Surface Flux & Colors of Stars

Section 2 of 8

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.