Stars — Nights of the Universe

A Luminous Blue Variable star named HD168625, located in our Milky Way Galaxy, is surrounded by a bipolar nebula that is similar to the one around SN1987A, a supernova that exploded in 1987 in the Large Magellanic Cloud and was the nearest supernova to Earth in 400 years. Image Left — NASA, JPL-Caltech, Nathan Smith/UC Berkeley

Image of HD 168625 and HD 168607 at right is from the Aladin Previewer website. In 2003, survey evidence showed that this pair of rare luminous late type B (early type A) hypergiants may indeed be a "physical pair", a determination made via spectroscopic monitoring. — E. L. Chentsov and E. S. Gorda Luminosity class, A Star's Temperature (T) & Radius (R)

Finishing our section on spectra, we have what is called a star's Luminosity Class and it is expressed using the Roman numerals I to V, as in our example from the previous page, HD 66171 which was a G2 V class star. This is the energy output by the star every second, and the properties used to determine this are:

E≈ σ; T_eff⁴
E is approximately equal to σ and T_eff is raised to the 4th power where:
T_eff is the effective temperature in Kelvins
E is the energy per unit surface area in erg/cm2 and
σ is the Stefan-Boltzmann constant, equal to 1.82849716 x 10^-23 MDays^-3 K^-4.

The effective temperature is the best measure of the actual temperature of the gases in a star's outer layers and this is equal to the temperature of a black body having the same radius and radiating the same amount of energy as the star. Our first task is to look at a star's temperature, remembering from our section II information that the hotter the star the greater the amount of radiation emitted or power output per unit of surface area. An analogy is the household light bulb and dimmer switch. At it's lowest setting the light bulb is dim and not very hot but as we rotate the dimmer switch the bulb begins to brighten, getting hotter, and this results in a greater amount of energy being emitted from it's surface area. The same principle may be applied to stars with the caveat that since the range of stellar luminosities is great our calculations need use of a constant that approximates the relationship between power and temperature. The formula used is:

l ≈ σ T⁴
l is approximately equal to σ T raised to the 4th power

Our next step in calculating luminosity is to determine the star's surface area or radius since this relation is such that as both the unit of area and surface temperature increase so too does the emitted radiation or power output per unit of area which is what we are now calculating. The formula used for determining surface area is:

4π R²
where R is equal to the Radius of the star
(4 pi R squared)

Our final step at this point is to combine the two above formulas in order to calculate the star's total luminosity:

L ≈ 4π R² σ T⁴

So now we have all we need in hand to calculate a star's luminosity class - except! Looking at the above calculation begs the question, what is the star's radius? All the above works great if we have a star's radius to begin with but as only some hundreds of star's have had a direct measurement made of their radii the above seems to be rather acemdemic. Not so, as we are able to measure the luminosity of a star via other means - spectroscopic comparison - and through this we can actually use the above equation to determine the radius of the star, thus having in hand the information required to make our calculations (also see: Spectroscopy of Variable Stars for more on calculating a star's absolute luminosity and density via this method).

Apparent Magnitude

Before ending this section, let us quickly look at Apparent Magnitude and how we calculate the difference between two stars. Remember, the higher the magnitude number the dimmer the star. So, let's say for example that Φ is a magnitude 11 star and Ω a magnitude 6 star. How many times is one star fainter than the other? Well, the relative magnitude formula is expressed as:

b_Φ = 2.512 ( M_Ω - M_Φ ) b_Ω
Where b = the apparent brightness of star Φ and star Ω
M = the apparent magnitude of star Φ and star Ω and
2.512 is raised to ( M _Ω - M_Φ ) x b_Ω

By way of example, let's say star Ω appears five magnitudes fainter than Star Φ. Keeping in mind that the scale is logarithmic, such that a difference of 5-magnitude translates into a factor of 100x, our solution would be as follows:

b_Φ / b_Ω = 2.512^{(M_Ω - M_Φ)} → b_Φ / b_Ω = 2.512⁽⁵⁾
or 100 times fainter.

Keep in mind that the information presented in this section on stars is not definitive, there is still much more involved in this aspect of stellar science. For those wishing a more thorough study and understanding see the following websites:

1. Australia Telescope Outreach and Education
   The Colors of Stars
2. Chapter 9: The Life of a Star
   AAVSO Variable Star Astronomy - Chapter 9
3. Luminosity Class and the HR diagram
   Michael Richmond's classes

The above image of star HD 66171 is from the Palomar Observatory Sky Survey (POSSII) and was obtained via the Aladin previewer utility.