equals the chemical potential µ at T = 0. All the lowest energy states are occupied. Highest occupied state has energy µ. For particles in a box:

Transcription

1 5 The Ideal Fermi Gas at Low Temperatures M 5, BS 3-4, KK p83-84) Applications: - Electrons in metal and semi-conductors - Liquid helium 3 - Gas of Potassium 4 atoms at T = 3µk - Electrons in a White Dwarf star - eutrons in a eutron star pulsar) 5 Ideal Fermi Gas at T = f!) = e! µ ) + As T!, f! ) = if! < µ, or f!) = if! > µ equals the chemical potential µ at T = All the lowest energy states are occupied Highest occupied state has energy µ For particles in a box:! =! k M, k = k x + k y + k z k x = l! L, k = m! y L, k = n! y L l m n umber of Running total particles up and down) ~ States form a cubic lattice in k-space of side! L States inside an octant of k-space of radius k f are filled 4 3!k 3 f = 8! L ) 3 where the is due to spin degeneracy, = s + for double occupancy of k space up and down)

2 The 8 is the positive octant k x, k y,k z > 4 3!k f! L 3 is the volume of the sphere ) 3 is the volume of k-space per k state k f = 3! 3 = Fermi wave number = L 3 = the volume of the box ) 3 where n = k f = 3! n = particle density The surface of a sphere is the boundary between occupied and unoccupied states = Fermi surface Energy of particles on the Fermi surfare = = Fermi energy =! k f M =! M 3 k) 3 = µ at T = = M f defines the Fermi velocity = the speed of the particles on the Fermi surface p f =!k f : defines the Fermi momentum 5 Alternative derivation of k f,! k)dk = k dk = the number of k states k! k + dk k f k k =! f k)dk = dk = k 3 3 ) = 3 k f 3 where the is once more due to spin degeneracy So: k f = 3! 3 as before Or work with energy: = g!)d! g!) = k)dk g!) = k) dk d!! =! k M d! dk =! k M k f

3 k! dk = M M!! h ) d g!) = = = = 3 M 4!! d M! g k)d! 3! = number of k states per unit energy range 3 M 4! )! d! = M! ) 3 * 3! 3 +, - / =! M 3 3 as before! ) = g ) = number of states allowing for spin degeneracy!) = 3 M! ) = for unit volume 53 Free Electron Theory of Metals Conduction electrons! ideal Fermi gas Approximations: ) Positive ions are roughly equal to a uniformly distributed positive charge distribution potential well of finite depth ) Ignore the Coulomb repulsion between electrons justification of approx difficult departure are visible from ideal gas behavior - Free electron theory is a good starting point for a more relativistic theory Typical metal = n =! 8 m 3 eg for Potassium 4! 8 atoms per metre cubed = 4! 8 electrons per metre cubed ) m for k, which is roughly / Angstrom k f =! n Wavelength = k f ~ 9 m ~ atomic spacings Therefore expect stray

4 quantum effects =! k m e = e Define = degeneracy or Fermi temperature = k B is the temperature at which classical gas of particles would have energies of! 5 4 k for K At T = the electrons near the Fermi surface have kinetic energies corresponding to classical temperatures of 4 k The Fermi velocity v f =! f =!k f 86 ) 5 ms * for k m e m e Large but still much less than the speed of light Relativistic corrections are only needed for heavy metals 54 Ideal Fermi Gas at T > If T << T F eg metal at room temperature) a few electrons in the neighbourhood of the Fermi surface are parallel to the state above What is µ? f!) = =! e! µ )! + - fixed µ Change sum to an integral f!)g!)d! = g!) ~!

5 Integral = everything below the line At T =, µ = T >, µ < µ =! T T ) + f ) Mandl problem If T <<, µ! f For metals at 3k, can take µ = Because the electrons are crammed into their lowest possible energy states and this dominates their behaviour, an ideal Fermi gas at low temperatures is said to be highly degenerate 54 Heat Capacity of Fermi gas T << Exact calculation: umber of electrons!! + d! = f!)g!)d! Energy of electrons!! + d! =!)g!)d! Total energy E of electrons E =!)g!)d!! C v = de dt v For T << ; C v =! k T B This is the exact result Approximate calculation ignore changes in µ ow looking at the regions between the solid line, and the vertical line at µ

6 Electrons from the left side of the line! < ) are thermally excited to the region to the right side of the line,! > The two regions are the same area Let them be triangles base! height = k BT ) g ) = 4 k BT )g ) Every electron that has moved has changed its energy by ~ k B T ) Therefore the increase in the energy!e 4 k BT ) g f C v = d!e dt What is g )? g ) = k B Tg f ) m! But =! m k f Eliminate : ) = 3 =! 3! f m g Hence C v! 3 4 k B T! 3 f 4 k T B 3! T This result C v ~ k B T solves a mystery f If electrons behave classically, E = 3 k B T 3 C v = 3 k B Early success of quantum theory of Fermi gas was to explain why such a large contribution to the heat capacity was not seen Quantum theory reduces classical value by a factor T Only a fraction of the electrons T are involved in the heat capacity Observation of Electronic Contribution to C v It is difficult at high temperature because contributions are ~ 3k b from lattice vibrations At low temperatures, lattice vibrations contribute ~ T 3 C v =! T + T 3

7 ! = electronic specific heat coefficient =! k B for ideal gas C v T =! + T Slope is! For potassium,! theory = 67mJ mol k,! exp eriment = 8mJ mol k Discrepancy due to the failure of the assumption of the free electron theory We ignored: a) Periodic potential of the positive ion cores these produce band structure effects b) Coulomb repulsion between electrons 543 Paramagnetic Susceptibility of Conduction Electrons T << T P - temperatures have a small effect Assume T = B = Previous diagram, split into up and down, and rotated through 9 degrees Spin ups have µ z = µ B Spin downs have µ z =!µ B Apply a field B along z: de =!µ B where µ is the magnetic moment of the electron de =!µ z B = ±µ B B Positive for up spins, negative for down spins

8 Spin up electrons in the red area move into the blue area, where the energy is reduced If µ B <<, one can approximate the shaded area by a rectangle For a metal, B << 3! 4 T - an enormous field energies of the highest occupied state are the same as before = the number of electrons moving is base times height = g!) µ B) Magnetization per unit volume =! )µ B Use g ) = 3 use if! = p m M = 3µ B B! p is the Pauli paramagnetic susceptibility = µ M B! p = 3µ µ B f ote that this is independent of temperature Cf the Curie law result for classical spins:! curie = µ µ B k B T! p! curie = 3k BT f = 3 T = f ) µ Bµ B Like the heat capacity, the magnetic susceptibility is reduced by a factor of order T T from the classical result f! p is comparable to the Landau diamagnetic susceptibility For free electrons,! L = 3! p makes! p hard to measure Metal 5! p experiment) 5! p theory) Li 5 a 4 83 K 67 Rb 63 Cs 58 Discrepancies are due to the neglection of band structure effects, as well as Coulomb repulsion of the electron 544 X-Ray Emission Spectrum

9 The lines are electron energy states in iron cores The curve/line is the conduction electrons Knock out electrons from the K-shell, then say) conduction electrons fall down into vacant states emitting photons Spectrum of emitted photons depends on the number of electrons of each energy This plot is the intensity of the emitted X-rays proportional to g!) ), vs the photon energy There is a sharp cut-off at the Fermi energy 55 Liquid 3 He This is the Fermi form of a He ion It is a byproduct of the nuclear weapons program Phase diagram for most substances: S = Solid L = Liquid G = Gas C = Critical Point T = Triple Point The substance is only crystal at T = For 3 He :

10 At lower T, the liquid acts as a degenerate Fermi liquid It turns into a solid at around 34 Atmosphere The minimum point of the solid curve is at around k S = Solid BCC ) System remains liquid down to T = at low pressures at least for pressures less than 34 atmospheres) Quantum effects cause this Solid: atoms localized positions are known large uncertainty in momentum large momentum large kinetic energy p This will encourage a liquid to form This can m overcome the energy level gained by going into a crystalline structure, as the liquid state now has lower energy ) Liquid solid phase boundary at low T has dp dt negative Clausius-Clapepron dp dt = S L! S S L! S S L and S S are the liquid and solid entropies per kg L and S are the liquid and solid volume per kg For most materials, dp dt >, S > S and > L S L S For water, dp dt <, S > S liquid more disordered than solid < L S L S liquid more dense than solid ice floats on water) For 3 He, L > S ie contracts on freezing like most materials But S L < S S the liquid is more ordered than the solid 3 He is a Fermi liquid at low temperatures, and is very ordered at low temperatures Particles in the Fermi sphere are filled by 3 He particles with up and down spins ie each k state is filled The particles are not so ordered) Each k state less than k f contains two atoms, one with spin up, and one with spin down Ordering is in k-space, not real space For solid, the spins which lie on the nuclei of the 3 He ) are disordered each spin can be up or down because the atoms are located on crystal sites This gives the contribution to the entropy of k B ln! = ) S = k ln! ) At mk, the solid undergoes a magnetic ordering transition to the antiferromagnetic phase, S S!

11 dp dt ~ L! S As T!, dp dt! What is the Fermi temperature T F of 3 He? At P =, n = 6! 8 m 3 = =! k B k B M 3 n ) 3 = 5k B: m = 3! 66! 7 kg For T <<, 3 He should behave like a Fermi system C =! T no lattice, therefore no T 3 term as in metals) D = Degenerate Fermi fluid T < T S, liquid makes a transition to the A-phase of superfluid 3 He 56 Electrons in Stars AC Phillips, The Physics of Stars ) What is the sun made of? Mostly Hydrogen Atomic hydrogen? o temperatures are too high The proton and the electron have been dis-associated, ie it is a plasma How hot is the sun? Central temperature 7 k Do the electrons form a degenerate Fermi gas, ie is T <<? Mass of the Sun is roughly! 3 kg The radius of the sun is 7! 8 m Assume that the sun has a uniform density not true) The electron density in the Sun is equal to the number of Hydrogen atoms, which is the same as the number of Protons So:

12 n = M! M H! =! 3 66! 7! 4 3! 7! 8 ) 3 = 8! 9 m 3 This is a little bit larger than in metals = =! k B k B m 3 n) 3 = 4 5 k Therefore the electrons in the sun are classical Why does the sun not collapse under its own gravity? Partly gas pressure, but also the outward pressure of the radiation Where do we find degenerate Fermi gas in stars? We need to increase the density So look at dense stars neutron stars, or white dwarfs 56 White Dwarf Stars Entering the white dwarf phase in the ultimate fate of light stars such as the sun H burns at 7 k we need the kinetic energy to overcome the coulomb repulsion) After all the H has burnt, the star shrinks and heats until He burns to Carbon at 8 k If the star is then heavy enough, it can progress to burn C at 5! 8 k, etc So the temperature needs to rise in order to set kinetic energies to overcome the larger coulomb barrier of the heavier nuclei But suppose that the star never gets hot enough for Helium to burn or for Carbon to burn? What steps come from shrinking the star? Inward gravitational pressure is balanced by pressure from the degenerate electron gas white dwarf electron gas White dwarf slowly cools and dies If mass = M s, what is its radius R? ) Gravitational pressure = f G, M s, R Pressure = force per unit area P = GM s R 3 4! R ~ GM s R 4 Assume the star is of constant density Gravitational potential energy: E p =! 3 5 GM s R GM s pa! R = = 3 R 5 R p = 3 GM s 5 R 4! R This inward gravitational pressure is counteracted by pressure being exerted outwards excited by electrons) What is the pressure of the electrons? Assume T <<, and take T = for simplicity

13 de = TdS! pd =! pd at T = ) p =! E at T = ) t f E =!g!)d! t = k dk! k f m = 3 m! )! 3 d! = m! ) = 3 5 =! m k f =! 3 m has been used to eliminate the volume term! = 3 5 p =! )! f! P = 5 3 =! 3 5 f ) = )! m 3* ) 3 = 3 k = 3 ) 3! 5 m = ) f 3 is the number of electrons in the star = M s, where M H is the mass of a H atom M H For either carbon or helium white dwarfs, these will be approximately one electron for every nucleons = 4 3! R3! = A! m A = 3! ) 3 M s M H 5 3 R ! 6 B: this is the pressure pushing outwards Hence GM s!! R 4 m M s M H 5 3 R 5

14 ! R! 5 mgm 3 s M 3 H the more massive the star, the smaller R gets) For example, 4EriB: M s! 3 kg R theory = 79! 6 m R actual = 87! 6 m ote that this is around the size of the Earth good agreement R! as M s! R!!, therefore QM effects work on enormous length scales Check assumptions that electron gas is degenerate: = =! k B k B m = 3 M s M H, = 4 3! R3 3 9 k But T ~ 7 k Therefore T <<, ie cannot be hot enough to burn He or C ow, P electrons! R 5, P grav ~ R 4 ie electron pressure increases faster at small R Electron pressure P electrons ) increases faster at small R It ought to be able to resist collapse of heavier: wrong Work out f for electrons at Fermi surface for = 9 k m f = k B = 87ke f = 8! 8 ms So what s the problem? electrons are relativistic and it is important to treat them as such for heavier stars non-relativistic approximation is OK for white dwarfs)! = p c + m c 4 We will work in the extreme relativistic limit, p c >> m c 4! = pc =!ck At T =, k E =!g!)d! = f k dk!ck k = f! k dk = 3! k 3 f k f = 3! 3

15 =!ck f E = 4 4!c p =! E ~ M s R ) =!c =!c 9) 3 8 ~ electron pressure Gravitational pressure M s therefore for heavy stars, gravity will always win if R 4 4 For heavy stars: 4! M! star becomes a neutron star 58! M! star becomes a black hole 55 eutron Stars When mass is > 4M!, then the star goes through all the stages of nuclear fusion to reach iron core Then get catastrophic collapse via a supernova explosion this releases some of the star s original mass As star collapses, electron density n rises Electron KE rises ~ relativistic) v 3 Therefore KE becomes so high that inverse beta decay occurs e! + p n + Also, atomic nuclei break up left with a giant lump of nuclear matter Predominantly made of neutrons only a few e and p left Gravitational pressure can now be resisted by the pressure of gas of degenerate neutrons with T ~ 7 k Star is initially at 4 k! 8 k in years neutron star or pulsar) Solid crystalline crust of nuclei and neutrons, with a neutron fluid inside p = 5 =! m n = 4 3! R3 = m s m n 5 p ~! 3 m n R 5 3 3! GM s R 4 gravitational pressure)

16 ! R! 5 m n Gm 3 n m 3 s For m s ~ 3! 3 kg ~ 5M!, R ~ km = 43! 44 m 3 = 8 J = Ge f = m n = 5 8 ms ie about ½c So it is non-relativistic only just For heavier stars, relativistic effects become more important as with white dwarfs and electrons Eventually, pressure of neutrons can no longer withstand gravity M max ~ 58 solar masses Collapse continues black holes