Density of states and Fermi energy.

1 Density of states and Fermi energy. Rohlf Ch. 12 Number of states up to E:N=V6 22m 2 3/2E3 of states : dNdE=Vm3/22 2 3E1/2 Fermi energy: EF=(3/ )2/3h28mne2/3 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=34034/34 2()2/3h2meR2Ne5/3=CNe5/3R2EG= 35GM2R= 35 GmN2NN2R= BNN2R Use the following data to find B and C for the 10 27 kg me= 1031 kgG= 10 11 m3kg-1s 2M =2 1030kg NN= 1057 Ne= 1057 h= 10 34 J-sC= 10 38 B= 10 64E=Ee+EG=CNe5/3R2 BNN2 RdEdR= 2 CNe5/3R3+BNN2R2=0 R=2 CNe5/3 BNN2= 103 kmEe=CN5/3R2EG= BNN2 RFind the equilibrium radius by mimimizing the energy with respect to R.

2 Density and gravitational energy of white dwarf M V EG= BNN2R Fermi energy and zero-point energy of electrons:Ee=CNe5/3R2EF=53 EeNeData:R= 106 mM =2 1030kg NN= 1057 Ne=NN/2 Exercises R= 106 mM =2 1030kg Ne=NN/2= 1057 Density of white dwarf 2 1030kg 43 106()3m3= 109kg-m-3= 106 gm-cm-3 Fermi Energy of electrons:EF=53 EeNeEe=CNe5/3R2= 1042J= 1061eVEF=53 CNe2/3R2= 10 38() 1057()2 106()2= 10 14J = .194 106eV non relativistic kinematics is becoming invalid. R=Ne5/3N2NR=Ne5/3N2NN2nWNeW5/3RW=105/310 2RW=10 1/3RW= Ee= Ne5/3RW2 NeW5/3R2 EeW =105 1061= 1063 eVEF= 53 EeNe= 1063()36 1057()= 106 eV 3mec2 What about more massive stars: assume M=10M Volume up to k vk= 6k3. Volume per state vs= of states up to k: N= vkvs =L36 of states up to E: k2=p2 2=E2 me2c4c2 2 NV=k36 2=16 2E2 me2c4()3/2c3 3At T=0, electrons fill all states , 2 per state, to the Fermi energy NeV=EF2 m2c4()3/23 2c3 3 Relativistic number of states up to k.

3 Fermi energy.! Density of states : 1 VdNdE=dndE=8 E2 m2c4()1/2Ec3h3 E>>m 8 3c3h3E2dnedE= 8 c3h3E2 Relativistic number of states up to k. Fermi energy.!Total energy = NeE=NeEdNedEdE0Ef dNedEdE0Ef =NeE3dE0Ef E2dE0Ef =Ne34 EFTotal 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. dNedE=8 Vc3h3E2 EF=3 2()1/3 cNeV 1/3 Total energy = NeE=Ne34EF Total zero point energy Ee=Ne34 EeFEeF=3 2()1/3 cNeV 1/3= 10 26 NeV 1/3Ee= 10 26()NeV 1/3= 10 26()43 R3 1/3Ne4/3= 10 26Ne4/3R= Ne4/3R=( )(.466)( 10 34)(3 108)2( )Ne4/3 REe= 10 26Ne4/3 RCompare Fermi and gravitational energies.

4 Gravitational energyEG= 35GM2R= 35 GmN2NN2R= 35 10 11() 10 27()2NN2R= 10 64NN2R= REe=Ne4/3R= Ne4/3R EG= 10 64NN2R= NN2RE=Ee+EG= Ne4/3R NN2R= Ne4/3 NN2 RMinimize the energy:dEdR= Ne4/3R2+ NN2R2=1R2 Ne4/3+ NN2()= is no stable minimum! dEdR=1R2 Ne4/3+ NN2()= NN2> Ne4/3the energy always decreases R becomes small. Gravity always wins out over the Fermi energy and the star collapses. 10 64NN2> 10 26Ne4/3 Ne=12 NNNN2/3> 1038 NN> 1057. N 1057M For a more accurate measure, should not assume E>> more careful calculation gives the Chandrasekhar mass M Compare a white dwarf s energy with a neutron star. decay: n p+e+ mn mp me MeVInverse decay (electron capture)e+p n+ requires minumum electron 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.

5 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=3h240mN3 V 2/3NN5/3=3h240mN94 2 2/3NN5/3R2=CNNN5/3R2EG= 35GM2R= 35 GmN2NN2R= BNNN2R Use the following data to find Bn and Cn for the neutron 10 27 kg me= 1031 kgG= 10 11 m3kg-1s 2M =2 1030kg NN= 1057 h= 10 34 J-sCn= 10 42 Bn= 10 64 R=2 CNNN5/3 BNNN2= 103 m= kmEnergetics of neutron star. Exercise Verify this result Schwartzschild solution to Einstein's gravitational field 2 GMc2r dt2 1 2 GMc2r 1dr2 r2d 2 r2sin2 d 2 For M 0, ds2=c2dt2 dr2 r2d 2 r2sin2 d 2=c2dt2 dx2 dy2 dz2 flat spacetime.

6 Radial motion: ds2=c21 2 GMc2r dt2 1 2 GMc2r 1dr2 Light: ds2=0,dr=c1 2 GMc2r dtdrdt=c1 RSr RS=2 GMc2RS=Schwartzschild r=RS drdt=0 When r<RS drdtis always negative. Black =2 GMc2=2 10 11 2 10303 108()2 3 103=3kmFor several solar masses, RS crosses neutron star dr1 2 GMc2r= dr1 RSrSchwartzschild radius: r=2 GMc2(event horizon)World line of object falling toward large mass. (Eddington-Finkelstein coords.) Radial motion: ds2=c21 2 GMc2r dt2 1 2 GMc2r 1dr2c2dt2=ds21 2 GMc2r 1+1 2 GMc2r 2dr2cdt= dr1 2 GMc2r= dr1 RSrcdt=t0t t t0= dr1 RSr=r0r r r0+RSlnr RSr0 RS ct= r+RSlnr RS+Const()