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Star brightness and magnitude

Apparent magnitude ranks how bright a star looks on a scale inherited from ancient astronomy, in which brighter means a smaller number. Add the parsec, a distance defined from the Earth's orbit, and one logarithmic equation converts between how bright a star looks and how bright it actually is.

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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, and convert between them.
  • Use p = 1/d with p in seconds of arc and d in parsecs, and say why the units are part of the equation.
  • Say what absolute magnitude means without reaching for the equation.
  • Convert between apparent and absolute magnitude with m − M = 5 log(d/10).

COMMON MISCONCEPTION

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, and the dimmest a naked eye can catch, sixth. The scale stuck. Modern astronomy is therefore saddled with a system in which brighter means a smaller number. Once telescopes and detectors made brightness measurable the ranks were pinned down, and apparent magnitude m became the measure of how bright an object looks from Earth. Objects brighter than first magnitude simply continue through zero into negative numbers.

The apparent magnitude scale runs backwards: brighter means smaller numbers, and each step of one magnitude divides the received light by 2.51Sun −26.7full Moon −12.7Venus −4.5Sirius −1.5eye limit +6.0brighter this wayapparent magnitude: the backwards scaleeach step of 1.0 divides the received light by 2.51
FIG. 1The scale in use: the Sun at −26.7, the full Moon at −12.7, Venus at −4.5, Sirius, the brightest star in the night sky, at −1.5, down to the naked-eye limit near +6.

The eye judges brightness on a squashed, roughly logarithmic scale, so Hipparchus's equal-feeling steps are really equal ratios. 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, the fifth root of 100. Brightness here means the intensity arriving at your detector. The unaided eye's judgement is subjective, and tying the scale to measured ratios is what removed the subjectivity.

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, so 2.513 = 16 times brighter (15.8 before rounding).

One check catches most sign errors. P has the smaller number, so P must be the brighter. Say that first, then multiply.

Luminosity and radiant flux

Underneath the magnitude language sit two physical quantities worth separating. A star's luminosity L is the total power it radiates, in watts. What reaches a detector a distance d away is the received flux, the power per square metre, strictly a flux density or irradiance; some boards' papers name it radiant flux intensity. Geometry alone dilutes it:

F=L4πd2F = \frac{L}{4\pi d^{2}}NOT ON THE AQA DATA SHEET: LEARN IT

the star's whole output spread over a sphere of radius d. Measure F, and the inverse square gives the distance the moment anything tells you L. Objects whose luminosity is known independently are called standard candles, and the usual examples are certain pulsating stars and, further out, type Ia supernovae. Spot one in a distant galaxy, compare its known L with its measured F, and the galaxy's distance follows. CIE frames the whole distance ladder in exactly these terms.

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.

The parsec defined: when the radius of Earth's orbit subtends an angle of one arcsecond at a star, the star is one parsec away1 AUEarthSunstard = 1 parsec = 3.26 light yearsone arcsecond of parallax across the Earth to Sun distancedefines the parsec; the angle is drawn enormously exaggerated
FIG. 2The definition at a glance: when one astronomical unit, the radius of the Earth's orbit, subtends an angle of one arcsecond at the star, the star's distance is one parsec.

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. The unit justifies itself twice over. Real parallax measurements arrive in arcseconds, and the magnitude equation below is built around the parsec.

The parallax angle p of a star is the angle that one AU subtends at it, which is half the total sideways shift the star makes over six months. Because the parsec was defined to make that triangle come out at one, the distance follows with no constant to carry. OCR sets the relation as an equation, and no other board does:

p=1dp = \frac{1}{d}

p in seconds of arc, d in parsecs, and those units are not a footnote to the equation, they are the equation. Written in radians and metres the same physics needs the awkward constant that the parsec was invented to absorb, so p = 1/d is simply false in any other pair of units. Quote them every time you quote it.

Put numbers through it and the scale of the sky comes out. A star with a parallax of 0.10 arcseconds lies at 1/0.10 = 10 pc; halve the parallax to 0.050 arcseconds and the star is twice as far, at 20 pc. Proxima Centauri, the nearest star, holds the record at 0.77 arcseconds, which puts it at 1.3 pc, or 4.2 light years. Nothing lies closer, so every stellar parallax ever measured is below one arcsecond, and they shrink fast: the method runs out of precision within a few thousand parsecs, and the standard candles of the next lessons take the distance ladder on from there.

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, a like-for-like measure of the star's own output. The two magnitudes are linked by the star's actual distance d, in parsecs:

mM=5logd10m − M = 5\,\text{log}\frac{d}{10}ON THE AQA DATA SHEET

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 gives d.

WORKED EXAMPLE

Sirius, ranked by absolute magnitude

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.

Now 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.

GUIDED PRACTICE

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.

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. That is comfortably visible, among the brighter stars in the sky. Distance was hiding a respectable star.

INDEPENDENT PRACTICE

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 that against the sky. Polaris is indeed a middling second-magnitude star. Its absolute magnitude of −3.6 says it is enormously luminous, and 133 parsecs of distance tames it.

ASSESSMENT FOCUS

  • 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, and 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. It is the distance at which one astronomical unit subtends one arcsecond. Have it word-perfect, with 1 pc = 3.26 ly.
  • OCR only: p = 1/d earns its second mark from the units, so write them down. p in seconds of arc, d in parsecs. Converting the angle to radians or the distance to metres first is the classic way to throw the question away, because the equation is only true in the units the parsec was built for.
  • In m − M = 5 log(d/10), d must be in parsecs. Feeding in light years is the standard trap, and it costs the whole calculation.
  • Watch the direction of the subtraction too. Writing M − m instead of m − M flips the sign of the log and moves the star to the wrong side of 10 pc, which is worth doing a sanity check on before you write the final line.
  • Absolute magnitude questions often ask what M means, which is the apparent magnitude the star would have at 10 pc. That 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, exactly as you should expect for a star nearer than 10 pc.

Sirius appears brighter, because −1.5 is smaller than +2.0.

The gap is 3.5 magnitudes, so the ratio is 2.513.525 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 gives 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.

20 questions on this topicAnswer them one at a time and mark yourself against the mark scheme.Practise this topic

Or read them with their mark schemes on the star brightness and magnitude questions page.

13 flashcards on this topicDefinitions, off-sheet equations and a spot-the-error card, scheduled by spaced repetition in your browser.Revise with flashcards

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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, and convert between them.
  • Use p = 1/d with p in seconds of arc and d in parsecs, and say why the units are part of the equation.
  • Say what absolute magnitude means without reaching for the equation.
  • 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.