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