Luminosity and Flux of Stars

How to calculate the total amount of energy radiated per second by a star

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In astronomy, luminosity is the total amount of energy radiated by a star, galaxy, or another astronomical object per unit time. It is related to the brightness, which is the luminosity of an object in a given spectral region. In SI units luminosity is measured in joules per second or watts.

Introduction to Astronomy Series

Luminosity, L, is the total outward flow of energy from a radiating body per unit time, in all directions and over all wavelengths. The SI units of luminosity are Watts (W) which quantify the rate of energy transfer in joules per second.

Flux, F, is defined as the total flow of light energy perpendicularly crossing a unit area per unit time, or the total energy from an object per unit area over time. Flux has units of J s-1 m-2 or W -2 (that's Joules per second per metre squared, or simply Watts per second)

Luminosity is the rate at which a star, or any other body, radiates its energy. It is the same as the classification of light bulbs. A 40W bulb radiates less energy than a 100W bulb. The same is true for stars, however, they radiate far greater quantities of energy. The energy output of our Sun is around 3.89x1026 W (that's 389 followed by 24 zeros!), and our Sun isn't even a bright star! As stars go, our Sun is a 20W bulb.

Because of the large quantities involved with luminosity, astronomers prefer to use a more convenient unit called a solar luminosity, given the symbol L. One solar luminosity is equal to the luminosity of our Sun, but even so, stars can be as high as 1x106L? so very large numbers cannot be avoided!

A star which has a luminosity of 2L is twice as luminous as our Sun, and a star of 0.5L is half as luminous.

Luminosity

Whereas flux is the energy received over a unit area, luminosity is the total energy output of the star. Since the star radiates in all directions (isotropically) we only receive a tiny fraction of the energy radiated which is how we observe flux and calculate apparent magnitude. It would be helpful to know the relationship between the flux observed at Earth and the star's luminosity (total energy output).

Imagine a star at a distance d radiating equally in all directions. Flux is measured with a detector (whatever type) and has a surface area of 1 m2 and is perpendicular to the star. From the diagram above we can see that as the star radiates in all directions and forms a sphere around the star.

From previous definitions, luminosity (L) is the total energy output per second and flux (F) is the total energy per second crossing a unit area of surface, so we can determine the relationship between flux and luminosity. The surface area of a sphere of radius d is:

Equation 21 - Sphere, Surface Area of

So the flux, F, measured at Earth by the detector of unit area is given by:

Equation 22 - Flux and Luminosity

We can see from the equation that flux decreases as distance increases and we can also see that distance is squared. It follows from this that light obeys the inverse square law - the observed flux from a star is inversely proportional to the square of the distance between it and an observer. This is more clearly illustrated in the diagram below.

The Inverse Square Law

The brightness of a star seen from the Earth depends its intrinsic luminosity and also on its distance from the Earth. The observed brightness of a given star decreases inversely proportionally to its distance away. The presence of interstellar gas will further decrease the observed brightness. From the diagram we can see that for every additional distance unit, r, the light is spread over an additional area, r2.

Flux

In the article about the magnitude scale we saw that Pogson devised a scale whereby a 1st magnitude star is 100 times brighter than a 5th magnitude star. This logarithmic scale states that a 1st magnitude star is 2.512 times brighter than a 2nd, which is 2.512 times brighter than a 3rd and so on...

We can use this constant ratio per magnitude to obtain a formula for the ratio of fluxes. Consider two stars that have apparent magnitudes m and n and measured fluxes of Fm and Fn, the ratio of the fluxes is given by:

Equation 18 - Ratio of Fluxes

If one star is 6th magnitude `(n = 6)` and another star is 1st magnitude `(m = 1)` then the magnitude difference is given by `(n-m) = 6-1 = 5` We can use Equation 18 to calculate the ratio of these fluxes.

Equation 19 - Ratio of Fluxes

We can see that this equation has shown that a difference in 5 magnitudes affects magnitude by a factor of 100 as per Pogson's rule. We can further reinforce this relationship between flux and magnitude by showing that the magnitude difference between two objects can be expressed in terms of the logarithm of the flux ratio. This form is known as Pogson's relation and one form or another one of the most useful equations in the astronomer's toolbox.

Equation 20 - Pogsons Relation

Last updated on: Wednesday 24th January 2018