<<

. 12
( 28)



>>

i j
can therefore rewrite equation (13.17) as
Since the i and j both go over the full range, 1
1 d2I
ab2 2K U (13.27)
to N, we could interchange the i and j on the
2 dt
right-hand side of equation (13.20) without really
If we take the time average of these quantities
changing anything. If we rewrite the right-hand
over a sufficiently long time, the left-hand side
side, interchanging i and j, and then adding it to
13 CLUSTERS OF STARS 227



If the virial theorem applies, then the total energy
approaches zero. This leaves
1051 erg.
is E U 2 1.2
2K U
0 (13.28)
We now look at the kinetic energy. In a cluster
where the represents the time average of the
of stars, the kinetic energy is in the random
enclosed quantity. Equation (13.28) is the sim-
motions of the stars. If the cluster has N stars,
plest form of the virial theorem.
each of mass m, the kinetic energy is
The virial theorem applies to any gravitation-
1N
a b a mv2
ally bound system that has had sufficient time to
K i
2i1
come to equilibrium. Even simple systems, like
binary stars, obey the virial theorem (see Problem N
a b m a v2
1
13.11). If the orbits are circular then K U 2 at i
2 i 1
all points. For elliptical orbits, r and v are chang-
The total mass of the cluster is M mN; so
ing, so K and U are changing, while their sum E is
fixed. This means that we have to average over a
1N
a b 1mN2 a b a v2
1
K (13.31)
whole orbit to get K U 2. i
Ni1
2
Remember, for any system, the total energy is
If we take the sum of N quantities and then divide
E K U (13.29)
by N, the result is the average of that quantity.
Therefore 11>N2 g v2 is the average of the quantity
So for a system to which the virial theorem, in 1
2
v . We write this average as v2 . Remembering that
the form of equation (13.28) applies, we set K
mN M, equation (13.31) becomes
” U /2, to give
11 2 2 M v2
K
E U2 (13.32)
(13.30)

(We don™t have to take the time average of E, since If we put this and the potential energy into the
E is constant.) Remember, the gravitational poten- virial theorem, we find
13 5 2GM2 R
tial energy, defined so that it is zero when the
M v2 (13.33)
particles are infinitely far apart, is negative.
Dividing both sides by M gives
Therefore, the total energy of a bound system is
13 5 2GM R
negative. This means that we have to put energy
v2 (13.34)
in to break up the system.
The quantity v2 is the mean (average) of the
13.3.2 Energies square of the velocity. If we take the square root
of this quantity, we have the root mean square veloc-
We now look at the kinetic and potential energies
ity or rms velocity. It is a measure of the internal
of a cluster. In Section 9.1, we saw that the gravi-
tational potential energy for a constant density motions in the cluster.
sphere of mass M and radius R is
Example 13.2 The rms velocity in a cluster
13 5 2GM2 R
U Find the rms velocity for the cluster used in the
previous example.
Example 13.1 Potential energy for a globular
cluster
SOLUTION
Find the gravitational potential energy for a spheri-
We use equation (13.34) with the given quantities:
cal cluster of stars with 106 stars each of 0.5 M .
10.6 2 16.67 dyn cm2 g2 2 10.5 2 1106 2 12
The radius of the cluster core is 5 pc. 8
1033 g2
10
v2
15 pc 2 13.18 1018 cm>pc 2

1012 1cm s 2 2
SOLUTION
2.5
We use the above equation to give
10.62 16.67 dyn cm 2 g2 2 3 10.5 2 12 1033 g 2 1106 2 4 2
Taking the square root gives
8
10
15 pc 2 13.18 1018cm pc 2
U
106 cm s
vrms 1.6
16 km/s
1051 erg
2.5
228 PART III STELLAR EVOLUTION



We can relate the gravitational potential energy
n stars/volume
to the escape velocity ve, the speed with which an
object must be launched from the surface to
escape permanently from the cluster. Consider a
r
particle of mass m, moving outward from the sur-
face at speed ve. If the object escapes, it must get
so far away that the potential energy is essen-
v trel
tially zero. Since the kinetic energy is always
greater than or equal to zero, the total energy of Fig 13.8. Calculation of relaxation time.
the object far away must be greater than or equal
to zero. Since the total energy is conserved, the
total energy for an escaping object must be zero
Earth. The highest speed molecules escape, leav-
or positive when it is launched. The kinetic
ing the water a little cooler.)
energy of the particle is
In discussing cluster dynamics, it is important
11 2 2 mv2
KE (13.35) to remember that we are studying the motions of
e
the stars in a cluster relative to the center of mass
Since it is at the surface of the sphere of mass M
of the cluster. We are not concerned with the
and radius R, its potential energy is just
overall motion of the cluster through space.
GmM R
PE (13.36) We call the time it takes to re-establish equi-
librium the relaxation time, trel. We can estimate
For the total energy to be zero (the condition that
trel by following a single star as it moves through
the particle barely escapes), KE PE, giving
the cluster (Fig. 13.8). We assume that there are n
v2 2GM R (13.37) stars per unit volume in the cluster. We would
e

like to know how long our star will go between
Note that the escape velocity is approximately
collisions with other stars. That depends, in part,
twice the rms speed. For a gravitationally bound
on how we define a collision. We would like to
system, we would expect ve 7 vrms. If it were the
define a distance r and say that if two stars pass
other way around, many particles would have
within this distance we will count it as a colli-
speeds greater than the escape velocity and would
sion. We define r so that the potential energy of
escape. The cluster would not be gravitationally
the star is equal in magnitude to the kinetic
bound.
energy of our star. If our star is moving with
speed v, this means that r is defined by
13.3.3 Relaxation time
Gm2 r mv2 2
In any given cluster, stars will be in orbits about (13.38)
the center of mass. Pairs of stars can exchange
We can think of our star as sweeping out a
energy and momentum via gravitational encoun-
cylinder in a given time trel. The radius of the
ters. (By exchange, we just mean that some energy
cylinder is r, and the length is vtrel. Therefore, the
and momentum is transferred, not that each one
volume swept out is r2vtrel. The number of stars
acquires the other™s energy and momentum.) As
in this volume is n multiplied by the volume. If
we have said, if there are enough collisions, an
we define trel so that it is the time for one colli-
equilibrium velocity distribution will be reached.
sion, we have the condition
Not all are moving at the average speed. Some
n 1 r2v trel 2
move faster and others move slower. In a cluster, 1 (13.39)
there will be some stars with speeds greater than
Solving for trel gives
ve. They will escape. This alters the velocity distri-
1 n r2v
bution by removing the highest velocity stars. The trel (13.40)
remaining stars must adjust, re-establishing the
Substituting for r from equation (13.38) gives
equilibrium velocity distribution. (This is equiva-
v3 4 G2m2n
lent to the evaporation of a puddle of water on the trel (13.41)
13 CLUSTERS OF STARS 229



approximate age of the universe is 15 Gyr. So, if we
The number of stars per unit volume is simply
don™t include the correction for distant encounters,
N
14 3 2R3
then the relaxation time would be longer than the
n
age of the universe, but, with that correction, that
(13.42)
is not a problem.
Mm
14 3 2R3 We can also define an evaporation time, which
is the time for a significant number of stars to
Substituting into equation (13.41) gives
leave the cluster. The evaporation time is approx-
v3R3 imately 100 times the relaxation time. For the
trel
3G2mM cluster in the above example, the evaporation
(13.43)
1R v 2 v4R2 1M m 2 time would be 200 Gyr, much more than the age
of the universe. So, that cluster has had enough
3G2M2
time to become relaxed, but not evaporate.
Using the virial theorem to eliminate two factors Once relaxation takes place, the velocity distri-
of v2, and ignoring numerical factors that are bution will evolve toward a Maxwell“Boltzmann
close to unity (since this is just an estimate), this distribution, given for a gas in equation (9.16). For
simplifies to a cluster of stars, we let kT become (1/3)mv2rms, the
1R v 2 1M m 2 number of stars in the velocity range v to v dv is
trel
given by:
NR v (13.44)
v2 3v2
n1v 2 dv r a b exp a b dv (13.45)
Equation (13.44) is just an estimate in which v3 2v2
rms rms
the effects of a few close encounters dominate.
Calculations show that a core denser than the
However, for every close encounter that a star has,
outer parts of the cluster will develop. It has been
it has many more distant encounters. This is
speculated that, at some point, the core can become
because the effective area for an encounter at dis-
so massive that it collapses to form a large black
tance r is proportional to r2. The effect of any one
hole. Recent observations of some globular clus-
distant encounter is small. However, because there
ters have revealed the existence of a luminous
are so many of them, they actually dominate the
extended object in the center. The size of these
relaxation process and speed it up. A more detailed
objects are in the 0.1 pc range. In each case the
calculation shows that the effect of many distant
object is bluer than the rest of the cluster, mean-
encounters is to reduce trel by a factor of 12 ln(N/2).
ing that it is not an unresolved group of red stars.
Example 13.3 Relaxation time
Estimate the relaxation time for the cluster dis-
13.3.4 Virial masses for clusters
cussed in the two previous examples.
For dynamically relaxed clusters, we can use the
virial theorem to estimate the mass of the cluster.
SOLUTION
For a uniform cluster, with N stars, each of mass
We use equation (13.44). For the speed v, we use the
m, and the total mass of the cluster M mN, the
calculated vrms:
cluster potential energy is
1106 2 15 pc 2 13.18 1018 cm pc 2
3 GM2
ab
trel 6
U
1.6 10 cm s
R
5
1019 s
1.0
where R is the radius of the cluster. The kinetic
1011 yr
3 energy is (equation 13.32)
11 2 2 M v2
300 Gyr K
If we apply the correction for distant encounters,
Substituting these into the virial theorem gives
12 ln(N/2) 1.6 102, trel is reduced to 2 Gyr. For
M v2 3 GM2 5R (13.46)
comparison, we will see in Chapter 21, that the
230 PART III STELLAR EVOLUTION



Solving for M, we have If we substitute this into the virial theorem (equa-
tion 13.47), the mass is given by
2
ab
5 vR
M (13.47)
5 v2 R
G
3 r
M (13.52)
G
It is important to understand which motions
we are talking about. The cluster has some overall Example 13.4 Virial mass of cluster
motion of its center of mass, shared by all the Find the virial mass of a cluster with vr 10 km/s
stars in the cluster. The stars have individual and R 5 pc.
motions within the cluster (with respect to the
center of mass of the cluster). The net motion of SOLUTION
each star is the vector sum of these two motions, From equation (13.52) we have

152 11.0 106 cm>s2 2 15 pc 2 13.18 1018 cm>pc 2
and that is what we observe. In equation (13.47)
the quantity v2 is the average of the square of the
dyn cm2>g2
M 8
6.67 10
star velocities with respect to the center of mass
1039 g
of the cluster. 1.2
The best way for us to measure the velocities of
105 M
6
individual stars is through their Doppler shifts.
When we talked about binary stars (Chapter
However, this only gives us the component of the
5), we noted that the best way to measure the
velocity along the line of sight. This means that
we are measuring v2 rather than v2 . However, if mass of an object, or a group of objects, is to
r
measure their gravitational effects on other
the internal motions of the stars are random, we
objects. The gravitational effects are independent
can relate these two quantities.
of how bright the objects are; they depend only
Suppose we resolve the motion of any star into
on how massive they are. Using virial masses is an
its components in an (x, y, z) coordinate system.
extension of these ideas. The more massive the
The velocity can then be written in terms of its
cluster, the greater the internal motions that we
components as
will observe. To determine vr , it is not even
vxˆ
x vy ˆ
y vz ˆ
z
v (13.48)
necessary to measure radial velocities for all the
where x, ˆ and ˆ are the unit vectors in the three
ˆy z stars in the cluster. We just need a representative
directions, respectively. To be definite, we can let sample.
the x-direction correspond to the line of sight. What are the limitations of this method? An
The square of v, which is v v, is simply the sum important one is that we don™t know if any par-
of the squares of the components, ticular cluster is dynamically relaxed, or even
gravitationally bound. We may measure large
v2 v2 v2 v2 (13.49)
x y z
internal motions in an unbound system and mis-
If we then take the average of both sides of the
take them for bound motions in a more massive
equation, we have
system. This can introduce errors that are off by
v2 v2 v2 v2 as much as a few orders of magnitude. Our calcu-
x y z

lation of the relaxation time suggests that all but
However, if the motions are random, the averages
the youngest clusters should be relaxed. We will
of the squares of the components should be the
talk about indicators of age of a cluster in the
same for all directions. This means that
next section. Another limitation can be from geo-
v2 v2 v2 (13.50)
x y z
metric effects. We may observe clusters that are
Using this, equation (13.49) becomes elliptical, rather than spherical. Or we may
observe clusters that don™t have a uniform den-
v2 3 v2
x
sity. The most likely variation is having a higher
and since the x-direction is the one corresponding density in the center. The effect of these geo-
to the line of sight, vr vx, so metric effects can be to produce errors of order
v2 3 v2 unity (see Problem 13.16). For most applications,
(13.51)
r
13 CLUSTERS OF STARS 231



astronomers use the virial theorem, knowing that the information from all of the main sequence
it is a technique that may be off by a factor of two stars in the cluster. This is more accurate than
or three. But this can be very useful for measur- studying a single star.
ing the masses of a variety of astronomical sys- The HR diagram for a group of galactic clus-
tems. We will look more at virial masses when we ters is shown schematically in Fig. 13.9. Note
talk about the masses of interstellar clouds that the lower (cooler or later) part of the main
(Chapter 14) and the masses of clusters of galaxies sequence is the same for all the clusters shown.
(Chapter 18). For each cluster, there is some point at which
the main sequence stops. Beyond that point, no
hotter stars are seen on the main sequence. The
13.4 HR diagrams for clusters hotter stars all appear to be above the main
sequence. The point at which this happens for a
given cluster is called the turn-off point. Stars of
By studying the HR diagram for a cluster, we are
studying a group of stars with a common dis- earlier spectral type (hotter than) the turn-off
tance. We can study their relative properties point appear above the main sequence, meaning
without knowing what their actual distance is. If that they are more luminous, and therefore
we do know the distance to the cluster, we plot larger than main sequence stars of the same
directly the absolute magnitudes on the HR dia- spectral type. Each cluster has its turn-off point
gram. If we don™t know the distance, we plot the at a different spectral type. Data for one cluster
apparent magnitudes. We then see how many are shown in Fig. 13.10, to see the scatter in the
magnitudes we would have to shift the diagram points.
up or down to calculate the right absolute mag- We interpret this behavior as representing stel-
nitudes for each spectral type. The amount of lar aging, in which stars use up their basic fuel
shift gives the distance modulus for the cluster, supply, as described in Chapter 10. Hotter, more
and therefore the distance. This procedure is massive, stars evolve faster than the cooler, low
known as main sequence fitting. It is like doing a mass stars, and leave the main sequence sooner.
spectroscopic parallax measurement, but it uses We assume that the stars in a cluster were formed

’8
h + χ Per
NGC2362
6
h + χ Per
’6


’4
8
Pleiades NGC6664
’2 M5
NGC6664 M41
M11
M41 K 10
MV
M11
0
Hyades
Coma
M67
Hyades
M5
NGC 12
NGC
2
752
752


4
14

6
0 1 2 3
J-K
8
’0.4 0 0.4 0.8 1.2 1.6 2.0 Fig 13.10. Color“magnitude diagram for a galactic cluster
B“V H and Chi Persei (the double cluster shown in Fig. 2.4).
Fig 13.9. Schematic HR diagrams for various galactic clusters. [Courtesy of 2MASS/UMASS/IPAC/NASA/JPL/Caltech]
232 PART III STELLAR EVOLUTION



at approximately the same time. As a cluster ages, clusters. For globular clusters, only the lower
later and later spectral types evolve away from (cooler) part of the main sequence is present.
the main sequence. This means the turn-off point All earlier spectral types have turned off the
shifts to later spectral types as the cluster ages. main sequence. This tells us that globular clus-
We can tell the relative age of two clusters by ters must be very old. Globular clusters contain
comparing their turn-off points. If we know how a large number of red giants. In Chapter 10 we
long different spectral type stars actually stay on saw that the red giant state is symptomatic of
the main sequence, we can tell the absolute age old age in a star.
of a cluster from its turn-off point.
We note that there are some galactic clusters,
13.5 The concept of populations
not shown in Fig. 13.9, that are missing the lower
(cooler) end of the main sequence. We think that
these clusters are very young. The lower mass There is another important difference between
stars are still in the process of collapse, and have the stars in galactic and globular clusters. It con-
not yet reached the main sequence. (We think cerns the abundances of “metals”, elements heav-
that lower mass stars take longer to collapse than ier than hydrogen and helium. Many globular
higher mass stars.) clusters have stars with very low metal abun-
An HR diagram for a composite of globular dances, while galactic cluster star are higher in
clusters is shown in Fig. 13.11. These appear to metal abundance. We refer to high metal stars as
population I stars and low metal stars as population
be different from the HR diagrams for galactic
II stars. We have a general sense that population I
stars represent younger, more recently formed
stars. We interpret the metal abundance differ-
ences as reflecting the conditions in our galaxy at
the time each type of star was formed. When the
older stars were formed, our galaxy had only
0
hydrogen and helium. When the newer stars were
formed, the galaxy had been enriched in the met-
MV als. This enriched material comes from nuclear
processing in stars, followed by spreading into the
5
interstellar medium, especially through supernova
explosions.
The differences between galactic and globular
clusters start us thinking about old and new
material in our galaxy. The globular clusters are
10
older, and form a spherical distribution, while
the galactic clusters are newer and are confined
to the galactic disk. This suggests that, a long
time ago, star formation took place in a large
15
spherical volume, but now it only takes place in
the disk. This is supported by the fact that globu-
0 1 2 3 lar clusters are free of interstellar gas and dust,
(V’I)0 the material out of which new stars can form,
while galactic clusters are sometimes associated
Fig 13.11. A composite color“magnitude diagram for a
with gas and dust.
“metal-poor” globular cluster, constructed from real photo-
The concept of stellar populations is impor-
metric data from several Milky Way globular clusters includ-
ing M3, M55, M68, NGC 6397 and NGC 2419. [William tant in our understanding of the evolution of our
Harris, McMaster University, STScI/NASA 10th Anniversary galaxy. This will be discussed farther in Chapters
Symposium Proceedings (STScI, Baltimore), May 2000]
14“16.
13 CLUSTERS OF STARS 233




Chapter summary
In this chapter we saw what could be learned This distribution is brought about by star“star grav-
from studying clusters of stars. Clusters are use- itational encounters, and the effects of many dis-
ful in studying a variety of astronomical prob- tant encounters are important. We saw that it is
lems because they provide us with groupings of possible for gravitationally bound clusters to evap-
stars all at the same distance. orate by losing the stars that are moving the fastest.
We looked at an important technique for We developed the idea of stellar populations,
determining the distances to nearby clusters. It is signifying old and new material in the galaxy.
free of any assumptions about luminosities, and Globular clusters seem associated with the old
serves as an important cornerstone in our dis- material, and galactic (open) clusters seem associ-
tance determination scheme. ated with the newer material. Some of these dif-
We looked at the dynamical properties of clus- ferences are very evident in comparing HR dia-
ters. Clusters have internal motions which eventu- grams, as well as in comparing the metal content
ally reach some equilibrium velocity distribution. of the stars in the two types of clusters.


Questions
13.1. Suppose you had photographs of a cluster 13.5. Construct a table, contrasting the properties
taken ten years apart. How would you use of globular clusters and galactic clusters.
those photographs to find the convergent 13.6. If a cluster is moving through space at
point of the cluster? 50 km/s, how should this motion be
included in v2 , which appears in equation
13.2. List the distance measurement techniques
that we have encountered so far in this book, (13.34)?
13.7. In a cluster for which ve vrms, some stars
and estimate the distance range over which
each is useful. can still escape. How does this happen?
13.3. What is the relationship between main 13.8. What are the advantantages and disadvan-
sequence fitting and spectroscopic parallax? tages of using the virial theorem to deter-
13.4. What is the significance of the main mine the mass of a cluster?
sequence turn-off point of a cluster?

Problems
13.1. The Hyades has a proper motion of 0.07 arc *13.4. In Chapter 4, we discussed the resolving
sec/yr and appears 26 from its convergent power of gratings. How is the minimum
point. The radial velocity is 35 km/s. (a) How radial velocity shift we can measure related
far away is the cluster? (b) What is the actual to this resolving power?
speed of the cluster? *13.5. For the globular cluster treated in the exam-
13.2. Suppose we can detect proper motions down ple in this chapter, estimate the average
to 0.1 arc sec/yr. How far away can we detect time between collisions in which the stars
a transverse velocity of 10 km/s? actually hit.
13.3. Suppose we have two photographs of a clus- 13.6. Compare the rms speeds in a typical globu-
ter, taken ten years apart. In the second pho- lar cluster with those in a typical galactic
tograph the cluster has moved over by cluster.
1.0 arc sec. Two stars that were originally 13.7. Verify that equation (13.44) can be obtained,
20.0 arc sec apart are now 19.5 arc sec apart. as outlined, from equation (13.43).
13.8. The crossing time for a cluster is the average
What is the angle between the current line
of sight to the cluster and the line of sight time for a cluster to move from one side of
to the convergent point? the cluster to the other. (a) What is the
234 PART III STELLAR EVOLUTION



relationship between the crossing time and 13.13. Find an expression that gives the virial mass
the relaxation time? (b) How would you in solar masses, when rms velocity is in km/s
explain that relationship? and the size is in pc.
*13.9. Calculate the relaxation time for the typical 13.14. (a) Find the virial mass of a cluster for which
galactic cluster discussed in this chapter, the rms radial velocity dispersion is 5 km/s
and the radius is 15 pc. (b) What are U, K and
and compare it with that for the typical
E for this cluster?
globular cluster.
*13.10. Show that the derivation of the virial theo- 13.15. Compare the evaporation times for typical
rem gives the same result if we allow the galactic and globular clusters discussed in
masses of the stars to be different. this chapter. Include the correction for the
13.11. Show that binary stars obey the virial effects of distant collisions.
theorem. *13.16. Find an expression for the gravitational
13.12. Suppose we have a gravitationally bound potential energy for a spherical cluster of
mass, M, and radius, R, where the density
object in virial equilibrium. Show that if
falls as 1/r2. (Hint: Integrate equation (9.6)
more than half of the mass of the system is
lost, with no change in the velocities of the with a variable density.) How does your
remaining material, then the system will be result compare with the case of a uniform
unbound. density sphere?


Computer problem

13.1. Find the virial masses, and U, K and E for clusters
with the properties shown in Table 13.1.


Table 13.1.
vrms (km/s) R (pc)

2 2
2 5
2 10
5 2
5 5
5 10
10 2
10 5
10 10
Part IV
The Milky Way
Most of the light we can see from our galaxy appears as a narrow band
around the sky. From its appearance, we think that we are in the plane of
a disk, and that this disk looks something like the Andromeda galaxy.
However, our location within our own galaxy makes its structure very
dif¬cult to study. In this part we will see both how we learn about our
galaxy and what we have learned about it so far.
Most of the light that we see comes directly from stars. Among all the
objects we can see, the stars provide most of the mass. Averaged over the
whole galaxy, the gas and dust between the stars “ the interstellar
medium “ contains only about 1% as much mass as the stars themselves.
Of the interstellar medium, 99% of the mass is in the form of gas, and
1% of the mass is in the form of dust. However, this small amount of dust
is very ef¬cient at blocking light, making optical observations of distant
objects dif¬cult.
We expect that stars form out of interstellar material. Since most of
the mass of the interstellar material is in the form of gas, it is the gas that
will provide the gravitational attraction for the star formation process. In
this part, we will ¬rst look at the contents of the interstellar medium.We
will then look at how stars are born. Finally, we will see how the stars and
interstellar medium are arranged in the galaxy as a whole.
Chapter 14




Contents of the interstellar medium

are dying, and we know that there is still a large
14.1 Overview number of stars in the galaxy. We therefore pre-
sume that stars are being created at a rate that
When we look at photographs of the Milky Way approximately offsets the rate at which they are
(see Fig. 16.1), we note large regions where no dying. This is not an airtight argument, because
light is seen. We think that these are due to dust it could be that many stars were formed early in
blocking the light between us and the stars. We the history of the galaxy and we are just seeing
can see the same effect on a smaller scale (Fig. 14.1). the ones that haven™t died yet. However, we know
that O stars live only about 107 years or less on
Note that there is a high density of stars near the
edges of the image. As one moves close to the cen- the main sequence. Since we see O stars today,
ter, the density of stars declines sharply. Near the there must have been O star formation in the last
107 years. We think that the galaxy is ten billion
center, no stars can be seen. This apparent hole in
the distribution of stars is really caused by a years old. Compared with this, ten million years
small dust cloud, called a globule. The more dust is almost like yesterday. If the conditions were
right for star formation in the last 107 years, they
there is in the globule, the fewer background
stars we can see through the globule. We can use must be right for star formation now. The actual
images like this to trace out the interstellar dust. star formation process will be discussed in
We find that it is not uniformly distributed. Chapter 15.
Rather, it is mostly confined to concentrations or
interstellar clouds.
14.2 Interstellar extinction
We detect the presence of the gas by observing
absorption or emission lines from the gas. By
If we want to see direct emission from the dust,
tracing these lines, we find that the gas also has
we have to look in the infrared, as we will discuss
an irregular distribution. Often the gas appears
in the next section. In the visible part of the spec-
along the same lines of sight as the dust clouds.
trum the dust is generally evidenced by its block-
From this apparent coincidence we form the idea
ing of starlight. The blocking arises from two
that the gas and dust are generally well mixed,
processes, scattering and absorption. In scattering,
with the gas having about 99% of the mass in a
the incoming photon is not destroyed, but its
given cloud. In this chapter, we will see how the
direction is changed. In absorption, the incoming
masses of different types of clouds are deter-
photon is destroyed, with its energy remaining in
mined.
the dust grain. The combined effects of scattering
One of the reasons that the interstellar
and absorption are called interstellar extinction. In
medium is so interesting is that it is the birth-
Fig. 14.2 these two processes are depicted schemat-
place of stars. How do we know that stars are still
ically. A dark nebula, in which background light
forming in our galaxy? We have seen that stars
238 PART IV THE MILKY WAY



Fig 14.1. An image of a globule
at various wavelengths.The
globule is the region with the
fewest number of stars per unit
area.The dust in the globule is
blocking the light from the
background stars. [ESO]




star would have an apparent magnitude m with-
is being blocked, and a reflection nebula, in which
out extinction, but its light passes through a cloud
scattered starlight is being sent in our direction,
with A magnitudes of extinction, then the star will
are depicted in Fig. 14.3.
be observed to have a magnitude m¿ m A.
(Remember, extinction dims the starlight, so the
14.2.1 The effect of extinction
magnitude increases.)
We quantify interstellar extinction as the number
We can relate the extinction, in magnitudes,
of magnitudes by which a cloud dims starlight
to the optical depth of the dust. This is useful,
passing through it. For example, if a particular
since it is the extinction in magnitudes that will
be directly measurable, but it is the optical depth
that is directly related to the dust properties. If we
Absorbed
Scattered have light of incident I0 passing through the cloud
of optical depth , and intensity I emerges, then
these are related by (as we saw in equation 6.18)
I I0 e (14.1)
From the definition of extinction and the magni-
tude scale
A m m

2.5 log10 (I0/I) (14.2)
Using equation (14.1), this becomes
A 2.5 log10 (e )
2.5 log10 (e)

(2.5)(0.4343)

(1.086) (14.3)
Fig 14.2. Scattering and absorption. Light is incident from
the right. Light rays that are absorbed stop inside the cloud.
This means that one magnitude of extinction cor-
Light rays that are scattered change direction. Rays that are
responds approximately to an optical depth of
scattered or absorbed are color coded as indicated.
one.
14 CONTENTS OF THE INTERSTELLAR MEDIUM 239



the apparent magnitude m. In the presence of A
magnitudes of extinction, the star will appear A
magnitudes fainter than without extinction, so
m M A
5 log (r/10 pc) (14.4)
Since we know m, M and r, we can find A.
Obviously the presence of interstellar extinction
will affect distance measurements by spectro-
scopic parallax. If we don™t correct for extinction,
then a star will appear to be farther away than it
actually is. You can see that if both r and A are
unknown, then equation (14.4) only gives us one
equation with two unknowns. We will see below
that there is a way of obtaining additional infor-
mation by observing at different wavelengths.
Example 14.1 Interstellar extinction
Suppose we observe a B5 (M 0.9) star to have an
apparent magnitude of 9.2. The star is in a cluster
(a)
whose distance is known to be 400 pc. What is the
extinction between us and the star?

SOLUTION
We solve equation (14.4) for A to give

A m M 5 log (r/10 pc)
9.2 0.9 5 log(40)

2.1 mag


14.2.2 Star counting
If we record an image of a field which has some
interstellar extinction, fewer stars will appear
than if the extinction were not present. This is
(b)
because the light from each star is dimmed by
Fig 14.3. The Horsehead Nebula, in Orion™s belt, is formed
the extinction. Some stars that would have
by the dust blocking the light from the glowing gas in the
appeared on the image if there were no extinc-
background. In this image, north is to the left. (a) A wider
tion are now too dim to appear with extinction.
view. The fuzzy blue patch at the lower left of the Horsehead
We can estimate the extinction in a cloud by com-
is a re¬‚ection nebula, where dust is scattering light from a
paring the number of stars we can see through
hidden background star towards us. Just off the left (north)
the cloud with the number we can see in an
of the image is the southwestern most star in Orion™s belt.
unobscured region of the same size. Suppose an
(b) A closer view from HST. This shows the intricacy in the
structure in both the glowing gas and the absorbing dust. image is exposed to a threshold magnitude m0.
[(a) NOAO/AURA/NSF; (b) STScI/NASA] All stars with apparent magnitude less than m0
(that is, stars that are brighter than m0) will
appear on the image. If the light from each star is
dimmed by A magnitudes, only stars that have
If we have a star of known distance and spec-
undimmed magnitudes of m0 A will appear.
tral type, we can determine the extinction
between the star and us. The spectral type gives There are two ways of applying this idea. In
us the absolute magnitude M. We can measure one, we measure the number of stars per unit area
240 PART IV THE MILKY WAY



If we now look at a region with extinction A,
only the stars that would have had magnitude
m0 A without extinction will show up. We
therefore count
log N'(m)




on
A2 N¿1m¿ 2 dm¿
m0 A
cti




es
N1m0 (14.7)




ud
tin
q




nit
Ex


A



ag
No




Therefore, if we know N (m), we can predict


M
N(m0 A) for various values of A. If we use plates

A
with a limiting magnitude of 20, then we can gen-
erally obtain good star count data for A in the
range 1 to 6 mag. For A much less than 1 mag, the
m
difference between an obscured and an unob-
Fig 14.4. Effect of extinction on star counts.The plot gives scured region is hard to detect. For A much greater
the number of stars per magnitude interval, as a function of than 6 mag, there are very few stars bright enough
magnitude m.The distribution is such that when the vertical
to shine through, and the obscured region will
axis is logarithmic, the curve is close to a straight line.The
appear blank, a situation in which 6 mag of extinc-
effect of a cloud with A magnitudes of extinction is to make
tion is indistinguishable from 20 mag.
each star A magnitudes fainter, shifting the curve to the right
by A magnitudes.
14.2.3 Reddening
If we measure interstellar extinction we find that
it is not the same at all wavelengths. In general,
in each magnitude range. In the more common
the shorter the wavelength is, the higher is the
way, we measure just the total number of stars per
extinction. This means that blue light from a star
unit area. We define the function N (m) such that
is more efficiently blocked than red light. In the
N (m) dm is the number of stars per unit area with
presence of extinction, the images of stars will
magnitudes between m and m dm in the
therefore appear redder than normal, as shown
absence of extinction. We measure N (m) for a
in Fig. 14.5. This is called interstellar reddening. You
region we think is partially obscured by dust and
for a nearby region we think is unobscured. If we
plot graphs of these two quantities, as shown in
Fig. 14.4, we see that the two curves look like each
other, except that one is shifted by a certain num-
ber of magnitudes. The amount of the shift is the
extinction in the partially obscured region.
Often we don™t have enough stars in each
magnitude range to obtain a good measure of
N (m), In that case we must use integrated star
counts. We let N(m) be the number of stars per
unit area brighter than magnitude m. This is
related to N (m) by

N1m 2 N¿1m¿ 2 dm¿
m
(14.5)
q


If a photographic plate has a limiting magnitude
m0, then the number of stars per unit area, with-
out extinction, is

N1m0 2 N¿1m¿ 2 dm¿
m0
Fig 14.5. Interstellar reddening. More blue light is
(14.6)
removed from the incoming beam than red.
q
14 CONTENTS OF THE INTERSTELLAR MEDIUM 241



can see the effect of reddening in the various Example 14.2 Spectroscopic parallax with
wavelength images of the globule in Fig. 14.1. extinction
More stars shine through at longer wavelengths. 0.9, B V
Suppose we observe a B5 star (MV
Suppose we measure the magnitude of a star 0.17) to have mB 11.0 and mv 10.0. What is the
in two different wavelength ranges, say those cor- visual extinction between us and the star, and how
responding to the B and V filters. Then, from far away is the star?
equation (14.4) we have
mV MV AV SOLUTION
5 log (r/10 pc) (14.8a)
From equation (14.9) we have
mB MB AB
5 log (r/10 pc) (14.8b)
AV) mV) MV)
(AB (mB (MB
If we take the difference mB mV, the distance
1.00 0.17
r drops out, giving
mV) MV ) AV )
(mB (MB (AB (14.9) 1.17
In equation (14.9), the quantity on the left- We can now use the ratio of total-to-selective
hand side is directly observed. The first quantity extinction to convert this to AV :
on the right-hand side depends only on the spec-
AV R (AB AV)
tral type of the star. It is simply the B V color of
the star. We know it because we can observe the 3.6 mag
star™s absorption line spectrum to determine its
We can now find the distance from
spectral type. Since the spectral type determina-
mV MV AV
5 log (r/10 pc)
tion depends on the presence of certain spectral
lines, it is not greatly influenced by interstellar
7.3
extinction. We can therefore determine the quan-
This gives
tity (AB AV ).
Since both AB and AV are proportional to the
r 280 pc
total dust column density ND, their difference is
also proportional to ND. If we define a quantity
14.2.4 Extinction curves
AV 1AB AV 2
R (14.10)
If we study how extinction varies with wavelength,
we can learn something about the properties of
it will not depend on ND since ND appears in both
interstellar dust grains. We try to measure the
the numerator and denominator. We call this
A( ) in the directions of several stars, to see the
quantity the ratio of total-to-selective extinction.
degree to which grain properties are the same or
Extensive observational studies have shown that,
different in different directions. Since the dust
in almost all regions, R has a value very close to
column densities are different in various direc-
3.1. (There are a few special regions where R is as
tions, we do not directly compare values of A.
high as 6.) This has a very important consequence.
Instead, we divide by AV or (AB AV), to get a quan-
It means that if we can measure (AB AV), we need
tity that is independent of the column density. It
only multiply by 3.1 to give AV. We have already
is conventional to plot the following function to
seen that the difference can be determined by
represent interstellar extinction curves:
knowing the spectral type of a star and measur-
A1 2
ing its B and V apparent magnitudes, and then
f1 2
using equation (14.9). Note that we have not made (14.11)
AV
use of knowing the distance to the star r. We can
A typical curve is shown in Fig. 14.6. One gen-
still go back to equation (14.8a) to find the dis-
eral feature is that in the visible part of the spec-
tance to the star. So, the method of spectroscopic
trum f1 2 is roughly proportional to 1 . In the
parallax works even in the presence of interstel-
ultraviolet there is a broad ˜hump™ in the curve.
lar extinction. We just need to do an extra obser-
The size of this hump varies from one line of
vation at a different wavelength.
242 PART IV THE MILKY WAY



brightness when the filter is rotated by 90 . The
greater the amount of polarization, the greater
the contrast between maximum and minimum.
When unpolarized starlight passes through
interstellar dust clouds, it can emerge with a
slight degree of linear polarization. This means
that the polarization must be caused by the dust
itself. The amount of polarization is very small,
only a few percent at most. We find that there is
a weak wavelength dependence on the amount of
polarization. We also find that the amount of
polarization generally depends on the visual
extinction AV. When AV is low, the polarization is
low. When AV is high, the polarization can be low
or high. Dust is necessary for the polarization, but
something else must be necessary. There must be
a mechanism of aligning non-spherical dust
grains (at least partially) to produce the polariza-
Fig 14.6. Average interstellar extinction curves.The vertical
axis is just 3 A1 2 4 AV 4 .This is plotted as a function of 1 , tion. We will discuss this in the next section.
since it is approximately linear in 1 near the visible.The V
14.2.6 Scattering vs. absorption
¬lter would be at 1 1.8.The solid line is for most
We have said that extinction is the combined
normal regions, with R (ratio of total-to-selective extinction,
de¬ned in equation 14.10) 3.1.The dashed line is for effect of scattering and absorption. The relative
regions with an unusually high value of R 5. [ John Mathis, importance of these two effects depends on the
University of Wisconsin]
physical properties of the grains and the wave-
length of the incoming light. The fraction of the
sight to another. In the infrared there are absorp- extinction that results from scattering is called
tion features of various strengths. We will see the albedo a of the dust grains. If a is the fraction
that these infrared features tell us a lot about the scattered, then 1 a must be the fraction
grain composition. absorbed.
The albedo is much harder to measure than
14.2.5 Polarization the extinction. Studies of reflection nebulae are
Sometimes the light we receive from celestial particularly useful, since they provide us with
objects is polarized. We can detect this polariza- light that we know is scattered by the dust. It
tion by placing a polarizing filter in front of our appears that the albedo is quite high, about
detector. Such a filter only passes radiation whose 50“70%, at most wavelengths. (The albedo is
electric field vector is parallel to the polarization lower in the range of the ultraviolet bump in the
direction of that filter. As we rotate the filter, dif- extinction curve, meaning that the bump is due
ferent polarizations of the incoming light are to a strong absorption feature.) The high albedo
passed. If the incoming light is unpolarized, then also means that a photon may be scattered a few
the amount of light coming through will not times before it is actually absorbed.
depend on the angle through which the filter has If a photon is scattered by a dust grain, we
been rotated. If the incoming light is completely would like to know the directions in which it is
linearly polarized, there will be one position of most likely to travel. Studies indicate that about
the filter for which we see the image at full half of the scattered photons move in almost the
brightness and another position, 90 away, for same direction as they were going when they
which we see no light. If the incoming light is struck the grain. The rest of the photons have
partially polarized, we will see a maximum almost an equal probability for being scattered in
brightness in one position and a minimum any direction.
14 CONTENTS OF THE INTERSTELLAR MEDIUM 243



would be no way of producing the polarization.
14.3 Physics of dust grains Therefore, some significant fraction of the grains
must either be elongated, like cigars, or flat-
In this section we will see how we use observa- tened, like disks.
tions and theory to deduce a number of proper- There must also be a mechanism for actually
ties of interstellar dust grains. We will also see aligning the asymmetric grains. This mechanism
how grains interact with their environment. probably involves the interstellar magnetic field.
Some of the things we would like to know about The grains are probably not ferromagnetic. This
interstellar dust are: means that a collection of grains cannot make a
permanent magnet. However, they may be para-
(1) size and shape;
magnetic. In a paramagnetic material the indi-
(2) alignment mechanism for polarization;
vidual particles have magnetic moments. These
(3) composition;
can be aligned by a magnetic field. The tendency
(4) temperature;
to align is offset by the random thermal motions
(5) electric charge;
of the grains. We think that a partial alignment
(6) formation and evolution.
arises from a combination of two effects: (1) the
14.3.1 Size and shape tendency of elongated grains, shaped like cigars,
We try to deduce grain sizes from the observed to rotate end-over-end rather than about the long
properties of the interstellar extinction curve, axis, since less energy is required for end-over-end
the variation of extinction with wavelength. If motion; and (2) the tendency for the magnetic
the grain size r is much greater than the wave- moment of the grain to align with the rotation
length , then we are close to the situation in axis.
which geometric optics applies. The wavelength
is unimportant, and A1 2 is roughly constant. If 14.3.2 Composition
r V , the waves are too large to ˜see™ the dust We can deduce the composition of the larger
grains, and A1 2 is very small (explaining the low grains from infrared absorption features. This is
extinction at long wavelengths). If r is compara- not as exact as using optical absorption spectra
ble to , then diffraction effects in the scattering to tell us about the compositions of stellar atmos-
process are important. Hence the wavelength pheres. Because the dust grains are solids, cer-
dependence is strongest in this range. tain motions of atoms within the grains are
We compare the observations with theoreti- inhibited by the close bonding to neighboring
cal calculations of scattering and absorption by atoms, so the spectra consist of a few smeared
grains of different sizes and composition to see out features instead of many sharp spectral lines.
which gives the best agreement. We find that We observe absorption features at 10 m and
interstellar grains are not all the same size. 12 m, which correspond to the wavelengths for
There is some spread about an average (just as vibrational transitions (stretching of bonds) in
people are not all the same height, but have silicates (SiO and SiO2) and water ice. Since sili-
some spread of heights about some average cates are an important component of normal dirt
value). In fact, the situation is more complicated on Earth, we sometimes talk of grains as being
than that. Observations of extinction curves ˜dirty ice™.
indicate that there are probably at least two dif- The extinction in the ultraviolet (including
ferent types of grains with distinctly different the hump) cannot be explained by dirty ice. For
average sizes (just as men and women are differ- that, carbon is probably needed. Therefore, inter-
ent types of people with different average stellar grains are probably a combination of large
heights). dirty ice grains and small graphite grains.
We can deduce something about the shapes of At the smallest end of the size distribution,
interstellar grains from their ability to polarize probably only 1 nm across, are grains consisting of
beams of light. If the grains were perfect spheres, 20 to 100 C atoms in an aromatic hydrocarbon form.
These are called polycyclic aromatic hydrocarbons,
there would be no preferred direction, and there
244 PART IV THE MILKY WAY



14.3.3 Electric charge
30
HR4049 We can deduce the electric charge for grains
20
from theoretical considerations. There are two
10
ways for grains to acquire charge: (1) Charged
0
particles (both positive and negative) from the
30 gas can strike the grains and stick to the sur-
IRAS21282+5050 face. (2) Photons striking the grain surface can
20
Flux [Jy]




eject electrons via the photoelectric effect. The
10
grain is left with one unit of positive charge for
0
every electron ejected. In an equilibrium situa-
tion the net charge on the grains must be
300 NGC7027 constant.
200 We first consider the situation in which the
photoelectric effect is not important. This would
100
be the case in regions of high extinction. For
0
particles striking the grains, the negative charges
8 10 12
10
» [µm] are mostly carried by electrons, and the positive
charges are mostly carried by protons. At any
Fig 14.7. Infrared spectra of dust, showing the main
given temperature the average speed of the elec-
spectral lines.The emission features at 3.3, 6.2, 7.7, 8.6 and
trons, which have lower mass, will be greater
11.3 m are probably from PAHs. [ESA/ISO, SWS, F. J.
than that of the protons. Therefore, electrons
Molster et al.]
will hit grains at a greater rate than protons or
C , ionized carbon. If the grains are initially
PAHs. They can be identified through some of
neutral, this will tend to build up negative
their emission lines, like those shown in Fig. 14.7.
charge. However, once the grains have a small
They are stable at high temperatures, and were
negative charge, the electrons will be slowed
originally proposed to explain diffuse emission
down as they approach the grains, while the pos-
from hot extended clouds.
itive charges will be accelerated. Therefore, if the
It also seems likely that many grains, especially
grains have a net negative charge it is possible to
the larger ones, are not of uniform composition.
have protons and electrons striking the grain at
Rather, they have layers, like the schematic grain
the same rate, keeping the charge on the grain
in Fig. 14.7. There could be a core of silicates and
constant. Note that, if the grains have a net neg-
carbon. Outside of that are various mantles, one
ative charge, the gas must have a net positive
with water and ammonia ice and another with O,
charge if the interstellar medium as a whole is to
N and CO (carbon monoxide) all in solid form.
be neutral.
There may even be a thin outer layer of hydrogen.
Example 14.3 Charge on dust grains
Estimate the electric charge (in multiples of e)
required to keep the charge on dust grains con-
stant. Take the radius of the grain to be 10 5 cm
and the gas kinetic temperature to be 100 K.
Core

SOLUTION
Inner Mantle
We estimate the grain charge for which the electric
Outer Mantle
potential energy of an electron at the grain surface
is equal in magnitude to the average kinetic energy
Silicates of the electrons in the gas. If the net charge on the
Oxygen
Carbon Water Carbon Monoxide grain is Ne, the potential energy for an electron
Ammonia Nitrogen on the surface, a distance r from the center, is
Fig 14.8. A multilayered interstellar grain.
Ne2 r
U
14 CONTENTS OF THE INTERSTELLAR MEDIUM 245



If a is the albedo, then (1 a) is the fraction of
The average kinetic energy is (3/2)kT. Equating
these and solving for N gives incoming radiation absorbed by the grain.
Therefore the rate P at which energy is being
ab2
3 kT r
absorbed by the dust grain is
N
2e
11 a 2 14 R2 2 1 T 4 2 1 r 2 2
13 2 11.38 erg K2 1100 K 2 110
g
* *
16 5
10 cm2
14 d2 2
Pabs (14.14)
12 2 14.8 10
esu2 2
10
11 a2 R2 T4 r2
* *g
0.9
d2
This says that each grain should have a net charge
The quantity R2 d2 is the solid angle sub-
of about e. However, the actual charge is about a *
tended by the star as seen from the dust grain. We
factor of ten larger because we have only consid-
say that the star acts like a dilute blackbody. It has
ered electrons of the average energy. Electrons mov-
the spectrum of a blackbody at a temperature T*,
ing faster than average contribute significantly to
but the intensity is down by a factor of (solid
the charge buildup. Also, the charge becomes more
angle/4 ).
negative at higher temperatures.
We now look at the rate at which the grain
radiates energy. Since it can only absorb 11 a 2
If the photoelectric effect is dominant, the
grains will have a positive charge. There must be a of the radiation striking it, it can only emit
11 a 2 of the radiation that a perfect blackbody
balance between the rate at which electrons are
being ejected and the rate at which they strike the would emit. (A perfect blackbody has an albedo of
grain. For positively charged grains, the electrons zero, by definition.) If the grain temperature is Tg,
in the gas are attracted, meaning the tendency for the power radiated is
the electrons to strike the grains at a greater rate
11 a 2 4 r2 T 4
Prad (14.15)
than positively charged particles is enhanced. g g

Equating the power radiated and the power
14.3.4 Temperature received, and solving for T, gives
The temperature of a large dust grain is deter-
T* (R*/2d)1/2
Tg (14.16)
mined by the fact that, on a time average, it must
emit radiation at the same rate as it receives radi- Note that the final result does not depend on
ation. This keeps the temperature constant. The the size of the grain or the albedo. That is because
temperature of a dust grain will therefore depend both enter into the emission and absorption
on its environment. If it is very close to a star it processes. (See Problem 14.10 for a discussion of
will be hot. If it is far from any one star it is cool, what happens if the albedo is a function of wave-
receiving energy only from the combined light of length.) This result is the same as that derived for
many distant stars. a planet in Section 23.2.
Let™s look at the case of a dust grain a distance
Example 14.4 Temperature of a dust grain near
d from a star whose radius and temperature are R*
a star
and T*. We will assume that the albedo is the same
What is the temperature of a dust grain a distance
at all wavelengths. (If the albedo varies with wave-
5000 stellar radii from a star whose temperature is
length, as in more realistic cases, the fraction of
104 K?
incoming radiation absorbed is different at differ-
ent wavelengths, and the calculation is harder. SOLUTION
See Problem 14.11.) The luminosity of the star is Using equation (14.16), we have
4 R2 T4
L* (14.12) 12
1104 K2 B
R*
* * R
12 2 15000 2R*
T
The fraction of this power striking the grain is
2
the projected area of the grain rg , divided by the 100 K
area of a sphere of radius d. That is
When dust is sufficiently warm (Tg 20 K), it
2 2
rg 4d
fraction striking grain (14.13) is a good emitter in the infrared, and we can
246 PART IV THE MILKY WAY



of processes. Sometimes molecules can simply
Planetary
Nebula sublime from the surface. (Sublimation is a phase
Nova change directly from the solid phase to the gas
Supernova phase.) Collisions with atoms in the gas can break
up grains. Collisions between grains can also
destroy the grains.




Molecular Cloud
14.4 Interstellar gas
Red
14.4.1 Optical and ultraviolet studies
Giant Dense
Core Early studies of cold interstellar gas utilized opti-
Star cal absorption lines. When light from a star
Formation
passes through a cloud, as shown in Fig. 14.10,
some energy is removed at wavelengths corre-
sponding to transitions in the atoms and mole-
cules in the cloud. These studies revealed the
existence of trace elements such as sodium or
calcium. (These elements happen to have conven-
ient spectral lines to study.) In addition to these
Fig 14.9. Life cycle of a dust grain.
atoms, some simple molecules were discovered:
CH (in 1937), CN (in 1940), and CH (in 1941).
determine its temperature directly from infrared
observations.

14.3.5 Evolution
We still know very little about the evolution of
dust grains. Fig. 14.9 indicates, schematically, the
life cycle of a typical dust grain. The densities in
Starlight passing
interstellar clouds are probably too low for the through cloud
grains to be formed directly where we see them.
We think that most dust grains are formed in the
envelopes of red giants undergoing mass loss. As
Background
material leaves the surface, it is hot enough to be
Star
gaseous. However, as it gets farther from the sur-
face, it cools. When the temperature is low
Interstellar Cloud with
enough, about 1000“2000 K, many of the materi-
Atoms & Molecules
als, such as silicates, can no longer exist as a gas.
They form small solid particles. These particles

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