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

1. Apr 6, 2012

### 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 havent understood is that since the cavity emits Eb watts per unit area, shouldnt 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?

2. Apr 7, 2012

### conquest

are emissivity and absorptivity extensive or intensive properties?

3. Apr 7, 2012

### 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.

4. Apr 7, 2012

### Staff: Mentor

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.