Energy balance in Kirchhoff's law of thermal radiation

Question

Kirchhoff's law of thermal radiation is usually motivated by an energy balance : at thermal equilibrium, the absorbed power should be exactly compensated by the radiated power, hence $$ \alpha(\Omega,\nu)\mathcal{K}(\Omega,\nu)=\mathcal{L}(\Omega,\nu)=\epsilon(\Omega,\nu) \mathcal{L}_{BB}(\Omega,\nu) $$

where $\alpha$ is the absorptivity (fraction of incident power absorbed by the body), $\mathcal{K}$ is the incident spectral radiance (incoming power per surface unit, steradian and Herz), $\mathcal{L}$ is the body spectral radiance (radiated power per surface unit, steradian and Herz) $\epsilon$ is the emissivity of the body (ratio between the body radiance and that of a black body).

If the body is illuminated by a thermal radiation, $\mathcal{K}=\mathcal{L}_{BB}$ and we conclude $$ \alpha(\Omega,\nu)=\epsilon(\Omega,\nu) $$

My question is the following : why is this relation true for all frequency and solid angles ? Why would energy conservation apply in each mode, and not only globally ? Why don't we rather consider an integrated form such as $$ \int d\nu d\Omega \, \alpha(\Omega,\nu) \mathcal{L}_{BB}(\Omega,\nu)= \int d\nu d\Omega \, \epsilon(\Omega,\nu) \mathcal{L}_{BB}(\Omega,\nu) $$

Complementary question : can Kirchhoff law be related to transition amplitude in a quantum description ?

Answer 1

I came up with a solution from the university of Arizona.

Let's consider a body with emissivity $\epsilon_{1}(\lambda)$ and absorptivity $\alpha_{1}(\lambda)$.

The energy balance for the body exposed to a thermal radiation of radiance $\mathcal{L}_{T}(\lambda)$ imposes $$ \int d\lambda\,\mathcal{L}_{T}(\lambda)\epsilon_{1}(\lambda)=\int d\lambda\,\mathcal{L}_{T}(\lambda)\alpha_{1}(\lambda) $$ which is not enough to conclude on the identity between emissivity and absorptivity at all wavelength.

Let's now put the body inside a cavity, the radiative behavior of which is characterized by $\epsilon_{2},\,\alpha_{2}$. We assume thermal equilibrium at temperature T between the body and the cavity.

The spectral irradiance arriving on the body is $$\phi_{in}(\lambda)=\mathcal{L}_{T}(\lambda)\epsilon_{2}(\lambda)+\left(1-\alpha_{2}(\lambda)\right)\phi_{out}(\lambda)$$ where the spectral irradiance reaching the cavity is expressed as $$\phi_{out}(\lambda)=\mathcal{L}_{T}(\lambda)\epsilon_{1}(\lambda)+\left(1-\alpha_{1}(\lambda)\right)\phi_{in}(\lambda).$$

Solving this simple system leads to $$ \phi_{in}(\lambda) =\mathcal{L}_{T}\frac{\epsilon_{2}+\epsilon_{1}(1-\alpha_{2})}{1-(1-\alpha_{1})(1-\alpha_{2})}\\ \phi_{out}(\lambda) =\mathcal{L}_{T}\frac{\epsilon_{1}+\epsilon_{2}(1-\alpha_{1})}{1-(1-\alpha_{1})(1-\alpha_{2})} $$ and the energy balance requires $$ \int d\lambda\,\phi_{in}(\lambda)=\int d\lambda\,\phi_{out}(\lambda)\\ \Rightarrow\int d\lambda\,\mathcal{L}_{T}\frac{\alpha_{1}\alpha_{2}}{1-(1-\alpha_{1})(1-\alpha_{2})}\left(\frac{\epsilon_{2}}{\alpha_{2}}-\frac{\epsilon_{1}}{\alpha_{1}}\right)=0 $$ The trick is that this relation holds for whatever cavity material, ie whatever $\epsilon_{2},\,\alpha_{2}$. This is possible if and only if $\epsilon\propto\alpha$, and the first equation imposes $\epsilon(\lambda)=\alpha(\lambda)$ for each wavelength, hence the Kirchhoff law of radiation.

This proof is written in spectral domain, but can be extended to the angular domain as well.

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I read that Kirchhoff law was only holding in time-reversal symmetric systems. Where is this assumption used here ?