Density of states and Fermi energy.
Rohlf Ch. 12 Sec.12.6
V ⎛ 2m ⎞
Number of states up to E : N =
⎜
⎟
6π 2 ⎝  2 ⎠
3/2
E 3/2 .
dN
Vm 3/2 1/2
Density of states: =
E
2
3
dE
2π 
2/3
(3 / π ) h 2 2/3
Fermi energy: EF =
ne
8m
Application of zero point energy
to astrophysics.
Some aspects of the structure of a star may be understood by considering the opposing forces of gravitational energy,
which makes the star become smaller and the Fermi energy,
which makes the star become larger.
3
Ee =
40
34/3
( 4π )
2 2/3
h2
CN e5/3
5/3
N
=
2 e
me R
R2
2 2
3 GM 2
3 GmN
NN
BN N2
EG = −
=−
=−
5 R
5
R
R
Use the following data to find B and C for the Sun.
mN = 1.67 × 10 −27 kg
M  = 2 × 10 30 kg
me = 9.1× 10 31 kg
N N = 1.2 × 10 57
G = 6.67 × 10 −11 m 3kg-1s −2
N e = 0.6 × 10 57 h = 6.63 × 10 −34 J-s
C = 1.36 × 10 −38 B = 1.12 × 10 −64
Find the equilibrium radius by mimimizing the energy with respect to R.
Ee =
CN 5/3
BN N2
EG = −
R
R2
E = Ee + EG =
CN e5/3
R2
BN N2
−
R
dE
2CN e5/3 BN N2
=−
+ 2 = 0
3
dR
R
R
R =
2CN e5/3
BN N2
= 7.2 × 10 3 km
Exercises
Density and gravitational energy of white dwarf
M
ρ
V
BN N2
EG = −
R
Fermi energy and zero-point energy of electrons:
Ee =
CN e5/3
R2
5 Ee
EF =
3 Ne
Data: R = 7.2 × 10 6 m
M  = 2 × 10 30 kg
N N = 1.2 × 10 57
N e = N N / 2 R = 7.2 × 10 6 m
M  = 2 × 10 30 kg
N e = N N / 2 = 0.6 × 10 57
Density of white dwarf
ρ
2 × 10 30 kg
(
4
π 7.2 × 10 6
3
)
3
m3
= 1.28 × 10 9 kg-m -3 = 1.28 × 10 6 gm-cm -3
Fermi Energy of electrons:
5 Ee
EF =
3 Ne
5
EF =
3
CN e2/3
2
R
Ee =
(
CN e5/3
R2
= 3.5 × 10 42 J = 2.2 × 10 61 eV
−38
)(
0.6 × 10
5 1.36 × 10
=
2
3
7.2 × 10 6
(
)
)
57 2/3
⇒ non relativistic kinematics is becoming invalid.
= 3.1× 10 −14 J = .194 × 10 6 eV
What about more massive stars: assume M = 10M 
R=
N e5/3
R=
N e5/3 N 2nW
N 2N
5/3
N 2N N eW
Ee =
CN e5/3
EF = R2
RW =
10 5/3
10 2
RW = 10 −1/3 RW = 0.46RW
2
N e5/3 RW
10 5/3
61
63
Ee = 5/3 2 EeW =
2.2
×
10
=
4.8
×
10
eV
2
0.46
N eW R
(
5 4.8 × 10 63
5 Ee
= 3 Ne
3 6 × 10 57
(
)
) = 1.3 × 10 eV  3m c
6
e
2
Relativistic number of states up to k. Fermi energy.!
π 3
π3
Volume up to k vk = k . Volume per state vs = 3 .
6
L
Number of states up to k :
vk
L3 3
N=
= 2k .
vs 6π
Number of states up to E :
2
k =
p2

2
=
E 2 − me2 c 4
c2 2
(
(
)
2 4 3/2
EF − m c
Ne
=
V
3π 2 c 3 3
)
2 4 3/2
N
k3
1 E − me c
=
=
2
V 6π
6π 2
c 3 3
At T=0, electrons fill all states, 2 per state, to the Fermi energy
2
2
Relativistic number of states up to k. Fermi energy.!
Density of states:
(
2
)
2 4 1/2
E −m c
1 dN dn
=
= 8π
V dE dE
c 3h 3
E
dne
8π
= 3 3 E 2
dE
c h

E >> m  8π
2
E
3c 3h 3
Total zero point electron energy:
Ef
Total energy = N e E = N
∫
dN e
E
dE
dE
0
e Ef
∫
0
dN e
dE
dE
Ef
=N
∫
0
e Ef
E 3 dE
∫E
0
2
dE
3
= N e EF
4
Repeat of previous analysis for relativistic electrons.
Homework: Verify the following. Read lecture notes
to be posted and compare with text for photons – p343-4. ( )
EF = 3π
2 1/3
⎛ Ne ⎞
c ⎜
⎝ V ⎟⎠
1/3
dN e 8π V 2
= 3 3E
dE c h
3
Total energy = N e E = N e EF
4
Compare Fermi and gravitational energies.
Total zero point energy Ee = N e
( )
EeF = 3π
2 1/3
⎛N ⎞
c ⎜ e ⎟
⎝ V ⎠
(
1/3
= 9.75 × 10
)
3
⎛N ⎞
Ee = N e 9.75 × 10 −26 ⎜ e ⎟
⎝ V ⎠
4
= 4.54 × 10
−26
3
E
4 eF
1/3
−26
(
(6.24)(.466)(6.6 × 10 −34 )(3 × 10 8 ) N e4/3
=
2(6.35)
R
Ee = 4.5 × 10
N e4/3
R
1/3
)
−26
3 9.75 × 10
4/3
=
N
e
1/3
4 ⎛4
3⎞
⎜⎝ π R ⎟⎠
3
N e4/3
N e4/3
=α
R
R
−26
⎛ Ne ⎞
⎜⎝
V ⎟⎠
Gravitational energy
3 GM 2
3 Gm N2 N N2
EG = −
= −
5 R
5
R
(
)(
⎛ 3⎞
= − ⎜ ⎟ 6.67 × 10 −11 1.67 × 10 −27
⎝ 5⎠
)
2
2
N N2
β
−64 N N
= − 1.1 × 10
= −
R
R
R
N e4/3
Ee =
R
N e4/3
= α
R
EG = −1.1 × 10
N e4/3
N N2
α N e4/3 − β N N2
E = Ee + EG = α
−β
=
R
R
R
Minimize the energy:
(
)
α N e4/3 β N N2
dE
1
4/3
2
=−
+
=
−
α
N
+
β
N
= 0.
e
N
2
2
2
dR
R
R
R
There is no stable minimum!
−64
N N2
N N2
= −β
R
R
(
)
dE
1
= 2 −α N e4/3 + β N N2 = 0.
dR R
IF β N N2 > α N e4/3 the energy always decreases R becomes small.
Gravity always wins out over the Fermi energy and the star collapses.
−1.1 × 10 −64 N N2 > 4.5 × 10 −26 N e4/3 N e =
1
NN
2
N N2/3 > 1.6 × 10 38 N N > 2.1 × 10 57. N  ≈ 1.2 × 10 57
M  1.7 M 
For a more accurate measure, should not assume E >> mc 2 .
A more careful calculation gives the Chandrasekhar mass M  1.4 M 
Compare a white dwarf’s energy with a neutron star.
β decay: n → p + e + ν
mn − m p − me ≈ 0.8 MeV
Inverse β decay (electron capture) e + p → n + ν requires minumum electron energy.
In a nucleus it may be energetically favorable for an inner atomic electron to be captured by a
proton and turn into a neutron, emitting a neutrino. Then, the number of electrons in the star may
begin to reduce, and this speeds up the process as the Fermi energy increases, until all the electrons
have been used up. With the reduction in electron Fermi pressure the star collapses under gravity
until balanced by the increasing Fermi pressure of the nucleons.
Energetics of neutron star.
2
3h ⎛ 3 ⎞
En =
⎜
⎟
40mN ⎝ π V ⎠
2/3
2
⎛ 9 ⎞
40mN ⎜⎝ 4π 2 ⎟⎠
3h
N N5/3 =
2/3
N N5/3
R
2
=
C N N N5/3
R2
2 2
3 GM 2
3 GmN
NN
BN N N2
EG = −
=−
=−
5 R
5
R
R
Use the following data to find Bn and Cn for the neutron star.
m N = 1.67 × 10 −27 kg
M  = 2 × 10 30 kg
me = 9.1× 10 31 kg
N N = 1.2 × 10 57
G = 6.67 × 10 −11 m 3kg-1s −2
h = 6.63 × 10 −34 J-s
Cn = 7.3 × 10 −42 Bn = 1.12 × 10 −64
R =
2C N N N5/3
BN N N2
= 12.3 × 10 3 m = 12.3 km
Exercise Verify this result
Schwartzschild solution to Einstein's gravitational field equation.
2GM ⎞ 2 ⎛
2GM ⎞
2
2⎛
ds = c ⎜ 1 − 2 ⎟ dt − ⎜ 1 − 2 ⎟
⎝
⎝
c r ⎠
c r ⎠
−1
dr 2 − r 2 dθ 2 − r 2 sin 2 θ dφ 2
For M → 0,
ds 2 = c 2 dt 2 − dr 2 − r 2 dθ 2 − r 2 sin 2 θ dφ 2
= c 2 dt 2 − dx 2 − dy 2 − dz 2 → flat spacetime.
2GM ⎞
2GM ⎞
⎛
⎛
Radial motion: ds 2 = c 2 ⎜ 1 − 2 ⎟ dt 2 − ⎜ 1 − 2 ⎟
⎝
⎝
c r ⎠
c r ⎠
−1
dr 2
Light: ds 2 = 0,
2GM ⎞
⎛
dr = c ⎜ 1 − 2 ⎟ dt
⎝
c r ⎠
R ⎞
dr
⎛
= c ⎜1 − S ⎟
⎝
dt
r ⎠
RS =
2GM
c2
RS = Schwartzschild radius.
dr
= 0 When r < RS
dt
When r = RS
RS =
2GM
c2
=
2 × 6.7 × 10 −11 × 2 × 10 30
(
3 × 10 8
)
2
dr
is always negative. → Black hole.
dt
≈ 3 × 10 3 = 3km
For several solar masses, RS crosses neutron star RN .
cdt = ±
1−
dr
=±
2GM
c2 r
dr
R
1− S
r
World line of object falling
toward large mass. (Eddington-Finkelstein coords.)
Schwartzschild radius: r =
(event horizon)
2GM
c2
2GM ⎞
2GM ⎞
⎛
⎛
Radial motion: ds = c ⎜ 1 − 2 ⎟ dt 2 − ⎜ 1 − 2 ⎟
⎝
⎝
c r ⎠
c r ⎠
2
2
2GM ⎞
⎛
c dt = ds ⎜ 1 − 2 ⎟
⎝
c r ⎠
2
2
2
−1
2GM ⎞
⎛
+ ⎜1 − 2 ⎟
⎝
c r ⎠
−2
dr 2
−1
dr 2
cdt = ±
t
1−
dr
=±
2GM
c2 r
dr
R
1− S
r
r
⎛
r − RS ⎞
dr
∫ cdt =t − t0 = ± ∫ RS = ± ⎜⎝ r − r0 + RS ln r0 − RS ⎟⎠
t0
r0 1 −
r
ct = ± ( r + RS ln r − RS + Const )