. 11
( 28)


When an iron core is built up, there is no nism. As pulsars give off energy, they slow down.
longer a source of energy. The core begins to col- If the mass of a neutron star is greater than
roughly 3 M , then neutron degeneracy pressure
lapse. When a high enough density is reached,
the star explodes in a supernova. We see the will not support it and it will become a black hole.
blown off material as a supernova remnant. Such objects must be detected in binary systems.

11.1. Why is the stability of 56Fe important in the 11.2. What are the distinguishing characteristics of
synchrotron radiation?
steps that lead to a supernova?

11.3. At a given density, for equal numbers of different parts of the cycle. Explain how you
electrons and neutrons, electrons will exert a would use a strobe light to obtain a sequence
greater degeneracy pressure. Why, then, are of photographs, each showing the drumhead
neutron stars supported by neutron degener- at a slightly different point in the cycle.
acy pressure? (b) How is this like the technique for studying
11.4. (a) Suppose you have a drumhead vibrating at optical pulses from pulsars?
103 Hz. You want to “freeze” its motion at

11.6. A uniform density sphere of mass M has
11.1. Suppose a supernova explosion throws off
5 M of material at an initial speed of initial radius r and an angular speed 0. It
103 km/s. (a) Calculate the initial kinetic collapses under its own gravity to a radius r,
energy of the shell and the sum of the mag- conserving angular momentum. (a) How do
nitudes of the momenta of all the pieces in the initial and final kinetic energies com-
the shell. (b) Suppose the shell slows down pare? (b) Account for any difference.
11.7. (a) What is the escape speed from a 1.5 M
by conservation of momentum in sweeping
up interstellar material. How much mass neutron star of radius 10 km? (b) How does
will be swept up before the shell slows to it compare with the speed of light?
10 km/s? (c) If the average density of inter- *11.8. Suppose we can measure the arrival time of
stellar material is 1 H atom/cm3, what is the pulses to within 10 8 s. (a) Explain how we
radius of the shell when it reaches 10 km/s? can measure a pulsar period more accu-
11.2. Estimate the pressure in a 1.5 M neutron rately than this by timing a large number of
star with a radius of 15 km. pulses. (b) How many pulses do you have to
11.3. We have already seen that hydrostatic measure to measure a period of 0.1 s with
an accuracy of 10 10 s?
equilibrium allows us to estimate the
central pressure of a star as PC GM2/R4. *11.9. Explain why the average brightness of an
(a) Show that this provides the following object is approximately equal to its duty
mass“radius relation for neutron stars: cycle multiplied by its peak brightness. Use
the fact that if b(t) is the brightness as a
M1/3 (2)(1/4)5/3(h2/4 2
function of time, then the average of b(t)
over some time interval T is given by
(b) Use this expression to find the radius of a
1.5 M neutron star.
b1t 2 dt
11.4. (a) If F(r) is the force on an object as a b
function of radius r, show that the “tidal”
11.10. (a) Examine the stability of a rotating object
force on an object of length r is
against centrifugal disruption. Show that,
F (dF/dr) r for a rotating object, the requirement that
the gravitational force must at least balance
(b) Calculate the tidal force on your body
the centrifugal force, produces an expres-
near the surface of a neutron star. (Do the
sion for the minimum radius like equation
calculations for standing and prone
(11.9). (b) Using this result, what is the mini-
mum rotation period for a white dwarf?
11.5. For a 1.5 M neutron star with R 15 km,
*11.11. Suppose we approximate the rate of change
rotating 100 times per second, compare the
of a magnetic field for a stationary observer
gravitational force on an object at the
near a neutron star as the field strength
surface with the force required to produce
divided by the rotation period. (a) Taking
the circular motion for that object (at the
the magnetic field to be 1012 gauss, calcu-
star™s equator). In other words, compare the
late the magnitude of the induced electric
weight of an object with the centripetal
field. (b) Over what distance will an electron
force on it.

*11.13. (a) For a pulsar of mass M, radius R, and
have to travel in this field to reach a speed
period P, slowing at a rate dP/dt, at what rate
of 0.1c?
*11.12. Suppose that, as a pulsar slows down, the is its angular momentum changing?
quantity (1/P)(dP/dt) stays constant (say at a (b) What provides the torque for this change
value of b, where b is a positive number). in angular momentum?
(a) If the initial period (t 0) of the pulsar is *11.14. Find an expression, analogous to that for
P0, find an expression for P(t), the period as a the Chandrasekhar limit, discussed in
function of time. (b) If the initial rotation Chapter 10, for the maximum mass star
energy is E0, find an expression for E(t), the that can be supported by neutron degener-
energy as a function of time. (c) If the pulsar acy pressure.
is formed with P0 10 3 s, how long will it 11.15. Compare the radius of a 1.4 M neutron star
take to reach P 3 s? with its Schwarzschild radius.

Computer problem

1. Construct a table showing neutron star sizes for surface, and the rate of change of that acceleration,
masses 1, 2, 3 and 10 M . Also include columns for dg/dr.
average density, the acceleration of gravity at the
Chapter 12

Evolution in close binaries

we look at the energy of a particle in this system,
12.1 Close binaries the rotation of the coordinate system introduces
a term in addition to the gravitational potential.
This term is equal to J2/2mr2, where J, m and r are
If the two stars in a binary system are very close
to each other, each has the effect of altering the the angular momentum, mass and distance,
structure of the other star. When this occurs we respectively, from the origin of some particle. (We
call the system a close binary system. The surface can think of it as the term in the potential energy
of a star can be distorted by the stronger gravita- which gives rise to the pseudo “centrifugal force”
tional force that the companion exerts on the in the rotating system.) When we add this term to
near side than on the far side. Remember, we said the gravitational potential, we have an effective
that any effect that depends on variations in the potential that can be used to describe the motions
gravitational force from one position to another of particles. We can draw surfaces of constant
is called a tidal effect. (A similar situation applies effective potential, as in Fig. 12.1. The effective
as the Sun and Moon distort the Earth™s ocean force (gravity plus “centrifugal”) at any point on
surface, raising the tides.) one of these surface is perpendicular to the sur-
The distortion of stars results in internal dis- face. (This is analogous to contour maps of gravi-
sipation of energy. As a star rotates, different tational potential “ elevation “ on Earth. The
material is incorporated in the bulge. Different gravitational force is perpendicular to the con-
layers of material rub against each other, in a tour lines, and you don™t have to do any work to
fluid friction. This lost energy has to come from move along an equipotential line.)
somewhere. It comes from both the orbital There are five points, called Lagrangian points,
energy and the rotational energy of the star. As a where the effective gravitational force is zero.
result, eventually the orbits circularize and the These points are designated L1, L2, L3, L4, L5. (Note
two stars always keep the same sides towards that the L5 Society wants to place a space station
each other. This is the lowest energy arrangement at the L5 point for the Earth/Moon system.) The
for the system (see Problem 12.1). We say that the point L1 lies between the two stars, at the inter-
spins are synchronized. (The Moon™s spin and section point of the “figure eight” shaped surface.
orbital motion around the Earth are synchro- L1 is the dividing point between material being
nized, and the Moon keeps the same side towards attracted to one star or the other. The two sides of
the Earth.) the figure eight are called Roche lobes (Fig. 12.2).
In certain situations, it is possible for material The equipotential surfaces in a fluid must be
from one star to be pulled off the surface onto the surfaces of constant pressure. If they were not,
other star. To see how this can happen, we look at there would be pressure differences forcing
a binary system from a coordinate system rotat- fluid along the surface, and these forces could
ing with the same period as that of the orbits. If not be balanced by any gravitational forces,

CYGX-1 33M

L3 L2

9M 6M
M1 M2

A0620-00 0.7M
Surfaces of constant
effective potential
Fig 12.2. The sizes of the two Roche lobes depends on
the mass and separations of the orbiting systems.This ¬gure
shows the Roche lobes for three mass combinations.The
masses are the best estimates for the mass in three
observed close binary systems.

In discussing the evolution of close binary sys-
tems, we divide them into three classes (Fig. 12.3):
(1) detached, in which each star is totally con-
tained within its own Roche lobe; (2) semidetached,
in which the photosphere of one star exactly fills
its side of the Roche lobe; and (3) contact binaries,
in which both stars are at or over the Roche lobe.
So far in this book, we have discussed the evo-
Equatorial lution of isolated stars. However, we have already
plane (b) seen that approximately half of all stars are in
binary systems. If the binaries are completely
Fig 12.1. Surfaces of constant effective potential.The
detached, and there is no mass transfer, then the
effective potential is the true gravitational potential plus that
resulting from the centrifugal force in the coordinate system evolution will not be altered by the presence of
rotating with the orbiting system. (a) We look from above and the companion. However, mass transfer in semi-
see the intersection of surfaces of constant potential and the detached or contact systems can influence stellar
plane of the orbit.The heavy ˜¬gure eight™ is the intersection
of the Roche lobe and the plane of the orbit. (b) A three-
In general, the more massive star in a binary
dimensional view of the Roche lobe. (In this case, M1 ≠ M2.)
system will evolve off the main sequence first.
When that star becomes a red giant, it may
become large enough to fill its Roche lobe. In that
which must be perpendicular to the surfaces. The
case mass transfer to the companion will take
equipotential surfaces must therefore also be sur-
place. This can alter the evolution of the compan-
faces of constant density. (Again, this is analo-
ion. The degree to which it alters the evolution
gous to the situation on Earth. By the equation of
depends on the nature of the companion. As the
hydrostatic equilibrium, we can only have a pres-
more massive star continues to lose mass, its
sure gradient in the direction of the gravitation
Roche lobe shrinks, but the Roche lobe for the
force. That would be perpendicular to the poten-
companion grows. This means that mass transfer
tial lines.)

Close Binaries
12.2 Systems with white dwarfs

We first consider systems in which the first star to
evolve off the main sequence becomes a white
dwarf. Eventually, the white dwarf™s companion
goes through its evolution off the main sequence.
The companion becomes a red giant and fills its
Roche lobe. At this point mass transfer starts back
in the other direction from the original mass
transfer. Now mass is falling in on the white dwarf.
Not all the infalling matter strikes the white
dwarf. Because of its angular momentum, some of
the material goes into orbit around the white
dwarf. This orbiting material forms a disk, called
an accretion disk (Fig. 12.4). The disk forms because
material can fall parallel to the axis of rotation but
not perpendicular to that axis. (We will discuss
Common envelope
disk formation in more detail in Chapter 15.)
As material falls in, its potential energy
Fig 12.3. Classi¬cation of close binary systems. Shaded
areas are ¬lled with material. decreases, so its kinetic energy increases. The
increase in kinetic energy will not equal the
potential energy decrease, because some energy
will take place until the masses of the stars are
will be radiated away. We can expect roughly half
about equal. Some slow mass transfer may con-
of the change in potential energy to show up as
tinue after that point.
kinetic energy. But this increase in kinetic energy
At some point the star that was losing mass
will not produce much radiation on its own. This
will become a white dwarf or some other col-
is where the accretion disk helps. As the faster
lapsed object. In the following sections we look at
moving gas strikes the accretion disk, it slows
examples of each type of collapsed object that we
down, but its temperature increases. The now
have discussed “ white dwarf, neutron star and
heated gas can then radiate.
black hole.

Fig 12.4. Artist™s conception of
mass transfer leading to an
accretion disk. [STScI/NASA]

We can estimate the energy available from the
change in potential energy as mass falls in. If the
mass starts at distance r1 from the white dwarf,
and ends up a distance r2 from it, the luminosity
is given by (see Problem 12.5)

a ba b
r2 r1

GM a ba b
dM 1 1
r2 r1

In this equation dM/dt is the rate at which
mass is falling in, and M is the mass of the white
dwarf. We should think of equation (12.1) as the
upper limit to the actual luminosity. That is
because all of the energy gained in the infall is
not converted into outgoing radiation. In fact, the
formation of the accretion disk is crucial in this
process. The accretion disk provides a place where
the energy from the infalling material can be con-
verted into heat. The heated disk then radiates.

Example 12.1 Mass accretion luminosity Fig 12.5. HST image of the shell around recurrent nova
T Pyxidis.This is at a distance of 2000 pc.The shell is a little
Calculate the luminosity for a mass infall rate of
less than 1 parsec across.This image shows that it is made
10 8 M /yr, onto a 1 M white dwarf. Assume that
up of a large number of small objects.This is the material
the material starts 1.0 1011 cm away from the
that has collected from the various nova outbursts. (Note
white dwarf, and ends up 1.0 109 cm away.
that Fig. 4.16 shows an HST image of the shell expanding
around Nova Cygni 1992, taken 467 days after that
SOLUTION outburst.) [STScI/NASA]
We first convert the mass loss rate into g/s:

11.0 M yr 2 12.0 1033 g M 2
13.1 107 s yr 2 years. There is evidence that mass is ejected in
the process. In some cases this material can be
1017 g/s
seen expanding away from the star. The amount
of mass ejected is about 10 5 M .
The luminosity is then
We think that novae are the result of ther-
(6.67 10 8 dyn cm2/g2 )(2.0 1033 g)
monuclear explosions on the surfaces of white
(6.5 1017 g/s)
dwarfs with mass falling in from a companion. The
{[1/(1.0 109 cm)] [1/(1.0 1011 cm)]}
mass falling in is from the envelope of a red giant,
1034 erg/s and therefore contains hydrogen. (Remember, the
white dwarf has used up all of the hydrogen in
This is approximately 20 times the luminosity of
the core and has expelled the rest in its planetary
the Sun.
nebula.) The surface of the white dwarf is hot
enough for fusion of the hydrogen to take place.
Occasionally we observe a star that suddenly
It takes place rapidly in a small explosion, which
brightens by 5 to 15 magnitudes. These objects
probably stops the mass transfer for a while.
are called novae (Fig. 12.5). The name suggests the
When the transfer resumes, another explosion
appearance of a new star where one was not pre-
can take place. Shells of material left around
viously seen. Some of these novae appear to be
novae are shown in Figs. 12.5 and 4.16.
recurrent, on time scales of up to hundreds of

SNI MV light curves
Fig 12.6. Light curves for type I
supernovae. [Craig Wheeler,
University of Texas, Austin]


0 40


-12 SN 1994I
SN 1993J

0 50 100 150 200 250 300 350 400
phase (days)

We think that mass transfer onto a white
12.3 Neutron stars in close
dwarf can also account for type I supernovae.
binary systems
Sometimes enough mass falls onto the white dwarf
to make its mass greater than the Chandrasekhar
limit, MCH. In this case, electron degeneracy no In the preceding section we saw how mass transfer
longer supports the star, and it collapses. The in semidetached systems can alter the evolution of
energy from the collapse drives nuclear reactions, a star. In this section we consider a neutron star
which eventually build up 56Ni (with an even and a normal star in orbit around their common
number of 4He nuclei). The 56Ni beta decays to center of mass. As the normal star evolves
form 56Co, which, in turn, beta decays to 56Fe. towards a red giant, it fills its Roche lobe and
Type I supernovae have light curves with double matter starts to fall onto the neutron star.
exponential behavior. The time scales of the two At first it was thought that this situation
exponentials turn out to be characteristic of these could not develop. It was not clear how such a
two beta decays. We think that this process system could form. The problem is that, for a
accounts for most of the iron in the universe, neutron star to be present, there must have been
since the iron created in more massive stars is a supernova explosion in the past. The first star
destroyed in the type II supernova, as discussed in to go supernova in a binary is the more massive
Chapter 11. Light curves for type I supernove are star. The supernova explosion drives away most
shown in Fig. 12.6. There is little variation in the of the mass of the more massive star, meaning
peak brightness of type I supernovae. We will see that more than 50% of the original mass of the
in Chapter 18 that this makes them very good system was blown away. If this happens far too
“standard candles” in distant galaxies. quickly for the system to adjust the orbit
The nuclear energy released in these reactions radius/period, it becomes unbound. To see this
is greater than the binding energy of the white for circular orbits, we assume that the system
dwarf, and the star is destroyed, leaving no rem- starts with stars of mass m1 and m2 and a separa-
nant. This explanation accounts for the light tion R, with speeds appropriate for a circular
curves of type I supernovae, their spectra and lumi- orbit, given by equation (5.15). Assume that star
nosities, and their occurrence in what are thought 1 explodes and is left with a mass m, but that R
to be old systems (as we will discuss in Chapter 13). and the orbital speeds don™t change. The new

energy is then given by equation (5.26), with m to general relativity, appear to exert the gravi-
substituted for m1. For the new system to be on tational force of a 1.3 M star. Thus, there may
the boundary of being bound, its energy would be stars we think are white dwarfs, but which
be zero, so we can find the value of m that makes are really neutron stars.)
E 0. It is (see Problem 12.6) We now suppose that we have a binary system
with a neutron star and a normal star, with mass
m1 c d
1m2 m1 2
m (12.2) being transferred from the normal star to the
neutron star. The mass falling in is heated and
Note that for m1 being much greater than m2 gives off irregular bursts of X-ray emission. To see
this approaches m1/2, meaning the massive star how this works, we look at the case of a well stud-
would have to lose half its mass. Note that, by the ied X-ray source, Her X-1 (Fig. 12.7a). (The name
discussion in Section 5.4, this result doesn™t implies the brightest X-ray source in the constel-
change for elliptical orbits, since the equation for lation Hercules.) It is also coincident with a vari-
the energy is the same, with R replaced by the able star HZ Her. The star is a binary with a period
semi-major axis. of 1.7 days. The mass of the unseen companion is
Therefore, the explosion should break up the estimated to be in the range 0.4 to 2.2 M . The
binary system. This scenario explains the exis- X-rays are observed to pulse with a period of 1.24 s.
tence of “runaway stars”. These are individual
stars moving through space with much higher
than average speeds.
X-ray observations have been made of sys- Her X“1 RXTE ASM
2“12 keV intensity (ASM cts/s)

tems in which it appears that a neutron star is
still in orbit with a normal star. This means that
there must be some way of forming such a sys-
tem, and eventually theoreticians have come up
with a number of plausible scenarios. Different
scenarios might work in different types of sys-
tems that are found. (1) Theoretical simulations 0

have shown that, while most systems in which 52000 52050 52100 52150 52200
the more massive star goes supernova first will Time (Modified Julian Date)

become unbound, there are some combinations (a)
of initial conditions that will lead to bound sys-
tems after the supernova explosion. (2) Before GRB010326A
the supernova explosion, the more massive star Bands
Sum of 4 Detectors
might have filled its Roche lobe and transferred (keV)
mass to the less massive star. If enough mass is 6-120

transferred before an explosion, the system can 6-30
stay bound. (3) An alternative explanation is
that the compact star may have originally been
a white dwarf, not a neutron star. However,
mass transfer from the companion may have
pushed the mass of the white dwarf beyond the 0
11500 11600 11700 11800 11900 12000
1.44 M limit. The electron degeneracy pressure
UT seconds on 2001-03-26
could no longer support the star and it would
collapse until it became a neutron star. So, the
Fig 12.7. (a) X-ray emission from HZ Her.The intensity
neutron star would have formed without a
is plotted as a function of time into the period. (b) X-ray
supernova explosion. (It is interesting to note
emission from a burster. [(a) Alan M. Levine, MIT Center for
that a 1.44 M white dwarf that suddenly col-
Space Research; (b) MIT Center for Space Research]
lapsed to form a neutron star would, according

The period changes regularly throughout the Binary Pulsar

1.7 day cycle. This can be interpreted as a Doppler
shift (see Problem 12.7). We can think of the X-
rays as a signal being emitted with a period of 100
1.24 s. When the source is moving away from us

…r (km/s)
the period appears longer, and when the source is 100
coming toward us the period appears shorter. The 200

X-ray source also appears to be eclipsed every 300
0.0 1.0
1.7 days.
Time (periods)
The mass transfer rate, estimated from the Shift in orbit
X-ray observations, is about 10 9 M yr. The lumi- in one year

nosity is about 1037 erg/s. The temperature is esti- (a) (b)
mated to be 108 K. At this temperature, we can
Fig 12.8. Binary pulsar orbit. (a) The radial velocity as a
estimate the frequency of a photon with energy
function of time within a period, expressed as a fraction of
kT. The frequency, , is kT/h, or 2 1018 Hz. This
the period. (b) The shift in the orbit over a one year period.
corresponds to a wavelength of 0.14 nm, clearly
[ Joseph Taylor, Princeton University]
in the X-ray part of the spectrum.
Theoreticians have speculated on the future
evolution of such a system. The mass transfer rate
may become so large that the X-ray emission is sar. Also, as long as there is mass transfer, the rota-
quenched. The outgoing X-rays are effectively tion rate can be kept very high, explaining why the
blocked by the infalling material. This system slowdown of millisecond pulsars is very small.
may eventually end up as two compact objects. While discussing neutron stars in binary sys-
Mass transfer in a system with a neutron star tems, we should mention one other interesting
can also explain the existence of pulsars with object. This is a radio pulsar discovered in 1974 by
Joseph Taylor and Russel Hulse , then of the University
very short periods, as short as a few milliseconds,
known as millisecond pulsars. One of the intrigu- of Massachusetts. This pulsar has a period of 0.059 s
ing features of these objects is that their periods but the period varies periodically, suggesting that
are not decreasing as rapidly as those for normal the pulsar is in a binary system. The variation in
pulsars. Some of them have values of P/(dP/dt) as the period is like a Doppler shift (Fig. 12.8), with
large as 1010 per year. That is, in one year, the the period appearing longer when the pulsar is
change in the period is only 10 10 (one-ten-bil- moving away from us and shorter when it is com-
lionth) of the period. This means that they are ing towards us. This system provides us with an
extraordinarily stable clocks (something which interesting test of a prediction of general relativ-
had originally been hoped for normal pulsars ity“gravitational radiation. For their studies of
this binary pulsar, Taylor and Hulse were awarded
until their period changes were observed, as dis-
cussed in Chapter 11). the Nobel Prize in physics in 1993.
To see how mass transfer can explain millisec- Since the gravitational radiation carries away
ond pulsars, remember that if transferred material energy, the total energy of the orbit should be
is not aimed directly at the center of the neutron decreasing. In hopes of seeing this, Taylor and his
star, it has a large amount of angular momentum. co-workers have monitored this binary pulsar for
This explains the formation of an accretion disk, as several years. Some results are shown in Fig. 12.9.
material goes into orbit rather than falling onto They have found that the orbital period is chang-
ing by 2.3 10 12 s/s. The change in energy of
the neutron star surface. As material leaks inward
from the accretion disk onto a (normal) pulsar, it the orbit corresponds to the energy that would be
transfers a lot of angular momentum to the pulsar, given off by gravitational radiation from the
causing a large increase in the rotation rate orbiting bodies. This may provide us with the
(decrease in the period). This explains how a nor- first indirect observational confirmation of grav-
mal pulsar can be “spun up” into a millisecond pul- itational radiation.

in would also emit X-rays (before crossing the event
GTR prediction
horizon). We could start searching for stellar black
holes by looking for irregular X-ray sources.
One interesting possibility is known as Cyg X-1
(Fig. 12.10), the brightest X-ray source in Cygnus.
phase shift ’1
The Uhuru satellite showed this to have both
short and long term variability. Until the Einstein
’2 observatory was launched in 1978, the positions
1975 1976 1977 1978 1979
of X-ray sources were not accurately determined.
Fig 12.9. A comparison of theory and experiment for the However, there is also a radio source associated
binary pulsar.The plotted quantity keeps track of energy lost with the X-ray source. We know that the X-ray and
in the orbit over the years that it has been observed.The
radio sources are associated because they have
predictions from general relativity include gravitational
the same pattern of variability. The position of
radiation as the mechanism for energy loss. [Joseph Taylor,
Princeton University]


12.4 Systems with black holes 1000

In Chapter 11, we saw that neutron stars are sup-
Counts per 64 ms

ported by neutron degeneracy pressure. However,
if the neutron star is too massive, neutron degen- 600
eracy pressure is insufficient to support the star.
We think this limit is about 3 M . We know of no
other source of pressure that will stop the col-
lapse of the star. It will collapse right through the
Schwarzschild radius for its mass, and will 200

become a black hole. Black holes would be a nor-
mal state of evolution for some stars. 0
0 20 40 60
How would we detect a stellar black hole? We
Time (seconds)
obviously could not see it directly. We could not
even see it in silhouette against a bright source
1996.0 1997.0 1998.0 1999.0 2000.0 2001.0 2002.0
since the area blocked would be only a few kilo-
meters across. We have to detect stellar black
2 “ 12 keV intensity (ASM cts/s)

holes indirectly. We hope to see their gravita-
tional effects on the surrounding environment.
This is not a hopeless task, since we might expect 4
to find a reasonable number of binary systems
with black holes. By studying a binary with a sus- 2
pected black hole, the problem would be to show
the existence of a very small object (as inferred
from orbits) with a mass in excess of 3 M .
How do we find a candidate binary to study? In 50000 50500 51000 51500 52000 52500
Time (Modified Julian Date)
Section 12.3, we saw that a neutron star in a binary
system can give rise to strong X-ray sources. The
importance of the neutron star is that its radius is Fig 12.10. X-ray light curves for suspect stellar black
so small that infalling material acquires enough holes in close binaries: (a) Cyg X-1, (b) LMCX-3. [(a) Edward
energy to give off X-rays. A stellar black hole would H. Morgan, MIT Center for Space Research; (b) Alan Levine,
be smaller than a neutron star, so material falling MIT Center for Space Research]

the radio source is determined accurately from it is like another O9.7Ib star. However, we know
radio interferometry. Once the radio position was that it is in a close binary system, and is tidally dis-
known, optical photographs were studied to see if torted. We are only now beginning to understand
there is an optical counterpart. This optical coun- stellar structure and evolution in close binary sys-
terpart was found to be a ninth magnitude star tems. It may be that the spectrum we classify as
HDE226868. A study of the star™s spectrum shows O9.7Ib is really produced by a star of a different
that is an O9.7Ib (blue supergiant) star. This places mass. We are also not sure of the inclination of the
orbit, and it enters in the mass function as sin3 i. It
its mass at about 15 M .
is also possible to avoid an 8 M companion if we
This star is also a spectroscopic binary. Its
orbital period is 5.6 days. The star also varies in postulate the triple-star system, but searches for a
brightness with this period. The amount of varia- third component have not been successful.
tion is 0.07 mag. The source of the variation is For ten years, Cyg X-1 was the only strong stel-
thought to be a distortion in the shape of the lar binary black hole candidate. Astronomers
blue supergiant due to thee strong tidal effects of have studied large numbers of possible candi-
the unseen companion. The distortion results in dates. The starting points are generally X-ray
the star appearing as different sizes at different sources for which optical counterparts can be
points in its orbit. From the amount of brightness identified. In the last few years, a few more solid
variation, it was concluded that the inclination possibilities have emerged.
of the orbit is 30 . One is an X-ray source in the Large Magellanic
We now look at what can be deduced about the Cloud (a companion to our own galaxy), known as
LMCX-3. Based on Einstein positional observations,
companion. From the analysis of the Doppler shift
data, the mass function (discussed in Chapter 5) an optical counterpart to the X-ray source was
can be found. Since we know the mass of the blue identified. It turns out to be a 17th magnitude B3V
supergiant and have an estimate of the inclination star. The star is a spectroscopic binary with an
angle of the orbit, we can derive the mass of the orbital period of 1.7 days, and an orbital Doppler
unseen companion from the mass function. The shift of 230 km/s. The system does not eclipse, so
best estimate for this is 8 M . There is some uncer- this puts some constraints on the orbital inclina-
tainty, but it seems very likely that the mass of the tion. An inital analysis put the most likely mass
unseen companion is a least greater than 4 M . range of the compact object as 4 to 11 M . A more
detailed analysis showed that it is more than 6 M .
The X-ray observations tell us that the com-
panion must be very small. This is because the One uncertainty in studies of this object is that it
X-ray emission varies significantly in intensity on seems that some of the optical light comes from
time scales of about 5 ms. This requires that the the accretion disk, making analysis of the optical
emitting region be less than 1500 km in extent variability more difficult. (This is not a problem in
(see Problem 12.9). Thus, we have an object that is Cyg X-1, where the optical light appears to all come
definitely smaller than a normal star, but which from the visible star.) Even with these uncertain-
has a mass greater than 4 M (and probably closer ties, it seems that 4 M is a reliable lower limit.
to 8 M ). It would therefore seem likely that the The third strong candidate that has emerged is
object is a black hole! different from Cyg X-1 or LMCX-3 in that it is a
It is interesting that such an astonishing con- transient X-ray source. Depending on what cata-
loge it is found in, it has numerous names: 0620-00,
clusion rests on the foundation of some “stan-
V616 Mon, Nova Mon 1975, 1917. It has been identi-
dard” observational techniques. These include
using the spectrum of HDE226868 to find its fied with an optically recurrent nova (with 1917
mass, and the classical analysis of spectroscopic and 1975 being the two most recent outbursts).
binary orbits, using the mass function. The identification of the X-ray and optical novae
Though most astronomers probably accept the allowed for the secure identification with the
black hole explanation for Cyg X-1, the argument X-ray source with an optical source that could be
is not airtight. For example, our knowledge of the studied in detail. The optical star is a K5/7 dwarf.
mass of the visible star comes from assuming that The properties of low mass dwarfs are better

known than the more massive stars found in Cyg .2

X-1 and LMCX-3. This K dwarf was found to be a
spectroscopic binary with a period of 7.8 hr, and
a large orbital Doppler shift (470 km/s). Analysis .1

of the mass function places a lower limit of 3 M

Doppler shift
on the mass of the compact companion. An analy-
sis of the mass ratio of the two objects placed the 0
mass of the compact object between 4 and 9 M .
Many objects have been studied in detail, but .05

there are always problems in the chain of reason-
ing. For example, a promising source, LMCX-1, has
two possible optical counterparts. Of the three .15
4000 4500 5000 5500 6000 6500
objects mentioned here, Nova Mon 1975 probably (Julian Day “ 2,440,000)
has the best evidence for the presence of a black (a)
hole. It is amazing to contemplate how far
astronomers have come when we can calmly say
that the most likely explanation for some observed SS 433
phenomenon is the presence of a black hole!

12.5 An unusual object: SS433

To form an idea of the fascinating range of phe-
nomena encountered in close binary systems, we
take a look at an object best known by its desig-
nation in a particular catalog, SS433. (It was cata-
loged long before is unusual nature was realized.)
The object is a binary at the center of a supernova
remnant. It is therefore not surprising that one of
the members of the binary system is a neutron
star or black hole. The period of the binary sys-
tem is 13.087 days. The system is also a periodic
6000 6500 7000 7500
X-ray source. °
Optical observations reveal absorption and
emission lines with very large Doppler shifts. The
required speeds are up to 0.26c. At any time, both Fig 12.11. Doppler shift data for SS433. (a) The vertical
axis shows the Doppler Shift, / , relative to the average
blueshifted and redshifted components are pres-
velocity of the system.The horizontal axis is in days.The
ent. The magnitude of the Doppler shift goes
dots represent the actual measurements, and the smooth
through a 164 day cycle, as shown in Fig. 12.11.
curves are ¬ts of models to the data. (b) Actual spectra of
There is an interesting asymmetry in the Doppler
SS433 at three different times, showing the shifts from the
shifts. The redshift is always larger than the velocities of the H emission. (The absorption features
blueshift. The maximum redshift corresponds to marked “ ” are from the Earth™s atmosphere.) [Bruce
about 50 000 km/s, and the maximum blueshift Margon (University of Washington)/Margon, B. Astrophys. J.
corresponds to about 30 000 km/s. If we take an Lett, 230, L41, 1979, Fig. 3]
average of the redshifted velocity and blueshifted
velocity at any instant (remembering that blueshift which one member is either a black hole or a neu-
corresponds to a negative radial velocity), we tron star. The period of the binary is 13.087 days.
obtain a fairly constant value of about 12 000 km/s. The compact object is a source of two jets, mov-
The basic model to explain this behavior is ing in opposite directions. The jets are emitted in
shown in Fig. 12.12. It involves a binary system in a cone with a half-angle of about 20 . The cone is

precession we mean the changing of the orienta-

tion of the axis of rotation caused by an external
164 day

precession torque. This is like the precession of the Earth,

under the combined effects of the Sun and Moon,
as discussed in Chapter 23.) This causes the pro-
jected angle of the jets to go through a 164 day
cycle, giving the variations in the Doppler shifts.

In this model there is a natural explanation

for the 12 000 km/s offset in the Doppler shifts. It

comes from a transverse Doppler shift arising
from the fact that the jet is moving at an appre-
ciable fraction of the speed of light. (We discussed
the transverse Doppler effect in Chapter 7; it is
essentially a time dilation effect.) At v/c 0.26,
1.036. According to equation (7.5) this pro-
duces a wavelength which is 1.036 times the rest
wavelength. If we interpret this shift as arising
from a normal Doppler shift, it corresponds to
v/c 0.036, or about 11 000 km/s. Since this is just
a time dilation effect, this is added in as a con-
Jet 82

stant to any radial Doppler shift. Since it always

20° increases the wavelengths, it make the redshifts

days l

larger and the blueshifts smaller.
In this general model, the X-rays come from

material falling into an accretion disk around the
compact object. The binary is an eclipsing binary,
so we can estimate the masses of the components.
The best estimate for the mass of the companion
is about 4 M . This would make it a black hole.
However, there are enough uncertainties in the
Fig 12.12. A model for SS433, showing the twin jet geom-
estimate to allow it to be either a black hole or a
etry. [Bruce Margon (University of Washington)]
neutron star. Theoreticians are still working on
the mechanisms for the collimating of the jets,
inclined by 80 to the line of sight. The compact and for generating the energy for getting the jets
object precesses with a period of 164 days. (By moving so fast.

Chapter summary
We saw in this chapter how the evolution of stars We saw how mass transfer onto neutron stars
is altered by placing them in close binary sys- can produce strong X-ray emission, such as from
tems. The change in evolution arises mainly from HZ Her. We also saw how mass transfer can spin
strong tidal effects and from mass transfer. up the pulsar, making it into a millisecond
If the component receiving mass is a compact pulsar. We also saw how studies of a binary pul-
object “ white dwarf, neutron star or black hole “ sar have provided evidence for gravitational
infalling material can acquire enough energy to radiation.
emit X-rays as it falls in to form an accretion disk. We saw how studies of Cyg X-1, LMCX-3 and
We saw how mass transfer onto white dwarfs Nova Mon 1975 have provided us with evidence
can account for novae. If mass transfer is large for the existence of a stellar black hole.
enough, it can account for type I supernovae.

12.1. Explain why surfaces of constant effective 12.5. How does mass transfer explain the existence
potential must be surfaces of constant density. and stability of millisecond pulsars?
12.2. Why is the formation of an accretion disk a 12.6. Why is the binary pulsar so important?
likely event when we have mass transfer onto 12.7. Explain the steps leading to the conclusion
a compact object in a binary system? that Cyg X-1 is a black hole. Which are the
12.3. Why is the first star in a binary to go super- most suspect? Compare the problems in the
nova usually the more massive one? reasoning for LMCX-3 and Nova Mon 1975.
12.4. Why is it unusual to find a neutron star in a
binary system?

12.1. Show that, for a fixed total angular momen- 12.3, what is the fractional change in the
orbital energy E/E per second?
tum, the “synchronized” spins situation is
the lowest energy state for an orbiting system. 12.5. Derive the first line of equation (12.1).
(Hint: Consider the sum of the orbital and 12.6. Derive equation (12.2), following the outline
rotation energies.) given in the discussion.
12.2. The radial force Fr is related to a potential 12.7. For the system HZ Her, how large a shift in
V(r) by the pulsar period would be observed if the
visible star has a mass of 3.0 M , the com-
Fr dV(r)/dr
pact object has a mass of 2.0 M , and the
Show that the “centrifugal” force can be orbital period is 1.7 days? The pulsar period
derived from the term J 2 /2mr 2 in the effective is 1.24 s.
potential. 12.8. For HZ Her, assuming that the material falls
*12.3. Suppose we have a pulsar in orbit around in from far away at the given rate, how close
another object. The pulse period, as emitted by must it come to the star to provide the given
the pulsar, is P0. The orbit is circular with a X-ray luminosity?
speed v. We are observing in the plane of the 12.9. For Cyg X-1, the most likely value of the mass
orbit. Find an expression for the observed pulse function is
period as a function of the position of the pul-
(Mx sin i)3/(Mx Mopt)2 0.25 M .
sar in the orbit. (Hint: Consider the advance
and delay in the arrival time of pulses as the For an inclination angle of 30 , and an optical
star mass of 33 M , find the mass of the com-
pulsar moves toward us and away from us.)
12.4. For the rate of change in the orbital period pact object.
for the binary pulsar discussed in Section

Computer problems

12.1. Draw a graph showing mass accretion luminosity the system, using equation (12.2), for combina-
tions involving m1 2, 5, 10, 20 and 50 M and
for material falling onto a neutron star (at a rate
of 10 8 M /yr) for neutron star masses ranging m2 1, 2 and 5 M .
from 1 to 3 M . For each neutron star mass, 12.3. Repeat Problem 12.9, showing the effects of
assume the radius as given by the mass“radius changing the inclination angle to 15 and 45 , and
the optical star mass to 28 M and 38 M .
relation discussed in Chapter 11.
12.2. Construct a table showing how much mass would
have to be lost in a supernova explosion to unbind
Chapter 13

Clusters of stars

106 stars, and are 20 to 100 pc across. They seem to
When we look at the spatial distribution of stars
in our galaxy, we find that most of the light is have no associated gas or dust. Some do have plan-
concentrated in a thin disk. We are inside this etary nebulae, though. Globular clusters are not
disk, so we see it as a band of light on the sky, confined to the disk of the galaxy. Harlow Shapley
called the Milky Way. We will discuss this farther used RR Lyrae stars and Cepheids to find the dis-
in Part III, but we will see in this chapter that tances to globular clusters. This placed the globu-
location of stars in the galaxy can tell us some- lar clusters in three dimensions. It was found that
thing about those stars. In particular, some stars the globular clusters form a spherical distribution
are confined to the thin disk of the Milky Way, with the Sun being about 10 kpc from the center.
while others form a more spherical distribution. (This is still one of the best techniques for finding
In this chapter, we will discuss groupings of stars, the distance to the galactic center.)
called clusters, and see how they vary in size, con- Before we look at the properties of the clusters
tent and galactic distribution. themselves, we will look at an important tech-
nique for determining distances to relatively
nearby galactic clusters.
13.1 Types of clusters
13.2 Distances to moving clusters
We distinguish between two types of star clusters “
galactic clusters and globular clusters.
Galactic clusters are named for their confine- Let™s assume that we have a star (or cluster) mov-
ment to the galactic disk. A selection of images of ing through space with a velocity v. The velocity
galactic clusters is in Fig. 13.1. A familiar galactic makes an angle A with the line of sight. We can
cluster, the Pleiades, is shown in Fig. 13.1(a). Note break the velocity into components parallel to
the open appearance in which individual stars the line of sight and perpendicular to the line of
can be seen. Because of this appearance, galactic sight. The component parallel to the line of sight
clusters are also called open clusters. Galactic clus- is the radial velocity vr and is responsible for the
ters typically contain 103 stars, and are less than Doppler shift we observe. The component per-
10 pc across. Recent sensitive near IR surveys are pendicular to the line of sight is the transverse
showing more members than we had previously velocity vT. It is responsible for the motion of the
thought in many clusters. In the photograph, we star across the sky, called the proper motion.
see some starlight reflected from interstellar From the right triangle in Fig. 13.3, we can see
dust. Galactic clusters are sometimes associated that these quantities are related by
with interstellar gas and dust.
v2 v r2 vT2 (13.1)
Globular clusters are named for their compact
spherical appearance (Fig. 13.2). They have 104 to vr v cos A (13.2)



Fig 13.1. Open clusters. (a) The Pleiades (M45) in Taurus.
ond, is just the transverse velocity divided by the
The nebulosity seen here is starlight re¬‚ected from interstel-
distance to the star d:
lar dust. (b) M6, also known as the Butter¬‚y Cluster. (c) M7
in Scorpius. Its distance is about 300 pc, and it is about 8 pc
(rad/s) (13.5)
across. (d) M37 in Aurigi, at a distance of 1.5 kpc.
The greater the transverse velocity, the faster the
[(a) Courtesy of 2MASS/UMASS/IPAC/NASA/JPL/Caltech;
(b)“(d) NOAO/AURA/NSF] star will appear to move across the sky. Also, the
closer the star is, the greater the motion across
vT v sin A (13.3) the sky. In general, proper motions are very small,
amounting to a few arc seconds per year, or less.
tan A vT/vr (13.4)
(The largest proper motion is 10.3 arc sec/yr for
Barnard™s star.) For this reason, we would like to
The relationship between proper motion and
rewrite equation (13.5), expressing in arc sec-
tangential velocity is shown in Fig. 13.4. The
onds per year, v in kilometers per second, and d in
proper motion , expressed in radians per sec-



Fig 13.2. Globular clusters. (a) M3, in Canes Venatici. (b) M5,
at a distance of 0.8 kpc, is one of the most massive clusters in
our galaxy. (c) M15 in Pegasus, at a distance of 1 kpc. (d) M80.
[(a), (b) NOAO/AURA/NSF; (c), (d) STScI/NASA]

parsecs. We can then rewrite equation (13.5) as
vT d
31 km>s 4 31 rad>s 4 3 1 km4
3 1 rad4
3 2.063 105 arc sec>rad4
33.086 1013 km4
1 yr
3 3.156 107s>yr 4 31 pc 4
vT(km/s) 4.74 (arc sec/yr)d(pc) (13.6)
In general, we can measure the radial velocity
(from the Doppler shift) and we can also measure
If, instead of vT, we know A, then we can use
the proper motion. If we know the transverse
equation (13.4) in equation (13.7) to give
velocity, we can find the distance from
d vr tan A/4.74 (13.8)
d vT/4.74 (13.7)

vr f
Li t #2
Star igh
of S
v #3
Lin Sight
ht f
Line o
Cluster A1
i A2 A3
fS v
eo A2
L A3

A Observer Convergent
Fig 13.5. Convergent point. As the cluster moves farther
Fig 13.3. Space velocity.The velocity of the star is v, which away, the angle A between v and the line of sight approaches
makes an angle A with the line of sight.The radial and tan- zero.That is A1 A2 A3. As this happens, the cluster
gential components of the velocity are vr and vT, respectively. approaches one line of sight, the convergent point. Note that
this ¬gure is exaggerated to show the effect. Real clusters do
not move that much over the times we could observe them.
With a cluster of stars, we can compare the
proper motion with the rate at which the angular
size of the cluster changes to find A. To see how
call this point the convergent point of the cluster.
this works, let us consider the case of a cluster
We can see from the figure that the angle
moving away from us (positive radial velocity). As
between the current line of sight and the line of
the cluster moves away (Fig. 13.5), A becomes
sight to the convergent point is the current value of
smaller and approaches zero. As the cluster moves
A. Similar reasoning applies to clusters that are
farther away, the proper motion approaches zero
moving toward us. They are moving away from
(by equation 13.5). Thus, the cluster appears to be
their convergent point, so we find it by extrapo-
heading toward a particular point in the sky. We
lating their motion backwards in time.
To apply these ideas, we take a series of
vT (∆t) Fig 13.4. The depend- images a number of years apart. From the proper
ence of the proper motion motion, and the change in angular size, we can
on the distance d and tan-
find the convergent point, and therefore we
gential speed vT.We follow
know A. This is shown schematically in Fig. 13.5.
the motion of the star for a
time t.

µ (∆t)

Fig 13.6. Schematic representation of motions of stars
within a cluster. Arrows represent total motions.

We measure the radial velocity vr and proper (See Problem 13.10.) We let the position of the ith
particle, relative to some origin, be ri. If we have
motion . Since we know A, vr and , we can find
d from equation (13.8). The best determination of two particles, i and j, the vector giving their sepa-
ration is rj ri, as shown in Fig. 13.7. We let Fi be
a distance to a moving cluster is the Hyades. This
the net force on the ith particle. We can therefore
determination is an important cornerstone in
our determination of distances to more distant write the equation of motion for this particle as
objects in our galaxy and in other galaxies.
Fi m (13.9)
dt 2
13.3 Clusters as dynamical entities We are looking for a relationship between vari-
ous types of energy. The vector dot product between
In this section we look at the internal dynamics force and distance will give us an energylike
of star clusters. If the gravitational forces between quantity. We therefore take the dot product of ri
the stars are sufficient to keep the cluster together, with both sides of equation (13.9) and then sum
we say the cluster is gravitationally bound. (We the resulting quantities over all the particles to
have already discussed the idea of gravitational give
binding when we talked about binary stars, in
a b
Chapter 5 and Chapter 12.) However, gravity does a ri F i m a ri (13.10)
dt 2
i i
1 1
more than assure the overall existence of the
cluster. As stars move around within the cluster, We can rearrange this by using the following:
d2 1ri2 2
pairs of stars will pass near each other. The gravi-
a 2ri b
tational attraction between the two stars in the 2
dt dt
pair will alter the motion of each star. The momen- (13.11)
tum and energy of each star will change in this d2 ri
dri 2
2a b a b
gravitational encounter. Thus, these encounters dt 2
alter the distribution of speeds, the number of
where we have used the fact that for any vector
stars traveling at a given speed. If there has been
sufficient time for many encounters to occur, the
distribution of speeds will reach some equilib- x2 x2
rium. For every star that suffers a collision chang-
This gives
ing its speed from v1, there is another collision in
which some other star has its speed changed to N N N
1 d2 dri 2
a b 2 a a mr2 b a ma b
a ri F i i
v1. (We refer to these gravitational encounters as 2 dt i 1 dt
i i
1 1
collisions, even though the stars never actually (13.12)
get close enough for their surfaces to touch.)
When a cluster has reached this stage, we say that
ri “ rj
it is dynamically relaxed.

13.3.1 The virial theorem rj
In a dynamically relaxed system, the kinetic and
potential energies are related in a very specific
way. This relationship is known as the virial theo-
rem. In this section we derive it.
We begin with a collection of N particles. (We ri
can think of each star in a cluster or each atom in
a gas cloud being represented by a particle.) To
simplify the calculation we assume that all parti-
Fig 13.7. Position vectors ri from the origin to the ith par-
cles have the same mass m. The final result would
ticle and rj from the origin to the jth particle.
be the same if we allowed for different masses.

We now introduce the velocity of the ith par- the right-hand side, we will have a quantity equal
ticle, vi, defined as to twice the original right-hand side. This gives us
ri rij rj rji
a b a a Gm2 a b (13.21)
vi dri/dt (13.13)
a ri F i rij 3 rji 3
2 i 1j 1
i 1
We also introduce the quantity I, defined as i j

ri 1rj ri 2 rj 1ri rj 2
a b a a Gm2 a b
I a mri (13.14)
rij 3
2 i 1j 1
i 1
i j
If ri were the distance of particle i from some (13.22)
axis, then I would be the moment of inertia about
where we have also used the fact that ’ rij ’ ’ rji ’.
that axis. However, an axis is a line, and ri is
This procedure actually allows for a simplifica-
measured from a point (the origin). Therefore, I is
tion. To see this, we first multiply out the numer-
a quantity similar to, but not the same as, the
ator on the right-hand side to obtain
familiar moment of inertia. (For most normal
a b a a Gm2
shaped objects it has a numerical value close to
a ri F i 2 i 1j 1
the moment of inertia.) We also set the total i 1
i j
kinetic energy K equal to the sum of the kinetic r2 r2
r i rj r j ri
a b (13.23)
i j
energies of the individual particles,
rij 3
a b a mv2 We also note that
K (13.15)
2 i1 i
1ri rj 2 1ri rj 2
’ rij ’ 2
In terms of these quantities, equation (13.12)
r2 r2 ri rj r j ri (13.24)
becomes i j

Substituting into equation (13.23) gives
1 d2I 2
a ri F i a mvi (13.16)
rij 2
2 dt2 N
a b a a Gm2 a b
i i
1 1
a ri F i (13.25)
rij 3
2 i 1j 1
2 i 1
1 dI i j
2K (13.17)
2 dt2
which simplifies to
We evaluate ri Fi for the Fi being the gravi- N
1 N N Gm2
a baaa b
tational forces between particles. We let a ri F i (13.26)
2 i 1 j 1 rij
i 1
i j
rij rj ri (13.18)
The term on the right-hand side is the sum of
be the vector distance between the two particles.
the gravitational potential energies of each pair
The net force on a given particle is the sum of the
of particles. Note that each pair appears twice in
forces exerted on it by all of the other particles.
the sum, since the energy of the pair is inde-
pendent of which particle in the pair we count
m2Grij first. For example, for particle 1 and 2, both the
Fi (13.19)
a r3 quantities Gm2 ’ r12 ’ and Gm2 ’ r21 ’ appear. This
means that the double sum on the right-hand

side of equation (13.26) gives us twice the gravita-
which gives us
tional potential energy, but there is also a factor
ri rij
Gm2 a b
of one-half in front, so the right-hand side is
a ri F i (13.20)
aa rij 3 equal to the gravitational potential energy U. We
i i 1j 1


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