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Introductory Astronomy
Magnitude and Distance Modulus Summary
At first the idea of magnitudes as brightness can be confusing. The scale runs backwards–larger
numbers are fainter. The negative part of the scale indicates the brightest objects. On top of that
quirk, the scale is logarithmic and people think in a linear fashion. On the magnitude scale stars
twice as bright are 0.752 magnitudes brighter. One magnitude brighter is 2.512 times brighter.
You used the magnitude scale outside during lab so you are a little familiar with it. You will see
how magnitudes make the math easy when calculating distance. That idea has to appeal to those
of you who have admitted math phobias.
Table 1 lists the apparent magnitudes of some common
objects. The range from brightest to faintest covers little
more than 50 magnitudes. That is the beauty of a
logarithmic scale–a wide range of values can be represented
within a limited and easily digested range of numbers.
Table 1 Apparent Visual Magnitude
of Selected Objects
Object
mv
Sun
-26.74
Apparent magnitude means these numbers are for the
objects as seen from Earth. Those were the magnitudes in
lab in the first semester. Note that Vega at a magnitude of 0
was the standard that originally defined the scale. Anything
brighter than Vega has a negative magnitude and anything
fainter has a positive magnitude.
Full Moon
-12.5
Table 2 shows the difference in magnitude and the
corresponding brightness ratio of the object. A one
magnitude difference between two objects means that the
brighter object is 2.5 times brighter. Note that it is the
magnitude difference and not the magnitude per se that we
are discussing here. Thus a zero magnitude star is 2.5 times
brighter than a first magnitude star just as a fifth magnitude
star is 2.5 times brighter than a sixth magnitude star.
Venus (at brightest)
-4
Sirius
-1.5
Vega
0
100 W bulb at 14 miles
4.3
Bolometric m for bulb above
1.0
Naked-eye limit
6.5
10 x 50 binoculars
~11
Pluto
13
Celestron/Meade 8-in limit
15
Keck (10-meter) limit
29
Hubble limit
30
How do we use this table? The magnitude difference between two stars along with Table 2 gives
us the brightness ratio, how many times brighter the star is. We can also work the other way.
Thus if star A is 40 times brighter than star B, the table tells us that A is 4 magnitudes brighter. It
is pretty easy once you get the feel for it.
Note that this table repeats itself in a way. The first six values in the table (for a magnitude
difference of 0 up to 5) are all one needs. Notice that a magnitude difference of 1 corresponds to
a brightness ratio of 2.5 and that a magnitude difference of 6 corresponds to a brightness ratio of
250 that is 100 times the ratio of a magnitude difference of 1. Likewise the value for a magnitude
difference of 7 is 100 times the magnitude difference of 2 and so on. Every fifth value is 100
times the previous one. Thus this table can be extended indefinitely to higher magnitude
differences as need.
As an example suppose that Sirius is 14 magnitudes brighter than another star. How many times
brighter is it? By extending the table for magnitude differences between 10 and 15, we find that a
14 magnitude difference tells us that Sirius is 400,000 times brighter than that other star. Here is
how we make that computation: A magnitude difference of 14 is 9 + 5. Thus the brightness ratio
for a magnitude difference of 14 is found by multiplying the brightness ratio for 9 magnitudes
(which is 4,000) by 100 arriving at 400,000 for the brightness ratio.
Table 2 allows us to determine the brightness ratio only if we have an
integer magnitude difference. In astronomy lecture the table will provide
all the information needed to analyze magnitudes and brightness ratios. It
is on the first page of every test. However, in the lab when there is a noninteger magnitude difference, a simple equation allows a calculation of the
brightness ratio:
bΔm = 2.512Δm
eq 1
Here Δm is the magnitude difference and b m is the brightness ratio that
corresponds to that difference. The inverse of this equation provides the
magnitude difference when we know the brightness ratio:
Table 2 Magnitude
Difference &
Brightness Ratio
Δm
b
0
1
1
2.5
2
6.3
3
16
4
40
5
100
6
250
7
630
8
1,600
9
4,000
10
10,000
15
1,000,000
Δ
Δm = 2.512 × log ( bΔm )
eq 2
Now consider a more useful measure, the absolute magnitude, M, which is
one of two methods for designating the absolute brightness of an object.
(The other is luminosity.) Absolute magnitude is the magnitude a star
would have if it were located 10 parsec from Earth. By using a standard
distance for all stars, we eliminate the effect of distance and have a
magnitude that allows us to compare directly the true brightness of various
stars.
Thus Sirius, the brightest star in our sky, is –1.5 in apparent magnitude but appears that bright
only because it is a mere 2.6 parsecs away. Were Sirius located at 10 parsecs, it would haves an
absolute magnitude of 1.5, a full three magnitudes fainter. If Sirius were actually at 10 parsecs,
then it would be outclassed by over 20 other stars in our night sky. On the other hand the star
Betelgeuse with an apparent magnitude of 0.5 (two magnitudes fainter than Sirius) is located 430
parsecs from the Earth. It would be a magnitude -7.6 star at 10 parsecs and would outshine the
first quarter moon!
There is a "simple" relationship between the distance (d) of a star, its apparent magnitude (m)
and its absolute magnitude (M). This relationship is contained in the equations below which are
not as formidable as they first appear:
⎛d⎞
m − M = 5log ⎜ ⎟
⎝ 10 ⎠
or if we solve for the distance,
d = 10
⎛ m-M+5 ⎞
⎜
⎟
⎝ 5 ⎠
eq 3
eq 4
Given any two of the quantities in these equations, we can calculate third. Note that the left side
of the equation 3, m - M, is determined by the right side of the equation which depends only on
d. The quantity “m - M” is actually a measure of the distance. We call that quantity the distance
modulus and you will hear me say it as "little m minus big M" to avoid confusion over which m.
Its value is the number of magnitudes by which a star would brighten if we moved it from where
it is located to a distance of 10 parsecs from the Earth. In the few cases where the star is closer
than 10 parsecs, it has a negative value.
Table 3 relates the distance modulus to the distance. We will use it on lecture tests where there is
little time for complex calculations. Its importance will be more obvious later in as we discuss
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how to find the large distances beyond our galaxy. As with the previous table, Table 3 repeats
itself after the first six values of the distance modulus.
After the first six rows of Table 3 it is possible to extend the table to any
higher values needed. Every time the distance modulus increase by 5, the
distance is 10 times greater. However, if you work with the table, you will
find that a distance modulus of 40 parsecs corresponds to a distance of one
billion parsecs or 3.26 billion light years. The most distant objects in the
universe are only a little farther than that, so this table runs out somewhere
between 40 and 45.
The importance of this table is that we use it to easily find a distance. If
we can see an object then we have its apparent magnitude, its little "m". If
we can in some way infer (guess if you want to be cruel to astronomers)
its big "M", then we have its distance modulus and thus its distance just by
looking it up in Table 3. There are a number of proven ways to estimate
values of big "M". In some case we can study an object's spectrum or
analyzing the way its light varies to obtain the value. Other times simply
recognizing the object as belonging to a group with a known standard
brightness is enough.
Table 3 Distance
Modulus &
Distance
m-M
d (pc)
0
10
1
16
2
25
3
40
4
63
5
100
6
160
7
250
8
400
9
630
10
1,000
15
10,000
Consider an example. Suppose we find a supergiant star in a star cluster in
our galaxy and we measure its apparent magnitude (any good astronomer
can do this in a few minutes) and obtain a value of 7. And suppose that the spectrum of this star
gives every indication that this star is virtually identical in every respect to similar stars whose
absolute magnitudes we have determined and they are all about -9. If we assume that our new
supergiant has an absolute magnitude of -9, we can easily obtain an estimate of its distance in the
following way. First see that
m - M = 7 - (-9) = 16.
Thus the distance modulus is 16. Table 3 shows the distance is somewhere over 10,000 parsecs.
How much? A distance modulus of 6 indicates a distance of 160 parsecs. By extending the table
to missing values we see a distance modulus of 11 (which is 6 + 5) must indicate a distance of
1,600 pc (160 X 10) and therefore a distance modulus of 16 (6 + 5 + 5) must indicate a distance
of 16,000 pc (160 X 10 X 10). I hope you agree that this method is more user-friendly than
equation 4.
This method has been one the foundations of distance measurement for objects outside our
galaxy. Indeed there are exploding stars called supernova that can be seen back when the
universe was very young. Since supernova of various types all have about the same absolute
magnitude, when we find one of these supernova and measure its apparent magnitude, we can
use this method to measure the distance to the supernova and hence the distance to objects back
near the beginning of the Universe.
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