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Star brightness and magnitude
Astronomy still ranks stars on a scale invented two thousand years ago, and it runs backwards: the brighter the star, the smaller the number. Add a definition of distance built from the Earth's orbit and one tidy logarithm connects how bright a star looks to how bright it really is.
Builds on Telescopes across the spectrum.
IN THIS TOPIC
- Use the apparent magnitude scale, including the 2.51 brightness ratio per step.
- Define the parsec, the light year and the astronomical unit.
- Convert between apparent and absolute magnitude with m − M = 5 log(d/10).
WHAT YOU PROBABLY THINK
The bigger a star's magnitude number, the brighter it shines.
The backwards scale
Around 130 BC the Greek astronomer Hipparchus sorted the visible stars into six ranks: the brightest he called first magnitude, the dimmest a naked eye can catch, sixth. The scale stuck, which is why modern astronomy is saddled with a system where brighter means a smaller number. Once telescopes and detectors made brightness measurable, the ranks were pinned down: apparent magnitude m measures how bright an object looks from Earth, and objects brighter than first magnitude simply continue through zero into negative numbers.
The eye judges brightness on a squashed, roughly logarithmic scale, so Hipparchus's equal-feeling steps are really equal ratios, and the modern definition makes that exact: five magnitude steps correspond to a brightness ratio of exactly 100, so one magnitude step is a ratio of 2.51, since 2.51 multiplied by itself five times is 100. Brightness here means the intensity arriving at your detector, and the judgement of the unaided eye is subjective, which is exactly why the scale had to be tied to measured ratios.
WORKED EXAMPLE
Three steps apart
Star P has apparent magnitude +1.0 and star Q has +4.0. How many times brighter does P appear?
Each step is a factor of 2.51, and there are three of them: 2.513 = 16 times brighter (15.8 before rounding).
The check that catches sign errors: P has the smaller number, so P must be the brighter. Say that first, then multiply.
Three ways to say how far
Astronomy keeps three units of distance, each built for its own scale. The astronomical unit (AU) is the mean Earth-Sun distance, 1.50 × 1011 m. The light year (ly) is the distance light travels in a year, 9.46 × 1015 m. The third is subtler and matters most for this unit: watch a nearby star from opposite ends of the Earth's orbit and it appears to shift slightly against the distant background, an effect called parallax. The smaller the shift, the further the star.
One parsec (pc) is the distance at which one AU subtends an angle of one arcsecond, a 3600th of a degree. The name is the definition compressed: parallax of one arcsecond. Work the triangle through and 1 pc = 3.08 × 1016 m, which is 3.26 light years. It earns its keep because real parallax measurements arrive in arcseconds, and because the magnitude equation below is built around it.
Absolute magnitude: the level playing field
Apparent magnitude mixes up two different things: how much light a star pours out, and how far away it happens to sit. A feeble nearby star can outshine a giant across the galaxy. To compare stars fairly, imagine picking every one up and placing it at the same distance, 10 parsecs. The apparent magnitude a star would have at 10 pc is its absolute magnitude M, and it is an honest measure of the star's own output. The two magnitudes are linked by the star's actual distance d, in parsecs:
The logic sits in the signs. A star nearer than 10 pc has d/10 below 1, the log goes negative, and m comes out smaller than M: moving a star closer than its standard shelf makes it look brighter. Further than 10 pc, the log is positive and the star looks dimmer than its absolute magnitude. The difference m − M is called the distance modulus, because knowing it hands you d.
WORKED EXAMPLE
Sirius, honestly ranked
Sirius has apparent magnitude −1.5 and sits 2.64 pc away. Find its absolute magnitude.
M = m − 5 log(d/10) = −1.5 − 5 log(0.264).
log(0.264) = −0.578, so M = −1.5 + 2.89 = +1.4.
Read the story in the numbers: at its true distance Sirius dazzles at −1.5, but parked at 10 pc it would be an ordinary-looking star of +1.4. It dominates our sky by being close, and only mildly by being bright.
YOUR TURN
A star at a round distance
A star of apparent magnitude +6.0, right at the naked-eye limit, lies 100 pc away. Find its absolute magnitude, and state whether it would be naked-eye visible from 10 pc, before opening the working.
Show the working
M = 6.0 − 5 log(100/10) = 6.0 − 5 log(10) = 6.0 − 5 = +1.0.
At 10 pc its apparent magnitude would equal its absolute magnitude, +1.0: comfortably visible, among the brighter stars in the sky. Distance was hiding a respectable star.
TRY IT UNSEEN
Polaris, the other way round
Polaris has absolute magnitude −3.6 and lies about 133 pc away. Predict its apparent magnitude.
Show the working
m = M + 5 log(d/10) = −3.6 + 5 log(13.3) = −3.6 + 5 × 1.12 = +2.0.
Check against the sky: Polaris is indeed a middling second-magnitude star. Its absolute magnitude of −3.6 says it is enormously luminous; 133 parsecs of distance tames it.
THE EXAM BIT
- State the direction of the scale before anything else: smaller and negative numbers are brighter, and the naked-eye limit is about +6. Most magnitude errors are direction errors.
- One step is a ratio of 2.51 in received intensity; five steps are exactly 100. For a ratio between two stars, raise 2.51 to the power of the magnitude difference.
- The parsec definition is one sentence: the distance at which one astronomical unit subtends one arcsecond. Have it word-perfect, with 1 pc = 3.26 ly.
- In m − M = 5 log(d/10), d must be in parsecs. Feeding in light years is the standard trap.
- Absolute magnitude questions often ask what M means: the apparent magnitude the star would have at 10 pc. The definition mark comes before any algebra.
CHECK YOURSELF
Vega has apparent magnitude 0.0 at a distance of 7.68 pc. Find its absolute magnitude. Then state which appears brighter from Earth, Sirius at m = −1.5 or Polaris at m = +2.0, and by what brightness factor.
Show a hint
Vega is nearer than the 10 pc shelf, so decide first whether M should be bigger or smaller than m.
Show the answer
M = 0.0 − 5 log(0.768) = 0.0 + 0.57 = +0.6: moved out to 10 pc, Vega would look slightly dimmer than it does now, as expected for a star nearer than 10 pc.
Sirius appears brighter: −1.5 is smaller than +2.0.
The gap is 3.5 magnitudes, so the ratio is 2.513.5 ≈ 25 times.
Magnitude runs backwards: smaller number, brighter star, 2.51 per step.
Absolute magnitude is the view from 10 parsecs; the distance modulus m − M hands you d.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
CHECK YOUR PROGRESS
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- Use the apparent magnitude scale, including the 2.51 brightness ratio per step.
- Define the parsec, the light year and the astronomical unit.
- Convert between apparent and absolute magnitude with m − M = 5 log(d/10).
Open the full revision checklist to track your progress across the whole unit.
No animated video for this topic yet; these notes stand alone. InkPhysics on YouTube.