Luminosity of Stars
- What Determines a Star's Luminosity?
- Comparing Luminosities and Brightness
The absolute magnitude of a star is simply a simple way of describing its luminosity. Luminosity, L, is a measure of the total amount of energy radiated by a star or other celestial object per second. This is therefore the power output of a star. A star's power output across all wavelengths is called its bolometric luminosity. Astronomers in practice also measure an object's luminosity in specific wavebands so that we can discuss an object's X-ray or visible luminosities for example. This is also used to measure a star's colour as described on the next page.
Our Sun has a luminosity of 3.84 × 1026 W or J.s-1 which can be denoted by the symbol Lsol (actually the subscript symbol is normally a dot inside a circle - the standard astrological symbol for the Sun but this cannot be shown in html). Rather than always use this exact value it is often more convenient to compare another star's luminosity L* to the Sun's as a fraction or multiple. Thus if a star is twice is luminous as the Sun, L*/Lsol = 2. This approach is convenient as the luminosity of stars varies over a huge range from less than 10-4 to about 106 times that of the Sun so an order of magnitude ratio is often sufficient.
As we have seen in the section on spectroscopy, we can approximate the behaviour of stars as black body radiators.
1. 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, Teff. So; the power output, l ∝ T4 or l = σT4 for a perfect black where σ is a constant called the Stefan-Boltzmann constant. It has value of σ = 5.67 × 10-8 W m-2 K-4 in SI units. 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.
2. 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 4.3 so the star with greater surface area must be intrinsically more luminous than the smaller one. This becomes apparent when we plot stars on an HR diagram.
Assuming stars are spherical then surface area is given by:
where R is the radius of the star.
To calculate the total luminosity of a star we can combine equations 4.4 and 4.5 to give:
Using equation 4.6 all we need in order to calculate the intrinsic luminosity of a star is its effective temperature and its radius. 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 4.6 to determine the radius of the star.
Let us imagine we have two stars, A and B that we wish to compare. If we can measure their respective apparent magnitudes, mA and mB how will they differ in brightness? The ratio of the brightnesses (or intensities) IA/IB corresponds to their difference in magnitude, mB - mA . Remember, as a difference of one magnitude means a brightness ratio of the fifth root of 100 or 1001/5, a difference of mB - mA magnitudes gives a ratio of (1001/5)mB - mA ∴
Note this equation is specified in the NSW HSC Physics Formula Sheet. If you are mathematically astute you should realise that this is in fact the same as equation 4.1 from the previous page, ie IA/IB = 2.512mB - mA.
On the previous page we used the distance modulus equation (4.2). How is this equation derived? It is simply an application of the luminosity ratio relationship (4.7).
The inverse-square law of light means that the flux, l (or intensity) of a star at a distance d can be related to its luminosity L at a distance D by the following relationship:
At distance of 10 parsecs, D is represented by absolute magnitude, M and the flux at distance d is represented by the apparent magnitude, m then the luminosity ratio is given by:
Example 1: Comparing brightness of two stars given apparent magnitudes.
α Car (Canopus) has an apparent magnitude of -0.62 whilst the nearby star Wolf 359 has an apparent magnitude of 13.44.
a) Which star appears brightest in the sky?
b) How many times brighter is it than the other star?
a) The answer to this part is really just checking your understanding of the concept of apparent magnitude. As Canopus has a lower value (-0.62) than Wolf 359 (+13.44) it appears brighter in the night sky. In fact Canopus is the second brightest star visible in the night sky after Sirius A whereas with an apparent magnitude of 13.44 Wolf 359 is far too faint to be visible to the naked or unaided eye.
b) How much brighter is Canopus than Wolf 359? For this we can use equation 4.7:
substituting in we get:
Example 2: Calculating brightness range for a variable star.
δ Cephei is a pulsating variable star that changes its apparent magnitude from 3.5 to 4.4 with a period of 5.366 days. It was the first such star discovered and has given its name to a class of variable stars. The importance of these is discussed in a later section. How much brighter is δ Cephei when at maximum brightness than at minimum?
Again, let us start with equation 4.6:
In this type of problem we simply substitute in the two values for the apparent magnitudes for the same star so;
Example 3: Comparing two stars' luminosities.
How much more luminous is Betelgeuse than our Sun?
The Sun has an absolute visual magnitude MS = 4.8 and Betelgeuse has an absolute magnitude MB = -5.14 so we can rewrite equation 4.7 to give us:
so substituting in: