Download The Magnitude System

The Magnitude System
The magnitude system is invented by Hipparchus in 120 BC, and is still used today, although the
magnitude scale itself is not a meaningful “physical quantity” (it is just a scale). Initially this system was
arbitrary because Hipparchus decided that all the brightest and most beautiful stars had a magnitude of 1
while the faintest ones had a magnitude of 6. On Staten Island we do not see 6th magnitude stars, mostly
because the sky is too bright due to light pollution from NYC. However, with the aid of telescopes we
can see much fainter stars and galaxies. The Hubble Space Telescope was able to push this limit to
close to 30.
Magnitudes have to be somehow related to the brightness of stars. And brightness is a measure
of the total number of photons of energy E that are emitted from the star per second and that pass
though the area of our detectors. The unit of brightness is ergs/sec/cm2, or if you use SI units, it is
Watts/m2. Magnitude does not have a unit - it is a scale. Since brightness and magnitudes express the
same thing, we have to figure out the relation between them.
It turns out that our eyes read out light signals logarithmically (this means that brightness and
magnitudes have some type of logarithmic relation). The relationship was originally determined
empirically and now it is used as a definition. It was found that a difference in 5 magnitudes corresponds
to a ratio of 100 in brightness. This means that if you have two stars, one that is really bright, say that
has a magnitude of 5, to one that you can barely see, one that has a magnitude of 6 (i.e. magnitude
difference of 5), then our detectors would measure100 times as many photons per second from the
brighter star.
Let’s compose a formula for this:
difference in magnitudes = −5 = m1 − m2
b
ratio in brightness = 100 = 1
b2
since our eyes read out signals logarithmically, we will have to take the logarithm of
b 
log  1  = log (100 ) = 2
 b2 
now this has to equal to the difference in magnitude, i.e., it has to be equal to -5
b 
so if we multiply log  1  by -2.5 we get
 b2 
b 
− 2.5 ⋅ log  1  = −2.5 ⋅ log (100) = −2.5 ⋅ 2 = −5 = m1 − m2
 b2 
brightness ratio
magnitude difference
So, a magnitude difference of -5 equals -2.5 times the logarithm of the brightness ratio.
b1
, i.e.
b2
Or, in general, using the first and the last part of the previous equation this is:
b 
m1 − m2 = −2.5 ⋅ log 1 
 b2 
Another way of writing this is in terms of the brightness ratio, rather than the magnitude difference. Thus
b
solving the above equation for 1 we get:
b2
b 
m1 − m2 = −2.5 ⋅ log 1 
 b2 
divide by -2.5
b 
m1 − m2
= log  1 
− 2.5
 b2 
1
= −0.4
− 2.5
b 
− 0.4 ⋅ (m1 − m2 ) = log 1 
 b2 
take the exponent
10
(
− 0. 4 ⋅ m1 − m 2
 b1 
) = 10 log  b 
solving this gives
2
b1
− 0. 4 ⋅ m − m
= 10 ( 1 2 )
b2
Both of the two formulae work. To test this try it with a few numbers. Some examples are given in the
Table below. First of all let us calculate the brightness ratio for a star that is one magnitude brighter than
the other.
b1
−0 .4 ⋅ (m1 − m 2 ) )
= 10
= 10− 0 .4 ⋅ ( −1) = 100 .4 = 2.512
b2
This means that a star that is brighter by 1 magnitude than another one emits 2.512 times a much light
(or expressing this mathematically: if the magnitude difference, i.e., m2 - m1 = 1, then the brightness ratio
is b1/b2 = 2.512). If the magnitude difference is 2, the brighter star emits 2.512 x 2.512 times (= 6) as
much light). If the magnitude difference is 3, the brighter star emits 2.512 x 2.512 x 2.512 times (=16)
as much light, and if magnitude difference 5 magnitudes, the brighter star emits 2.512 x 2.512 x 2.512 x
2.512 x 2.152 = 100 times as much light.
m1 - m2
0
1
2
3
4
5
b1/b2
1
2.512
6
16
40
100
2
Correlating magnitudes, brightness, luminosity and distance
Brightness, b, is what we measure on earth. It is related to the luminosity, L, via
L = 4πd 2 b or b =
L
4πd
2
.
So the brightness of an object decreases with the inverse square law of the distance. Since the
b 
brightness is related to the magnitude, via m1 − m2 = −2.5 log  1  , we ought to be able to determine a
 b2 
relationship of how the magnitude of a star changes with its distance. Therefore, we need to combine
 b1 
L
these equations and substitute b =
 . We get:
2 into m1 − m2 = −2.5 log 
4πd
 b2 
 L 


4πd1 2 

m1 − m2 = −2.5 ⋅ log
 L 
 4πd 2 

2 
Since the luminosity of the star is the same regardless at what distance it is, the L’s cancel out.
 L 
 1 

 2
2 
 1 d 2
4πd1 
d
m1 − m2 = −2.5 ⋅ log
= −2.5 ⋅ log  1  = −2.5 ⋅ log 2 ⋅ 2 
 L 
 1 
1 
 d1
 4πd 2 
d 2

 2 
2 
 d2 2 
d 
d 
m1 − m2 = −2.5 ⋅ log 2  = −2.5 ⋅ log 2  = −5 ⋅ log  2 
 d1 
 d1 
 d1 
2
And here is the formula relating magnitude to distance:
d 
m1 − m2 = −5 ⋅ log  2 
 d1 
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The Absolute Magnitude and the Distance Modulus
If we measure the brightness of a star and if we know the distance to the star, then we can
L
calculate its luminosity via b =
. Now, let’s do a thought experiment and put all the stars in the
4πd 2
sky at the same distance. Let’s take a distance of 10 parsec. The brightness of a star then is
L
B=
2 . Next we want to compare the real brightness of the star to the brightness the star
4π (10 pc)
b
would have if placed at a distance of 10 parsec. So we need to evaluate the ratio of , i.e.
B
L
b
4πd 2
=
L
B
4π (10 pc)
2
Substituting that ratio into the first “framed” formula, we get:



L
1 



2
b


 d2 
m − M = −2.5 ⋅ log = −2.5 ⋅ log  4πd
=
−
2
.
5
⋅
log


L
1 
B


2 
2
 4π (10 pc ) 
 (10 pc) 
2
 (10 pc )2 
 d 
 10 pc 
 10 pc 

m − M = −2.5 ⋅ log
=
−
2
.
5
⋅
log


=
−
5
⋅
log


=
5
⋅
log


2

 d 
 d 
 10 pc 
 d

m − M = 5 ⋅ log d − 5 log(10 pc) = 5 ⋅ log d − 5
So here is a final famous formula:
m − M = 5 ⋅ log d − 5
The absolute magnitude, M, is defined as the magnitude the star would have if it was placed at a
distance of 10 parsec. To distinguish it from the apparent magnitude we denote it with a capital M, and
the apparent magnitude with a small m.
Comparing the absolute to the apparent magnitudes, we have obtained an expression for the
distance. So if we know how to determine the absolute magnitude (and there are methods) and if we
measure the apparent magnitude with a photometer, we would be able to figure out how far that star is.
Therefore m − M is actually a measure is distance and astronomers call it the “distance modulus”.
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Sample Problems
1. Star A has a visual magnitude of +4.0 and star B has a visual magnitude of +10.0 Which star is
brighter? How much brighter is it?
Star A is brighter because its magnitude is a smaller number.
b1/b2 = 10 -0.4(4-10) = 10 2.4 = 250
2. Star C has an apparent visual magnitude of 5.0 and an absolute visual magnitude of 3.0. Is star C
closer or farther than 10 pc? How far away is Star C?
Since the absolute magnitude is brighter than the apparent magnitude, the star must be farther away
than 10 pc.
mv - Mv = 5 log d - 5 ;
5 log d = 5 + 5 - 3 = 7; log d = 7/5 = 1.4; D = 25 pc.
3. Sirius is 2.7 pc away and has an apparent brightness of -1.47 magnitudes. What is its absolute
visual magnitude?
Mv = mv - 5 log d + 5;
Mv = -1.47 - 2.2 + 5 = 1.3 mag
4. Lets assume that the brightest stars we see at Wellesley have an apparent magnitude of +4.0.
However if we use the 8” Meade telescopes, we can see stars that have apparent magnitudes of
+10.0. How much dimmer are the stars that we can observe through the telescope?
b1/b2 = 10 -0.410-4) = 10 -2.4 = 1 / 250
5. When a star goes supernova, it emits 100,000,000 times as much light. How much brighter will its
magnitude be?
b 
m1 − m2 = −2.5 ⋅ log 1  ; m1 - m2 = -2.5 log (108); m1 - m2 = -2.5 x 8 log 10 = -20
 b2 
The star will appear 20 magnitudes brighter, and it is well possible that a star in a distant galaxy,
suddenly “shows up” if it goes supernova. In fact, in 1989, the supernova in the Small Magellanic
Cloud (seen in the southern hemisphere) could be seen with the naked eye.
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