2 Okay, why should we want to discuss a Fermi-Dirac gas? Gosh, its probably because it is necessary for theunderstanding of some important problems.OK, what problems?Some major problems that can be tackled with thisformalism are free electrons in conductors, free electrons inwhite dwarf stars and neutrons in a neutron star.Fine, lets press on.Before we do, I would like to mention the Third Law ofThermodynamics.One expression of this law is that as(There are exceptions to this statement, such as with glasses.)Another expression is: It is impossible to reduce the temperatureof a system to absolute zero with a finite number of processes.
3 We now consider particles with half-integer spin (s=1/2, 3/2, ….) called fermions. These particles obey the Pauli Exclusion Principle.They are indistinguishable in a gas and so obey Fermi-Dirac Statistics.We haveis called the fermi function.gives the probability that a single state will be occupied.At any T, ifWe define the fermi energy by(not fermi level!)At T=0Therefore[The ground state istaken to be at 0eVand so μ(0)>0]
4 The PEP permits only one particle per state so the N particles are crowded into the N lowest energy states. At T=0 only onemicrostate is possible, so w=1 and S=kln(w)=0. Hence S=0 at T=0.This is in accordance with the 3rd law of thermodynamics.T=01The density of states is the same as before except that there are twopossible values for the quantum number ofHence we have 2 particles per spatial state.Taking this intoconsideration gives, from our previous calculation of the density ofstates:
5 At T=0 the fermi function is unity up to the fermi energy The energy of an ordinary gas molecule is of order kT. We define afermi temperature by
6 This is analogous to the bose temperature defined earlier. The fermi temperature is a reference temperature, it is not the temperature ofthe gas. We will calculate the fermi temperature a little later: it is inthe rangeThe more difficult situation in when the temperature isgreater than zero. Now states above the fermi energy are occupied.The 1 in the denominator makes the integration difficult. Numericalcalculations can be made to determine the chemical potential as afunction of the temperature. However approximations can be made,valid for ,givingNotice that in contrast to bosons of which
7 A knowledge of the chemical potential permits the calculation of the fermi distribution function. T=0KT=0.2 TF
8 Next we are going to treat the electrons in a solid as a gas. This may seem unreasonable given the strong coulomb interactionbetween electrons and the presence of the positively chargedlattice sites. The many-body theory of solids shows that theseeffects can be reasonably ignored, at least to a good 1stapproximation.
9 Free electrons in a metal Free electrons in a metal. (This is relevant to conduction in materials,white dwarfs, He-3, and nuclear matter.)To a good first approximation, the electrons confined in the interiorof a metal are similar to molecules in a gas, so we speak of an electrongas. Of course this is a simplification because it ignores, among otherthings, the periodic ionic potential, which leads to band structure soimportant in semiconductor physics. (An effective mass is oftenintroduced to partially compensate for the simplifications.)The metal will be at a fixed temperature and there will be acorresponding chemical potential, called the fermi level. The potentialenergy for the electron gas is as follows:!fermi level=work function (3-4eV)
10 An electron near the surface of the metal feels a strong attractive force due to the positive metal ions in its vicinity. To free an electron froma metal one must give an electron at the fermi level a certain energy ,called the work function. In the photoelectric effect, this energy issupplied by absorbing a photon.Note that the fermi level is not the same as the fermi energybut, forAs shown in your textbook, for Ag, the fermi energy is 5.54eV. Thisis high and the most striking property of a fermi gas. Contrast thiswith a gas molecule at room temperature, which has an energy ofabout 0.025eV. The energy of the electrons at the fermi energy, atabsolute zero, is about that of a gas molecule at a temperature ofabout 64,000K.The energy of an electron gas at T=0K is called the zero pointenergy.At room T (Ag)
11 When the gas is said to be in the degenerate region. This is not to be confused with the degeneracy of an energy level.Using equation (1)This is why is often confused withFermi level Fermi energyWe now plotT=01T
12 T=0It is the electrons in the tail of this distribution that can be mosteasily extracted from a metal by various processes.Internal energy of the gas:Again we can make approximations to obtain a series solutionto this equation.
13 At T=0This is a large energy. It comes about bythe PEP.For example, for Ag at T=0,For an ordinary gas at T=300KThe electrons in an electron gas have much more energy at 0K than the molecules of a gas at room temperature.
14 Specific heat of metals (one of the great accomplishments of the theory).The law of Dulong and Petit: for all elementary solidsat room temperature. (This is an experimental result.)This law has a simple explanation based on the principle ofequipartition of energy. Each atom of the solid is considered to be alinear oscillator with 6 degrees of freedom (vibrating in 3 dimensionsand the oscillator has both kinetic and potential energy).At lower temperatures the specific heat decreases and , at thesetemperatures, a more sophisticated quantum-mechanical approachis required.
15 Continuing with the electron gas, we associate a specific heat with the free electrons by defining:Using equation (2)SinceSince we are considering low temperatures it is certainly true thatT is very much smaller than the fermi temperature and so the secondterm is negligible.
16 (Notice the linear dependence on T) At room temperature and for AgThis small value explains a puzzle regarding the specific heat ofmetals. It was expected that the free electrons, having 3 translationaldegrees of freedom, would contribute an additional (3/2)R to thespecific heat (from equipartition theorem). This is obviouslynot in agreement with experiment. Our calculation shows that, indeed,the electronic contribution is small.The reason that it is small is as follows. Even though thekinetic energies of the free electrons are much greater than the thermalenergy of the particles in a gas, the change in energy (dU/dT) of theelectrons is small. Only the electrons near the fermi level canincrease their energies because of the availability of unoccupiedstates, and only a small fraction of the free electrons are near
17 đ đ The entropy of the electron gas: For the free electrons S=0 at T=0 in accordance with the 3rd law of thermodynamics. Thisis also true for a Bose gas.
19 The pressure of an electron gas: Now that we have the Helmholtz function, we can calculate the pressure. (potentials yield properties)Our first chore is then to write F explicitly in termsof the volume!!!Rewriting F
20 Comparing with equation (2) (slide 15) Example: For AgSince
21 In spite of this tremendous pressure, the potential barrier at the surface of the metal keeps the free electrons from evaporating fromthe surface.This pressure tends to increase the volume. This is balancedby the interaction between the electrons and the ions.If we were to continue with a description of solids we wouldthen choose a more realistic potential. The regular spacing of theions is described by a periodic potential. This would lead to anenergy level diagram which has bands of energy states separatedby gaps in which there are no states. This structure permits us tounderstand electrical conduction in semiconductors.
22 Here is how you can coax MAPLE to give you a numerical answer to an integral.> assume(x>=0); > int((x^2)/(exp(x)-1),x=0..infinity); 2*Zeta(3) > evalf(%);
23 White Dwarf Stars. These stars have masses comparable to the mass of the sun and radii comparable to the radius of the Earth. Thereforethey are extremely dense. The core temperature is of the order ofUnder these conditions the atoms are completely ionized so wehave nuclei and an electron gas. At the white dwarf stage the H hasbeen used up in thermonuclear reactions, fusion is greatly reduced, thestar cools and begins to collapse. This collapse is stopped by thepressure of the electron gas.In an elementary discussion of white dwarfs, severalapproximations are made. Relativistic effects are not usually dominantso it is not too egregious to make a non-relativistic calculation. Alldensities are assumed uniform (perhaps the most seriousapproximation).The first white dwarf discovered was Sirius B, so we shall useit as an example.
24 The white dwarf consists of light elements and hence Let Nnuc be the number of nuclei in the star and A the averagemass number. The total mass, M, is then M=NnucAMHSince the atoms are completely ionized the number of electrons is:N=ZNnucThe fermi temperature can now be calculated.
25 The temperature of the star is much less than the fermi temperature and so we have a degenerate electron gas.Now we can calculate the pressure of this degenerate gas.From slide 20Is this pressure, due to the degenerate electron gas, sufficient toprevent future collapse?Condition for stability: U=minimumU has two terms: electron gas and gravitational
26 We will write U=U( R ) and determine the radius at equilibrium The gravitational U can be obtained by building up the starby bringing in mass elements from infinity. We obtainSince and using the expression for the energy on slide 15
28 This is equal to the observed radius, to within the accuracy of the calculation. Hence it is reasonable to assume that Sirius B is nowa stable white dwarf. However it will eventually become invisible.In most known white dwarfs the core contains C and O withan outer layer which consists of H or He or both. The star coolsslowly. It is estimated that white dwarfs take about yearsto become invisible. Since the age of the universe is aboutyears, most white dwarfs are still visible.
29 If the mass of a star is sufficiently large so that the electrons are relativistic, a stable equilibrium is not possible. The largestpossible mass is called the Chandrasekhar limit. This limit isabout 1.44 solar masses. A star exceeding this mass collapses,the density approaches that of nuclear matter and the electronscombine with the protons (inverse β-decay) to form a neutronstar. Now a degenerate neutron gas stabilizes the star. Again thereis a limiting mass, which is about three solar masses.More massive stars collapse to form black holes.
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