5 Stars and Stellar Structure

Home , List of smallest stars, Polytrope, Stellar structure

5.1 Phenomenology of stars

Almost all the lightwe see in the Universe comes from stars, either directly,

or else indirectly from the fact that stars heat the surrounding gas and dust.

Almost the only exceptions to this rule are

non-thermal radiation (synchrotron radiation, for example) from high

energy particles spiraling around in magnetic elds

emission from quasars, thought to b e from the gravitational p otential

energy released as matter spirals, via an accretion disk, into a black

hole.

Wehave already seen that stars can b e classi ed, purely empirically,on

the basis of their sp ectra: O B A F G K M R N S. This is prettymucha

classi cation by surface temp erature, but not entirely. More sophisticated

application of physics to a sp ectrum allows one to determine separately

the temp erature (which lines, and which ionization states are present)

the surface gravity (at a given temp erature, a line will b e more pressure-

broadened if the surface gravity is larger)

chemistry or elemental abundances (relative strength of lines from dif-

ferent elements)

5.1.1 Elemental abundances, p opulations I and I I

The last classi cation ab ove, elemental abundances, reveals things not only

ab out stars in isolation, but also things ab out the history of the Universe, i.e. 113

cosmology.To refresh your memory ab out the elements, here is the Perio dic

Table.

The numb ers shown on the p erio dic table are the atomic numb ers or num-

b ers of protons, Z , in the nucleus. This of course determines the number

of electrons and hence the chemistry. For nuclear pro cesses, however, we

also care ab out the total number of nucleons (protons plus neutrons), A,in

the nucleus. For all the elements from He (not H!) through S (sulfur), the

relationship is

A 2Z

within one AMU. That is, there are close to equal numb ers of protons and

neutrons. Heavier elements get slightly richer in neutrons so A b ecomes

somewhat larger than 2Z (but not much). Iron, for example has Z =26;A

56.

Studying their sp ectra, it is found that there are two distinct classes of

stars, called Population I and Population I I (usually read as \Pop One"

and \Pop Two"). Pop I stars are young stars, like the Sun, created by 114

ongoing star-formation pro cesses within our Galaxy (or in other galaxies).

Pop I I stars are old stars and are thought to fossilize the initial epoch of star

formation after the Big Bang.

When this classi cation was rst devised, the connection with age was

not immediatelyobvious, so the numb ering scheme (\I" and \I I") is p erhaps

illogical. A go o d way to rememb er it is to recall that a I I-year-old child is

older than a I-year-old child.

The observational distinction b etween Pop I and Pop I I is that Pop I stars

(e.g., the Sun) are relatively rich in \heavier" elements. For astronomers,

\heavier" means anything b eyond H and He in the p erio dic table, that is,

Li, Be, B, C, N, O, etc. Chemists would consider these to b e light elements!

Another astronomer's habit of whichyou should b e aware is that the \heavy"

(though not really heavy!) elements are often referred to lo osely as \metals",

even though they're not. So, to astronomers, the whole p erio dic table is

reduced to H, He, and \metals".

We know quite a lot ab out the elemental abundance of Pop I stars, b e-

cause wehave one right in our neighb orho o d (the Sun), and also b ecause,

except for H and He which escap ed into space, the Earth itself has in most

resp ects the exact elemental abundances of a typical Pop I star. Here is

a table, and also graph (note logarithmic scale!) of these \Solar System

abundances". 115

Pop I Abundances

Mass

Atomic Number (main isotop e) Relative Number Mass Fraction

H 1 1 1 0.77

He 2 4 7 10 0.21

4 3

C 6 12 4 10 4 10

5 3

N 7 14 9 10 1 10

4 3

O 8 16 7 10 9 10

\metals"

4 3

Ne 10 20 1 10 1 10

total 0.02.

5 4

Mg 12 24 4 10 8 10

5 4

Si 14 28 4 10 8 10

;

5 3

Fe 26 56 3 10 1 10

H 1 He 10-1 10-2 10-3 -4 C O Ne Mg Si S Fe 10 N -5 Al Ca Ni 10 Na Ar -6 Cr 10 P Cl K Ti Mn -7 Co Zn 10 F V Cu -8 10 Sc -9 10 B 10-10 Li Be 10-11 -12 10 1234567 8 9 101112131415161718192021222324252627282930

Atomic Number

Hydrogen and Helium are primordial to the big bang (in fact, the Helium

is pro duced during the rst few minutes of the big bang). All the elements

from Carb on on are pro duced solely in stars. The combined mass fraction

of \metals" (Carb on and heavier) is usually denoted Z ,soPop I stars have

typically Z 0:02. (Don't confuse this use of the letter Z with its other use,

denoting atomic numb er. They are unrelated.)

Note the very low abundances of Li, Be, B. These are hardly pro duced

in the big bang at all, and they are in fact destroyed by stars { they get

\co oked" into heavier elements. Also note that the even elements tend to b e

more abundant(by a factor of order 10) than their o dd neighb ors. This is 116

b ecause they get \built" eciently out of -particles (Helium nucleii) that

contain 2 protons + 2 neutrons.

So, Pop I stars are made of material that has already been processed

through an earlier generation of stars, either Pop I I stars or else older Pop

I stars (whose creation and destruction is a continuous pro cess). In fact,

many astrophysicists study what amounts to the \ecology" of stars, that is,

pro cesses by which they form, evolve, are disrupted, and return their now-

pro cessed material to the interstellar medium for formation in a subsequent

generation of stars.

Since Pop I I stars were (as far as we know) the original generation of

stars, you would exp ect them to havemuchlower metal abundances. Indeed

that is true. Typically they mighthave Z 0:002, ab out 1=10 of the Solar

abundances. Some haveeven smaller Z values.

In our Galaxy,Pop I stars are lo cated in the disk, while Pop I I stars are in

the so-called bulge and halo. Thus, we know that these parts of the Galaxy,

b eing p opulated with primordial stars, are older.

5.1.2 Nuclear reactions

Stars are p owered bynuclear reactions that transmute (we sometimes lo osely

say \burn") lighter elements into heavier ones. Main-sequence stars are all

nucleii (protons) into one Helium nucleus.

The reason that burning H to He pro duces energy is of course the fact

that the He nucleus weighs slightly less than the 4 H's. In atomic mass units

(AMUs): 117

4 M =41:008 ! 1 M =4:0027 :

H He

That is

M 4:032 4:0027

= =0:007 :

4M 4:032

So, 0:7% of the mass of each proton is converted to energy (E = mc ).

In practice, b ecause of the various rules governing nuclear reactions and

their probabilities, the reaction is not simply

1 1 1 1 4

H+ H+ H+ H ! He [not!]

(The leading sup erscripts are used to indicate mass numb er.) For one thing,

the ab ove do es not conservecharge! Instead, the reaction pro ceeds by a series

of 2-b o dy interactions. In lower mass stars (< 1:5M ) the so-called p p

cycle dominates:

1 1 2 +

H+ H ! D + e +

2 1 3

D+ H ! He +

3 3 4 1 1

He + He ! He + H+ H

Higher mass stars (> 1:5M ) \burn" via the CNO cycle:

12 1 13

C+ H ! N+

decays

13 13 +

N ! C+e +

13 1 14

C+ H ! N+

14 1 15

N+ H ! O+

decays

15 15 +

O ! N+e +

15 1 12 4

N+ H ! C+ He : 118

Notice that the Cisacatalyst in this case: it comes into the cycle at the

top and is regenerated at the b ottom. The net reaction involves only the

particles shown ab ove in b oldface, namely

1 4 +

4 H ! He + 2e +2:

(The 's are just energy coming o .)

The key fact ab out nuclear reactions is that they are extremely temp er-

ature sensitive. That is, ab ove a certain threshold, a small increase in tem-

p erature makes a huge increase in reaction rate. This fact makes most stars

thermally very constant: across a wide range of stellar masses, the central

temp eratures of stars (where the nuclear burning takes place) are b etween

7 7

1 10 K and 2 10 K. That is, this small temp erature range translates

into the entire range of luminosity needed to \hold up" against gravity b oth

massive stars (high luminosity) and light stars (low luminosity). A good

empirical approximation for the central temp erature of main-sequence stars

1=3

T 1:5 10 K :

5.2 Stellar structure

5.2.1 Order-of-magnitude stellar structure

The virial theorem, whichwe previously derived for a collection of gravitating

b o dies, applies equally well to any combination of inverse-square-law forces

between b o dies. Since the nucleii and electrons in a star interact almost

exclusively by Coulomb electromagnetic forces (and gravity), the virial the-

orem applies to them. Thus, in a star, the total kinetic energy of particles is 119

exactly half the gravitational p otential energy. (There is no electromagnetic

p otential energy to sp eak of, b ecause of bulk charge neutrality).

We can estimate these two terms as follows:

Gdm dm GM

1 2

P:E: =

r R

3 M

kT : K:E: = N kT

particles

2 m

Here M is the star's mass, R its characteristic size (radius, say), T its char-

acteristic (central T ,say) temp erature. Twiddles mean equality in order of

magnitude, i.e. ignoring numerical constants like2 or .

Equating (at twiddle accuracy) P.E. and K.E., and using the empirical law

for T given previously (whichwas motivated by the temp erature sensitivity

of nuclear reactions), we get a mass-radius relation for stars,

2=3

GM m GM m M

p p

kT k (15 10 K) M

(6:67 10 ) M

cm =

23 16 6

(6 10 )(1:38 10 )(15 10 ) M

2=3

=1:010 cm :

The actual solar radius is 7 10 , so our twiddle calculation is actually

pretty go o d (the neglected numerical factors cancel to near-unity)! Howdo

we do for other stars on the main sequence? Here is some actual data: 120

Physical Prop erties of Main-Sequence Stars

log (M=M ) Sp ectral log (L=L ) M M log (R=R )

bol V

class

1:0 M6 2:9 12.1 15.5 0:9

0:8 M5 2:5 10.9 13.9 0:7

0:6 M4 2:0 9.7 12.2 0:5

0:4 M2 1:5 8.4 10.2 0:3

0:2 K5 0:8 6.6 7.5 0:14

0.0 G2 0.0 4.7 4.8 0.00

0.2 F0 0.8 2.7 2.7 0.10

0.4 A2 1.6 0.7 1.1 0.32

0.6 B8 2.3 1:1 0:2 0.49

0.8 B5 3.0 2:9 1:1 0.58

1.0 B3 3.7 4:6 2:2 0.72

1.2 B0 4.4 6:3 3:4 0.86

1.4 O8 4.9 7:6 4:6 1.00

1.6 O5 5.4 8:9 5:6 1.15

1.8 O4 6.0 10:2 6:3 1:3

Here is the comparison with our twiddle mo del:

log (R/R )

1 our prediction

"actual data" 0.5

−1 −0.5 0.51 1.5 log (M/M )

−0.5

Mass-Radius Relation for Stars

What ab out the luminosity of stars? Can we predict that with twiddle

calculations? Yes, but we need to know something ab out the opacity of stellar

material, that is, howmuch resistance it gives to the outward di usion of 121

photons. I will write down the calculation, even though it is b eyond the scop e

of this course. You can study it for extra credit (or intellectual curiosity).

" # " # !

L c d(aT )

Di usion Energy Density

= Flux = = :

Co ecient Gradient

4r 3 dr

Here is the opacity, which comes from atomic physics. Let us assume that

this is a constant (as it very nearly is for highly ionized matter). Then,

neglecting constants, wehave

M=R

T M=R (virial theorem) :

5 4 4 2

R T (M=R) R

M : L /

R M R

Notice that the R's cancel, so we never had to use the mass-radius relation,

or the empirical formula for central temp erature, but only the virial theorem.

How go o d is this? Since we do not at this stage knowanumerical value for

,we cannot check the constant, but only the scaling from the solar value:

log(L/L ) 6

prediction 4 data normalized here 2 prediction

−1 −0.5 0.51 1.5 log(M/M )

−2

Mass-Luminosity Relation for Stars 122

Pretty good! So it lo oks like stars really do ob ey the laws of physics.

This motivates us to do a more careful job of writing down their governing

\equations of stellar structure," whichwe will do next.

5.2.2 Quantities describing the stellar interior

What is a star? A self-gravitating gaseous system. Why call it a gas? It is so

hot that all the original material inside has long since ionized to nuclei and

electrons | a plasma. But for our purp oses, it will b e sucient to describ e

this as a fully ionized gas of electrons plus ions (plus photons | in sucha

gas, \radiation pressure" may b e imp ortant), whichisentirely neutral |

there are no signi cantinternal dynamics of the plasma.

The simplest description of stellar structure is that the stars are spherical

and static (no rotation, magnetic eld, no pulsation, oscillation,:::), and

we will only deal with such ob jects here. In other words, stars involve the

interplay of gravitation, gas dynamics, and radiation.

The basic idea is to write down quantities that describ e the stellar interior

as a function of radius r , then to write down relations b etween them, either

algebraic, or else di erential equations, until we get a closed set of equations

(as many equations as unknowns). Then we can think ab out solving them.

Here is a list of such quantities:

(r ) density: At the center of the star this will have some value and

decrease to zero at the surface of the star.

M (r ) mass: This is the mass interior to a radius r . It comes into the

calculation of the gravitational force as a function of r . It is zero at

r = 0, and equal to the star's total mass M at the stellar surface.

0 123

P (r ) pressure: The pressure at any r is of course the total weight(per

square centimeter) of all the overyling mass.

T (r ) temp erature: Gas and radiation are in lo cal thermo dynamic equi-

librium.

L(r ) luminosity: This is the net outward total energy ow at each radius.

It grows from zero to the star's total luminosityaswe pass from

r = 0 through the star's energy generating region, and thereafter

(as wemove outward) is constant at the star's total luminosity.

There are also various quantities describing the microscopic lo cal prop er-

ties of the gas:

(r ) mean particle mass (also called mean molecular weight): This comes

into the p erfect gas law P =(=)kT . It is the mean counting

b oth electrons and nucleons as particles (since b oth contribute to

pressure).

(r ) opacity or \mass absorption co ecient": This has units of cm = g

and is the total cross sectional area for absorbing or scattering pho-

tons p er gram of material. It controls the rate at which photons

di use outward to transp ort the luminosity.

(r ) The nuclear generation rate, in ergs p er cm p er sec.

Now, if you are lazy, here is the go o d news: In this course we are going

to use various tricks or approximations that close the set of equations with

only the rst three variables ab ove: ; M ; P . This \closure" is not exactly 124

correct for all cases, but it will allow us to learn some interesting things ab out

some real stars, and will let us defer the whole sub ject of \radiative transfer"

(involving T ; L; ; K , and E ) to later courses, Astronomy 145 and 150.

5.2.3 Equations of stellar structure

Equation of Hydrostatic Equilibrium

Consider an element of gas in equilibrium in the star

density ρ

r + dr

r g

The pressure P (r ) is larger than the pressure P (r + dr )by just the weight

p er unit area of the material b etween r and r + dr in the lo cal gravitational

acceleration g . If the element is of area dA,wehave

mass of

= dAdr

element

! "

GM (r )

weightof

= g dA dr = (r ) dA dr

element

G(r )M (r )

weight of element

= dr :

p er unit area

So,

dP P (r + dr ) P (r ) G(r )M (r )

= = :

dr dr r

The minus sign is b ecause pressure increases as r gets smaller (downward

direction).

Mass Equation 125

Mass (interior to radius r ) is just the integral of the density in spherical

co ordinates:

0 02 0

M (r )= (r)4r dr :

We usually prefer to write this as a di erential equation. Taking the deriva-

tive with resp ect to the upp er limit of integration gives trivially

=4r :

Equation of State

If we could just nd an algebraic relation b etween pressure and density

P = P ()

wewould b e done: 3 equations for the 3 unknowns P ; ; M as a function

of the indep endentvariable r . In real life, however, pressure dep ends not

only on density but also on temperature and composition.For a mixture of a

p erfect gas and radiation, wehave

kT + P =P + P = aT

gas radiation

(rememb er our calculation of radiation pressure in 3.3.4, and of radiation

energy density in 3.5.4?). Luckily, it is often true that P P ,so

radiation gas

that the second term can b e neglected, and that we can either (i) derive from

other physics, or (ii) make a go o d guess, ab out how T in the rst term varies

with .Further, in many cases of interest, is constant. Then, we will have

arrived at a so-called barytropic equation of state P = P (). Let us now

make that assumption. Later, we will catalog the actual cases for whichit

o ccurs. 126

5.2.4 Polytrop es: The Lane-Emden equation

A polytropic equation of state is a sp ecial case of a barytropic equation of

state where the relation b etween P and is a pure p ower law

P = K :

Here n (which need not b e an integer) is the so-called p olytropic index. (The

weird notation \1 + " can b e thought of as arising from the p erfect gas law

P / T

along with an assumed p ower law relating T and ,

1=n n

T / or / T :

But that is just notational history.)

It turns out that main sequence stars are prettywell mo deled by n =3

p olytrop es. That is, the run of temp erature and density in the star roughly

follows

1=3

T / :

So the value n = 3 is a go o d one to keep in mind as we pro ceed, although

we will meet other values later.

Hydrostatic equilibrium:

dP GM

dr r

Mass:

=4r

Thus, eliminating M (r ),

" #

r dP d 1

= 4G

r dr dr 127

P = K

Change to dimensionless units:

= dimensionless length, , where a = some scale distance

= dimensionless density function (actually,very temperature-like)

= , where = central density, and (0)=1

c c

n+1

P =K

Substitute for P :

" #

2 2

1 1 1 d a d

n 1+n

K = 4G

2 2 n

a a d a d

" #

1 d d

n n

K = 4G (1 + n)

2 2 n

a d d

0 1

1 d 4Ga d

2 n

@ A

d d

(1 + n) K

Wenow can see that an \inspired" choice for the length scale a would b e

(1 + n) K

a =

and we obtain the canonical form of the Lane{Emden equation:

d d 1

2 n

= :

d d

It is a common pro cedure in physics to convert equations to \mathematical"

form like this! 128

5.2.5 Boundary conditions and Lane{Emden functions

The ab ove Lane{Emden equation can only b e integrated numerically. Con-

ceptually,we rst rewrite the equation as the equivalent form

" #

00 0 n

= +()

where prime denotes d=d . Then starting at = 0 with known boundary

conditions on (0) and (0), we start \stepping" in .For each (theoretically

in nitesimal) step, we up date by its Taylor series

( + d )=()+()d

and also up date by its Taylor series

0 0 00

( + d ) = ( )+ ( ) d

0 0 n

= ( ) + d :

In practice there are b etter numerical metho ds than this, but they reduce to

this one conceptually.

What are the b oundary values (0) and (0)? First, (0)=1by de -

nition of (see ab ove). Second, we know that dP =dr =0atr = 0, since

GM =r (lo cal acceleration of gravity) go es to zero there. Thus from

1+ n+1

P / /

we get

d d dP 1

n n 0

0= / 1+ / (n +1) / (n +1) :

dr n dr dr

n 0

Since (0) is nite (actually = 1), (0) must vanish. 129

With these b oundary conditions, you might enjoy deriving the rst terms

in the p ower series expansion for ( ), namely

1 n

2 4

=1 + :

6 120

Numerical integration of the kind just describ ed can give these so-called Lane-

Emden functions to any desired accuracy. Here is a graph for n =0;1:5;3:0,

and 3:5.

.8 φ

n = n = n = .2 3.5 0

Lane-Emden function 1.5 n = 3 0 02 46810

scaled radius ξ

5.2.6 Physical prop erties of p olytrop es

For n<5, the Lane-Emden function go es to zero at a nite value of (and

therefore r ) which is called . This is the surface of the star! The stellar

radius in physical units is therefore

(1 + n) t K

R = a = :

1 1

4G 130

For the mass M (r ), wehave

Z Z

=r=a r

2 3 2 n

M(r)= 4r dr =4a d :

0 0

However (here is a great trick!) if wemultiply the Lane-Emden equation by

and integrate from 0 to we get

2 2 n

= d :

So we can immediately read o

3 2

M (r ) = 4a

=r=a

" # !

3=2

d 1 (n +1)K

: =

G d

=r=a

The total mass is obtained by setting = (stellar surface). To get phys-

ical masses and radii wethus need tabulated numerical values for and

(d=d ) (you could in principle read these o the ab ove graph of the

1 1

Lane-Emden functions)

n =

1 c

0 2:4494 4:8988 1:0000

0:5 2:7528 3:7871 1:8361

1:0 3:14159 3.14159 3:28987

1:5 3:65375 2.71406 5:99071

2:0 4:35287 2.41105 11:40254

2:5 5:35528 2.18720 23:40646

3:0 6:89685 2.01824 54:1825

3:5 9:53581 1.89056 152:884

4:0 14.97155 1.79723 622:408

4:5 31.83646 1.73780 6189:47 131

You mightwonder what happ ens if you integrate the Lane-Emden equation

beyond its rst ro ot ? Don't even think ab out it! If is negative then the

density is negative, which is completely unphysical. The equation is p erfectly

\happy" to b e truncated at = , since = 0 implies P = 0, which is the

correct surface b oundary condition.

Recapping, the p olytrop e stellar mo del gives 2 algebraic relations among

the 4 quantities K; ;M, and R.Nowadays we might view M and K as the

indep endentvariables (K , which puts a scale on how hot the star is, deriving

from nuclear theory) and use the mo del to determine and R. Historically,

b efore nuclear pro cesses were understo o d, p eople used measured values for

M and R and then derived and K , whichwere otherwise unknown. From

either viewp oint, once these 4 quantities are known, we can go on to calculate

how all quantities of interest vary with radius, for example by using the

graphs of the Lane-Emden functions given ab ove. Also, one can easily derive

relations like

3 d

mean density of p olytrop e =

center 1

1 GM

P = central pressure

center

4 2

4 (n + 1)( R )

m P

center p center

T = central temp erature

center

(using the p erfect gas law)

Example: n = 3 p olytrop e: M =1 M ; R =1 R (use these measured val-

ues to get and K ).

c 132

2 0 0

From the numerical solution, =6:90 , =2:02 , and = 0:0424 .

1 1 1

Thus,

3 31

a = =7:910 g ;

2 0

1 1

and the scale length

a =1:01 10 cm :

Therefore, if the Sun can b e represented as a n = 3 p olytrop e, it has

central density ; =76:7g cm

mean density= 0:0184 =1:41gcm

central pressure;P =1:25 10 dyne cm

central temp erature;T =1:97 10 K

14 4=3

We also get the constant K =3:85 10 cgs (in P = K ).

5.3 Sp eci c cases of p olytropic mo dels

5.3.1 Adiabatic indices for a p erfect gas

If you compress a gas element whose thermal conductivity is small, so that

there is no heat conducted into or out of the element, then its pressure

is said to increase adiabatical ly. In this case we can use the 1 Lawof

Thermo dynamics to get a relation b etween P and for the gas element.

Consider a volume V containing N particles. Then the total energy E is

given by the rule \1=2kT p er degree of freedom," namely

E = NkT

where is the numb er of degrees of freedom. For a monatomic, fully ionized,

gas (the usual case in stars) wehave = 3 corresp onding to x; y , and z

translational motions. 133

The p erfect gas lawinvolves only the numb er density of particles N=V ,

not , and is

P = kT

which gives a relation b etween P and E for p erfect gasses,

E = PV :

Now the 1 Law of Thermo dynamics, which is really just conservation of

energy,says that when you squeeze a gas from volume V to volume V dV

(smaller volume), the increase in its internal energy is just the work you have

done squeezing it:

dE = PdV :

(The minus sign is b ecause the internal energy increases with decreasing

volume.) Combining the last two equations (taking a di erential of the rst):

(PdV + VdP) = PdV

VdP = 1+ PdV

2 2

dP 2+ dV

= :

P V

Integrating, we get

( )

P = constant V :

But since the density of a xed quantity of gas (N particles) varies inversely

with its volume, this is just

)

(

P /

which is p olytropic with index n = =2. The most common case, =3,

5=3

gives n =1:5;P / . Notice that as the numb er of degrees of freedom 134

increases, the p olytropic index increases. In fact, the limiting case of an

isothermal gas (P / T , T constant, so P / ) corresp onds to !1.

This is b ecause one can view the work compressing the gas as b eing spread

over an in nite number of \internal" degrees of freedom, resulting in no

increase in temp erature.

5.3.2 Fully convective stars

Convection is the buoyancy-driven pro cess of dynamical circulation that car-

ries heat upward in gas or liquid in a gravitational eld. You see it when

you heat a p ot of water on the stove, or when the Sun heats the ground (and

hence the nearby air) on the Earth. In essence convection is nothing more

than \hot air rises and co ol air sinks."

In general, convection transp orts heat much faster than conduction do es.

Thus, a uid element in the convective owisvery nearly adiabatically com-

pressed (as it sinks) or decompressed (as it rises). Since convection is also

generally turbulent, the mixing of di erent uid elements is ecient. Thus, a

gas in turbulent convection is quite accurately all on the same adiabat. That

is, if it is monatomic and fully ionized ( =3,above), it satis es

5=3

P = constant

where all uid elements have the same constant. This is just what is needed

for the validity of a p olytropic mo del with n =3=2.

The Sun is not fully convective; most of it is stably strati ed with the

deep er, denser material b eing on a \lower adiabat" (smaller value of constant

in ab ove equation). That is why the run of density and pressure in the Sun

4=3 5=3

is b etter describ ed by P / (p olytropic index n = 3) than by P / 135

(p olytropic index n =1:5), even though the material in the Sun is monatomic

and fully ionized. The Sun has an outer convectiveenvelop e only in the last

1/6 or so of its radius.

Low mass stars, 0:3M , are almost completely convective, so they are

good n =1:5 p olytrop es. Also, stars of all masses go through an initial

convective phase (called the Hayashi phase) b efore they settle down to the

main sequence. This phase can typically last several million years. It, to o,

is well describ ed by the Lane-Emden equations for an n =1:5 p olytrop e.

5.3.3 Equation of state for degenerate matter

Degenerate matter is matter whose resistance to compression comes not from

thermal, the kinetic energy of its particles, but rather from the fact that its

electrons (or, more generally, fermions) ob ey the Pauli exclusion principle

and can't o ccupy the same quantum state. Thus, degenerate matter resists

compression even at zero temp erature.

Normal terrestrial solids and liquids (ro cks, water, etc.) are, roughly

sp eaking, degenerate in this sense, although the fact that their electrons are

b ound into atoms is an additional complication. In astrophysics, cold (or

nearly cold) dead stars known as white dwarf stars are comp osed of degen-

erate matter (nucleii plus electrons). Neutron stars, comp osed of virtually

100% neutrons, are also degenerate. One says that these ob jects are \sup-

p orted by degeneracy pressure."

Since temp erature do esn't enter into it, degenerate matter is barytropic,

with pressure a function of density only, P = P (). Let's calculate this

equation of state in detail.

In Section 5.3.1, when we applied the 1 Law of Thermo dynamics, the 136

temp erature T entered only as an intermediate step in getting a relation

between P and E=V (energy p er volume), namely

2 E

P = (p erfect gas).

3 V

Supp ose we had gotten some di erent constant

P = (generalization).

Then, using this, plus the 1 Law:

(PdV + VdP) PdV = dE =

1 1 dV dP

1+ = :

0 0

V P

Integrating gives

0 0

( +1) +1

P = constant V / :

So we see that we get a p olytropic equation of state

P /

with

= 1= :

In other words, a p olytropic gas P / has, in general,

PV =( 1)E:

Now back to degenerate electrons. In our discussion of quantum phase

space density in Section 3.5.2, we already saw that fermions, b ecause of

the Pauli exclusion principle, have a maximum phase space densityof1per

quantum unit h of phase space. Actually it's 1 for each so-called spin state, 137

just as, when for photons, we counted each p olarization separately. Electrons

1 1

havetwo spin states, called + and .

2 2

Recall that phase space volume is the pro duct of actual space volume

3 3

and momentum space volume d xd p. Because of the limitation on phase

space density,ifwe force some electrons into a smaller volume d x, they must

push out into a larger momentum volume d p. The lowest energy state of an

electron gas (the state it will take on in practice at zero or low temp erature) is

when the electrons ll all momentum states in a sphere out to some radius p

in momentum space (called the Fermi momentum) and none b eyond. Thus,

for N electrons in a volume V wehave the phase space density

N 2

N = : =

V p

Since N=V is n , the numb er density of electrons, wehave

3 3

p = h n :

Now to get the energy p er volume, E=V ,weintegrate over the density distri-

bution in momentum space, putting in E = p =(2m ) as the energy of each

e e

electron. (Note that this is valid for non-relativistic electrons)

Z Z

2 5

E 4 p 2 p

2 3

4p dp = = E N d : p =

3 3

V 2m h 5 h m

e e

Now what ab out the relation b etween pressure P and energy density

E=V ? Actually, degeneracy never comes into this. It is just the same cal-

culation we did in Section 3.3.4, but for a nonrelativistic particle moving at

velo city V . Go back and lo ok at the equations around these gures 138 θ A θ

and you should then understand the equations,

" # " #

p 2 E 1 2p 1 2p 1

= P = = =

3 A(2L=v ) 3 A(2L=[p=m ]) 3 ALm 3 V

e e e

2 E

P = :

3 V

0 5=3

So from the previous discussion of we know rightaway that P / .

Another waytoverify this and also get the constant in the relationship,

is by substituting the ab ove expression for E=V in terms of p , and also

substituting the earlier expression for p in terms of n . The result is the

amazing universal formula

2=3

1 3 h

5=3

P = n (non-relativistic electrons).

20 m

Something di erent happ ens if the electron gas is so compressed that the

Fermi momentum p starts approaching m c (i.e. the electron velo city starts

F e

approaching c). Then, the ab ove assumptions E = p =(2m ) and v = p=m

e e e e

start breaking down! In the extreme relativistic limit, the ab ove calculation

of P go es over to b eing exactly the same as we previously did for photons,

E 1 4 1

0 4=3

) = ) = ) P / : P =

3 V 3 3 139

The calculation of E=V now uses E = pc, and is

Z Z

E 2c 2

3 2 4

= E N d p = 4p dp = p : pc

3 3

V h h

So the same two substitutions as b efore give

1=3

1 3

4=3

P = hc n (relativistic electrons).

The crossover b etween these two limits can b e roughly taken to b e where

they are equal,

1=3 2=3

3 1 3 h 1

4=3 5=3

hc n = n

e e

8 20 m

h 4 2

: ) n =

3 5 m c

The length h=m c =2:42 10 cm is called the Compton wavelength of

the electron. An electron forced into a b ox smaller than ab out 2=5 Compton

wavelengths on a side, we see, b ecomes relativistic. This is of course a conse-

quence of the Heisenb erg uncertainty principle relating p ositional lo calization

to momentum de-lo calization.

A nal detail for our astrophysical applications is to relate n , the electron

density,to , the total mass density. Except in the case of hydrogen (which

is not present in white dwarf stars anyway) each electron is accompanied by

exactly one proton and byvery nearly one neutron. Thus

= n (m + m + m ) 2n m :

e e p n e p

We can then substitute n = =(2m ) in the ab ove equations for P , getting

e p

the actual equation state P = P (). More precisely we might de ne as

the mean molecular weight p er electron, and use m n , with =1

e p e e 140

for hydrogen, 2 for everything else. If a gas contains a mass fraction X of

hydrogen,

X =

then you can readily work out that

1+X

Elsewhere, you might see (not ), which is de ned as the mean molec-

ular weight per particle (not p er electron). If Y = is the helium mass

fraction, and Z =(1XY) is the mass fraction of everything heavier than

He, then a go o d approximation is

1+3X + Y

Here is a handy table of values for mixtures of H and He:

Pure H gas: =1; =

Hydrogenless gas: =2; =

1+ Y

\Cosmic gas": X =0:74 ; Y =0:26 ) =1:15 ; =0:60

5.3.4 White dwarf stars

White dwarfs are stars supp orted entirely by electron degeneracy pressure,

so we can immediately apply the results of the previous section:

# "

2=3

3 h 1

5=3

(nonrelativistic) P =

5=3 5=3

m m

p e

" #

1=3

hc 1 3

4=3

P = (extreme relativistic).

4=3 4=3

p e 141

Write these as

P = K ;

e e

with n = for the nonrelativistic case and n = 3 for the very relativistic case.

Then, from p olytrop e theory (Section 5.2.6), a white dwarf radius and mass

are

" #

1=2

(n +1)K

(1n)=(2n)

R =

# "

3=2

(n +1)K

(3n)=(2n) 2 0

M = : 4

c 1 1

Nonrelativistic White Dwarf

Take n = ; then

= 3:654

2 0

= 2:714

1 1

1=6

5=6

4 6 3

(=2) R = 1:122 10 km =10 gcm

e c

1=2

5=2

6 3

( =2) : M = (0:4964 M ) =10 gcm

e c

Eliminate the unknown central density :

M =(0:7011 M ) R=10 km ( =2) :

3 7 3

So M / R , with 10 gcm when M 1:5 M , and the more mas-

sive a white dwarf is, the smaller it is!

Relativistic White Dwarf

Take n = 3; then

= 6:897

1 142

2 0

= 2:018

1 1

1=3

2=3

4 6 3

R = 3:347 10 km =10 gcm (=2)

c e

M = (1:457 M )(2= ) :

Isn't it p eculiar that M is indep endentof ! It has a xed value, 1:457M

(if = 2), that is called the \Chandrasekhar mass," M . What do es

e Ch

this mean? Recall that the extreme relativistic case is a limiting case as

!1. Thus, the meaning is that as !1we get driven to the

c c

relativistic case with M ! M . That is the maximum mass that electron

degeneracy pressure can p ossibly supp ort. For higher masses there are simply

no solutions!

Another way of understanding this is to lo ok at the mass-radius diagram,

plotting b oth the p olytrop es n =3=2 and n = 3, and also a more complicated

mo del that go es smo othly from the non-relativistic to extreme relativistic

limits: 143 from Schatzman (1958) -1.5

Van Maanen 2 o2 Eri B AC 70 8247 n = 3/2 -2.0 polytrope Sirius B W 219 . Ross 627 10

log (R/R ) -2.5 L 930-80

n = 3 polytrope "Chandrasekhar Mass" -3.0

-3.5 -0.6 -0.4 -0.2 0 0.2 0.4 0.6

log 10 (M/M . )

So, over this range the radius for a given mass decreases; and at R =0

(M =M ) the relativistic electrons can no longer supp ort the star. The

p oints plotted on the curve are observational measurements of actual white

dwarf stars, and demonstrate that our theory is basically correct, even accu-

rate quantitatively!

Note that = 2 is appropriate b ecause X ' 0: the star must have

burned all its hydrogen en route (used up all nuclear fuel).

In terms of fundamental constants, we can write

3=2

hc 1

M =3:10 ;

2 2

G m

p e 144

where

1=2

m :

Planck

This so-called Planck mass has the numerical value 0.22 g. So M de-

p ends only on m and m ,even though it is electrons that supp ort the

Planck p

star. Nowhere do es m enter the formula. In the relativistic limit, the rest

mass does not enter the E=(PV ) relation. Thus, m do es not enter into the

calculation of electron pressure.

5.3.5 Stellar structure virial theorem

Wehave seen that as ! 4=3 from ab ove (e.g. from 5=3 ! 4=3 for relativis-

tic electrons), something \go es wrong" with the stellar structure equations.

We might ask as a p oint of principle whywe can't build a star out of ma-

terial with an even smal ler .For example, a molecule a likeNH has not

only the 6 obvious degrees of freedom (3 translational and 3 rotational) but

also 9 vibrational mo des. So a star made of convecting ammonia would b e a

p olytrop e with

=1+ < 4=3 :

(This is of course fanciful, since stars are always so hot that molecules are

destroyed. But there is a p oint of principle to understand here.)

We can use thermo dynamic results and the pressure equation to deduce

a virial theorem relating to stellar structure. Start from the equation of

hydrostatic equilibrium

dP GM (r ) (r )

dr r

r , so that and de ne V (r )= volume o ccupied by gas inside radius r =

dV =volume of dr -shell, containing mass (r )4r dr . 145

Multiply the pressure equation by V (r ) dr :

dP GM (r ) (r )

V (r ) dr = V (r ) dr ;

dr r

i.e.,

4 1 1 1

V (r ) dP = GM (r ) (r ) r dr = GM (r ) dM (r ) :

3 r 3 r

Integrate over the star

r =R r =R

Z Z

1 GM (r ) dM (r ) 1

V (r ) dP = = U ;

3 r 3

r =0 r =0

where U is the total gravitational p otential energy. Hence, integrating by

parts,

r =R

r=R

U =[PV ] P (r ) dV :

r=0

r =0

At r =0, V =0.At r =R, P '0. (The surface of a star is approximately

avacuum.) Therefore,

r=R

[PV ] =0 :

r=0

Hence,

r=R

U +3 PdV =0:

r=0

This is the most general form of the stellar structure virial theorem.

For an ideal gas, the energy p er volume u is

: u =

(We formerly wrote this as u = E=V when it was assumed constantover the

volume V .) Thus,

Z Z

PdV =( 1) udV =( 1) E; 146

where E is the total internal energy of gas. Therefore,

U +3( 1)E =0:

Note that for a p erfect, monatomic gas, = , and the total internal

energy/unit mass is just the total kinetic energy of the gas particles, i.e.,

E T .SoU+3( 1)E = 0 is equivalentto U +2T = 0, the same as self-

gravitating particle virial theorem. (Theorems are consistent { fortunately!)

The total energy of a star is

E = E + U

tot

E = E 3( 1)E = (3 4)E

tot

(3 4)

= + U :

3( 1)

Wenow see that

if > ; E < 0 (therefore; b ound star)

tot

if < ; E > 0 (therefore; unb ound \star" ):

tot

. Oth- We see that stars are stable only if their adiabatic index exceeds

erwise, they are unstable to converting their internal energy into expansion

velo city | they blow themselves apart!

5.4 Beyond the Chandrasekhar mass

What happ ens as M approaches or exceeds M ?Further physical pro cesses

come into play as the central density, , increases.

c 147

5.4.1 Inverse decay

Normally, neutrons decaybyn!p+e + with energy release 1 MeV.

Therefore, when the Fermi energy p er electron b ecomes 1 MeV,an inverse

reaction e + p ! n + can go. As the radius decreases, the density rises and

E rises until E ! 1 MeV. Then, the electrons disapp ear by combining

F F

with protons to pro duce neutrons. This means that the electron pressure

drops, and the electrons b ecome less able to supp ort the star. The star

collapses further and faster until all electrons combine with protons. Thus,

the star b ecomes a mass of neutrons, a neutron star.

The collapse maygoeven further { there is a limiting mass for a neutron

star, to o. (We can calculate that in the same way as for electrons.)

5.4.2 Neutron stars and pulsars

Neutron stars are exactly like white dwarf stars in the theory of their struc-

ture, but with supp ort from degenerate neutrons instead of electrons. Sup-

p ose these neutrons form a gas like the electrons (actually, they are thought

to \solidify" or \liquify" in some parts of the star).

In the case of relativistic neutrons, the neutron degeneracy pressure is

" #

1=3

1 3 hc

4=3

P = ;

4=3 4=3

p n

where is the mean molecular weight p er neutron in a.m.u. ' 1 (since there

is ab out 1 m p er neutron). So we get the same n = 3 p olytrop e solution as

electrons with

5:83 M ; M =(1:457 M )

and the limiting mass of a neutron star is 5:83 M . The only change in

the argument from the white dwarf case is to replace by .

e n 148

Actually, this estimate is rather inaccurate: (1) b ecause at this density

nuclear forces (strong force) b etween neutrons are appreciable, and this aids

gravity,thus decreasing the limiting mass; and (2) b ecause the Newtonian

gravitational p otential =c 1, so Newtonian gravity is a p o or approx-

surface

imation, and we should use General Relativity. The physics overall is rather

uncertain, but the limiting mass of a neutron star is probably 3 M . Un-

til we know the equation of state of a neutron uid, this mass will remain

uncertain.

What is the radius of a neutron star? Neutron stars are denser than

white dwarf stars. In the limit, the central density should corresp ond to the

neutrons almost touching. The neutron{neutron separation is then 1fm

13 39 3 15 3

10 cm, so the neutron numb er density 10 cm , and 2 10 gcm .

Hence, if M 1 M , R 10 km.

We know that neutron stars exist b ecause of evidence of pulsars. (We

know that white dwarfs exist, to o, b ecause we can see them optically; e.g.,

Sirius B.) Pulsars are distinguished by the regular arrival of radio pulses

separated byintervals of a few ms (0.5 ms = shortest) to a few seconds

( 5 sec = longest).

0 2468

Time (s)

Chart record of individual pulses from one of the rst pulsars discovered, PSF 0329 + 54. They were

recorded at a frequency of 410 MHz and with an instrumental time constant of 20 ms. The pulses o ccur

at regular intervals of ab out 0.714 s. 149

The p erio d is quite regular (but slowly decreasing, due to \spindown" of

the neutron star). The ux and pulse shap e are somewhat variable, but the

long-time averages are stable. Some pulsars put out optical and x-ray pulses

as well as radio pulses.

What makes pulsars pulse? The answer is thought to b e that the pulses

are from electrons trapp ed in the magnetic eld of the rotating neutron star.

The idea is that the pulses are thus emitted in a cone rotating with the

neutron star. Supp ose the p erio d is P (= 2= ). Then we can derive a limit

r per to the neutron star radius r .At the equator, the centrifugal force =

unit mass had b etter not exceed the gravitational force = GM =r p er unit mass.

to +

rotation axis

emission cone sweeps across once per rotation period neutron

star charge

2 2

Therefore, r < GM =r for stability. So for a given P ,we require

1=3

2=3

GM P P

r< 1500 km :

4 sec

Crab pulsar: P =33ms r<150 km white dwarf radius

white dwarf radius Pulsar 1937+21: P =2 ms r<24 km

100

These ob jects can't b e white dwarf stars (p erio ds to o short), and they can't

b e planets (or the rate of loss of energy would slow them to o quickly), so we

infer that they are neutron stars. 150

We see neutron star slowdown at a rate that is consistent with energy loss

by electromagnetic radiation due to rotating magnetic dipole. This requires

very strong magnetic elds, 10 Gauss (b elieved to b e present).

5.4.3 Black holes

What happ ens at M>6M (certainly greater than limiting mass of a neu-

tron star)? Then matter collapses to a black hole; the star vanishes inside

its Schwarzschild radius r .

2GM

r =

A classical derivation of r (equate escap e velo city to the sp eed of light)

is p ossible, but wrong: this is a result of General Relativity. Putting in

numerical values gives

r =3 km :

This is not a lot smaller than a neutron star: neutron stars are almost to o

small and dense to supp ort themselves. Nothing escap es from the region

s s

gets out from there either; the strong eld causes infall of gas.

This infalling gas heats up as it falls in and may b e seen by the x-rays it

emits. The direct evidence for black holes is meager: they are hard to see!

We require unseen ob ject of high density and M 4 M (theoretical neutron

star limit) in a binary system. A few such ob jects are known. (The evidence

6 8

for massive black holes, 10 10 M , at the center of quasars is a lot more

certain.) 151