Radiation Transport

Section 2 of 7

The Mean Free Path

Part 2: The Mean Free Path — How Far a Photon Travels

Start microscopically. If a photon moves through targets of number density $n$ , each with cross section $\sigma$ , the mean free path is $\ell = 1/(n\sigma)$ . Since opacity is cross-section per unit mass, $\kappa = n\sigma/\rho$ , so $n\sigma = \kappa\rho$ and:

Higher density means more targets per volume; higher opacity means more effective targets per gram. Both shorten the distance a photon travels before interacting.

Mean free path

The average distance a photon travels between interactions, $\ell = 1/(\kappa\rho)$ — about $0.02~\text{cm}$ in the solar core. When $\ell \ll R$ , radiation cannot stream out; it diffuses.

Worked Example 1Mean Free Path in the Solar Core

Problem

Use one-zone solar-core values $\rho_c \approx 150~\text{g cm}^{-3}$ and $\kappa_\text{es} \approx 0.34~\text{cm}^2\,\text{g}^{-1}$ to find the mean free path, and compare it to the solar radius.

StepCompute the mean free path

$\kappa\rho_c = (0.34)(150) = 51~\text{cm}^{-1} \Rightarrow \ell = \frac{1}{\kappa\rho_c} \approx 1.96 \times 10^{-2}~\text{cm} \approx 0.20~\text{mm}$

StepCount mean free paths across the Sun

$\frac{R_\odot}{\ell} \approx \frac{7.0 \times 10^{10}}{2.0 \times 10^{-2}} \approx 3.5 \times 10^{12} = \tau$

Dimensional check

$[\kappa\rho] = (\text{cm}^2\text{g}^{-1})(\text{g cm}^{-3}) = \text{cm}^{-1}$ , so $[\ell] = 1/[\kappa\rho] = \text{cm}$ ✓.

Result

A photon travels only ~0.2 mm between interactions, and the Sun is $\sim 3.5 \times 10^{12}$ mean free paths thick — not just opaque, but enormously optically thick. The crucial result is $\ell \ll R_\odot$ : the diffusion regime.

Quick check

Suppose $\kappa$ doubles while $\rho$ stays fixed. (1) What happens to $\ell$ ? (2) To $\tau$ across the same star? (3) Does diffusion become faster or slower?