# 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

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.