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108 cm,
from r1 out to r2, conservation of energy would For r 7
give us
2
hc GMh hc GMh 4
1 2.12 10
(8.2) 1
r1c 1 r2c 2
1 2

Solving for the ratio of the wavelengths gives The shift for spectral lines in the Sun is very
small. The shift for white dwarfs is measurable.
c1 d
GM
10 4)
The two best cases so far are Sirius B (3
r2c2
2
and 40 Eridani (6 10 5).
(8.3)
c1 d
GM
1
There is an interesting way to measure the
r1c2
gravitational redshift on Earth. It utilizes a phe-
As we have said, this derivation should not be nomenon known as the MГ¶ssbauer effect (Fig. 8.10).
considered rigorous вЂ“ it is more of a dimensional This involves the emission of a gamma-ray by a
analysis. However, it gives a result that agrees nucleus held firmly in place by a solid crystal. In
with the full general relativistic calculation for a free nucleus, the gamma-ray would lose a little
shifts that are not too large. The actual result is
2GM 1 2 Nucleus
Before
1
r2c2
в‰Ґ ВҐ
2
(8.4)
2GM
1
1
r1c2 Photon
After
(Оі ray)
If we use the fact that (1 x)1/2 (1 x/2) for
x V 1, equation (8.4) gives the same result as
equation (8.3). Recoiling
If we use equation (8.4) and take r2 в€ћ, and nucleus
use the approximation for small shifts, we obtain (a)
GM
2
1 (8.5)
r1c2 Before
1

Nucleus
If we compute the wavelength shift, , we find
in atom
GM
Вў
in crystal
(8.6)
rc2
Example 8.1 Gravitational redshift After
Find the gravitational redshift for radiation emit-
ted from the surface of the Sun and for radiation
Recoiling
emitted from a 1 M white dwarf, whose radius is
Crystal
1% that of the Sun.
(b)
SOLUTION
Fig 8.10. The MГ¶ssbauer effect. (a) Emission of a gamma-
1033 g we use equation (8.5) to get
For M 2.0
ray by a free nucleus.To conserve momentum, the nucleus
16.67 dyn cm2>g2 2 12 1033 g 2
8 recoils.The recoiling nucleus carries away some energy.
10
2
r 13
1 Therefore, the energy of the gamma-ray is less than the
1010 cm>s2 2
1
energy difference between the two levels involved in the
105 cm
1.48 particular transition. (b) If the nucleus is part of an atom,
1
which is, in turn, part of a crystal, the whole crystal must
r
recoil. Since the crystal is much more massive than the
1010 cm,
For r 7 nucleus, its recoil is negligible.This means that the energy of
the gamma-ray is always equal to the difference between the
2 6
1 2.12 10 two levels involved in the transition.
1
8 GENERAL RELATIVITY 147

energy due to the recoil of the nucleus. (The recoil From this we see that t2 t1. This effect has
is to conserve momentum.) When the nucleus is been tested by taking identical clocks, leaving one
in a crystal, the whole crystal takes up the recoil. on the ground and placing the other in an airplane.
It moves very little because of its large mass, and (Of course, you must first correct for the special rel-
the energy loss by the gamma-ray is small. This ativistic effect due to the motion of the airplane.)
means that the gamma-ray energy is well defined. The airplane experiments have yielded results that
If the gamma-ray is emitted by a nucleus in agree with theory. Even more recently, tests on
one crystal, it can be absorbed by a nucleus in an rockets have yielded even more accurate results.
identical crystal, as long as there is no wave-
length shift while the photon is in motion. A
group of physicists tried an arrangement in
Just as the classical theory of electricity and mag-
which the gamma-rays were emitted in the base-
netism predicts that accelerating charges will give
ment and absorbed on the roof. The small gravi-
tational redshift was enough for the gamma-rays
predicts that certain types of systems should give
to arrive at the roof with the wrong wavelength
to be absorbed. The gamma-rays could be
is more complicated than electromagnetic radia-
blueshifted back to the right wavelength by
tion. When a gravity wave passes by the geometry
moving the crystal on the roof towards that in
of space-time briefly distorts. The types of systems
the basement. By seeing what Doppler shift is
that might produce gravitational radiation are
necessary to offset the gravitational redshift,
orbiting systems with objects close together or
the size of the gravitational redshift can be
collapsing objects.
measured. The result agrees with the theoretical
Some groups have attempted to detect gravi-
prediction.
A phenomenon related to the gravitational
sources, hoping to detect small changes in very
redshift is gravitational time dilation. All oscillators
large detectors. None of these has been successful
or clocks run slower in a strong gravitational
yet. More recently, an ambitious program involv-
field than they do in a weaker field. If we have
ing laser interferometers to detect small changes
two clocks at r1 and r2, the times they keep will be
is being developed (Fig. 8.11).
related by the same expression as the gravita-
The best evidence for gravitational radiation at
tional redshift. That is
the time of writing has been indirect. Astronomers
2GM 1>2 have been studying a binary pulsar system (to be dis-
1
r2c2
в‰Ґ ВҐ
t2 cussed in Chapter 12), in which the system appears
(8.7)
to be losing orbital energy at exactly the rate that
t1 2GM
1
r1c2 would be predicted for gravitational radiation.

Fig 8.11. To detect gravity
waves, physicists are hoping to
measure very small changes in
very large objects.This shows a
view of one station of the Laser
Interferometer Gravity
Observatory (LIGO), in Hanford,
WA. A passing gravity wave would
cause a small shift in the interfer-
ence patterns in the laser. [Photo
courtesy of Caltech/LIGO]
148 PART II RELATIVITY

8.3.5 Competing theories If we set r1 2GM/c2 (the Schwarzschild
One of the reasons that there has been so much radius), we find that 2 is infinite, even if r2 is
interest in testing general relativity as accurately only slightly greater than r1. This means that no
as possible is that there are some competing the- electromagnetic energy can escape from within
ories to EinsteinвЂ™s. These theories generally have the Schwarzschild radius. We call an object that
the same starting point, but differ in their is contained within its Schwarzschild radius a
details. The result of the tests of these competing black hole.
theories is that the experimental foundations of
EinsteinвЂ™s theory are now much stronger than
Find the Schwarzschild radius for an object of one
they were when the theory was initially studied.
solar mass.

8.4 Black holes SOLUTION
From equation (8.8) we have

122 16.67 dyn cm2>g2 2 12 1033 g 2
8
10
One of the exciting aspects of astronomy is the
13 1010 cm>s 2 2
Rs
possibility of studying a variety of fascinating
objects. By our earthly standards, even a normal
105 cm
3.0
star contains extreme conditions. However, there
are other objects that make the conditions on 3.0 km
stars seem commonplace. Among these objects
Since the Schwarzschild radius varies linearly
are black holes, objects from which no light can
with mass and has a value 3 km for a 1 M object,
escape.
we can write an expression for RS for an object of
any mass. It is
13.0 km2 1M>M 2
RS (8.10)
Shortly after Einstein published his general the-
ory of relativity, Karl Schwarzschild worked out the
Remember, every object has its Schwarzschild
solution for the curvature of space-time around a
radius. However, it can only be a black hole if it is
point mass. He found that there is a critical
contained within this radius. For example, the Sun
radius at which a singularity occurs. A singularity
is much larger than 3 km, so it is not a black hole.
is a place where some quantity becomes infinite.
The density of a 1 M black hole would be
This critical radius is called the Schwarzschild
quite high, almost 1017 g/cm3. It is higher than
the density of the nucleus of an atom. However, as
2GM/c2 we consider more massive black holes, the density
RS (8.8)
goes down. This is because the radius is propor-
Real objects are not pointlike, but have some
tional to the mass, but the volume is proportional
finite extent. An interpretation of SchwarzschildвЂ™s
to the radius cubed (and therefore to the mass
result is that if an object is completely contained within
cubed). This means that the density will be pro-
its Schwarzschild radius, the singularity will occur.
portional to 1/M2. Since we know the density for a
We can understand the significance of this
1 M black hole, we can write the density for any
critical radius by recalling the discussion of grav-
other mass black hole as
itational redshift in Section 8.2. We saw that if a
11 1017 g>cm3 2 1M>M 2 2
photon is emitted at a wavelength 1 at a distance (8.11)
r1 from a mass M, and is detected at r2, its wave-
By the time the mass reaches 108 M , the den-
length 2 is given by
sity is only a few grams per centimeter cubed,
2GM 1 2 just a few times the density of water.
1
r2c2
в‰Ґ ВҐ
We would expect the region just outside a
2
(8.9)
2GM black hole to be characterized by a large change
1
1
r1c2 in gravitational force over a small distance.
8 GENERAL RELATIVITY 149

8.4.2 Approaching a black hole
On the Earth the tides result from the changing
of the gravitational forces exerted by the Sun and What is it like to fall into a black hole? We con-
Moon from one side of the Earth to the other. By sider two astronauts. One approaches the black
extension, we refer to any effect of the variation hole, and the other stays a safe distance away. The
of a gravitational force as a tidal effect. Near a various steps are indicated in Fig. 8.12.
black hole, the gravitational force should fall off We assume that the astronaut approaching
very quickly with small changes in distance from the black hole can send out signals in various
the surface. We write the acceleration of gravity directions, including back to the other astronaut.
as a function of radius As the first astronaut approaches the black hole,
the first thing the distant astronaut would notice
GM/r2
g(r) (8.12)
is the redshift in the signals received. The magni-
Differentiating with respect to r gives tude of the redshift increases as the first astro-
naut becomes closer to the Schwarzschild radius.
2GM/r3
dg(r)/dr (8.13)
Before the Schwarzschild radius is reached,
Though the gravitational force falls off as 1/r2, another effect becomes noticeable. The paths of
the tidal effects fall off as 1/r3, meaning that they photons sent out by the first astronaut are not
are most important for small values of r. straight lines. They bend. The only direction in
which the astronaut can aim a beam and not have
Example 8.3 Black hole tidal forces
Find the difference between the acceleration of
gravity at the feet and head of an astronaut just 2 2

outside a 1M black hole.
1
SOLUTION
The change in g, g, is given by 1

g (dg/dr) r
RS RS
where r is the distance over which the change is

1.
to be found. In this case, r is the height of the

5
RS
astronaut, which we will take to be 2m. We find
dg/dr from equation (8.13) to be
(a) (b)
8
dyn cm2/g2)(2.0 1033 g)/
dg/dr (2)(6.67 10
105 cm)3
(3
2 2
10 2
1 10 cm/s /cm
Exit
(Note that the minus sign means that gravity is
Cone
stronger at the feet than at the head.) For r 2
102 cm, we have 1
1
1012 cm/s2
g 2
RS RS
This is two billion times the acceleration of gravity
at the surface of the Earth. The astronaut would be
pulled apart with a force of over a billion times the
astronautвЂ™s weight!
(c) (d)
The tidal force, dg/dr, is proportional to M/r3,
just as is the density. Therefore, the tidal force Fig 8.12. Approaching a black hole. Astronaut 1
approaches the black hole while astronaut 2 stays behind. In
will be less for more massive black holes, falling
each frame, the emitted and received wave correspond to a
to more tolerable values for very massive black
beam sent from 1 to 2.
holes (see Problem 8.9.)
150 PART II RELATIVITY

Plane perpendicular to
it bend is straight up. If the beam is not aimed axis of rotation through
Ergosphere
sufficiently close to the vertical, the bending will rotating mass

be so great that the light will not escape. Only
light aimed into a cone about the vertical, called
the exit cone, will escape. As the first astronaut
moves closer to the Schwarzschild radius, the exit
cone becomes smaller. At a distance equal to
(3/2)RS, photons aimed horizontally go into orbit
Disk singularity
around the black hole. The sphere of orbiting
Infinite redshift
photons is called the photon sphere. If you were to surfaces Event Horizons
look straight out, along the horizon, you would
The second astronaut never actually sees the Side
View
first astronaut reach the Schwarzschild radius. Side
View
The gravitational time dilation is so great that, as
RS is approached, the second astronaut thinks that
it takes the first astronaut an infinite amount of
time to reach RS. The time dilation makes the first
astronaut appear to slow down as RS is approached. Fig 8.13. Rotating black hole.The structure of the surface
From the point of view of the first astronaut, is complicated, so we show two cuts. In the upper п¬Ѓgure, we
there is no such respite. The Schwarzschild radius see the intersection of various surfaces with the plane per-
pendicular to the axis of rotation. At the center is a disk sin-
is reached very quickly. If the black hole is of suf-
gularity. (This is just a disk and doesnвЂ™t extend above or
ficiently small mass, the tidal forces would tear
below the plane.) There are two inп¬Ѓnite redshift surfaces and
the first astronaut apart. However, if the black
two event horizons. Between the event horizons, the roles
hole is massive enough, the tidal forces might be
of space and time are reversed.The region between the
survived and the astronaut crosses RS. When this outer inп¬Ѓnite redshift surface and the outer event horizon is
happens, we say that the astronaut has crossed called the ergosphere.The lower п¬Ѓgure shows a side view.
the event horizon. If the black hole is massive
enough, the astronaut might not notice anything
unusual, except that escape is impossible! is somewhat more complicated than that of a
Once inside the black hole, the inevitable jour- non-rotating black hole, and is depicted schemat-
ney to the center continues. The gravitational time ically in Fig. 8.13. The situation shown is for the
dilation is so great that time passes slowly. case in which the angular momentum per unit
However, the headlong rush through space con- mass, J/M, is less than GM/c. For the case shown,
tinues. Outside the black hole, it is time that there are two infinite redshift surfaces instead of
rushes on while distance is covered slowly. It is as a single event horizon. Between the two surfaces,
if crossing the event horizon has interchanged to the roles of space and time are reversed, just as
roles of space and time. inside the event horizon in the non-rotating case.
The second astronaut can tell nothing about The region between the outer infinite redshift
what is going on inside the black hole. In fact, the surface and the event horizon is called the ergos-
only properties of a black hole that can be deduced phere. The name results from the fact that there is
are its mass, radius, electric charge and angular a way to extract energy from the black hole by
momentum. (So far, we have assumed zero angular moving particles through the ergosphere in the
momentum. We will discuss rotating black holes correct trajectory.
below.) The external simplicity of black holes is
8.4.3 Stellar black holes
summarized in a theorem that states that black
holes have no hair. In Chapter 11 we will see that some types of stars
So far we have been discussing non-rotating evolve to a point were nothing can support them.
black holes. The structure of a rotating black hole Such a star will collapse right through the
8 GENERAL RELATIVITY 151

Schwarzschild radius for its mass, and will
-
become a black hole. Black holes would be a nor- Travel Briefly
+
mal state for the evolution of some stars. How Annihilated
would we detect a stellar black hole? We obvi-
Electron-Positron
ously could not see it directly. We could not even Pair Created
see it in silhouette against a bright source, since (a)
the area blocked would be only a few kilometers
across. We have to detect stellar black holes indi-
Free
rectly. We hope to see their gravitational effects -
on their surrounding environment. This is not a +
hopeless task, since we might expect to find a rea-
Event
sonable number in binary systems. We will dis- Horizon
cuss the probable detection of black holes in
binary systems in Chapter 12. Captured

8.4.4 Non-stellar black holes (b)
Black holes that have masses much less than a
Fig 8.14. Pair production. (a) The process in free space. An
solar mass are called mini black holes. We think
electron and positron are created out of nothing, but quickly
that mini black holes might have formed when come back together to annihilate. (b) Near a black hole, one
the density of the universe was much higher than of the particles can be captured before they can annihilate,
it is now. (The conditions in the early universe and the other escapes, carrying energy away from the black
will be discussed in Chapter 21.) These may still hole.
exist. The British physicist Stephen Hawking has
found that there is a mechanism by which mini
As a result of the uncertainty principle, a quan-
black holes could actually evaporate. Hawking is
tum mechanical vacuum is a very busy place. At any
studying the relationship between gravity and
place it is possible to create a particleвЂ“antiparticle
quantum mechanics, and the process he has pro-
pair (Fig. 8.14). (We will discuss antiparitcles in
posed is a quantum mechanical one.
Chapter 21.) It requires an energy equal to 2mc2,
This mechanism involves a different concept
where m is the mass of the particle (and the
of a vacuum than we are accustomed to seeing. In
antiparticle). The pair can exist for at most a
classical physics, a vacuum is simply nothing. In
time h/[(2 )mc2]. Before the time is up, they must
quantum mechanics it is possible to make some-
find each other and annihilate. Since electrons
thing out of nothing, if you donвЂ™t do it for long. It
have masses that are much less than protons, an
amounts to borrowing energy for a brief time
electronвЂ“positron (antielectron) pair will live
interval. The more energy you borrow, the less
longer than a protonвЂ“antiproton pair. We can
time you can borrow it for. It is related to the
therefore think of a vacuum as being made up of
uncertainty principle (which we discussed in
continuously appearing and disappearing elec-
Chapter 3). We have talked about the uncertainty
tronвЂ“positron pairs (with a small contribution
principle as it relates to momentum and position.
from heavier particleвЂ“antiparticle pairs). The phe-
However, it also relates to energy and the lifetime
nomenon is called vacuum polarization.
of a state. It says that if the state has a lifetime t,
When an electronвЂ“positron pair is created
then the energy of the state is uncertain by an
amount E, given by just outside a black hole, it is possible for one of
the particles to be pulled into the black hole
tE h/2 (8.14)
before the two recombine. The other particle will
The longer lived a state, the more accurately its continue moving away from the black hole. The
energy can be determined. Since the energy of a two cannot recombine. The particles then exist
state is uncertain by E, it is possible for us to have for much longer than the time limit for violating
this extra amount of energy and not detect it. conservation of energy. We must therefore make
152 PART II RELATIVITY

up the energy from somewhere. This process verse. In the lifetime of the universe, black holes
actually reduces the mass of the black hole. The smaller than some given mass should have disap-
black hole shrinks slightly. For mini black holes peared. Those at that mass should just be dying
this energy loss can be a significant fraction of now. Some physicists have suggested that when
the mass of the black hole. Eventually, the black this happens we should be able to see the burst of
hole shrinks to the point where it disappears in a gamma radiation.
small burst of gamma radiation. The more mas- At the other end of the mass scale, much larger
than stellar black holes, are maxi black holes. They
sive a black hole is when it starts out, the longer
it will live. An estimate for the lifetime of a black probably result from large amounts of material
hole of mass M (in grams) is (10 26 s)(M3). So, a gathering together in a small region. In Chapter 19,
black hole of about 1014 g would have a liftime of we will see evidence for 108 to 109 M black holes
about 1010 yr, a little less than the age of the uni- being present in the centers of many galaxies.

Chapter summary
In this chapter we saw how the general theory of We saw that there are several effects of general
relativity has changed our thinking about the relativity that can be tested. These include the
nature of space and time. advancement of the periastron of orbiting bodies,
We then saw how the ideas of space-time carry the bending of electromagnetic radiation, gravita-
over to a theory of gravitation вЂ“ general relativity. tional redshift and time dilation, and gravitational
The interpretation of gravitational fields is that radiation.
they alter the geometry of space-time, causing it We also saw how the gravitational redshift
to behave like that on a curved surface. The start- leads to a concept of black holes, objects from
ing point for general relativity is the principle of which nothing can escape.
equivalence, which tells us that inertial and grav-
itational masses are the same.

Questions
8.1. What do we mean when we say that gravity would you вЂњmeasureвЂќ the radius of a black
alters the geometry of space-time? hole? (Hint: Think in terms of a measurement
8.2. Re-do the analysis of the person on a scale in that doesnвЂ™t involve crossing the event
the elevator (all three cases) explicitly noting horizon.)
the uses of inertial mass and gravitational 8.6. (a) What is the exit cone? (b) When the exit
mass. cone closes, what happens to photons aimed
8.3. Why donвЂ™t you need a solar eclipse to measure straight up?
the bending of radio waves past the edge of 8.7. Is there a place near a black hole where you
the Sun? could look straight ahead and see the back of
discussed in this chapter. 8.8. If the Schwarzschild radius of the Sun is 3 km,
8.5. Since we cannot run a rule from the center of does that mean that the inner 3 km of the Sun
a black hole to the Schwarzschild radius, how is a black hole?

Problems
8.1. Consider an object with the same density as of starlight past the edge of this object as a
the Sun. Find an expression for the bending function of the size of the object.
8 GENERAL RELATIVITY 153

8.2. For a neutron star (to be discussed in 8.6. (a) Compute your Schwarzschild radius.
Chapter 11) with 1 M and a radius of 10 km, (b) What would the density be for a black hole
through what angle is light bent as it passes of your mass?
the edge? 8.7. For what mass black hole does the density
equal 1 g/cm3 ?
8.3. For a white dwarf with 1 M and a radius of
5 103 km, find the wavelength to which the 8.8. For what mass black hole does the difference
H line will be shifted by the time it is seen between the acceleration of gravity at an
by a distant observer. astronautвЂ™s feet and head equal the accelera-
tion of gravity on the Earth (1000 cm/s2 )?
8.4. (a) For the test of the gravitational redshift
involving the MГ¶ssbauer effect, calculate 8.9. Find an expression for dg/dr at the surface of
the shift in going from the EarthвЂ™s surface a black hole as a function of the mass of a
to 50 m above the EarthвЂ™s surface. (b) How black hole. Your expression should be a scal-
fast would the receiver have to move toward ing relationship as in equation (8.10).
8.10. How does the rate of a clock 1.5 RS from a
the source to compensate for the redshift?
3M black hole compare with the rate of a
speed that an object falling from the roof clock far from the black hole?
8.11. How close must you be (in terms of RS) to a
would acquire just before striking the
3M black hole to find that a clock runs at
ground.
8.5. Show that the Schwarzschild radius can also 10% the rate it runs when it is far away?
be found by taking the escape velocity from 8.12. If an electronвЂ“positron pair forms from the
an object of mass M and radius R, and setting vacuum, how long can they live before they
it equal to c. must annihilate?

Computer problems

8.2. For gravitational redshift, make a graph of Вў >
8.1. For stars in the mid-range of each spectral type
vs. r, for an interesting range of r (assuming 1 M ).
(O5, B5, . . .), make a table showing the angle of
bending of starlight just passing the limb of that 8.3. Make a table showing the Schwarzschild radii and
average density for black holes of 10n M where n
star. Also include in your table a white dwarf
which has 1 solar mass in an object the size of the goes from 0 to 10. Also include a column showing
Earth. the acceleration of gravity at the surface.
Part III
Stellar evolution
Now that we know the basic properties of stars, we look at how the laws
of physics determine those properties, and then how stars change with
time вЂ“ how they evolve. Stars go through a recurring full life cycle.They
are born, they live through middle age, and they die. In their death, they
distribute material into interstellar space to be incorporated into the next
generation of stars.
In describing the life cycle, we can start anywhere in the process. In
Chapter 9, we discuss the most stable part of their life cycle, life on the
main sequence вЂ“ stellar middle age. In Chapters 10, 11 and 12 we will look
at the deaths of different types of stars. After discussing the interstellar
medium in Chapter 14, we will look at star formation in Chapter 15.
Chapter 9

The main sequence

In this chapter we look at the inner workings of a The gravitational potential energy is the work
star once it has settled into a main sequence exis- required to bring all of the material from far
tence. We start by looking at the sources of stellar away (infinity) to the final configuration. The
energy, and then we look at the physical processes final result does not depend on the order in
that govern stellar structure. which the various parts of the sphere are assem-
bled, so we do the calculation in the easiest way
that we can envision. We can think of the sphere
9.1 Stellar energy sources as being made up of shells (Fig. 9.1). We can
assemble the sphere one shell at a time, starting
with the smallest.
When material collapses to form a star, there is
LetвЂ™s assume that we have already assembled
gravitational potential energy stored from the (neg-
shells through radius r. We now want to calculate
ative) work done by gravity in bringing the mate-
the work done to bring in the next shell. The
rial together. We might wonder how long this
thickness of the shell is dr. The volume of the
energy supply will last. We start by calculating the
(thin) shell is its surface area multiplied by its
gravitational potential energy in a uniform sphere.
thickness:
9.1.1 Gravitational potential energy 4 r2 dr
dV (9.2)
of a sphere
The mass contained in the shell is the volume
We are dealing with systems of large numbers of
multiplied by the density:
particles. Therefore, rather than thinking in
terms of individual particles of mass m, we can 4 r2 dr
dM (9.3)
think of a fluid of average density . (The density
The total mass of material already assembled is
is simply the mass per particle, multiplied by the
14 3 2R3
number of particles per unit volume.) In this sec- M1r2 (9.4)
tion, we will evaluate the potential energy for a
The quantity M(r) is important since the shell that
uniform (constant density) sphere. Even though
ends up at radius r will only feel a net force from
real objects might not be exactly uniform or
material inside it. Even after we bring in more
spherical, the results will generally only change
material outside this shell, the net force exerted
by numerical factors of order unity.
on any particle in the shell by any matter outside
We begin by calculating the gravitational poten-
radius r is zero. Also, for mass in the shell, the
tial energy of a uniform sphere of mass M, radius R,
mass M(r) exerts a force equal to that which
and density . These quantities are related by
would be exerted by the same mass all located at
14 3 2 R3 the center of the sphere.
M (9.1)
158 PART III STELLAR EVOLUTION

or
13 2 2 1M m 2kT
K (9.10)

9.1.2 Gravitational lifetime for a star
r It is possible for stars to use their stored gravita-
tional potential energy to power the star. In this
section, we calculate how long this process can go
on. For example, could the Sun be powered in this
dr
R way even now? To see, we estimate the gravita-
energy divided by the rate at which the energy is
Fig 9.1. We model stars by studying spherical shells.
being lost. The rate at which the energy is being
lost is the luminosity. That is
For any two point masses, remember the grav-
E
itational potential energy (relative to infinity) is tg
dE dt
given by
E
U Gm1m2 r (9.5) (9.11)
L
We let m1 M(r) and m2 equal the mass of the
where L is the luminosity.
shell, dM. The work to bring this shell is
We estimate E as the negative of the SunвЂ™s cur-
GM1r2 dM r
dU1r2 rent gravitational potential energy:
3 2 14 2 2r4 dr 13 5 2GM2
G14 (9.6)
E
R
To find the effect of all the shells, we integrate
10.6 2 16.67 dyn cm2 g2 2 12 1033g 2 2
the quantity dU from r 0 to r R: 8
10
17
R
1010 cm2
U dU1r2
0
1048 erg
2 (9.12)
3 2 14 2
R
2
r4 dr This gives a lifetime, called the Kelvin time,
G14

12 1048 erg 2
0

3 2 14 2 11 5 2 R
14 1033 erg s 2
2 5
tg
G14 (9.7)

13 5 2G3 14 32 R3 4 2 R
1014 s
5
3 GM2
ab 107 yr
2
5R
This is a lifetime of 20 million years. It may sound
For other shaped objects, the gravitational
like a long time. However, we know from geological
potential energy is generally proportional to
evidence that the Earth has been around for over
GM2/R, where R is some average length. The con-
four billion years. This means that the Sun must be
stant of proportionality is generally close to unity.
at least that old. Therefore, the Sun (and presum-
The thermal energy is (3/2)kT per particle. The
ably other stars) cannot exist in a stable configura-
total thermal energy is then
tion on stored gravitational energy.
13 2 2NkT
K (9.8)
9.1.3 Other energy sources
where N is the total number of particles in the
An alternative source of energy could be chemical
cloud. If the mass per particle is m, and the total
reactions. After all, we use chemical reactions to
mass of the cloud is M, then
make automobiles go on Earth. We can estimate
the amount of energy stored in the Sun, capable
N Mm (9.9)
9 THE MAIN SEQUENCE 159

of being released in chemical reactions. Typical
9.2 Nuclear physics
energies of these chemical reactions should be
equivalent to some fraction of the binding energy
9.2.1 Nuclear building blocks
of molecules that might be formed and destroyed
We have already seen that the positive charge in
in these reactions. This means that we might
atoms is confined to the small nucleus (10 13 cm
expect something like 1 eV per atom in the Sun.
across). The nucleus is composed of protons and
The total chemical energy is then 1 eV multiplied
neutrons. Because they are the building blocks of
by the total number of atoms in the Sun, M/m.
the nucleus, we call them nucleons. The proton
The energy available is then
has charge e (where e is the charge on the elec-
11 eV2 11.6 erg eV2 12 1033 g 2
12
10 tron). Since atoms are neutral, the number of
11.67 g2
E 24
10 electrons orbiting the nucleus equals the number
of protons in the nucleus. The chemical proper-
1045 erg
2
ties of an atom depend on the number of orbiting
electrons. Therefore, the number of protons in
This is much less than is stored in gravitational
the nucleus ultimately determines the identity of
potential energy, so chemical reactions clearly
the element. We designate the number of protons
cannot provide a longer term energy source for
by the symbol Z, called the atomic number. The
the Sun.
charge on the nucleus is Ze. The highest natu-
You might wonder if we have not been too
rally occurring value of Z is 92 (uranium), with
casual in our dismissal of chemical reactions as a
possible energy source for the Sun. After all, we
with higher values of Z.
didnвЂ™t even say what chemical reactions might be
Neutrons are electrically neutral. The mass of
involved. The important point is that any chemi-
the neutron is slightly greater than the mass of
cal reaction involves moving electrons from atom
the proton. Nuclei with the same numbers of pro-
to atom, and the energies associated with this are
tons can have different numbers of neutrons N.
a few eV, independent of what the reaction is.
The total number of nucleons in the nucleus is
Suppose there were some very energetic reaction
called the mass number, A Z N. The mass of a
that produced 10 eV per atom, that would
given nucleus is approximately Amp. Two nuclei
increase the chemical energy by a factor of ten
with the same number of protons but different
but it would still be many orders of magnitude
numbers of neutrons are called isotopes of the
short of the required amount. If our estimate had
same element. We generally designate an ele-
shown that chemical reactions might work, then
ment by a letter symbol (e.g. H for hydrogen), pre-
we would have to worry about the details, figure
ceded by a superscript giving A, followed by a sub-
out what reactions were important and then do a
script giving Z. (This subscript is redundant since
more accurate calculation of the stored energy.
the symbol tells us Z, and is sometimes left out.)
However, our estimate tells us that it is not worth
For example, the three known isotopes of hydro-
wasting our time on the details. These types of
gen are 1H1, 2H1, 3H1 or simply 1H, 2H, 3H.
calculations, called order of magnitude calculations,
For any given number of protons, we do not
are very important in astronomy. They help us
find an arbitrary number of neutrons. In general,
eliminate processes that obviously donвЂ™t work
we find that the stable elements have approxi-
and allow us to focus our attention on possibili-
mately equal numbers of protons and neutrons.
ties that might.
For larger values of Z the stable nuclei have a
The answer to the problem of stellar energy
slightly larger number of neutrons than protons.
sources is nuclear reactions. The typical energies
We now look at the force that holds nuclei
available in nuclear reactions are about 1 MeV
together. We generally refer to four forces in
per atom, instead of 1 eV. This is an improvement
of a factor of 106. To see how nuclear reactions nature. In order of decreasing strength, they are
strong nuclear, electromagnetic, weak nuclear and
provide energy for the Sun, we look at some of
gravity. The nuclear forces are shorter ranged, as
the elements of nuclear physics.
160 PART III STELLAR EVOLUTION

keep the protons farther apart.) This explains why
FN
FN
n
n a nucleus needs approximately as many neutrons
as protons.
FN
FN
p 9.2.2 Binding energy
n
Since the nuclear force is attractive, we must do
FN
FN work to move two nucleons apart. The work
p
p
FE
FE required to disassemble a nucleus is the binding
energy of the nucleus (Fig. 9.3). The binding
FE FE
energy is analogous to the binding energy of an
p
e
atom вЂ“ the energy required to separate an elec-
(a) tron from the rest of the atom, doing work
against the electrical attraction. Since the
FN FN nuclear force is so strong, nuclear binding ener-
p p
gies are greater than atomic binding energies by
FE
a factor of about 106. We measure nuclear bind-
ing energies in MeV, rather than eV. The greater
the binding energy of a nucleus, the more work
p p we must do to get the nucleus apart. Therefore, a
FE FN
larger binding energy means a more stable
(b) nucleus.
The binding energies of nuclei are so large
Fig 9.2. Properties of the nuclear and electrical forces
that we can measure them directly by comparing
between neutrons and protons. (a) The nuclear force, FN , is
the mass of a nucleus with the masses of its com-
the same for a proton and a proton, a neutron and a
neutron, or a proton and a neutron.The electrical force only ponents. This follows from EinsteinвЂ™s relationship
between mass and energy (E mc2) and conserva-
acts between the proton and the proton.The magnitude of
the electrical force is the same as that between a proton tion of energy
and an electron. (b) How the forces vary with distance. As
Mnucleusc2 Zmpc2 Nmnc2
BE (9.13)
the two protons are brought closer together, the electric
repulsion becomes stronger, but the nuclear attraction
becomes stronger faster.
Binding energy/nucleon (MeV)

opposed to the electromagnetic and gravitational
forces that can be felt at very large distances. For 8.5
example, we look at the force between two pro-
tons. When they are far apart, the electric force
dominates, and the protons repel each other. 8.0
When they are close together, the attractive
nuclear force dominates. The strong nuclear
7.5
force between two protons is the same as between
two neutrons and is the same as between a neu- 0 50 100 150 200 250
tron and a proton (Fig. 9.2). A
This gives us an idea of the role of a neutron Fig 9.3. We must do work to break a nucleus apart.This
in a nucleus. If we just have protons, the electric work is the binding energy. Shown here are nuclear binding
repulsion will be appreciable, and the nucleus energies for nuclei that are found in nature.The horizontal
will not be stable. If we add neutrons, we have axis is the mass number A, and the vertical axis the binding
energy, divided by the number of nucleons in the nucleus
additional binding from the nuclear force but no
(BE/A). Nuclei with higher binding energies per nucleon are
more stable.
repulsion is actually reduced, since the neutrons
9 THE MAIN SEQUENCE 161

that energy. If we hope to power stars by liberat-
where BE is the binding energy. This means that
ing nuclear energy, a nuclear reaction in which
the mass of a nucleus is less than the masses of its
components by BE/c2. This is also true for atoms. the products have a greater binding energy than
However, the binding energy for atoms is so small the reactants must be found. Then, the difference
that the mass difference is negligible (about 10 6 in energy will be available to heat the surround-
of the nuclear binding energy). ings. Before we discuss the particular nuclear
reactions that work in stars, we look at the types
Example 9.1 Proton rest energy
of nuclear reactions that can take place.
Compute the rest energy of a proton and express
The first type of reaction is a decay. In a decay,
the result in MeV.
a nucleus emits a particle. Depending on the type
of particle that is emitted, the result is either a
SOLUTION
different nucleus, or the same nucleus in a lower
11.67 g 2 13.0 1010 cm s 2 2
24
energy state. We identify three types of decay,
E 10
depending on the type of particle that is emitted.
3
1.5 10 erg
The three types are called alpha ( ), beta ( ), and
3
1.5 10 erg gamma ( ).
12
An alpha particle is the nucleus of the most
1.6 10 erg eV
common form of helium, 4He. This means that an
9.4 108 eV
alpha particle has two protons and two neutrons.
940 MeV
This particular combination is very stable and
can be emitted as a single group. The final
Example 9.2 Deuteron binding energy
nucleus has two fewer protons and two fewer neu-
Calculate the binding energy of the deuteron, an
trons than the original nucleus. This means that
isotope of hydrogen containing one proton and one
the element changes as the atomic number Z
neutron, given the following data:
decreases by two. The mass number A decreases
24
mp 1.6726 10 g by four. This process can only take place if the
resulting nucleus is more stable than the original
24
mn 1.6749 10 g
nucleus. This means that the binding energy of
24
md 3.3436 10 g the final nucleus must be greater than that of the
original nucleus. Once the alpha particle is far
enough from the nucleus, it no longer feels
SOLUTION
The binding energy is given by the nuclear attraction. However, since it has a
1mp md 2 c2
charge of 2e, it feels the Coulomb repulsion of
mn
BE
the remaining nucleus. The alpha particle accel-
13.9 g 2 12.9979 1010 cm s 2 2
27
10 erates away from the nucleus. The energy liber-
ated in the reaction is carried away in the form of
6
3.6 10 erg
the kinetic energy of the alpha particle.
2.2 MeV A gamma-ray is simply a high energy photon.
Since it carries away neither mass nor charge, nei-
By comparing the result of this example with the
ther A nor Z change. The identity of the nucleus
previous example, we see that the nuclear binding
does not change in gamma decay. The photon
energy is a few tenths of a percent of the total rest
does carry away energy. The protons and neu-
energy. This mass difference is relatively easy to
trons in a nucleus must also be treated quantum
measure.
mechanically as being waves. This means that
there are allowed states for the motions of the
9.2.3 Nuclear reactions nucleons within the nucleus. A gamma-ray is
emitted when a transition is made from a higher
We have seen that there are large amounts of
energy state to a lower energy state. The energy of
energy involved in holding a nucleus together.
the gamma-ray is equal to the energy difference
However, we must still have a way of liberating
162 PART III STELLAR EVOLUTION

between the states. We can learn a lot about the take place is 11 min. In nuclei, the reaction can
internal structure of nuclei by studying the ener- take place only if the resulting proton ends up in
gies of gamma-rays that are emitted, just as opti- a lower energy state than the original neutron.
cal spectroscopy tells us about the structure of This places an upper limit on the number of neu-
trons that a nucleus with a given Z can have and
atoms.
A beta particle is an electron or a positron. A still be stable. If we have too many neutrons,
positron is the antiparticle to the electron. Every some will beta decay and become protons in
particle has a corresponding antiparticle. A parti- lower energy states.
Another type of nuclear reaction is called fis-
cle and its antiparticle have identical masses, but
sion, in which a nucleus breaks into smaller
all other properties, such as charge, are the nega-
tive of each other. When a particle and an parts. The controlled fission of uranium provides
antiparticle come together, they can convert all the energy in current nuclear power plants.
of their mass into energy вЂ“ in other words they Generally, in fission, a very heavy nucleus breaks
annihilate each other вЂ“ without violating any into a form that is of more stable, middle mass
conservation laws. In a beta decay, an electron or nuclei. From Fig. 9.3, we see that the most stable
a positron is emitted. The charge of the nucleus nuclei are the ones with intermediate masses,
increases by e (or e if a positron is emitted), with iron being the most stable. The fission of
but the mass number A remains unchanged. The uranium or plutonium has been used in вЂњatomic
net result is to change a neutron into a proton. bombsвЂќ.
(The reverse, changing a proton into a neutron is The final type of nuclear reaction we will con-
sider is fusion, in which lighter nuclei can come
also possible, but requires a source of energy,
because the neutron is more massive than the together to build heavy nuclei. Fusion is impor-
proton.) When a neutron changes to a proton, Z tant in stellar energy generation, because there is
increases by one, but N deceases by one, again an ample supply of lighter elements. One prob-
leaving A unchanged. lem with fusion is that it is hard to start. When
In beta decays, an additional particle, called a two nuclei are far apart the nuclear force
neutrino, is emitted. Neutrinos were originally between them vanishes. They feel only the elec-
postulated because an analysis of beta decays trical repulsion. To bring nuclei close enough
indicated that some energy and angular momen- together for the nuclear force to take over, we
tum were being lost in the process. It was there- must do work against the electrical repulsion.
fore assumed that a massless neutral particle was This requires accelerating the particles to high
carrying away this energy and angular momen- energies and letting them collide. Since they have
tum. If neutrinos are truly massless (and we will high energies, they will be slowed by the electri-
have more to say on this point in Part VI of this cal repulsion, but not stopped. On the Earth, we
book), then they travel at the speed of light. The can accomplish this acceleration for small quan-
existence of the neutrino was verified experi- tities of matter in particle accelerators, but this is
mentally in the 1950s. Neutrinos do not interact impractical for large quantities of material. This is
with matter via the strong nuclear or electro- the major problem that must be overcome before
magnetic forces. This means that a reaction in we can realistically think about using fusion as an
which a neutrino is involved must proceed by the economical source of energy on the Earth.
weak nuclear force. The weak force is so weak
9.2.4 Overcoming the fusion barrier
that neutrinos rarely interact with matter. Weak
One way to increase the energies of particles is to
decays also proceed at much slower rates than do
raise their temperature. If the temperature is high
strong decays.
enough, particles will be moving fast enough to
The basic beta decay reaction is
overcome the electrical repulsion, and fusion
nSp e (9.14)
reactions can take place. On the Earth this poses
The bar over the indicates an antineutrino. In problems, since we have trouble containing a gas
free space, the average time for this reaction to at the required temperature, tens of millions of
9 THE MAIN SEQUENCE 163

kelvin. However, in stars the material is confined
by gravity. If sufficiently high temperatures are
reached, the nuclear reactions can take place. If
the temperatures are not high enough to support
nuclear reactions, then we donвЂ™t have a star. Electrical
repulsion
We can estimate the temperature required for
nuclear fusion to take place. LetвЂ™s consider the
case of two protons. Their electric potential
energy when they are a distance r apart is

Energy
e2 r
U (9.15)
Distance
so we can take their potential energy to be zero.
We let K be the kinetic energy of the particles at
this point. We would like the particles to come to
rest (zero kinetic energy) a distance rp, the radius
of the proton, apart. From conservation of energy,
the kinetic energy when the protons are far apart
Nuclear
must be
attraction
e2 rp
K Fig 9.4. The potential energy for two protons as a func-
14.8 tion of distance.This includes the electrical force and an esti-
10 2
10 esu2
11
mate of the nuclear force.
13
10 cm2
6
2 10 erg
There is another effect which allows nuclear
If we divide this by the Boltzmann constant, k,
reactions to take place at a lower temperature
we have an estimate of the temperature at which
than we have calculated. Fig. 9.4 shows the poten-
the average kinetic energy of the particles in the
tial energy of two protons as a function of dis-
gas is equal to this energy. This gives a tempera-
tance between them. For the most part, the force
ture of 2 1010 K, a very high temperature. If we
is repulsive (electrical), but if the protons are
had considered the case of nuclei with charges
brought close enough together the force would
Z1e and Z2e, the potential energy becomes
be attractive. Suppose E0 is the minimum energy
Z1Z2e2/r. This means that even higher temperatures
for a particle to be able to overcome the Coulomb
are needed for the fusion of higher charged nuclei.
repulsion. We can see that this energy E0 is the
Actually, the temperature doesnвЂ™t have to be as
вЂњheightвЂќ of the potential energy barrier. Particles
high as our calculation suggests. If the tempera-
with energies greater than E0 will pass through
ture is as high as we calculate, then the energy of
the barrier. However, according to classical
the average particle will be high enough for it to
physics, particles with energies less than E0 will
participate in fusion. We have already seen that
reach a point at which their energy equals the
not all particles in a gas have the average energy.
height of the barrier at some distance r0. At this
Even at lower temperatures there will be some
point all of the energy is in the form of potential,
particles with a high enough energy to undergo
so the kinetic energy is zero. The particle has
fusion. For particles with a MaxwellвЂ“Boltzmann
velocity distribution, the probability of finding a
direction. The point at which the particle turns
particle with an energy between E and E + dE is
back is called the turning point.
proportional to a term like the exp(вЂ”E/kT) in the
Quantum mechanically, we should talk about
Boltzmann equation. That is
the probability of finding a particle in various
P1E 2dE r E1 2 e E kT
dE (9.16) places. LetвЂ™s look again at the particle with energy
164 PART III STELLAR EVOLUTION

1
less than E0. Initially, the particle has a high prob-
ability of being found farther from the proton
than the classical turning point. As the classical
turning point is approached, the probability can-
0.8
not suddenly go from some finite value to zero,
because that probability is related to a wave phe-
nomenon. So, as we go into the barrier, the prob-
eв€’E/kT
ability of finding the particle falls off gradually. 0.6
This means that there is some probability of find-
eв€’b/E
1/2
ing the particles closer together than the classi-
cal turning point. This phenomenon is called bar-
Product 0.4
rier penetration or tunneling.
In general, a particle can penetrate a distance
approximately equal to its wavelength, h/mv. More
precisely, the probability of penetrating a distance 0.2
x is related to the wavelength by
ax
P1x2 dx r e dx
0
axmv h
e dx (9.17)
0 1 2 3 4 5
where a is some constant. Suppose we have to tun-
E/kT
nel a distance x, equal to the classical turning point
Fig 9.5. The probability of a nuclear fusion, as a function
r0, defined by the initial kinetic energy being equal
of the particle energy E, at a given gas temperature T.This
to the electrostatic potential energy at r0, or
shows the combined effects of the location of the classical
Z1 Z2 e2
1 turning point and the quantum-mechanical tunneling.
mv2
r0
2
Substituting into equation (9.17), we have the
temperature. That is, small changes in T will pro-
probability of penetrating r0:
duce large changes in the reaction rates.
P1r0 2 r e aZ1Z2e2 2hv

9.3 Nuclear energy for stars
b E1 2
re
where b is a constant. The effective area of the
nucleus for a reaction is approximately 2 and is When a star is on the main sequence, its basic
source of energy is the conversion of hydrogen
proportional to 1/E. When we combine this with
the MaxwellвЂ“Boltzmann velocity distribution, we
up with one 4He nucleus. However, it is unlikely
find the probability of a reaction by nuclei of
that four protons will get close enough to directly
energy E is proportional to
form a 4He nucleus in a single reaction. There are
e1 b>E1>2 2
E>kT
different series of reactions that achieve this net
This means that at a given T there will be a most result, and they will be discussed below.
likely value of E for reactions to take place. This is We can calculate the energy released by con-
verting four protons to one 4He by comparing
known as the Gamow peak shown in Fig. 9.5. It
turns out that most reactions involve particles their masses. We find that
m1 4He 2 0.00714mp 2
that are on the low energy side of the velocity dis-
4mp (9.18)
tribution. These lower energy particles have
This means that 0.007 of the mass of each pro-
longer wavelengths and can penetrate farther
ton is converted into energy. (The 007 will be
into the barrier. The exponential behavior of the
familiar to James Bond fans.)
reaction rates also makes them very sensitive to
9 THE MAIN SEQUENCE 165

Once the deuteron has been created, it can
Example 9.3 Lifetime of the Sun
quickly react with another proton:
Estimate the lifetime of the Sun for producing
energy at its current rate from nuclear fusion. p S 3He
d (9.20)

which we can also write as
SOLUTION
2 1
H S 3He
If 0.007 of the mass of each proton in the Sun is H
converted into energy, and if we assume that most
If each of these reactions takes place twice,
of the mass of the Sun was originally in the form
then we have started with six protons, and we
of protons, then 0.007 of the SunвЂ™s total mass is
now have two 3He nuclei. (3He is an isotope of
available for conversion into energy. The total
helium with two protons and one neutron. It is
energy available is therefore
one of the few stable nuclei with fewer neutrons
than protons.) The 3He nuclei combine to give
0.007 M c2
E

10.0072 12.0 1033 g 2 13.0 1010 cm s 2 2 3 3
He S 4He 1 1
He H H (9.21)

1052 erg Note that we have two of the protons back. The
1.3
net result is that we have converted four protons
The lifetime is this energy divided by the into a 4He nucleus, along with two positrons, two
luminosity: gamma-rays and two neutrinos. The energy given
off in this chain is carried away by the positrons,
tn EL
gamma-rays and neutrinos. The positrons will
11.3 1052 erg 2 immediately scatter off (or annihilate) particles in
14 1033 erg s2 the gas, resulting in heating of the gas. The gamma-
ray will travel a small distance before being
1018 s
3.2
absorbed, also heating the gas. The hot gas will
1011 yr
1 emit new photons, which are absorbed. This
process of absorption and re-emission of the pho-
However, we think that only 10% of the mass of
tons takes place until photons are close enough to
the Sun is in a region hot enough for nuclear
the surface to escape. The neutrino interacts so
reactions вЂ“ the core, so we must lower our estimate
weakly that it will escape from the star completely.
by a factor of ten. This leaves us with a lifetime of
Stars more massive than the Sun have higher
ten billion years. We think that the Sun has already
central temperatures than the Sun. This means
lived half of this time.
reactions involving higher charged nuclei are
We now look at actual nuclear reactions, the possible. One such process is responsible for the
net result of which is to convert four protons in a buildup of elements heavier than helium. It is
4
He nucleus, plus energy. The basic series of called the triple-alpha process, because the net
reactions in stars like the Sun is called the protonвЂ“ result is to convert three alpha particles into a 12C
proton chain, because it starts with the direct com- nucleus. The repulsion between two alpha parti-
bination of two protons: cles is four times that between two protons, so
higher temperatures, about 108 K, are needed.
p pSd e (9.19)
The first step in the chain is
which we can also write as 4 4
He S 8Be
He (9.22a)
1 1 2
H HS H e
We should note that the binding energy of two
4
He nuclei exceeds that of the 8Be because the
7
This process requires a temperature of 10 K.
4
In this reaction the e+ is a positron, the antiparti- He is such a stable nucleus. This tells us that the
8
Be should be unstable and break up. However,
cle to the electron. We can see by the presence of
there will be so many alpha particles around
the neutrino that this process is a weak interac-
that some of the 8Be nuclei will capture an alpha
tion, and therefore goes very slowly.
166 PART III STELLAR EVOLUTION

Оі catalyst for this cycle. (A catalyst helps something
13N
12 C 1H happen, but is not itself changed in the process.)
)
ay
dec There are more complicated versions of this cycle
ta
(be
ОЅ involving even heavier elements, but the basic
e+ +
ideas are the same.
Оі
As we try to build up heavier and heavier ele-
14N
13C 1H
ments through fusion, the electrical repulsion
becomes stronger. This becomes an effective barrier
Оі to the formation of heavier elements. However, no
1H 15O
14N
matter how high the Z a nucleus has, we can always
ay)
dec get a neutron near it with no electrical repulsion. If
eta
ОЅ (b the neutron is moving slowly, it can be captured by
Оі
e
the nucleus. This can be important in stars,
12C 4He
1H
15N
because some reactions provide free neutrons, so
there are generally some free neutrons available.
We can schematically represent what happens
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