Surface Flux & Colors of Stars

Section 5 of 8

Wien's Law: Color to Temperature

Part 5: Wien’s Law — Color to Temperature

To use Stefan-Boltzmann we need an independent temperature. Wien’s displacement law supplies it: measure a star’s color (where its spectrum peaks) and read off temperature directly.

conceptcolor-temperature

verbal

A star’s color is its temperature. Hotter stars peak at shorter (bluer) wavelengths; cooler stars peak at longer (redder) wavelengths. The relationship is inverse — double the temperature, halve the peak wavelength.

equation

seewien-displacement(T)

figure

see Fig.energy-wavelength-connection(T)

Each Planck curve has a single peak, and Wien’s law quantifies the pattern: $\lambda_{\text{peak}} \propto T^{-1}$ . Unlike distance, temperature can be read directly from color. In practice astronomers compare flux through standardized filters (a color index) as a proxy for the peak — the same Wien physics.

Wien's law

$\lambda_{\text{peak}} = b/T$ with $b = 0.2898\,\mathrm{cm\,K}$ — the peak wavelength of a blackbody’s per-wavelength Planck curve is inversely proportional to temperature. Hotter is bluer.

Two-part diagram. Top: 'The Energy-Wavelength Connection' with equation E = hc/lambda and a wave transitioning from red (long wavelength) to blue (short wavelength). Bottom: 'The Temperature Signature' as a color gradient from a cool red star (3,000 K) to a hot blue star (30,000 K), noting Wien's law allows temperature calculation from peak color. — Figure 5E = hc/lambda means shorter wavelength = higher energy. Wien's law reads temperature from color: cool stars are red (~3,000 K), hot stars blue (~30,000 K).Course illustration (A. Rosen)

Problem

From $\lambda_{\text{peak}} = b/T$ : (1) $T$ doubles → peak wavelength? (2) A star twice as hot as the Sun peaks where, relative to the Sun? (3) One star peaks red (700 nm), another blue (400 nm) — what’s the temperature ratio?

Worked Example 1The Sun's temperature from its color

Problem

The Sun’s spectrum peaks at $\lambda_{\text{peak}} \approx 500\,\mathrm{nm}$ . Calculate its surface temperature using Wien’s law, $T = b/\lambda_{\text{peak}}$ , with $b = 2.898 \times 10^6\,\mathrm{nm\,K}$ .

StepApply Wien's law

$T_\odot = \dfrac{2.898 \times 10^6\,\mathrm{nm\,K}}{500\,\mathrm{nm}} = 5796\,\mathrm{K}$ .

Dimensional check

$[T] = \mathrm{nm\,K}/\mathrm{nm} = \mathrm{K}$ ✓ — the nm cancels, leaving kelvin.

Result

$T_\odot \approx 5800\,\mathrm{K}$ , matching the directly measured solar surface temperature — a validation that the Sun radiates approximately as a blackbody.

Worked Example 2Rigel and Betelgeuse (temperature contrast)

Problem

Rigel (blue) peaks at $\lambda_{\text{peak}} \approx 240\,\mathrm{nm}$ ; Betelgeuse (red) at $\approx 830\,\mathrm{nm}$ . Find both temperatures via Wien’s law.

StepRigel

$T_{\text{Rigel}} = \dfrac{2.898 \times 10^6\,\mathrm{nm\,K}}{240\,\mathrm{nm}} \approx 12{,}100\,\mathrm{K}$ .

StepBetelgeuse

$T_{\text{Betelgeuse}} = \dfrac{2.898 \times 10^6\,\mathrm{nm\,K}}{830\,\mathrm{nm}} \approx 3{,}500\,\mathrm{K}$ .

Dimensional check

$\mathrm{nm\,K}/\mathrm{nm} = \mathrm{K}$ ✓ for both.

Result

$T_{\text{Rigel}}/T_{\text{Betelgeuse}} \approx 3.5$ — Rigel is about 3.5 times hotter. Their visible colors directly reflect this: hot is blue, cool is red. This color–temperature relationship anchors the HR diagram’s horizontal axis.

Vertical electromagnetic spectrum showing wavelength bands with corresponding temperatures: Gamma/X-Ray at top for million-degree plasma and black holes, UV/Visible in middle for stars (3,000K-50,000K), Infrared/Radio at bottom for dust (100K) and cold gas (10K). Rainbow colors shown in the visible band. — Figure 6The EM spectrum is a temperature ladder: gamma/X-ray = million-degree plasma; UV/visible = stellar surfaces (3,000-50,000 K); infrared/radio = dust and cold gas (10-100 K).Course illustration (A. Rosen)