The absorption coefficient is usually defined as

$$ \alpha(\nu)=\frac{1}{d}\ln\frac{I_0(\nu)}{I(\nu)}, $$ where $d$ is the thickness of the sample, $\nu$ is the light frequency, and $I_0$ and $I$ are the initial and transmitted light intensities, respectively. Assume there is a sharp line in the spectrum at a frequency $\nu_{if}$, caused by a transition of an active center from a state $i$ to a state $f$.

I have found in some sources (unfortunately without further references) the following expression (sometimes the factor $1/n$ is missing): $$ \int\alpha(\nu)d\nu\sim \frac{N\nu_{if}}{n}|M_{if}|^2 $$ where $N$ is the concentration of the active centers in the sample, $n$ is the refractive index of the sample (probably assumed to be frequency independent) and $M_{if}$ is the matrix element for the light induced transition $i\rightarrow f$. The proportionality coefficient presumably depends only on fundamental physical constants. The integration shall be performed over the "line contribution".

How can the expression be derived? Does the result depend on the shape of the line? The proportionality to $N$ and $|M_{if}|^2$ is obvious, but is there a clear physical interpretation for the factors $\nu_{if}$ and $n$?

  • $\begingroup$ I will provide the answer if my comments aren't helpful. Considering that the absorption coefficient may be written as $\alpha(\nu)=\sigma(\nu)N$ where $\sigma(\nu)$ is the photoabsorption cross section, you'll end up with $\int \sigma(\nu)d\nu\propto\frac{\nu_{if}}{n}|M_{if}|^{2}$. To make things clearer from here I would recommend Googling optical oscillator strength and specifically this paper: ui.adsabs.harvard.edu/abs/1994ApJS...95..301M/abstract $\endgroup$
    – Newbie
    Jan 13, 2022 at 18:20

1 Answer 1


It will be a good if you bring the context where you found this equation. Been then i→f represents the induced frequency (wavelength) transition, then I am lean to conclude that this equation is related to the emission event proportional to absorption events . Very much like the phase and source functions as represented in the diffusion approximation of the Radiative Transfer Equation. I have to add that it is confusing the fact that you have a Index or reflection term. I.R have little meaning in a diffuse scattering heterogeneous system.

How can the expression be derived? There will be no straight answer to this but if the radiation is in the UV or illvuminecense region, then look at laws governing Raleigh and Compton.

Does the result depend on the shape of the line? I am not sure what line you are looking at. But I believe that this integral equation and the proportionality shown is not related to the The Beer-Lambert Law.

  • $\begingroup$ All these questions here read so natural, don't they? Hahaha! As if a graduate student is hired to formulate them ... in the interest of this site too, of course. Many of us still like to think rather naive, don't we? Or am I alone? $\endgroup$ Dec 11, 2023 at 20:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.