- Apparent magnitude

- Absolute magnitude

- Distance modulus

- Interstellar reddening

There were no binoculars or telescopes in the time of Hipparchus. The original classification system was based on naked-eye observations, and was fairly simple. Under the best observing conditions, the average person can discern stars as faint as sixth magnitude. The light pollution in Chicagoland limits our view of the night sky at Northwestern. On a typical clear night in Evanston, you can see 3rd to 4th magnitude stars. On the very best nights, you might be able to see 5th magnitude stars. Although the loss of one or two magnitudes does not seem like much, consider that most of the stars in the sky are dimmer than 5th magnitude. The difference between a light polluted sky and dark sky is astonishing.

As instruments were developed which could measure light levels more
accurately than the human eye, and telescopes revealed successively
dimmer stars, the magnitude system was refined. A subscript may be
added to the apparent magnitude to signify how the magnitude was
obtained. The most common magnitude is the **V** magnitude,
denoted **m _{V}**, which is obtained instrumentally using
an astronomical

Object |
m_{V} |

Sun | -26.8 |

Full Moon | -12.5 |

Venus at brightest | -4.4 |

Jupiter at brightest | -2.7 |

Sirius | -1.47 |

Vega | 0.04 |

Betelgeuse | 0.41 |

Polaris | 1.99 |

Naked eye limit | 6 |

Pluto | 15.1 |

Hubble Space Telescope | 31 |

While you may perceive one star to be only a few times brighter than another, the intensity of the two stars may differ by orders of magnitude. (Light intensity is defined as the amount of light energy striking each square cm of surface per second.) The eye is a logarithmic detector. While the eye is perceiving linear steps in brightness, the light intensity is changing by multiplicative factors. This is fortunate; if the eye responded linearly instead of logarithmically to light intensity, you would be able to distinguish objects in bright sunlight, but would be nearly blind in the shade! If logarithms are a faint memory, you should peruse a refresher on logs and logarithmic scales before continuing.

Given that the eye is a logarithmic detector, and the magnitude system
is based on the response of the human eye, it follows that the
magnitude system is a logarithmic scale. In the original magnitude
system, a difference of 5 magnitudes corresponded to a factor of
roughly 100 in light intensity. The magnitude system was formalized
to assume that a factor of 100 in intensity corresponds exactly to a
difference of 5 magnitudes. Since a logarithmic scale is based on
multiplicative factors, each magnitude corresponds to a factor of the
5th root of 100, or 2.512, in intensity. The magnitude scale is thus
a logarithmic scale in base 100^{1/5} = 2.512. The following
table illustrates the point.

Magnitude difference | Relative intensity |

0 | 1 |

1 | 2.51 |

2 | 6.31 |

3 | 15.8 |

4 | 39.8 |

5 | 100 |

10 | 10^{4} |

15 | 10^{6} |

To reiterate, each magnitude corresponds to a factor of 2.512 in intensity, and each 5 magnitudes corresponds to a difference in 100 in intensity. A few examples will help clarify the point.

**Q:**Star A has an apparent visual magnitude of 7, and its light intensity is 100 times dimmer than that of star B. What is the apparent visual magnitude of star B?**A:**The intensities differ by a factor of 100, which means that the difference in magnitudes must be 5. Since A is dimmer than B, and brighter stars have lower magnitudes, the apparent visual magnitude of B must be 7 - 5 = 2. Star B is thus a second magnitude star.**Q:**Star A has a magnitude of 2.5, and star B has a magnitude of 4.5. What is the ratio of the intensities I_{A}/I_{B}, where I_{A}is the intensity of star A, and I_{B}is the intensity of star B?**A:**The difference in magnitudes between the two stars is 4.5 - 2.5 = 2, so the intensities must differ by a factor of 2.512 squared, or 6.31 (see Table 2). Star A is brighter, because its magnitude is a smaller number. I_{A}/I_{B}should thus be larger than 1. It follows that I_{A}/I_{B}= 6.31.

Sometimes the numbers are not this simple, and we need general
equations. How can we mathematically describe the relationship
between intensity ratios and magnitude differences? While the
magnitudes are increasing linearly, the intensity ratios are
increasing logarithmically in base 2.512. Denoting magnitudes by
**m**, intensities by **I**, and using subscripts **A** and
**B** to denote stars A and B, we can express the relationship
between intensity and magnitudes as follows:

Convince yourself that this equation describes the numbers in Table 2. This form of the relationship is best when you know the relative magnitudes, and want to calculate the intensity ratio. We can manipulate the equation to put it in a more convenient form for the case when we know the intensities, but wish to find the relative magnitudes. Taking the log of both sides we get

Recalling from the general rules for
logs that log_{10} M^{P} = p log_{10} M,
we can rewrite this as

or

The result is most commonly expressed in the form

Let's do one more example of the magnitude system, this time using the equations.

**Q:**The intensity of star B is a factor of 10 higher than that of star A. Star A has a magnitude of 2.4. What is the magnitude of star A?**A:**First of all, think through the problem intuitively. Which star is brighter? Estimate by how many magnitudes the stars should differ. Should B have a higher or a lower magnitude?Star B is brighter, so it should have a lower magnitude than A. If the intensities differ by a factor of 10, Table 2 shows that the difference in magnitudes should be somewhere between 2 and 3. Since m

_{A}= 2.4, the magnitude of B must therefore lie between 0.4 and -1.4 magnitudes. Now let's use the equations.m Our answer lies within the predicted range._{B}- m_{A}= 2.5 log_{10}(I_{A}/ I_{B}).m

_{B}= m_{A}+ 2.5 log_{10}(I_{A}/ I_{B})m

_{B}= 2.4 + [ 2.5 log_{10}(0.1) ]m

_{B}= 2.4 + [ 2.5 (-1) ]m

_{B}= -0.1

Whew!

If we want to compare the intrinsic brightness of stars using the
magnitude system, we have to level the playing field. We will have to
imagine that all the stars are at the same distance, and then measure
their apparent brightness. We define the **absolute magnitude** of
an object as the apparent magnitude one would measure if the object
was viewed from a distance of 10 parsecs (10 pc, where 1 pc = 3.26 light years). We denote absolute
magnitude by an upper case **M**. As before, we denote such
magnitudes measured through a V filter by the subscript V. The
absolute magnitude is thus a measure of the intrinsic brightness of
the object. Let's look at the apparent and absolute magnitudes of the
stars in Table 2.

Object |
m_{V} |
M_{V} |

Sun | -26.8 | 4.83 |

Sirius | -1.47 | 1.41 |

Vega | 0.04 | 0.5 |

Betelgeuse | 0.41 | -5.6 |

Polaris | 1.99 | -3.2 |

Which of these stars is intrinsically the brightest? The supergiant Betelgeuse is the brightest, with an absolute magnitude of -5.6. The sun is intrinsically the dimmest!

How can we calculate the absolute magnitude of a star? There are two general ways. One method is to determine the distance to the star, measure the apparent magnitude, and scale the apparent magnitude to a distance of 10 pc. Secondly, if we know the spectral type and luminosity class of the star in question, we can estimate the star's luminosity, which is closely related to absolute magnitude. (Spectral types and luminosity classes are topics beyond the scope of this lab.) In this writeup we will only examine the first method. The relationship between distance, apparent magnitude and absolute magnitude will be derived in the next section.

Imagine a point source that flashes on for one second, then off again.
The light pulse spreads out in all directions, traveling at the speed
of light. You can think of the light pulse as an expanding shell of
light energy. The original light energy in that 1-second pulse is
spread over an ever enlarging shell. Intensity is the light energy
that strikes 1 square cm in one second. As the shell expands, there
is less light energy available for each square cm on the shell. At a
distance **d** from the original source, the shell has a surface
area of (4 x pi x d^{2}). Thus, the intensity **I** decreases
as 1/d^{2}. Although this is easiest to picture with a pulse
of light, the relationship between distance and intensity is just as
true for any source of (incoherent) light. Light intensity decreases
as the distance squared. All other things being equal, a light source
3 times more distant is 9 times as faint. For two identical light
sources **A** and **B** at distances **d _{A}** and

Now that we know how distance and intensity are related, and how intensity and magnitudes are related, we can determine how distances and magnitudes are related.

In the first section on apparent magnitudes we saw that

Instead of comparing the intensities and magnitudes of two different
stars, we will compare the intensities and magnitudes of the same star
at two different distances. Substituting (d_{B} /
d_{A})^{2} for (I_{A} / I_{B}), we get

which can be rewritten as

Now let's look at the special case when d_{A} = 10 pc, so that
m_{A} = M_{A}. We can then drop the subscripts A and
B, and simply write

If we require that **d** be specified in
pc, we can rewrite the equation as

The quantity **(m - M)** is called the **distance modulus**.
Once we know the distance modulus, we can easily calculate the
distance to the object. Rewriting the equation as

and exponentiating both sides, we find that

where **d** is in pc. Assuming no other factors are
involved, we now have the explicit relationship between apparent
magnitude, absolute magnitude, and distance. Knowing any two allows
us to solve for the third. The distance modulus automatically
provides two of the unknowns. Table 4 adds distance modulus and
approximate distance (in both pc and light years) to the values in
Table 3.

Object |
m_{V} |
M_{V} |
m_{V}-M_{V} |
d(pc) |
d(ly) |

Sun | -26.8 | 4.83 | -31.6 | 4.72 x 10^{-6} |
1.53 x 10^{-5} |

Sirius | -1.47 | 1.41 | -2.88 | 2.65 | 8.65 |

Vega | 0.04 | 0.5 | -0.46 | 8.1 | 26 |

Betelgeuse | 0.41 | -5.6 | 6.0 | 160 | 520 |

Polaris | 1.99 | -3.2 | 5.2 | 110 | 360 |

What else might affect the apparent brightness of a star? As we will see in the last section, interstellar dust dims starlight. If a star is far enough away, we must take this dimming into account.

Extinction is stronger at shorter wavelengths, as shorter wavelengths
interact more strongly with dust particles. Red light passes through
gas and dust more easily than blue light. The more gas and dust
between you and the source, the stronger the reddening. You observe
this effect daily! When the sun and moon are near the horizon, you
are viewing them through more atmosphere than when they are overhead.
That is why the sun and moon look reddish when they rise and set. The
reddening of starlight due to the interstellar extinction is known as
**interstellar reddening**. Astronomers often used the terms
extinction and reddening interchangeably.

The extinction or reddening to an object is usually given in
magnitudes, and denoted by an upper case **A**. Since extinction
is a function of wavelength, a subscript specifies the wavelength for
the stated value. A star whose light is dimmed by 1.2 magnitudes when
viewed through a V filter would have an extinction of A_{V} =
1.2.

How do we correct the equation for distance when accounting for extinction? Without extinction, we have

Can you guess how this equation should be altered? We need to
brighten the apparent magnitude to correct for the extinction. Lower
magnitudes are brighter, so we want to subtract A_{V} from the
apparent magnitude. The revised equation is thus

where d is in pc.

How can you determine A_{V} if you don't know the distance to
the star? One trick is to realize that extinction is very low at
longer wavelengths. Distances calculated from apparent and absolute
magnitudes at long wavelengths will be relatively accurate. The
spectral type and luminosity class of the star determine its absolute
magnitude, and also how the stellar emission changes with wavelength.
If you can obtain the spectral type and luminosity class of the star,
you can then compare the calculated emission at various wavelengths
with the corresponding observed emission. The difference between the
expected and observed emission at a given wavelength gives you the
extinction at that wavelength. An explanation of spectral types and
luminosity classes is beyond the scope of this lab.

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