How do you calculate luminosity

Temperature : A black body radiates power at a rate related to its temperature - the hotter the black body, the greater its power output per unit surface area. An incandescent or filament light bulb is an everyday example. As it gets hotter it gets brighter and emits more energy from its surface. The relationship between power and temperature is not a simple linear one though.

The power radiated by a black body per unit surface area is given varies with the fourth power of the black body's effective temperature, T eff.

As a star is not a perfect black body we can approximate this relationship as:. This relationship helps account for the huge range of stellar luminosities. A small increase in effective temperature can significantly increase the energy emitted per second from each square metre of a star's surface. Size radius : If two stars have the same effective temperature but one is larger than the other it has more surface area.

The power output per unit surface area is fixed by equation 4. This becomes apparent when we plot stars on an HR diagram. Using equation 4. In practice this equation is not used to determine the luminosity of most stars as only a few hundred stars have had their radii directly measured. If however, the luminosity of a star can be measured or inferred from other means eg by spectroscopic comparison then we can actually use equation 4.

Let us imagine we have two stars, A and B that we wish to compare. If we can measure their respective apparent magnitudes, m A and m B how will they differ in brightness?

If you are mathematically astute you should realise that this is in fact the same as equation 4. On the previous page we used the distance modulus equation 4. How is this equation derived? It is simply an application of the luminosity ratio relationship 4. Absolute magnitude is a different way to measure the luminosity. Instead of expressing it in watts, it can be shown on a logarithmic scale. The lower the absolute magnitude, the more luminous the star is - some very bright stars can even have negative magnitudes!

For example, the absolute magnitude of the Sun is equal to 4. Apparent magnitude, on the other hand, is a measure of brightness when the star is seen from Earth - hence, it takes into account the distance between the star and the Earth.

You can find it with the apparent magnitude calculator, using the following equation:. The absolute magnitude is defined as the apparent magnitude of an object seen from the distance of 10 parsecs. Let's analyze Sun with this luminosity calculator to investigate its absolute and apparent magnitude. Input the radius and temperature of the Sun into the calculator. The luminosity calculator will automatically find the luminosity of the Sun.

It is equal to 3. Embed Share via. Luminosity Calculator By Bogna Szyk. Table of contents: What is luminosity?

Luminosity equation Absolute and apparent magnitude Calculating luminosity: an example. Why do light sources appear fainter as a function of distance? The reason is that as light travels towards you, it is spreading out and covering a larger area. This idea is illustrated in this figure:. Again, think of the luminosity—the energy emitted per second by the star—as an intrinsic property of the star. As that energy gets emitted, you can picture it passing through spherical shells centered on the star.

In the above image, the entire spherical shell isn't illustrated, just a small section. Each shell should receive the same total amount of energy per second from the star, but since each successive sphere is larger, the light hitting an individual section of a more distant sphere will be diluted compared to the amount of light hitting an individual section of a nearby sphere.

The amount of dilution is related to the surface area of the spheres, which is given by:. See Technical Requirements in the Orientation for a list of compatible browsers. How bright will the same light source appear to observers fixed to a spherical shell with a radius twice as large as the first shell? Since the radius of the first sphere is d, and the radius of the second sphere would be 2 x d This equation is not rendering properly due to an incompatible browser. Since the same total amount of light is illuminating each spherical shell, the light has to spread out to cover 4 times as much area for a shell twice as large in radius.

The light has to spread out to cover 9 times as much area for a shell three times as large in radius. So, a light source will appear four times fainter if you are twice as far away from it as someone else, and it will appear nine times fainter if you are three times as far away from it as someone else.