# Einstein Coefficients

Einstein coefficients describe the absorption and emission of photons via electronic transitions in atoms. There are three coefficients:

Left: Photon absorption is described by ${\displaystyle B_{12}}$. Center: Spontaneous photon emission is described by ${\displaystyle A_{21}}$. Right: Stimulated photon emission is described by ${\displaystyle B_{21}}$.

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

There are 3 coefficients:

• ${\displaystyle {A_{21}}}$ 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 ${\displaystyle {A_{21}}}$. So ${\displaystyle {A_{21}}^{-1}}$ is the mean lifetime of the excited state. e.g. For ${\displaystyle H_{\alpha }}$ (n=3 to n=2): ${\displaystyle {A_{21}}\approx 10^{9}s^{-1}}$.
• ${\displaystyle {B_{12}}}$ governs absorptions causing transitions ${\displaystyle 1\to 2}$. The transition probability per unit time is ${\displaystyle {B_{12}}J_{\nu }}$, where ${\displaystyle {B_{12}}}$ is the probability constant, and ${\displaystyle J_{\nu }}$ is:
${\displaystyle J_{\nu }\equiv {\int {I_{\nu }d\Omega } \over 4\pi }\,\!}$

It depends on ${\displaystyle I_{\nu }}$ (the intensity), but it does not depend on direction, so we integrate over all angles. The ${\displaystyle 4\pi }$ is a normalization constant which makes ${\displaystyle J_{\nu }}$ the mean intensity, instead of the total intensity. However, we have to remember that there are uncertainties in the energy-level separations. ${\displaystyle \phi (\nu )\equiv }$ is called the line profile function. It describes some (maybe gaussian) distribution of absorption around ${\displaystyle \nu _{0}}$ (the absorption frequency), and is subject to the requirement that:

${\displaystyle \int _{0}^{\infty }{\phi (\nu )d\nu }=1\,\!}$

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

${\displaystyle R_{ex}^{-1}={B_{12}}\int _{0}^{\infty }{J_{\nu }\phi (\nu )d\nu }\approx {B_{12}}{\bar {J}}\,\!}$
• ${\displaystyle {B_{21}}}$ 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 ${\displaystyle {B_{21}}{\bar {J}}}$.

## 1 Einstein Relations among 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 that ${\displaystyle {n_{2} \over n_{1}}={g_{2} \over g_{1}}e^{-h\nu _{0} \over kT}}$:

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

In thermal equilibrium ${\displaystyle J_{\nu }=B_{\nu }}$:

{\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

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

## 2 Rewriting ${\displaystyle j_{\nu },\alpha _{\nu }}$ in terms of Einstein coeffs

In a small volume ${\displaystyle dV}$:

{\displaystyle {\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 ${\displaystyle \alpha _{\nu }}$ in terms of the Einstein coefficients. The excitation probability per time is ${\displaystyle n_{1}B_{12}{\bar {J}}}$, and the energy lost in crossing the small volume ${\displaystyle \propto n_{1}{B_{12}}{I_{\nu }d\Omega \over 4\pi }\phi (\nu )d\nu }$ (it is the probability per time per volume of going ${\displaystyle 1\to 2}$ by absorbing ${\displaystyle I_{\nu }}$ from a cone of solid angle ${\displaystyle d\Omega }$ and frequency range ${\displaystyle [\nu ,\nu +d\nu ]}$). Thus, the energy is given by:

{\displaystyle {\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 ${\displaystyle dV=dA\,ds}$:

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

Correcting for stimulated emission, we get:

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

## 3 Estimating Cross-Sections

The absorption coefficient, written in terms of Einstein constants is:

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

Thus, the cross-section of an atom for absorption of a photon is:

${\displaystyle \sigma _{12}={{B_{12}}\phi (\nu )h\nu \over 4\pi }\,\!}$

To estimate ${\displaystyle {B_{12}}}$, we use the fact that, ignoring g’s, ${\displaystyle {B_{12}}\sim {B_{21}}}$, and ${\displaystyle {{A_{21}} \over {B_{21}}}={2h\nu ^{3} \over c^{2}}}$. Then using the approximation that that ${\displaystyle \phi (\nu )\sim {\frac {1}{\Delta \nu }}}$, we get:

${\displaystyle \sigma _{12}\sim {{A_{21}} \over \left({2h\nu ^{3} \over c^{2}}\right)}{\frac {1}{\Delta \nu }}\,\!}$
${\displaystyle {\sigma _{12}\sim {\lambda ^{2} \over 8\pi }{{A_{21}} \over \Delta \nu }}\,\!}$

In a single atom, ${\displaystyle \Delta \nu \sim {A_{21}}}$, so ${\displaystyle \sigma _{12}\sim {\lambda ^{2} \over 8\pi }}$.