# How are absorption cross sections calculated?

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?

• Not quite clear what you want. Cross-sections are usually "calculated" by the application of quantum mechanics to the interaction in question. I'm not sure how far down that rabbit-hole you want to go. Apr 4, 2022 at 17:10

## 1 Answer

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.

• Thanks for your answer. I think what I'm trying to understand is the process described in your first paragraph. If we have, say, a single molecular species, how do we go from knowledge of its energy levels and transitions, to getting cross sections? If this is the 'application of the relevant quantum mechanics' that you mention, is there a simple way to describe the steps taken? Apr 4, 2022 at 21:30