need a surface integral to make sense. Hence $\vec F\neq0$ only indicates an imbalance in directional fluxes, not the presence of a source. That would be a job for $div F$, consistent with Maxwell
AtmosphericPrisonEscape
<observational-astronomy><exoplanet><photometry>
How to understand radiation 'flux' as compared to the flux in an electric field?
Splinter_X
# How to understand radiation 'flux' as compared to the flux in an electric field?

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

I'm working on Section 2.3 from Sara Seager's *Exoplanets Atmosphere*, and went across this definition for radiation flux:
$\mathbf{F}(\mathbf x, \nu,t) = \int\_{\Omega}I(\mathbf{x},\mathbf{\hat{n}},\nu,t)\mathbf{\hat{n}} d\Omega$. Notice how here the specific intensity $I(\mathbf{x},\mathbf{\hat{n}},\nu,t)$ is a scalar and that the 'flux' $\mathbf{F}(\mathbf x, \nu,t)$ is a vector.

This differs from the 'flux' I was used to in electromagnetism and vector analysis, which is defined by a surface integral like $\Phi = \oint \mathbf{E}d\mathbf{a}$. And the RHS would be a point multiplication which produces a scalar flux $\Phi$, instead of a scalar.

Now this difference in definition produces an interesting result: *the net flux for an isotropic radiation surface element is zero*. I was very surprised to find it at first, but this conclusion is confirmed by various sources, including exercise problem 2 in the same chapter in *Exoplanets Atmosphere*, section 5 in this Max Planck Institute lecture note <https://www.mpi-hd.mpg.de/personalhomes/frieger/HEA3.pdf>, and page 4 in this book <https://crossfield.ku.edu/A391_2020A/Rybicki_Ch1-firsthalf_Radiation.pdf>. These sources often interpret the result as 'there is as much energy crossing surface element $dA$ in its normal direction $\mathbf{\hat n}$ as in the $-\mathbf{\hat n}$ direction'.

It's hard to comprehend the conclusion if you borrow intuition from the electric field. *The electric flux surrounding an isolated charge that produces isotropic line fields is definitely not zero*, and a non-zero flux means that something is giving off electric field in all directions. That's what the Gaussian theorem is all about. Similarly, for an isotropic radiation source, I can't stop feeling that 'something is giving off energy in this surface element'. Yet the net radiation flux is zero, and the sources' interpretation suggests that energy is neither lost nor gained.
[![enter image description here](https://i.sstatic.net/pZtTmFfg.jpg)](https://i.sstatic.net/pZtTmFfg.jpg)
So the analogy between the radiation field and the electric field is broken... but why? Mathematically speaking, how are these two fields different? (Why is the Gaussian theorem not applicable to radiation?) And physically speaking, how do you understand the statement that an isotropic radiating source is not losing energy? Is there a better intuition than the electric field to relate the radiation field to?

For the physical side, this discussion <https://www.physicsforums.com/threads/is-the-net-flux-from-an-isotropic-source-zero.483418/> differentiates an 'isotropic emitter' from 'a field point in isotropic radiation'. Nevertheless, it's still hard to convince myself that even 'a field point in isotropic radiation' is not losing energy.

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

> **Comments:**
> **username_1**: Your conclusion that the surface integral around a homogenous source equals to 0 is mathematically wrong; it yields the total flux of the sources enclosed. (Your ignorance is not physics, but math). I suggest to study calculus: https://en.wikipedia.org/wiki/Surface_integral  
> **username_0**: You have mistaken the definition of flux here for that in a standard calculus course.  Eq 2.3 integrates a scalar (specific intenisty) multiplied with a normal vector, which is why we end up with F as a vecotr, while the more common flux for a vector field integrates the dot product between a field vector and a vectored surface.  
> **username_0**: I think you're trying to indicate that the flux defined here should not be understood as the more common scalar energy, which would be a valid point.  
> **username_2**: I think there are two subtleties here that you missed: 1. The presence of charges is defined by the $div E$ - from which you have to integrate over the projected contributions onto a cube. That situation is fundamentally different from 2. the specific intensity, which can be multi-valued at the same location $\vec x$, due to it's dependence on the (additional dimension) direction $\vec n$. It doesn't need a surface integral to make sense. Hence $\vec F\neq0$ only indicates an imbalance in directional fluxes, not the presence of a source. That would be a job for $div F$, consistent with Maxwell  

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## 1 answers

### Answer by username_3 on 2024-12-09. Score: 0

If the specific intensity is isotropic, then indeed the net flux through a surface is zero.

I am not sure what you mean by a perfectly radiating source, but the radiation field say at the photosphere of a star is not isotropic and there is a net flux outwards.

> **Comments:**
> **username_0**: Oh so surface elements at photospheres actually pour more energy outward than inward? This makes more sense. Thanks!  

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