A doubt regarding the proof of kirchoff's law of thermal radiation

AI Thread Summary
The discussion centers on the behavior of a small opaque body inside an isothermal cavity that acts as a black body, focusing on the relationship between absorptivity (a) and emissivity (e). It highlights that the small body absorbs energy based on its absorptivity and emits energy based on its emissivity, leading to the conclusion that a equals e. However, there is confusion regarding whether to equate energy flow in watts by considering the internal area of the cavity, as the areas of the small body and cavity differ. The conversation also touches on the complexities of calculating how much emitted radiation from the cavity interacts with the small body, suggesting that geometric factors can complicate the analysis. Ultimately, the principles of Kirchhoff's Law are invoked to clarify the relationship between emissivity and absorptivity.
ashutoshd
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In the proof that I have studied, there is a Isothermal cavity that behaves as a black body. there is another small opaque body inside the cavity at the same temperature with emissivity e, absorptivity a and area s. now the irradiation on the small body is Eb = σ times T raised to 4. now the small body absorbs "a" times Eb. And emissive power is e times Eb. Then by conservation of energy we have a*Eb= e* Eb. therefore a = e. What i haven't understood is that since the cavity emits Eb watts per unit area, shouldn't we multiply it by the internal area of the cavity to get energy flow in watts and then equate it with the emissive power of the small body also multiplied by its own area. This way we will be equating the rates of energy streaming rather than the radiation fluxes as has been done in the proof. But since the areas of the small body and the internal area of the cavity are obviously not equal, will a not be equal to e then?
 
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are emissivity and absorptivity extensive or intensive properties?
 
I thought that Kirchoff's Law claims that:

<br /> e/a = f(\nu, T)<br />

where f is a function function which coincides with the emissivity of a black body.
 
ashutoshd said:
What i haven't understood is that since the cavity emits Eb watts per unit area, shouldn't we multiply it by the internal area of the cavity to get energy flow in watts [...]
Some parts of this energy flow will hit the cavity again, as long as your body in the cavity is not really close to the cavity walls (and in that case, the areas are equal).
In the general case, it can be tricky to calculate how much of the emitted radiation of the cavity hits the cavity again. But you can use the surface of your (convex?) body to determine this. Or just ignore the geometric details and assume a constant energy density in the volume.
 
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