Stellar Radii and the Stefan-Boltzmann Law
Stars radiate roughly like blackbody radiators.
I have already used this information to show how the
surface temperatures of stars may be inferred. We may also
use this information to show how the
radii of stars may be estimated. For this exercise, we use
the Stefan-Boltzmann law.
The Stefan-Boltzmann law
tells you how much energy a blackbody radiator of a given
temperature radiates per unit area
of its surface (i.e., it tells you the flux of radiation which comes off
the surface of a blackbody as a function of its temperature).
We have
Flux of energy = σ x Temperature4
The constant σ depends upon the unit system in which we choose to work.
In this course, we will are only concerned
with comparisons of objects (ratios) and so σ is
not needed.
Example:
Take two objects, one of temperature 2,000 K and one of temperature
4,000 K. How much more energy will the hotter object radiate (if they have
the same diameter)?
The ratio of the two energy fluxes is,
Flux 1/Flux 2 = σ (4,000 K)4/ σ(2,000 K)4 = 16
The hotter object is 16 times brighter than the cooler object.
Luminosity
Take a star of temperature T. How luminous will this
star be?
Since the star is roughly a blackbody radiator, it produces a
flux of energy at its surface given by σ x T4
(according to the Stefan-Boltzmann law
). To
figure out the total energy radiated, one needs to figure out how many
square meters there are on the star's surface.
This is
straightforward for a spherical star. The surface area of a sphere
is 4 pi R2. Combining the above two relations, we have that
Luminosity = surface area x flux = 12.6 R2 σ T4
Suppose that I have two stars with the same L.
One star is 6,000 K
and the other is 3,000 K, which star is larger and by how much?
Well, do roughly the same thing as above.
L = 12.6 R2 σ (6,000 K)4
L = 12.6 r2 σ (3,000 K)4
and so
L / L = 1 = (R/r)2 x (6,000/3,000)4 = (R/r)2 x 16
===> the hotter star is 1/4 the size of the cooler star!
