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Presentation on theme: "CHAPTER 19 FERMI-DIRAC GASES."— Presentation transcript:


2 Okay, why should we want to discuss a Fermi-Dirac gas?
Gosh, its probably because it is necessary for the understanding of some important problems. OK, what problems? Some major problems that can be tackled with this formalism are free electrons in conductors, free electrons in white dwarf stars and neutrons in a neutron star. Fine, lets press on. Before we do, I would like to mention the Third Law of Thermodynamics. 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 temperature of 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 have is called the fermi function. gives the probability that a single state will be occupied. At any T, if We define the fermi energy by (not fermi level!) At T=0 Therefore [The ground state is taken to be at 0eV and 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 one microstate 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=0 1 The density of states is the same as before except that there are two possible values for the quantum number of Hence we have 2 particles per spatial state.Taking this into consideration gives, from our previous calculation of the density of states:

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 a fermi temperature by

6 This is analogous to the bose temperature defined earlier. The fermi
temperature is a reference temperature, it is not the temperature of the gas. We will calculate the fermi temperature a little later: it is in the range The more difficult situation in when the temperature is greater than zero. Now states above the fermi energy are occupied. The 1 in the denominator makes the integration difficult. Numerical calculations can be made to determine the chemical potential as a function of the temperature. However approximations can be made, valid for ,giving Notice that in contrast to bosons of which

7 A knowledge of the chemical potential permits the calculation of the fermi distribution function.
T=0K T=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 interaction between electrons and the presence of the positively charged lattice sites. The many-body theory of solids shows that these effects can be reasonably ignored, at least to a good 1st approximation.

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 interior of a metal are similar to molecules in a gas, so we speak of an electron gas. Of course this is a simplification because it ignores, among other things, the periodic ionic potential, which leads to band structure so important in semiconductor physics. (An effective mass is often introduced to partially compensate for the simplifications.) The metal will be at a fixed temperature and there will be a corresponding chemical potential, called the fermi level. The potential energy 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 from a metal one must give an electron at the fermi level a certain energy , called the work function. In the photoelectric effect, this energy is supplied by absorbing a photon. Note that the fermi level is not the same as the fermi energy but, for As shown in your textbook, for Ag, the fermi energy is 5.54eV. This is high and the most striking property of a fermi gas. Contrast this with a gas molecule at room temperature, which has an energy of about 0.025eV. The energy of the electrons at the fermi energy, at absolute zero, is about that of a gas molecule at a temperature of about 64,000K. The energy of an electron gas at T=0K is called the zero point energy. 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 with Fermi level Fermi energy We now plot T=0 1 T

12 T=0 It is the electrons in the tail of this distribution that can be most easily extracted from a metal by various processes. Internal energy of the gas: Again we can make approximations to obtain a series solution to this equation.

13 At T=0 This is a large energy. It comes about by the PEP. For example, for Ag at T=0, For an ordinary gas at T=300K The 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 solids at room temperature. (This is an experimental result.) This law has a simple explanation based on the principle of equipartition of energy. Each atom of the solid is considered to be a linear oscillator with 6 degrees of freedom (vibrating in 3 dimensions and the oscillator has both kinetic and potential energy). At lower temperatures the specific heat decreases and , at these temperatures, a more sophisticated quantum-mechanical approach is required.

15 Continuing with the electron gas, we associate a specific heat with
the free electrons by defining: Using equation (2) Since Since we are considering low temperatures it is certainly true that T is very much smaller than the fermi temperature and so the second term is negligible.

16 (Notice the linear dependence on T)
At room temperature and for Ag This small value explains a puzzle regarding the specific heat of metals. It was expected that the free electrons, having 3 translational degrees of freedom, would contribute an additional (3/2)R to the specific heat (from equipartition theorem). This is obviously not 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 the kinetic energies of the free electrons are much greater than the thermal energy of the particles in a gas, the change in energy (dU/dT) of the electrons is small. Only the electrons near the fermi level can increase their energies because of the availability of unoccupied states, 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. This is also true for a Bose gas.

18 Helmholtz thermodynamic potential.
F=U-TS Eqn(2) –T[eqn(3)]

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 terms of the volume!!! Rewriting F

20 Comparing with equation (2) (slide 15)
Example: For Ag Since

21 In spite of this tremendous pressure, the potential barrier at the
surface of the metal keeps the free electrons from evaporating from the surface. This pressure tends to increase the volume. This is balanced by the interaction between the electrons and the ions. If we were to continue with a description of solids we would then choose a more realistic potential. The regular spacing of the ions is described by a periodic potential. This would lead to an energy level diagram which has bands of energy states separated by gaps in which there are no states. This structure permits us to understand 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. Therefore they are extremely dense. The core temperature is of the order of Under these conditions the atoms are completely ionized so we have nuclei and an electron gas. At the white dwarf stage the H has been used up in thermonuclear reactions, fusion is greatly reduced, the star cools and begins to collapse. This collapse is stopped by the pressure of the electron gas. In an elementary discussion of white dwarfs, several approximations are made. Relativistic effects are not usually dominant so it is not too egregious to make a non-relativistic calculation. All densities are assumed uniform (perhaps the most serious approximation). The first white dwarf discovered was Sirius B, so we shall use it 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 average mass number. The total mass, M, is then M=NnucAMH Since the atoms are completely ionized the number of electrons is: N=ZNnuc The 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 20 Is this pressure, due to the degenerate electron gas, sufficient to prevent future collapse? Condition for stability: U=minimum U 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 star by bringing in mass elements from infinity. We obtain Since and using the expression for the energy on slide 15

27 For equilibrium

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 now a stable white dwarf. However it will eventually become invisible. In most known white dwarfs the core contains C and O with an outer layer which consists of H or He or both. The star cools slowly. It is estimated that white dwarfs take about years to become invisible. Since the age of the universe is about years, 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 largest possible mass is called the Chandrasekhar limit. This limit is about 1.44 solar masses. A star exceeding this mass collapses, the density approaches that of nuclear matter and the electrons combine with the protons (inverse β-decay) to form a neutron star. Now a degenerate neutron gas stabilizes the star. Again there is a limiting mass, which is about three solar masses. More massive stars collapse to form black holes.


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