Physics › Astrophysics › The apparent magnitude scale
The apparent magnitude scale
Apparent magnitude m ranks how bright a star looks from Earth: smaller numbers are brighter, and one step is a brightness ratio of 2.51. Behind the ranking sit two physical quantities: luminosity, the power a star radiates, and flux, the power per square metre a detector receives, related by F = L/4πd².
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Star brightness and magnitude, part 1 of 2. Part 2 is Stellar distances and absolute magnitude.
Builds on Telescopes beyond the visible and Alpha, beta and gamma radiation.
The maths behind it: Logarithms and their laws on InkMaths.
IN THIS TOPIC
- Use the apparent magnitude scale, including the 2.51 brightness ratio per step.
COMMON MISCONCEPTION
The bigger a star's magnitude number, the brighter it shines.
The magnitude scale runs backwards: smaller numbers mean brighter stars, and the brightest objects have negative magnitudes.
The apparent magnitude 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. On the apparent-magnitude scale, a lower number means a brighter object. 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 eye's response to brightness is approximately logarithmic, so Hipparchus's equally spaced ranks correspond to equal ratios of received intensity rather than equal differences. 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; tying the scale to measured ratios removed that.
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:
the star's whole output spread over a sphere of radius d. Measure F, and the inverse square gives the distance whenever L is already known. 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.
Describe the drawing
A star at the left, with the edge rays of one cone of its light drawn out to the right. At distance r the cone fills a small square patch; at 2r the same cone fills a square of twice the side, divided into four cells each the size of the near patch. The same power crosses both, so the far patch receives a quarter of the flux per unit area.
Three ways to say how far has moved to the next part of this lesson.
Absolute magnitude: the level playing field has moved to the next part of this lesson.
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.
- 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.
- Keep luminosity and flux apart by their units. Luminosity is the star's total output in watts; flux is what arrives, in watts per square metre, diluted by 4πd2. A standard candle answer runs known L, measured F, inverse square law, distance.
CHECK YOURSELF
Star A has apparent magnitude +1.2 and star B has +4.7. State which appears brighter and find the brightness ratio. Then a standard candle of known luminosity 4.0 × 1028 W gives a measured flux of 2.5 × 10−12 W m−2; find its distance.
Show a hint
Direction first, then 2.51 to the power of the difference. The candle is the inverse square law rearranged.
Show the answer
A appears brighter, because 1.2 is the smaller number.
The gap is 3.5 magnitudes, so the ratio is 2.513.5 ≈ 25 times.
F = L/4πd2 gives d = = √(4.0 × 1028 / (4π × 2.5 × 10−12)) = 3.6 × 1019 m, about 1200 parsecs.
Magnitude runs backwards: smaller number, brighter star, 2.51 per step.
Luminosity is the power a star emits; flux is what arrives, diluted by 4πd², and a known luminosity turns a measured flux into a distance.
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.
Or read them with their mark schemes on the apparent magnitude scale questions page.
WHERE TO GO NEXT
- Exponentials and logarithms is the maths this lesson leans on, worked through step by step.
USED LATER IN
CHECK YOUR PROGRESS
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- Use the apparent magnitude scale, including the 2.51 brightness ratio per step.
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