Stellar Evolution in General and in Special Effects:

Core Collapse, C-Deflagration, Dredge-up Episodes

Cesare Chiosi

Department of Astronomy

University of Padova, Italy

PART AGENERALITIES

Elementary Theory of Nuclear reactions

2

p n Nuc

2

B

B

Why nuclear energy generation ? A source able to survive f or

about 10 Gyr.

Basic Principle ΔE = ΔM c with ΔM = [ZM +(A - Z)M - M ]

Binding Energy E =ΔM c

Binding Energy per nucleon E / A

56

B

7 Mev (average)

Maximum E at A =56 : esonergetic reactions up Fe, endoenergetic

reactions above.

2[ ( ) ]

931.48 [ ( ) ] Mev

= 931.48 [2 x 1.00782 + 2 x 1.00866 - 4.0026] = 28.3 Mev

B p n

B p n

Zm A Z m M c

E Zm A Z m M

E

4 H He

The case of p+n D & 4 H He2[ ( ) ]

931.48 [ ( ) ] Mev

= 931.48 [2 x 1.00782 + 2 x 1.00866 - 4.0026] = 28.3 Mev

B p n

B p n

Zm A Z m M c

E Zm A Z m M

E

2

2

( )

'p n

d

E m m c

E m c

Energy at separationEnergy in bound state

p+n D

Binding energy per nucleon

Kinetic vs Binding Energies

K

K B

I n a star the kinetic energies of the themal

motions of nuclei are

1E » KT » 1 - - 100 Kev

2theref ore

1 1E » - - E

1000 100Very low!!!!!

Typical dimensions of the Nuclei

-13 1/ 3

-13 3 3

A nucleus can be considered as a sphere of

radius R » 1.4 10 A cm

and

4πvolume V » (1.4 10 )

3 A cm

Standard notation

• A particle a impinges on a nucleus X, producing a nucleus Y and liberating a particle b

• a + X Y + b or X(a,b)Y• Q-value (liberated energy) Q=[Ma + Mx – My - Mb]c^2

Schematic Potential

How does a reaction happen? Consider two nuclei of charge Za and Zb and

masses Ma and Mb at the ever changing distance r (they are in motion).

The repulsive coulomb potential is

for r > rN (the nuclear radius), whereas it is

strongly actractive for r < rN .

The minimum relative distance between the two nuclei for energy E of the relative motion at infinite distance is

2

0

4

a bZ Z e

r

2

04a b

c

Z Z er

E

Barrier Thickness

The coulombian potential at the border of the nucleus is

the particle sees a very thick potential barrier.

2a b a b

rN0 N N,F

3c rN

N

Z Z e Z ZΦ = » 1.4 Mev

4πε r r

it f ollows that

r Φ Mev= » »10

r E Kev

How many nuclei with E=RN?

-E/ KT -1000

Maxwell - Boltzmann distribution

e e 0.

BUT QUANTUM MECHANI CS !!!!

Barrier crossing probability

1/ 2 -1/ 2G

4 42

G G a b2

The probability that the incident particle may cross the Coulomb Barrier is

E exp[-( ) ] E

E16π e μ

where E is the Gamow energy E = (Z Z ) 2h

a b

a b

7G

M Mwith μ the reduced mass μ = .

M +M

For T =10 K and E =493 Mev

exp(-22)

very small but not zero!!!!

However…… This is only part of the story because we must

evaluate

1. The probability that two nuclei may encounter;2. The probability that once the encounter has

occurred and the barrier is crossed the final result is Y+b;

3. Take into account that nuclei are moving with different relative velocities.

To this aim we introduce two simple but powerful concepts

A) The Bohr model of a nuclear reactionB) The Cross Section

Bohr Model

• Formation of a composite nucleus in an excited state

a + X C* Y+b• The two steps are fully independent; the

second step depends only on the energy and angular momentum of C* and not on the first step a + X (i.e. the energy of the incident particle).

• A reaction is favoured if the first step brings C* in (or close to) a quasi stationary state.

Quasi-Stationary States

Quasi-stationary states

Typical of composite nucleus.

Explains resonances.

Cross Section

• Medium with number density of targets n

• The probability that an incoming particle interacts is PDX=nDx

• The probability of no interaction is

Pno,Dx=1-nDx

. . . . . . . . . . . . . . . . . . . . . . . . . .

Dividing a finite distance x into x/N intervals we get Pno,x=lim[1-nx/N]^N exp(-nx)

Mean Free Path & s

G BP N R

G

BP

N

P P P F

where

P Probability that two particles encounter (geometrical, depends on E)

P Probability of Barrier Penetration (depends on E)

P Probability

R

that the nucleus C* decays into Y+b (intrinsic, does not

depend on E)

F Dimensionless factor that accounts for resonances

The cros section is an effective area proportional to the probability of the reaction occurring in a collision

0

1exp( )l x nx ndx

n

G BP N R

G

BP

N

σ P P P F

where

P Probability that two particles encounter (geometrical, depends on E)

P Probability of Barrier Penetration (depends on E)

P Probabil

R

ity that the nucleus C* decays into Y +b (intrinsic, does not

depend on E)

F Dimensionless f actor that accounts f or resonances

PG: De Broglie Term

Let a be a particle in relative motion with respect to X, with relative velocity

hv, energy E, angular momentum L = l(l +1) , reduced mass μ, and De

2π

h 1Broglie wave length = to

2π 2π 2μE

2

G

which we associate the area

1 1 1 σ =(2l +1)π( ) P (2l +1)

2π E E E

PBP: Barrier Penetration

l l

The penetration probability of particle a into nucleus X is expressed by the

Transmission Factor

κ T P(E)

Kwhere κ is the wave number outside the well (depend

l

1/ 2

l 0

s on E) and K is the wave

number in the well (depends on nucleus and matching of eigenf unctions), P(E)

is the penetration f actor. Note that κ 1/ λ E .

For l =0 P(E) P (E)

-1/ 2 1/ 2G0

E P (E) E exp[-( ) ]

E

PN: Nuclear Term

• The nuclear term is proportional to the probability that the composite nucleus C* decays into the end product Y+b.• To describe the energy situation of C* we make use of the so-called Shell Model in analogy to the classical description of electrons. The nucleus C* has a set of energy levels to disposal each of which has an intrinsic width

i ii

i

bN

h 1ΔE =Γ =

2π t

where t is the lif etime of the level.

The nuclear term can be expressed as the ratio

ΓP =

Γwhere Γ is the sum of all possible levels.

FR: Correction for Resonances

R

R 2R

R R

Suitable f actor enhancing the cross section at resonances when E =E

A F (E) =

(E - E ) +B

where A and B are constants.

AFor E E , F maximum eff ect

BFor

R R E E , F 0 the cross section is decreased

Nuclear Reaction RatesConsider two type of nuclei, A and B, with number densities nA and nB. Suppose B at rest and A moving with velocity v. The mean free path of particles B is l=1/nB and mean lifetime tB=l/v=1/nBv.

Therefore in the unit volume, nA nuclei react with nB nuclei at the rate RAB=nA nB v per s

In reality both A and B are moving with relative velocity v which obeysthe probability distribution P(v) expressed by the Mawell-BoltzmannLaw, therefore the product v has to be weighed on P(v)

AB

0

( ) R A Bv vP v dv n n v

and

Maxwell-Boltzmann

23/ 2 2

¥1/ 2 3/ 2

0

1/ 2 3/ 2AB A B

Maxwell - Boltzmann distibution of velocities

μ μvP(v)dv =( ) exp[- ]4πv dv

2πKT 2KTSubstitute v with E and insert in <σv >

8 1 E<σv >=( ) ( ) Eσ(E)exp(- )dE

πμ KT KT

and fi nally

8 1R =n n [ ] ( ) S(E)exp[-

πμ KT ¥

1/ 2G

0

21/ 3 1/ 6 5/ 6G

0 G3

EE- ( ) ]dE

KT E

The exponential term is signifi cantly diff erent f rom zero in

a narrow interval centered at

E (KT) 4E =[ ] and large Δ = E (KT)

4 3 2

Approximate Expression

22 2

G A B

0

-24 3/ 2 1/ 3 -3 -1A B G GAB 0

AB μ A B

2/ 3A B

e 2π 1E =(απZ Z ) 2μc with α=

4πε c h 137

n n E E1R =6.48 10 ( )( )S(E )( ) exp[-3( ) ] m s

1+δ A Z Z 4KT 4KT

as in the exponential term we have (Z Z ) the fi rst reactions to occ

=

2 BAo o

AB A B

ur

are those with the lowest Zs.

Finally the energy generation rate of a reaction per unit mass and time is

XX1ε = N ρ <σv >Q where N is the Avogadro number

1+δ A A

The PP Chains

Lifetimes in the pp-chain

The CN-CNO cycle

CN Cycle

OpFFpO

NpO

eOF

FpO

NpO

eOF

FpOOpN

16191918

1518

1818

1817

1417

1717

17161615

),( ),(

CN refuels ),(

),(

CN refuels ),(

),( ),(

CNO Cycle

He-burning 3C

* of levelsenergy of Diagram 12C

8 4

4

Be He

2 He

0

4.43

7.65

Pair emission

Rest Mass Energies (Mev)

0.282

0.099

C12

Reactions in He-burning…..

4 8

8 4 12

12 12

12

2 He+(99±6) Kev Be

Be+ He+(282±6) Kev C*

C* C+(2γ)+(7.656±0.008) Mev

Lif etime of Be is longer than the encounter lif etime of two

The lif etime of C* is long enough f or the exis

He

H

tence of two decay channels

Energy liberated per nucleon 7.275/ 12 =0.61 Mev about 1/ 12 of the energy

liberated by pp chain and CN - CNO.

tThe ratio of »0.1t

Companion reactions

Mev 9.31

Mev 4.73

Mev 7.16

24420

20416

16412

MgHeNe

NeHeO

OHeC

Heavy elements start being synthetized !

C-burning

Mev 0.11-

Mev 4.68

Mev 2.60-

Mev 2.23

Mev 13.93

24

24

23

23

241212

Mg

Mg

nMg

pNa

MgCC

New, p, n and particles are created…..very important

Ne-photodissociation (burning)

Mev 4.73-

2824

2420

1620

SiMg

MgNe

αOγNe

For the first time photo-dissociation becomes important

Oxygen-burning

Mev 0.424 2

Mev 0.393- 2

Mev 9.593

Mev 1.459

Mev 7.676

Mev 16.539

30

24

28

31

31

321616

pSi

Mg

Si

nS

pP

SOO

SSi

SiMg

SipP

3228

2824

2831

Followed by ….

Si-burning

NiSiSiSi 56282828 7

In reality Si is fused in heavier and heavier nuclei by means of many reactions in which p, n, and emitted by photo-dissociations are rapidly captured. A sort of equilibrium condition is established in which Si is converted to elements of the Iron-group (for which the binding energy per nucleon is maximum). The end of the nuclear exo-energetic history of a star.

THE END NUCLEAR REACTIONS AS LARGE SCALE ENERGY SOURCES

Burnings in Tc vs c plane

Electron Screening

• The coulomb interaction energy has been evaluated considering barenuclei, i.e. neglecting the effect of the electron gas in which they areimmmersed.• At high densities the nuclei tend to (locally) attract the electronswhich form a cloud of negative charge around and shield the charge the nuclear charge.• This will lower the effective nuclear charge and the coulomb barriersin turn thus favouring the nuclear reactions.• Polarization of the charge induced by a nucleus of charge Ze: thenumber density of electrons ne in vicinity of a nucleus is slightly higherthan the mean value <ne>. The other nuclei are pushed away and the local number density of nuclei ni is lower than the mean value <ni>.

Electron screening: continue

For a non degenerate gas, the number density of particles with charge q in presence of a potential is modified according to

In most cases |q|<< KT so that the exponential can be approximated to1-q/KT, therefore

which show the increase and decrease of electron and nuclei densities.

)1(

)1(

KT

enn

KT

eZnn

ee

iii

)/exp( KTqnn

Electron screening: continue

Let us consider all types of nucleus and derive the total charge density. In absence of =0. In presence of we have

We may write the above relations in the following way

2

21

1

i i ei

i ii

i i

en

KT

( Z n n ) n

Z ( Z )χ μ X

A

e iwith n=n + n and

introducing the mass abundances we get

with μ the molecular weight

2 2

0i i ei

ii e

i

( Z e ) n e n

0

(Z e ) e- n n

KT KT

whereas f or we have

Electron screening: continue

Charge density and potential are related by the Poisson equation(in spherical symmetry)

RD is the radius of the electron cloud effectively shielding the nucleus

decreasedstrongly is potential theR rfor

exp

issolution the

distance Debye the 4

with

)(or 4

D

2

2

222

)r/R(r

Ze

ne

KTR

dr

rd

r

R

D

D

D

Comparison between RO and RD

21 2

O OO

1/ 2

D DO 7

O 1 2 O

Z Z eR = distance of maximum approach at energy E

E

Ratio between the two radii

R RE T200 f or typical values 50÷100

R Z Z ςρ R

The particle may cross the whole cloud

The lower barrier makes the ractions easier. This occurs throughthe term exp(-2) in calculation of PB, which in fact depends on thequantity EC-E, where EC is the coulomb interaction energy

Final Correction

2 2 21 2 1 2 1 2

C DD

21 2

DD

Expanding the exponential term we get

Z Z e Z Z e Z Z eE - E = exp(-r/ R )- E = - - E

r r R

which is equivalent to increase the energy of the particle in absence of

electron screening

Z Z eE =E+ =E+E

R

Example: d

D

1/ 2

(E / KT -E/ KT -2πη)-1 -2πη 1/ 2 1/ 2 -E KT D

DD

S

erive <σv>

Substitute E with E and η with η =(E/ E)

Eσ(E)vf (v) =(E e )E (E e ) (1- )e

EE

Assume E / KT <<1, neglecting (1- ) and integratingE

over E instead of E we get

<σv>= D

S

E<σv>exp( )

KT<σv> is larger than <σv>

Nuclear Statistical Equilibrium

Starting from C-burning in all major nuclear steps p, n, a particles together with light nuclei rare produced . For instance

The generated photons may dissociate another Ne. We get an equilibrium state in which the abundances of are regulated by law similar to the Saha relationship between ionic species and electrons

ONe 1620

α and ONe,1620

3/ 2O α O α O α3

Ne Ne Ne

2O α Ne

iin

n n 2π m m KT G G1= ( ) exp(-Q/ KT)

n m Gh

where

Q =(m +m - m )c

together with the condition n ρ and n the initial abundances

In general ……

• In reality this is only one of the many possible reactions. The problem becomes identical to that of simultaneous ionization

and recombination of many atomic species. The processes are

are mutually correlated as they produce electrons that affect

the recombination rate.• At high T many nuclear species are present whose

abundances are each other correlated. As 56Fe is the most stable

nucleus (maximum binding energy per nucleon), it plays the pivotal role

in establishing the equilibrium state among the nuclear species (Nuclear Statistical Equilibrium, NSE).

Iron photo-dissociationLet us consider the case of Iron photo-dissociation

we want to find the ratio .

4n13αFeγ 56 nnFe /

56

Consider the ensemble of reactions given by

γ+(Z,A) (Z - 2,A - 4)+α (a)

γ+(Z,A) (Z,A -1)+n (b)

Starting f rom Fe, we have 13 reactions of ty

13 4 13 4 13 424α n α n α n

2Fe Fe Fe

2α n Fe

pe (a) and 4 of type (b).

The ratios of abundances are regulated by

n n G G m m2π KT= ( ) ( )exp(-Q/ KT) (c)

n G h m

Q =(13m +4m - m )c . The occurrence of reactions (a) in eq

4 17n α α Fe

uilibrium implies

n / n = 4/ 13 and theref ore the lef t - hand side of (c) is (4/ 17) (n / n )

Consequences…

Together with the condition

For given T , and the ratio nn/na, the above relations constitute a system in nFe and na

From these considerations we get the important result that the equilibrium conditions a low T require the formation of Fe, whereas the same at high T require the break-down of Fe in P, n, a

unFeiiu )mn4n(56nAnm

Nuclear Reactions at High Densities

• Nuclei organized in a crystal lattice• Nuclei oscillate about an equilibrium position• Reactions can occur even at T=0 due to the very high density

Ions separated by 2Ro

RN radius of nuclei

Nuclear reactions at high densities

1-3O

3/2

1/2

OO

OO

3O

222O

22O

O2OO

22

NOO

22

O

22

O

22

)r(πψ is mode lfundamenta the to associated

function wave the and )MΩ2

(r within oscillates particle The

Ω21

E is level lfundamenta the of energy The

MRe4Z

Ω and mass ion the is M where xMΩ21

-V(x)

oscillator

harmonic the becomes it Rx for ])x

R[(

1R

e2Z

R-R|x| where R

e2ZxR

eZxR

eZV(x)

potential The

Nuclear reactions at high densities

)rR

2exp(rR

τ

)MR

()2Ze

(r e2Z

REα

Rx

u where

duαu1

u)

e4MZ2(exp[τ

R-Rb and ra where dx] o]E-[2MV(x)2

exp[-τ

crossed are barriers that yprobabilit

some is there 0T at even 0 is level lfundamenta the As

2O

2o

O

O

1/43O1/2

22OO

O

1/2/RR1

/Rr2

21/2

2

22

NOO1/2

b

a

NN

OO

Nuclear reactions at high densities

O

1/322

21/3A

22

2

44

1/25/442A

O

A

1/2O

5/4O

2

3/4221/2

3

2

R1

)π3

(e2MZ

)2n

(e2MZ

2.85ε ,101.1γ weight, atomic A

where )εexp(SγZ)AAρ

(WnR

is second and volume unit per

reactions of number the therefore /2,R radius of sphere each

to assigned is ione one when density number ion the be n Let

])(MR

4Zeexp[)R(

M)e(ZS(E))

24(W

velocity relative the 2E/Mv with S(E)E1

|ψ| vW

is nuclei of pair a for rate The

τ

Nuclear reactions at high densities

....)ignition..-(C

WD a of stages final the in yinstabilit of source be may It

years).10n(Rt scale

time short very a on Mg in C convert may reaction This

.g/cm 106ρ be can density typical the

and b Mev 108.83S(E) C,C reaction the For

S(E). assigne to now have We

5A

39

16

Ignition masses (of cores)

1/ 3 2/ 3 4/ 3c c

H

Let start with a simple usef ul representation of hydrostatic

equilibrium π P »[ ] GM ρ36

Kinsert the equation of state f or a perf ect gas P= ρTμm

K c1/ 3 2/ 3 1/ 3

HπT =[ ]36

The temperture increases with density by contraction until

an important quantity of nuclear energy is produced thus

halting f urther contraction or electrons become denegerate

and

Gμm M ρ

5/ 3e ei iNR

EOS changes.

P= P +P=K n + nKT

Ignition masses 2

NR i Hi c5/ 3 5/ 3

e H

5/ 3i e

1/ 3 2/ 3 1/ 3 2/ 3c cH

2 8/ 32/ 3 4/ 3H

c maxNR

K μmρ

μ m

μ

The new relation f or temperature is π KT =[ ] Gμm M ρ -36

which is maximum at

G (μ )mπ [KT ] =[ ] M36 4K

I nverting this relation we get th

1/ 2 3/ 4 3/ 4NRig ig2 8/ 3

H6

ig ig Θ

8ig ig Θ

e minimum mass f orignition

4K36 M [ ] [ ] [KT ]π G (μm )

- - Main sequence H-burning, T »1.5x10 K, M 0.05 M

- -He-burning, fl ash, T »1.0x10 K, M 0.5 M

- - Case of C- ignition in degenerate conditions is sligthly more

complicate

Ignition masses 3: the CO core

γe γ

e H e

Highly degenerate CO core: conditions f or C- ignition.

Pressure is approximated to that of sole electrons and in turn

it is supposed to be made of two terms

K ρ P P = ρT +K ( )

μ m μ

where γ

2/ 3 4/ 3c c c

1/ 3

=5/ 3 f or non relativistic electrons, and γ=4/ 3 f or

relativistic electrons. Start now f rom the usual

P =f G M ρ

where f =[π/ 36] . Substituting we get

2/ 3 1/ 3 γ-1 -γc c c γ c e

e H

1/ 3 2/ 3 χ -(4/ 3+χ)c c c γ c e

e H

KT =f G M ρ - K ρ μ

μ m

At increasing density γ 4/ 3. Let us pose γ=4/ 3+χ

KT =ρ [f G M - K ρ μ ]

μ m

f rom which we see that T increases with ρ i

3/ 2 -24/ 3c crit e

f

K M M =( ) μ

f Gwhich is of the order of the Chandrasekhar mass.

Thermal runawayI n a sphere of mass M and Radius R, the variation in P and ρ due to variations

of R by r +dr =r(1+x) - - defi nition of x - - are expressed as

dρ dP =-3x and =-4x

ρ P

f rom w

α -δ

hich we have

1dρ 1dr 1dP 1dr =-3 and =-4

ρ dt r dt P dt r dt

With the aid of EOS ρ P T we get

1 dT 4 - 3 1dr =

T dt δ r dt

I ntroduce auxiliary va

4(1 )

4 3A

P A P

dP dT 4δriables p= , θ = and p= θ

P T 4 - 3f rom the fi rst law of thermodynamics we derive

dq=du+PdV = c T(θ - p) =c*Tθ c* =c

where c * is the gravothermal specifi c

A

heat.

For a monoatomic perf ect gas =δ=1, =2/ 3 c* <0 f or

dq >0 dT < 0. Thermostat.

For a non relativistic degenerate gas δ 0, α 3/ 5 c * > 0 f or

dq > 0 dT > 0. Thermal runaway.

Ι gnition of a f uel in a degenerate gas leads to an explosion that can

tear the star apart.

Grouping stars according to their evolutionary history

– Low mass stars ( ) H and He burnings, He degenerate core, He flash, AGB mass loss, CO White Dwarf .

– Intermediate mass stars ( ), H and He burnings, no degenerate He core, no He-flash, AGB, mass loss , CO White Dwarfs of larger mass.– Massive stars ( ) complete the whole

sequence of nuclear burnings form a Fe degenerate core sourrounded by many layers and burning shells,

mass loss by stellar wind that does not affect evolution of inner core.

6 8HeF upM M M M

MM 98

MMM HeF 2.27.1

CO WD and Fe cores much similar: both electron degenerate

Compact Stars (cores)These considerations apply to both isolated White Dwarfs

and thecentral cores of stars in some evolutionary stages.

3

31/ 3F e

F

The equation of state of f ully degenerate electrons in any realitivistic state is given by

P Af (x) Bx where

p 3h nx p ( )

mc 8π

2 1/ 2 2 2 1/ 2o

e

4 5 3 322 -2 5 -3

e e3 3o

N ρ n f (x) x(x 1) (2x -3) 3ln[(1 x ) x]

μ

πm c 8πmcA 6.002 10 dynes cm B μ 9.736 10 μ g cm

3h 3h N

f or x 0 non relativistic, x f ully re

22 2 1/ 2 3

2

2 2 F2

lativistic

Combining the equation of hydrostatic equilibrium and mass conservation we get

1 d d πGB [r (x 1) ] x

dr dr 2ArE

I ntroducing z x 1 where z ( ) 1 mc

c

1/ 2

c

and the variables r ας z z φ

2A 1with α ( ) ( ) we obtain

πG Bz

Basic equation For White Dwarfs

2 2 3/ 2c2 2

c

1 d dφ 1ς (φ ) parametric in z

dς dςς z

dφwith boundary conditions ς 0 φ 1 0

dς

I t becomes the politrope with n 3 f or z (x ) f ully relativ

c c

istic

the politrope with n 3/ 2 f or z 1 (x 0) non relativistic

The solution φ(ς) is f ully determined by z i.e. ρ .

Consequence: the structure of the White Dwarf (degenerate c

22A(

G

c

1/ 2 3/ 21 1 12

c

ore) is f ully

determined by its central density.

I n general the radius and mass are univocally determine by ρ . The radius and

mass are given by

1 2A 1R ) ( ) M 4 ( ) ( )[- ' ]

Bz G B

2e

Constant radius

The case of n 3 (f ully relativistic) the total mass becomes

5.836 M M the Chandrasekhar Mass ..... but Radius 0

396 g/cm 1010 rangeDensity

Evolution of white dwarfs

Two important corrections for White Dwarfs

Ch

For M 0 we expect R 0 and ρ const instead of R and ρ 0.

For M M , ρ and R 0 we expect R 0.

Two corrections are required:

Low mass end: this is "simple". Taking into account the ele

o c zp 0 C

zp

ctrostatic interactions between

nuclei and electrons, the energy per electron is given by

E E E E where E is the usual energy in a Fermi gas, E is the coulombian correction

( 0) and E is

0 C

0

the zero point energy of the ion oscillation in the lattice (negligible).

Theref ore

E EEP - P usual pressure radius is decreased

(1/ n) (1/ n) (1/ n)

High mass end: more complicate (require a new concept : neutronization of matter and

change of type of star)

Mass-Radius Relationship

More on WDs: GR Effects

The structure of WDs has been examined in Newtonian description ofGravity (NG). However at very high densities General Relativity (GR)intervenes often leading to instability.

In NG an equilibrium configuration may exist if the gravitational forceand pressure gradient balance each other. In principle at any a pressure P() can be found to secure balance.

In GR the pressure itself is an effective source of mass therefore increasing P will also increase the gravitational force stability is not granted.

The structure of a WD is determined by its c the total energy isalso a function of the density E(c).

Equilibrium & Stability

0cd

dE

At equilibrium

Stability is determined by

stability marginal 0

maximum) (local unstable 0

minimum) (local stable 0

2

2

2

2

2

2

c

c

c

d

Ed

d

Ed

d

Ed

Looking for stability

Derive the total energy E(c) of a politrope with index n, takingGR effects into account (at first order) and analyze stability.

PVEi Total Energy = Internal + Gravitational

R

GMEg

2

KPEquation of State of politrope

A simple case

)3

4(

03

4

ln

ln

.

3/53/5

)3

4(

2

3

3/13/53

12

2

11

0

2

10

GME

yieldsstabilityforconditionThe

d

Md

densitywithincreasetohasmassthestabilitygetTo

M

yieldsmequilibriuforconditionThe

knownareKAll

GMKMKE

R

GMKPMKE

R

GMKPVKE

c

Two well known results

Correction for self-mass

In GR the gravitational energy is equivalent to an effective mass

3/23/724

3/23/72

3

21

33c

2

3

2

2

212

21

2

11

)())((

eliminate and introduce

therefore

toscorrespond

MM

GKM

M

GE

RR

MR

M

c

G

R

GMmE

c

||Em

R

GME

corr

effcorr

eff

New condition for stability

21 5 3 1 3 7 3 2 3

2 3 4 2

4 32

3

2

3

/ / / /c c c

Insert this in the total energy

GE K M K GM K M

celiminate the density and derive the condition for stability

K4 GM

3 K c R

The interval for instability is increased. For inst

2

ance with n=3

GMthe variation of is 2.25( )

c R

Now we need to specify for the the physical

environment we are dealing with

for a degenerate gas of electrons

cm 10

or Eliminate

requires above) (seestability for condition The

3

2

3

4

3

2

3

4

ln

ln

ln3

4 ln4ln

kp n via torelated isdensity The

)1(3

2get we

211 c/p)(m of seriesin v(p)expand

3

8 fromStart

92958

32223

35

22

2

223/2

2

22

3/22

22

2

22

1/33e

2

224

3

2

2221

2

22

e

0

23

μ)M

M(R

)m(μ)h

cm(

cR

GM p

)p

cm(

Rc

GM

p

cm

k

cm

d

Pd

constk

cmconst

p

cmpP

pp

p

cmp

h

cP

)p

cmc()

p

cmc(v

dppv(p)ph

πP

/-e

/

Θ

/pe

e/

F

F

e

F

eee

e

F

eFe

FF

F

eFe

e/e

p

e

F

The last espressiongives the radius below which GR effects leads to instability

Compare with theradius of typical WDs

Comparing critical densities

The radius of WD scales with density as cm 10)( 93/1

relR

For > 600 rel , this is smaller than the radius we have previously determined, therefore a WD may easily go unstable for GR.

Note that the density above which GR effects are important is largerthan the density at which neutronization begins in Fe cores……see below…

Effects of rotation on limit mass

2222 1

MRJRMT

Suppose the WD (or the core of a massive star) is rotating withangular velocity and angular momentum J (conserved). The kinetic rotational energy is

Apply the Virial theorem (W is the gravitational energy)

01

32

2 MR

JP

MW

Applying the Virial Theorem

3/43/5 PPEquation of state non relativistic relativistic

icrelativist 0

icrelativistnon 0

232

2

3

2

3

22/32

2

2/3

2

2/3

R

M

MR

J

R

GM

R

M

MR

J

R

GM

The non relativistic case is trivial and leads to the well known M(R)

From Virial

The new limit massNon rotating case: the relation can be satisfied only by a certain valueof M as the remaining two terms have the same dependence on R.

Rotating case: there is a term 1/R^2 for each value of M an equilibriumconfiguration is possible for R sufficiently small. As a consequence of itthe limit mass will increase

2 4 3

3 3

3 2

3 2

20 1

2 21 1

/

/

3 Ch

/3

Ch

GM T M( )

R |W | R

invert and get the mass

MM

T TG( ) ( )

|W | |W |

for typical values of T/|W| (0.14--0.26) we obtain

M 2M

Neutron Stars (generated by Fe cores)

Neutronization of matter occurs due to inverse -decays betweennuclei and electrons at the top of the Fermi distribution. In brief

(Z,A) e (Z 1,A) ν

(Z 1,A) (Z,A) e ν

The second reaction cannot occur as all states f or

electrons are fi lled by degeneracy. The number of

f ree electrons decreases whereas that of f ree

neutrons increases.

At increasing density the neutronization processes become more and more important until the binding energy of nuclei become positive and they melt into lighter nuclei and free neutrons.

The result is that matter is converted into free neutrons at the expenses of the gravitational energy of the system.

Hydrostatic equilibrium in general relativity

This is the first important change to be made with the neutron stars

At very high densities and in relativistic conditions the mass densityo=mnnn (expressed by means of the rest mass density o) must be replaced by the total mass density =+u/c^2

3-1r r

2 2 2 2

The equation of hydrostatic equilibrium in general relativity is

GM 2GMdP P 4πr P =- ρ(1+ )(1+ )(1- )

dr r ρc mc rc

the Tolman- Oppenheimer -Volkoff equation, which in the so- called

post - New3

r r2 2 2 2

tonian approximation becomes

GM 2GMdP P 4πr P =- ρ(1+ + + )

dr r ρc mc rc

New equation of state: 1• Neutronization and liberation of free neutrons must change the EOS.• How does it occur?• We must derive the EOS by determining the minimum –energy confi-guration for a system made of about 10^57 baryons possibly made ofseparate nuclei in -equilibrium with a relativistic gas of electrons. From standard nuclear physics for systems with A<90 the lowest energystate consists of the Iron nucleus with the tightest binding energy. For A>90, the lowest energy state will again be made of several such iron nuclei. The situation gets modified when electrons can combine with protons and incresase the number of neutrons in the nucleus.• To see this, let us remember that the tight binding for A=56 arises the balance between coulomb repulsion of protons and actractive nuclearforces. When neutrons increase, the coulomb repulsion decreases andenergetic nuclei with A>56 may exist. The critical n/p ratio is reached

New equation of state: 2at approximately 4x10^11 g/cm^3. Any further increase in density leads to a two-phase system in which electrons, nuclei and free neutrons coexist.The number of free neutrons increases with density.

'N e e n n

The energy density of a mixture of nuclei, f ree electrons and f ree neutrons can be written

ε=n M(A,Z)+ε (n )+ε (n ) where N ref ers to nuclei, n to neutrons etc... and M(Z,A)

is the energy of th

2 2N p e b

22 2/ 3 2 5

N 1 2 3 4 1/ 3

e nucleus (A,Z)

M(Z,A) =[(A - Z)m c +Z(m +m )c - AE ] which can be analytically represented by

b Z1 Z=m c b +b A -b Z +b A( - ) +

2 A A

b =0.991749, b =0.01911, b =0.000840, b =0.10175, b =0

'e N n

.000763

I t is conventional to include the rest energy of nucleons and electrons in M(Z,A) and hence

ε does not have the rest energy. Denoting with n=n A +n the total density of baryons and

the neutr n non f raction as Y =n / n we recast the energy as

Explanation

New equation of state: 3

'n e e n n e n n

n n

2 2 2F,e e F,e e

M(Z,A) Zε n(1 Y ) ε (n ) ε (n ) n n(1 Y) n nY

A AWe must minimize ε(n,A,Z,Y ) with respect to A, Z, Y at constant n. We get

M M M(ε mc ) A Z(ε mc ) ε

Z A A A

(1)

(2)

F,e

2 1/ 2 n3 4 5 e1/ 3

N

1/ 2 1/ 2 1/ 22

5

21/ 3 252

1 4 2 4

or

m2Z Zb b (1- )-2b [(1 x ) 1]

A mA

bZ ( ) A 3.54A

2b

b Z2b A 1 Zb b ( )

3 4 A 3A (3)

2 1/ 2 nn/ 3

N

F

m(1 x )

m

where x (p / mc) f or each species.

New equation of state: 4

2n

These equations can be used to determine the pressure- density

relation (EOS) as f ollows :

we start with A 56, eqn (2) yields a value f or Z; f or this value we check

whether x 0, if so then f ree

e e e e.

neutron exist and their density and

pressure can be derived f rom standard relationships f or degenerate

neutrons; eqn (1) yields x and hence we can calculate ε ' , P , and n

Known these quant

e e ne n e n2 2

ities, we derive the P( ) f rom the implicit relations

n M(Z,A)/ Z ε ' εε Aρ , P P P, n n n

Zc c

New equation of state: 5

neutrons of made is star the PP When

P PP 102.5 ρ

P108

P 101.5 ρ

P21

P 101 ρ

P PP 104 ρ

sequence pressure - density The

N

N14

eN13

eN12

Ne11

Mass bounds for Neutron Stars

• The simplest model is based on ideal degenerate EOS for non relativi- stic neutrons and Newtonian Gravity.• Identical to a politrope with n=3/2 that immediately yields

1/ 2 3 -1/ 6c cΘ Θ15 3 15 3

12 3min

min

ρ ρ15.12 km M 1.102 ( ) M ( ) M and R 14.64( )

R10 g/ cm 10 g/ cm

which is valid only if neutrons are stable against β- decay ρ 4.5 10 g/ cm

or equivalently at ρ ρ R 36 Θ km and M 0.1 M

These values are not correct because real EOS is not the ideal non relativistic neutron gas!

However the concept of minimum density above which neutron stars can exist is very

importan 10 3

10 12 3

t. Remember that WDs have maximum densities of 10 g/ cm above which they

become unstable to General Relativity ef f ects. I n the density interval 10 - - 10 g/ cm

no stable compact star can exist.

F NS WDMin Maxinally note that M M

• Important question: Can a WD become a NS? Answer NO WD have not reached the lowest energy state as NS

Corrections for General Relativity

In Neutron Stars the term GM/Rc^2 is not small compared withunity, must include proper treatment and corrections.

SCHEME: Given P(), the structure of the star and associated M(R) must derived from solving the TOV equation for hydro-static equilibrium. Complicate!

15

ΘMax

13ΘMin

105.8ρ at km 8.8R M 0.72M

102.6ρ at km 300R M 0.18M

get we discussed already have we EOS the Using

The existence of a Maximum Mass can be understood in terms of GR

Why a maximum mass?Start from the case of non relativistic, degenerate neutrons describedby the politrope with n=3/2

2/ 3int 1 c 1

1/ 3 5/ 3grav 2 c 2

Calculate the contribution to the Total Energy f rom

I nternal Energy E =kKρ M with k =0.795873

Gravitational Energy E =-k Gρ M with k =02

7/ 3 2/ 3GR 4 c 42

int

F n

.760777

GGR Corrections ΔE =-k M ρ with k =0.6807

cTo compute ΔE we start f rom a suitable expression of the internal energy

expanded in power of x =p / mc

2 2 4int

2 4int

3n

O n nn

3 3 u=c ( x - x ) where the fi rst term is E

10 56and the second yields the correction

3 ΔE =- c x dm

56mx

Using the relation ρ = mn =3π

44/ 3

int 3 c3 16/ 3 2n

3

we obtain ΔE =-k Mρm c

with k =1.1651

Adding up…..

2/ 3 5/ 3 1/ 3 4/ 3 7/ 3 2/ 3c c c c

4 23

1 2 4 216/ 3 2n

c

1/ 3 2/ 3 -2/ 3 1/c c c

The total energy is

E AMρ BM ρ CMρ DM ρ

k Gwith A kK B k G C D k

cm c

EApplying the condition f or equilibrium 0

ρ

2Aρ BM ρ 4Cρ

3 4/ 3 1/ 3c

2

2c

1/ 3 2/ 3 -2/ 3 1/ 3 4/ 3 1/ 3c c c c

c

15 3Θ c

2DM ρ 0

Eand f or stability 0

ρ

2Aρ 2BM ρ 4Cρ 2DM ρ 0 which

can be solved f or M and ρ to get

M 1.11 M and ρ 7.43 10 g/ cm .

The total energy is

2Θ

2Tot Θ

E -0.08 M c and the total mass is

M M E/ c 1.03 M which are f airly consistent with

previous estimates.

Is there a rigorous upper limit?

• Neutron stars are the most compact configurations that can withstand gravity. Therefore it is very important to establish their maximum mass which would provide the limit at which stablelowest-energy configuration of matter can exixt in its most compact form.• Determining the Mmax is a cumbersome affair because of our poor knwoledge of the EOS at supernuclear densities. Indirect estimate• Suppose we have an EOS up to a certain density o. For > o we would like to extend the EOS in such a way that it provides the maximum pressure at any .• Because the speed of sound must be Vs < c, ANY EOS must obey the constraint

2cd

dP

It follows that ….

.M -5-3 range the in limit upper rigorous a be must

there that clarifies reasoning refined more slightly A

M )g/cm104.6

ρ( 3.2M

get we equationTOV the into expression this Inserting

.constraint above the satisfies

ρρ for )cρ-(ρPP

expression The

Θ

Θ1/2-

3140

02

00

The Mass-Radius relationship again

Basic concepts on Black Holes

General Relativity sets the maximum mass for a neutron star to a value that

depends on the EOS in use.

If a collapsing nucleus has a mass in excess of this value the collapse cannot

halted and a Black Hole (BH) is formed.

A BH is a region of space-time enclosed by the event-horizon, a region whose

gravitational field is so strong that no matter no radiation can escape from this

surface.

A BH can be detected only by its gravitational effects on nearby objects.

Generalities: 1

The existence of BH has been predicted long ago by Laplace with very simple

considerations: suppose a test particle with mass m in the gravitational field of

an object with mass M, and assume that the velocity of the test particle is zero

at infinite distance.

km 3 M 1Mfor 2

thatfollowsit light) of (speed assume usLet

2 with 0

2

1 energy Total

0

2BH

22

BHS

esc

esc

Rc

GMRR

cv

R

GMv

R

GMmmv

v

Generalities: 2Rotation and charge remain to a BH. The charge can be later easily neutralized

by accretion of matter.

Any other properties of the material collapsing to a BH is definitly lost.

To describe a BH General Relativity is required.

Let be a non rotationg, highly concentrated object with mass M, the gravitational

field around is governed by the Einstein solutions. Each line-element ds

(distance between two events in the four-dimensional space) is given by

tensormetric theis where2ij

jiij gdxdxgds

In brief

densityenergy

theanddensity massrest theof sum the and

pressure theis P wherePTTT ,T

termsdiagonal only the null-non has T gas, idealan For

tensorMomentum-Energy : T

(scalar) curvatureRieman : R

tensormetric : g

tensorRicci :R

and 8

where2

1-

equations

Einstein by the described is field nalgravitatio The

20

3322112

00

k i

ik

ik

ik

22

c

U

c

c

GT

cRgR ikikik

Three basic equations

Spherical, symmetric and static distribution of matter. Assume spherical coordinates r,

2 2 2 2 2 2 2 2

2 21 1

( r ) ( r )ds e c dt e dr r ( d sin d )

where (r) and (r) are two suitable functions

1. Get the field equations

2. Get the equation for the hydrostatic equilibrium in GR.

dP Gm P- ρ( )(

dr r ρc

3

12 2

3

2 2 2 2

4 21

4 21

πr P Gm)( )

mc rc

which in Post Newtonian approximation becomes

dP Gm P πr P Gm- ρ( )

dr r ρc mc rc

and for c 0 becomes the classical newtonian equation.

3. Get the metric in the so-called vacuum so

2 2 2 1 2 2 2 2 2 2

2 2 2 2

1 1

1

-s s

ss

lution of Schwarzschild

r rds ( - )c dt - ( - ) dr - r d - r sin dφ

r rr

ds ( - )c dt - dσ where r is the Schwarzschild radius.r

Singularity & Proper time

Note that the term is singular at r=rs. The singularity

is not physical as it can be eliminated by a change of coordinates.

1)/1( rrs

The proper time measured by an observer is related to the line element by

dt=ds/c.

A stationary observer (dr = d= d= 0) at r = measures =t

For two stationary observers one at r, , and the other one at the

proper times are related by

2/1)1(r

r

d

d s

Gravitational redshift

Suppose the first emits a signal at regular interval d (an atom emitting a

frequency =1/d), the other receives the signals and measures the intervals

doo i.e. =1/doo. The redshift caused by the gravitational field is

z rr for

1)1(1

s

2/100

r

r

d

dz s

Curvature in the 3D space

The components of the metric tensor show that the 4D space as well as the

3D space are curved.

At the surface of a star with mass M and radius R the curvature of the space

of positions is

2

232

1 BH afor

1

2

1

-

s

R -K

RR

r

Rc

GMK

Motion of a test particle: 1Suppose a test particle freely moving a gravitational field between the

points A and B. The Universe line is a geodesic, i.e. a line whose length

SAB satisfies the condition

0 B

A

AB dsS

If a particle locally moves with velocity v over the spatial distance d the interval

of proper time d decreases at increasing velocity. One has d = d = 0 for v = c.

E.g. photons travels along geodesic of length d0.

For particles with mass > 0, v < c and d ^2 > 0 ds^2 < 0, the separation is said Time-Like. The Universe lines of material particles are always time-like.

Separations with ds^2 <0, d^2 <0 would require v> c and are named Space-Like.

E.g. The distance between two simultaneous events.

Motion of a test particle: 2

Null geodesic (ds^2=0) yield the propagation of photons and describe the

so-called Light-Cones in the space-time.

Eliminate the singularity

|1|ln s

s

r

r

c

rtt

To study the properties near r = rs let us introduce the new temporal coordinate

The relation for ds becomes

222222222 sin)1( 2 )1( drdrdrr

rtddrc

r

rtdc

r

rds sss

Which is no longer singular at r = rs

Light Cones of radial photons

Consider only photons emitted radially (d= d= 0) for which we derive

r

rr

r

cc

r

r

cdr

td

cr

r

dr

td)(

r

r-(

ds

s

s

sss

1

11)

dr

td( and

1)

dr

td(

solutionswith

0)1(12

)1

or 0

21

22

2

These are the slopes of the radial limits of the light cones at r = rs.

The two solutions: meaning

The first solution represents photons moving inward with speed c.

The second solution changes sign at r = rs

• It is > 0 for r > rs: the photons can be emitted outward (dr >0)

• For r rs the cone rotates inwards

• For r = rs: no photons are emitted ouwards

• For r < rs: the solution becomes negative, all photons are emitted

inwards and no photon can be emitted outward (leave the star)

The motion of a test particle

Assume that all varaibles depend on t which monotonically changes along

the Universe line d= ds/c and indicate the temporal derivatives with

Introducing in the basic equation we get the identity

d

dxx

dxdx

)sin()1()1( 222221222 rrr

rt

r

rcxxgc ssji

ij

The condition that the Universe lines are geodesic implies s==0 leading to

the Eulero Lagrange equations

Eulero-Lagrange equation

Where L is the lagrangian

For the Eulero-Lagrange equation becomes

0)(

x

L

x

L

d

d

1 22 i j /ijcL [ g x x ] and by comparison L 1/2

ctx 0

)1( 0])1[( constAtr

rt

r

r

d

d ss

Consider for simplicity only radial motion and assume some initial distance ro with

v = 0 and = 0. Performing some substitutions and algebric manipulations we get

Final result

2/12 ]1[r

rAcr s

r

rA s 12

with and integrate

0 0

0

1 21

2 s

r r r(sin ) with arc - cos( )

c r r

The function (r)

Nothing happens in the proper time when the particle arrives at rs. The

total proper time is

!small!Very

s 108.9c

r M MFor

c

r 17.5 r 5 rfor

c

r 49.7 r 10rfor

)(2

6s

ss0

ss0

2/30

s

os

r

r

c

r

The observer at roo

2/10

0

s

s

)1(

)]sin(2

[]))2

tg/()2

tg(abs[ ln(/r

t

(r)function t the toleadswhich

r r as 0 /)1(

s

s

s

r

r

r

r

c

Ar

r

dt

d

An observer at roo sees a different story. In fact the relation between and t is

The fact that such an observer sees the -clock to slow

down as r rs means that t(r) will reach r = rs only at too.

Events inside rs are fully masked for such an observer

Consequences

• For an observer at rs the collapse proceeds quickly but smoothly through

the surface.

• Once the surface of the stars has fallen inside rs no static solution exists

and the collapse towards the central singularity cannot be opposed.

• The singularity at r = 0 is real but the physical conditions are not known.

• For a distant observer the scene is very different. At his t-clock

the collapse of the star surface slows down as r rs which indeed

can be reached only for too

• The surface of the star will appear as at rest and the emitted light more

and more shifted toward the red (z 00 )

• For this observer the stars will disappear from his view and will be

detectable only via the gravitational interaction with nearby objects.