3.9.2.1, 3.9.2.2AstrophysicsCore

Magnitudes and Stellar Distances

A nearby faint star can look brighter than a distant luminous one. The magnitude system separates apparent brightness from a comparison at a standard distance. This lesson is part of the Astrophysics option; study it if this is your centre’s chosen Paper 3 Section B topic.

40 min 3.9 Astrophysics

What you'll be able to do

Interpret the reversed magnitude scale
Use magnitude differences to compare received intensities
Calculate distance with the distance modulus

A logarithmic, reversed scale

Apparent magnitude describes how bright a star appears from Earth. A smaller or more negative magnitude means brighter. On the historical Hipparcos scale the faintest stars visible to an unaided eye in good conditions are around magnitude 6.

A difference of five magnitudes corresponds to a factor of 100 in received intensity; one magnitude corresponds to approximately 2.51. Comparing two sources requires a ratio of intensities, not a subtraction of intensities. A star with magnitude 1 is brighter than one with magnitude 4 by about $2.51^3$ .

\frac{I_1}{I_2}=10^{0.4(m_2-m_1)}

The sign ensures that a smaller magnitude corresponds to a larger intensity.

Absolute magnitude and the parsec

Absolute magnitude is the apparent magnitude a star would have if placed 10 parsecs away, with no intervening absorption. It lets you compare intrinsic optical output. A parsec is the distance at which 1 astronomical unit subtends 1 arcsecond; 1 pc is about $3.09\times10^{16}$ m or 3.26 light years. A light year is a distance, not a time.

The distance modulus assumes that light spreads without significant absorption between star and observer. Put distance in parsecs inside the logarithm. If apparent magnitude is numerically larger than absolute magnitude, the star is farther than 10 pc.

m-M=5\log_{10}\!\left(\frac{d}{10}\right),\qquad d=10\times10^{(m-M)/5}\ \mathrm{pc}

Here the numerical value of d is measured in parsecs.

What the observations can tell you

Received intensity also follows the inverse-square law for an isotropic source in transparent space. Moving a star twice as far away reduces received intensity to one quarter, while its intrinsic luminosity and absolute magnitude stay the same.

Dust makes a star appear dimmer. Ignoring extinction can make a magnitude-based distance too large. Explain this chain: dust raises apparent magnitude, raises the distance modulus, and therefore raises the inferred distance.

Work through the reasoning

Try each problem before opening the worked solution. State the physical model, show substitutions with consistent units, and explain what your answer means.

Example 1A star has

m=8.0

and

M=3.0

. Find its distance.

m-M=5.0

, so

\log_{10}(d/10)=1.0

d/10=10

, hence

d=100

pc.

Answer100 pc, approximately 326 light years.

Example 2Star A has magnitude 2 and star B magnitude 7. Compare their received intensities.

1B has the larger magnitude and is therefore fainter.

I_A/I_B=10^{0.4(7-2)}=10^2

AnswerA delivers 100 times the intensity of B.

Equation recap

\frac{I_1}{I_2}=10^{0.4(m_2-m_1)}

The sign ensures that a smaller magnitude corresponds to a larger intensity.

m-M=5\log_{10}\!\left(\frac{d}{10}\right),\qquad d=10\times10^{(m-M)/5}\ \mathrm{pc}

Here the numerical value of d is measured in parsecs.

Common mistakes to avoid

Putting metres into the distance-modulus formula.

Use the numerical distance in parsecs.

Calling apparent magnitude a direct measure of luminosity.

It depends on distance and intervening absorption as well as source output.

Key takeaways

Five magnitudes correspond to a factor of 100 in intensity.
Absolute magnitude compares stars at 10 pc.
The distance modulus needs parsecs and an extinction assumption.

Test yourself

Answer a few questions, then review anything you missed.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

Browse practice topics

Related lessons

All 8 lessons