The Magnitude Scale

by Roger Venable, Summer 2000

We measure the brightnesses of stars, and sometimes of star clusters and of galaxies, by the scale of magnitudes. This system has been used for 2000 years. Originally, the brightest stars were said to be of the first magnitude, while stars in the next tier of brightness were said to be of the second magnitude, etc., down to the sixth magnitude. The brighter the object, the lower the magnitude number. We now have it formalized so that a difference in brightness of five magnitudes is a factor of 100 in brightness. Thus, a difference of one magnitude is a factor of approximately 2.512 in brightness (2.512 is the fifth root of 100). Here are some proportions on this scale:

Magnitude
00 01 02 03 04 05 06 07 08 09 10 15 20 |
Inverse of brightness compared to Mag. 0
1 2.512 6.31 15.9 39.8 100 251.2 (or, 100 x 2.512) 631 (or, 100 x 6.31) 1585 (do you sense a pattern here?) 3980 (etc.) 10,000 (10,000 = 100 x 100) 1,000,000 100,000,000 |

The magnitude scale is an exponential scale, and computations in it can be done using logarithms. In the magnitude scale, the base is 2.512 and the exponent is the negative of the magnitude number. That is, the actual brightness is inversely proportional to 2.512 raised to the power of the magnitude number. But many magnitude calculations do not actually involve calculations using logarithms -- they’re just basic arithmetic.

Here is one such problem to solve using only arithmetic:

Consider the galaxy M32, a small eliptical galaxy in orbit around the great Andromeda galaxy, M31. As seen from earth, M32 is magnitude 9.0. If it contains 3,000,000,000 stars, what is the apparent magnitude of the average star in it? To find out, without using detailed logarithm calculations, think of the problem as follows:

We want to know the magnitude of magnitude 9.0 divided by 3 x 109.

We already know the relative real brightness of the average star: it is mag 9.0 divided by 3 x 109. Thus we need to know the number of magnitudes represented by a factor of 3 x 109 in brightness, and we can simply add this number to 9.0 to get the answer. 3 x 109 can be thought of as 30 x 108. And 108 is 100 x 100 x 100 x 100. Each factor of 100 is five magnitudes, so 108 is 20 magnitudes.

From the table of magnitudes versus relative real brightnesses above, it can be seen that a factor of 30 in brightness is approximately 3.7 magnitudes. 9.0 + 20 + 3.7 = 32.7 magnitude. That’s the answer. By the way, the Hubble Space Telescope can see down to about 28th mag, so it can’t see the average star in M32 -- it can only discern the brightest ones.

Astronomers do this sort of magnitude calculation in theirs heads frequently. There are two keys to understanding it: (1) Such problems ask you to divide or multiply. With exponents, of which magnitudes are an example, you add them when multiplying or subtract them when dividing. (2) Ratios of real brightnesses can be broken down into factors of 100 and a single number between 1 and 100. E.g., 3,000,000.000 is 30 x 100 x 100 x 100 x 100. Each 100 is five mags, and the other number can be interpolated from your recollection of the table above.

Here’s another problem: Deneb (Alpha Cygni) is 1600 light years away. It is magnitude 1.26. The sun is mag -26.5. If Deneb were the same distance from us as our sun is, how would it compare to the sun in brightness? Solution:

First, we’ll compare the distances of the two stars. Light travels 186,000 miles per second, so in a year it goes 186,000 x 60 x 60 x 24 x 365.25 miles, or 5.87 trillion miles (5.87 x 1012).

So, Deneb’s 1600 light years are 1600 * (5.87 x 1012 ) miles, or 9.39 x 1015 miles away.

The sun is 9.3 x 107 miles away, so Deneb is 1.0 x 108 times farther. That’s because (9.39 x 1015 ) / ( 9.3 x 107) = 1.0 x 108 .

Next, we note the effect of the inverse square law on Deneb’s apparent brightness. By the inverse square law, Deneb appears 1.0 x 1016 times fainter than it would if it were at the sun’s distance, because (1.0 x 108) * (1.0 x 108) = 1.0 x 1016.

Now we need to convert the factor of 1.0 x 1016 to magnitudes. It’s easy this time: there are 8 of the “100-factors” in 1016 -- that is, 100 x 100 x 100 x 100 x 100 x 100 x 100 x 100. Each 100 factor is worth 5 magnitudes, so the brightness difference of 1.0 x 1016 is 40 magnitudes.

Subtracting 40 from Deneb’s apparent magnitude of 1.26, we get -38.7 for the magnitude of Deneb when it’s repositioned to the distance of the sun. So Deneb would appear 12.2 magnitudes brighter than the sun -- that is, -38.7 is 12.2 magnitudes brighter than -26.5. That 12.2 magnitudes is two factors of 100 for the first 10 magnitudes and a factor of about 8 for the remaining 2.2 magnitudes. 100 * 100 * 8 is 80,000. That’s right -- Deneb is about 80,000 times as bright as the sun! Our collective goose would be cooked.

(Note: Deneb is a “yellow” star like the sun, so it has about the same brightness per unit of surface area. Therefore, it must have a much greater surface area -- specifically, 80,000 times as great. So its diameter is about 283 times that of the sun (the square root of 80,000 is about 283). If its center were at the sun’s location, Earth would be inside its photosphere! And Pluto would receive enough heat to be vaporized! And a note for the purists among the readers: in doing this calculation I rounded off a few figures, and this caused the final result to be mistaken by 25%. Deneb is actually ‘only’ 60,000 times as bright as the sun.)

Absolute magnitude. It is useful to compare the brightnesses of objects as though they were all at the same distance from you. But calculating the brightnesses as though they were all at the sun’s distance, as we did with Deneb, is a bit cumbersome, because the sun is so very close, astronomically speaking. Consequently, astronomers calculate the brightnesses that objects would have were they 32.6 light years away. They call this brightness the absolute magnitude of an object. The sun, at that distance, would be of magnitude 4.8, and Deneb would be 12.2 mags brighter at -7.4. Comparing absolute magnitudes is the usual way we compare brightnesses. So there are two fundamental types of magnitude, absolute magnitude and regular magnitude. Sometimes we can confuse the two, so to be very clear we often refer to regular magnitude as apparent magnitude.

The distance of 32.6 light years was chosen because it is 10 parsecs, and it happens to convey to almost all stars absolute magnitudes between -8 and +15, which are close to zero and thus easy to manipulate in your head. A parsec is the distance at which the earth’s six-month motion across its orbit gives an object a yearly parallax of one second of arc. At a distance of ten parsecs, an object would have a parallax of 0.1 second. A parsec is about 3.26 light years.

Distance modulus. Imagine that you are tracking the magnitude changes of supernova 1987A in the Large Magellanic Cloud, 160,000 light years away. You note sequential daily apparent magnitudes of 3.5, 3.4, 3.3, 3.2, 3.1, etc, and you want to know the real brightness -- the absolute magnitude -- of the star on each day. You don’t have to do the distance calculation repeatedly. You do it only once, and find that the star is 18.5 magnitudes fainter than it would be if it were at a distance of 10 parsecs. This same factor of 18.5 magnitudes applies to the supernova every day, and it also applies to every other object in the LMC. It is related to distance, and is called the distance modulus of the LMC. It is the apparent magnitude minus the absolute magnitude. Deneb’s distance modulus is 1.26 - (-7.4), or about 8.66. Sol’s is -26.5 - 4.8, or -31.3. The Andromeda galaxy, M31, has a distance modulus of about 24.5, as does every object in it. With the supernova, we can do an inverse calculation: we know its apparent magnitude, 3.5, and its distance modulus, 18.5, so we can calculate its absolute magnitude. D.M. = Mapp - Mabs, so Mabs = Mapp - D.M. Its absolute magnitude is -15.0.

Algebra with logarithms. Lastly, for the engineers in the club, I’ll show a magnitude calculation in which skill in manipulating logarithms is necessary. Imagine that you have measured the brightness of Castor (Alpha Geminorum) and find it to be of visual magnitude 1.59. Looking through a telescope, you find it to be a double star, and you measure the brightness of the dimmer of the two to be magnitude 2.85. What is the brightness of the brighter of the two? It equals the brightness of the pair minus the brightness of the dimmer one. But you can’t subtract magnitudes, because the scale is exponential. Here’s one way to solve it:

Let x = the magnitude of the brighter of the two stars of Castor.

Then 2.512-1.59 = 2.512-2.85 + 2.512-x or 2.512-x = 2.512-1.59 - 2.512-2.85 and we need to solve for x.

The right side of the lower equation can’t be evaluated as a whole but its two components can each be evaluated separately:

2.512-1.59 : log(2.512-1.59 ) = -1.59 log 2.512 = -1.59 * .4 = -0.636 = 9.364 - 10 antilog (9.364 - 10) = 0.2312 (so a star of mag 1.59 has 0.2312 of the brightness of a star of mag 0)

2.512-2.85 : log(2.512-2.85) = -2.85 log 2.512 = -2.85 * .4 = -1.14 = 8.86 - 10 antilog (8.86 -10) = 0.0724 (so a star of mag 2.85 has 0.0724 of the brightness of a star of mag 0)

Then 2.512-x = 0.2312 - 0.0724 = 0.1588 (so the brighter of the pair has 0.1588 of the brightness of a star of mag 0)

Now we can solve for x: log (2.512-x ) = log 0.1588 -x log 2.512 = log 0.1588 - 0.4x = 9.2009 - 10 = -0.7991 x = 2.0 (this is the answer, and it is the magnitude of the brighter of the pair)