that the shells are expanding at velocities of a few

tens of kilometers per second.

Blueshifted V

V The physical conditions in planetary nebulae

are determined from observations of various spec-

tral lines. Different lines are sensitive to different

temperature and density ranges. For example, Fig.

Lines of Sight Observer 10.10 shows photographs of a planetary nebula,

(a) (b) taken at wavelengths of certain emission lines.

Different lines reveal different aspects of the neb-

Fig 10.9. Lines of sight through a planetary nebula.

ula structure. Information is also obtained from

(a) Appearance.The shaded regions represent the places

studies of radio waves emitted by the nebulae.

where the lines of sight pass though the shell.The line of

sight near the edge passes through more material than that From these studies, we find that masses of

through the center.This is responsible for the ringlike planetary nebulae are of the order of 0.1 M . The

appearance. (b) Doppler shifts. Material on the near side is

temperatures are about 104 K. The mass tells us

moving toward the observer, producing a blueshift, and

that up to 10% of the stellar mass is returned to

material on the far side is moving away, producing a redshift.

the interstellar medium in the ejection of the

some planetary nebulae have a ringlike appear- nebula. This material will be included in the next

ance. However, they are spherical shells. We see generation of stars to form out of the interstellar

them as rings because our line of sight through medium. (This is in addition to mass lost through

the edge of the shell passes through more mate- stellar winds.)

Fig 10.10. The Ring Nebula,

shown in Fig. 10.8(a), is shown

here in four different colors of

light, highlighting gas with different

physical conditions.The wave-

lengths are indicated below each

frame. [NOAO/AURA/NSF]

186 PART III STELLAR EVOLUTION

We can apply the same idea to a solid, which

10.4 White dwarfs we can think of as a structure with many energy

levels. The electrons fill the lowest energy levels

10.4.1 Electron degeneracy first, but as they fill the electrons end up in

The material left behind after the planetary neb- higher and higher levels, as shown in Fig. 10.11.

ula is ejected is the remnant of the core of the The average energy of the electrons is, again,

star. It is mostly carbon or oxygen, and its tem- much greater than kT. In fact, the distribution of

perature is not high enough for further nuclear energies of electrons in a solid at room tempera-

fusion to take place. The gas pressure is not high ture is negligibly different from that in a solid

enough to support the star against gravitational at absolute zero. We call an electron gas in

collapse. This collapse would continue forever if which all of the electrons are in the lowest

not for an additional source of pressure when a energy states allowed by the exclusion principle

high enough density is reached. This pressure a degenerate gas.

arises from electron degeneracy. In a degenerate gas, most of the electrons

Electron degeneracy arises from the Pauli will have energies much greater than they

exclusion principle, which states that no two elec- would in an ordinary gas. These high energy

trons can be in the same state. For two electrons electrons also have high momenta. They can

to be in the same state, all of the quantum num- therefore exert a pressure considerably in excess

bers describing that state must be the same. For of the pressure exerted by an ideal gas at the

example, in an atom, there is a quantum number same temperature. The higher pressure is called

describing which orbit the electron is in, and degeneracy pressure. We have everyday examples

another describing how that orbit is oriented (by of this pressure. For example, it is responsible

giving the component of the angular momentum for the hardness of metals. (Metals consist of a

along some axis). In addition, we must take into regular arrangement of positive ions, held

account the fact that the electron has intrinsic together by many shared electrons. The exclu-

angular momentum, called “spin”. The spin can sion principle results in those shared electrons

have two opposite orientations. For convenience,

we call them “up” and “down” (depending on the

direction of the angular momentum vector). An

up electron and a down electron in the same

energy level are considered to be in different

states. However, two is the limit. We can only put

...

two electrons into each energy level.

We can see how this affects the properties of

E E5

atoms with many electrons. Suppose we build the

E4

atom by adding electrons one at a time. The first

E3

electron goes into the lowest energy level. The kT

E2

second also goes into the lowest level, but with

E1

the opposite spin orientation. The first level is

E0

now full. The third electron must go into the next

level. After we have added all of the electrons, we

E = 2{E0 E1 E2 E3 E4 . . .} >> NkT

can add up the excitation energies of all of the

electrons. We will find that the average excita- Fig 10.11. Energy levels in a degenerate gas.The energies

tion energy of the electrons in the atom is much of the levels are indicated on the right. In each level an

greater than kT. This means that electrons are in upward arrow represents an electron with its spin in one

higher levels than we would guess by just consid- sense, and a downward arrow represents an electron with

its spin in the opposite sense.The dashed line indicates the

ering the thermal energy available. The problem

average thermal energy per particle.The total energy is the

is that the electron cannot jump into the filled

sum of the energies of the individual electrons.

lower states.

10 STELLAR OLD AGE 187

x-component of the momentum. The momentum

being in higher states than one would expect

just based on temperature.) per second per unit surface area is just the pres-

We can also describe this pressure in terms of sure exerted by the gas on the wall:

the uncertainty principle. In Chapter 3, we saw

P n vx px (10.3)

that we must think of electrons as having wave

properties. We can only talk about the probability If we have ne electrons per unit volume, then

of finding an electron in a given place, or moving there is one electron per box with volume 1/ne.

The side of such a box is (1/ne)1/3, so the average

with a given speed. As a result of this wave prop-

erty, we cannot simultaneously describe the posi- spacing between electrons is

11>ne 2 1>3

tion and momentum of the electron. If we can

(10.4)

¢x

determine the momentum with an uncertainty

p, and the position with an uncertainty x, the If we say that the average momentum is of the

uncertainty principle tells us that order of the momentum uncertainty, then

h>2 (10.1) px h>2 ¢x

¢p ¢x

1h>2 2 ne 1>3

For a given x, the uncertainty in the momentum (10.5)

is

The speed of each electron is its momentum

h>2 ¢x (10.2)

¢p divided by its mass, so

px>me

When the density becomes very high, we are vx (10.6)

trying to force the electrons close together. This

This gives a pressure, using equation (10.3), of

means that we are trying to confine them to a

1h>2 2 2 n5>3>me

small x. Therefore, the uncertainty principle tells P (10.7)

e

us that the uncertainty in the momentum is large.

This is just an estimate of the pressure. A more

This means that large momenta are possible. These

detailed calculation yields a pressure that is a fac-

high momentum electrons are responsible for the

tor of about two higher than that given in equa-

increased pressure. Fig. 10.12 shows a container

tion (10.7).

with density n and particles moving with speed vx

Equation (10.7) gives the pressure in terms of

in the x-direction. The number of particles hitting

electron density. We would like to express it in

a wall per second per unit surface area is n vx.

terms of the total mass density . If the density of

The momentum per second per unit surface area

positive ions with charge Ze is nZ, then in a neu-

delivered to the wall is then n vx px, where px is the

tral gas the density of electrons must be

ne ZnZ (10.8)

y

Each positive ion has a mass of Amp, if we ignore

the difference between the proton and neutron

masses. The total density of the gas is then

x

Amp nZ m en e

z

vx t Amp nZ (10.9)

A

In going to the second line we have ignored the

mass of the electrons relative to the mass of the

vx nucleons. Using equations (10.8) and (10.9), the

electron density is related to the total density by

1Z>A 2 1 >mp 2

Fig 10.12. Pressure in a degenerate gas.We consider the

ne (10.10)

force on the section of area A of the right-hand wall of the

box, due to the x-component of the motions of the Substituting this into equation (10.7) and adding

particles.

the factor of two to account for the difference

188 PART III STELLAR EVOLUTION

For an ideal gas, the pressure is given by

between our estimate and the detailed calculation,

1ne nZ 2 kT

we have

P

21h>2 2 1Z>A 2 1 >mp 2 >me

2 5>3 5>3

P (10.11)

where ne nZ represents the total density.

Note that, in a degenerate gas, the pressure However, each atom of atomic number Z con-

depends on the density, but not on the tempera- tributes Z electrons, so

ture. We have already seen that this point is

ne ZnZ

important in deciding whether the triple-alpha

process will take place in a controlled way (nor- We therefore have

1Z 1 2ne kT

mal gas) or in a flash (degenerate gas).

P

10.4.2 Properties of white dwarfs Z ne kT

A star supported by electron degeneracy pressure

We can now relate this to the density . If A is

will be quite small, since it must collapse to a

the mass number of the nuclei, then (ignoring the

high density before the degeneracy pressure is

difference between proton and neutron masses)

high enough to stop the collapse. These objects

are quite hot, being the remnant of the core of a A n Z mp

star. These objects are the stars that appear on

This gives

the HR diagram as white dwarfs.

1Z>A 2 1 >mp 2kT

P

Example 10.2 White dwarf density

Estimate the density of a white dwarf if it has a 106 g>cm

10.5 2 a b 11.38

1.0 16

10 erg>K2

solar mass packed into a sphere with approxi- 24

1.67 10 g

11.0 107 K 2

mately 10 2 R (approximately the size of the

Earth) as found in Section 3.5.

1020 dyn>cm2

4.1

The degeneracy pressure is a factor of about 100

SOLUTION

We find the density by dividing the mass by the higher than the normal thermal pressure, even at

volume: this very high temperature.

12 1033 g 2 We can also use the expression for degeneracy

14 >32 17 pressure (equation 10.11) to relate the mass and

108 cm2 3

radius of a white dwarf. We saw in Example 9.4

106 g>cm3

1

that we could use hydrostatic equilibrium to

(Remember, the density of water is only 1 g/cm3, so approximate the central pressure by

GM2>R4

a white dwarf is very dense.)

P (10.12)

Example 10.3 White dwarf degeneracy pressure If we put this into equation (10.11), and substitute

M/4R3 for the density, we find

6 3

For a white dwarf of density 1.0 10 g/cm , and

Z/A 0.5, estimate the degeneracy pressure and

M2 h2 Z 5>3 M 5>3

2a ba b a 3b

compare it with the thermal pressure of a gas at a G (10.13)

R4 4 me Amp 4R

temperature of 1.0 107 K.

Rearranging gives

SOLUTION

h2 5>3

2a ba b

Z

GM1>3R

We find the pressure from equation (10.11): (10.14)

4 me 4Amp

11.05 erg s 2 2

27

2c d

10

The right-hand side is all constants, so for a given

P 28

9.11 10 g

10.5 2 11.0 106 g>cm3 2

R a mass can be calculated (Problem 10.12).

5>3

c d A degenerate gas has a very low opacity to

24

1.67 10 g

radiation. For a photon to be absorbed, an elec-

1022 dyn>cm2

3.2

tron would have to jump to a higher energy state.

10 STELLAR OLD AGE 189

times detect their presence in binary systems by

Path of White Dwarf

measuring their gravitational influence on a star

that we can see. We deduce the mass of the white

dwarf from its influence on the visible compan-

ion™s orbit (Fig. 10.13).

10.4.3 Relativistic effects

Path of The treatment of degeneracy pressure must be

Center modified by considerations introduced in the

of Mass special theory of relativity (Chapter 7). This was

Path of

first realized by S. Chandrasekhar (who shared the

Visible Star

1983 Nobel Prize in physics for his work on stellar

Fig 10.13. Even if we cannot see a white dwarf in a binary

structure). Chandrasekhar found that these cor-

system, we can detect its presence.The visible star will

rections reduce the degeneracy pressure. This

appear to wobble as it and the white dwarf orbit their

provides an upper limit to the mass that can be

common center of mass.

supported by electron degeneracy pressure.

The modification arises from the fact that

However, such a transition would have to be an electrons cannot travel faster than the speed of

already empty state, and may require more light. In using equation (10.5), we can still say

energy than is carried by an optical photon. In that px is (h/2 )(ne)1/3. However, we can no longer

addition, degenerate gases are good heat con- say that vx px/me. We have to use the correct rel-

ductors, which explains why metals are good ativistic expression, as discussed in Chapter 7. To

heat conductors. The low opacity and high ther- find the maximum degeneracy pressure, we take

mal conductivity mean that a degenerate gas vx c. This gives

cannot support a large temperature difference.

1h>2 2 c 1ne 2 4>3

Pmax (10.15)

The internal temperature of a white dwarf is

approximately constant across that star, at about A more detailed calculation gives approximately 0.8

107 K. The outermost 1% is not degenerate, and it times this. Using this, the expression analogous

is in that thin layer that the temperature falls to equation (10.11) is

from 107 K to roughly the 104 K indicated by its

10.82 1h>2 2 c 1Z>A 2 4>3 1 >mp 2 4>3

P (10.16)

spectral type. These conditions make a white

dwarf very different from a normal star. In addi- If we use the same assumptions to find the

tion, Zeeman splitting measurements suggest mass“radius relation (equation 10.14), we find

very strong magnetic fields, about 107 gauss in that the radius drops out, and we simply have an

some cases. expression for the mass (Problem 10.14):

10.05 2 1hc>2 G2 3>2 1Z>A mp 2 2

What happens to a star after it becomes a

M (10.17)

white dwarf? As it radiates it must get cooler. This

This mass corresponds to the maximum

is because it is giving off energy, but has no

pressure, so it is the maximum mass that can be

source of new energy. The degeneracy pressure

supported by electron degeneracy pressure. A

does not depend on the temperature, so the star

more accurate calculation, which takes into

will maintain its size as it cools. Eventually, it will

account variations in pressure and density with

be too cool to see. It will take tens of billions of

distance from the center of the star, gives a

years for a star that is now visible as a white

mass that is a factor of about 60 higher. The

dwarf to become that cool. We can understand

resulting mass, called the Chandrasekhar limit, is

the long lifetime when we realize that a white

dwarf radiates like a 104 K blackbody, but has 1.44 M . A star whose nuclear processes have

more energy than a thermal reservoir at 107 K stopped, and whose mass is greater than 1.44 M ,

will continue to collapse beyond the white

(see Problem 10.13).

dwarf phase. The fate of such stars will be dis-

Even at their current temperatures, their small

cussed in Chapter 11.

sizes make white dwarfs difficult to see. We some-

190 PART III STELLAR EVOLUTION

Chapter summary

In this chapter we saw what happens to stars that In the red giant or supergiant phase, the outer

use up their supply of hydrogen fuel in their cores. layers are loosely bound to the star. Mass loss

Low mass stars evolve into red giants. Higher mass takes place. One form of mass loss is ejection of a

stars evolve into red supergiants. planetary nebula.

We saw how some stars at this stage are unsta- We saw how the remnant of a low mass star is

ble to pulsations and become Cepheid variables. a white dwarf, a star supported by electron degen-

The Cepheids are particularly important because eracy pressure. There is a limit, the Chandrasekhar

of a period“luminosity relation. This relation limit, on the mass that can be supported by elec-

allows us to use Cepheids as important distance tron degeneracy pressure. Best estimates of this

limit are 1.44 M .

indicators.

Questions

10.1. In which aspects of post-main sequence 10.4. How does electron degeneracy pressure sup-

evolution does the mass enter into port a white dwarf?

consideration? How are low mass stars really 10.5. (a) If we try to confine an electron to within an

atom, x 1 …, what is p, the uncertainty in

different from high mass stars?

10.2. Why is there hydrogen left in a shell around its momentum? (b) What velocity would the

the core after it has been used up in the core? electron need to have this momentum?

10.3. For the surface oscillations of a Cepheid, 10.6. List the distance measurement techniques

show how the energy goes from kinetic to that we have encountered so far in this book,

potential throughout the stellar oscillation and estimate the distance range over which

cycle, as in the analogy of a swing. each is useful.

Problems

*10.5. The momentum of a photon is E/c. (a) Calculate

10.1. Suppose we observe a classical Cepheid with

a period of 10 days and an apparent magni- the momentum per second delivered to the

outer layers of a 102 L star if all the photons

tude of 6. How far away is it?

10.2. Suppose we observe a classical Cepheid with are absorbed in that layer. (b) How does the

a period of 8 days and m 7. What is the force on the layer compare with the gravita-

period of a type II Cepheid at the same dis- tional force on the layer if the layer has a

radius R 100 R and a mass M 0.1 M and

tance with the same apparent magnitude?

the rest of the star has a mass of 1 M .

*10.3. Suppose a classical Cepheid has a period

P 10 days, and assume that the surface 10.6. (a) What is the kinetic energy in a 0.1 M

planetary nebula, expanding at 103 km/s?

oscillates sinusoidally with a maximum

amplitude Rmax 103 km. (a) Write an (b) Compare that with the gravitational

expression for R vs. t, assuming that potential energy of the shell when it was at

R 0 at t 0, and that the star is R 100 R , assuming that there is 1 M of

expanding at t 0. (b) Find an expression material left behind.

for the speed of the surface, v(t). (c) Find 10.7. Suppose a planetary nebula is a spherical

the wavelength observed for the H line shell whose thickness is 10% of its radius.

vs. t. Compare the length of the longest line of

1/2

M/R3 and

*10.4. Use the fact that P , sight with that through the center, and

L R2T4 to derive a relationship between relate your answer to the appearance of the

period and luminosity. planetary nebula.

10 STELLAR OLD AGE 191

10.8. Suppose a planetary nebula expands at a is their momentum uncertainty p?

speed v, determined from Doppler shift (c) What is the velocity corresponding to

observations. The angular size is observed that momentum?

to be increasing at the rate d /dt. In terms of 10.12. (a) Use the mass“radius relationship for

these quantities, find an expression for the white dwarfs to calculate the radius of a

distance to the planetary nebula. 1 M white dwarf. (b) Use this result to

10.9. Suppose we observe the H line in a planetary rewrite equation (10.14) in a form that gives

nebula. The maximum wavelength is R in kilometers when M is expressed in solar

656.32 nm, and the minimum wavelength is masses.

656.24 nm. How fast is the nebula expanding? 10.13. (a) What is the thermal energy stored in a

107 K white dwarf of 1 M ? (b) Assuming

10.10. Calculate the escape speed from a 1 M

that it radiates like a 104 K blackbody, esti-

white dwarf. Compare it with that of a main

sequence star of the same mass. mate its lifetime as a luminous object.

10.11. (a) At white dwarf densities, what is the aver- 10.14. Derive equation (10.17).

age separation between electrons? (b) What

Computer problems

masses ranging from 0.1 M to the Chandrasekhar

10.1. Suppose a planetary nebula is a spherical shell

whose thickness is 10% of its radius. Find the limit, and make a graph of size vs. mass over that

length of a line of sight through the material as a range.

function of position across the nebula. 10.3. Tabulate the radii of white dwarfs with masses

10.2. Use the mass“radius relationship for white corresponding to the mid-range of each spectral

dwarfs to calculate the radii of white dwarfs with type (O5, B5, . . .).

Chapter 11

The death of high mass stars

In Chapter 10 we saw how stars evolve to the red less than the Chandrasekhar limit, the core can

giant or red supergiant stages, and how low mass be supported by electron degeneracy pressure.

stars (less than 5 M ) lose enough mass to leave However, once the core goes beyond that limit,

behind a white dwarf as the final stellar remnant. there is nothing to support it, and it collapses. In

We also saw that electron degeneracy pressure the collapse, some energy, previously in the form

can only support a 1.44 M remnant. In this chap- of gravitational potential energy, is liberated.

Since that energy is available, the 56Fe can react

ter we will see what happens to higher mass stars.

It is important to remember that stars lose mass by using up the energy. This means that the core

as they evolve. This mass loss can be through winds, does not get any hotter. It continues to collapse.

or the ejection of planetary nebulae. (In the next A runaway situation develops in which the iron

chapter, we will see that stars in close binary sys- and nickel consume liberated energy. As the iron

tems can transfer mass to a companion.) Though is destroyed, protons are liberated from nuclei.

we only have estimates for the total amount of The electrons in the star can combine with these

mass loss, it seems likely that massive stars can lose protons to form neutrons and neutrinos. This

more than half of their mass by the time they pass reaction can be written

through the red supergiant phase. A star™s evolu-

e pSn (11.1)

tion will depend on how much mass it starts with,

and how much mass it loses along the way. The core is driven to a very dense state in a short

time, about one second. What happens next is not

completely understood, but the collapse results in

11.1 Supernovae an explosion in which most of the mass of the star

is blown away. The neutrons created in reaction

11.1.1 Core evolution of high mass stars (11.1) probably play a role in this. They also obey the

In the core of a high mass star the buildup of exclusion principle, and exert a degeneracy pres-

heavier elements continues. If we look at nuclear sure (the details of which we will discuss in the next

binding energies (Fig. 9.3) we see that the isotope section). This pressure can stop the collapse and

of iron 56Fe has the highest binding energy per cause the material to bounce back. In addition, so

nucleon. This makes it the most stable nucleus. many neutrinos are created, and the material is so

This means that any reaction involving 56Fe, be it dense, that a sufficient number of neutrinos inter-

fission or fusion, requires an input of energy. act with the matter forcing the material outward.

Such an exploding star is called a supernova.

When all of the mass of the core of the star is con-

verted to 56Fe (and other stable elements, such as This type of supernova is actually called a type II

supernova. Another type of supernova, type I, seems

nickel), nuclear reactions in the core will stop.

At this stage, the core will start to cool and the to be associated with older objects in our galaxy.

thermal pressure will not be sufficient to support (The mechanism for type I supernovae probably

the core. As long as the mass of iron in the core is involves white dwarfs in close binary systems,

194 PART III STELLAR EVOLUTION

Fig 11.1. A supernova in

another galaxy can be almost as

bright as the whole galaxy.This

shows a supernova in the spiral

galaxy NGC4603. [STScI/NASA]

11.1.2 Supernova remnants

and is discussed in Chapter 12.) During the explo-

sion, nuclear reactions take place very rapidly, The material thrown out in a supernova explosion

and elements much heavier than iron are cre- is called a supernova remnant. It contains most of the

ated. This material is then spread out into inter- material that was once the star. In young super-

stellar space, along with the results of the normal nova remnants we can actually see the expansion

nucleosynthesis during the main sequence life of of the ejected material. These remnants are

the star. This enriched material is then incorpo- important because they spread the products of

rated into the next generation of stars. nucleosynthesis in stars throughout the interstel-

The light from a supernova explosion can lar medium. There, this material enriched in

exceed that of an entire galaxy (Fig. 11.1). The “metals” will be incorporated into the next gen-

energy output in a type II supernova is about eration of stars. This explains why stars that

1053 erg. About 1% of this shows up as kinetic formed relatively recently in the history of our

energy of the shell, and 0.1% as light. (Most of the galaxy have a higher metal abundance than the

energy is in the escaping neutrinos.) After a rapid older stars. In the later stages of a supernova rem-

increase in brightness, the supernova fades grad- nant™s expansion, we still see a glowing shell, like

ually, on a time scale of several months (Fig. 11.2). those in Fig. 11.3. These shells also serve to stir up

SNII MV light curves

Fig 11.2. Light curves for type

II supernovae. [Craig Wheeler,

University of Texas, Austin]

-18 -17

-16

-15

SN 1993J

SN 1987A

SN II-L

SN II-P

0

-14

MV

-12

-10

56

Co

-8

0 100 200 300 400 500

phase (days)

11 THE DEATH OF HIGH MASS STARS 195

(c)

(a)

(d)

(e)

(b)

Fig 11.3. Views of supernova remnants. (a) An optical image of a portion of the Veil Nebula supernova remnant. (b) A (false

color) radio image (6 cm wavelength) of the source known as Cas A (the brightest radio source in Cassiopeia), taken with the VLA,

providing an angular resolution of 0.1 arc sec. (c) A far infrared image of Cas A, taken with the ISO satellite. (d) An X-ray image of

Cas A, made with the Chandra Observatory. (e) HST image of the region of SN1987A in the LMC.The small bright ring shows the

interaction of the expanding supernova remnant with the surrounding medium. [(a) NOAO/AURA/NSF; (b) NRAO/AUI/NSF;

(c) ESA/ISO,ISOCAM/CEA and P. Lagage et al. (d) NASA; (e) STScI/NASA]

196 PART III STELLAR EVOLUTION

e-

B

(a)

v

B

(b)

Fig 11.4. Synchrotron radiation. (a) Electrons spiral

around magnetic ¬eld lines. (b) In their circular motion, they

are constantly accelerating, and therefore give off radiation.

At any instant, most of the radiation is emitted in a small

(a)

cone centered on the velocity of the electron.The beaming

of the radiation is a relativistic effect.

the interstellar medium. This point will be dis-

cussed further in Chapter 15.

Supernova remnants generally have mag-

netic fields that are strong by interstellar stan-

dards, and a supply of high energy electrons. In

Chapter 6, we discussed the motion of charged

particles in a magnetic field. The component of

the electron™s velocity along the field lines is not

changed, since the force is perpendicular to the

field direction and to the electron™s velocity. The

electron traces out a helix as it circles around

the field lines, and drifts along the direction of

the field, as shown in Fig. 11.4. Since the velocity

of the electron is changing (in direction), the

electron must be accelerating. Classical electro-

magnetic theory tells us that accelerating

charges must give off radiation. This radiation is

called synchrotron radiation. (b)

One of the best studied supernova remnants is

Fig 11.5. The Crab Nebula, a young supernova remnant.

the Crab Nebula in the constellation of Taurus.

(a) In normal light from the ground. (b) From HST showing

(This supernova was observed as a bright object in

great detail. For comparison, an X-ray image is shown in

1054 AD.) Some optical images are shown in Fig. Fig. 4.33(c). [(a) NOAO/AURA/NSF; (b) STScI/NASA]

11.5. An X-ray image is shown in Fig. 4.33(c). All of

the radiation is synchrotron radiation.

tion, and we change the orientation of the detec-

Synchrotron radiation has a number of distin-

tor, we see an intensity of radiation that varies

guishing characteristics. One is that the radiation

with the angle of the detector. This means that the

is polarized. If we have a radio telescope receiver

electric field vector of the incoming radiation is

that detects only one direction of linear polariza-

11 THE DEATH OF HIGH MASS STARS 197

that if the mass is more than 1.44 M electron

’22

10

degeneracy pressure cannot support it. The col-

’24

10 lapse of the core continues beyond even the high

densities associated with a white dwarf. As the

’26

10

density increases, electrons and protons are

’28

forced together to make neutrons. The resulting

W/m / Hz

10

2

object is called a neutron star.

’30

Crab

10

Nebula

11.2.1 Neutron degeneracy pressure

’32

10

Neutrons have spin properties similar to those of

’34

electrons. They therefore also obey the Pauli

10

exclusion principle. (We have already seen how

’36

10

this affects the energy levels of neutrons in a

’2 ’4 ’6

2

10 m 1 10 10 10 m

nucleus, in Chapter 9.) Neutrons are therefore

1 µm

capable of exerting a degeneracy pressure if the

’2 ’4

2

10,000 10 1 10 10

density is high enough. We can estimate the neu-

»

tron degeneracy pressure as we did the electron

Fig 11.6. Spectrum of the Crab Nebula, from radio to

degeneracy pressure in the preceding chapter.

gamma-rays.

The result corresponding to equation (10.7),

including the factor of two, is

preferentially aligned along some direction. Even 2(h/2 )2nn5/3/mn

P (11.2a)

the visible light is polarized.

nnmn, so

Since the star is all neutrons,

The spectrum of synchrotron radiation is also

quite distinctive. Most of the radiation is given off 2(h/2 )2 5/3

/mn8/3

P (11.2b)

at long wavelengths. The intensity of radiation

In comparing this with equation (10.7), we see

falls off as the wavelength raised to some power.

that at a given density electron degeneracy pres-

This is called a power law spectrum. Fig. 11.6 shows

sure will be greater by a factor of mn/me, or about

the spectrum of the Crab Nebula. The intensity of

2000. However, in a neutron star there are no free

radiation is much greater in the radio part of the

electrons, so neutron degeneracy pressure is all we

spectrum than in other parts of the spectrum,

have to support the star. Because the density of a

but the source is so strong that we are able to

neutron star is so much higher than that of a

detect synchrotron radiation in all parts of the

white dwarf, the neutron degeneracy pressure in a

spectrum, even at gamma-ray energies. So much

neutron star is greater than the electron degener-

energy is given off in synchrotron radiation that

acy pressure in a white dwarf. The neutron degen-

the high energy electrons should be losing their

eracy pressure will halt the collapse of the core.

energy over a very short time. However, they are

Let™s consider some properties of neutron stars.

still radiating strongly some 1000 years after the

We can estimate the radius from a mass“radius

supernova explosion. Until recently, it was a prob-

relationship, like that for white dwarfs (Problem

lem for astronomers to explain where the energy

11.3). Neutron stars will be smaller than white

comes from to maintain the high energy elec-

dwarfs by a factor of about me/mn. We find the

trons in the Crab Nebula.

radius to be about 15 km for a mass of 1 M . This

means that a neutron star concentrates more

11.2 Neutron stars than a solar mass in a sphere smaller than the

island of Manhattan.

Example 11.1 Density of a neutron star

In this section we look at what happens to the

core that is left behind in the supernova explo- Estimate the density of a neutron star and compare

sion. The core is compressed so that normal gas it with that of a neutron. Take the mass of the star

to be 1.4 M .

pressure cannot support it. We have already seen

198 PART III STELLAR EVOLUTION

Another interesting effect comes from the

SOLUTION

fact that g changes very quickly with radius R.

The density is the mass divided by the volume:

11.4 2 12 1033 g 2

Differentiating the expression for g gives

14 32 11.5 2GM/R3

106 cm2 3 dg/dr

1014 g>cm3 If we use the numbers in the above example,

2

1.1 108 (cm/s2)/cm. This is

we find dg/dR

The density of a neutron is

equal to a change of 105 times the acceleration of

11.7 g2

24

10 gravity on the Earth per centimeter. If you were

14 32 11.0

n 13

cm2 3 floating near the surface, your feet would be

10

pulled in with a much greater acceleration than

1014 g>cm3

4

your head. Your body would be pulled apart by

So we see that the density of a neutron star is very these tidal forces. By tidal forces, we mean effects

close to that of a neutron. This means that the neu- that depend on the difference between forces on

trons in a neutron star must be packed very close opposite sides of an object. Some astrophysicists

together, with very little empty space. have jokingly noted that if an astronaut visits a

neutron star, it should be in a prone position to

Example 11.2 Neutron degeneracy pressure minimize the tidal effects.

Compare the neutron degeneracy pressure in a The large acceleration of gravity also has

neutron star with the electron degeneracy pressure another interesting effect. The equation of hydro-

in a white dwarf. static equilibrium (equation 9.44) tells us that the

rate at which the pressure in the atmosphere of a

SOLUTION

neutron star falls off, dP/dR, is proportional to g.

Using equations (11.2b) and (10.11), The atmospheric pressure on a neutron star thus

53

a b a b

Pns me falls off very quickly. This leads to an atmosphere

ns

that is only about 1 cm thick. (The thin atmos-

Pwd mn

wd

phere is another reason for an astronaut to stay

1014 5 3

a 6b a b

2 1

in a prone position.)

2000

2 10

1010 11.2.2 Rotation of neutron stars

1

In the process of the collapse of a core to become

Example 11.3 Acceleration of gravity on a a neutron star, any original rotation of the core

neutron star will be amplified. If the angular momentum of

For the star in the above example, find the acceler- the core is conserved, the core must rotate faster

ation of gravity at the surface as it becomes smaller. Since the core shrinks by a

large amount, the rotation speed is increased by

SOLUTION a large amount.

The angular momentum is given by

GM

g

R2 J I (11.3)

16.67 dyn cm g 2 11.42 12

8 2 2 33

10 10 g2

where I is the rotational inertia and is the angu-

11.5 106 cm2 2 lar speed. If we put in the rotational inertia for a

uniform sphere, we find

1013 cm>s2

8.3

(2/5)MR2

This is almost 1011 times the acceleration of gravity J (11.4)

at the surface of the Earth! (With such strong gravi-

To get a feel for how fast a neutron star can

tational fields, we should really use general relativ-

rotate, let™s assume that the angular momentum

ity to calculate particle motions.) You can calculate

of a neutron star is equal to that of the Sun. (This

your weight on the surface of a neutron star.

is probably a conservative estimate since the Sun

11 THE DEATH OF HIGH MASS STARS 199

is not rotating very rapidly compared with other Crust

104

stars we see.) Using equation (11.4), we have

1011

2

R2

R (11.5)

Solving for 1 2 , we have

ρ (g/cm3)

1 2 1R R2 2 (11.6)

1010)/(1.5 106)]2

[(7

109

2

The rotation period of the Sun is about 30 days.

4 x 1014

The period of a neutron star is therefore Superfluid

Neutrons

106 s)/(2 109) 1 km

Pns (2.6

Fig 11.7. Structure of a neutron star.

3

1.3 10 s

In addition to its other extreme properties, a 2 109 times that of the Sun. It should be noted

neutron star might be rotating 1000 times per that, even though flux conservation arguments

second! give us what might be an order of magnitude for

B, there is some disagreement over the actual

11.2.3 Magnetic ¬elds of neutron stars

mechanism for the buildup of large fields in neu-

We might also expect a neutron star to have a

tron stars. Some theories involve dynamo mecha-

strong magnetic field. This is a consequence of

nisms, in which rotation plays an important role.

Faraday™s law, which can be written as

Most of the interior of a neutron star is

AE dl d B / dt thought to be a fluid. Because of certain quan-

tum mechanical properties, this fluid can flow

In this expression B is the magnetic flux

with no frictional resistance or viscosity. A fluid

through a surface. (For a small surface element,

with no viscosity is called a superfluid. (This is

the flux is the dot product of the magnetic field

analogous to a superconductor which allows the

and a vector whose magnitude is the surface area

flow of electricity with no resistance.) Actually,

and whose direction is perpendicular to the sur-

the existence of vortices (such as whirlpools) in

face.) The integral on the left is performed around

the rapidly rotating fluid leads to some viscosity.

a closed path that forms the boundary of the sur-

This is what couples the fluid to the outside layers.

face. E is the electric field induced by the change

Outside the fluid is a solid crust made up of heavy

in flux. The integral on the left is the induced

elements, such as iron. The crust is probably less

electromotive force (emf) around the closed path.

than 1 km thick (Fig. 11.7). We have already seen

If there is an induced emf around the path, and if

that the atmosphere is even thinner.

the material has some conductivity, currents will

The possibility of the existence of neutron

flow to oppose the change in flux. When the flux

stars was first realized in the 1930s. However,

through the surface stays constant, we say that

they were objects that existed in theory only. It

the flux is frozen into the material.

was felt that their small size would make them

The flux is proportional to the strength of the

hard to see. In 1967 an accidental discovery in

magnetic field and the surface area. This means

radio astronomy revived interest in neutron stars.

that the quantity

BR2 constant (11.7)

11.3 Pulsars

The magnetic field should therefore be propor-

tional to 1/R2, just as the rotation rate is. If the 11.3.1 Discovery

In 1967, Antony Hewish and his graduate student,

neutron star started off with a solar magnetic

Jocelyn Bell Burnell, were looking for the twinkling

field, it will end up with a magnetic field of about

200 PART III STELLAR EVOLUTION

1855+09

1913+16 Fig 11.8. Distribution of pulsars

0950+08

1953+29 on a galactic coordinate system.

0820+02

Their tendency to concentrate

0655+64 near the galactic plane indicates

that they are within our own

galaxy.We will talk more about

galactic structure in Chapter 16.

Crab

0329+54

2303+46

Vela

1937+21

0529’66

1831’00 1509 ’58

1641’45

of radio sources. This twinkling, called scintillation, they noted that the signal was a string of pulses,

is analogous to the twinkling of stars. However, it with a very regular separation. By counting a large

is not caused by the Earth™s atmosphere. It is number of pulses, they determined the period to

be 1.337 301 1 s. A few other pulsars were found,

caused by charged particles in interplanetary

space. The observations required very good time suggesting a general phenomenon. Since then,

resolution. For most radio observations, the hundreds of pulsars have been discovered. Most

sources are so weak that we have to observe for a of them are close to the plane of our galaxy, as

long time to detect a signal. There is generally no shown in Fig. 11.8. This suggests that they are in

call to see how fast things are changing. However, the galaxy. If they were extragalactic, their distri-

Hewish and Burnell were looking for changes on bution would not correlate with anything within

time scales that were much shorter than for typi- our galaxy.

cal radio observations. This good time resolution We see some properties of pulsar signals in

is what allowed them to make an unusual acci- Fig. 11.9. Note that each pulse is not exactly the

dental discovery. Hewish received the Nobel Prize same as the previous one. The period may even

in physics for this work. vary slightly from one pulse to the next. However,

Hewish and Burnell found a rapidly varying if we take the average of 100 pulses and compare

source. When they looked at it in more detail, it with the average of the next 100 pulses we find

Fig 11.9. Selection of pulsar

pulse pro¬les. Each curve is an

average over many cycles.They

are plotted as intensity vs. frac-

tion of a cycle. Periods in sec-

onds: (a) 0.94295 s; (b) 0.71452 s;

(c) 0.0057575 s; (d) 3.74552 s;

(e) 0.0052557 s; (f) 0.29011 s;

(g) 0.0015578 s; (h) 0.0049311 s;

(i) 0.016052 s. [Dunc Lorimer

& Michael Kramer (Jodrell Bank

Obs.)]

11 THE DEATH OF HIGH MASS STARS 201

that the two averages agree quite well. We can see pulsar periods, the orbital radii would have to be

that the pulsar is actually “off” most of the time. smaller than a normal star, or even a white dwarf.

The pulse is on for only a small fraction of the If we have orbiting objects, we would need two

period. We define the duty cycle as the fraction of neutron stars.

the pulse period in which the pulse is actually on. There is a problem with orbiting systems. The

For most pulsars the duty cycle is about 5%. This energy of the orbit is (by equation 5.41)

means that the peak brightness is about 20 times

E Gm1m2/2R

the average brightness (Problem 11.9). Notice that

there may also be a small pulse somewhere in the As the pulsar gives off radiation, to conserve

middle of the cycle. This is called an interpulse, energy, the orbital energy must decrease. This

means that E becomes more negative, or the

and fewer than 1% of pulsars have one.

absolute value of E becomes larger. For this to

11.3.2 What are pulsars? happen, R must become smaller. As R becomes

smaller the period must decrease, since P R3/2

When the discovery of pulsars was announced,

astronomers immediately attacked the problem (equation 11.8). However, we observe pulsar peri-

of what they are. Initial guesses even included ods to increase, rather than decrease (see below).

the possibility that they were signals sent by There is an additional problem with orbiting neu-

intelligent civilizations (called LGM theories, for tron stars. We saw in Chapter 8 that the general

little green men). However, as more pulsars were theory of relativity predicts that such a system

discovered in different parts of the galaxy, it would lose energy very quickly by giving off grav-

seemed that a more natural explanation was itational radiation. We would be able to detect this

needed. Efforts were concentrated on trying to as a significant decrease in the orbital period.

find the clock mechanism. Three basic mecha- (We will discuss a binary pulsar system, in which

nisms were tried: (1) pulsation, (2) orbital motion, this effect has been studied, in the next chapter.)

and (3) rotation. Three different types of objects This leaves rotation as the mechanism for

were considered: (1) normal stars, (2) white producing the regularity in the pulses. For any

dwarfs, and (3) neutron stars. rotating star the rotation must not be so fast that

Pulsation of stars is reasonably well under- objects on the surface lose contact with the sur-

stood. In Chapter 10, we saw that, for radial oscil- face. The gravitational force must be greater than

lations, the period is roughly proportional to the force required to keep a point moving in a

(G ) 1/2, where is the average density of the star. circular path. If not, an object on the surface will

Pulsations of normal stars have periods of hours go into orbit just above the surface. This gives the

to days, and would not explain pulsars. For a same constraint on the size of an object as equa-

period of 0.1s, a density of about 109 g/cm3 is tion (11.9), with the sum of the masses replaced

required. This is about 103 times the density of a by the mass of the single rotating star. This again

white dwarf, but only 10 5 of the density of a rules out a rotating normal star and a rotating

neutron star. No stellar object has a density even white dwarf. (White dwarfs might actually work

close to the right range. Therefore, the radial pul- for some of the slower pulsars, but not for the

sations in stars are ruled out. fastest.) This leaves rotating neutron stars as the

We next consider orbital motion. The period best candidates for pulsars.

and radius of an orbit are related by (equation 5.20) In describing the emission from pulsars, an

analogy has been drawn with a lighthouse. The

23

m2)P2

R /G

4 (m1 (11.8)

light in the lighthouse is always on, but you can

Solving for the radius gives only see it when the beam points in your direc-

tion. If you stay in one place, you will see the light

2

m2)P2]1/3

R [(G/4 ) (m1 (11.9)

appear to flash on briefly once per cycle. In this

For a period of 1 s, the radius is 2000 km; for model of neutron stars, there is some emission

a period of 0.1 s, the radius is 400 km; for a period point or “hot spot” on the surface of the star, pro-

of 0.01 s, the radius is 100 km. For the range of ducing a beam of radiation, like a lighthouse beam.

202 PART III STELLAR EVOLUTION

We can only see the radiation when the beam is on that cone, the pulsar will never be visible. This

pointed in our direction. is also true for a lighthouse. If you are directly

The emission mechanism may be related to the above the lighthouse, you will never see the light.

strong magnetic field that we think many neutron Any given pulsar will only be seen by about 20%

stars must have (Fig. 11.10). The magnetic axis of of the potential observers. This means that our

the star is probably not aligned with the rotation galaxy contains many more pulsars than the few

axis. (This is not unusual; it also happens on the hundred that we actually observe. This is espe-

Earth.) If the beam of radiation is somehow colli- cially true since we cannot survey the whole

mated along the magnetic axis, we only see it when galaxy for pulsars. Current estimates place the

the beam points in our direction. We may even see number of pulsars in the galaxy at about 200 000.

two pulses as opposite magnetic poles pass by. Probably the most extensively studied pulsar

The details of the mechanism are not clear. One is one in the center of the Crab Nebula, the rem-

possibility is that the rotation and strong magnetic nant of the supernova explosion observed in

field result in an observer near the surface seeing a 1054. We have already seen that the emission

rapidly changing magnetic field. Faraday™s law tells from the Crab Nebula is polarized, suggesting

us that a rapidly changing magnetic field can accel- synchrotron radiation, and that it gives off most

erate charged particles to large energies. These par- of its energy in the radio part of the spectrum.

ticles then spiral around the magnetic field lines, We have also seen that the energy loss via syn-

producing synchrotron radiation. This is only a chrotron radiation is so great that the nebula

very general outline of what might be taking place. should have faded considerably. Something is

A considerable effort is still going on to understand replacing the energy that is radiated away. The

the emission of radiation under the extreme con- total energy loss rate of the nebula is about 3

1038 erg/s, or about 105 L . It is known that there

ditions that exist near the surface of a neutron star.

The lighthouse nature of the pulsar mecha- is a star with an unusual spectrum at the center

of the nebula. This star is known as Baade™s star

nism has another consequence. The beam only

after Walter Baade who first noted its peculiarity.

traces out a cone on the sky. If the observer is not

A very rapid pulsar was discovered at the posi-

tion of this star. Its period is 0.033 s. This pulsar

Rotation

was very important in ruling out rotating white

Axis

dwarfs as the source of pulsars. After the radio

Radiation

pulsar was discovered, it was suggested that it

Beam

might be possible to observe optical pulses from

the star. However, to catch the star in different

parts of the cycle would require exposures of

about 10 3 s, much too short to see anything. An

interesting technique was used to get around this

problem. The star was observed for many cycles,

but the image was recorded only during a small

part of each cycle. The part of the cycle was the

same cycle after cycle. For example, we expose the

same 10 3 s every period until we can see a good

image. We then shift our exposure by 10 3 s and

repeat the process. We use the radio signals to

synchronize our observing with the pulsing star.

Radiation

Beam Optical pulses from the Crab pulsar are shown

in Fig. 11.11. The optical pulsations are clearly vis-

Fig 11.10. Role of a magnetic ¬eld in a pulsar emission

mechanism. If the magnetic axis is not aligned with the ible. When the star is “on” it is the brightest star

rotation axis, then the magnetic axis can act like a searchlight in the field. When it is “off ” we cannot see it at

beam. [NRAO/AUI/NSF]

all. If we just take a normal photograph of the

11 THE DEATH OF HIGH MASS STARS 203

Fig 11.11. The Crab Pulsar (NP

0532) in visible light. Each frame

shows an image of the ¬eld at

equally spaced points in the cycle.

When the pulsar is on, it is the

brightest object in the ¬eld.When

it is off, we cannot see it.

[NOAO/AURA/NSF]

nebula, we see just the average brightness of the that the fastest ones must be younger. The increase

star. If the duty cycle is 5%, then the average in pulse period means that the star is not rotating as

brightness is about 5% of the peak brightness. fast as it once was. This means that the kinetic

energy of the star is decreasing. This is not surpris-

11.3.3 Period changes ing, since the star is giving off energy in the pulses.

Just as astronomers were becoming used to the idea The energy of a rotating sphere is given by

of pulsars as dependable clocks in the sky, it was dis- 2

E (1/2)I (11.10)

covered that all pulsars are slowing down. Some

The rate of change of the energy is found by dif-

sample data are shown in Fig. 11.12. It was also found

ferentiating both sides with respect to time:

that the fastest pulsars were slowing down with the

greatest rate of change of the period. This suggests dE/dt (I ) (d /dt) (11.11)

We can relate the fractional change in energy,

using equation (11.10):

2

(1/E)(dE/dt) (2/I )(I ) (d /dt)

2(1/ ) (d /dt) (11.12)

The fractional change in energy is therefore equal

to twice the fractional change in the frequency.

We can relate the change in frequency to the

change in period:

2 /P (11.13)

Differentiating, we obtain

( 2 /P2)(dP/dt)

d /dt

Fig 11.12. Period changes for the Crab pulsar.The general

Using equation (11.13) to eliminate one of the

slowdown is clear. Glitches, brief period increases, are indicated

factors of P on the right-hand side gives

by the locations of the arrows. [Michael Kramer/Lyne & Smith,

Pulsar Astronomy, 2nd edn, CUP]

(1/ ) (d /dt) (1/P)(dP/dt) (11.14)

204 PART III STELLAR EVOLUTION

The fractional rate of energy change then The lifetime of the pulsar, t, is just the energy,

becomes divided by the rate at which it is being lost:

t E/(dE/dt)

(1/E)(dE/dt) 2(1/P)(dP/dt) (11.15)

12

1/(2 10 /s)

Thus, when we observe the rate at which the pul-

sar is slowing down, we can directly relate it to 1011 s

5

the energy loss.

104 yr

1.7

Example 11.4 Pulsar energy loss

This means that pulsars have very short lifetimes

Consider a rotating neutron star with a mass M 2

by astronomical standards. (Actually, the loss rate is

M , and a radius R 15 km, a period P 0.1 s, and

not constant, so the above calculation actually

a rate of change of the period dP/dt 3 10 6 s/yr.

gives an upper limit to the lifetime.)

Find (a) the kinetic energy, (b) the rate at which the

kinetic energy is decreasing, and (c) the lifetime of We should note that the rate at which the pul-

the pulsar if it loses energy at this rate. sar in the Crab Nebula is losing energy is equal to

the rate at which the nebula itself is losing

energy via synchrotron radiation. This solves the

SOLUTION

We find the energy from equation (11.10), remem- long-standing problem of the source of energy for

bering that the rotational inertia for a sphere is the nebula. As the pulsar loses energy, somehow

(2/5)MR2. We also use the fact that the frequency is that energy ends up in the nebula. The connection

2 /P: between the pulsar and the nebula is probably the

strong magnetic field. When we see radiation

(1/2)(2/5)MR2 2

E

from the Crab Nebula, it is being indirectly fueled

1033 g)[(1.5 106 cm)(2 /0.1 s)]2

(1/5)(4 by the slowing down of the Crab pulsar.

Pulsars show another variation in their peri-

1048 erg

7

ods. These are sudden decreases in the period,

To find the rate at which is energy is changing, we called glitches (also shown in Fig. 11.12). After each

find the fractional rate at which the period is glitch, the normal slowdown resumes. If the angu-

changing. We first convert dP/dt into more useful lar momentum is conserved, the rotation of the

units: star can only speed up if the size of the star

decreases. We think that the fast rotation and

6

3 10 s yr

dP

extreme conditions at the surface put a tremen-

7

dt 3 10 s yr

dous strain on the solid crust. Periodically, this

strain is relieved by a quake, and the crust then

13

1 10

settles into a more stable configuration. (Actually,

this explanation of glitches may work for the Crab

The fractional rate of change is

pulsar, which has small glitches, but it cannot

13

(1/P)(dP/dt) (1/0.1 s)(1 10 )

explain the larger glitches seen in some other pul-

12

sars. It is thought that these larger glitches may

1 10 /s

involve a transfer of angular momentum from the

This means that each second the period increases

superfluid interior to the crust.)

by 10 12 of a period. We use equation (11.15) to find

We can relate the change in period to the

the rate of change of the energy:

change in radius. The angular momentum is given

by

(1/E)(dE/dt) 2(1/P)(dP/dt)

12

J I

2 10 /s

P2

I12

Multiplying both sides by E gives

12>52 MR2 12 P2

12

dE/dt (2 10 /s)E

14 5 2MR2 P (11.16)

1037 erg/s

1.4

11 THE DEATH OF HIGH MASS STARS 205

Solving for P gives

11.4 Pulsars as probes of

14 5 2 MR J

2

P

interstellar space

Differentiating with respect to R gives

14 5 2 2MR J Let™s look at what happens to pulsar signals as

dP dR

14 5 2 2MR 15>4 2P

they propagate through interstellar space. The

speed of the radiation depends on the index of

21P R2 (11.17) refraction n, which is a function of wavelength .

The speed of the radiation is given by

This means that a small change in period, dP, will

c1n 2

be caused by a small change in radius, cn (11.19)

11 2 2 1dP P 2

dR R (11.18) where c is the speed of light in a vacuum (where

n 1). We can write the speed as a function of

The fractional change in radius is one-half the

wavelength

fractional change in period.

c1 2 c n1 2 (11.20)

Example 11.5 Pulsar glitch

For the pulsar in the above example find the Suppose we have two signals at wavelengths

change in radius for a period change of 10 7 s. 1 and 2. Their speeds will be different. Their

times t1 and t2 to travel a distance d are

12

SOLUTION

t1 dn1 c

We find dR by multiplying both sides of equation

22

(11.18) by R to give t2 dn1 c (11.21)

dR (1/2)(dP/P)R The difference between the two times is

1d c 2 3n1 12 22 4

7

106 cm)

(1/2)(1 10 /0.1)(1.5 n1 (11.22)

¢t

0.75 cm This time delay for signals at different wavelengths

is called dispersion (Fig. 11.14). If we know n( ), then

This is not a very large change.

we can measure t and find out the distance d that

On top of slowdown and glitches, there are

the signals have traveled. If d is known, measuring

additional irregular variations, such as those

t tells us about the index of refraction.

shown in Fig. 11.13. Understanding these tran-

For radio waves passing through interstellar

sients is a current problem in pulsar studies.

space, the index of refraction results from the

interaction of the radio waves with electrons, and

Received

Emitted

∆x

∆x = c ∆t

Fig 11.13. Variations in the Crab period when the effects Fig 11.14. Dispersion. Pulses are emitted at the same time

of the slowdown and glitches have been removed. [Michael at two different wavelengths on the left.They are received at

Kramer/Lyne & Smith, Pulsar Astronomy, 2nd edn, CUP] different times on the right.

206 PART III STELLAR EVOLUTION

is proportional to the interstellar electron density degeneracy, they put a limit on the mass that can

ne. That is, be supported by neutron degeneracy pressure.

n1 2

There is some uncertainty about this limit.

1 r ne (11.23)

Neutron degeneracy alone can support about 2

Then the time delay must be given by M . Neutron“neutron interactions raise this to

about 3 M , and this is the number that has been

¢t r d ne (11.24)

regularly used. However, some calculations show

When we observe a pulsar at two wavelengths, that this mass might be as large as 8 M . Whatever

the signals will not arrive at the same time. the limit, if a neutron star is more massive than

However, we know they left the pulsar at the same this limit, it cannot be stable. We know of no

time. We can attribute the time difference to inter- other source of pressure that will halt the col-

stellar dispersion. By measuring the time delay t, lapse. The star will eventually be small enough to

we can derive the quantity dne. If we observe be contained within its Schwarzschild radius. It

objects for which we have another estimate of dis- will become a black hole, as described in Chapter

tance, we can derive the average interstellar elec- 8. Therefore, unless all neutron stars are formed

tron density. This turns out to be 0.03 cm 3 on with masses less than about 3 M , black holes

average. Once we know this average ne, we can appear to be a normal stellar final state. Though

measure t for other pulsars, which gives us d ne. this can happen for individual stars, or stars in

Since we know ne we can estimate d. This gives us binary systems, it is not likely that we can detect

a way of estimating the distances to pulsars that isolated stellar black holes. We have to detect its

we cannot otherwise measure distances to. presence by its gravitational effects on another

star. We will therefore discuss attempts to detect

stellar black holes in the next chapter, where we

11.5 Stellar black holes discuss close binary systems.

Just as special relativistic effects put a limit on

the mass that can be supported by electron

Chapter summary

In this chapter we saw what happens to stars of The left-over core of the star becomes a neu-

higher mass after they leave the main sequence. tron star, supported by neutron degeneracy pres-

The higher mass of the core means a higher tem- sure. Neutron stars are observed as pulsars. We

perature. This means that nuclear reactions can discussed the accidental discovery of pulsars, and

proceed farther (formation of heavier elements) the chain of reasoning to make the connection

in high mass stars than in low mass stars. A shell between pulsars and neutron stars. The condi-

structure develops, with each layer closer to the tions on neutron stars are quite extreme, often

center fusing heavier nuclei than the layer out- involving enormous magnetic fields. These strong

side it. fields are involved in the pulse emission mecha-