21 Apparent and Absolute Magnitude

Warning: Math Ahead. Don’t Panic.

Our goal is to find the luminosity of and distance to a star. We know we can’t find both, or even either, of these properties by observing only the brightness of a star. We also know that stars come in a wide range of luminosity and vary greatly in their distances from the Sun. What appears to be a bright star might be very close (Rigil Kentaurus A) or intrinsically very luminous (Rigel) or a little bit of both (Sirius).

In the magnitude system, we approach this problem by talking about apparent magnitude and absolute magnitude. The apparent in apparent magnitude is “what you see”, what is apparent. Apparent magnitude is what we measure with our telescope/camera.

The absolute in absolute magnitude implies something that is fixed and unchanging. If we could board an interstellar spaceship and travel through the stars, their apparent magnitudes would change – getting brighter as they got closer and dimmer as they got farther away. Their absolute magnitudes would not. If we don’t know how far away a star is, then we can’t figure out its absolute magnitude, we can only measure its apparent magnitude. But if we can figure out how far away a star is, then we can calculate its absolute magnitude.

The basic principle behind this is the inverse square law of light.  The inverse in the inverse square law means that as a star gets further away, it gets dimmer. Increasing distance means decreasing brightness. The square in the inverse square law means that this drop off happens as the square of the distance. If you double the distance, brightness doesn’t just halve, it quarters. (If you halve the distance, brightness doesn’t double, it quadruples.)

The magnitude system works well with the inverse square law for light. Imagine two stars with equal luminosity (equal absolute magnitudes), one of which is twice as far away. The further star…

• Is dimmer
• Has a higher apparent magnitude
• Has a brightness that is a quarter that of the closer star
• Has an apparent magnitude that is about 1.5 larger than the closer star

If the closer star has an apparent magnitude of 10, the further star has an apparent magnitude of about 11.5.  If the closer star has an apparent magnitude of 12, the further star has an apparent magnitude of about 13.5.

The magnitude system is set up to work perfectly when you multiply the distance by 10. Ten squared (10*10) is 100. So a star that is 10 times further away is 100 times fainter. If a star is 100 times fainter, its apparent magnitude increases by exactly 5.

For example if star A is 10 parsecs (a measure of distance) away and has an apparent magnitude of 12, then an equally luminous star 100 parsecs away would have an apparent magnitude of 17, and an equally luminous star 1000 parsecs away would have an apparent magnitude of 22. Every time the distance increases by a factor of 10 (is multiplied by 10), the apparent magnitude increases by 5.

Comparing apparent magnitudes can be useful, but we also want a measurement of the intrinsic luminosity of an object, that is, its absolute magnitude. Absolute magnitude is defined as the (apparent) magnitude that a star would have if it were exactly 10 parsecs away. In the example above, the absolute magnitude of the three identical stars would be 12 (the apparent magnitude of the star that was 10 parsecs away).

The equation relating apparent magnitude, absolute magnitude, and the distance in parsecs is

[latex]m-M=5\log\left(\dfrac{d}{10 pc}\right)[/latex]

In this equation, [latex]m[/latex] is the apparent magnitude, [latex]M[/latex] is the absolute magnitude, and [latex]d[/latex] is the distance in parsecs.

The logarithm tells us how many powers of 10 greater (or less) than 10 pc the distance to the star is. If the star is 100 parsecs away, then 100 pc/10 pc is 10 and [latex]\log\left(10\right)=1[/latex] (100 pc is one power of ten greater than 10 pc).  Similarly log₁₀ of (10 pc/10 pc) is 0 and log₁₀ of (1000 pc/10 pc) is 2.  The factor of 5 in front of the logarithm represent the apparent magnitude increasing by 5 (corresponding to 100 times fainter) every time the  distance is multiplied by 10.  A star 100 pc away has an apparent magnitude +5 greater than its absolute magnitude, and for a star 1,000 pc away [latex]m=M+10[/latex].

Note that magnitudes do not have to be whole numbers.  If a star was 20 pc away, then

[latex]m-M=5\log\left(\frac{20 pc}{10 pc}\right)=1.5[/latex]

If the star’s absolute magnitude were 12, its apparent magnitude would be 13.5.  (The absolute and apparent magnitudes of most stars are not whole numbers.)

Solving the equation for apparent magnitude…

[latex]m=M+5\left(\log_{10}\left(d\right)-1\right)[/latex]

The plus sign after [latex]M[/latex] lets you know that apparent magnitude ([latex]m[/latex]) is going to increase as the distance gets larger. (The [latex]-1[/latex] comes from simplifying the logarithm.)

The equation to find the absolute magnitude if you know the apparent magnitude and the distance is…

[latex]M=m-5\left(\log_{10}\left(d\right)-1\right)[/latex]

Now there’s a negative sign between [latex]m[/latex] and the log term. If you have two stars with equal brightness (equal [latex]m[/latex]), the star that is further away has a higher luminosity. Which means, because magnitude is strange, that is has a smaller absolute magnitude ([latex]M[/latex]) and hence the negative sign.