AST-17

Saha ionisation snapshot

n_{i+1} n_e / n_i = (kT / λ_th³) exp(−χ / kT) (g-factor = 1 sketch).

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StarsSaha

Governing equation

\dfrac{n_{i+1}n_e}{n_i}=\left(\dfrac{2\pi m kT}{h^2}\right)^{3/2}e^{-\chi/kT}

$T \in \left[3000, 30000\right]$

where

T: Temperature (K)
\chi: Ionisation (eV)
n_e: Electron density (10¹⁸/m³)
n_{i+1}/n_i: Ion ratio (—)

The figure, part by part

Each numbered marker wears the color of its card. Hover a card to light the marker. Tap a marker to jump to that card. The slider walks a red point along the curve; Animate drawing traces it like a pen.

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①

Governing law — red

\dfrac{n_{i+1}n_e}{n_i}=\left(\dfrac{2\pi m kT}{h^2}\right)^{3/2}e^{-\chi/kT}

$T \in \left[3000, 30000\right]$

$T = 3000$ $n_{i+1}/n_i = 5.64\times 10^{-15}$

Along the curve

②

Domain of the variable — blue

T_{\min} = 3000,\quad T_{\max} = 30000

$T \in \left[3000, 30000\right]$

$T = 3000$ $T = 3000$

Along the curve

③

Response curve — green

\dfrac{n_{i+1}n_e}{n_i}=\left(\dfrac{2\pi m kT}{h^2}\right)^{3/2}e^{-\chi/kT}

$T \in \left[3000, 30000\right]$

$T = 3000$ $n_{i+1}/n_i = 5.64\times 10^{-15}$

Along the curve

④

Rate of change — amber

\frac{\mathrm{d}n_{i+1}/n_i}{\mathrm{d}T}

$T \in \left[3000, 30000\right]$

$T = 3169.8$ $n_{i+1}/n_i^{\prime} = 4.09\times 10^{-15}$

Along the curve

Lecture brief

Historical brief

Hubble expansion, Jeans collapse, Eddington luminosity, Bondi accretion and Stefan–Boltzmann stars are the first astrophysical budgets. The sheets scale a star, a cloud and a horizon. This sheet (AST-17 — Saha ionisation snapshot) is the form associated with Saha. Working symbols: $T$ , $\chi$ , $n_e$ $\rightarrow$ $n_{i+1}/n_i$ . Why Balmer lines peak around A0: hydrogen ionises, then the n=2 population falls.

Purpose

Purpose: compute

n_{i+1}/n_i

from

T

\chi

n_e

in Astrophysics via

\dfrac{n_{i+1}n_e}{n_i}=\left(\dfrac{2\pi m kT}{h^2}\right)^{3/2}e^{-\chi/kT}

n_{i+1} n_e / n_i = (kT / λ_th³) exp(−χ / kT) (g-factor = 1 sketch). Use it when a real astrophysics question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given

T = 10000.000\,\mathrm{K}

\chi = 13.600\,\mathrm{eV}

n_e = 1.000\,\mathrm{10¹⁸/m^{3}}

, the governing relation

\dfrac{n_{i+1}n_e}{n_i}=\left(\dfrac{2\pi m kT}{h^2}\right)^{3/2}e^{-\chi/kT}

yields

n_{i+1}/n_i = 337.8795\,\mathrm{—}

. A star atmosphere, two ion stages, a ratio. Move a slider: the numbers are this situation, not a canned story.

Calculator

Inputs

Temperature T10000 KIonisation \chi13.6 eVElectron density n_e1 10¹⁸/m³

Outputs

Ion ratio n_{i+1}/n_i337.8795 —

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Narration of this film

A star atmosphere, two ion stages, a ratio.

Why Balmer lines peak around A0: hydrogen ionises, then the n=2 population falls.

Reading speed

Watch on YouTube