<<

. 15
( 28)



>>

image of a region, containing a
bipolar ¬‚ow and a number of
Herbig“Haro objects.The solid
contours are the CO emission
that is blueshifted with respect to
the average velocity.The dashed
contours show the redshifted
CO.The plus ( ) marked B is
the location of the suspected
source for the ¬‚ow. [Snell, R. A.
et al., Astrophys. J. Lett., 239, L17,
Figs. 2 & 5]




mass. It has been suggested that the combined
this wind is relatively gentle, and can clear some
effects of winds in OB associations can drive off
of the debris from around the forming star, leav-
enough mass to unbind the association, explain-
ing any planetary system intact. For massive (O
ing why associations are not gravitationally
and B) stars the strong winds drive away a large
bound. Winds can also carry away some of the
angular momentum in a cloud, allowing the col-
lapse of the remaining material to continue.
Jet
Material
flowing
away from
Observer



Disk
(Cutaway View)




Material
flowing
towards
Jet
Observer

Fig 15.20. How an isotropic ¬‚ow can be converted into
two jets.The source of the isotropic out¬‚ow is in a hole in a
disk. In most directions the disk can block the ¬‚ow, but not
Fig 15.21. IR spectrum of Herbig“Haro object number
in the direction where the disk is thinnest. Jets emerge in
212. Notice the bright emission lines. [ESO]
that direction.
15 STAR FORMATION 285



15.6.4 T Tauri stars and related objects
Another group of pre-main sequence objects are
called T Tauri stars (Fig. 15.22). T Tauri is a vari-
able star in the constellation Taurus, and T
Tauri stars are variables with properties like
those of T Tauri. Light curves are shown in Fig.
15.23, and a spectrum is shown in Fig. 15.24.
These stars are spectral class K, and appear above
the main sequence on the HR diagram. These
show an irregular variability. Their spectra are
also characterized by the presence of emission
lines.
There are three possible sources of the vari-
ability we see:
(1) The variability could arise in a photosphere.
One model for this involves star spots. These
are dark areas, like sunspots, only larger. As
the star rotates, a different fraction of the



(a)




(b)
Fig 15.22. Images of two T Tauri stars. (a) T Tauri.The star
is unremarkable, but there is a small nebula to the right.The
star is in a dust cloud.The presence of the dust can be
deduced from the fact that it blocks light from the back-
ground stars.There are therefore fewer background stars in Fig 15.23. Light curves for a selection of T Tauri stars in
the center of the photograph than around the edges. (b) HL the Orion Nebula cluster.The horizontal axis is fractions of
Tau. Again, there is some nebulosity near the star and the a period (which is given in days in the upper right-hand
star is in a dust cloud. [(a) Courtesy of 2MASS/UMASS/ corner of each curve).Variability is caused by rotation of
IPAC/NASA/JPL/Caltech; (b) Laird Close, (University of stars with large, cool spots. [William Herbst (Wesleyan
Arizona)/Close, L., Astrophys. J., 486, 766, 1997] University)/Herbst,W. et al., Astrophys. J. Lett, 554, L197, 2001]
286 PART IV THE MILKY WAY



12500


10000


7500


5000


2500



6550 6560 6570 6580
Wavelength (angstroms)
Fig 15.24. Strong H emission line from a T Tauri star.
[George Herbig, IFA, Hawaii]



observed surface is covered by the spots, and
the brightness changes. Fig 15.25. Far IR image of protostellar core.This is a
(2) The variations may arise in the chromosphere. ground-based image from Mauna Kea, at 850 m.The beam
size is shown in the red circle to the lower right, so you can
(3) The variations may actually result from
see that the sources are barely resolved. Notice the separation
changes in the opacity of the dust shell sur-
into two sources.The irregular edges of the image are due to
rounding the star.
problems with the detectors near the edge.This is an example
The emission lines show Doppler-shifted of what is called a Class 0 protostar, which is thought to be the
youngest stage, where there is a strong out¬‚ow but the sur-
absorption wings, like those in Fig. 15.24. This sug-
rounding cloud has not been driven away. [Yancy Shirley,
gests material both falling into the star and mate-
University of Texas, Austin, made with SCUBA on the JCMT]
rial coming off the star. The infall may be close to
the star as the final stage of collapse, while the out-
flow is a wind (like the solar wind, but stronger) far- From studies of the spectral lines, we think
ther away from the star. Alternatively, the infall that the winds may have speeds of about
may be in the form of a disk around the star™s equa- 200 km/s. The mass loss in the wind, dM/dt, is
about 10 7 M /yr. The total luminosity in the
tor, while the outflow is along the polar axes.

Fig 15.26. HST images of
infrared emission from selected
disks around forming stars. All
six objects are in Taurus, at a
distance of 150 pc. [STScI/NASA]
15 STAR FORMATION 287



wind is that rate at which kinetic energy is car- contains the Orion Nebula and several smaller HII
ried away in the wind, regions. The Orion Nebula is actually on the front
side of the molecular cloud. Behind the HII region
(1/2)(dM/dt)v2
Lw (15.28)
is a dense molecular core. It is totally invisible in
Using the numbers given, we find a wind the optical part of the spectrum because of the
luminosity of about 1 L . That is, the star gives off foreground material. However, it can be studied
as much energy per second in its wind as the Sun in detail using radio observations of molecules. In
gives off at all wavelengths. However, the wind addition, it is a source of infrared emission.
phase is a short lived one. The wind does sweep
away some of the dust that has collected around
the star. We think that a similar wind from the
Sun was important in clearing debris out of the
early Solar System.
It is now possible to make far IR images of pro-
tostellar cores, like that in Fig. 15.25. Far IR emis-
sion HST and IR satellites have provided us with
some images, such as those in Fig. 15.26, which may
be the result of infrared emission from dust shells
around recently formed stars. Molecular spectral
line observations of these disks will require reso-
lutions achievable using interferometers.

15.7 Picture of a star forming
region: Orion
(a)
The Orion region (Fig. 15.27) is one of the most
extensively studied star forming regions. It is rel-
atively nearby, only about 500 pc from the Sun. It
is away from the plane of the Milky Way, so there
is little confusion with foreground and back-
ground stars. There is also an interesting variety
of activity in this region.
The region contains a large OB association.
There are four distinct subgroups. The two oldest
are near Orion™s belt, and the two youngest are
near the Orion Nebula (Fig. 15.28a) in Orion™s
sword. The Orion Nebula is an HII region powered
by the brightest stars in the youngest subgroup.
Images of the Orion Nebula and its cluster are in
Figs. 15.28(b)“(e).
The region also contains two giant molecular
cloud complexes, also shown in Fig. 15.4. A far IR
image of the GMCs is shown in Fig. 15.27(a). The
northern complex is associated with the belt
region, which contains the two oldest subgroups,
(b)
and a strong HII region just north of the
Fig 15.27. Large-scale images of the Orion region, mostly
Horsehead Nebula. The southern complex is asso-
at a distance of 500 pc. (a) Far IR image from IRAS. (b) X-ray
ciated with the sword region, which contains the
image from ROSAT. [(a) NASA, (b) NASA/MPI]
two youngest subgroups. The southern complex
288 PART IV THE MILKY WAY




(c)




(a)




(d)
Fig 15.28. Images of the Orion region, mostly at a distance
of 500 pc. (a) Large scale view of the belt and sword regions.
Two of the three belt stars appear at the top of the image.The
lower left belt star illuminates an HII region that has the
Horsehead Nebula superimposed.The sword region is near the
bottom.The Orion Nebula is overexposed in this image, and
surrounds the central sword star. (b) The Orion Nebula.There
is a small cluster of O and B stars that ionize this nebula.A
dense dust cloud provides over 100 mag of visual extinction in
the central part.The object near the top of the photo is a sep-
arate small HII region illuminated by a single star. Compare its
simple appearance to the complex structure of the big nebula.
(b)
15 STAR FORMATION 289




(e)
Fig 15.28. (Continued) (c) Central part of the Orion Nebula.This gives a better view of the O and B stars.The brightest four
stars in the central part of the cloud are called the Trapezium, because of their arrangement. Notice the dense dark cloud
intruding from the upper left. (d) A larger ¬eld HST image of the Orion Nebula, assembled from 15 smaller ¬eld images.This
image is approximately 1 pc across. (e) HST image of ¬laments in the nebula.The diagonal length of this image is 0.5 pc.The
plume of gas at the lower left is the result of a wind from a newly formed star. Red light depicts emission in nitrogen; green is
hydrogen; blue is oxygen. [(a) © Anglo-Australian Observatory/Royal Observator, Edinburgh. Photograph from ULK Schmidt
plates by David Malin; (b) Courtesy of 2MASS/UMASS/IPAC/NASA/JPL/Caltech; (c) ESO; (d), (e) STScI/NASA]



The general picture of the southern complex of gas heated to high temperatures in the regions
that has emerged has the HII region expanding where the wind strikes the surrounding cloud.
into the molecular cloud, compressing it. This Infrared line emission from H2 is seen from this
compression probably triggered a new generation 2000 K gas. We can also study the proper motions
of star formation. This new cluster of stars now of the maser and see that the region is expanding.
appears as a series of small infrared sources and (The best measurement of the distances to this
masers. There is evidence for an energetic flow. region comes from studying the proper motions
We see very broad (over 100 km/s) lines in CO and of masers, as discussed in the preceding section.)
other molecules. These wings have the character- It is likely that this dense region will appear as
istics of the bipolar flows discussed in the preced- another OB subgroup when there has been suffi-
ing section. We also see evidence for small regions cient time to clear the interstellar material away.


Chapter summary
In this chapter we saw how the various compo- saw how rotation can slow or stop the collapse,
nents of the interstellar medium, discussed in and lead to the formation of binaries, planetary
Chapter 14, are involved in the process of star for- systems and disks.
mation. We first looked at the conditions for grav- We saw how the magnetic field in a cloud
itational binding, and found that molecular can affect its collapse if the magnetic energy is
clouds are the likely sites of star formation. comparable to the gravitational potential
Current problems in star formation include how energy. As a cloud collapses, we expect that flux
the collapse is actually initiated, what fraction of freezing will lead to an increase in the magnetic
the mass is converted into stars, how planetary field strength within the cloud. We saw that
systems form, and how OB associations form. We there are two ways to overcome the support the
290 PART IV THE MILKY WAY



magnetic field provides. One produces a steady Once an O or early B star forms in a molecular
stream of low mass stars, and the other pro- cloud, the gas around the cloud is ionized, pro-
duces waves of high mass (as well as low mass) ducing an HII region. The size of the HII region is
star formation. determined by a balance between ionizations and
We saw how the molecular clouds, being recombinations. The continuous radiation from
cool and dense, are the most likely sites of star HII regions comes from free“free scattering of
formation. The most massive stars seem to be electrons. Recombinations lead to electrons pass-
born in the giant molecular clouds. These ing through many energy levels, giving off recom-
clouds have masses in excess of 105 M and are bination line radiation.
found in complexes with masses in excess of Other indicators of recent star formation are
106 M . masers and energetic flows. The bipolar flows pro-
We looked at how collapsing clouds eventu- vide indirect evidence for the existence of disks
ally become protostars, glowing mostly in the that collimate the flows.
near infrared. Their radiation is provided by the We looked at the Orion region as an nearby
decrease in the gravitational potential energy as example of ongoing high and low mass star for-
the cloud collapses. mation.


Questions
15.10. How is it possible for a gravitationally bound
15.1. Explain why cool dense regions are the most
cloud to give birth to a gravitationally
likely sites of star formation.
unbound association?
15.2. Suppose we have a cloud of 2 pc radius,
15.11. (a) What is the difference between ioniza-
formed of a material whose Jeans length is
tion bounded and density bounded HII
10 km. Will the cloud collapse?
regions? (b) Why are density bounded HII
15.3. As a cloud collapses, what happens to the
regions important?
Jeans mass of the new individual fragments?
15.12. Explain why the CII zone around a star is
15.4. If a cloud has a higher density in the center
much larger than the HII region. Consider
than at the edge when its collapse starts,
(a) the relative ionization energies, and
explain what happens to the density con-
(b) the relative abundances.
trast between the center and the outer part
15.13. Explain why almost all carbon ionizations
as the collapse continues. (Hint: Think about
result from photons with energies
free-fall time.)
between the ionization energy of carbon
*15.5. As a cloud collapses, the acceleration of the
and that of hydrogen, even though pho-
particles increases. Therefore, the free-fall
tons that can ionize hydrogen can also
time will be less than we would calculate on
ionize carbon.
the basis on constant acceleration. However,
15.14. Why is the temperature of an HII region so
we still use the initial free-fall time as an
high?
estimate of the total free-fall time. Why does
15.15. (a) How can an expanding HII region trig-
this work? (Hint: Think in terms of the time
ger star formation? (b) Why is the rate at
for the radius to halve, and halve again, and
which the gas can cool important to the
so on.)
process?
15.6. Explain why a rotating cloud flattens as it
15.16. Why is coherence important for a maser?
collapses.
15.17. How do we know that masers are small?
15.7. How does the formation of multiple star
15.18. What features of T Tauri stars lead us to
systems help in the star formation process?
believe that they are still in the formative
15.8. As a cloud collapses, is it likely that it will
rotate as a rigid body ( independent of r)? process?
15.9. What do we mean by a “trigger” for star
formation?
15 STAR FORMATION 291



Problems
15.1. Compare the Jeans mass and radius for ity if most of the energy is liberated over the
typical HI and molecular clouds. last 50 years of the collapse.
15.2. Suppose we made an interferometer with 15.14. From the evolutionary track in Fig. 15.6,
calculate the radius of a 3 M protostar
one telescope on the Earth and the other on
when it has reached a temperature of 104 K.
the Moon. What would be its angular
15.15. (a) For a star of radius R, emitting like a
resolution at a wavelength of 1 cm?
blackbody of temperature T, find an expres-
15.3. Suppose that a cloud contracts to one-tenth
of its initial size. How do the Jeans length sion for the number of photons per second
and mass compare with those in the original emitted capable of ionizing hydrogen. (b) For
h kT, evaluate the integral. (c) Why is
cloud?
15.4. Compare the density and pressure of a dense the approximation in (b) valid for an
interstellar cloud with those of the gas in ordinary star?
your room. 15.16. Compare the radii of HII regions around
15.5. (a) Find the Jeans length for a cloud with a an O5 and an O7 star if the density is
density 106 H2 molecules/cm3 and T 100 K. nH 103 cm 3.
(b) What angle does this subtend at a 15.17. What is the Doppler width of the H110 line
distance of 1 kpc from the Earth? at a wavelength of 6 cm, for a temperature
of 104 K?
15.6. Use the results of Example 15.2 to give the
free-fall time (in years) for objects whose 15.18. What is the shift from the H110 to the
density is given as a fraction of 105 atoms/cm3. C110 line wavelength, if it just depends on
15.7. An interstellar cloud is found to be rotating the nucleus“electron reduced mass?
such that a velocity shift of 1 km/s/pc is 15.19. Suppose an HII region is formed in a cloud
with a density of 104 H2 molecules/cm3. If the
observed across it. (a) What is the angular
temperature in the HII region is 104 K and the
speed (rad/s)? (b) What is the rotation period?
15.8. For an object on the surface of the Sun, temperature of the surrounding molecular
cloud is 102 K, how much will the HII region
compare the two terms on the right-hand
side of equation (15.14b). expand before the pressures equalize?
15.9. For a spherical interstellar cloud of mass *15.20. (a) Derive an expression, analogous to equa-
1000 M and radius 10 pc, rotating once tion (15.27), for the radius rS (He) of an ion-
every 107 years, compare the two terms on ized helium region around a star in terms of
Nuv (He), the number of photons per second
the right-hand side of equation (15.14b).
15.10. For a spherical interstellar cloud of mass emitted by the star capable of ionizing
1000 M and radius 10 pc, rotating once helium. Give your answer in terms of np and
every 107 years, compare the gravitational nHe . Assume that is the same for H and
potential energy with the kinetic energy of He. Assume that anywhere that He is ion-
ized, all of the H is ionized, so that ne np.
rotation.
15.11. For a uniform sphere, what does it mean for (b) Repeat the calculation for carbon, giving
d2I/dt2 to be small compared with the gravi- your answer in terms of Nuv (C), np and nC.
tational potential energy? For carbon, assume that it is ionized over a
15.12. For the protostar considered in Example 15.6, much larger region than H because of its
lower ionization energy, so that ne nC.
how much more energy will be released as
the star contracts by another factor of five? 15.21. From observations of the masers in Orion,
If this takes another 100 years, what is the we find an average radial velocity of
average luminosity over that period? 24.0 km/s, and proper motions of 0.01 arc
15.13. For a 10 M protostar that has collapsed to a sec/yr. How far away are these masers?
radius of 1000 R , (a) calculate the energy 15.22. (a) For a maser, show that a population
that has been liberated to this point, and inversion corresponds to a negative excita-
(b) use this to calculate the average luminos- tion temperature. (b) The negative excitation
292 PART IV THE MILKY WAY



temperature gives a negative optical depth. 15.27. (a) For the typical T Tauri wind described in
Show from the radiative transfer equation this chapter, what is the momentum per sec-
that this implies amplification. ond carried away by the wind? (b) If the
15.23. If the Sun turns off, how long will it take wind drives away dust, and slows by conser-
the light to go from full intensity to zero? vation of momentum, and the wind is effec-
15.24. (a) If an object dims in a day, how large can tive at driving dust away until it slows to
it be? (b) What angle would it subtend at a 5 km/s, what is the rate at which dust can be
distance of 1 kpc? driven away?
15.25. Calculate the rate at which energy is deliv- 15.28. By how much is the H line shifted by the
ered to a cloud by a 10 8 M /yr flow with a 200 km/s wind in T Tauri?
speed of 100 km/s. 15.29. Calculate the energy used up in ionizing
1 M of atomic H. To what radius must a
15.26. For the flow in the previous question, what is
1 M protostar collapse for this much energy
the rate at which interstellar material is
being swept up if the sweeping occurs as long to be released in the change in gravitational
as the wind speed is greater than 5 km/s? potential energy?


Computer problems

15.1. Make a table showing the Jeans mass and length 15.3. Suppose we start with an interstellar cloud with a
for clouds with densities of 1, 103 and 105 H density of 105 H atoms/cm3. Using the variable
atoms/cm3, and kinetic temperatures of 10, 30 and acceleration in equation (15.6), find the actual
100 K. collapse time, and compare it with the estimate
15.2. Add a column to Table 15.2, giving the Stromgren for the free-fall time, given in equation (15.8).
radius in each case. Assume that np is 104 cm 3. 15.4. Reproduce Table 15.2, using the radii and temper-
2.6 10 13 cm3/s.
Take a constant value of atures of the spectral types given.
Chapter 16




The Milky Way galaxy

around them. Population I stars are confined to
16.1 Overview the galactic plane.
Population II stars are thought of as being the
Throughout this book we have discussed the com- “old” component of the galaxy. They are found in
ponents of our galaxy: stars, clusters of stars, globular clusters and are characterized by low
interstellar gas and dust. We now look at how metalicity. They have no gas and dust around
these components are arranged in the galaxy. The them. Their galactic distribution is very different
study of the large scale structure of our galaxy is from that of population I stars. The population II
difficult from our particular viewing point. We stars form a spherical distribution, as opposed to
are in the plane of the galaxy, so all we see is a a disk. This spherical distribution is sometimes
called the halo. When we talk about a spherical
band of light (Fig. 16.1). The interstellar dust pre-
vents us from seeing very far into the galaxy. We distribution, we do not mean just a spherical
see a distorted view. shell around the galaxy. Instead, we mean the
The first evidence on our true position in the spherically symmetric distribution whose density
galaxy came from the work of Harlow Shapley, falls off with increasing distance from the galac-
who studied the distribution of globular clusters tic center. Population II objects also seem to have
(Fig. 16.2). He found the distances to the clusters a larger velocity spread in their motions than do
from observations of Cepheids and RR Lyrae stars. population I objects. Table 16.1 shows the charac-
Shapley found that the globular clusters form a teristic thicknesses and velocity dispersions for
spherical distribution. The center of this distribu- some components of the galaxy.
tion is some 10 kpc from the Sun. Presumably, the The schematic arrangement of these compo-
center of the globular cluster distribution is the nents is shown in Fig. 16.3. First we see the disk
center of the galaxy. This means that we are and the halo. Note that the halo has a spherical
about 10 kpc from the galactic center. distribution with a density of material that falls
In Chapter 13, when we studied HR diagrams off radially. We then look at the disk. First there is
for clusters, we introduced the concept of stellar an overhead view, showing the location of the Sun.
populations I and II. The distribution of these The best estimates, which we will discuss later in
populations in the galaxy can help us understand this chapter, place the Sun 8.5 kpc from the galac-
how the galaxy has evolved. Population I material tic center, about halfway out to the edge of the disk.
is loosely thought of as being the young material We then look at a side view of the disk. The inner
in the galaxy. Population I stars are found in part of the disk (closer to the center than the Sun)
galactic clusters, and are characterized by high is relatively thin and flat. The outer part of the disk
is warped and also gets thicker, that is, it flares. We
metalicity. Some are also associated with inter-
stellar gas and dust, suggesting that they are will also discuss the evidence for this in this
chapter. Surrounding the center region is a bulge.
young enough to have some of their parent cloud
294 PART IV THE MILKY WAY




(a)

(b)

(c)

(d)

(e)

(f)

(g)

(h)

(i)

(j)
Fig 16.1. Images of the Milky Way at different wavelengths. (a) Radio continuum (408 MHz), from Bonn, Jodrell Bank and Parkes.
(b) HI column density, derived from 21 cm emission (Leiden“Dwingeloo survey). (c) Radio continuum (2.4“2.7 GHz), from Bonn and
Parkes. (d) H2 column densities based on CO emission at 115 GHz (CFA at Harvard and in Chile). (e) Composite mid and far infrared
from IRAS (1.25, 2.2 and 3.5 m). (f) Mid infrared (6.8“10.8 m), data from the SPIRIT III instrument on the Midcourse Space
Experiment (MSX) satellite. (g) Near infrared from COBE (12, 60 and 100 m). (h) Optical from wide ¬eld photographs (compiled at
ESO). (i) X-ray (0.25, 0.75 and 1.5 keV) from ROSAT. (j) Gamma-ray (CGRO).All of the images (except the optical) have angular reso-
lutions of between one-half and two degrees.All images are centered on the galactic center, and cover 10 above and below the galac-
tic plane. [Courtesy of NASA, provided by the GSFC Astrophysics Data Facility (ADF) (a), (e), (g), (i), (j) ADF; (b) Dap Hartmann,“The
Leider/Dwingeloo Survey of Galactic Neutral Hydrogen”, Ph.D.Thesis, University of Leider, 1974; (c) A. R. Duncan, University of
Queensland,Australia; (d) Thonas Dame, CFA; (f) Stephan D. Price, Hanscom AFB; (h) Axel Mellinger, University of Potsdam, Germany]




16.2 Differential galactic rotation

All of the material in the galaxy orbits the galactic
center. If the galaxy were a rigid body, all of the gas
and dust would orbit with the same period.
However, material closer to the galactic center
orbits with a shorter period than material farther
out. This is not an unusual situation. After all, the
planets in the Solar System exhibit the same behav-
ior: Mercury takes less time to orbit the Sun than
does the Earth, and so on. When the orbital period
depends on the distance from the center, we say
that the material is exhibiting differential rotation.
(a)
16 THE MILKY WAY GALAXY 295




Disk



Halo



(a)

(b)



Sun



8.5 kpc
Galactic
Center




(b)

(c)
Fig 16.2. Three-dimensional distribution of globular rp
Bulge
Wa Flare
clusters. (X, Y, Z) are linear coordinates (kpc) centered at
the galactic center.The galactic plane is Z 0.The Sun is
Disk
Flare
located at ( 8.5, 0, 0). (a) Overhead view.The dashed circle rp
Wa
shows the extent of the galactic disk. (b) Side view, with the
(c)
solid line indicating the extent of the disk. (c) The other side
view, with the solid line indicating the extent of the disk. Fig 16.3. Schematic arrangements of major components
[William Harris, McMaster University] of the galaxy. (a) The relationship between the halo and the
disk.The lighter shading as one moves farther out in the
halo is to suggest a decrease in density with distance from
16.2.1 Rotation and mass distribution the center. (b) Top view of the disk, showing the location of
the Sun. (c) Side view of the disk, showing the warp and ¬‚are
The orbital period of any particle will depend
in the outer part of the disk, and the bulge near the galactic
on the mass about which it is orbiting and the
center.
radius of the orbit. Just as we use the period
296 PART IV THE MILKY WAY



Therefore, if we can measure v(R), we can deduce
Table 16.1. Scale heightsa and velocity
M(R), the mass distribution in the galaxy.
dispersions.
Equivalently, we can use (R), since
1R2
Scale Velocity
v1R2>R (16.4)
Constituent height (pc) dispersion (km/s)
Substituting into equation (16.3) gives
O Stars 50 5
1R2R3>G
2
M1R2 (16.5)
GMCs 60 7
Galactic clusters 80 If all of the mass is, indeed, concentrated at
HI clouds 120 the center of the galaxy, then M(R) is a constant,
F stars 190 20
so equation (16.5) gives Kepler™s third law (men-
Planetary nebulae 260 20
tioned in Chapter 5). (We speak of the orbits as
M stars 350
being “Keplerian”.) The function v(R), or (R), is
RR Lyrae stars:
called the rotation curve for the galaxy.
Short period 900 25
Long period 2000 30
16.2.2 Rotation curve and Doppler shift
Globular clusters 3000
The differential galactic rotation produces
a
Scale height is the distance over which a quantity Doppler shifts in spectral lines that we observe
falls to 1/e of its maximum value. from gas at different distances from the galactic
center than the Sun. This is illustrated in Fig.
16.4. In Fig. 16.4(a), we look at five test particles at
different distances along the line of sight. In each
and size of the Earth™s orbit to tell us the mass case the Doppler shift depends on the relative
of the Sun, we can use the orbital periods at radial velocity of the test particle and the Sun.
different distances from the galactic center to That is, we take the line of sight component of
tell us about the distribution of mass in the the particle™s motion and subtract the line of
galaxy. sight component of the Sun™s motion.
To see how this works, we assume that all mat- In Fig. 16.4(b) we look at the Doppler shifts
ter follows circular orbits. At a distance R from for each test particle. Point 1 is slightly closer to
the center, the orbital speed is v(R), and the angu- the center than the Sun. It is moving slightly
lar speed is (R). The mass interior to radius R is faster than we are, so there will be a small
Doppler shift. It is moving away from us so that
1r2 dV
R
M1R2 (16.1) the shift will be to longer wavelength (redshift).
0
Point 2 is where our line of sight crosses the
where (r) is the density at radius r, and dV is a same circle. The speed is the same as at point 1,
volume element. For a spherical mass distribu- and the angle with the line of sight is the nega-
tion, the motion of an object at R depends only tive of that at point 1. Since the line of sight
on M(R). Furthermore, the mass M(R) behaves as if component depends on the cosine of that angle,
it were all concentrated at the center. (This also that component is the same. The Doppler shift
works for particles in the plane of a thin disk.) for point 2 is therefore the same as for point 1.
For a particle of mass m orbiting at a radius R, Point 3 is where the line of sight passes closest
the gravitational force is GM(R)m/R2, and this must to the center. Material is moving fastest around
provide the acceleration for circular motion, so that circle. It is also moving directly away from
mv2 1R2 us, so that point has the largest Doppler shift to
GM1R2m
(16.2) the red. Point 4 is in the same orbit as the Sun.
R2 R
Our distance from that object is constant, so our
Solving for M(R) gives relative radial velocity must be zero, meaning
v2 1R2R that the Doppler shift is zero. Point 5 is farther
M1R2 (16.3) from the center than the Sun, and is moving
G
16 THE MILKY WAY GALAXY 297



measured around the galactic plane, starting in
the direction of the galactic center.
We also define a convenient reference frame
Sun
for measuring velocities, called the local stan-
1 dard of rest, LSR. If the only motion the Sun had
were its orbital motion about the galactic cen-
ter, then the local standard of rest would coin-
3
GC cide with the Sun™s motion. That is, we could
2
simply measure motions with respect to the
4
Sun (so-called heliocentric velocities). However,
because of gravitational interactions with its
5
nearest neighbors, the Sun has a small motion
superimposed on its circular orbital motion, so
Line of
it is not a convenient reference point for veloci-
Sight
ties. There are actually two ways of defining the
(a)
LSR:
(1) Dynamical. The origin of the coordinate sys-
12 3
4 tem orbits at a distance R0 from the galactic
5
center, and R0 is the distance of the Sun from
the galactic center. The coordinate system
moves with a velocity v(R0) v0, or (R0) 0,
appropriate for circular motion at R0. Defined



Doppler Shift
Longer
Shorter
Wavelength
Wavelength Sun
(b)
Fig 16.4. Doppler shift produced by galactic differential
l
rotation by material at different locations along a given line
of sight. (a) The locations of ¬ve test particles in an overhead
Galactic
view. Arrows for each particle indicate the velocity (magni-
Center
tude and direction). For each particle the Doppler shift will
depend on the relative radial velocity of that particle and the
Sun. (b) For each particle, the position of the arrow shows
the amount of Doppler shift.



slower than us. We are therefore overtaking it,
so there is a Doppler shift to shorter wavelength
(blueshift).
In determining the rotation curve for our
galaxy, it is convenient to introduce a set of coor-
b
dinates, known as galactic coordinates, measured
Disk
from our viewing point at the Sun. They are
Galactic
Sun
shown in Fig. 16.5. The galactic latitude, b, meas- Center
ures the angle above or below the galactic plane
Fig 16.5. Galactic coordinates.
of an object. The galactic longitude, /, is an angle
298 PART IV THE MILKY WAY



this way, the motion of the LSR depends only Using the relationship between sines and
on M(R0). cosines, gives
(2) Kinematic. The origin of the coordinate system
vr v(R) sin v0 sin
moves with the average velocity of all the
R (R) sin R0 sin (16.7)
stars in the vicinity of the Sun. This averages 0
out the effects of the random motions of
We can measure , but not , so we must elimi-
these stars.
nate it using the law of sines:
The two definitions should result in the same sin(180 )/R0 sin /R (16.8)
velocity system. However, there are small differ-
Simplifying the left-hand side gives
ences, which we will ignore. (The difference tells
us about the dynamical properties of the galaxy.) sin / R
sin /R0 (16.9)
With respect to the LSR, the Sun is moving at
Substituting into equation (16.7) gives
about 20 km/s towards a right ascension of 18 h,
and a declination of 30 . (In galactic coordinates, vr R0 (R)sin R0 sin (16.10)
0
56 , b 23 .) Once we know the Sun™s
this is
Factoring out the R0 sin gives
motion, we can use it to correct Doppler shift
measurements to give us the radial velocity of the vr [ (R) 0]R0sin (16.11)
object with respect to the LSR.
In tracing the behavior of vr, it is convenient
We now look at the Doppler shifts we will
to divide the galaxy into quadrants, based on the
observe for some material a distance R from the
value of . This is illustrated in Fig. 16.7.
galactic center, moving in a circular orbit with a
Let™s look along a line of sight at some galac-
speed v(R). The situation is shown in Fig. 16.6 for
tic longitude and see how the radial velocity
R R0, but the result holds for R R0 (see Problem
changes with increasing distance d from the Sun.
16.12). The relative radial velocity is given by
This is illustrated in Fig. 16.8. We first look at the
vr v(R) cos (90 v0 cos (90
) ) (16.6) case for 90 (first quadrant). As we look at
material closer to the galactic center, the quantity
0] becomes larger. This means that vr
[ (R)
Sun
increases. We see that this line of sight has a
v0
d
l
R0
P

III II
v
GC Rmin Sun

IV I
Galactic
Center


Fig 16.6. Differential rotation and radial velocities.The sun
is a distance R0 from the galactic center.We observe an
object at point P, along a line making an angle with the line
of sight from us to the galactic center. P is a distance R from
Fig 16.7. Galactic quadrants.
the galactic center.
16 THE MILKY WAY GALAXY 299



when it is negative in the first quadrant it is pos-
100
itive in the fourth quadrant.
In the second quadrant (90 180 ), all
terminal velocity
lines of sight pass only through material out-
side the Sun™s orbit. There is no maximum vr; it
45°
50
just increases with d. The behavior in the third
quadrant is the negative of that in the second
quadrant.
vr (km/s)




We can also find an expression for the relative
0
transverse velocity. This will produce proper
motions. The relative velocity is given by
vT v(R) sin (90 ) v0 sin (90 )
’50
v(R) cos v0 cos
135°

R (R) cos R0 cos (16.12)
0

From Fig. 16.6 we see that
’100
0 5 10 15 20 R0 cos d R cos (16.13)
radius (kpc)
This gives
Fig 16.8. Radial velocity as a function of distance from
R cos R0 cos d (16.14)
the Sun, d, along a given line of sight.The upper curve is
typical of between 0 and 90 .The maximum vr corresponds
Substituting this into equation (16.12) gives
to the point at which the line of sight passes closest to the
galactic center.The close point with vr 0 corresponds to vT d] R0
(R)[R0 cos cos (16.15)
0
local material, and the far point with vr 0 corresponds to
Grouping the terms with cos gives
our line of sight crossing the Sun™s orbit. Inside the Sun™s orbit,
each vr (except for the maximum) occurs twice.The lower
vT (R) d
[ (R) 0]R0 cos (16.16)
curve is typical of values of between 90 and 180 . All of
The quantity R0, our distance from the galac-
these points are outside the Sun™s orbit, so each circle is
crossed only once, and each vr is reached only once. tic center, is determined from studies of the dis-
[F rancoise Combes, Observatoire de Meudon] tribution of globular clusters, and more recently
from the studies of clusters of masers near the
galactic center. For approximately 20 years prior
to 1985, the generally accepted value was
10.0 kpc. However, data accumulated by 1985 sug-
point of closest approach to the galactic center.
gest a smaller value. As of 1985, the International
This point is called the subcentral point. Of all the
Astronomical Union started recommending the
material along this line of sight, the material at
value
the subcentral point produces the largest vr. As we
go beyond the subcentral point, we are recrossing R0 8.5 kpc
orbits, and vr becomes smaller. Eventually, vr
By using an agreed upon value, astronomers
reaches zero, when the Sun™s orbit is crossed again.
can be sure that they are using the same values
For points beyond the Sun™s orbit, (R) 0. This
when they compare their studies of various
means that vr is negative and increases in
aspects of galactic structure. Prior to 1985, the
absolute value as d increases.
adopted value for v0, the orbital speed about the
For 270 (the fourth quadrant), the behav-
galactic center, was 250 km/s. The value recom-
ior of vr is similar to that in the first quadrant,
mended in 1985, to go with the new R0, is
except that when vr is positive in the first quad-
v0 220 km/s
rant it is negative in the fourth quadrant, and
300 PART IV THE MILKY WAY



maximum Doppler shift. We then assign that
Example 16.1 Galactic rotation
Doppler shift to material at the subcentral point
For the values of the galactic rotation constants,
(the point of closest approach to the galactic cen-
find the time it takes the Sun to orbit the galactic
ter) for that particular longitude. We can see
center and the mass interior to the Sun™s orbit.
from Fig. 16.6 that the distance of the subcentral
point to the galactic center, Rmin, is
SOLUTION
The time for the Sun to orbit is simply the circum- Rmin R0 sin (16.17)
ference, 2 R0, divided by the speed v0:
From equation (16.11), we see that if vmax is
2 R0 the maximum radial velocity along a given line of
t
v0 sight, then the angular speed (Rmin) for that line
2 18.5 103 pc 2 13.1 1013 km>pc 2
of sight is given by

1220 km>s2 (R0 sin ) (vmax/R0 sin ) (16.18)
0


1015 s By studying lines of sight with longitudes
7.5
ranging from 0 to 90 , the corresponding value
108 yr
2.4
of Rmin will range from zero to R0. This means
that we measure (R) once for each value of R
The Sun orbits the galactic center in 240 million
from zero to R0. However, we have already said
years. We find the mass interior to R0 from equa-
that, if the material is moving in circular orbits,
tion (16.3):
one measurement per orbit is sufficient to deter-
v2R0
0 mine the rotation curve.
M1R2
G There are some limitations to this technique.
1220 105 cm>s 2 2 18.5 103 pc2 13.1 1018 cm>pc 2 We have already seen that the distribution of
16.67 dyn cm2>g2 2 interstellar gas is irregular. If there happens to be
8
10
no atomic hydrogen at the subcentral point for
1044 g
2.0 some line of sight, we will see a vmax which is less
than the value that we would see if there were
1011 M
1.0
material at the subcentral point. There are also
problems arising from non-circular orbits. The
effect of both of these problems can be reduced
16.3 Determination of the
by repeating the procedure for the fourth quad-
rotation curve rant. Because of the inclination of the galactic
plane relative to the celestial equator, the fourth
The rotation curve for material within the Sun™s quadrant studies must be performed in the south-
orbit can be determined reasonably well from ern hemisphere, and have been done in Australia.
21 cm line observations. In determining the rota- For a number of years it seemed that there were
tion curve, we make two important assumptions: disagreements between the first and fourth quad-
(1) The orbits are circular. This means that we rant rotation curves, but we now think we under-
need to determine v(R) at only one point for each stand them in terms of non-circular motions. There
value of R. (2) There is some atomic hydrogen all is further evidence for such motions in the large
along any given line of sight. It is especially radial velocities observed for close to 0 and 180 .
important that there be some hydrogen at the Another problem is that we cannot really cover
subcentral point of each line of sight. the full range from zero to R0. For close to zero,
The method takes advantage of the fact that, sin is close to zero, and the Doppler shift is very
for lines of sight through the part of the galaxy small. Random motions of clouds are much larger
interior to the Sun™s orbit, there is a maximum than the radial velocity due to galactic rotation.
Doppler shift. It is easy to inspect the 21 cm Similarly, for near 90 , is close to 0, providing
spectrum at each longitude and determine the a small radial velocity due to galactic rotation.
16 THE MILKY WAY GALAXY 301



parallax. This gives a reliable rotation curve at
least out to about 20 kpc. The combined rotation
curve (using HI inside the solar circle and CO out-
side, is shown in Fig. 16.9. There is no falloff in
v(R) out to 20 kpc, and there may even be a slight
rise. This means that there is much more mass
outside the Sun™s orbit than previously thought!
We can see from equation (16.3) that if v(R) is
constant from 8 to 16 kpc, then M(16 kpc) will be
twice M(8 kpc). This means that there is as much
mass between 8 and 16 kpc as there is out to 8 kpc.
However, the luminosity of our galaxy is falling
very fast as R increases. Since the luminous part
of the matter is mostly in the disk, it would seem
Fig 16.9. Rotation curve for the Milky Way.The curve
that this extra mass cannot be part of the disk.
inside the Sun™s orbit is well determined from 21 cm studies.
Current thinking places the extra mass in the
Outside the Sun™s orbit, we use HII regions in molecular
halo of the galaxy. We still have little idea of what
clouds.We use spectroscopic parallax on the exciting stars
to give the distance to the HII regions, and then CO form this matter takes. It has been suggested that
observations to measure the Doppler shift. [Daniel Clemens it can be in the form of faint red stars, but recent
(Boston University) Clemens, D., Astrophys. J., 295, 422, 1985]
results make this seen unlikely. This is our first
encounter with dark matter, matter whose gravi-
tational effects are felt, but which is not very
When we eliminate these ends, a reasonable rota-
luminous. (Astronomers used to call this “miss-
tion curve has been derived for R in the range 3 to
ing” matter, but it is not missing. We can tell that
8 kpc. This is shown as part of the curve in Fig. 16.9.
it is there by its gravitational effects. We just
Using HI to determine the rotation curve out-
can™t see it.) We will see that there is strong evi-
side the Sun™s orbit is more difficult. There is no
dence for dark matter in other galaxies, and we
maximum Doppler shift along any line of sight. It
will discuss it farther in Chapter 17.
is therefore necessary to measure independently v
and d, the distance from the Sun to the material
Example 16.2 Galactic mass distribution
being studied. From d and , we can deduce R (see
For what (spherically symmetric) mass distribution
Problem 16.6). Until molecular observations, there
is v(R) constant?
was no reliable rotation curve for R R0. The best
astronomers could do was to derive the mass dis-
SOLUTION
tribution for R R0, and then make some assump-
From equation (16.3), we know how M(R) is related
tions about how the mass distribution would
to v(R), and from equation (16.1) we know how (r)
continue for R R0. From the assumed mass dis-
is related to M(R).
tribution, a rotation curve could be derived. It
For a spherical coordinate system, the volume
was assumed that there was relatively little mass
element is 4 r2dr, so equation (16.1) becomes
outside the Sun™s orbit, so the rotation curve was
1r2 4 r 2 dr
R
characterized by a falloff in v(R) that was close to M1R2
that predicted by Kepler™s third law. 0
However, recent observations of molecular
Differentiating both sides tells us that
clouds have provided a direct method of measur-
1r2 4 r2
dM1r2
ing the rotation curve outside the Sun™s orbit.
Molecular clouds associated with HII regions were dr
studied. The radial velocities were determined from
or
observations of the carbon monoxide (CO) emission
1r2
1 dM1r2
from the clouds. The distance to the stars exciting
4 r2 dr
the HII regions were determined by spectroscopic
302 PART IV THE MILKY WAY



If v(R) is a constant, say v0, then equation (16.3) tells ambiguity. Remember, there is no distance ambi-
us that guity for material outside the Sun™s orbit, making
this an interesting part of the galaxy to study.
v2r
0
M1r2
G
16.4 Average gas distribution
v2
dM1r2 0
To understand star formation on a galactic scale,
G
dr
we must know how the interstellar gas, out of
Equating the two expressions for dM(r)/dr gives
which the stars will be formed, is distributed in the
v2 galaxy. We are interested in the average distribu-
1r2
0
tions of various constituents. By “average” we mean
4 r2G
that we are interested only in the large-scale struc-
This means that the density falls as 1/r2. While this
ture. We would like to know the radial distribution
sounds like a rapid falloff, remember that the
of interstellar gas. (Remember, this is not the same
volume of a shell of thickness dr and radius r is
as M(R), which includes mass in all forms.) We
4 r2 dr. So, in calculating the mass of each shell,
would also like to know the degree to which the gas
the r2 factors cancel. This means that, as we go
is confined to the disk. We can express this as a
farther out, the mass of each shell stays constant.
thickness of the disk, as determined from various
So, as far as we continue to add shells with this 1/r2
constituents. We would also like to know whether
density falloff, the mass of the galaxy grows by the
the thickness is constant, or whether it varies with
same amount with every shell we add.
position in the galaxy. Finally, we would like to
Once we have a rotation curve for our galaxy, know if the plane of the galaxy is truly flat, or if it
it is possible to use measured Doppler shifts to has some large-scale bumps and wiggles.
determine distances to objects. Since these dis- We first look at HI. The amount of HI doesn™t
tances are determined from the motions of the fall off very quickly as one goes to larger R. For
objects, they are called kinematic distances. For any example, the mass of HI interior to R0 is about
particular object, we measure vr and . We then 1 109 M , and the mass exterior to R0 is about
determine the angular speed from
1vr>R0 sin /2 (16.19)
0

It is assumed that we know R0 and 0. Once
we know we can use the known rotation curve
to find the value of R to which the corresponds.
There are a few limitations to this technique.
It does not work for material whose radial veloc-
ity due to galactic rotation is less than that due to
the random motions of the clouds. This rules out
material near longitudes of 0° and 180°, as well
as material close to the Sun.
Another problem arises for material inside
the Sun™s orbit. There are two points along the
same line of sight that produce the same radial
Fig 16.10. Radial distribution of H2, with the H2 deduced
velocity. (The one exception is the subcentral
from CO observations, assuming a constant conversion from
point.) Both of these points are the same distance
CO luminosity to mass.There is growing evidence that this
from the galactic center, but they are different conversion factor actually changes with environment, and
distances from us. This problem is called the dis- this curve may underestimate the mass in the outer galaxy
tance ambiguity. We can use the rotation curve to by a factor of three. Also, remember that there is a larger
say that the object is in one of two places, and we volume, so even a lower density of material can still translate
must then use other information to resolve the into a signi¬cant mass. [Thomas Dame, CFA]
16 THE MILKY WAY GALAXY 303



2 109 M . (Of course, this larger mass is spread 1.2

out over a larger volume. See Problem 16.5.) Note
1.0
that the mass of HI is only about 1% of the total
mass interior to a given radius. This means that 0.8
the gas does not provide most of the large-scale




h (kpc)
gravitational force in the galaxy. It just responds 0.6
to the gravitational effects of the stars, and what-
ever dark matter there is in the halo. 0.4

The abundance of H2 (Fig. 16.10) falls off more
0.2
rapidly with R than does that of HI. Inside the
Sun™s orbit, the mass of H2 is approximately equal 0.0
to that of HI, about 1 109 M . Outside the Sun™s 15 20 25 30
10
R (kpc)
orbit, the mass of H2 is about 5 108 M , about
one-quarter that of HI. There appears to be a peak Fig 16.11. Thickness of the plane of the Milky Way as
determined from 21 cm observations.The plot shows the
in the H2 distribution about 6 kpc from the galac-
layer thickness (measured as a gaussian width), with dotted
tic center. This is sometimes called the molecular
lines for range 20 160 , the dashed for 200 340 ,
ring. It appears that most of the H2 seems to be
and the solid line the average of both ranges. is galactocen-
concentrated into a few thousand giant molecular
tric azimuth, where 0 is the direction 0 . [Butler
clouds, rather than a large number of small
Burton, Sterrewacht, Leiden University]
clouds. The distribution of molecular hydrogen
H2 is generally deduced indirectly from observa-
confined to the plane. Fig. 16.13 shows the thick-
tions of CO. There are still disagreements over
ness of the plane as measured by various tracers. In
how to derive the H2 abundance from the inten-
general, we think that components that are more
sity of the CO emission. There is growing evi-
closely tied to the plane are younger. This means
dence that the conversion factor changes with
that HII regions and molecular clouds have formed
galactic environment, and that the mass of the
more recently than HI clouds or globular clusters.
outer galaxy (like the derived mass distribution
Molecular hydrogen is more closely confined
in Fig. 16.10) are underestimated by as much as a
to the galactic plane than atomic hydrogen. The
factor of three.
We generally express the thickness of the disk
4
by finding the separation between the two
points, one above and one below, at which the HI
3
density falls to half of its value in the middle of
the plane. This is called the full-width at half-
z (kpc)




maximum or FWHM. At the orbit of the Sun, the 2

thickness of the HI layer is about 300 pc. At R
15 kpc, the thickness is about 1 kpc. This trend is 1
shown in Fig. 16.11. This means that the plane
becomes thicker as one goes farther out from the 0
center of the galaxy (as depicted schematically in
Fig. 16.3c). This is called the “flaring” of the galac- -1
10 15 20 25 30
tic disk. In addition, we find that the disk isn™t
R (kpc)
flat. It has a warp to it, like the brim of a hat (Fig.
16.12). This is also shown schematically in Fig. Fig 16.12. Height of the plane of the Milky Way, as deter-
16.3(c). The warp is most prominent outside the mined from 21 cm observations.The solid contours are
Sun™s orbit. The bend is upward in the first and heights (in kpc) above the plane, and dashed contours are
second quadrants and downward in the third and heights below the plane, where the solid lines trace the warp
heights in the ¬rst and second quadrants (upper) and in the
fourth quadrants.
third and fourth (lower).The dotted lines are ¬ts used in
It is also interesting to see the degree to which
modelling. [Butler Burton, Sterrewacht, Leiden University]
various constitents (atomic, molecular, ionized) are
304 PART IV THE MILKY WAY



from our own, rather than part of the interstellar
1
medium in our galaxy. Some examples of spiral
l = 320° ’ 348° 100 m
galaxies are shown in Fig. 17.3. Given that many
HI
.8 CO other galaxies seem to have a spiral appearance,
Normalized Intensity




5GHz
it is reasonable to wonder whether our galaxy is
.6 also a spiral. Unfortunately, from our vantage
point within the galaxy, it is very difficult to see
.4 the overall pattern. In other galaxies, the spiral
pattern appears in the disk.
However, we can detect certain similarities
.2
between our galaxy and spiral galaxies. In partic-
ular, spiral galaxies have a significant amount of
0
gas and dust. The amount of gas and dust in our
’4 ’2 0 2 4 galaxy is comparable to that in other spirals. For
Galactic Latitude (Deg.) this reason, astronomers have been encouraged
in trying to unravel the spiral structure of our
Fig 16.13. Thickness of the plane as measure by four trac-
galaxy.
ers. In general, components that are more concentrated to
the plane are thought of as being younger, in the sense that
16.5.1 Optical tracers of spiral structure
they have not had as much time to spread out. [Eli Dwek,
NASA/Sodroski,T. et al., Astrophys. J., 322, 101, 1987] When we look at spiral galaxies we see that the
spiral arms are not continuous bands of light.
Rather, they appear to contain knots of bright
thickness of the plane in H2 is about half that in stars and glowing gas. For example, it appears
HI. We have seen that one feature of population I that HII regions and OB associations trace out
material is its confinement to the galactic plane. spiral arms in other galaxies. For this reason, we
We therefore think of molecular clouds as repre- have tried to see if the HII regions we can see
senting a more extreme population I than that optically in our galaxy form any distinct pat-
represented by the HI clouds. This might indicate tern. By using optical observations we can rely
that molecular clouds, as a whole, are more on distances determined by spectroscopic paral-
recently formed than the HI clouds. Finally, the lax for the stars exciting the HII regions.
H2 shows the same flaring and warp as the HI. Similarly, we can also look at the distribution of
Additional information on the total gas distri- OB associations. A drawback is that, with optical
bution (HI plus H2) comes from observations of observations, we cannot see very far along the
gamma-rays. These gamma-rays are created when plane of the galaxy. When we study the distribu-
cosmic rays strike protons. It doesn™t matter what tion of HII regions and OB associations, it is
types of atoms or molecules the protons are in. clear that the placement is not entirely random.
The results are somewhat uncertain, but provide We seem to see at least pieces of connected
an additional constraint. The general conclusion chains of HII regions and OB associations. These
from such observations supports the conclusions pieces have been identified as a series of named
from the CO observations. “arms”, identified by the constellation in which
they are most prominent. This is all a tantaliz-
ing hint of spiral structure, but is not a defini-
16.5 Spiral structure in
tive picture.
the Milky Way
16.5.2 Radio tracers of spiral structure
Early photographic surveys revealed a number of We can view the situation on a larger scale by
nebulae with a spiral appearance. In Chapter 17 using radio observations to look at the distribution
we will discuss the reasons for believing that of interstellar gas. This allows us to see across the
these spiral nebulae are distant galaxies, distinct whole galaxy. We can utilize kinematic distances,
16 THE MILKY WAY GALAXY 305



studying the spiral structure in inner parts of our
own galaxy, especially with the distance ambigu-
ity. Or, it may be telling us that the spiral pattern
in the inner galaxy is not that well defined. We
will see in Chapter 17 that, in many other spirals,
the pattern gets stronger as one moves farther out.
Outside the Sun™s orbit, the approach is more
direct. Since we have a rotation curve for the outer
part of the galaxy and there is no distance ambi-
guity, it is easier to trace out the large-scale struc-
ture. It is in the regions outside the Sun™s orbit
that we see the best evidence for spiral structure,
with some features being traced over at least a
quarter the circumference of the galaxy (Fig. 16.15).
There is a growing confidence that the outer part
of our galaxy is a four arm spiral. This work is still
going on.
Much of our understanding of spiral struc-
ture comes from comparing our galaxy to other

Locations of Mapped Clouds
Fig 16.14. Spiral structure of the galaxy as determined
from giant molecular cloud complexes inside the solar circle.
The sizes of the circles indicate the masses of the complexes,
as indicated in the upper right.The 4 kpc and Scutum arms
are drawn from the 21 cm maps.The Sagittarius arm is
drawn as it would best ¬t the CO data. [Thomas Dame,
CFA/Dame,T.M. et al., Astrophys. J., 305, 892, 1986, Fig. 9]


but must still deal with the distance ambiguity.
Initial radio studies of the interstellar gas and spi-
ral structure involved the 21 cm line. Again, long
connected features have been identified.
With the discovery of molecular clouds, it was
hoped that they would reveal the spiral structure
GC
of our galaxy. This is because the optical tracers
of spiral structure we see in other galaxies “ OB
M /M
associations, HII regions, dust lanes “ are all asso-
> 105
ciated with giant molecular clouds. A number of Ro
3 — 104 ’ 105
groups have carried out large-scale surveys of R = 13 kpc
104 ’ 3 — 104
emission from CO throughout the galaxy. We
3 — 103 ’ 104
look at some results in Fig. 16.14. Most of the
work involved material inside the Sun™s orbit. The
problem is that the distance ambiguity makes it
difficult to place uniquely all of the emitting
(a)
regions. One approach has been to take specific
models for spiral arms and predict the outcome Fig 16.15. Molecular clouds outside the Sun™s orbit and
spiral structure. (a) First and second galactic quadrants.The
of CO observations. Again, pieces of arms have
cloud masses are denoted by the symbols, shown at the
been identified, and again we see pieces of arms.
lower left.The circle at 13 kpc is drawn in for reference.
This may be telling us about the difficulties of
306 PART IV THE MILKY WAY



Visual extinction in the galactic plane makes
3 x 104 - 1 x 105 Ma
optical studies of the galactic center virtually
1 x 105 - 3 x 105 Ma
3 x 105 - 1 x 106 Ma impossible. We are able to observe the galactic
1 x 106 - 3 x 106 Ma
center in the radio and infrared parts of the
solar circle
spectrum. In the radio, we detect continuum
emission from ionized gas and line emission
VLSR =
from molecular clouds. In the infrared, contin-
15 10 km/s
uum observations tell us about the dust temper-
ature and opacity, and spectral line observations
tell us about the neutral and ionized gas. Near
infrared observations can also be used to study
rich star clusters near the center. Millimeter

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