Understanding Emission and Absorption Power in Three-Body Systems

[LagrangeEuler]

Three bodies 1,2,3 are in closed region. Region is at temperature $T$.
$e_{\lambda,T}$ - spectral emission power
$a_{\lambda,T}$ - spectral absorption power

In experiment
$(\frac{e_{\lambda,T}}{a_{\lambda,T}})_1=(\frac{e_{\lambda,T}}{a_{\lambda,T}})_2=(\frac{e_{\lambda,T}}{a_{\lambda,T}})_3$

I am confused. Does it perhaps $(\frac{e_{\lambda,T}}{a_{\lambda,T}})_1=1$? How much body absorbs and emits so much. Right?

 

[dauto]

Yes, emission power and absorption power must be equal.

 

[LagrangeEuler]

Why then in next step
$(\frac{e_{\lambda,T}}{a_{\lambda,T}})=E_{\lambda,T}$

 

[Philip Wood]

I can make sense of this only by assuming that…

a_{\lambda, T} means the spectral absorptivity of a surface at temperature T, that is the fraction of radiation of wavelength \lambda which it absorbs. A number (≤ 1) without units.

E_{\lambda, T} means the spectral emissive power of a black body at temperature T, that is the power it emits between wavelength \lambda and wavelength \lambda + \Delta \lambda per unit area, divided by \Delta \lambda, as \Delta \lambda approaches zero.

e_{\lambda, T} means the spectral emissive power of the surface at temperature T, that is the power it emits between wavelength \lambda and wavelength \lambda + \Delta \lambda per unit area, divided by \Delta \lambda, as \Delta \lambda approaches zero.

The relationship you've just quoted would then express Kirchhoff's Law, crudely summarised as 'good absorbers are good emitters', since it states that a_{\lambda, T} is equal to the ratio of the spectral emissive power of the surface in question to that of a black body (which has the greatest spectral emissive power, at all wavelengths, of ANY surface at the same temperature).

 

[Philip Wood]

LE: Are you clear now?