Spark image

Luminosity and brightness


Luminosity is a measure of the total energy given output by a star at all wavelengths from gamma radiation to radio waves. For example the Sun gives out about 500 million million million MJ of energy every second so its luminosity is 500 million million million MJ.

The luminosity depends on:
(a) the size of the star
(b) the temperature of the star

(a) for a star with a certain surface temperature the bigger the star the more energy it gives out. A star with double the radius of another one will have an area four times as great and so have a luminosity four times greater than the first star

(b) for a star of a certain size the hotter the star the more energy it gives out and so the greater its luminosity. A star with a temperature of double another one will have a luminosity sixteen times greater.


The brightness is how bright a star appears when seen from the Earth. This depends on:
(a) the actual luminosity of the star
(b) the distance of the star from the observer on the Earth

If we have two stars of the same luminosity with one star double the distance of the other from the Earth the closer star will look four times brighter. It obeys the inverse square law.

The photograph shows the Pleiades star cluster. The brighter stars look about the same brightness in fact they are. They are all part of the cluster and about the same distance from the Earth. However some of the background stars may be just as bright they don't look it because they are much further away.

I am very grateful to Marcos Mataratzis and Vivek Hira for the permission to use their photograph.
To see the full collection please visit: Astrophotography

Luminosity theory

The luminosity depends on the surface area of the star. The surface area (A) of a star of radius R is 4pR2. A star with a radius of 2R will therefore have four times the luminosity of a star of radius R if the stars are at the same temperature.

The luminosity of a star also depends on its temperature. If we consider a star to be a perfect black body then the radiation emitted per second (E) is given by the Stefan-Boltzmann law.

Energy emitted per second (E) = sAT4 where sis Stefan's constant (5.7x10-8 Wm- 2K-4) and T is the absolute temperature of the star. (Strictly we should write the difference in the temperature between the star and its surroundings but since the temperature of a star is very large, as least a few thousand K and the temperature of space is very low about 3 K, this slight inaccuracy can be ignored).

We can calculate this energy output for a star of the same size as our Sun. R = 6.96x108 m and T = 6000K.
Therefore E = 5.7x10-8x6.08x1018x60004 = 4.5x1026 J.
A star with a temperature twice that of our Sun would have an energy output sixteen (24) times as great.

Consider a star like Rigel in Orion (on the right in the photo). The surface temperature of Rigel is 11600 K (1.9 times that of the Sun) and its radius is seventy times that of the Sun. The energy output of Rigel per second is therefore 702x1.94 = over 68000 times that of the Sun. In other words an enormous 3.08x1031 J. Altair in Aquila at nearly 17 light years away has an output of just less than ten times that of the Sun.

I am grateful to David Darling for permission to use this photo. To see his full collection visit: Astronomy encyclopedia

Wien's displacement law and the temperature of a star

If we assume that stars are perfectly black bodies then we can use Wien's displacement to calculate their temperature.

The law states that:

Wavelength (lmaxA) X Temperature (TA) = Wavelength (lmaxB) X Temperature (TB)

where lmaxA ,TA and lmaxB ,TB are the peaks of the energy wavelength curve and the absolute temperature of two black bodies, in this case two stars.

Consider two stars A and B whose energy-wavelength curves are shown in the diagram.

Star A has an energy peak nearer the blue end of the spectrum and is at a higher temperature than star B that has an energy peak towards the red end of the spectrum.

Example problems
A black body (the Sun) with a surface temperature of 6000 K emits radiation with lmax = 420 nm. Calculate the temperature of Sirius (the brightest star in the northern skies) if lmax for Sirius is 72 nm.
lmax1T1 = lmax2T2
Therefore 420x10-9x6000 = 72x10-9xT2 Therefore: T2 = 35 000 K

Inverse square law

The brightness of a star seen from the Earth depends its intrinsic luminosity but also on its distance from the Earth. The observed brightness for a given star decreases inversely proportionally to its distance away. The presence of interstellar gas will further decrease the observed brightness.

© Keith Gibbs 2012