Download 5 Elements of nuclear physics 5.1 Strong interaction and structure of atomic nuclei

5
5.1
Elements of nuclear physics
Strong interaction and structure of atomic nuclei
Strong interaction is characterized by a short range, ∼ 10−13 cm. Only hadrons (baryons and
mesons) take part in strong interactions, neutrons and protons interacting identically. The strong
force is attractive but has a strong repulsive core so that the nuclear matter could be considered
as incompressible fluid with an effective surface tension. The radius of the nucleus is estimated as
r0 = 1.2 · 10−13 cm,
R = r0 A1/3 ;
(5.1)
where A is the mass number.
The nuclear binding energy is defined as the rest mass energy of all protons and neutrons in
the nucleus minus the rest mass energy of the nucleus:
B(Z, A) = [Zmp + (A − Z)mn − m(Z, A)]c2 .
(5.2)
Because of the short range of nuclear forces, the binding energy of the nucleus is proportional, in
the first approximation, to the number of nucleons, B1 ∝ A. Inasmuch as nuclei in the surface
layer are at special conditions, one has to introduce an effective surface tension, which contributes
to the binding energy a negative term proportional to the surface area of the nucleus, B2 ∝
−A2/3 . The electrostatic repulsion of protons contributes to the binding energy one more term,
B3 ∝ −Q2 /R ∝ −Z 2 /A1/3 . At last according to the Pauli principle, all particles could not be at
rest; both protons and neutrons acquire the kinetic energy up to the corresponding Fermi energy,
εF ∝ N 2/3 , where N is the density of the corresponding specie; Np ∝ Z/A; Nn ∝ (A − Z)/A.
Therefore one has to add to the binding energy a negative term (the kinetic energy contributes
negatively to the binding energy)
B4 ∝ −
Z 5/3 + (A − Z)5/3
.
A2/3
(5.3)
This energy has a maximum at Z = A/2, when the number of protons and neutrons are equal.
Expanding around the maximum in small (Z − A/2) yields
−B4 ∝ 2−2/3 A +
20 · 21/3 (Z − A/2)2
.
9
A
(5.4)
The first term could be merged with B1 whereas the second one describes the asymmetry energy.
Summing up all contributions, one can finally write the binding energy of the nucleus in the form
called Weiszäcker formula:
B(Z, A) = aV A − as A2/3 − aC
Z2
(Z − A/2)2
−
a
,
a
A1/3
A
(5.5)
where the empirical constants are
aV = 15.85 MeV;
aS = 18.34 MeV;
aC = 0.71 MeV;
aa = 92.86 MeV.
(5.6)
In light nuclei, the Coulomb energy is small. Then due to the asymmetry term, the stable
isotopes typically have equal numbers of protons and neutrons (there are exceptions among very
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light nuclei, like 7 Li). In heavy nuclei, the role of the Coulomb repulsion becomes significant
(because of the long range of the interaction), which is compensated by a neutron excess. For a
fixed A, the maximal binding energy is achieved at
Z=
A
.
2 + 0.015A2/3
(5.7)
The binding energy per nucleon, B(Z, A)/A, grows with A but eventually becomes to decrease
due to the Coulomb repulsion. The most tightly bound is 56
26 Fe.
Since hadrons are heavy, new particles could not be produced in strong interactions at small
and moderate energies. In this case, strong interaction reactions just redistribute neutrons and
protons between the states. For example,
2
D + 2 D → 3 He + n.
(5.8)
If high energy baryons collide, new baryons are produced provided the energy and momentum
conservation laws are fulfilled. The lightest baryon is π 0 -meson; his mass is 135 MeV. Therefore
π 0 production reactions, e.g.
p + p → p + p + π0,
(5.9)
have the smallest energy threshold among strong-interaction reactions with particle production.
5.2
Electro-magnetic and weak interactions
As it was demonstrated in the previous section, the electro-magnetic interaction (in fact the
electrostatic repulsion) plays a major role in intermediate and heavy nuclei. In addition, electromagnetic interaction opens new channels for nuclear reactions because emission/absorption of
photons permits to conserve energy. For example, a deuteron could not be formed from free
proton and neutron via strong interaction because there is no way to remove the binding energy
of the deuteron, 2.2 MeV, from the system. Due to the electro-magnetic interaction, the reaction
becomes possible:
p + n → 2 D + γ.
(5.10)
Due to electro-magnetic interaction, the neutral pion decays,
π 0 → γ + γ,
(5.11)
with the mean life time 8.4 · 10−17 s. Therefore collisions of high energy protons could lead, via
the reactions (5.9) and (5.11), to the gamma-emission. In particular, interaction of the cosmic
rays with the interstellar gas produces the observable gamma-emission.
The electro-magnetic interaction could be also responsible for transitions between the excited
states of nuclei. The typical excitation energy is ∼ 1 MeV therefore these transitions may be
accompanied by emission of gamma-ray lines.
The weak interaction opens the way for mutual transformations of protons and neutrons. For
example, a free neutron decays as
n → p + e− + ν̃
(5.12)
with the mean life time 880 s. This occurs spontaneously because mn − mp − me = 0.8 MeV > 0.
Reverse reactions, e.g.,
p + e− → n + ν or p → n + e+ + ν
(5.13)
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are possible only at special conditions. The first reaction occurs if the electron energy is high
enough, e.g., in a very hot or strongly degenerate plasma. The second reaction could occur in
bound states, e.g., 15 O → 15 N+e+ + ν.
Note that in all reactions, the baryon number (=1 for baryons, -1 for antibaryons and 0 for
mesons) and the lepton number (=number of leptons - number of antileptons) are conserved.
5.3
Thermonuclear reactions
Hydrogen burning into helium is possible because 4mp − mHe − 2me = 27 MeV > 0. Inasmuch
as two protons have to be converted to neutrons, the weak interaction should be involved, which
results in a very slow reaction rate. In the p − p cycle, the conversion occurs in the reaction
p + p → D + e+ + ν
(+1.44 MeV),
(5.14)
p + D → 3 He + γ (+5.49 MeV);
3
He +3 He → 4 He + 2p (+12.85MeV).
(5.15)
(5.16)
which is then followed by the chain:
The overall reaction rate is determined by the first reaction. It becomes possible if protons
approach the nuclear distance, ∼ 10−13 cm. At this distance, the Coulomb barrier reaches e2 /r ∼ 1
MeV. Within the stars, the temperature is only ∼ 107 K ∼ 1 keV therefore the proton fusion occurs
via quantum tunneling through the barrier.
The reaction (5.14) is reverse to the reaction (5.12) but could occur only within the nucleus
because a free neutron is heavier than a free proton. Therefore the reaction rate per pp pair could
be roughly presented as
Wpp ∼ P Wn ,
(5.17)
1
s−1 is the neutron decay rate and P the probability for two protons to find
where Wn = 880
themselves within the distance ∼ rD ≈ 2 · 10−13 cm, where rD is the radius of the deuterium
nucleus. Introducing the wave function of the pair, one can write
Z
|ψ|2 dV
(5.18)
P ∼
r<rD
Let us estimate this quantity for a pair within a normalization volume V .
At distances much larger than the classical turning radius, r rcl = e2 /E, the wave function
of the pp pair is presented as
!
X
~ sin pr
1
1 ip·r/~
~
(5.19)
ψext = √ e
=√
+
... ,
pr
V
V
l≥1
where the last equality presents the expansion in angular momenta. The distance r ∼ rD could be
achieved only by l = 0 pairs therefore it is the first term in the rhs that should be matched with
the wave function within the Coulomb barrier. The last may be roughly presented in the WKB
approximation as (recall that for l = 0, the wave function could be presented as ψ = u/r, where
u is a solution to the one-dimensional Shrödinger equation)
Z
C
1 rcl
ψint ∼ √ exp −
pdr ,
(5.20)
r p
~ r
23
p
p
where p = 2m[(e2 /r) − E] = 2mE[(rcl /r) − 1], m = mp /2 is the reduced mass. The rough
matching of (5.19) and (5.20) at r ∼ rcl yields
C∼p
~
V pr (rcl )
∼
~
V 1/2 (mp E)1/4
(5.21)
Even in order to get a rough estimate, one has to evaluate the integral in the exponent with high
accuracy:
r
p
r
Z
Z r
Z
mp E rcl rcl
e2 mp 1
1
1 rcl
− 1dr =
− 1 dx
(5.22)
pdr =
~ rD
~
r
~
E rD /rcl x
rD
r
r
Z r
mp c2 1 1
π
mp c2
≈α
− 1 dx = α
.
(5.23)
E
x
2
E
0
Here we take into account that rD ≈ re and E ∼ kT ∼ 1 keV so that rD /rcl ≈ E/me c2 ≤ 0.01.
Now the wave function within the nucleus could be estimated as
#
"
r
~
π
mp c2
ψ ∼ ψint (rD ) ∼
exp − α
(5.24)
rD V 1/2 (m2p Ee2 /rD )1/4
2
E
" r #
3/4 1/4 1/2
re
me c2
me
1 E0
1
exp −
,
(5.25)
=
1/2
αV
rD
E
mp
2 E
where E0 = π 2 α2 mp c2 = 450 keV.
Substituting the above result to (5.18) and then to (5.17) yields the reaction rate per pair
" r #
3/2 1/2
3
1
re
me c2
me
E0
4 rD
exp −
.
(5.26)
Wpp ∼ Wn π
2
3 V α
rD
E
mp
E
Now one has to average this quantity over the Maxwell distribution. Taking into account an
extremely strong dependence of Wpp on the particle energy, one cannot just substitute E by the
thermal energy but has to perform averaging explicitly:
Z
Z
√
1
2
−E/kT 3
−E/kT
p
W
e
d
p
=
W
e
EdE.
(5.27)
hWpp i =
pp
pp
(2πmkT )3/2
π(kT )3
One can calculate the integral
Z
I=
0
∞
E
exp −
−
kT
r
E0
E
!
dE
(5.28)
taking into account that the function in the exponent has a sharp minimum at
Emin =
E0 k 2 T 2
4
24
1/3
.
(5.29)
Expanding around the minimum one gets
s
r
Emin 3
E0
E0
E
+
=3
+
(E − Emin )2 .
5
kT
E
kT
8 Emin
Substituting this expression into the exponent, one finds
" s
#
Z ∞
E0
Emin
3
2
I = exp −3
exp −
(E − Emin ) dE
5
kT
8 Emin
−∞
" r
1/3 #
π 1/6
27E0
2/3
5/6
=2
E (kT ) exp −
.
3 0
4kT
(5.30)
(5.31)
(5.32)
The reaction rate per unit volume is now finally found by substituting V = 1 into hWpp i and
multiplying by the number of pairs in the unit volume, N 2 /2:
" 3/2
1/3 #
1/6
1
27E0
211/3 π 2 3
re
me (me c2 )1/2 E0
Q = 3/2 N rD Wn 2
exp −
(5.33)
3
α
rD
mp
(kT )2/3
4kT
!
2
14.5
N
(5.34)
= 6 · 10−38 2/3 exp − 1/3 cm−3 · s−1 .
TkeV
TkeV
A more detailed calculation yields the coefficient 1.3 · 10−37 .
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6.1
Stars
Mechanical and thermal equilibrium in stars
The mechanical equilibrium of a star is described by the hydrostatic equation
dp
GM (r)ρ
=−
;
dr
r2
supplemented by the equation of state,
p=
dM (r)
= 4πr2 ρ;
dr
ρ
kT.
µmp
(6.1)
(6.2)
The thermal equilibrium is described by the radiation diffusion equation
1 d 2
r F = q(T, ρ);
r2 dr
F =−
c dU
16σT 3 dT
=−
;
3κR dr
3κR (T, ρ) dr
(6.3)
where q(T, ρ) is the thermonuclear energy release rate. In some cases, the convection energy
transfer should also be taken into account.
The structure of the star may be calculated by solving numerically the above set of equations.
Simple scalings could be obtained by replacing d/dr by 1/R, where R is the star’s radius. Then
eqs. (6.1) and (6.2) yield an estimate for the central temperature of the star
kTc ∼
µmp GM
.
R
25
(6.4)
For M = MJ ; R = RJ , one gets an estimate Tc ∼ 2 · 107 K.
The virial theorem provides a useful relation between the average kinetic and potential energies for a bound system of particles with binary interaction of the form U ∝ rn ; namely,
2hkinetic energyi = nhpotential energyi. For stars, n = −1 and the kinetic energy is in fact
the thermal energy, which yields
Etot = Eth + Ep = −Eth ,
(6.5)
R
R
where Eth = 23 (ρ/µmp )kT · 4πr2 dr; Ep = ρGM (r) · 4πrdr. Note that in this specific case, the
virial theorem could be obtained by multiplying eq. (6.1) by 4πr3 and integrating the lhs by parts.
It follows immediately from eq. (6.5) that the total energy of the star is negative, as it should be
for a bound body. When the star looses energy (e.g. for radiation), the thermal energy increases
so that the heat capacity of the star is negative. This implies stability of the star.
If the gravitational and pressure forces are out of balance, the star contracts or expands. The
characteristic dynamical time-scale is the free-fall time
s
r
R
R3
=
.
(6.6)
td =
GM/R2
GM
p
Any density perturbation propagates with the sound velocity, vs ≈
kT /µmp . Taking into
account the estimate (6.4), one finds that the sound crossing time, R/vs , is reduced to td . For the
Sun, td = 20 min.
The ratio of the total energy of a star to the rate of loss of energy from its surface is called the
thermal (Kelvin-Helmholtz) time-scale,
tth =
GM 2
.
RL
(6.7)
For the Sun, tth = 3 · 107 y.
Star’s luminosity may be estimated from eq. (6.3) as
L = 4πR2 F ∼ R2
σTc4
RTc4
∝
.
κR R
κR
(6.8)
If opacity is dominated by the Thomson scattering, κR = σT ρ/mp , eqs. (6.4) and (6.8) yield the
mass-luminosity relation in the form
L∝
M4
∝ M 3.
R3 ρ
(6.9)
If opacity is dominated by the bound-free or free-free absorption (3.39, 4.21), κ ∝ ρ2 /Tc3.5 , one
finds
15/2
RTc
R7 (M/R)15/2
M 11/2
L∝
∼
=
∝ M 5 Tc1/2 .
(6.10)
ρ2
M2
R1/2
A strong dependence of the thermonuclear energy release rate on the temperature implies that
the central temperature of stars does not vary much; therefore one can take L ∝ M 5 .
26
The ratio of the Thomson to the bound/free-free opacity goes as
3
R
T 3.5
T 3.5 R3
κT
3.5
∝
∼
=T
M 2 ∝ T 1/2 M 2
κbf
ρ
M
M
(6.11)
therefore in massive, high luminosity stars, opacity is dominated by the Thomson scattering.
The above estimates of the mass-luminosity relations show that the luminosity grows rapidly
with the mass of the star. The amount of nuclear fuel available is ∝ M therefore the life time of
the star, t ∝ M/L, decreases with the star mass. The life time of the Sun is about 10 Gyr. The
life time of very massive stars could be as small as a few Myr.
The radiation pressure is pr = U/3 therefore the luminosity may be presented as
L = 4πr2 F =
4πc 2 dpr
r
.
κR dr
(6.12)
If the hydrostatic equilibrium (6.1) is maintained by the radiation pressure, star’s luminosity is
L=
4πcGM ρ
.
κR
(6.13)
No equilibrium is possible at higher luminosities. In high-luminosity stars, κR = ρσT therefore the
limiting luminosity, which is called the Eddington limit, is
LEdd =
M
M
4πcGM
= 1.3 · 1038 J erg/s = 3 · 104 J LJ .
σT
M
M
(6.14)
Comparing with the mass-luminosity relation (6.9), one sees that there is a maximal mass of star;
numerically, Mmax ≈ 100MJ .
6.2
Degenerate stars
Degenerate gas. The Fermi momentum is pF = ~(3π 2 n)1/3 . For non-relativistic or highly
relativistic particles we have, E = p2 /2m or E = pc, respectively. This yields the Fermi energy
~2
(3π 2 n)2/3 ;
EF mc2 ;
2m
EF =
(6.15)
2
1/3
c~(3π n) ;
EF mc2
The degenerate gas becomes relativistic when pF = mc, which occurs at the number density
1 mc 3
.
(6.16)
nr = 2
3π
~
For the electron gas, nr ≈ 1030 , cm−3 .The corresponding mass density is ρr ≈ 106 g/cm3 .
The pressure is estimated as p ∼ vF pF n. The exact calculation yields
(
(3π 2 )2/3 ~2 5/3
n ;
EF mc2 ;
5
m
p=
2
1/3
(3π )
c~n4/3 ;
EF mc2 .
4
Minimum mass of the star
27
(6.17)
In the course of contraction of the protostellar object, the internal temperature increases. At
the late stage, the star is opaque and the radiation cooling time exceeds the free-fall time; then
the contraction is quasistatic so that one can estimate the central temperature from eq. (6.4):
1/3
G 3ρ
GM mp
∼
M 2/3 mp .
(6.18)
kTc ∼
2R
2 4π
The contraction stops either when the thermonuclear reactions start or when the gravity is balanced by the pressure of degenerate electrons. In the last case, no star is formed. Let us estimate
the minimal mass of the star. In the non-relativistic case, the Fermi energy (6.15) grows with the
density faster than Tc ,
ρ1/3
EF
∝ 2/3 .
(6.19)
kTc
M
Therefore the protostar becomes degenerate when
2/3
(3π 2 )2/3 ~2
ρ
.
(6.20)
kTc = EF =
2me
mp
Eliminating ρ from eqs. (6.18) and (6.20), one finds the temperature when the matter becomes
degenerate
8/3
mp me G2
kTc ∼ 7/3 2 2 M 4/3 .
(6.21)
2 π ~
If this temperature is not sufficient for the thermonuclear hydrogen burning, a brown dwarf is
formed, which only radiates away the energy stored in the course of gravitational contraction1 .
The minimal temperature necessary for pp reaction may be estimated as (see eq. (5.34)) kTmin ∼
10−3 E0 = 0.45 keV = k(5 · 106 K) . Substituting this quantity instead of kTc into eq. (6.21) yields
an estimate for the minimum mass of the star
7/3 4 4 3/4
7/3 4 2 2 2 3/4
2 π e
2 π α c~
mp =
mp = 3 · 1032 g = 0.15MJ .
(6.22)
Mmin ∼
3
3
2
10 mp me G
103 m3p me G2
According to detailed calculations, Mmin = 0.08MJ .
White dwarfs and neutron stars.
One sees that in degenerate stars, one can use the polytropic equation of state, p = KρΓ , where
Γ = 5/3 at EF mc2 and Γ = 4/3 at EF mc2 . Let us introduce the dimensionless variables
x = r/R∗ ;
w=
ρR∗3
;
M∗
u = M/M∗ ;
(6.23)
where R∗ and M∗ are the radius and the mass of the star, respectively. Then the hydrostatic
equations (6.1) are written as
dwΓ
awu
=− 2 ;
dx
x
du
= 4πwx2 ;
dx
(6.24)
(6.25)
1
In massive brown dwarfs, some small amount of energy is released due to fusion of lithium and deuterium
initially presented in the matter. Taking into account that fusion of these nuclei does not require weak interaction,
it could occur at relatively low temperatures
28
where a = GM∗2−Γ R∗3Γ−4 /K is the dimensionless coefficient. The boundary conditions are
u(0) = 0;
u(1) = 1;
w(1) = 0.
(6.26)
The second order set of equations could satisfy three boundary conditions only for a specially
chosen constant a0 (Γ), which could be found numerically. Then one finds the mass-radius relation
R∗ = [a0 (Γ)K/G]1/(3Γ−4) M −(2−Γ)/(3Γ−4) .
(6.27)
In white dwarfs, the equilibrium is maintained by the pressure of the degenerate electron gas.
Since they are formed from already evolved star cores composed from intermediate mass elements,
like C and O, ρ = (A/Z)mp n ≈ 2mp n. For non-relativistic electrons, the mass-radius relation
(6.27) is written as
32/3 π 4/3 a0 (5/3)~2 −1/3
M
.
(6.28)
R∗ =
5(2mp )5/3 me G
One sees that the radius of the white dwarf decreases with the mass so that the density grows as
m5p G3 2
M∗
ρ∼ 3 ∼
M .
R∗
~6
(6.29)
Eventually the degenerate gas becomes relativistic. The transition occurs at the density (6.16)
when the mass of the star reaches
(~c)3/2
Mr ∼ 3/2 2 .
(6.30)
G mp
When the gas becomes relativistic, the equation of state changes, Γ = 4/3. The mass-radius
relation now reduces to (cf. with Mr )
MCh =
31/2 π[a0 (4/3)~c]3/2
.
32G3/2 m2p
(6.31)
This is the Chandrasekhar mass, the maximal possible mass of the white dwarf. For larger
masses, the pressure of degenerate electrons is unable to balance the gravity force therefore the
star collapses into the neutron star or black hole. Numerically MCh = 1.4MJ .
In neutron stars, the gravity force is balanced by the pressure of degenerate neutrons. If the
neutron matter could be considered as an ideal gas, one would get a similar expression for the
critical mass (note that the expression for Chandrasekhar mass is independent of the electron
mass, i.e. of the mass of particles providing the pressure. However, the neutron matter is a fluid.
Depending on the equation of state of the neutron fluid, this mass is estimated as 2 − 3MJ . A
stellar remnant of a larger mass inevitably collapses into a black hole.
29