Density of states and Fermi energy. Rohlf Ch. 12 Sec.12.6

Number of states up to E : N =V

6π 22m2

⎛⎝⎜

⎞⎠⎟

3/2E3/2 .

Density of states:            dNdE

=Vm3/2

2π 23 E1/2

Fermi energy:                     EF =(3 /π )

2/3h2

8mne

2/3

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.

Ee =340

34/3

4π 2( )2/3h2

meR2Ne5/3 = CNe

5/3

R2EG = − 3

5GM 2

R= − 3

5GmN

2 NN2

R= − BNN

2

R

Use the following data to find B and C for the Sun.

mN = 1.67 ×10−27 kg me = 9.1×1031 kg G = 6.67 ×10−11 m3kg-1s−2

M = 2 ×1030 kg NN = 1.2 ×1057 Ne = 0.6 ×1057     h = 6.63×10−34 J-s

C = 1.36 ×10−38                  B = 1.12 ×10−64

E = Ee + EG = CNe5/3

R2 − BNN2

R

dEdR

= − 2CNe5/3

R3 + BNN2

R2 = 0

  R = 2CNe5/3

BNN2 = 7.2 ×103 km

Ee = CN 5/3

R2EG = − BNN

2

R

Find the equilibrium radius by mimimizing the energy with respect to R.

Density and gravitational energy of white dwarf

ρ MV

        

EG = − BNN2

R      

Fermi energy and zero-point energy of electrons:

Ee = CNe5/3

R2

EF = 53EeN e

Data: R = 7.2 ×106 m M = 2 ×1030 kg NN = 1.2 ×1057 Ne = NN / 2     

Exercises

R = 7.2 ×106 m M = 2 ×1030 kg Ne = NN / 2 = 0.6 ×1057

Density of white dwarf

ρ 2 ×1030 kg 43π 7.2 ×106( )3 m3

= 1.28 ×109 kg-m-3 = 1.28 ×106 gm-cm-3

Fermi Energy of electrons:

EF = 53EeN e

Ee =CNe

5/3

R2 = 3.5×1042 J = 2.2 ×1061 eV

EF = 53CNe

2/3

R2 = 53

1.36 ×10−38( ) 0.6 ×1057( )2/3

7.2 ×106( )2= 3.1×10−14 J = .194 ×106 eV

⇒non relativistic kinematics is becoming invalid.

R = Ne5/3

N 2N

R = Ne5/3

N 2N

N 2nWNeW5/3 RW = 10

5/3

102RW = 10−1/3RW = 0.46RW

Ee = CNe5/3

R2        Ee =   Ne

5/3RW2

NeW5/3R2

EeW   =105/3

0.4622.2 ×1061 = 4.8 ×1063  eV

EF = 53EeN e

= 5 4.8 ×1063( )3 6 ×1057( ) = 1.3×106  eV   3mec

2

What about more massive stars: assume M = 10M

Volume up to k vk =π6k3.     Volume per state vs=π 3

L3 .

Number of states up to k : N= vkvs

= L3

6π 2 k3.

Number of states up to E :

k2 =p2

2 =E2 − me

2c4

c22                NV

=k3

6π 2 =1

6π 2E2 − me

2c4( )3/2

c33

At T=0, electrons fill all states, 2 per state, to the Fermi energy

NeV

=EF

2 − m2c4( )3/2

3π 2c33

Relativistic number of states up to k. Fermi energy.!

Density of states:

1VdNdE

=dndE

=8πE2 − m2c4( )1/2

E

c3h3     E >> m    8π

3c3h3 E2

dnedE

=   8πc3h3 E

2

Relativistic number of states up to k. Fermi energy.!

Total energy = Ne E = Ne

E dNe

dEdE

0

E f

∫dNe

dEdE

0

E f

∫= Ne

E3 dE0

E f

∫

E2dE

0

E f

∫= Ne

34EF

Total zero point electron energy:

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.

dNe

dE=8πVc3h3

E2

EF = 3π 2( )1/3 c Ne

V⎛⎝⎜

⎞⎠⎟1/3

Total energy = Ne E = Ne34EF

Total zero point energy Ee = Ne34EeF

EeF = 3π 2( )1/3c Ne

V⎛⎝⎜

⎞⎠⎟1/3

= 9.75 ×10−26 NeV

⎛⎝⎜

⎞⎠⎟1/3

Ee = Ne34

9.75 ×10−26( ) NeV

⎛⎝⎜

⎞⎠⎟1/3

=34

9.75 ×10−26( )43πR3⎛

⎝⎜⎞⎠⎟

1/3 Ne4/3

= 4.54 ×10−26 Ne4/3

R= α Ne

4/3

R

=(6.24)(.466)(6.6 ×10−34 )(3×108 )

2(6.35)Ne

4/3

R

Ee = 4.5 ×10−26 Ne4/3

R

Compare Fermi and gravitational energies.

Gravitational energy

EG = −35GM 2

R= −

35GmN

2 NN2

R

= −35

⎛⎝⎜

⎞⎠⎟

6.67 ×10−11( ) 1.67 ×10−27( )2 NN2

R= −1.1×10−64 NN

2

R= −

βR

Ee =Ne

4/3

R= α Ne

4/3

R EG = −1.1×10−64 NN

2

R= −β NN

2

R

E = Ee + EG = α Ne4/3

R− β NN

2

R=

αNe4/3 − βNN

2

R

Minimize the energy:

dEdR

= −αNe

4/3

R2 +βNN

2

R2 =1R2 −αNe

4/3 + βNN2( ) = 0.

There is no stable minimum!

dEdR

=1R2 −αNe

4/3 + βNN2( ) = 0.

IF βNN2 >αNe

4/3 the energy always decreases R becomes small.

Gravity always wins out over the Fermi energy and the star collapses.

−1.1×10−64 NN2 >4.5 ×10−26Ne

4/3            Ne =12NN

NN2/3 > 1.6 ×1038      NN > 2.1×1057. 

N ≈ 1.2 ×1057 M 1.7 MFor a more accurate measure, should not assume E >> mc2 .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 −mp −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 aproton and turn into a neutron, emitting a neutrino. Then, the number of electrons in the star maybegin to reduce, and this speeds up the process as the Fermi energy increases, until all the electronshave been used up. With the reduction in electron Fermi pressure the star collapses under gravityuntil balanced by the increasing Fermi pressure of the nucleons.

En =3h2

40mN

3πV

⎛⎝⎜

⎞⎠⎟

2/3NN5/3 = 3h2

40mN

94π 2

⎛⎝⎜

⎞⎠⎟

2/3NN5/3

R2= CNNN

5/3

R2

EG = − 35GM 2

R= − 3

5GmN

2 NN2

R= − BNNN

2

R

Use the following data to find Bn and Cn for the neutron star.

mN = 1.67 ×10−27 kg me = 9.1×1031 kg G = 6.67 ×10−11 m3kg-1s−2

M = 2 ×1030 kg NN = 1.2 ×1057 h = 6.63×10−34 J-s

Cn = 7.3×10−42                  Bn = 1.12 ×10−64

 R = 2CNNN5/3

BNNN2 = 12.3×103 m = 12.3 km

Energetics of neutron star.

Exercise Verify this result

Schwartzschild solution to Einstein's gravitational field equation.

ds2 = c2 1− 2GMc2r

⎛⎝⎜

⎞⎠⎟dt2 − 1− 2GM

c2r⎛⎝⎜

⎞⎠⎟−1dr2 − r2dθ2 − r2 sin2 θdφ2

For M → 0,

ds2 = c2dt2 − dr2 − r2dθ2 − r2 sin2 θdφ2

= c2dt2 − dx2 − dy2 − dz2     →       flat spacetime.

Radial motion: ds2 = c2 1− 2GMc2r

⎛⎝⎜

⎞⎠⎟dt2 − 1− 2GM

c2r⎛⎝⎜

⎞⎠⎟−1dr2

Light: ds2 = 0,

dr = c 1− 2GMc2r

⎛⎝⎜

⎞⎠⎟dt dr

dt= c 1− RS

r⎛⎝⎜

⎞⎠⎟

RS =2GMc2

RS = Schwartzschild radius.

When r = RS drdt

= 0     When r < RS drdt

is always negative. → Black hole.

RS =2GMc2 =

2 × 6.7 ×10−11 × 2 ×1030

3×108( )2≈ 3×103 = 3km

For several solar masses, RS crosses neutron star RN .

cdt = ± dr

1− 2GMc2r

= ±dr

1− RSr

Schwartzschild radius: r = 2GMc2

(event horizon)

World line of object falling toward large mass. (Eddington-Finkelstein coords.)

Radial motion: ds2 = c2 1− 2GMc2r

⎛⎝⎜

⎞⎠⎟dt2 − 1− 2GM

c2r⎛⎝⎜

⎞⎠⎟−1dr2

c2dt2 = ds2 1− 2GMc2r

⎛⎝⎜

⎞⎠⎟−1

+ 1− 2GMc2r

⎛⎝⎜

⎞⎠⎟−2dr2

cdt = ± dr

1− 2GMc2r

= ±dr

1− RSr

cdt =t0

t

∫ t − t0 = ±dr

1− RSr

=r0

r

∫ ± r − r0 + RS lnr − RSr0 − RS

⎛

⎝⎜⎞

⎠⎟

ct = ± r + RS ln r − RS + Const( )