Radiation and Convection Inside Stars

Energy made in the core has to get out. In the Sun's inner region it travels as light, but not in a straight line: the gas is so dense that a photon is absorbed or scattered after a tiny distance, the mean free path $\ell$, and sent off again in a random direction. Building on the conditions at the center, we can estimate how long the trip takes.

Why it works

Each step is a vector $\vec{s}_i$ of length $\ell$ in a random direction. After $N$ steps the displacement is $\vec{D} = \sum \vec{s}_i$, so $$D^2 = \sum_i s_i^2 + \sum_{i \ne j}\vec{s}_i \cdot \vec{s}_j.$$ The cross terms are as often negative as positive and average to zero, leaving $D^2 \approx N\ell^2$, or $D \approx \sqrt{N}\,\ell$. To cover a distance $R$ takes $N = (R/\ell)^2$ steps, a total path of $N\ell = R^2/\ell$, and a time $t \approx R^2/(\ell c)$.

Key idea

A random walk of $N$ steps covers only about $\sqrt{N}\,\ell$. Crossing a star of radius $R$ takes $t \approx R^2/(\ell c)$, which is $R/\ell$ times longer than a straight flight. For the Sun the answer is on the order of $10^{5}$ years.

A photon's random walk. After $N$ steps it is typically only $\sqrt{N}$ step lengths from where it began.

Radiation and convection

In the Sun, energy diffuses outward by radiation from the core to about 0.7 of the radius. Farther out the gas is cooler and far more opaque, and radiation would need a very steep temperature drop to carry the energy. Instead the gas churns: hot blobs rise, cool and sink, in the convective zone, whose tops you can see as granules on the surface. Stars more massive than about 1.3 Suns have convective cores (their energy is made in a small, very hot region); stars below about 0.35 solar masses are convective all the way through.

Worked examples

Worked example: Counting steps

How many random steps of 1 cm does it take to get 1 m from the start?

$N = (D/\ell)^2 = (100\text{ cm}/1\text{ cm})^2$.

Answer: $N = 10{,}000$ steps, not 100.

Worked example: Escaping the Sun

With an average mean free path of 1 mm, how long does energy take to random-walk out of the Sun? ($R = 6.96 \times 10^{8}$ m.)

$t = R^2/(\ell c) = (6.96 \times 10^{8})^2/(10^{-3} \times 3.00 \times 10^{8})$.
$t = 1.6 \times 10^{12}$ s.

Answer: About 51,000 years. With $\ell = 0.5$ mm, about 100,000 years. A straight flight would take 2.3 seconds.

Worked example: Where the mean free path comes from

Photons scatter off free electrons with cross-section $\sigma_T = 6.65 \times 10^{-29}$ m². What is $\ell$ where $n_e = 1.0 \times 10^{30}$ m⁻³?

$\ell = 1/(n_e\sigma_T)$.
$= 1/(6.65 \times 10^{1})$ m.

Answer: $\ell \approx 1.5$ cm. Deeper in, densities are higher and absorption by ions adds to the scattering, so the average step in the Sun is much shorter.

Common mistake

Thinking light made in the core reaches the surface in a few seconds because it travels at the speed of light. Each photon goes only about a millimeter or so before it is absorbed and re-emitted in a random direction, so the energy wanders outward slowly; estimates for the Sun range from tens of thousands to a few hundred thousand years, depending on the model. Check with $D = \sqrt{N}\,\ell$: doubling the distance takes four times as many steps.

Radiation and Convection Inside Stars

Radiation and convection

Worked examples

Practice

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Sources and further reading