More on Black Bodies

For a sphere to be a black body, it requires that $f_{abs}=f_{emis}=1$ . However it can still be Planckean if $0 .

Local Thermodynamic Equilibrium (LTE)

Local Thermodynamic Equilibrium means

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

$dI_{\nu }=-\alpha _{\nu }dsI_{\nu }\,\!$ But those photons also pick up some intensity:

$dI_{\nu }=+j_{\nu }ds\,\!$ But since the photons should not pick up energy going through the box (everything is at the same temperature),

$S_{\nu }\equiv {j_{\nu } \over \alpha _{\nu }}=I_{\nu }=B_{\nu }\,\!$ 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 $S_{\nu }=B_{\nu }$ ) because the source function is a property of the matter alone (the absorptive and emissive properties of it). However, $I_{\nu }\neq B_{\nu }$ .

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

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

Derivation identical to Rybicki’s. We should memorize these.

These coefficients govern the interaction of radiation with discrete energy levels. Say we have 2 energy levels with a difference $\Delta E=h\nu _{0}$ . There is some uncertainty associated with $\nu$ , but we’ll say it’s small for now.

There are 3 coefficients:

• ${A_{21}}$ governs decay from 2 to 1, and is the transition probability per unit time. The probability of spontaneous de-excitation and release of photon is Poisson-distributed with mean rate ${A_{21}}$ . So ${A_{21}}^{-1}$ is the mean lifetime of the excited state. e.g. For $H_{\alpha }$ (n=3 to n=2): ${A_{21}}\approx 10^{9}s^{-1}$ .
• ${B_{12}}$ governs absorptions causing transitions $1\to 2$ . The transition probability per unit time is ${B_{12}}J_{\nu }$ , where ${B_{12}}$ is the probability constant, and $J_{\nu }$ is:
$J_{\nu }\equiv {\int {I_{\nu }d\Omega } \over 4\pi }\,\!$ It depends on $I_{\nu }$ (the intensity), but it does not depend on direction, so we integrate over all angles. The $4\pi$ is a normalization constant which makes $J_{\nu }$ the mean intensity, instead of the total intensity. However, we have to remember that there are uncertainties in the energy-level separations. $\phi (\nu )\equiv$ is called the line profile function. It describes some (maybe gaussian) distribution of absorption around $\nu _{0}$ (the absorption frequency), and is subject to the requirement that:

$\int _{0}^{\infty }{\phi (\nu )d\nu }=1\,\!$ Say that $\Delta \nu$ is the width of the distribution around $\nu _{0}$ . $\Delta \nu$ is affected by many factors: ${A_{21}}$ (the natural, uncertainty-based broadening of at atom in isolation), $\nu _{0}{V_{T} \over c}$ (the thermal, Doppler-based broadening), and $n_{coll}\sigma _{coll}v_{rel}$ (collisional broadening, a.k.a. pressure broadening). So really, the transition probability per unit time is:

$R_{ex}^{-1}={B_{12}}\int _{0}^{\infty }{J_{\nu }\phi (\nu )d\nu }\approx {B_{12}}\_J\,\!$ • ${B_{21}}$ governs stimulated emission. In this example, we are in energy state 2, and an incoming photon causes a transition to energy level 1 and the emission of 2 photons. The transition per unit time is ${B_{21}}\_J$ .

Einstein Relations among coefficients

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

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

$\_J={{{A_{21}} \over {B_{21}}} \over {g_{1}{B_{12}} \over g_{2}{B_{21}}}e^{-h\nu _{0} \over kT}-1}\,\!$ In thermal equilibrium $J_{\nu }=B_{\nu }$ :

{\begin{aligned}\_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 $\_J$ earlier, we get:

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

${{A_{21}} \over {B_{21}}}={2h\nu ^{3} \over c^{2}}\,\!$ Rewriting $j_{\nu },\alpha _{\nu }$ in terms of Einstein coeffs

In a small volume $dV$ :

{\begin{aligned}j_{\nu }&\equiv {dE \over dt\,dV\,d\nu \,d\Omega }\\&={h\nu _{0}{A_{21}}n_{2}\phi (\nu ) \over 4\pi }\\\end{aligned}}\,\! We can express $\alpha _{\nu }$ in terms of the Einstein coefficients. The excitation probability per time is $n_{1}B_{12}\_J$ , and the energy lost in crossing the small volume $\propto n_{1}{B_{12}}{I_{\nu }d\Omega \over 4\pi }\phi (\nu )d\nu$ (it is the probability per time per volume of going $1\to 2$ by absorbing $I_{\nu }$ from a cone of solid angle $d\Omega$ and frequency range $[\nu ,\nu +d\nu ]$ ). Thus, the energy is given by:

{\begin{aligned}E&=n_{1}{B_{12}}{I_{\nu }d\Omega \over 4\pi }\phi (\nu )d\nu h\nu dt\,dV\\&=\alpha _{\nu }I_{\nu }ds\,dt\,d\Omega \,dA\,d\nu \\\end{aligned}}\,\! Recognizing that $dV=dA\,ds$ :

$\alpha _{\nu }={n_{1}{B_{12}}\phi (\nu ) \over 4\pi }h\nu \,\!$ Correcting for stimulated emission, we get:

${\alpha _{\nu }={(n_{1}{B_{12}}-n_{2}{B_{21}})\phi (\nu )h\nu \over 4\pi }}\,\!$ 