How are absorption cross sections calculated?

Question

I would like to develop a more intuitive understanding of cross sections, in the context of radiative transfer.

I understand that a cross section, $\sigma_\nu$, is a measure of a given atom/molecule's ability to absorb radiation, and is related to the mass absorption coefficient, $\kappa_\nu^m$, in the following way:

$\displaystyle \kappa_\nu^m = \frac{\sigma_\nu n}{\rho}$

(where n is the number density and $\rho$ is the mass density)

Cross sections are given in units of cm$^2$, but my understanding is that this should be thought of as a 'probability of absorption' rather than a 'physical area'.

How are cross sections calculated? How does this lead to the determination of a value in units of area?

Answer 1

In the case of the solar interior, where very often the physical conditions cannot be experimentally reproduced, the cross-sections are often calculated by the application of the relevant quantum mechanics to the interction in question. Usually, those calculations are tricky enough that they can only be accomplished for a grid of energies, densities and temperatures, and the the required results are interpolated from these so-called opacity tables. Sometimes these tables are the sum over relevant mixtures of chemical elements to give a "mean opacity".

The cross-section can be thought of as the equivalent opaque area that is contributed by a single absorber. If you imagine a cylinder of volume $V$ containing $n V$ of these absorbers (symbols as per your question), then if $\sigma_\nu$ is small, the effective total opaque area presented by the cylinder is $n V \sigma$.

Opacity $\kappa_\nu$ is defined as cross-sectional area per unit mass of absorber, but the mass in the cylinder equals $\rho V$. Hence the equation in your question.

In terms of probability, the likelihood of absorption is given by $\exp(-x/l_\nu)$, where $x$ is the path length through the absorbing medium and $l_\nu$ is the mean free path at frequency $\nu$. In the notation given in the question, $l_\nu=(n \sigma_\nu)^{-1} = (\kappa_\nu \rho)^{-1}$, which you can see has units of length.