If we think of a beam of light as a stream of particles (like a stream of water, for example) we can speak of a flux , or flow of light. Flux is the amount of light that comes from a certain area (usually one square meter) in a certain amount of time (usually one second). The amount of flux given off by an object depends only on its temperature, according to the Stefan-Boltzmann Law:

Flux = (Sigma)T4

where T is the temperature (in K) and the Greek letter Sigma is a constant term called the Stefan-Boltzmann Constant. Its value is unimportant for this class, as we shall see. Flux is not the true measure of an object's energy output. For example, a flashlight and a searchlight have similar temperatures, therefore similar fluxes. But from a distance of 100 yards, the searchlight is the brighter of the two. Why? Because the searchlight is bigger!

We call the total energy output per second of an object its luminosity. It depends not only on Flux (temperature) but also on size (or, more accurately, surface area). Stars are for the most part spherical, so we can compute their surface areas easily, using A = 4(pi)R2, where R is the radius of the sphere. Therefore

Luminosity = (Flux)(Surface Area) = (SigmaT4) (4(pi)R2)

While it is possible to compute the exact values of luminosities, it requires that we know the value of Sigma. We can get around this by comparing the luminosities of two objects, either two different objects, or the same object before or after some great change in temperature, radius, or both:

L1 / L2 = (SigmaT14) (4(pi)R12) / (SigmaT24) (4(pi)R22)

Sigma and 4 and pi all drop out, leaving us with:

L1 / L2 = (T14 R12) / (T24 R22)

Or, to arrange it another way:

L1 / L2 = (T14 / T24) (R12 / R22)

Often we will use the Sun as object #2, comparing its luminosity to that of another star.

Sample Calculations

To see some sample calculations with Luminosity, click on the examples below



When you finish with these examples, read on about "brightness."

But even luminosity is not the true measure of how bright an object appears. A flashlight at 2 feet is blinding, but a searchlight at 10 miles is not. How can this be if the searchlight has a higher luminosity?

As light leaves a source, it spreads out in a spherical pattern. As the photons get farther away from the light source, they spread out and become less and less concentrated (there are only so many photons to go around, after all). Thus the light source appears dimmer the farther away it is. This is expressed mathematically using the inverse-square relation:

B = L / 4 (pi) D2

B = (SigmaT4) (4(pi)R2) / (4 (pi) D2)

B = (Sigma T4) (R2 / D2)

B = (Sigma T4) ( R / D )2

Remember that R is the actual radius of the light source, and D is the distance of the light source. The units for R and D don't really matter, as long as they are the same (both R and D in km, for example). We can get around using Sigma here the same way we did above, by comparing the brightnesses of two objects.

Apparent brightness decreases as the square of the distance. Thus, if I moved a light source from 1 foot to 5 feet away from me, it would appear 25 times dimmer.


Updated 8/27/99
By James E. Heath

Copyright Ó 1999 Austin Community College