Luminosities and Absolute Magnitudes

\[\frac{L_1}{L_2} = 2.512^{(M_2-M_1)}\quad\text{[Eq. (15.5)]}\]

See also: Table 15.1 (below)

Note: If you feel uncomfortable with working with fractions, exponents, variables, or scientific notation, you should review the following tutorials before working through this tutorial:

You may also want to study or review the tutorials Brightnesses and Magnitudes and Parallaxes and Distances.

Sections

Introduction

In the Brightnesses and Magnitudes tutorial, the modernized version of Hipparchus’s magnitude scale was presented. Although not expressly mentioned in that tutorial, the magnitudes that were discussed there are referred to as apparent magnitudes. They are the magnitudes measured for astronomical objects without any concern for how far away the objects are; they are simply the magnitudes apparent in the night sky.

There are several reasons for why a celestial object may appear dimmer than other stars: (a) the object is simply intrinsically dimmer (it doesn’t give off as much light), (b) the object is farther away than other objects it is being compared to (recall the inverse square law of light tutorial), (c) intervening material (dust and gas) lies between the observer and the object being observed, or (d) the object is giving off much of its light in wavelength regions not being seen or detected with instruments. For now we won’t worry about (c) or (d); they are simply mentioned for completeness (and academic honesty ?).

In order to directly compare two objects, such as stars, it is important to compare “apples to apples.” Comparing one star that is 2 pc away with another that is 200 pc away is an unequal comparison if you want to know which star intrinsically produces more light. It is certainly possible that the star 200 pc away produces more light than the one that is 2 pc away even though they have the same apparent magnitude \(m\). The brightnesses in Equations (15.1) and (15.2) are the amount of light energy that reaches the observer or detector every second spread across the same amount of surface area [technically the units are watts per square meter (\(\text{W}/\text{m}^2)\)]. If two stars have the same apparent magnitudes (\(m_2 = m_1)\), then they must also have the same brightnesses (\(b_2 = b_1)\).

In order to “level the playing field” (mixing metaphors) astronomers have chosen to define a standard distance that allows stars to be virtually compared side-by-side. The standard distance is 10 pc, which is nothing special other than it is a nice round number. When a star is moved mathematically to 10 pc from the observer or detector, the apparent magnitude it would have if it really is 10 pc away is referred to as the star’s absolute magnitude \(M\). Now if two stars really do have the same absolute magnitudes (\(M_2 = M_1)\), then they really do produce the same amount of light energy every second. Since the stars are all mathematically 10 pc away, the “per square meter” in the last paragraph no longer matters because it is the same “per square meter” for all of the stars. This means that we can directly compare the amount of light energy per second (in units of watts, \(\text{W}\)). The amount of light energy per second is also known as power output, which astronomers refer to as luminosity \((L)\). Luminosity is the total intrinsic power output of a star or other astronomical object and absolute magnitude is a direct conversion to/from luminosity. Equation (15.5) provides the direct conversion between the two quantities.

If you compare Equation (15.2) with Equation (15.5) you will see that they look exactly the same except that the observed brightness ratio (\(b_1/b_2\)) is replaced by the intrinsic luminosity ratio (\(L_1/L_2\)), and the apparent magnitude difference (\(m_2 – m_1\)) is replaced by the intrinsic absolute magnitude difference (\(M_2 – M_1\)). This means that the two equations, (15.2) and (15.5), behave the same way mathematically, and Table 15.1 can be used for Equation (15.5) if \(m_2-m_1\) is replaced by \(M_2-M_1\) and \(b_1/b_2\) is replaced by \(L_1/L_2\).

Table 15.1

Examples

  1. Star 1 has an absolute magnitude of 2 and star 2 has an absolute magnitude of 5. Which star is more luminous? How many times more luminous?
    The magnitude system works “backwards” relative to what we usually think of. In the magnitude system, the smaller the absolute magnitude, the more luminous the star is (recall Hipparchus’s magnitude 1 versus his magnitude 6). This implies that star 1 with the smaller numerical value of absolute magnitude is the more luminous star. Since the absolute magnitude difference is \(M_2 – M_1 = 5-2 = 3\), according to Equation (15.5), \(\frac{L_1}{L_2} = 2.512^3 = 2.512\times 2.512\times 2.512 = 15.58\). This is also listed in Table 15.1 if \(m_2-m_1\) is replaced by \(M_2-M_1\) and \(b_1/b_2\) is replaced by \(L_1/L_2\).
  2. Star 1 has an absolute magnitude of 8 and star 2 has an absolute magnitude of 13. Which star has the greater luminosity? How many times more luminous?
    Since star 1 has the smaller numerical absolute magnitude, it is the more luminous star (it produces more light energy per second). The difference in absolute magnitudes between the two stars is \(M_2 – M_1 = 13-8 = 5\). By definition of the magnitude system, an magnitude difference of \(5\) corresponds exactly to a luminosity ratio of \(100\). This is also evident from \(2.512^5 = 2.512\times2.512\times2.512\times2.512\times2.512=100\). (Note: If you carry out the calculation you will find that the answer is not exactly \(100\). This is because \(2.512\) is a rounded-off value of the fifth root of \(100\).)
  3. Suppose a star is 5 pc from Earth and its apparent magnitude is \(m = 3\). Is its absolute magnitude numerically greater than 3 or less than 3?
    Since the star is closer to Earth than 10 pc, it would appear dimmer if it could be moved farther away from Earth. Because a dimmer star has a numerically larger magnitude, the apparent magnitude of the star would become numerically greater if it were to be moved out to 10 pc. Given that the absolute magnitude of a star is its apparent magnitude if the star was 10 pc away, the star’s absolute magnitude would be numerically greater than 3 (i.e., \(M > 3\)).
  4. The apparent magnitude of Sirius A is \(m = -1.46\) and its absolute magnitude is \(M = 1.42\). Is Sirius A closer to Earth or farther from Earth than 10 pc?
    Since the apparent magnitude of Sirius A is numerically smaller than its absolute magnitude (a negative number is smaller than a positive number), in order to get a dimmer star at 10 pc, Sirius A would need to be moved farther away. Therefore Sirius A is closer to Earth than 10 pc.
  5. Betelgeuse has an apparent magnitude of \(0.5\) and an absolute magnitude of \(-5.63\). Is Betelgeuse closer to Earth than 10 pc or is it farther away.
    The absolute magnitude of Betelgeuse is very negative, meaning that it is inherently a very luminous star. It’s apparent magnitude is also numerically much greater than its absolute magnitude. At 10 pc Betelgeuse would greatly outshine its apparent brightness in the sky. Betelgeuse must be much farther from Earth than 10 pc (Betelgeuse is actually about 131 pc from Earth).
  6. Star 1 has an absolute magnitude of \(16\) and star 2 has an absolute magnitude of \(0\). Which star is more luminous and by how many times?
    Since star 1 has the larger numerical value of absolute magnitude, it is less luminous by a lot! Since the difference in absolute magnitudes is \(16\), star 2 is more luminous by a factor of \(2.512^{16}\). By studying Table 15.1, you should notice that \(2.512^{16} = 2.512 \times 2.512^{15} = 2.512 \times 10^6\). Star 2 is 2.5 million times more luminous than star 1.

Quiz Yourself

Try the following exercises to make sure that you really do understand the material just presented. By the way, computers and many calculators write scientific notation in “computer-speak,” where \(10\) raised to an integer (either positive or negative) replaces \(\times 10^\text{exponent}\) with \({\rm E(exponent)}\) or \({\rm e(exponent)}\); in other words, \(3.14159 \times 10^9\) is represented by \(3.14159\text{E}9\). If your calculator has the \(\Large{\hat{ }}\) symbol, then you would enter the number as \(3.14159 \times 10{\Large{\hat{ }}}9\). If your calculator has a \(10^x\) button then you would enter the number as \(3.14159\times9\,\,10^x\). If you have having problems with entering numbers in scientific notation, read the manual or help screens, or ask your instructor.

(Answers to the problems are given below)

  1. Star 1 is \(2\) magnitudes more luminous than star 2. The absolute magnitude of star 2 is \(4\). (a) What is the absolute magnitude of star 1? (b) How many times more luminous is star 1 than star 2?
  2. There are two stars in the Procyon system, Procyon A and Procyon B. Their apparent magnitudes are: \(m_A = +0.40\) and \(m_B = +10.7\). (a) Since the two stars are essentially the same distance from Earth, which star appears to be brighter? How many times brighter? (b) Which star produces more light energy every second? How many times more energy per second?
  3. Pollux has an apparent magnitude of \(m = +1.16\) and an absolute magnitude of \(M = +1.09\). (a) Is Pollux closer to \(5\text{pc}\), \(10\text{ pc}\), \(20\text{ pc}\), or \(40\text{ pc}\) from Earth? (b) Is Pollux closer or farther away from Earth than the answer you gave in part (a)?
  4. (a) The Sun’s absolute magnitude is \(+4.74\) and the absolute magnitude of Betelgeuse is \(-5.63\). Which star has a greater luminosity? How many times more luminous? (b) The Sun’s apparent magnitude is \(m = -26.83\) and Betelgeuse’s apparent magnitude is \(m=0.50\). Which star appears brighter in the sky, and by what factor?
Answers
  1. (a) \(2\), (b) \(6.31\)
  2. (a) Procyon A is much brighter than Procyon B. Roughly \(\text{10,000}\) times more, or more precisely, \(2.512^{10.3} = \text{13,200}\) brighter. (b) Procyon A; \(\text{13,200}\) times more.
  3. (a) \(10\text{ pc}\) (b) farther away
  4. (a) Betelgeuse is more luminous by about \(\text{14,000}\). (b) The Sun appears brighter in the sky (obviously!) by a factor of 1.25 billion!
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