. 2
( 28)


theoretically. The T 4 dependence means that E stant (not to be confused with the Stefan“
Boltzmann constant). Its value is 1.38 10 16 erg/K.
depends strongly on T. If we double the temperature
(The quantity kT is proportional to the kinetic
of an object, the rate at which it gives off energy
energy per particle in the gas.) The Rayleigh“Jeans
goes up by a factor of 16. If we change the tempera-
law agrees with experimental results at low fre-
ture by a factor of ten (say from 300 K to 3000 K), the
energy radiated goes up by a factor of 104. quencies (long wavelengths), but disagrees at high
frequencies. In fact, you can see from equation
For a star, we are interested in the total lumi-
(2.8) that, as we go to higher and higher frequen-
nosity. The luminosity is the total energy per sec-
cies, the energy given off becomes arbitrarily large.
ond (i.e. the power) given off by the star. The

However, in looking at the blackbody curves (as in I( , T) d I( , T) d
Fig. 2.5) you see that there is a peak and then there
Solving for I( , T) gives
is less energy at higher frequencies. (The classical
prediction of arbitrarily large energies at high fre- I( , T) I( , T) (d /d )
quencies was sometimes referred to as the “ultra-
To find I( , T) we must be able to evaluate
violet catastrophe”.)
c/ , so
d /d . We do this by remembering that
The first step in solving the problem of theo-
retically predicting blackbody curves was to
deduce an empirical formula for the observed 2
d /d (2.10a)
spectra. By “empirical” we mean a formula that is
We don™t care about the minus sign, which
arbitrarily put together to describe the observa-
just tells us that frequency increases when wave-
tions, but which is not derived from any theory.
length decreases. Using this result gives
Once that is done, then we try to find a theory
that can be used to derive the formula. I( , T) (c/ 2)
I( , T) (2.10b)
In 1900, Max Planck, a German physicist, pro-
Now we can substitute c/ for to obtain the
duced an empirical formula that accurately
final result:
describes the experimental blackbody spectra:

2h 3>c2 2hc2 5
I1 , T 2
I1 , T 2 (2.10c)
>kT ehc kT
eh 1
Remember, the Planck function accurately
In this equation, h is called Planck™s constant,
10 27 erg s. describes blackbody spectra, but it was origi-
and has the numerical value 6.63
nally presented as an empirical formula. There
This value was determined to provide the best
was still no theoretical understanding of the ori-
agreement with observed blackbody spectra.
gin of the formula. Planck continued his work
Since the Rayleigh“Jeans law adequately
in an effort to derive the formula from some
describes blackbody spectra at low frequencies,
theory. Planck found that he could derive the
the Planck law must reduce to the Rayleigh“Jeans
formula from classical physics if he inserted a
law in the limit of low frequencies. We can see
mathematical trick. The trick amounted to tak-
this if we take low frequencies to mean that h
ing a sum rather than an integral. The trick cor-
V kT (or equivalently h /kT V 1). In that case, we
responds to the physical statement that a black-
can take advantage of the fact that for x V 1,
ex 1 x. The Planck function then becomes body can only emit radiation at a frequency in
multiples of the quantity h . That is, the energy
2h 3 kT
I1 , T 2 could only be emitted in small bundles or quanta
c2 h (singular quantum). The quanta have energy h .
Even though Planck was able to derive the black-
body formula correctly, he was still not satisfied.
There was no justification for the restriction
which is the Rayleigh“Jeans law. that energy must be quantized.
Equation (2.9) gives the Planck function in
terms of frequency. How do we find it as a func- 2.4.2 Photons
tion of wavelength? Your first guess might be sim- An explanation for why energy must be quan-
ply to substitute c/ for each occurrence of in tized was proposed by Albert Einstein, in 1905. (It
equation (2.9). However, we must remember that was for this explanation that Einstein was later
I( , T) gives the energy per second per frequency awarded the Nobel Prize in physics.) Einstein
interval, whereas I( , T) gives the energy per sec- was trying to explain a phenomenon known as
ond per wavelength interval. The functions must the photoelectric effect, in which electrons can be
reflect that difference (especially since they will ejected from a metal surface if light falls on the
need different units). We therefore require that surface. (This is the basis for photocells, which

are used in many applications.) Laboratory stud-
2.5 Stellar colors
ies had shown that increasing the intensity of
the light falling on the surface increased the
We have seen that the color of a star can tell us
number of electrons ejected from the surface,
about the star™s temperature. However, we now
but not their energy. Einstein said that all radia-
need a way of quantifying a color, rather than
tion (whether from a blackbody or otherwise)
just saying something is red, green or blue. For
must come in small bundles, called photons. The
example, if we compare two blue stars, how do we
energy, E, of a photon with a frequency is given
decide which one is bluer?
We define two standard wavelength ranges,
E h (2.11)
centered at 1 and 2 and take the ratio of the
observed brightnesses, b( 1)/b( 2). We then con-
This explains the observed properties of the
vert that brightness ratio into a magnitude dif-
photoelectric effect by stating that each electron
ference (using equation (2.3)), giving
is ejected by a single photon striking the surface.
Increasing the intensity of light increases the
m2 m1 2.5 log10[b( 1)/b( 2)] (2.12)
number of photons striking the surface per sec-
ond, and therefore increases the rate at which We define the quantity m2 m1 as the color,
electrons are ejected. Increasing the frequency of measured in magnitudes, corresponding to the
the light increases the energy at which the elec- wavelength pair, 1, 2. For definiteness, let™s
trons are ejected. (This latter prediction was assume that 2 1. As we increase the tempera-
finally tested by Robert Millikan in 1916.) A further ture, b( 1)/b( 2) increases. This means that the
quantity m2 m1 decreases, since the magnitude
test of the photon hypothesis came in an analysis
of collisions between light (photons) and elec- scale runs backwards. If we know that an object is
trons, by A. H. Compton. The fact that light is com- radiating exactly like a blackbody, we need only
posed of photons explains why Planck had to take the ratio of brightnesses at any two wave-
assume that energy is quantized in deriving the lengths to determine the temperature.
formula for blackbody spectra. As we have said, we don™t really measure the
The assertion that light is essentially a parti- intensity of radiation at a wavelength. Instead, we
cle went against the then accepted ideas about measure the amount of energy received in some
light. The question of whether light is a particle wavelength interval. We can control that wave-
length interval by using a filter that only passes
or a wave had been going on for centuries. For
example, Newton believed that it is a particle, light in that wavelength range. When we use a
and he worked out a theory of refraction “ the filter, we are actually measuring the integral of
I( , T) over some wavelength range. Actually, the
bending of light when it passes, for example,
from air to glass “ on the basis of light speeding situation is more complicated. The transmission of
up when it enters the glass. (We now know, how- any real filter is not 100% over the selected range,
ever, that as a wave it slows down.) The wave the- and this must be factored in (see Problem 2.23).
ory became dominant with the demonstration of Another complication is that continuous spec-
interference effects by Young and the explana- tra of stars do not exactly follow blackbody
tion of electromagnetic waves by Maxwell. In curves. Therefore, observations through two fil-
explaining the photoelectric effect, Einstein was ters are not generally sufficient to tell us the tem-
saying that the particle picture must be revived. perature of the star. Over the years, a system of
The explanation that, somehow, light can exhibit standard filters has been developed, so that
both wave and particle properties is referred to as astronomers at various observatories can com-
the wave“particle duality. This concept is the pare their results. The wavelength ranges of the
foundation of what we refer to as the quantum rev- various filters are shown in Table 2.2. The most
olution, since it was such a radical departure from commonly discussed filters are U (for ultraviolet)
B (for blue) and V (for visible, meaning the center
previous theories. We will discuss this point far-
ther in the next chapter. of the visible part of the spectrum). More recently,

still L, but it is spread over an area of 4 d2 so the
Filter systems.
Table 2.2.
energy flux, f, is
Width (nm)a
Filter Peak wavelength (nm)
L/4 d2
f (2.13)
U 350 70
The received flux falls off inversely as the
B 435 100
square of the distance.
V 555 80
Unfortunately, distances to astronomical
R 680 150
objects are generally hard to determine. There is
I 800 150
a direct method for determining distances to
Full width at half maximum. nearby stars. It is called trigonometric parallax, and
amounts to triangulation from two different
observing points. You can demonstrate parallax
R (for red), and I (for infrared) have been added.
for yourself by holding out a finger at arm™s
(There are actually a couple of filters in different
length and viewing it against a distant back-
parts of the infrared.)
ground. Look at the finger alternately using your
For example, the B V color is defined by
left and right eye. The finger appears to shift
B V 2.5 log10 [I( V) / I( B)] constant
against the distant background. Bring the finger
where I( V) and I( B) are the intensities averaged closer and repeat the experiment. The shift now
over the filter ranges. (The constant is adjusted so appears larger. If you could move your eyes far-
that B V is zero for a particular temperature ther apart, the effect would be even greater.
star, designated A0. These designations will be Even the closest stars are too far away to
discussed in the next chapter.) As the tempera- demonstrate parallax when we just use our eyes.
ture of an object increases, the ratio of blue to vis- However, we can take advantage of the fact that
ible increases. This means that the B V color the Earth orbits the Sun at a distance defined to be
decreases (again because the magnitude scale runs one astronomical unit (AU). Therefore, if we observe
backwards.) a star and then observe it again six months later,
we have viewing points separated by 2 AU. The sit-
uation is illustrated in Fig. 2.6. We note the posi-
2.6 Stellar distances
tion of the star against the background of distant
stars, and then six months later we note the angle
So far we have discussed how bright stars appear by which the position has shifted. If we take half
as seen from Earth. However, the apparent bright- of the value of this angle, we have the parallax
ness depends on two quantities: the intrinsic angle, p.
luminosity of the star and its distance from us. Once we know the value of p, we can construct
(As we will see in Chapter 14, starlight is also a right triangle with a base of 1 AU and the other
dimmed when it passes through clouds of inter- leg being the length, d, the unknown distance to
stellar dust.) Two identical stars at different dis- the star. From the right triangle, we can see that
tances will have different apparent brightnesses.
tan p 1 AU/d (2.14)
If we want to understand how stars work, we
must know their total luminosities. This requires Since p is small, tan(p) p (rad), which is the
correcting the apparent brightness for the dis- value of p, measured in radians. Equation (2.14)
tance to the star. then gives us
If we have a star of luminosity L, we can cal-
p(rad) 1 AU/d (2.15)
culate the observed energy flux at a distance d. If
no radiation is absorbed along the way, all the
It is not very convenient measuring such
energy per second leaving the surface of the star
small angles in radians, so we convert to arc sec-
will cross a sphere at a distance d in the same
onds (see Box 2.1):
time. It will just be spread over a larger area.
Therefore, the energy per second reaching d is p( ) 2.06

Box 2.1. Angular measure

The natural unit for measuring angles is the radian. If we
have a circle of radius R, and two lines from the center
making an angle with each other, then the length of the
arc bounded by the two lines is
L (rad)R
where (rad) is the value of measured in radians. Since
the full circumference of a circle is 2 R, the angle corre-
sponding to a full circle must be 2 radians. This tells us
that a full circle, that is 360 , is equal to 2 radians, or
180 is equal to radians. In astronomy, we often deal
with very small angles, and measurements in arc seconds
( ) are convenient. We can convert measurements by

1“ 2 1rad2
180° 60¿ 60“
rad 1° 1¿
1 AU
105 (rad)
When we take the derivatives of trigonometric func-
Fig 2.6. Geometry for parallax measurements.The ¬gure
tions (for example, d(sin )/d cos ), it is assumed
is not to scale. In reality the distance to the star, d, is much
that the angles are in radians. If not, a conversion factor
greater than 1 AU, so the parallax angle, p, would normally
must be carried through the differentiation.
be very small.
When angles expressed in radians have a value that
is much less than unity, we can use a Taylor series to
approximate them:
d1sin ¿ 2
where p( ) is the parallax angle measured in arc
sin1 2 sin 10 2
seconds. Substituting this into equation (2.15) 0
105/p( )
d/1 AU 2.06 (2.16)

d1tan ¿ 2
This gives the distance to the star in AU (1 AU
tan1 2 tan 10 2
1.50 108 km). 0
This method suggests a convenient unit for
measuring distances. We define the parsec (abbre-
viated pc) as the distance of a star that produces
d1cos ¿ 2
a parallax angle p of 1 arc sec. From equation
cos1 2 cos 10 2
105 AU (or
(2.16), we can see that 1 pc 2.06 0
1013 km, or 3.26 light years). We rewrite
equation (2.16) as 1 sin(0)
d(pc) 1/p( ) (2.17)
Note that for small , sin and tan are both approxi-
Remember, as an object moves farther away,
mately equal to , so they must be equal to each other.
the parallax angle decreases. Therefore, a star at
(Remember, in each of the above expressions, must be
a distance of 2 pc will have a parallax angle of
expressed in radians.)
0.5 arc sec.

Using the fact that log(x2 ) 2 log(x) gives
Example 2.3 Distance to the nearest star
The nearest star (Proxima Centauri) has a parallax
m M 5 log10(d/10 pc) (2.18)
p 0.76 arc sec. Find its distance from Earth in
The quantity 5 log10(d/10 pc), which is equal to
(m M), is called the distance modulus of the star.
It indicates the amount (in magnitudes) by which
distance has dimmed the starlight. If you know
We use equation (2.17) to give
any two of the quantities (m, M or d) you can use
d(pc) 1/(0.76)
equation (2.18) to find the third. For any star that
d 1.32 pc we can observe, we can always measure m, its
apparent magnitude. Therefore, we are generally
With current ground-based equipment, we can
faced with knowing M and finding d or knowing
measure parallax to within a few hundredths of
d and finding M. In the next chapter we will look
an arc second. Parallax measurements are there-
at some ways of determining M.
fore useful for the few thousand nearest stars.
They are a starting point for a very complex system Example 2.4 Absolute magnitude
of determining distances to astronomical objects. A star is at a distance of 100 pc, and its apparent
We will encounter a variety of distance determi- magnitude is 5. What is its absolute magnitude?
nation methods throughout this book. The
trigonometric parallax method is the only one SOLUTION
that is direct and free of any assumptions. For this We use equation (2.18) to find
reason, astronomers would like to extend their
M m 5 log (d/10 pc)
capability for measuring parallax. The Hipparcos
satellite measures parallaxes to 10 3 arc sec. 5 5 log (100 pc/10 pc)

5 5 log (10)
2.7 Absolute magnitudes
We should note that changing the distance of
The magnitudes discussed in Section 2.1, based on
a star changes its apparent magnitude, but it
observed energy fluxes, are called apparent magni-
does not change any of its colors. Because colors
tudes. In order to compare intrinsic luminosities of
are defined to be differences in magnitudes, each
stars, we define a system of absolute magnitudes. The
is changed by the distance modulus. For example,
absolute magnitude of a star is that magnitude
using equation (2.18)
that it would appear to have as viewed from a stan-
dard distance, d0. This standard distance is chosen mB MB 5 log (d/10 pc)
to be 10 pc. From this definition, you can see that if
mV MV 5 log (d/10 pc)
a star is actually at a distance of 10 pc, the absolute
and apparent magnitudes will be the same. Taking the difference gives
To see how this system works, consider two
identical stars, one at a distance d and the other
at the standard distance d0. We let m be the appar-
Therefore, the distance modulus never appears in
ent magnitude of the star at distance d, and M
the colors.
that of the star at distance d0. (Of course, M will
When we talk about determining an absolute
be the absolute magnitude for both stars.) The
magnitude, we are really only determining it over
energy flux falls off inversely as the square of the
some wavelength range, corresponding to the
distance, therefore the ratio of the flux of the star
wavelength range of the observations. We would
at d to that from the star at d0 is (d0/d)2 . Equation
like to have an absolute magnitude that corre-
(2.3) then gives us
sponds to the total luminosity of the star. This
2.5 log10(d/d0)2 magnitude is called the bolometric magnitude of
m M

the star. (As we will see in Chapter 4, a bolometer ated BC), which relates the bolometric magnitude
to the absolute visual magnitude MV. Therefore
is a device for measuring the total energy received
from an object.) For any type of star, we can define
MBOL MV BC (2.19)
a number, called the bolometric correction (abbrevi-

Chapter summary
We saw in this chapter what can be learned from face area than cooler ones (as described by the
the brightness and spectrum of the continuous Stefan“Boltzmann law), and also have their spec-
radiation from stars. tra peaking at shorter wavelengths (as described
We introduced a logarithmic scale, the mag- by Wien™s displacement law). We saw how
nitude scale, for keeping track of brightness. attempts to understand the details of blackbody
Apparent magnitude is related to the observed spectra (Planck™s law) contributed to the idea of
energy flux from the star, and the absolute mag- light coming in bundles, called photons, with
nitude is related to the intrinsic luminosity of specific energies. With a knowledge of blackbody
the star. spectra, we saw how stellar colors can be used to
We saw how, even though stars are obvious to deduce stellar temperatures.
us in the visible part of the spectrum, they, and We saw how finding the distances to astro-
other astronomical objects, give off radiation in nomical objects is very important, but can be
other parts of the spectrum. The richness of infor- quite difficult. If we don™t know the distance to
mation in other parts of the spectrum is a theme an object, we cannot convert its apparent bright-
that we will come back to throughout the book. ness into a luminosity. We introduced one
We introduced the concept of a blackbody, method of measuring distances “ trigonometric
which is useful because the continuous spectrum parallax. It is the most direct method, but only
of a star closely resembles that of a blackbody. works for nearby stars. The problem of distance
Hotter bodies give off more power per unit sur- determination will come up throughout the book.

2.1. Why is the magnitude scale logarithmic? 2.8. What are the different ways in which the
2.2. Are there any other types of measurements word “spectrum” is used in this chapter?
that we encounter in the everyday world that 2.9. Give some examples of objects whose spectra
are logarithmic? (Hint: Think of sound.) are close to that of blackbodies.
2.3. Why are astronomical observations potentially 2.10. How can we determine the temperature of a
useful in measuring the speed of light? blackbody?
2.4. What are the factors that have resulted in 2.11. If the peak of a blackbody spectrum shifts to
early astronomical observations being in the shorter wavelengths as we reach higher tem-
“visible” part of the spectrum? peratures, how can it be that a hotter black-
2.5. What do we mean by “atmospheric window”? body gives off more energy at all wavelengths
2.6. Why was Maxwell™s realization that a varying than a cooler one?
electric field can create a magnetic field 2.12. What is the evidence for the existence of
important in understanding electromagnetic photons?
waves? 2.13. Explain how we quantify the concept of color.
2.7. (a) Estimate the number of people on Earth 2.14. What is the value of using standard filters in
who are exactly 2 m tall. (By “exactly” we looking at stellar spectra?
mean to an arbitrary number of decimal 2.15. Suppose you could communicate with an
places.) (b) How does this relate to the way we astronomer on a planet orbiting a nearby
define the intensity function I( )? star. (The astronomer is native to that planet,

2.16. How would parallax measurement improve if
rather than having traveled from Earth.) You
we could do our observations from Mars?
determine the distance to the star (by
2.17. As we determine the astronomical unit more
trigonometric parallax) to be 2 pc. The distant
accurately, how does the relationship
astronomer says that you are wrong; the dis-
between the AU and the parsec change?
tance is only 1 pc. What is the problem?

2.1. What magnitude difference corresponds to a wavelength being 501.00 nm and 510.00 nm.
factor of ten change in energy flux? What do you conclude?
2.2. One star is observed to have m 1 and *2.10. (a) Use equation (2.9) to derive max, the fre-
another has m quency at which I( , T) peaks. Convert this
1. What is the ratio of
energy fluxes from the two stars? max into a wavelength max. (b) Use equation
2.3. The apparent magnitude of the Sun is 26.8. (2.10c) to find the wavelength at which it
How much brighter does the Sun appear than peaks. (c) How do the results in (a) and (b)
the brightest star, which has m 1? compare?
2.4. (a) What is the distance modulus of the Sun? 2.11. For a 300 K blackbody, over what wavelength
(b) What is the Sun™s absolute magnitude? range would you expect the Rayleigh“Jeans
2.5. Suppose two objects have energy fluxes, f and law to be a good approximation?
f f, where f V f. Derive an approximate 2.12. Derive an approximation for the Planck func-
expression for the magnitude difference m tion valid for high frequencies (h W kT).
between these objects. Your expression should 2.13. As we will see in Chapter 21, the universe is
have m proportional to f. (Hint: Use the filled with blackbody radiation at a tempera-
fact that ln (1 x) … x when x V 1.) ture of 2.7 K. (a) At what wavelength does the
2.6. Show that our definition of magnitudes has spectrum of that radiation peak? (b) What
the following property: If we have three stars part of the electromagnetic spectrum is this?
with energy fluxes, f1 , f2 and f3 , and we 2.14. (a) We observe the blackbody spectrum from
define a star to peak at 400 nm. What is the temper-
ature of the star? (b) What about one that
m2 m1 2.5 log10( f1/f2)
peaks at 450 nm?
m3 m2 2.5 log10( f2/f3) 2.15. Derive an expression for the shift in the
peak wavelength of the Planck function for a
blackbody of temperature T, corresponding to
m3 m1 2.5 log10(f1/f3) a small shift in temperature, T.
2.16. Calculate the energy per square centimeter
2.7. Suppose we measure the speed of light in a
per second reaching the Earth from the Sun.
laboratory, with the light traveling a path of
2.17. How does the absolute magnitude of a star
10 m. How accurately do you have to time the
vary with the size of the star (assuming the
light travel time to measure c to eight signifi-
temperature stays constant)?
cant figures?
2.18. (a) What is the energy of a photon in the
2.8. Let 1 and 2( 1, 2) be the wavelength (fre-
middle of the visible spectrum ( 550 nm)?
quency) limits of the visible part of the spec-
(b) Approximately how many photons per sec-
trum. Compare ( 1 2)/( 1 2) with ( 1
ond are emitted by (i) a 100 W light bulb,
2)/( 1 2). Comment on the significance.
(ii) the Sun?
2.9. (a) Calculate the frequencies corresponding
2.19. If we double the temperature of a blackbody,
to the wavelengths 500.00 nm and 500.10 nm.
by how much must we decrease the surface
Use these to check the accuracy of equation
area to keep the luminosity constant?
(2.10a). (b) Repeat the process for the second

*An asterisk denotes a harder Problem or Question. The convention continues throughout the book.

2.20. (a) How does the absolute bolometric magni- can be measured using the method of
tude vary with the temperature of a star trigonometric parallaxes? (b) By what factor
(assuming the radius stays constant)? (b) Does will the volume of space over which we can
the absolute visual magnitude vary in the measure parallax change if we can measure
to 0.001 arc sec? (c) Why is the volume of
same way?
*2.21. For a star of radius R, whose radiation follows space important?
a blackbody spectrum at temperature T, derive 2.27. If we lived on Mars instead of the Earth, how
an expression for the bolometric correction. large would the parsec be?
2.22. Suppose we observe the intensity of a black- 2.28. Suppose we discover a planet orbiting a
body, I0, in a narrow frequency range cen- nearby star. The distance to the star is 3 pc.
tered at 0. Find an expression for T, the tem- We observe the angular radius of the planet™s
orbit to be 0.1 arc sec. How many AU from
perature of the blackbody in terms of I0 and
0. (a) First do it in the Rayleigh“Jeans limit the star is the planet? (Hint: You can solve
and (b) in the general case. this problem by “brute force”, converting all
*2.23.Suppose we receive light from a star for the units. For an easier solution, think about
which the received energy flux is given by the what the answer would be if the star were 1
function f( ). Suppose we observe the star pc from us and the angular radius of the
orbit were 1 arc sec, and then scale the result
through a filter for which the fraction of
light transmitted is t( ). Derive an expression accordingly.)
for the total energy detected from the star. 2.29. Derive an expression for the distance to a star
(Hint: Start by thinking of the energy in terms of its distance modulus.
detected in a small wavelength range.) 2.30. If we make a 0.05 magnitude error in measur-
2.24. What is the distance to a star whose parallax ing the apparent magnitude of a star, what
is 0.1 arc sec? error does that introduce in our distance
2.25. Derive an expression for the distance of an determination (assuming its absolute magni-
object as a function of the parallax angle seen tude is known exactly)?
by your eyes?
2.26. (a) If we can measure parallaxes as small as
0.1 arc sec, what is the greatest distance that

Computer problems

2.1. Make a fourth column for Table 2.1, showing the lation you may assume that magnitudes are deter-
range of photon frequencies for each part of the mined in a narrow range of wavelengths around
spectrum. Make a fifth column showing the range the peak of each filter.
of photon energies for each part of the spectrum. 2.3. For the Sun, plot the difference between the
Make a sixth column showing the temperatures Rayleigh“Jeans approximation and the Planck for-
that blackbodies would have to peak at the wave- mula, as a function of wavelength, for wavelengths
lengths corresponding to the boundaries between in the visible part of the spectrum.
the parts of the spectrum 2.4. For the Sun, calculate the energy given off over the
2.2. Make a graph of the magnitude difference MB MV wavelength bands that correspond to the U, B and
V filters. Use this to estimate the colors U B and
as a function of temperature for a temperature
B V.
range of 3000 K to 30 000 K. To simplify the calcu-
Chapter 3

Spectral lines in stars

In Chapter 2 we discussed the continuous spectra gave the strongest lines letter designations that
of stars and saw that they could be closely we still use today.
described by blackbody spectra. In this chapter, The origin of these lines was a mystery for
some time. In 1859, the German chemist Gustav
we will discuss the situations in which the spec-
Robert Kirchhoff noticed a similar phenomenon in
trum shows an increase or decrease in intensity
over a very narrow wavelength range. the laboratory. He found that when a beam of
white light was passed through a tube containing
some gas, the spectrum showed dark lines. The
gas was absorbing energy in a few specific narrow
3.1 Spectral lines wavelength bands. In this situation, we refer to
the lines as absorption lines. When the white light
We know that if we pass white light through a was removed, the spectrum showed bright lines,
or emission lines, the wavelengths where absorp-
prism, light of different colors (wavelengths) will
emerge at different angles with respect to the ini- tion lines had previously appeared. The gas could
tial beam of light. If we pass white light through emit or absorb energy only in certain wavelength
a slit before it strikes the prism (Fig. 3.1), and bands.
then let the spread-out light fall on the screen, at Kirchhoff found that the wavelengths of the
each position on the screen we get the image of emission or absorption lines depend only on the
the slit at a particular wavelength. type of gas that is used. Each element or com-
Both William Hyde Wollaston (1804) and Josef von pound has it own set of special wavelengths. If
Fraunhofer (1811) used this method to examine two elements which don™t react chemically are
sunlight. They found that the normal spectrum mixed, the spectrum shows the lines of both ele-
was crossed by dark lines. These lines represent ments. Thus, the emission or absorption spec-
wavelengths where there is less radiation than at trum of an element identifies that element as
nearby wavelengths. (The lines are only dark in uniquely as fingerprints identify a person. This
comparison with the nearby bright regions.) The identification can be carried out without under-
linelike appearance comes from the fact that, at standing why it works.
each wavelength, we are seeing the image of the Whether we see absorption or emission depends
slit. It is this linelike appearance that leads us to in part on whether or not there is a strong enough
call these features spectral lines. If we were to make background source providing energy to be absorbed
a graph of intensity vs. wavelength, we would (Fig. 3.2). The strength of the spectral lines also
find narrow dips superimposed on the continuum. depends on how much gas is present and on the
The solar spectrum with dark lines is sometimes temperature of the gas. Sample emission and
referred to as the Fraunhofer spectrum. Fraunhofer absorption spectra of stars are shown in Fig. 3.3.



Gas with
Atoms & Molecules


Fig 3.1. If we allow white light to fall all over a prism, the

red from one part will overlap the blue from another part,
and we can™t see a clear spectrum. Instead, we pass white
light through a slit ¬rst.The beam of light is then spread out Source
as it passes through the prism. On the screen, we are seeing
a succession of images of the slit in different colors. If there
is a color missing from the white light, this will show up as a
gap on the screen in the shape of the slit.

Fig 3.2. Conditions for the formation of emission and
absorption lines. (a) We look at a cloud of gas with the
3.2 Spectral types atoms or molecules capable of producing spectral lines. Since
there is no continuum radiation to absorb, we can only have
When spectra were taken of stars other than emission. (b) We now look through the gas at a background
continuum source.This can produce absorption lines.
the Sun, they also showed absorption spectra.
Presumably, the continuous radiation produced
in a star passes through an atmosphere in which
the absorption lines are produced. Not all stars B stars the next strongest, and so on. These letter
have absorption lines at the same wavelength. designations were called spectral classes or spectral
Astronomers began to classify and catalog the types. We now know that the different spectral
spectra, even though they still did not understand types correspond to different surface tempera-
the mechanism for producing the lines. This points tures. However, the sequence A, B, . . . is not a
out an important general technique in astronomy “ temperature-ordered sequence. For reasons we
studying large numbers of objects to look for gen- will discuss below, hydrogen lines are strongest in
eral trends. In one very important study, over intermediate temperature stars.
200 000 stars were classified by Annie Jump Cannon The spectral classes we use, in order of decreas-
at the Harvard College Observatory. The benefactor ing temperature, are O, B, A, F, G, K, M. We break
of that study was Henry Draper, and the catalog of each of these classes into ten subclasses, identified
stellar spectra was named after him. The stars in by a number from zero to nine; for example, the
this catalog are still known by their HD numbers. sequence O7, O8, O9, B0, B1, B2, . . . , B9, A0, A1, . . . .
One set of spectral lines common to many (For O stars the few hottest subclasses are not
stars was recognized as belonging to the element used.) For some of the hotter spectral types, we
hydrogen. The stars were classified according to even use half subclasses, for example, B1.5. It was
the strongest hydrogen absorption lines. In this originally thought that stars became cooler as
system, A stars have the strongest hydrogen lines, they evolved, so that the temperature sequence

Fig 3.3. Samples of stellar spec-
tra.These are high resolution
spectra, with the visible part of
the spectrum (400 to 700 nm)
broken into 50 slices.Wavelength
increases from left to right along
each strip and from bottom to
top. (a) Procyon, also known as
Alpha Canis Majoris (the brightest
star in Canis Major). It has spec-
tral type F5 (see Section 3.2),
making it a little warmer than the
Sun. (b) Arcturus, also known as
Alpha Bootes. It is spectral type
K1, being cooler than the Sun.



was really an evolutionary sequence. Therefore, are tied to the nature of matter and light. In
the hotter spectral types were called early and the Chapter 2, we saw the beginnings of the quantum
cooler spectral types were called late. We now revolution with the realization that light exhibits
know that these evolutionary ideas are not cor- both particle and wave properties. We now see
rect. However, the nomenclature still remains. We how the ideas of quantization apply to the struc-
even talk about a B0 or B1 star being ˜early B™ and ture of the atom.
a B8 or B9 as being a ˜late B™. The modern picture of the atom begins with
the experiments of Ernest Rutherford, who studied
the scattering of alpha particles (helium nuclei)
3.3 The origin of spectral lines off gold atoms. Most of the alpha particles passed
through the gold atoms without being deflected,
The processes that result in atoms being able to suggesting that most of the atom is empty space!
emit or absorb radiation at certain wavelengths Some alpha particles were deflected through

index of refraction of air, 1.000 29, giving
large angles, suggesting a concentration of posi-
656.28 nm. (It is interesting to note that when spec-
tive charge at the center of each atom. This con-
centration is called the nucleus. A sufficient num- troscopists tabulate wavelengths, those longer than
200 nm are given as they would be in air, since that
ber of electrons orbit the nucleus to keep the
is how they usually will be measured. Radiation
atom electrically neutral.
with wavelengths less than 200 nm doesn™t pene-
There were still some problems with this pic-
trate through air, and its wavelengths are usually
ture. It did not explain why electron orbits were
measured in a vacuum, so the vacuum values are
stable. Classical electricity and magnetism tells
us that an accelerating charge gives off radiation.
An electron going in a circular orbit is accelerat-
3.3.1 The Bohr atom
ing, since its direction of motion is always chang-
The next advancement was by the Danish physi-
ing. Therefore, as the electrons orbit, they should
cist Neils Bohr who tried to understand hydrogen
give off radiation, lose energy and spiral into the
(the simplest atom), illustrated in Fig. 3.4. He pos-
nucleus. This is obviously not happening. The sec-
tulated the existence of certain stationary states. If
ond problem concerns the origin of spectral
the electron is orbiting in one of these states, the
lines. There is nothing in the Rutherford model
atom is stable. Each of these states has a particu-
of the atom that allows for spectral lines.
lar energy. We can let the energy of the nth state
The arrangement of spectral lines in a partic-
be En and the energy of the mth state be Em. For
ular element is not random. For example, in 1885,
definiteness, let En Em.
Johann Jakob Balmer, a Swiss teacher, realized that
Under the right conditions, transitions between
there was a regularity in the wavelengths of the
states can take place. If the electron is in the
spectral lines of hydrogen. They obeyed a simple
higher energy state, it can drop down to the lower
relationship which became known as the Balmer
energy state, as long as a photon is emitted with
an energy equal to the energy difference between
R(1/22 1/n2)
The constant R is called the Rydberg constant,
and its value is given by 1/R 91.17636 nm. The
4th Orbit
quantity n is any integer greater than two. By set-
ting n to 3, 4, . . . , we obtain the wavelengths for
the visible hydrogen lines (also known as the
Balmer series). Of course, this was just an empirical 3rd Orbit
formula, with no theoretical justification.
Example 3.1 First Balmer line 2nd Orbit
Calculate the wavelength of the longest wavelength
Balmer line. This line is known as the Balmer- 1st Orbit
alpha, or simply H .
We let n 3 in equation (3.1) to obtain
R(1/22 1/32)
Substituting for R and inverting gives
656.47 nm
This is the wavelength as measured in a vacuum.
Fig 3.4. The Bohr atom. Electrons orbit the nucleus in
We generally refer to the wavelength in air, since
allowed orbits.The relative sizes of the orbits are correct,
that is how we measure it at a telescope. The wave-
but on this scale the nucleus should be much smaller.
length in air is that in vacuum divided by the

the two states. If the frequency of the photon is , We put this into equation (3.2) to find the total
energy in terms of r:
this means that
E (3.4)
h En Em
The minus sign indicates that the total energy is
If the electron is in the lower energy state, it
negative. To see what this means, remember that
can make a transition to the upper state if the
we have defined the potential energy such that it
atom absorbs a photon with exactly the right
is zero when the electron and proton are infi-
energy. This explanation incorporated Einstein™s
nitely far apart. The electron and proton being far
idea of photons.
apart with no motion is the minimal condition
Bohr pointed out that one could calculate the
for the electron being free of the proton. So, if the
energies of the allowed states by assuming that
electron is barely free of the proton, the total
the angular momentum J of the orbiting elec-
energy would be zero. So, if the total energy is
trons is quantized in integer multiples of h/2 .
negative, as in equation (3.4), then we must add
The combination h/2 appears so often we give it
energy [(1/2)e2/r] if we want to bring it up to zero,
its own symbol, h (spoken as ˜h-bar™). We apply
which would free the electron. So the negative
this to a hydrogen atom, with an electron, with
energy means that the electron is not free. In this
charge e, orbiting a distance r from a nucleus
case we say that the system is bound.
with charge e. We assume that the nucleus is
We still have to find the allowed values of r.
much more massive than the electron, so we can
The angular momentum is J mvr. The quantiza-
ignore the small motion of the nucleus, since
tion condition becomes
both the nucleus and electron orbit their com-
mon center of mass. mvr nh/2
We first look at the kinetic energy
Solving for v gives
v nh/2 mr
The potential energy, relative to the potential
energy being zero when the electron is infinitely Squaring and multiplying by m gives
far from the nucleus, is given by mv2 n2h2/4 2 mr2
PE By equation (3.2), we have
(By writing the potential energy in this form, e2/r n2h2/4 2
rather than with a factor of 1/4 0, we are using
We now solve for r, giving the radius of the nth
cgs units. This means that charges are expressed
in electrostatic units, esu, with the charge on the
electron being 4.8 10 10 esu.) The total energy n2h2/4 2
rn (3.5)
is the sum of kinetic and potential:
Substituting into equation (3.4) gives the energy
(1/2)mv2 e2/r
E (3.2) of the nth state:

We can relate v and r by noting that the elec- En (1/2)e4m(4 2)/n2h2 (3.6)
trical force between the electron and the nucleus, 2
e2/r2, must provide the acceleration to keep the Note that this has the 1/n dependence that we
electron in the circular orbit, v2/r. This tells us would expect from the Balmer formula.
One modification that we should make is to
account for the motion of the nucleus (since it is
2 22
mv /r e /r not infinitely massive). We should replace the
mass of the electron, m, in equations (3.5) and
Multiplying both sides of the equation by r
(3.6) by the reduced mass of the electron and pro-
ton. The reduced mass, mr, is defined such that
2 2
(3.3) the motion of the electron, as viewed from the
mv e /r

(moving) proton, is as if the proton were fixed It is then very easy to calculate the energies of
and the electron™s mass is reduced to mr. An the other levels. For example, E2 3.4 eV.
expression for mr is (see Problem 3.2) Therefore the energy difference, E2 E1 is equal
to 10.2 eV. The energy levels are shown in Fig. 3.5.
This diagram is a convenient graphical representa-
me mp
tion of energy levels, called an energy level diagram.
In this diagram the levels are plotted as horizon-
0.9995 me
tal lines with vertical locations proportional to
where me and mp are the masses of the electron the energy. We can draw vertical arrows indicat-
and proton, respectively. ing possible transitions between the levels. The
length of the arrow would then indicate the
Example 3.2 Hydrogen atom energy
energy change associated with that transition.
Compute the energy of the lowest (ground) energy
Note that the levels are closer together as one
level in a hydrogen atom. Also, find the radius of
goes to higher values of n.
the orbit of the electron in that state.
Since the zero of potential energy is arbi-
trarily defined, we sometimes choose to shift
the energy scale by the binding energy (13.6 eV
We use equation (3.6) with n 1 to give
for hydrogen). This would make the energy of
11>22 14.8 esu2 4 19.11 g 2 14 2
10 28 2
10 10 the ground state (n 1) zero, and for the n 2
11 2 2 16.63 erg s 2 2
E1 27
10 state, 10.2 eV. A free electron would then have
2.2 10 erg
An erg is not a convenient unit to use to keep
track of such small energies, so we convert to 0 5
electron volts, eV (1 eV 1.6 10 12 erg, is the 4
energy acquired by an electron in being acceler-
ated through a potential difference of 1 volt),
Energy (eV)

E1 13.6 eV
The radius is given by equation (3.5):
16.23 erg s 2 2
19.11 g 2 14.8 esu2 2 14 2
r1 28 10 2
10 10
5.25 10 cm

0.0525 nm
Note that if we take n in equation (3.6) we
get E 0. However, n corresponds to a free 1
electron. Therefore, to move the electron far from Lyman
the nucleus, we must add 13.6 eV. The energy that
Fig 3.5. Hydrogen energy levels.The right hand column
we must add to an atom to break it apart is called
gives the principal quantum number, n.The energies are rela-
the binding energy. The energy goes to do work
tive to the state in which the electron and proton are in¬-
against the electrical attraction between the elec- nitely far apart, so the ground state energy is 13.6 eV.
tron and the nucleus as you try to pull the elec- Transitions (which can be either emission or absorption) are
tron away. grouped according to the lower level of the transition. For
Now that we have evaluated E1, we can rewrite example, the Balmer series consists of emissions with the
equation (3.6) as electrons ending in state n 2, and absorptions starting in
the n 2 state.
13.6 eV/n2
En (3.7a)

an alpha ( ) transition; if n m 2, we have a
an energy of 13.6 eV, or greater. The values of
beta ( ) transition. The first Balmer line is then
the energy differences between these states are
designated H2 . (Note that for the Balmer series
unaffected by this shift in the zero point of the
of hydrogen only, we sometimes drop the 2 and
just say H , H , etc.)
We can use equation (3.6) to derive the Balmer
formula. First, we rewrite the equation as
3.3.2 Quantum mechanics
En (3.7b)
The Bohr model of the atom allowed physicists to
where understand the organization of energy levels.
However, it was far from a complete theory. One
(1/2)e4m(4 2
R (3.7c)
shortcoming was that it did not explain why
The energy of an emitted or absorbed photon some spectral lines are stronger than others.
must equal the energy difference between the More fundamentally, it was an ad hoc theory. Bohr
two states: had no explanation of why stationary states exist,
or why angular momentum must be quantized in
En Em h
some particular way. These were just postulates.
hc/ A much deeper understanding was needed.
An important step was made by Louis de
Taking the energies from equation (3.7b) gives
Broglie, who proposed the revolutionary idea that
R(1/m2 1/n2)
1/ (3.8)
if light could exhibit a wave“particle duality,
then maybe all matter could. That is, an electron
which looks very similar to the Balmer formula,
orbiting a nucleus has certain wavelike proper-
except that the Balmer formula has a 2 instead of
the m. This means that the Balmer series all have ties, and it is those properties that determine the
states that are stable. One could think of the elec-
the second energy level as their lower level.
tron as having a certain wavelength. Stationary
We can use equation (3.8) to divide the hydro-
states could be those whose circumference con-
gen spectrum into different series. A given series
tained an integral number of wavelengths, pro-
is characterized by having the same lower energy
ducing a pattern that reinforced during each
state. For example, the Balmer series consists of
orbit (like a standing wave). It was necessary to
absorptions accompanying transitions from level
have expressions for the wavelength and fre-
2 to any higher levels, and emissions accompany-
quency of a particle, and de Broglie noted that if
ing transitions from higher levels down to level 2.
the wavelength was taken as h/p (where p is the
The first Balmer transition (involving levels 2 and
momentum of the particle) and the frequency as
3) has the smallest energy difference of the series.
E/h, then the orbits allowed by the standing wave
(Clearly the energy difference between levels 2
idea were the same as the orbits that Bohr found
and 3 is less than the energy difference between
from his postulates (see Problem 3.8).
levels 2 and 4, or between levels 2 and 5, and so
This is clearly a departure from our normal
on.) The Balmer series is important because the
way of looking at matter around us, and we can-
first few transitions fall in the visible part of the
not go through all of the ramifications here. To
spectrum. The series with the lower energy level
being level 1 is called the Lyman series. Even the this point, we have gone far enough to under-
stand stellar spectra. The picture as presented by
lowest transition in the Lyman series is in the
Bohr and de Broglie is quantum theory in its
most naive form. It was realized that if particles
We have developed a labeling system for vari-
behave, in some fashion, like waves then the
ous transitions. First we give the chemical symbol
description of particle motions (mechanics) must
for the element (e.g. H for hydrogen). Then we
give the m for the lowest level that characterizes be changed from Newton™s laws of motion to laws
of motion involving waves. (Of course, in the
the series (1 for Lyman, 2 for Balmer, etc.). Finally,
limit of large objects, such as apples falling to
we give a Greek letter denoting the number of
levels jumped. For example, if n m 1, we have Earth, these new laws of motion must reduce to

Newton™s laws, because we know that Newton™s tion. The number of atoms per unit volume in a
given state is called the population of that state. In
laws work quite well for apples and planets.)
Theories that describe the mechanics of waves this section we look at the factors that determine
are called wave mechanics or quantum mechanics. the populations of the various states. We refer to
processes that can alter the populations as excita-
One such theory was presented in 1925 by the
German physicist Erwin Schrödinger. In his theory tion processes. We have already seen one type of
the information about the motion of a particle is excitation process “ the emission and absorption
contained in a function, called a wave function. of photons. Electrons can jump to a higher level
Schrodinger™s interpretation of the wave func- when a photon is absorbed or they can jump to a
tion was that it is related to the probability of find- lower level when a photon is emitted.
ing a particle in a particular place with a partic- Populations can also be changed by collisions
ular momentum. This replaced the absolute with other atoms, as illustrated in Fig. 3.6. For
example, atom 1 can be in state i. It then under-
determinism of classical physics, with the state-
ment that we can only predict where a particle is goes a collision with atom 2, and makes a transi-
likely to be, but not exactly where it will be. tion to a higher state, j. In the process the kinetic
However, we can predict the average positions energy of atom 2 is decreased by the difference
and momenta of a large group of particles, and it between the energies of the two states in atom 1,
Ej Ei. The reverse process is also possible, with
is these average properties that we see (and meas-
ure) in our everyday world. Many physicists atom 2 gaining kinetic energy and atom 1 drop-
ping from state j to state i.
(including Einstein) were not comfortable with
this probabilistic interpretation, but quantum
theory has been very successful in predicting the
outcome of a wide variety of experiments. We (a)
will pick up on some of the threads of the quan-
tum revolution later in this book.

3.4 Formation of spectral lines

Now that we have some idea of how atoms can
emit or absorb radiation, we can return to stellar
spectra. The first point to realize is that in a star
we are not talking about the radiation from a sin-
gle hydrogen atom, but from a large number of (b)
them. We see a strong H absorption line in stars
because many photons are removed from the con-
tinuum by this process. It is clear, however, that
having a lot of hydrogen does not assure us of a
strong H absorption. In order for such absorp-
tion to take place, a significant number of atoms
must be in level 2, ready to absorb a photon. If all
the hydrogen is in level 1, you will not see the
Balmer series, no matter how much hydrogen is Fig 3.6. Collisional excitation. In each case, the left frame
present. shows the atoms before the collision and the right frame
shows them after. In each frame, the occupied level is indi-
3.4.1 Excitation cated by a heavier line. (a) To a lower state. After the colli-
sion, atom 1 is in a lower state and atom 2 is moving faster.
In general, the strength of a particular transition
(b) To a higher state. After the collision atom, 1 is in a higher
(emission or absorption) will depend on the num-
state and atom 2 is moving slower.
ber of atoms in the initial state for that transi-

The collisional excitation rates will depend on the atoms are in the ground state, so the ratio is
the kinetic temperature of the gas. The higher zero. As the temperature increases, the quantity
the temperature the faster the atoms are moving. in square brackets gets smaller, so the exponent
For atoms of kinetic temperature Tk the average becomes less negative, and the ratio increases. If
we let Tk go to infinity the ratio of populations
kinetic energy per atom is (3/2)kTk. As the tem-
perature increases more energy is available for approaches the ratio of statistical weights. For a
collisions. This makes higher energy states easier given temperature, increasing the energy separa-
to reach. Also, since the particles are moving tion between the two levels makes the exponent
faster, they spend less time between collisions. more negative, lowering the ratio. This makes
There are more collisions per second. sense, since the greater the energy separation, the
When a gas is in thermodynamic equilibrium harder it is to excite the atom to the higher level.
(which we discussed in the previous chapter), The Boltzmann distribution provides us with a
with a kinetic temperature Tk, the ratios of the convenient reference point, even for a system that
level populations are given by a Boltzmann distri- is not in thermodynamic equilibrium. For any
bution. If we let ni and nj be the populations of lev- given population ratio nj/ni, we can always find
els i and j, respectively, their ratio is given by some value of T to plug into equation (3.9) to make
the equation correct. We call such a temperature
nj gj
31Ej Ei 2>kTk 4 the excitation temperature. When they are not in
e (3.9)
ni gi
equilibrium, each pair of levels can have a differ-
ent excitation temperature. In thermodynamic
In this equation gi and gj are called statistical
equilibrium all excitation temperatures are equal
weights. They are needed because certain energy
to each other and to the kinetic temperature.
levels are actually groupings of sublevels that
have the same energy. The statistical weight of a
3.4.2 Ionization
level is just a count of the number of sublevels in
If we know the temperature in the atmosphere of
that level. Typically, g are small integers.
a star, we can use the Boltzmann equation to pre-
To help us understand the Boltzmann distri-
dict how many atoms will be in each state, i, and
bution, Fig. 3.7 shows how the ratio of popula-
predict the strengths of various spectral lines.
tions for an atom with just two levels depends on
However, there is still an additional effect that we
temperature. When the temperature is zero, all
have not taken into account “ ionization. If the tem-
perature is very high, some of the colliding parti-
cles will have kinetic energies greater than the
ionization energy of the atom, so the electron
Population Ratio

1.5 will be torn away in the collision. Once a hydro-
gen atom is ionized, it can no longer participate
in line emission or absorption.
When the gas is ionized, electrons and posi-
tive ions will sometimes collide and recombine.
When the total rate of ionizations is equal to the
total rate of recombinations, we say that the gas
is in ionization equilibrium. If the gas is in thermal
0 equilibrium and ionization equilibrium, then the
1 — 105 2 — 105
Saha equation tells us the relative abundances of
various ions. We let n(Xr) and n(Xr 1) be the densi-
Temperature (K)
ties of the r and r 1 ionization states, respec-
tively, of element X. (For example, if r 0, then
Fig 3.7. Level populations as a function of temperature for
a two-level system. In this case we have put in energies and we are comparing the neutral species and the
statistical weights (3, 5) for the n 2 and n 3 states of first ionized state.) The ionization energy to go
hydrogen (¬rst Balmer transition).
from r to r 1 is Eion. The electron density is ne,

and the kinetic temperature is Tk. Finally, gr and
Ionization energies (eV).
Table 3.1.
gr 1 are the statistical weights of the ground elec-
Atom Singly ionized Doubly ionized
tronic states of Xr and Xr 1 (assuming that most
of each species is in the ground electronic state).
H 13.6 “
The Saha equation tells us that
He 24.6 54.4
12 2 gr 2 mr kTk 3>2
a be
ne n1X r 3Ei>kTk 4 C 11.3 24.4
n1X r 2
gr 1 N 14.5 29.6
O 13.6 35.1
The Saha equation has the same exponential
Na 5.1 47.3
energy dependence as the Boltzmann distribution.
K 4.3 31.8
However, there is an additional factor of T k . This
Ca 6.1 11.9
comes from the fact that a free electron has more
Fe 7.9 16.2
states available to it at higher Tk than at lower Tk.
In addition there is a factor of ne on the left. This
is because a higher abundance of electrons leads to
so the left side of equation (3.10) simplifies to
a higher rate of recombinations, driving down the
n2>n0, where n0 is the number of neutrals. This
fraction of atoms that are ionized. Just as we did e
extra factor of ne makes even this simpler form of
with the excitation temperature in the Boltzmann
the Saha equation harder to solve for (ne/n0 ) than
equation, we can define an ionization temperature Ti,
the Boltzmann equation is to solve for the ratio of
which makes the Saha equation correct, even if
level populations. In Fig. 3.8, we show the ratio
the gas is not in thermodynamic equilibrium.
ne/n0 as a function of temperature, for a value of
In this equation ne is the number of electrons
ne reasonable for stars like the Sun.
from all sources, since any electron can combine
The ionization energies of some common
with a hydrogen ion (for example) no matter
atoms are given in Table 3.1. This table is useful in
where that electron came from (hydrogen,
deciding which ions you are likely to encounter
helium, etc.) In many situations, virtually all of
at various temperatures. In designating ionized
the ions are hydrogen. That is because hydrogen
atoms, there is a shorthand that has been adopted.
is by far the most abundant element, and because
The roman numeral I is used to designate the
the next most abundant element, helium, is very
neutral species, II the singly ionized species, III
hard to ionize. In that case, the number of elec-
the doubly ionized species, and so on. For exam-
trons is equal to the number of positive ions, n ,
ple, neutral hydrogen is H(I), ionized hydrogen
(H ) is H(II), doubly ionized carbon is C(III).
Ionization Fraction

3.4.3 Intensities of spectral lines
We are now in a position to discuss the intensities
of various absorption lines in stars. We will take
H as an example to see the combined effects of
0.5 excitation and ionization. At low temperatures,
essentially all of hydrogen is neutral, and most of
it is in the ground state. Since little H will be in
the second state, there will be few chances for H
absorption. The H line will be weak.
As we go to moderate temperatures, most of
1 — 104 2 — 104
the hydrogen is still neutral. However, more of
Temperature (K) the hydrogen is in excited states, meaning that a
Fig 3.8. The ratio of electrons to the total number of reasonable amount will be in level 2. H absorp-
hydrogen atoms (neutral plus ion), for an electron density tion is possible. As the temperature increases, the
appropriate to stars like the Sun.
H absorption becomes stronger.

At very high temperatures, the hydrogen
becomes ionized. Since there is less neutral hydro-
gen, the H line becomes weaker. This explains
why the H line is strongest in middle-temperature
stars, and why the original scheme of classifying
by hydrogen line strengths did not produce a
sequence ordered in temperature.
We can apply a similar analysis to other ele-
ments. The details will differ because of different
energy level structures and different ionization
Fig 3.10. The relative strengths of spectral lines from
energies. It should be noted that, after hydrogen
important species as a function of spectral type. Each species
and helium, the abundances of the elements fall
shows the effects of excitation and ionization. For example,
off drastically (see Appendix F for the abundances
the increase in H line strengths from K to A stars occurs
of the elements). In fact, astronomers often refer
because the increasing temperature results in more hydro-
to hydrogen, helium and ˜everything else™. The gen in the n 2 (and higher) levels. However, the higher
˜everything else™ are collectively called metals, temperatures of the B and O stars ionize much of the
even though many of the elements don™t fit our hydrogen and the lines get much weaker.
common definition of a metal.
We now look at the properties of different star has strong ZrO (zirconium oxide) lines
spectral types, in order of increasing tempera- as opposed to TiO lines, we call it an S-type.
ture. Sample spectra are shown in Fig. 3.9, and K Temperatures range from 3500 to 5000 K. There
the behaviors of a few spectral lines are shown in are many lines from neutral metals. The H
Fig. 3.10. lines are stronger than in M stars but most
of the H is still in the ground state.
M Temperatures in M stars are below 3500 K,
G Temperatures in the range 5000“6000 K. The
explaining their red color. The temperature
Sun is a G2 star. The H lines are stronger
is not high enough to produce strong H
than in K stars, as more atoms are in excited
absorption, but some lines from neutral
states. The temperature is high enough for
metals are seen. The stars are cool enough
metals with low ionization energies to be
for simple molecules to form, and many
partially ionized. Two prominent lines are
lines are seen from molecules such as CN
from Ca(II). When Fraunhofer studied the
(cyanogen) and TiO (titanium oxide). If cool
solar spectrum, he gave the strongest lines
stars show strong CH lines, we designate
letter designations. These Ca(II) lines are the
them as C-type or ˜carbon stars™. If any M
H and K lines in his sequence.
F Temperatures range from 6000 to 7500 K. The H
lines are a little stronger than in G stars. The
ionized metal lines are also stronger.
A Temperatures range from 7500 to 10 000 K.
These stars are white“blue in color. They
have the strongest H lines. Lines of ionized
metals are still present.
B Temperatures are in the range 10 000“30 000 K,
and the stars appear blue. The H lines are
beginning to weaken because the tempera-
tures are high enough to ionize a significant
Fig 3.9. Samples of spectra from stars of different spectral
fraction of the hydrogen. The lines of neutral
types.The name of the star appears on the right of each
spectrum, and the spectral type appears on the left. In each and singly ionized helium begin to appear.
spectrum, the wavelength increases from left to right. Hotter Otherwise there are relatively few lines in
stars are at the top. [NOAO/AURA/NSF]
the spectrum.

O Temperatures range from 30 000 to over magnitude. However, if we find a group of stars
60 000 K, and the stars appear blue. The ear- all at the same distance, we can plot their appar-
liest spectral types that have been seen are ent magnitudes, since the distance modulus


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