AstronomyMeasuring the Stars

The Magnitude Scale

Smaller magnitudes are brighter, and 5 magnitudes means 100 times the brightness.

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Astronomers still use a brightness scale invented over 2000 years ago. The Greek astronomer Hipparchus sorted the stars into six classes: the brightest were "first magnitude," the faintest he could see "sixth magnitude." So on this scale, smaller numbers mean brighter stars. It measures apparent brightness in a way that matches how our eyes judge it: by ratios.

Why it works

Our eyes respond to brightness by ratios, not differences: each step down the old scale looked about equally dimmer, and it turned out each step was the same factor. In 1856 Norman Pogson made this exact: $5$ magnitudes is a factor of exactly $100$. So one magnitude is a factor of $\sqrt[5]{100} = 2.512$, and a difference of $\Delta m$ magnitudes means a brightness ratio of $$\frac{b_1}{b_2} = 100^{\Delta m/5} = 2.512^{\Delta m}.$$ This is an exponential function: adding magnitudes multiplies brightness. Going the other way, $\Delta m = 2.5 \log_{10}\frac{b_1}{b_2}$. The scale extends past zero for very bright objects: Sirius is $-1.46$, Venus up to $-4.6$, the full Moon $-12.7$ and the Sun $-26.7$.

Key idea

Smaller (and negative) magnitudes are brighter. A difference of $5$ magnitudes is a factor of $100$ in brightness; $1$ magnitude is a factor of $2.512$.
Apparent magnitudes from the Sun ($-26.7$) to the faintest stars seen in a dark sky (about $6.5$). Brighter is up.

Worked examples

Worked example: First against sixth

How much brighter is a first-magnitude star than a sixth-magnitude star?
  1. $\Delta m = 6 - 1 = 5$.
  2. Five magnitudes is a factor of exactly $100$.

Answer: $100$ times brighter.

Worked example: Sirius and Polaris

Sirius is magnitude $-1.46$ and Polaris magnitude $1.98$. How many times brighter does Sirius look?
  1. $\Delta m = 1.98 - (-1.46) = 3.44$.
  2. Ratio $= 2.512^{3.44} = 100^{3.44/5} = 10^{1.376} = 23.8$.
  3. Sirius looks about $24$ times brighter.

Answer: About $24$ times.

Worked example: From a ratio to magnitudes

A telescope can see stars $10{,}000$ times fainter than the eye can. How many magnitudes deeper does it reach?
  1. $\Delta m = 2.5\log_{10}(10{,}000) = 2.5 \times 4 = 10$.
  2. If your eye reaches magnitude $6$, the telescope reaches magnitude $16$.

Answer: $10$ magnitudes.

Common mistake

On most scales, bigger numbers mean more. On this one a magnitude-6 star is fainter than a magnitude-1 star, by a factor of $100$. Check with the brightest things in the sky: the Sun is $-26.7$, the most negative magnitude of all.
Apparent magnitude depends on distance. To compare stars fairly, astronomers also use absolute magnitude: how bright a star would look from a standard distance of $10$ parsecs.

Practice

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Sources and further reading

  • OpenStax, Astronomy 2e, section 17.1, The Brightness of Stars
  • MIT OpenCourseWare 8.282, Introduction to Astronomy: magnitudes
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