cesare chiosi department of astronomy university of padova, italy
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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.
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