Surface Flux & Colors of Stars

Section 6 of 8

Inferring Stellar Radii

Part 6: Inferring Stellar Radii

We trust radii inferred from light because the chain is cross-validated: parallax distances match photometric luminosities, interferometry measures some angular diameters directly, and eclipsing binaries give independent sizes. The chain is model-based but repeatedly tested.

Solving the Stefan-Boltzmann law for radius: isolate $R^2 = L/(4\pi\sigma T^4)$ , then raise to the $1/2$ power, $R = (L/(4\pi\sigma T^4))^{1/2}$ . In solar units, $R/R_\odot = ((L/L_\odot)/(T/T_\odot)^4)^{1/2}$ .

Worked Example 3Sirius A

Problem

Sirius A has $L \approx 25\,L_\odot$ and $T \approx 9{,}900\,\mathrm{K}$ (from color), with $T_\odot = 5800\,\mathrm{K}$ , $R_\odot = 6.96 \times 10^{10}\,\mathrm{cm}$ . Find its radius.

StepTemperature ratio, fourth power

$T/T_\odot = 9{,}900/5{,}800 = 1.707$ , so $(T/T_\odot)^4 \approx 8.49$ .

StepSolve in solar units

$(R/R_\odot)^2 = 25/8.49 = 2.94$ , so $R/R_\odot = (2.94)^{1/2} \approx 1.71$ .

Dimensional check

Solar-unit ratios are dimensionless; the result is a pure multiple of $R_\odot$ ✓. In CGS, $R \approx 1.71 \times 6.96 \times 10^{10} = 1.19 \times 10^{11}\,\mathrm{cm}$ .

Result

$R_{\text{Sirius}} \approx 1.7\,R_\odot \approx 1{,}190{,}000\,\mathrm{km}$ . Hot stars don’t need to be big to be bright: at $1.7\,T_\odot$ , each cm² radiates $(1.707)^4 \approx 8.5\times$ the Sun’s surface power, so 25× the luminosity needs only ~70% more radius.

Predict first

Betelgeuse is 10^5 times more luminous than the Sun but only 60% as hot. Will its radius be closer to 10x, 100x, or 1000x the Sun's? Commit before checking the calculation.

Which order of magnitude — 10x, 100x, or 1000x R_sun — and why?

Then work Example 4 below.

Worked Example 4Betelgeuse (a supergiant)

Problem

Betelgeuse has $L \approx 10^5\,L_\odot$ and $T \approx 3{,}500\,\mathrm{K}$ . Find its radius and compare it to the Sun.

StepTemperature ratio, fourth power

$T/T_\odot = 3{,}500/5{,}800 = 0.603$ , so $(T/T_\odot)^4 \approx 0.133$ .

StepSolve in solar units

$(R/R_\odot)^2 = 10^5/0.133 = 7.54 \times 10^5$ , so $R/R_\odot = (7.54 \times 10^5)^{1/2} \approx 868$ .

Dimensional check

Dimensionless ratios ✓. In physical units, $R \approx 870 \times 6.96 \times 10^{10}\,\mathrm{cm} = 6.06 \times 10^{13}\,\mathrm{cm} \approx 4.1\,\mathrm{AU}$ .

Result

$R_{\text{Betelgeuse}} \approx 870\,R_\odot \approx 4.1\,\mathrm{AU}$ — its surface would reach past Mars and approach Jupiter. It is enormous because it is cool: each cm² radiates only $(0.603)^4 \approx 0.13$ of the Sun’s surface power, so producing $10^5 L_\odot$ demands ~760,000× the Sun’s surface area.