Brightnesses and Magnitudes

\[\text{if }\ \frac{b_1}{b_2} = 100\ \text{ then }\ m_2-m_1 = 5\quad\text{[Eq. (15.1)]}\]

and

\[\frac{b_1}{b_2} = 2.512^{(m_2-m_1)}\quad\text{[Eq. (15.2)]}\]

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:

Sections

Introduction

The magnitude scale was created by Hipparchus, one of the great, ancient Greek philosopher-scientists as a way to record the relative brightnesses of stars before any knowledge of distances or technological means to measurement. Hipparchus simply assigned 1 to the very brightest stars visible to the naked eye and 6 to those stars that were very dim and just barely discernable. The difference between the dimmest classification in Hipparchus’s system and the brightest is \(6-1=5\). At the time, Hipparchus also didn’t understand that the human eye and brain don’t perceive light linearly, but rather approximately logarithmically. This means that if we see with the naked eye that one star appears to be twice as another star, in reality, our eye receives more than twice as much light energy every second from the brighter star; in fact, the amount of energy received every second is exponentially greater. (Mathematically, logarithms and exponential functions are inverse functions of one another; they undo each other’s operations on numbers and variables.)

The modern magnitude scale provides a precise definition of the relationship between actual brightness ratios (brightness being the true amount of energy received every second for a specific amount of surface area at our eye, a photographic plate, or an electronic detector), and our perception of differences in brightnesses as magnitudes. What scientists are really interested in is the actual brightness of a star, rather than a value perceived by the human eye and brain. This is because actual brightness relates directly to a star’s physical characteristics (in other words, what the star is really like). The formal definition of that relationship is that a magnitude difference of \(5\) corresponds exactly to a brightness ratio of \(100\). The number in Equation (15.2), \(2.512\) is the fifth root of \(100\), so that \(2.512^5= 100\) as required by Equation (15.1); see Table 15.1 for example.

The reason that astronomy still uses magnitudes is largely historical; the magnitude system has been around for over 2000 years. That’s a lot of tradition and recorded data before the development of a more modern system of measurement. Magnitudes also have the advantage of keeping recorded values in a convenient range of roughly \(-30\) to \(+30\) rather than very large numbers that require scientific notion to display as suggested in Table 15.1. Equation (15.1) defines how brightness ratios are related to magnitude differences and Equation (15.2) specifies how to convert any magnitude difference to its equivalent brightness ratio. (Point of trivia: The full definition of the magnitude scale requires a specification of when \(m = 0\) for a particular star and its brightness. That defining star is Vega, or more technically, the averages of a group of stars. that are very close to Vega’s values.)

Table 15.1

Examples

  1. Star 1 has a magnitude of 2 and star 2 has a magnitude of 5. Which star is brighter? How many times brighter?
    The magnitude system works “backwards” relative to what we usually think of. In the magnitude system, the smaller the magnitude, the brighter the star is (recall Hipparchus’s magnitude 1 versus his magnitude 6). This implies that star 1 with the smaller numerical value of magnitude is the brighter star. Since the magnitude difference is \(m_2 – m_1 = 5-2 = 3\), according to Equation (15.2), \(\frac{b_1}{b_2} = 2.512^3 = 2.512\times 2.512\times 2.512 = 15.58\). This is also listed in Table 15.1.
  2. Star 1 has a magnitude of 8 and star 2 has a magnitude of 13. Which star is brighter? How many times brighter?
    Since star 1 has the smaller numerical magnitude, it is the brighter star. The difference in magnitudes between the two stars is \(m_2 – m_1 = 13-8 = 5\). By definition of the magnitude system, a magnitude difference of \(5\) corresponds exactly to a brightness 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. Star 1 has a magnitude of \(16\) and star 2 has a magnitude of \(0\). Which star is brighter and by how many times?
    Since star 1 has the larger numerical value of magnitude, it is the dimmer star by a lot! Since the difference in magnitudes is \(16\), star 2 is brighter by a factor of \(2.512^{16}\). By studying Table 15.1, you should notice that \(2.512^{16} = 2.512 \times 2.512^{15} = 2.512 \times 10^6\).
  4. The difference in magnitudes of the Sun and Sirius, the brightest star in the sky other than the Sun),is roughly \(28\). What is the ratio of brightnesses?
    From Equation (15.2), \(\frac{b_\text{Sun}}{b_\text{Sirius}} = 2.512^{28}.\) From the laws of exponents, \(x^{(a+b)} = x^ax^b\). This means that \(2.512^{28} = 2.512^{25+3} = 2.512^{25}2.512^3 .\) Again from the laws of exponents, \(x^{a\times b} = \left(x^a\right)^b\). Since \(2.512^{25} = 2.512^{5\times5} = \left(2.512^5\right)^5\) and \(2.512^5 = 100\), \(2.512^{25} = 100^5 = \text{10,000,000,000} = 1 \times 10^{10}.\) Also, \(2.512^3=15.85\). Finally \(2.512^{28} = \left(1\times10^{10}\right)\times15.85 = 1.585\times10^{11}\), or the Sun is 158 billion times brighter than Sirius. (Of course you can do this more quickly on a calculator using the \(x^y\) button, but it’s not as challenging or fun!)

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 brighter than star 2. The magnitude of star 2 is \(4\). (a) What is the magnitude of star 1? (b) How many times brighter is star 1 than star 2?
  2. The brightest star in the sky other than the Sun, Sirius, is actually one member of a binary star system and is formally known as Sirius A. Its companion is Sirius B. The magnitude of Sirius A is \(-1.46\) and the magnitude of Sirius B is only \(+8.44\). (a) What is the difference in magnitudes of the two stars? (b) Rounding off your answer to part (a) to the nearest whole number, use that result to estimate the brightness ratio of the two stars.
  3. Star 1 is \(631\) times brighter than star 2. (a) What is the difference in magnitudes of stars 1 and 2? (b) If star 2 has a magnitude of \(12\), what is the magnitude of star 1?
  4. Star 1 has a magnitude of \(15\) and star 2 has a magnitude of \(2\). Which star is brighter and by what factor?
Answers
  1. (a) The brighter the star, the smaller the magnitude which means that star 1 has a magnitude of \(4-2 = 2\). (b) \(\frac{b_1}{b_2} = 2.512^2 = 2.512\times2.512 = 6.31\).
  2. (a) \(8.44 – (-1.46) = 8.44 + 1.46 = 9.90\). (b) Rounding off to \(10\), the brightness ratio is \(2.512^{10} = 2.512^5 \times 2.512^5 = 100 \times 100 = \text{10,000}\). Sirius A is approximately ten thousand times brighter than Sirius B!
  3. If you use Table 15.1 in reverse, notice that \(b_1/b_2 = 631\) corresponds to a magnitude difference of \(m_2-m_1=7\); \(2.512^7 = 631\). Since star 1 is brighter than star 2, star 1 must have the smaller numerical value for its magnitude. Therefore, the magnitude of star 1 is \(12-7 = 5\). Star 2 is not visible to the naked eye, but star 1 is if the sky is dark enough.
  4. The magnitude of star 1 is numerically larger than that of star 2, and so is dimmer than star 2. The difference in magnitudes is \(15-2 = 13\). This means that star 2 is brighter than star 1 by a factor of \(2.512^{13}\). This can be calculated directly using your calculator’s \(x^y\) key (or equivalent) or you can use laws of exponents: \(2.512^{13} = 2.512^{10 + 3} = 2.512^{10}2.512^3 = 2.512^3\times2.512^5\times2.512^5 = 2.512^3\times100\times100 = 15.8\times10^4 = 1.58\times10^5\) or \(\text{158,000}\) times brighter.
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