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10 km/s; observer moving towards source at day of Autumn to the first day of Spring (in
5 km/s. the northern hemisphere).
5 BINARY STARS AND STELLAR MASSES 99



5.17. What is the relationship between Kepler™s sec- temperatures and the range of
ond law and angular momentum? luminosities?
*5.18. As an object approaches periastron in an 5.21. What do we mean when we talk about a
elliptical orbit, it is moving faster and faster. selection effect?
This means that there must be a force in the 5.22. If high mass stars are more massive than low
direction of motion (in addition to the force mass stars, how can most of the mass in our
that is perpendicular to the direction of Galaxy be in the form of low mass stars? If
motion, which is responsible for the curving most of the mass is in low mass stars, how
of the path). What is the source of the force can most of the luminosity come from high
that makes the object go faster? Use a dia- mass stars?
gram, showing the forces and components, to 5.23. When we study stellar structure, in Chapter
illustrate your answer. 9, we will see that once we know the mass of
5.19. At apastron the objects in elliptical orbits are a main sequence star all of its other proper-
moving slower than at periastron, so their ties (size, temperature, luminosity) are deter-
kinetic energy must be lower. What happens mined. How does that show up in the obser-
to the ˜missing™ energy? vational results of this chapter?
5.20. How does the range of masses for main 5.24. Why is it difficult to measure stellar sizes?
sequence stars compare with the range of



Problems
5.7. You are standing a distance d from a railroad
5.1. A police radar, operating at a frequency of
track. A train comes past you at a constant
5 GHz, detects a Doppler shift of 2.5 MHz as
speed v, passing you at time t 0. (a) What is
you approach. How fast are you going?
the radial velocity of the train as a function
5.2. (a) A source is moving away from an observer
of time? (b) Draw a graph of your result for
with a speed of 10 km/s, along a path that
both positive and negative times.
makes a 30 angle with the line of sight. At
5.8. By how much can the H line in some object
what wavelength will the H line be
be shifted as a result of the Earth™s motion
observed? (b) If the observer is also moving
around the Sun?
directly away from the source at a speed of
5.9. Show that, if we had taken the force on the
20 km/s, at what wavelength will the H line
second star in equation (5.13), we would still
be observed?
have obtained the result in equation (5.20).
5.3. Some source is moving away from an observer
5.10. In deriving the law of gravitation, Newton
along a line making a 30 angle with the line
actually started with Kepler™s third law as
of sight at a speed of 80 km/s. The observer is
being observationally given and worked
moving towards the source along a line mak-
backwards (from the derivation in this chap-
ing a 20 angle with the line of sight, at a
ter). Show how that derivation would be
speed of 10 km/s. (a) What is the radial veloc-
done.
ity? (b) By how much is the H line shifted?
5.11. For a binary system with stars of equal masses
(c) How would the answer be changed if the
m, in circular orbits, with a total separation r,
observer were moving in the opposite
what is the orbital speed v?
direction?
5.12. For a binary system with stars of masses 5
5.4. We observe a source with a radial velocity of
and 10 M , in circular orbits with a period of
25 km/s. By how much is the separation
3 yr, what is the total energy of the system?
between the H and H lines changed?
5.13. For a binary system with stars of masses m1
5.5. If the H line in a source is shifted by
and m2, in circular orbits, with a total separa-
0.01 nm, by how much is the H line shifted?
tion r, find an expression for the ratios of the
5.6. If the H line in a source is shifted by 0.10 nm,
kinetic energies of the two stars.
what is the radial velocity of the source?
100 PART I PROPERTIES OF ORDINARY STARS



5.14. Let M m1 m2 , and x m1 m2. (a) Find of the system is 20 yr. What are the masses of
expressions for m1 and m2 in terms of M and the two stars (a) if i 90 (b) if i 30 ?
x. (b) What is the significance of your result? 5.20. An astrometric binary is 10 pc from the Sun.
5.15. We observe a binary system in which the two The visible star orbits 2 arc sec from the cen-
stars are 1 and 2 arc sec, respectively, from ter of mass with a period of 30 yr. What is the
the center of mass. The system is 10 pc from mass of the unseen companion?
us. The period is 33 yr. What are the masses 5.21. An eclipsing binary system has a period of
of the two stars, assuming that i 90 ? days. One star has a Doppler shift of 100 km/s.
5.16. Suppose we can measure the positions of What is the mass of the companion?
stars to 0.01 arc sec. How far away could we 5.22. Derive equations (5.40a) and (5.40b).
detect an astrometric binary where the sepa- *5.23.A star moves in an elliptical orbit of eccentric-
ity e. The plane of the orbit makes an angle i
ration is 100 AU?
5.17. A star in a circular orbit has a speed of with the plane of the sky. The orbit is ori-
30 km/s. The period is 10 yr. The star is 2 arc ented so that the line joining the foci is per-
sec from the center of mass. How far away is pendicular to the line formed by the intersec-
this star from us? tion of the plane of the orbit and the plane of
5.18. Derive a form of equation (5.23) that relates the sky. Show that the projected orbit is an
mass, in solar masses, period, in years, and ellipse.
velocity, in kilometers per second. 5.24. What is the luminosity (in solar luminosities)
of (a) a 0.5 M and (b) a 5.0 M star?
5.19. The H lines from two stars in a binary sys-
tem are observed to have Doppler shifts of 5.25. For an elliptical orbit, calculate the angular
momentum L in terms of G, m1, m2, a and e.
0.022 and 0.044 nm, respectively. The period


Computer problems

5.1. Plot ellipses with the same major axes, but with (a) Draw a graph showing the relative probabilities
of finding values of sin3i in various small ranges
eccentricities from 0.1 to 1.0 in steps of 0.1.
5.2. Consider a binary system with stars of masses 5 and from zero to one. (b) The average value of some
10 M in elliptical orbits, with eccentricity e 0.8. function f(x) over the interval 0 to L is
The period is 8 yr. Assume that the more massive
a b f 1x2dx
1L
star is at periastron at angle 0. (a) Plot the speed f1x2
L0
of the more massive star as a function of position in
the orbit, . (b) If this system is viewed from along What is the average value of sin3i over the angle
the major axis, such that the more massive star is range 0 to /2 radians? (c) What does this tell you
closer to the observer when it is at periastron, plot about the lower limits on masses and the actual
the radial velocity vr as a function of . *(c) Plot vr as masses of binary systems?
a function of time t, taking t 0 when 0. 5.5. For stars in the middle range of each spectral type
5.3. For the system in the previous problem, draw the (05, B5, etc.), calculate the average density, and
orbit and draw an arrow showing the force on the express it as a ratio to that for the Sun.
more massive star at positions every 45 around 5.6. For stars in the middle range of each spectral type
the orbit. (05, B5, etc.), calculate the luminosity, and express
5.4. Suppose that many binary star systems, with ran- it as a ratio to that for the Sun.
domly distributed inclinations, are observed.
Chapter 6




The Sun: a typical star

The Sun (Fig. 6.1) is the only star we can study in structure from theories of stellar structure
any detail. It therefore serves as a guide to our pic- (which we will discuss in Chapter 9). The basic
tures of other stars. Any theory of stellar struc- structure of the Sun is shown in Fig. 6.2. The cen-
ter is the core. It is the source of the Sun™s energy.
ture must first be able to explain the Sun before
explaining other stars. As we have seen, the Sun™s Its radius is about 10% of the full solar radius. The
spectral type places it in the mid-range of main outermost layers form the atmosphere. We divide
sequence stars. If we understand the Sun, we have the atmosphere into three parts. Most of the light
we see comes from the photosphere, the bottom
the hope of being able to understand a signifi-
cant number of other types of stars. layer of the atmosphere. Above the photosphere
is the chromosphere, named because it is the source
of red light seen briefly during total eclipses of
6.1 Basic structure the Sun. The chromosphere is about 104 km thick.
The outermost layer is the corona, which extends
We have already seen that the mass of the Sun is far into space. It is very faint, and, for most of us,
2.0 1033 g and that its radius is 7 1010 cm (7 can only be seen during total solar eclipses.
105 km). Its average density is Beyond the corona, we have the solar wind, not
strictly part of the Sun, but a stream of particles
M
14 >3 2R3
from the Sun into interplanetary space.

1033 g
2 6.2 Elements of radiation
14 >3 2 17 10 3
10 cm2
transport theory
1.4 g>cm3

For comparison, the density of water is 1 g/cm3. Radiation is being emitted and absorbed in the
The Sun is composed mostly of hydrogen (94% by Sun in all layers. However, we see radiation
the number of atoms), with some helium (6% by mostly from the surface. Most radiation from
number of atoms), and only 0.1% other elements. below is absorbed before it reaches the surface.
The abundances of the elements are given in To understand what we are seeing when we look
Appendix G. Our best measurement of the effective at the Sun, we need to understand about the
temperature of the Sun is 5762 K. The solar lumi- interaction between radiation and matter. For
nosity is 3.8 1033 erg/s. (The effective temperature example, much of what we know about the solar
is the temperature that we use in the Stefan“ atmosphere comes from studying spectral lines
Boltzmann law to give the solar luminosity.) as well as the continuum. In studying how radi-
When we look at the Sun, we see only the out- ation interacts with matter, known as radiation
ermost layers. We have to deduce the internal transport theory, we see how to use spectral lines
102 PART I PROPERTIES OF ORDINARY STARS




Corona


Convection
Zone
Photosphere


Chromosphere
Core




Fig 6.2. Basic structure of the Sun.


Fig 6.1. The Sun. [NOAO/AURA/NSF]
N (6.2)
tot

nAl (6.3)
to extract detailed information about the solar tot

atmosphere. In making this definition, we have assumed
We first look at the absorption of radiation by that the incoming beam can “see” all the spheres.
atoms in matter. We can think of the atoms as act- No sphere blocks or shadows another. We are
ing like small spheres, each of radius r (Fig. 6.3). assured of little shadowing if the spheres occupy
Each sphere absorbs any radiation that strikes it. a small fraction of the area, as viewed from the
To any beam of radiation, a sphere looks like a cir- end. That is
cle of projected area r2. If the beam is within that
1tot2 VA (6.4)
circle, it will strike the sphere and be absorbed. We
say that the cross section for striking a sphere is Under these conditions, the fraction of the
r2. The concept of a cross section carries over into incoming radiation that will be absorbed, f, is
quantum mechanics. Instead of the actual size of just that fraction of the total area A that is cov-
an atom, we use the effective area over which some ered by the spheres. That is
process (such as absorption) takes place. So then r
1tot2
f A (6.5)
would be how close the photon would have to be to
the atom in order to be absorbed. Using equation (6.3), this becomes
We consider a cylinder of these spheres, with
f nl (6.6)
the radiation entering the cylinder from one end.
We would like to know how much radiation is
absorbed, and how much passes through to the
r
far end. We let n be the number of spheres per
unit volume. The cylinder has length l and area A,
so the volume is Al. The number of spheres in the A
cylinder is
N n Al (6.1)
l
We define the total cross section of all the
Fig 6.3. Absorption of radiation. Radiation enters from the
spheres as the number of spheres multiplied by
left. Any beam striking a sphere is absorbed.
the cross section per sphere:
6 THE SUN: A TYPICAL STAR 103



We define the optical depth to be this quantity: In terms of these quantities, the optical depth is
given by
nl (6.7)
l
Our requirement in equation (6.4) reduces to (6.12a)
L
V 1. Under this restriction, the optical depth of
any section of material is simply the fraction of l (6.12b)
incoming radiation that is absorbed when the
In the above discussion, we required that the
radiation passes through that material. (For
optical depth be much less than unity. Our inter-
example, if the optical depth is 0.01, then 1% of
pretation of as the fraction of radiation
the incoming radiation is absorbed.)
absorbed only holds for V 1. What if that is not
In general, will be a function of wavelength.
the case? We then have to divide our cylinder into
For example, we know that at a wavelength corre-
several layers. If we make the layers thin enough,
sponding to a spectral line, a particular atom will
we can be assured that the optical depth for each
have a very large cross section for absorption. At a
layer will be very small. We then follow the radi-
wavelength not corresponding to a spectral line,
ation through, layer by layer, looking at the frac-
the cross section will be very small. To remind us
tion absorbed in each layer (Fig. 6.4).
that is a function of (or ), we write it as (or
). This means that the optical depth is also a
function of (or ), so we rewrite equation (6.7) as
nl (6.8)
In our discussions, the quantity nl occurs
often. It is the product of a number density and a
length, so its units are measured in number per
I1 I2
unit area. It is the number of particles along the
...
I0 In
full length, l, of the cylinder per unit surface
area. For example, if we are measuring lengths in
centimeters, it is the number of particles in a col-
umn whose face surface area is 1 cm2, and whose
length is l, the full length of the cylinder. We call
d„ d„ d„ d„ d„
this quantity the column density.
We can see that the optical depth depends on
the properties of the material “ e.g. cross section
and density of particles “ and on the overall size
of the absorbing region. It is sometimes conven-
ient to separate these two dependencies by defin-
ing the absorption coefficient, which is the optical
depth per unit length through the material,

Iin
(6.9a) Iout
l
n (6.9b)

If gives the number of absorptions per unit
d„
length, then its inverse gives the mean distance
between absorptions. This quantity is called the
mean free path, and is given by
Fig 6.4. Radiation passing through several layers. Each layer
L 1 (6.10) has an optical depth d .The bottom of the ¬gure is for
calculating the effect of each layer.
1n (6.11)
104 PART I PROPERTIES OF ORDINARY STARS



1
Let™s look at the radiation passing through
some layer with optical depth d . Since d V 1,
it is the fraction of this radiation that is 0.8
absorbed. The amount of radiation absorbed in
this layer is I d . The amount of radiation passing
0.6
through to the next layer is I(1 d ). The change
e’„
in intensity, dI, while passing through the layer,
is 0.4
Iout Iin
dI (6.13)
0.2
Id (6.14)

Notice that dI is negative, since the intensity
0
is decreased in passing through the layer. 2 3 4 5
0 1

Now that we know how to treat each layer, we
must add up the effect of all the layers to find the Fig 6.5. e” vs. , showing the fall-off in transmitted
effect on the whole sample of material. We can radiation as the optical depth increases. Note that the curve
see that we have formulated the problem so that looks almost linear for small . For large , it approaches
we are following I as a function of . We let be zero asymptotically.
the optical depth through which the radiation
has passed by the time it reaches a particular
ex 1 x, for x V 1. In this case, equation (6.18)
layer, and I be the intensity reaching that layer.
becomes
Then ranges from zero, at the point where the
radiation enters the material, to , the full optical I I0 (1 ) (6.19)
depth where the radiation leaves the material.
This is the expected result for small optical
Over that range, I varies from I0, the incident
depths, where again becomes the fraction of
intensity, to I, the final intensity. Using equation
radiation absorbed.
(6.14),
As shown in Fig. 6.5, e” falls off very quickly
with . This means that to escape from the Sun,
dI /I ”d (6.15)
radiation must come from within approximately
In this form, all of the I dependence is on the
one optical depth of the surface. This explains
left, and the dependence is on the right. To add
why we only see the outermost layers. Since the
up the effect of the layers, we integrate equation
absorption coefficient is a function of wave-
(6.15) between the limits given above:
length, we can see to different depths at different
wavelengths. At a wavelength where is large,
I dI¿
d¿
we don™t see very far into the material. At wave-
I
I0 0
lengths where is small, it takes a lot of mate-
ln(I) ” ln(I0) (6.16)
rial to make 1. We take advantage of this to
study conditions at different depths below the
Using the fact that ln(a/b) ln(a) ln(b), this
surface.
becomes
So far we have only looked at the absorption
ln(I/I0 ) (6.17)
of radiation passing through each layer. However,
radiation can also be emitted in each layer, and
Raising e to the value on each side, remembering
that eln x x, and multiplying both sides by I0 the amount of emission also depends on the
optical depth. In general, we must carry out
gives
complicated radiative transfer calculations to
I0 e”
I (6.18)
take all effects into account. To solve these prob-
lems, we use powerful computers to make math-
We can check this result in the limit V 1,
ematical models of stellar atmospheres. In these
called the optically thin limit, using the fact that
6 THE SUN: A TYPICAL STAR 105



calculations, we input the distribution of tem- there are a large number of free electrons to col-
perature, density and composition and predict lide with those atoms.
the spectrum that we will see, including emis- If we have an H ion and a photon ( ) strikes
sion and absorption lines. We vary the input it, the photon can be absorbed, and the electron
parameters until we find models that produce set free:
predictions that agree with the observations.
’H
H e
The results of these calculations are not unique,
The H ion is a bound system. The final state
but they do give us a feel for what processes are
has an H atom and a free electron. We call this
important in stellar atmospheres. The more
process a bound“free process. In such a process,
observational data we can predict with the mod-
the wavelength of the incoming photon is not
els, the more confident we can be that the tem-
restricted, as long as the photon has enough
peratures, densities and compositions we derive
energy to remove the electron. The electron in
are close to the actual ones.
the final state can have any kinetic energy, so a
continuous range of photon energies is possible.
6.3 The photosphere This process then provides most of the contin-
uum opacity of the photosphere. The continuum
emission results from the inverse process.
Most of the visible photons we receive from the
Sun originate in the photosphere. One question
6.3.1 Appearance of the photosphere
you might ask is why we see continuum radiation
We have said that the Sun is the one star that we
at all. We have already seen how atoms can emit
can study in great detail. To do this, we try to
or absorb energy at particular wavelengths, pro-
observe the photosphere with the best resolution
ducing spectral lines. However, we have not dis-
possible. When we observe the Sun, the light-
cussed the source of the emission and absorption
gathering power of our telescope is not usually a
of the continuum. It turns out that the contin-
problem. Therefore, we can try to spread the
uum opacity in the Sun at optical wavelengths
image out over as large an area as possible, mak-
comes from the presence of H ions. An H ion is
ing it easier to see detailed structure. We there-
an H atom to which an extra electron has been
fore want a telescope with a long focal length to
added. As you might guess, this extra electron is
give us a large image scale. The solar telescope
held only very weakly to the atom. Very little
shown in Fig. 6.6 provides this type of detailed
energy is required to remove it again. H ions are
picture.
present because there is so much hydrogen, and

Fig 6.6. Solar telescope on Kitt
Peak (operated by NOAO).The
telescope has a very long focal
length, so that two can produce a
large image and study the detailed
appearance. Since the tube is so
long, it is not reasonable to move
it. Instead, the large mirror at the
top (the objective) is moved to
keep the sunlight directed down
the tube. [NOAO/AURA/NSF]
106 PART I PROPERTIES OF ORDINARY STARS




Bright
Bright
25




Dark
Dark
20
[x 1000 km]




15


10


5

0 Fig 6.8. Granulation and convection zone.This is a side
0 5 10 15 20 25 view to show what is happening below the surface. Hotter
[x 1000 km] gas is being brought up from below, producing the bright
regions.The cooler gas, which produces the darker regions,
Fig 6.7. Granulation in the Sun. Remember, the darker is carried down to replace the gas that was brought up.
areas are not really dark.They are only a little cooler than
the bright areas. [NOAO/AURA/NSF]
from sound waves in the upper layers of the con-
vection zone. This type of oscillation is one of many
that are studied for clues to the Sun™s interior
When we look on a scale of a few arc seconds,
structure. This area of research is called solar seis-
we see that the surface of the Sun does not have
mology. (On the Earth, seismologists study motions
a smooth appearance (Fig. 6.7). We see a struc-
near the surface to learn about the interior.)
ture, called granulation, in which lighter areas are
One interesting question about the photos-
surrounded by darker areas. The darker areas are
phere concerns the sharpness of the solar limb. The
not really dark. They are just a little cooler than
Sun is a ball of gas whose density falls off contin-
the lighter areas, and only appear dark in compari-
uously as one moves farther from the center.
son to the light areas. The granules are typically
There is no sharp boundary (like the surface of
about 1000 km across. The pattern of granulation
the Earth), yet we see a definite edge on the Sun.
also changes with time, with a new pattern appear-
In Fig. 6.9, we see some lines of sight through the
ing every 5 to 10 min.
We interpret this granulation as telling us
Fig 6.9. Lines of sight through the
about the underlying structure we cannot see
solar limb. For clarity, we think of the
directly. The granulation can be explained by cir-
Sun as being composed of a series of
culating cells of material, called convection zones
spherical shells.The density in each shell
(Fig. 6.8). (Convection is the form of energy trans-
decreases as we move farther from the
port in which matter actually moves from one center.This decreasing density is
place to another. Strong convection on the Earth indicated by the shading; two lines of
is responsible for the updrafts that produce thun- sight are indicated. Note that most of
derstorms. A pot of boiling water also has energy each line of sight is in the densest layer
through which that particular line passes.
transport by convection.) The brighter regions are
Even though line 2 is not shifted very far
warmer gas rising up from below. The dark
from line 1, line 1 passes through much
regions are cooler gas falling back down.
more material.
In addition to the granulation variations,
there is also a variation called the five minute oscil-
lation, in which parts of the photosphere are mov-
12
ing up and down. We think this convection results
6 THE SUN: A TYPICAL STAR 107



photosphere. As the line of sight passes farther the center of the Sun. In each line of sight, we can
from the center of the Sun, the opacity decreases only see down to 1. Line of sight 1 is looking
because (1) the path length through the Sun is straight down into the atmosphere, so it gets
less, and (2) it passes through less dense regions. closer to the center of the Sun before an optical
Since the amount of light getting through is pro- depth of unity is reached than does line of sight
portional to e , the effect of changing from 2, which has a longer path through any layer. We
line of sight to line of sight is enhanced by the see deeper into the Sun on line of sight 1 than we
exponential behavior. Therefore the transition do on line of sight 2. If the temperature decreases
from the Sun being mostly opaque to being with increasing height in the photosphere, we
mostly transparent takes place over a region that are seeing hotter material on line of sight 1 than
is small compared with our resolution, and the on 2, so line of sight 1 appears brighter than line
edge looks sharp. of sight 2. Since line of sight 1 takes us the deep-
est into the Sun, we define the point at which
6.3.2 Temperature distribution reaches unity on this line as being the base of the
photosphere. When we talk about the temperature
Another interesting phenomenon near the solar
limb can be seen in the photograph in Fig. 6.1. of the Sun, we are talking about the temperature
The Sun does not appear as bright near the limb at the base of the photosphere.
as near the center. This limb darkening is also an When we look at the Sun, it appears brighter
optical depth effect, as shown in Fig. 6.10. We along line of sight 1 than it does along line of
compare two lines of sight: (1) toward the center sight 2. This means that the Sun is hotter at the
of the Sun from the observer, and (2) offset from end of 1 than at the end of 2. From this, we con-
clude that the photosphere cools as one moves
farther from the center of the Sun. If the photos-
phere became hotter as one moves farther from
the center, we would see limb brightening.
We obtain more useful information about the
Base of
photosphere by studying its spectral lines (Fig. 6.11).
Photosphere
The spectrum shows a few strong absorption lines
and a myriad of weaker ones. The stronger lines
were labeled A through K by Fraunhofer in 1814.
These lines have since been identified. For exam-
„=1 ple, the C line is the first Balmer line (H ); the D
line is a pair of lines belonging to neutral sodium
„=1 (NaI); and the H and K lines belong to singly ion-
ized calcium (CaII). Sodium and calcium are much



1 2
Fig 6.10. Limb darkening. Line of sight 1 is directed from
the observer toward the center of the Sun, so it takes the
shortest path through the atmosphere.This line allows us to
see the deepest into the photosphere, and the base of the
photosphere is de¬ned to be where the optical depth
along this line reaches unity. Line of sight 2 is closer to the
edge, so it doesn™t allow us to see as deep into the photos-
phere. If the temperature decreases with increasing height in
the photosphere, then line of sight 1 allows us to see hotter
material than does line of sight 2, and the edge of the Sun
Fig 6.11. The solar spectrum. [NOAO/AURA/NSF]
appears darker than the center.
108 PART I PROPERTIES OF ORDINARY STARS



less abundant than hydrogen but their absorption
lines are as strong as H . We have already seen in
Chapter 3 that this can result from the combined
effects of excitation and ionization.

6.3.3 Doppler broadening of spectral lines
P(x)
When we study the lines with good spectral reso-
lution, we can look at the details of the line pro-
file. The lines are broadened by Doppler shifts due
to the random motions of the atoms and ions in
the gas (Figs. 6.12 and 6.13). If all the atoms were at
rest, all the photons from a given transition would
emerge with a very small spread in wavelength. x
However, the atoms are moving with random
Fig 6.13. Line pro¬le.We plot intensity as a function of
speeds in random directions. We therefore see a wavelength.
spread in the Doppler shifts, and the line is broad-
ened. This process is called Doppler broadening. If
containing the particles have some overall motion
the gas is hotter, the spread in speeds is greater,
that just shifts the center wavelength of the line,
and thus the Doppler broadening is also greater. If
the broadening would still be the same.
in addition to these random motions all the objects
We can estimate the broadening as a function
of temperature. If v 2 is the average of the square
of the random velocities in a gas, and m is the
2 mass per particle, the average kinetic energy per
3
4
particle is (1/2)m v 2 . If we have an ideal
1
monatomic gas, this should equal (3 2)kT, giving
5
(1/2)m v 2 (3/2)kT (6.20)
7
6
11 Solving for v 2 gives
8
12
9
v2 3kT/m (6.21)
10
Taking the square root gives the root mean square
(rms) speed
3kT 1>2
a b
vrms (6.22)
m
This gives us an estimate of the range of speeds
we will encounter. (The range will be larger, since
we can have atoms coming toward us or away
62 4 9 11 12 8 7 53 1
10 from us, and since some atoms will be moving
faster than this average, but the radial velocity is
Radial Velocity
reduced since only the component of motion
Fig 6.12. Doppler broadening.The top shows the (ran- along the line of sight contributes to the Doppler
dom) motions of a group of particles.The purple vectors are
broadening.) To find the actual wavelength range
the actual velocities; the green vectors are the radial com-
over which the line is spread out, we would use
ponents, which produce the Doppler shift. For each particle,
the Doppler shift expression in equation (5.4).
identi¬ed by a number, the radial velocities are plotted
below.The line pro¬le is the sum of all the individual Example 6.1 Doppler broadening
Doppler shifted signals, and with many more particles it Estimate the wavelength broadening in the H line
would have a smooth appearance.
in a gas composed of hydrogen atoms at T 5500 K.
6 THE SUN: A TYPICAL STAR 109



including the relative strengths of various lines
SOLUTION
and the details of certain line profiles.
Using equation (6.22) gives

13 2 11.38 erg>K2 15.5
From observations of various spectral lines, the
16
103 K2 1>2
c d
10
temperature profile for the photosphere has been
vrms 24
1.7 10 g derived. It is shown in Fig. 6.14. Note that the tem-
106 cm/s perature falls as one goes up from the base of the
1.2
photosphere. This is what we might expect, since
From equation (5.4), we estimate the linewidth as
we are moving farther from the heating source.
However, an interesting phenomenon is observed.
a b
vrms
¢
The temperature reaches a minimum at 500 km
c
above the base of the photosphere, and then begins
1656.28 nm2 11.2 106 cm s2 12
a b to rise with altitude. We will see below that this
13 1010 cm>s2 temperature rise continues into the higher layers.
0.03 nm
6.4 The chromosphere
In any spectral line, smaller Doppler shifts
relative to the line center are more likely than
larger ones. Therefore, the optical depth is great- At most wavelengths the chromosphere is optically
est in the line center, and falls off to either side. thin, so we can see right through it to the photos-
At different Doppler shifts away from the line phere. Under normal conditions the continuum
center, we will see different distances into the radiation from the photosphere overwhelms that
Sun. The farther we are from the line center, the from the chromosphere. However, during a total
deeper we see. By studying line profiles in detail eclipse of the Sun, just before and after totality, the
we learn about physical conditions at different Moon blocks the light from the photosphere, but
depths. Also, each spectral line has a different not from the chromosphere. For that brief
optical depth in the line center, so different lines moment, we see the red glow of the chromosphere.
allow us to see down to different depths in the The red glow comes from H emission. The optical
photosphere. When we perform model stellar depth of the H line is sufficiently large that we
atmosphere calculations, we try to predict as can study the chromosphere by studying that line.
many of the features of the spectrum as possible, At the center of the H line we see down only to
1500 km above the base of the photosphere.
800 We can also study the large scale structure of
the chromosphere by taking photographs through
filters that only pass light from one line. One such
photograph is shown in Fig. 6.15. In this picture
Height above base (km)




600
we see granulation on an even larger scale than
in the photosphere. This is called supergranulation.
The supergranules are some 30 000 km across. As
400
with the granules in the photosphere, the matter
in the center of the supergranules is moving up
and the matter at the edges is moving down.
200 These motions can be determined from Doppler
shifts of spectral lines. We also see smaller scale
irregularities in the chromosphere, called spicules.
These are protrusions from the surface some
0
700 km across and 7000 km high. An H image of
4000 5000 6000
the chromosphere is shown in Fig. 6.16.
T(K )
When the chromosphere is visible just before
Fig 6.14. Temperature vs. height in the photosphere.
and after totality, we see only a thin sliver of light.
110 PART I PROPERTIES OF ORDINARY STARS



spectrum shows emission lines. This is because
there is no strong continuum to be absorbed.
When we study the spectra of the chromos-
phere we find that it is hotter than the photos-
phere. The chromospheric temperature is about
15 000 K. (The Sun doesn™t appear this hot
because the chromosphere is optically thin and
doesn™t contribute much to the total radiation we
see.) We are faced with trying to explain how the
temperature rises as we move farther from the
center of the Sun. We will discuss this point in
the next section, when we discuss the corona.


6.5 The corona
6.5.1 Parts of the corona
The corona is most apparent during total solar
Fig 6.15. This spectroheliogram shows large scale motions
at the solar surface based on the H emission. eclipses (Fig. 6.17), when the much brighter light
[NOAO/AURA/NSF]


The effect is the same as if the light had been
passed through a curved slit. If we then use a prism
or grating to spread out the different wavelengths,
we will obtain a line spectrum, though each line
will appear curved. This spectrum is called a flash
spectrum because it is only visible for the brief
instant that the Moon is covering all of the photos-
phere but not the chromosphere. Note that the




(a)




(b)
Fig 6.17. Two views of total solar eclipses, showing the
Fig 6.16. H image showing the chromosphere. [NASA] corona. [(a) NOAO/AURA/NSF; (b) NASA]
6 THE SUN: A TYPICAL STAR 111



were in the same plane as the Earth™s orbit
from the photosphere and chromosphere is
around the Sun. However, the Moon™s orbit is
blocked out. The corona is simply too faint to be
inclined by about 5 , so total eclipses of the Sun
seen when any photospheric light is present. You
are rare events. The average time between total
might think that we can simulate the effect of an
solar eclipses is about one and a half years. Even
eclipse by holding a disk over the Sun. If you try
when one occurs, the total eclipse is observable
this, light that would come directly from the pho-
from a band not more than 300 km wide, and
tosphere to your eye will be blocked out. However,
totality lasts only a few minutes.
some photospheric light that is originally headed
Solar astronomers take advantage of eclipses
in a direction other than directly at you will scat-
whenever possible, but also look for other ways to
ter off the atoms and molecules in the Earth™s
study the corona. It turns out that some ground-
atmosphere, and reach you anyway (Fig. 6.18).
based non-eclipse observations are possible.
This scattered light is only a small fraction of the
Telescopes, called coronagraphs, have disks to
total photospheric light, but it is still enough to
block out the photospheric light, and reduce as
overwhelm the faint corona. This is not a prob-
much scattered light as possible. They are placed
lem during solar eclipses, because the Moon is
at high altitude sites, with very clear skies. For
outside the atmosphere, so there is nothing to
example, the Haleakala Crater on the island of
scatter light around it.
Maui (operated by the University of Hawaii) is at
Therefore, solar eclipses still provide us with
an altitude of 3000 m (10 000 ft).
unique opportunities to study the corona. For this
One way to get around the scattering in the
purpose, we are fortunate that the Moon subtends
Earth™s atmosphere is to put a coronagraph in
almost the same angle as the Sun, as viewed from
space. This is not quite as good as a solar eclipse,
the Earth. The Moon can exactly block the photo-
since space probes are not totally free of escaping
sphere and chromosphere, but not the corona.
gases. Some photospheric light is scattered by
Unfortunately, we do not have an eclipse of the
these gases. However, the results are much better
Sun every month. We would if the Moon™s orbit
than for a ground-based telescope. They also have
an advantage over eclipse studies in that they
allow for continuous study of the corona. For
example, the Orbiting Solar Observatory 7 (OSO
7) provided observations of the corona for the
period 1971“4.
There is an additional technique for studying
the corona. It involves studying radio waves.
Radio waves pass through the Earth™s atmosphere
and are not appreciably scattered. We therefore
Direct don™t have to worry about the radio waves behav-
Scattered
Light ing like the visible light in Fig. 6.18.
Light
In discussing what we have learned so far about
the corona, we divide it into three parts:
Earth's
Occulting
Atmosphere
Disk
(1) The E-corona is a source of emission lines
Observer directly from material in the corona. These
lines come from highly ionized species, such
Fig 6.18. Effects of scattered light on corona studies.We
as Fe XIV (13 times ionize iron). If we look at
can use an occulting disk, but sunlight can scatter off
the Saha equation, discussed in Chapter 3, we
particles in the Earth™s atmosphere and reach our telescope.
see that highly ionized states are favored
Since the corona is very faint, and the Earth™s atmosphere is
very ef¬cient at scattering, especially for blue light, this under conditions of high temperature and
scattered light can overwhelm the direct light from the low density. The high temperature provides
corona.
the energy necessary for the ionization.
112 PART I PROPERTIES OF ORDINARY STARS



portional to 2. Of course the amount of gas to be
The low density means that collisions leading
to recombinations are rare. cooled is proportional to . We can define a cool-
(2) The K-corona (from the German Kontinuierlich, ing time which is the ratio of the gas volume to be
for continuous) is the result of photospheric cooled (which is proportional to ) to the gas cool-
ing rate (which is proportional to 2). The cooling
light scattered from electrons in the corona.
(3) The F-corona (for Fraunhofer) is not really part time is then proportional to 1/ . Therefore, low
of the Sun. It comes from photospheric light density gases take longer to get rid of their stored
scattered by interplanetary dust. Since we heat.
are just seeing reflected photospheric light, Further, if the density is very low, a very high
the light still has the Fraunhofer spectrum. temperature doesn™t necessarily mean a lot of
Both the F- and K-coronas appear at approxi- stored energy. The energy is (3/2)kT per particle.
mately the same angular distance from the Even if the quantity kT is very large, if the total
center of the Sun, but there are experimental number of particles is very small, the energy
ways of separating their contributions to the stored is not as large as if the density were much
light we see. higher.
Example 6.2 Energy density in the corona and
6.5.2 Temperature of the corona
the Earth™s atmosphere
When we analyze the abundance of highly ion-
Compare the energy density (energy per unit
ized states, the Doppler broadening of lines and
volume) in the corona with that in the Earth™s
the strength of the radio emission, we find that
atmosphere.
the corona is very hot, about 2 106 K. As we have
stated, the density is very low, approximately 10 9
SOLUTION
times the density of the Earth™s atmosphere.
For each, the energy density is proportional to the
Again, we must explain why a part of the
density of particles n and the temperature T. (All
atmosphere farther from the center of the Sun is
other constants will drop out when we take the
hotter than a part closer to the center. We should
ratio.) Therefore
note that it is not necessarily hard to keep some-
energy density 1corona2 n1corona 2 T1corona 2
c dc d
thing hot if it can™t lose heat efficiently. For exam-
energy density 1Earth 2 n1Earth 2 T1Earth 2
ple, in a well insulated oven, once the required
temperature has been reached, the heat source can
106
31 4c d
2
9
be turned off and the temperature of the oven still 10
102
3
stays high. An ionized gas (plasma) like the corona
6
can lose energy only through collisions between 7 10
the particles. For example, an electron and an ion
Even though the corona is so hot, its very low den-
could collide, with some of the kinetic energy
sity gives it a lower energy density.
going to excite the ion to a higher state. The ion
We must still come up with some explanation
can then emit a photon and return to its lower
for getting energy into the corona. Some have
energy state. The emitted photon escapes and its
suggested mechanisms in which oscillations near
energy is lost to the gas. If the lost energy is not
the surface of the Sun send supersonic sound
replaced, the gas will cool.
waves (shock waves) into the Sun™s upper atmos-
Since collisions play an important role in the
phere. In addition, there are mechanisms for
above process, the rate at which the gas can lose
heating that involve the Sun™s magnetic field.
energy will depend on the rate at which colli-
These theories are still under study, and we still
sions occur. The rate of collisions should be pro-
do not have a definitive picture of the energy bal-
portional to the product of the densities of the
ance in the corona.
two colliding species. However, the density of
When we study photographs of the corona,
each species is roughly proportional to the total
we find its structure is very irregular. We often
gas density (each being some fraction of the
see long streamers whose appearance varies with
total). This means that the collision rate is pro-
6 THE SUN: A TYPICAL STAR 113



time. We think that these phenomena are related dark. They are just not as bright as the surround-
to the Sun™s magnetic field, as are other aspects ing areas. The gas in these darker areas is probably
of solar activity (to be discussed in the following at a temperature of about 3800 K. A closeup of
these sunspots shows that they have a darker inner
section).
region, the umbra, surrounded by a lighter region,
the penumbra.
6.6 Solar activity The number of sunspots on the Sun is not
even approximately constant. It varies in a regu-
6.6.1 Sunspots lar way, as shown in Fig. 6.20. It was realized in
When we look at photographs of the Sun (Figs. 6.1
the mid-19th century that sunspot numbers fol-
and 6.19) we note a pattern of darker areas. As
low an 11-year cycle. The number of sunspots in a
with the granulation, the dark areas are not really
peak year is not the same as in another peak year.
However, the peaks are easily noticeable.
We see more regularity when we plot the
height of each sunspot group above or below the
Sun™s equator as a function of when in the
sunspot cycle they appear. This was done in 1904
by E. Walter Maunder. An example of such a dia-
gram is shown in Fig. 6.21. Early in an 11-year
cycle, sunspots appear far from the Sun™s equator.
Later in the cycle, they appear closer to the equa-
tor. This results in a butterfly-like pattern, and in
fact these diagrams are sometimes called “but-
terfly diagrams”.
When Maunder investigated records of past
sunspot activity, he found that there was an
extended period when no sunspots were observed.
This period, from 1645 to 1715 is known as the
Maunder minimum. This minimum has recently
been reinvestigated by John A. Eddy, who found
records to indicate that no aurorae were observed
(a)
for many years during this period. Also during
this time, weak coronae were reported during
total solar eclipses. An unusual correlation was
also found in the growth rings in trees, suggest-
ing that altered solar activity had some effect on
growth on Earth. The mechanism by which this
takes place is poorly understood, but we can use
the growth rings as an indicator of solar activity
farther into the past than other records. A study
of growth rings in old trees indicates that the
Maunder minimum is not unique. There may have
been several periods in the past with extended
reduced solar activity.
(b)
Sunspots appear to be regions where the mag-
netic field is higher than on the rest of the Sun. In
Fig 6.19. Sunspot images.The darker inner area of each
spot is the umbra and the lighter outer area of each spot is discussing sunspots, we should briefly review prop-
the penumbra.The spots appear dark because the surround- erties of magnetic fields and their interactions
ing areas are brighter. [(a) NOAO/AURA/NSF; (b) NASA]
with matter. Magnetic fields arise from moving
114 PART I PROPERTIES OF ORDINARY STARS



200
Fig 6.20. Numbers of sunspots
Sunspot Number


per year as a function of year.The
11-year pattern is evident. Note
that the number in a peak year is
not the same from cycle to cycle.
100
[NASA]



0
1600 1700 1800 1900 2000
Date

(including spinning) charges. Magnetic field lines on a charge moving parallel to the magnetic field
form closed loops (Fig. 6.22). This is equivalent to lines. The force is maximum when the velocity is
saying that there are no point sources of magnetic perpendicular to the magnetic field. The force is
charge, magnetic monopoles. (We will discuss the perpendicular to both the direction of motion
implications of the possible existence of magnetic and to the magnetic field, and its direction is
monopoles in Chapter 21.) We call the closed loop given by the so-called “right hand rule”. This
pattern in Fig. 6.22 a magnetic dipole. means that the component of motion parallel to
The magnetic field strength B is defined such the field lines is not altered, but the component
that the magnetic force on a charge q, moving perpendicular to the field lines is. So, charged
with velocity v, is (in cgs units) particles will move so that they spiral around
magnetic field lines.
F = q v x B/c (6.23)
Through the forces that it exerts on moving
where the x indicates a vector cross product. charges, a magnetic field will exert a torque on a
There is no magnetic force on a charge at rest, or current loop. We can think of a current loop as
having a magnetic dipole moment (as in Fig.
6.22), and the torque will cause the dipole to line
up with the field lines. In this way, the dipole
40
moment of a compass needle lines up with the
Earth™s magnetic field.
20
How do we measure the Sun™s magnetic field?
Latitude




For certain atoms placed in a magnetic field, energy
0


’20


’40

1990 1995 2000
Year
Fig 6.21. Butter¬‚y diagram.This shows the distribution of
sunspots in solar latitude as a function of time in the
sunspot cycle. Early in a cycle, spots appear at higher lati-
tudes; late in a cycle, they appear close to the equator. Note
that a new cycle starts before the old cycle completely ends.
[Roger Ulrich, UCLA, Mt Wilson.This study includes data
from the synoptic program at the 150 ft Solar Tower of the
Mt Wilson Observatory.The Mt Wilson 150 ft Solar Tower is
operated by UCLA, with funding from NASA, ONR and NSF,
Fig 6.22. Field lines for a dipole magnetic ¬eld.
under agreement with the Mt Wilson Institute]
6 THE SUN: A TYPICAL STAR 115



Fig 6.23. Zeeman effect in
sunspots. (a) The placement of the
spectrometer slit across a
sunspot. (b) The spectrum at vari-
ous positions along the slit. In the
spectrum, away from the spot, the
spectral lines are unsplit. In the
center, near the spot, some of the
spectral lines are split into three.
The stronger the magnetic ¬eld
the greater the splitting.
[NOAO/AURA/NSF]



(a) (b)

levels will shift. This is known as the Zeeman effect think that the Sun™s magnetic field arises below
(Fig. 6.23). Different energy levels shift by differ- the surface, rather than in the core, as the Earth™s
ent amounts. Some transitions that normally does. We also know that the Sun does not rotate
appear as one line split into a group of lines. The as a rigid body. By following sunspot groups (Figs.
amount of splitting is proportional to the strength 6.25 and 6.26), we see that material at the equator
of the magnetic field. (We can check the amount takes less time to go around than material at
of splitting in different atoms in various mag- higher latitudes. For example, it takes 25 days for
netic fields in the laboratory.) An image of the
magnetic field strength over the whole Sun is
shown in Fig. 6.24.
Measurements by George Ellery Hale in 1908
first showed that the magnetic fields are stronger
in sunspots. He also found that sunspots occur in
pairs, with one corresponding to the north pole
of a magnet and the other to the south pole. In
each sunspot pair, we can identify the one that
“leads” as the sun rotates. In a given solar hemi-
sphere, the polarity (i.e. N or S) of the leading spot
of all pairs is the same. The polarity is different in
the two hemispheres.
This polarity reverses in successive 11-year
cycles. During one cycle all of the leading sunspots
in the northern hemisphere will be magnetic
north, while those in the south will be magnetic
south. In the next cycle all the leading sunspots
in the northern hemisphere will be magnetic
south. The Sun™s magnetic field reverses every 11
years! This means that the sunspot cycle is really
a 22-year cycle in the Sun™s magnetic field.
If the Sun™s magnetic field arose in the core,
as the Earth™s does, we would expect it to be quite
Fig 6.24. Solar magnetic ¬eld.This shows the result of
stable. (There is geological evidence that the Earth™s
observations of the Zeeman effect. Brighter areas corre-
magnetic field reverses periodically, but on geo-
spond to stronger ¬elds. [NOAO/AURA/NSF]
logical time scales, not every 11 years.) We now
116 PART I PROPERTIES OF ORDINARY STARS



material at the equator to make one circuit, while
North
it takes 28 days at 40 latitude.
Pole
As the Sun rotates differentially, the magnetic
field lines become distorted. This is because the
charged particles in the matter cannot move
across field lines, so the field lines are carried
along with the material. We say that the mag-
netic field is frozen into the material. The devel-
Equator
opment of the magnetic field is shown in Fig.
6.27. As the field lines wind up, the field becomes
very strong in places. Kinks in the field lines
break through the surface. The sunspots appar-
ently arise through some, as yet poorly under-
stood, dynamo motion, involving convective
motion and the magnetic fields.
After
6.6.2 Other activity
One
Sunspots are just one manifestation of solar activ-
Rotation
ity related to the Sun™s magnetic field. Another
form of activity is the solar flare, shown in Fig.
6.28. A flare involves a large ejection of particles.
Fig 6.25. Differential rotation of the Sun, as traced out by Flares develop very quickly and last tens of min-
sunspots. Since the Sun rotates faster at the equator than at
utes to a few hours. Temperatures in flares are
higher latitudes, sunspots at the equator take less time to
high, up to 5 106 K. They also give off strong H
make one rotation than do those at higher latitudes. In this
emission, and flares are seen when the Sun is
schematic diagram, a selection of sunspots starts at the same
photographed through an H filter. Flares have
meridian, but, after one rotation of the Sun, they are on dif-
been detected to give off energy in all parts of the
ferent meridians.




MARCH 7 MARCH 8 MARCH 9 MARCH 10 MARCH 13 MARCH 14 MARCH 15 MARCH 16 MARCH 17, 1989




Fig 6.26. Differential rotation of the Sun, as traced out by sunspots. This sequence of images shows this effect on the Sun.
[NOAO/AURA/NSF]
6 THE SUN: A TYPICAL STAR 117



solar wind originates in lower density areas of the
corona, called coronal holes.
The solar wind can have an effect on the
Earth. Most of the solar wind particles directed
at Earth never reach the surface of the Earth. The
Earth™s magnetic field serves as an effective
shield, since the charged particles cannot travel
across the field lines. Some of the particles, how-
ever, travel along the field lines and come closest
to the Earth near the magnetic poles. These
charged particles are responsible for the aurora
Fig 6.27. This shows how the solar magnetic ¬eld is
displays. This explains why aurorae are seen pri-
twisted and kinked by the differential rotation of the Sun. It
marily near the magnetic poles. When solar
takes about 8 months for the ¬eld to wrap around once.The
activity is increased, the aurorae become more
kinked parts of the ¬eld lines represent places where these
widespread. The increased abundance of charged
loops come outside the Sun, allowing charged particles to
particles in our atmosphere also creates radio
follow those looped paths. [NASA]
interference.

electromagnetic spectrum. The cause of flares is
not understood, but they appear to be related to
strong magnetic fields and the flow of particles
along field lines.
Solar activity is also manifested in plages (from
the French word for “beach”). Plages are bright
regions around sunspots. They show up in H
images of the Sun. They remain after sunspots
disappear. Plages are apparently chromospheric
brightening caused by strong magnetic fields.
Filaments are dark bands near sunspot regions.
They can be up to 105 km long. Filaments appear
(a)
to be boundaries between regions of opposite
magnetic polarity. When filaments are projected
into space at the limb of the Sun, they appear as
prominences (Fig. 6.29). Some prominences vary on
short times scales while others evolve more slowly.
The solar wind is a stream of particles that are
emitted from the Sun into interplanetary space.
We can see the effects of the solar wind when we
look at a comet that is passing near the Sun
(Chapter 26). The tail of the comet always points
away from the Sun. This is because the material
in the tail is driven out of the head of the comet
by the solar wind. The rate at which the Sun is
(b)
losing mass is 10 14 M /yr. The wind is still accel-
erating in its five- to ten-day trip from the Sun to Fig 6.28. Images of solar ¬‚ares. (a) Looking obliquely as
Earth. At the Earth™s orbit, the speed is about the ¬‚are stands out from the surface. (b) In this photograph,
taken through an H ¬lter, we are looking down on a large
400“450 km/s, and the density of particles is 5 to
solar ¬‚are from the side.This ¬‚are appears brighter than its
10 cm 3. The particles in the solar wind are posi-
surroundings in the H line. [NOAO/AURA/NSF]
tive ions and electrons. It is thought that the
118 PART I PROPERTIES OF ORDINARY STARS



Fig 6.29. Images of large prominences. (a) This shows the
large loop structure. (b) This H image shows an eruptive
prominence, where some of the material may actually escape
from the Sun. (c) The X-ray image of the Sun gives another
view of regions of activity. [(a), (b) NOAO/AURA/NSF;
(c) NASA]




(a)




(b)




(c)
6 THE SUN: A TYPICAL STAR 119




Chapter summary
We looked in this chapter at the one star we can The chromosphere is difficult to study. In the
study in detail, the Sun. chromosphere the temperature begins to increase
Most of what we see in the Sun is a relatively as one moves farther from the center of the Sun.
thin layer, the photosphere. In studying radiative This trend continues dramatically into the
transfer, we saw how we can see to different corona. The best studies of the corona have come
depths in the photosphere by looking at different from total solar eclipses or from space.
wavelengths. The distance we can see corresponds We saw how sunspots appear in place of inten-
to about one optical depth. sified magnetic fields, as evidenced by the
The photosphere doesn™t have a smooth Zeeman effect. The number of spots goes through
appearance. Instead, it has a granular appearance, a 22-year cycle, which includes a complete rever-
with the granular pattern changing on a time sal in the Sun™s magnetic field. The structure of
scale of several minutes. This suggests convection the magnetic field is also related to other mani-
currents below the surface. Supergranulation sug- festations of solar activity, such as prominences.
gests even deeper convection currents.


Questions
6.1. In Fig. 6.3, what would be the effect on the 6.10. Explain why we see a range of Doppler shifts
total cross section of (a) doubling the density over a spectral line.
of particles, (b) doubling the length of the 6.11. How does Doppler broadening affect the sepa-
tube, (c) doubling the radius of each particle? ration between the centers of the H line and
*6.2. (a) What does it mean to say that some sphere the H line in a star?
has a geometric cross section of 10 16 cm2? 6.12. What do granulation and supergranulation
(b) Suppose we do a quantum mechanical cal- tell us about the Sun?
6.13. If the corona has T 2 106 K, why don™t
culation and find that at some wavelength
some atom has a cross section of 10 16 cm2. we see the Sun as a blackbody at this
What does that mean? temperature?
6.3. In studying radiative transfer effects, we let 6.14. Why can™t you see the corona when you cover
be a measure of where we are in a given sam- the Sun with your hand?
ple, rather than position x. Explain why we 6.15. What advantages would a coronagraph on
can do this. the Moon have over one on the Earth?
6.4. What does it mean when we say that some- 6.16. (a) Why does the F-corona still show the
thing is optically thin? Fraunhofer spectrum? (b) Would you expect
6.5. Suppose we have a gas that has a large optical light from the Moon to show the Fraunhofer
depth. We now double the amount of gas. spectrum?
How does that affect (a) the optical depth, 6.17. Why is the low density in the corona favor-
(b) the amount of absorption? able for high levels of ionization?
6.6. Why do we need the “mean” in “mean free 6.18. (a) Why are collisions important in cooling a
path”? gas? (b) Why does the cooling rate depend on
6.7. (a) Explain how we can use two different opti- the square of the density? (c) How would you
cal depth spectral lines to see different dis- expect the heating rate to depend on the
tances into the Sun. (b) Why is H particu- density?
larly useful in studying the chromosphere? 6.19. Explain why charged particles drift parallel
6.8. What other situations have you encountered to magnetic field lines.
that have exponential fall-offs? 6.20. How are the various forms of solar activity
6.9. Explain how absorption and emission by the related to the Sun™s magnetic field?
H ion can produce a continuum, rather
than spectral lines?
120 PART I PROPERTIES OF ORDINARY STARS



Problems
*6.9. Suppose we have a uniform sphere of radius R
6.1. Appendix G gives the composition of the Sun,
measured by the fraction of the number of and absorption coefficient . We look along
various paths, passing different distances p
nuclei in the form of each element. Express
the entries in this table as the fraction of the from the center of the sphere at their points
mass that is in each element. (Do this for the of closest approach to the center. (a) Find an
ten most abundant elements.) expression for the optical depth as a func-

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