Does Radiation Intensity depend on [itex]\theta [/itex]?

[bobfei]

Hi,

https://www.physicsforums.com/attachment.php?attachmentid=60066&stc=1&d=1372909704
I would like to ask a question on thermal radiation. In the attached image, if all three surfaces $\{ dA,d{S_1},d{S_2}\}$ are black body surface, and $d{S_2} = d{S_1} = S$, then does $d{S_2}$ receive the same thermal from $dA$ as $d{S_1}$? In other words, are the view factor ${F_{A \to 1}}$ and ${F_{A \to 2}}$ equal?


I find this problem perplexing:

1. Straightforward intuition suggests that $d{S_1}$ should receive more energy since it is facing $dA$, and $d{S_2}$ might get less because it is at an oblique angle with dA.
2. In heat transfer there is a reciprocity relationship which states that for two diffuse emitters: $Are{a_1} \cdot {F_{1 \to 2}} = Are{a_2} \cdot {F_{2 \to 1}}$. For the pair of $dA$ and $d{S_2}$, if we stand at the position of $d{S_2}$ and look at $dA$, clearly we see only a stretched elliptical projection of $dA$, and the area times view factor is $d{S_2} \cdot {\rm{Solidangle}}(dA\cos \theta ) = S \cdot \frac{{S\cos \theta }}{{4\pi {R^2}}} = \frac{{{S^2}}}{{4\pi {R^2}}}\cos \theta$; because of the reciprocity relationship this equals $dA \cdot {F_{A \to 2}}$, and we get $S \cdot {F_{A \to 2}} = \frac{{{S^2}}}{{4\pi {R^2}}}\cos \theta \;\; \Rightarrow {F_{A \to 2}} = \frac{{S\cos \theta }}{{4\pi {R^2}}}$, so that the view factor from dA is dependent on $\theta$. This would suggest that $d{S_2}$, because of larger $\theta$, would receive smaller faction of $dA$’s energy than $d{S_1}$.
3. But from a microscopic perspective, if we consider each molecule of $dA$’s radiation, we see that for a small molecule, because $d{S_2} = d{S_1} = S$, the two will subtend the same solid angle and hence receive the same fraction of the molecule’s radiation. Then if we sum up the contribution of all molecules on $dA$, this seems suggest that the fraction of energy $d{S_1}$ and $d{S_2}$ received are equal.
4. Yet another question is the radiation’s interference. At the position of $d{S_2}$, particularly if we push $d{S_2}$ to the extreme that its center lies on the plane of $dA$, intuitively much of the energy from the left side of $dA$ traveling to $d{S_2}$ will be blocked by molecules on its right side, or cancelled. I have tried to sketch a proof by assuming some distribution of each molecule’s phase, but haven’t yet arrived at any result. Does interference really has a role here?

To me it is a difficult question. Please help!


Bob

 

[mfb]

The microscopic perspective is problematic - those molecules are not independent. If you want to imagine them as spheres (or other 3-dimensional objects on the surface), for the emission towards S2, the radiation gets (partially) shielded by other molecules.
You don't need interference (between what, for incoherent emission?) to see that effect.

If you have coherent emission, the radiation does not have to be isotropic, and you can get whatever you like - but in general, that is not thermal radiation any more.

 

[bobfei]

Quantum Mechanics needed to resolve the contradiction?

Could you give more details on the inter-molecule shielding effect? If we draw the projected cone as in the image, although it is oblique and has a smaller cross-section than when $\theta = 90^\circ$, geometrically no molecule blocks each other; the geometric complete blocking only happens when $\theta = 0^\circ$.

From this I found it counter-intuitive. How can real object has such a abrupt transition property in its radiation? Also, if my microscopic view that the molecules do not shield each other is true, this implies that as $\theta \uparrow$, more and more energy is concentrated to a narrower ellipse, which in the extreme leads to ${\lim _{\theta \to 90^\circ }}{\rm{(concentration) = }}\infty$, which is apparently unacceptable.

However as for the "partially shielding" argument, could you give some more detail on its derivation such has how do you assume the phase distribution among molecules? Can the molecules be treated completely using classical argument (basic calculus, etc.) or to reconcile the contradiction it in fact entails quantum mechanics? Maybe it is a difficult problem because Thermodynamics including the blackbody radiation theory was among the precursors of quantum mechanics?Bob

 

[mfb]

Here is a sketch:

Note that this is a very classical picture, but it gives the correct results.

 

[bobfei]

Awesome, I got it!

 

[bobfei]

Density interpretation from "ray counting"

Add for completion:

https://www.physicsforums.com/attachment.php?attachmentid=60066&stc=1&d=1372909704

1. In heat transfer there is a reciprocity relationship which states that for two diffuse emitters: $Are{a_1} \cdot {F_{1 \to 2}} = Are{a_2} \cdot {F_{2 \to 1}}$. For the pair of $dA$ and $d{S_2}$, if we stand at the position of $d{S_2}$ and look at $dA$, clearly we see only a stretched elliptical projection of $dA$, and the area times view factor is $d{S_2} \cdot {\rm{Solidangle}}(dA\cos \theta ) = S \cdot \frac{{S\cos \theta }}{{4\pi {R^2}}} = \frac{{{S^2}}}{{4\pi {R^2}}}\cos \theta$; because of the reciprocity relationship this equals $dA \cdot {F_{A \to 2}}$, and we get $S \cdot {F_{A \to 2}} = \frac{{{S^2}}}{{4\pi {R^2}}}\cos \theta \;\; \Rightarrow {F_{A \to 2}} = \frac{{S\cos \theta }}{{4\pi {R^2}}}$, so that the view factor from dA is dependent on $\theta$. This would suggest that $d{S_2}$, because of larger $\theta$, would receive smaller faction of $dA$’s energy than $d{S_1}$.

The reciprocity relationship essentially is based on ray-counting. The reason that counting from $d{S_2}$ to $d{S_1}$ gives the correct result, but not vice versa, it because from $d{S_2} \to d{S_1}$ the ray's "density" is always the same regardless of $\theta$, but from $d{S_1} \to d{S_2}$ we are actually increasing the ray density as $\theta \downarrow$.

I think the concept of "ray", as a geometric construct representing the direction of wavefront propagation, is valid here.

If anyone could add more insight I would sincerely appreciate that!


Bob