Surface Flux & Colors of Stars

Complete lesson

Two Stars with the Same Luminosity

After completing this reading, you should be able to:

Distinguish
received flux (what an observer measures) from surface flux (a stellar property).
Explain
why surface brightness is independent of distance using inverse-square scaling.
Apply
the Stefan-Boltzmann law $L = 4\pi R^2 \sigma T^4$ to connect luminosity, temperature, and radius.
Use
Wien’s displacement law to infer stellar surface temperature from observed color.
Calculate
stellar radius given measurements of luminosity and temperature.
Connect
the color sequence to the temperature sequence and preview the HR diagram’s organization.
Evaluate
the blackbody and spherical-symmetry assumptions and their impact on inferred properties.

What 'color' means here

When we say “color is temperature,” we do not mean only the color your eyes perceive. We mean electromagnetic color — the peak wavelength of a star’s Planck curve across the entire spectrum. Rigel’s “color” peaks in the ultraviolet (~240 nm); Betelgeuse’s in the near-infrared (~830 nm). Neither peak is visible, yet both encode temperature through Wien’s law.

Two complications matter in practice: interstellar dust scatters blue light preferentially, making stars look redder and cooler than they are; and astronomers rarely measure the exact peak — they compare flux through standardized filters and infer an effective temperature. “Color” here means “where the Planck curve peaks,” and the temperature we infer is always an effective temperature requiring dust corrections before it is reliable.

Part 1: Two Stars with the Same Luminosity

Look toward Orion. Blue-white Rigel at the foot and red Betelgeuse at the shoulder both radiate roughly 100,000 times the Sun’s power — yet one is a compact blue supergiant and the other an enormous red supergiant whose surface would reach past Mars’s orbit. How can two stars with the same luminosity differ so much in size?

A hot (blue) object radiates energy much more efficiently per square centimeter than a cool (red) object. So to produce the same total luminosity, a hot star can be smaller; a cool star must be huge. This lecture quantifies that intuition with the Stefan-Boltzmann law — the second major step in stellar inference.

Surface Flux and Received Flux

Part 2: Surface Flux vs. Received Flux

In Lecture 1 you learned that the flux received at Earth falls off with distance.

This received flux depends on the observer’s distance — the farther away, the dimmer it appears, even though the star’s intrinsic power is unchanged. The same luminosity $L$ is spread over a sphere of area $4\pi d^2$ that grows with distance.

Figure 1Light intensity decreases with the square of distance: the same luminosity is spread over a sphere of area 4-pi-d-squared that grows with distance.Course illustration (A. Rosen)

Received flux

The power per unit area an observer measures at distance $d$ , $F = L/(4\pi d^2)$ . Distance-dependent — it falls as $1/d^2$ .

Now consider a different quantity: if you could stand on the star’s surface, how much power per unit area would you receive? That is the surface flux:

$F_* = \frac{L}{4\pi R^2}$

Surface flux

The power per unit area radiated at the star’s own surface, $F_* = L/(4\pi R^2)$ . It depends only on the star’s intrinsic properties ( $L$ and $R$ ), not on the observer’s distance.

Surface flux depends only on the star’s intrinsic properties, not on where the observer is.

Quantity	Equation	What it measures	Depends on
Received flux ( $F$ )	$F = L/(4\pi d^2)$	Brightness we measure	Star’s distance; our location
Surface flux ( $F_*$ )	$F_* = L/(4\pi R^2)$	Radiation at the star’s surface	Star’s luminosity and radius only

NoteSymbol legend — quantities used in this reading

Symbol	Name	Units (CGS)
$L$	Luminosity	erg/s
$F$	Received flux	erg s⁻¹ cm⁻²
$F_*$	Surface flux	erg s⁻¹ cm⁻²
$R$	Stellar radius	cm (or $R_\odot$ )
$d$	Distance	cm (or pc)
$T$	Effective temperature	K
$\sigma$	Stefan-Boltzmann constant	erg cm⁻² s⁻¹ K⁻⁴
$b$	Wien’s constant	cm·K
$\lambda_{\text{peak}}$	Peak wavelength	nm or cm

Quick check

If a star’s distance doubles, received flux does what?
If a star’s radius doubles (with $L$ fixed), surface flux does what?
Two stars have the same received flux at Earth. Must they have the same surface flux?

Derivation graphic comparing a star viewed at two distances d1 and d2 with angular radii alpha1 and alpha2. Equations show angular radius, solid angle, flux F equals L over 4 pi d squared, and intensity I equals F over Omega simplifying to L over 4 pi squared R squared equals F-star over pi. — Figure 2Surface brightness is distance-independent because both received flux and apparent solid angle scale as 1/d-squared, so their ratio stays constant.Fundamentals of Astrophysics (Owocki)

Spherical coordinate diagram highlighting a small surface patch between theta and theta plus delta-theta and across delta-phi. From the sphere center, dashed rays outline the patch and the formula delta-Omega equals sin theta times delta-theta times delta-phi is shown. — Figure 3A small patch on the sky has solid angle delta-Omega = sin(theta) delta-theta delta-phi, the 2D angular area element used in flux and intensity integrals.Fundamentals of Astrophysics (Owocki)

Common student confusions

Watch out for these pitfalls before moving on.

The Stefan–Boltzmann Law

Part 3: The Stefan-Boltzmann Law

Every blackbody is described by its temperature. The Stefan-Boltzmann law quantifies how much power a blackbody radiates per unit area: $F_* = \sigma T^4$ , where $\sigma = 5.67 \times 10^{-5}\,\mathrm{erg\,cm^{-2}\,s^{-1}\,K^{-4}}$ . Multiplying by the spherical surface area $4\pi R^2$ gives the total luminosity.

conceptluminosity-temperature-radius

verbal

Temperature dominates and size amplifies. A blackbody’s output rises steeply with temperature — as $T^4$ per unit area — while total luminosity scales with surface area as $R^2$ . So a hot star can be small and still bright, and a cool star must be huge to match it.

equation

seestefan-boltzmann(L)

figure

see Fig.blackbody-stellar-spectra(T)

The Stefan-Boltzmann law connects three quantities — $L$ , $T$ , $R$ — so knowing any two gives the third. Temperature enters as $T^4$ : a steep dependence that makes even small temperature differences produce large effects.

Stefan-Boltzmann law

$L = 4\pi R^2 \sigma T^4$ — the total power a spherical blackbody of radius $R$ and effective temperature $T$ radiates. Surface flux $\sigma T^4$ times area $4\pi R^2$ .

Graph showing three Planck curves for stars at 8000 K (blue, peaks in the near-UV around 360 nm), 5000 K (yellow, peaks near 580 nm), and 3000 K (red, peaks in the near-IR around 970 nm). The visible band (about 400-700 nm) is marked. Y-axis is brightness; X-axis is wavelength. — Figure 4Hotter stars peak at shorter (bluer) wavelengths and emit more total light: an 8000 K star peaks near 360 nm, a 3000 K star near 970 nm. Wien's law: lambda_peak = b/T with b = 0.2898 cm K.JWST/STScI

Scaling with the Sun

Part 4: Dimensionless Form — Scaling with the Sun

Working in absolute CGS units means juggling $\sigma = 5.67 \times 10^{-5}$ and $R_\odot = 6.96 \times 10^{10}\,\mathrm{cm}$ — easy places for arithmetic errors. Express everything relative to the Sun instead: dividing the Stefan-Boltzmann law for any star by the same law for the Sun cancels the constants.

$\frac{L}{L_\odot} = \left(\frac{R}{R_\odot}\right)^2 \left(\frac{T}{T_\odot}\right)^4$

The $4\pi$ and $\sigma$ appear in both numerator and denominator, so they cancel — leaving only ratios. Solving for radius in solar units:

$\frac{R}{R_\odot} = \left(\frac{L/L_\odot}{(T/T_\odot)^4}\right)^{1/2}$

No physical constants appear — just ratios. Solar values: $T_\odot \approx 5800\,\mathrm{K}$ , $R_\odot = 6.96 \times 10^{10}\,\mathrm{cm} \approx 696{,}000\,\mathrm{km}$ .

Problem

From $L \propto R^2 T^4$ : (1) radius doubles, $T$ fixed → $L$ does what? (2) $T$ doubles, radius fixed → $L$ ? (3) both double → $L$ ?

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)

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.

ESO image of Betelgeuse with solar system orbits overlaid for scale. The star's disk extends past Mars's orbit and approaches Jupiter's orbit. Inner planets (Mercury, Venus, Earth, Mars) would be inside the star. Angular scale bar shows 0.015 arcseconds. — Figure 7Betelgeuse is enormous — it would engulf Mercury through Mars and extend to about 4 AU. Red supergiants are cool (~3,500 K) but luminous because of their vast surface area.ESO/L. Calcada

Color and the HR Diagram

Part 7: Color and the HR Diagram (Preview)

Hot stars are blue (short peak, high $T$ ); cool stars are red (long peak, low $T$ ); the relationship is quantitative, $\lambda_{\text{peak}} = b/T$ . Astronomers use color indices — brightness in two filter bands — to estimate temperature without finding the exact peak.

Hertzsprung-Russell diagram with logarithmic luminosity on the vertical axis (relative to the Sun) and surface temperature on the horizontal axis decreasing from about 30000 K at left to 3000 K at right. Colored stellar points and shaded regions mark the main sequence, giants, supergiants, and white dwarfs, with the Sun labeled near luminosity 1 and temperature about 5800 K. — Figure 8The H-R diagram separates stars into main sequence, giants/supergiants, and white dwarfs; temperature decreases left-to-right while luminosity increases upward.ESO

The Hertzsprung-Russell diagram plots luminosity (vertical) against temperature/color (horizontal). The key insight: the horizontal axis is a temperature axis — hot blue stars on the left, cool red stars on the right (historical convention). Since $L \propto R^2 T^4$ , lines of constant radius are diagonal curves across it; every star lies on some constant- $R$ curve.

The More You Know: Enrichment: why OBAFGKM is a temperature sequence

The spectral sequence O B A F G K M, defined by prominent spectral lines, is fundamentally sorted by temperature:

Type	Temperature	Color	Example
O	30,000–50,000 K	Blue-UV	θ Orionis
B	10,000–30,000 K	Blue-white	Rigel, Spica
A	7,000–10,000 K	White	Vega, Sirius
F	6,000–7,500 K	Yellow-white	Procyon A
G	5,200–6,000 K	Yellow	Sun
K	3,700–5,200 K	Orange	Aldebaran
M	<3,700 K	Red	Betelgeuse

Later spectral type = cooler temperature. Which lines appear is set by ionization equilibrium (covered in later spectroscopy lectures), but the ordering is temperature.

When you plot many stars, they cluster: the main sequence (hydrogen-burning, where the Sun lives), the giant branch (cool but luminous — large radii compensate), and the white dwarf sequence (hot but dim — tiny radii, ~ $0.01\,R_\odot$ ). Lines of constant radius separate these populations.

Assumptions and Synthesis

Part 8: Assumptions and Limitations

The Stefan-Boltzmann law assumes:

Spherical symmetry — uniform radiation from a sphere of radius $R$ . Reality: rapid rotators are oblate, binaries tidally distorted. Small error for most stars.
Uniform surface temperature — Reality: gradients and starspots exist, so $T$ is an effective temperature: the blackbody temperature reproducing the total flux.
Blackbody radiation — Reality: real spectra have absorption lines and non-thermal emission, but the overall Planck shape is a good approximation; lines are second-order.

Astronomers determine effective temperature from color (Wien’s law), spectral-energy-distribution fits, temperature-sensitive spectral lines, or parallax + photometry — methods that typically agree to within 100–200 K.

Top: rainbow spectrum image of star Altair showing dark absorption lines. Bottom: graph of brightness vs wavelength (about 400-700 nm) showing a smooth blackbody-like curve with sharp dips at absorption line wavelengths. Hydrogen Balmer lines labeled. — Figure 9A real stellar spectrum combines a continuous blackbody shape with absorption lines: the curve gives temperature (Wien), the lines give composition (spectroscopy).JWST/STScI

Infographic titled 'Comparing Density Through Spectra' comparing a blue giant and a white dwarf. Two rainbow spectra are shown with dark absorption lines, where the white dwarf's lines are visibly broader to illustrate pressure broadening at higher density. — Figure 10Broader absorption lines indicate higher pressure and density, so white dwarfs show pressure-broadened spectra compared with low-density giants.JWST/STScI

Effective temperature

The temperature a uniform blackbody would need to radiate the same total surface flux as the star, $F_* = \sigma T_{\text{eff}}^4$ . It is the " $T$ " used throughout the Stefan-Boltzmann and Wien relations.

Diagram showing interstellar reddening: a hot blue star emits light through a dust cloud (grains about 0.1 micrometers). Blue rays scatter away while red rays pass through, so the observer sees a red star. Labels indicate Source (Hot/Blue), Dust Grain size, and Observer sees Red Star. — Figure 11Dust preferentially scatters blue light, making distant stars appear redder than they truly are; this reddening must be corrected before inferring temperatures.Course illustration (A. Rosen)

Summary: Observable → Model → Inference

Circular flowchart titled 'The Astronomer's Decoder Ring' with Inference (Reality Revealed) at center. Four stages around the circle: Signal (photons arrive from distant objects), Measurement (flux and wavelength quantified through instruments), Model (apply physics like L = 4-pi-R-squared-sigma-T-to-the-fourth), Correction (account for dust and distance). — Figure 12The cycle that makes astronomy a science: Signal, Measurement, Model, Inference, Correction, and back to Model. Failed predictions drive model revision.Course illustration (A. Rosen)

We observe	We use	We infer
Received flux $F$ at distance $d$	$L = 4\pi d^2 F$	Luminosity
Peak wavelength $\lambda_{\text{peak}}$	Wien: $T = b/\lambda_{\text{peak}}$	Temperature
Luminosity $L$ and temperature $T$	Stefan-Boltzmann: $R = (L/(4\pi\sigma T^4))^{1/2}$	Radius

Each star yields at least three fundamental properties from two measurements (brightness + color) and one distance. That is the power of multi-wavelength, multi-technique astronomy.

Glossary

Effective temperature: The temperature a uniform blackbody would need to radiate the same total surface flux as the star, $F_* = \sigma T_{\text{eff}}^4$ . It is the " $T$ " used throughout the Stefan-Boltzmann and Wien relations.
Received flux: The power per unit area an observer measures at distance $d$ , $F = L/(4\pi d^2)$ . Distance-dependent — it falls as $1/d^2$ .
Stefan-Boltzmann law: $L = 4\pi R^2 \sigma T^4$ — the total power a spherical blackbody of radius $R$ and effective temperature $T$ radiates. Surface flux $\sigma T^4$ times area $4\pi R^2$ .
Surface flux: The power per unit area radiated at the star’s own surface, $F_* = L/(4\pi R^2)$ . It depends only on the star’s intrinsic properties ( $L$ and $R$ ), not on the observer’s distance.
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.