Transcript Fermi Gases

Nuclei
• Protons and neutrons in nuclei separately fill their
energy levels: 1s, 1p, 1d, 2s, 2p, 2d,
3s…………… (we’ll see in 461 their ordering and
split by total J)
• often easier to analyze as 2 Fermi gases of
(mostly) non-interacting particles
• density ~1/fm3 slightly higher for neutrons large A
p  # p 4
3 R
3
Nuc
n  # n 4
3

R
Nuc
3
• proton shifted higher due to Coulomb repulsion.
Both p,n fill to top with p<->n coupled by Weak
interactions so both at ~same level (Fermi energy
for p impacted by n)
n  p  e  
p  n  e  
p  e  n 
n
p
U 238  92 p,146n
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Nuclei
• gives Fermi momentum ~same for all except H
pF  2mEF  200 MeV / c
• if p,n were motionless, then the energy thresholds
for some neutrino interactions are:
e  p  n  e 
E  1.8 MeV
   n     p E  120 MeV
• but Fermi momentum allows reactions to occur at
lower neutrino energy.
  p  n  e
Ethrsh  1.8MeV
mp
Ep  pp
 1.4MeV
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Nuclei: Fermi Suppression
• But also have filled energy levels and need to give
enough energy to p/n so that there is an unfilled
state available. Simplest to say “above” Fermi
Energy
• similar effect in solids. Superconductivity mostly
involves electrons at the “top” of the Fermi well
e  p  n  e 
  n     p
  e   e
•
at low energy transfers (<40 MeV) only some p/n
will be able to change states. Those at “top” of
well.
• Gives different cross section off free protons than
off of bound protons.
• In SN1987, most observed events were from
antineutrinos (or off electrons) even though (I
think) 1000 times more neutrinos. Detectors were
water…..
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Fermi Gases in Stars
• Equilibrium: balance between gravitational
pressure and “gas” (either normal or degenerate)
pressure
• total gravitational Energy:
GM (r )  Mass (r  r )
E 

r
R
r
 (r )
2
E G
4r  dr  4r 2  (r )dr
0
0
r
E
3 M 2G
pressure  

  cons tan t
V
5 R
• density varies in normal stars (in Sun: average is 1
g/cm3 but at r=0 is 100 g/cm3). More of a constant
in white dwarves or neutron stars
• will have either “normal” gas pressure of P=nkT
(P=n<E>) or pressure due to degenerate particles.
Normal depends on T, degenerate (mostly) doesn’t
• n = particle density in this case
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Degenerate Fermi Gas Pressure
• Start with p = n<E>
• non-relativistic
relativistic
density states : D( p )  p 2 both cases
D ( E )  E 1/ 2
N 
EF
0
 E2
1/ 2
AE dE
 EF  n 2 / 3
E 

EF
0
AE 2 dE
 n1/ 3
1/ 2
E

E
dE

1/ 2
E
 dE
E  53 E F  K1n 2 / 3
P  K1 n  n 2 / 3  n 5 / 3
2
E

E
dE

2
E
 dE
 34 E F  K 2 n1/ 3
 K 2 n  n1/ 3  n 4 / 3
• P depends ONLY on density
• Pressure decreases if, for a given density, particles
become relativistic
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Older Sun-like Stars
• Density of core increases as H-->He. He inert (no
fusion yet). Core contracts
• electrons become degenerate. 4 e per He nuclei.
Electrons have longer wavelength than He
Ee  EHe thermal equilibriu m
 pe 
me
mHe
pHe  e 
mHe
mH
He
• electrons move to higher energy due to Pauli
exclusion/degeneracy. No longer in thermal
equilibrium with p, He nuclei
• pressure becomes dominated by electrons. No
longer depends on T
Ptotal  Pe  PH  PHe
Pe  PH  PHe
• allows T of p,He to increase rapidly without
“normal” increase in pressure and change in star’s
equilibrium.
• Onset of 3He->C fusion and Red Giant phase
(helium flash when T = 100,000,000 K)
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White Dwarves
• Leftover cores of Red Giants made (usually) from
C + O nuclei and degenerate electrons
• cores of very massive stars are Fe nuclei plus
degenerate electrons and have similar properties
• gravitational pressure balanced by electrons’
pressure which increases as radius decreases --->
radius depends on Mass of star
• Determine approximate Fermi Energy. Assume
electron density = 0.5(p+n) density
N M Sun 1
1

 5  1035 e / m 3
V
m p 2 volume Earth
E F (non  rel ) 
E F (rel )  hc
 
h2 N 3 2 / 3
8m V 

N 3 1/ 3
V 8
 0.3 MeV
 0.8 MeV
• electrons are in this range and often not completely
relativistic or non-relativistic---> need to use the
correct E2 = p2 + m2 relationship
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White Dwarves + Collapse
• If the electron energy is > about 1.4 MeV can have:
e  p    n EThreshold  1.4MeV
• any electrons > ET “disappear”. The electron
energy distribution depends on T (average E)
#e’s
EF
ET
• the “lost” electrons cause the pressure from the
degenerate electrons to decrease
• the energy of the neutrinos is also lost as they
escape ---> “cools” the star
• as the mass increases, radius decreases, and
number of electrons above threshold increases
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White Dwarves+Supernovas
• another process - photodisentegration - also abosrbs
energy “cooling” star. Similar energy loss as e+p
combination
  56Fe  134 He  4n
  He  2 p  2n
4
• At some point the not very stable equilibrium
between gravity and (mostly) electron pressure
doesn’t hold
• White Dwarf collapses and some fraction (20-50%
??) of the protons convert to neutrons during the
collapse
• gives Supernovas
LSN (light )  109  LSun (light )
LSN (neutrinos )  100(1000 ?)  LSN (light )
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Neutron Stars-approx. numbers
• Supernovas can produce neutron stars
- radius ~ 10 km
- mass about that of Sun. always < 3 mass Sun
- relative n:p:e ~ 99:1:1
• gravity supported by degenerate neutrons
N 2 M Sun

V
mn
1
44
3
3

6

10
/
m

.
6
/
F
4  (10 km) 3
3
separation  1.2 F
•
plug into non-relativistic formula for Fermi Energy
-----> 140 MeV (as mass =940 MeV, non-rel OK)
• look at wavelength
h
h
1240MeVF
 

 2F
p
2mE
2  940MeV 140MeV
• can determine radius vs mass (like WD)
• can collapse into black hole
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Neutron Stars
• 3 separate Fermi gases: n:p:e p+n are in the same
potential well due to strong nuclear force
• assume independent and that p/n = 0.01 (depends
on star’s mass)
EF ( p)  EF (n) VN 
2/3
EF (e)  EF (n) VN 
 14525MeV  6MeV
2 / 3 mn
me
 14525MeV
940
.5
 10,000MeV ( wrong )
• need to use relativistic for electrons
EF (e)  

N 3 1/ 3
V 8
hc  500MeV
• but not independent as p <---> n
n  p  e 
p  n  e  
p  e  n 
n  e  p  
• plus reactions with virtual particles
• free neutrons decay. But in a neutron star they can
only do so if there is an available unfilled electron
state. So suppresses decay
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Neutron Stars
• Will end up with an equilibrium between n-p-e
which can best be seen by matching up the Fermi
energy of the neutrons with the e-p system
• neutrons with E > EF can then decay to p-e-nu
(which raises electron density and its Fermi energy
thus the balance)
E F (e)  E F ( p )  E F ( n )
 3n p 
 3ne 
2


 hc  m p c  
 8 
 8 
1/ 3
 3nn 


 8 
2/3
2/3
h2
2m p
h2
 mn c 2
2mn
• need to include rest mass energies. Also density of
electrons is equal to that of protons
• can then solve for p/n ratio (we’ll skip algebra)
• gives for typical neutron star:
  2 1017 kg / m 3 
nn  110 44 / m 3
ne  n p  nn / 200
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