# Difference between revisions of "Radiative Equilibrium"

### Local Thermodynamic Equilibrium (LTE)

${\displaystyle S_{\nu }=B_{\nu }(T)\,\!}$

where ${\displaystyle T}$ is local. Take a sphere at surface temp ${\displaystyle T}$, with some gas (matter) inside it. Allow to come to thermal equilibrium. Inside the sphere, you see a Planckian spectrum (${\displaystyle I_{\nu }=B_{\nu }}$). Now suppose you take a box of gas (it is both emissive and absorbent) and shoot rays of photons through that box. According to the Radiative Transfer Equation, these photons suffer some absorption:

${\displaystyle dI_{\nu }=-\alpha _{\nu }dsI_{\nu }\,\!}$

But those photons also pick up some intensity:

${\displaystyle dI_{\nu }=+j_{\nu }ds\,\!}$

But since the photons should not pick up energy going through the box (everything is at the same temperature),

${\displaystyle S_{\nu }\equiv {j_{\nu } \over \alpha _{\nu }}=I_{\nu }=B_{\nu }\,\!}$

Now suppose you let all the photons out of this sphere. Those photons will no longer be in thermal equilibrium with the gas. Suppose we magically fix the temperature of the gas at this point. The source function will remain the same (it will still be the case that ${\displaystyle S_{\nu }=B_{\nu }}$) because the source function is a property of the matter alone (the absorptive and emissive properties of it). However, ${\displaystyle I_{\nu }\neq B_{\nu }}$.

Suppose we look along a column of this gas. At each point, the source function will be: ${\displaystyle S_{\nu }=B_{\nu }(T)}$. Therefore, ${\displaystyle I_{\nu }}$ along that column will be:

${\displaystyle I_{\nu }=\int _{0}^{T_{\nu }}{S_{\nu }(T_{\nu ^{\prime }})e^{-(T_{\nu ^{\prime }}-T_{\nu })}dT_{\nu ^{\prime }}}\,\!}$

### Radiative Equilibrium and Relations among Einstein Coefficients

Assume we have many atoms with 2 energy states, and ${\displaystyle n_{1}}$ is the # density in state 1, ditto for ${\displaystyle n_{2}}$. Assume we are in thermal, steady-state equilibrium, so:

${\displaystyle n_{1}{B_{12}}{\bar {J}}=n_{2}{A_{21}}+n_{2}{B_{21}}{\bar {J}}\,\!}$

This is because as many atoms need to be going from energy state 1 to 2 as visa versa. A second relation is: ${\displaystyle {\bar {J}}={n_{2}{A_{21}} \over n_{1}{B_{12}}-n_{2}{B_{21}}}}$. Using the Boltzmann distribution, ${\displaystyle {n_{2} \over n_{1}}={g_{2} \over g_{1}}e^{-h\nu _{0} \over kT}}$:

${\displaystyle {\bar {J}}={{{A_{21}} \over {B_{21}}} \over {g_{1}{B_{12}} \over g_{2}{B_{21}}}e^{-h\nu _{0} \over kT}-1}\,\!}$

In thermal equilibrium ${\displaystyle J_{\nu }}$ is given by the Planck Function:

{\displaystyle {\begin{aligned}{\bar {J}}&\equiv \int _{0}^{\infty }{J_{\nu }\phi (\nu )d\nu }\\&=\int _{0}^{\infty }{B_{\nu }\phi (\nu )d\nu }\\&\approx B_{\nu }(\nu _{0})\\&={2h\nu _{0}^{3} \over c^{2}(e^{-h\nu _{0} \over kT}-1)}\\\end{aligned}}\,\!}

Combining this with ${\displaystyle {\bar {J}}}$ earlier, we get:

${\displaystyle {g_{1}{B_{12}}=g_{2}{B_{21}}}\,\!}$

and

${\displaystyle {{A_{21}} \over {B_{21}}}={2h\nu ^{3} \over c^{2}}\,\!}$