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

[ashutoshd]

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?

 

[conquest]

are emissivity and absorptivity extensive or intensive properties?

 

[Dickfore]

I thought that Kirchoff's Law claims that:

$$e/a = f(\nu, T)$$

where f is a function function which coincides with the emissivity of a black body.

 

[mfb]

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.

 

[sardar]

Thank you for bringing up your doubt regarding the proof of Kirchoff's Law of Thermal Radiation. You are correct in pointing out that the areas of the small body and the internal area of the cavity are not equal, so the irradiation and emissive power should be multiplied by their respective areas. However, the key concept in Kirchoff's Law is that the emissivity and absorptivity of a material are equal at thermal equilibrium, regardless of their individual areas.

To understand this concept, it is important to remember that the emissivity and absorptivity of a material are properties that describe its ability to emit and absorb thermal radiation, respectively. At thermal equilibrium, the small body and the cavity are in thermal equilibrium, meaning they are at the same temperature. This also means that they are exchanging thermal radiation at the same rate. Therefore, the emissive power of the small body, which is e times the irradiation, must be equal to the irradiation of the cavity, which is σT^4 multiplied by the internal area of the cavity.

In other words, at thermal equilibrium, the rate of energy streaming (emissive power) from the small body must be equal to the rate of energy streaming into the cavity (irradiation), which is multiplied by the internal area of the cavity. This is why we equate the emissive power of the small body (e*Eb) with the irradiation of the cavity (σT^4*A), and not just the radiation fluxes as you have mentioned.

I hope this explanation helps clarify your doubt. If you have any further questions, please do not hesitate to ask. Science is all about questioning and seeking understanding, and it's great that you are thinking critically about this proof. Keep up the good work!