CHAPTER 9 Statistical Physics

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1 CHAPTER 9 Statistical Physics 9.1 Historical Overview 9.2 Maxwell Velocity Distribution 9.3 Equipartition Theorem 9.4 Maxwell Speed Distribution 9.5 Classical and Quantum Statistics 9.6 Fermi-Dirac Statistics 9.7 Bose-Einstein Statistics Ludwig Boltzmann, who spent much of his life studying statistical mechanics, died in 1906 by his own hand. Paul Ehrenfest, carrying on his work, died similarly in Now it is our turn to study statistical mechanics. Perhaps it will be wise to approach the subject cautiously. - David L. Goldstein (States of Matter, Mineola, New York: Dover, 1985)

Note: How to calculate n(e) of an ideal gas The number of particles with energies between E and E + de E 1 p 2 2m 2 2 mv p 2mE

energy range We may consider the energy distribution to be continuous.

The density of state between E and E + de can be defined by the number of state with momenta between p and p + dp: 2 2 ge ( )

2 Note: How to calculate n(e) of an ideal gas The number of particles with energies between E and E + de E 1 p 2 2m 2 2 mv p 2mE p x py pz : Maxwell-Boltzmann factor for classical system Density of state : the number of states available per unit energy range We may consider the energy distribution to be continuous. Therefore, the momentum also would be continuously distributed in 3-dim. Space. The density of state between E and E + de can be defined by the number of state with momenta between p and p + dp: 2 2 ge ( ) g( p) pdpg( p) Bpdp 2 2 p 2mE 2pdp 2mdE p dp 2 me mde 3 3 g( E) B 2m E de ne ( ) B 2mEdE Aexp( E) C EdE If the total number of particles is N, the normalization of n(e) gives. E 2N 2N 3 N n E de C Ee dec nede ( ) Eexp( EdE ) 3/2 0 0

3 9.6: Fermi-Dirac Statistics for Fermions B FD : normalization factor g(e): density of state First we consider the role of the factor B FD : Because the parameter (= 1/kT ) is contained in F FD, B FD is temperature dependent. Possible to express this temperature dependence as : E F is called the Fermi energy F FD = ½ when E = E F Or. Fermi energy can be defined as the energy at which F FD = ½ In the limit as T 0, At T = 0, fermions occupy the lowest energy levels available to them. Near T = 0, there is little chance that thermal agitation will kick a fermion to an energy greater than E F.

4 Fermi-Dirac Statistics In the limit as T 0, At T = 0, fermions occupy the lowest energy levels available to them (Pauli principle). Near T = 0, there is little chance that thermal agitation will kick a fermion to an energy greater than E F. As the temperature increases from 0 (T >0), the Fermi-Dirac factor smears out T = 0 T > 0 Fermi temperature, defined as T F E F / k When T >> T F, F FD approaches a simple decaying exponential T = T F T >> T F

5 Classical Theory of Electrical Conduction Fermi-Dirac Statistics: Now we can apply the Fermi-Dirac theory to the problem of understanding electrical conduction in metals. For comparison, first briefly review the classical theory of the electrical conductivity. Drude Model of metal conductivity: In 1900 Paul Drude developed a theory of electrical conduction Assumed that the electrons in a metal existed as a gas of free particles. the current in a conductor should be linearly proportional to the applied electric field that is consistent with Ohm s law. J E E J F ee Acceleration of e by E: a m m ee Drift velocity during the average time between electron-ion collisions: v d a m 2 J ee m 1 ne J ( ne) v d v d E J 2 ne m e n m (where n is the number density of conduction electrons) The mean time between collisions: 2 ne mv (l : the mean free path, : the mean speed)

6 Classical Theory (Drude model) of Electrical Conduction 2 ne mv From the Maxwell speed distribution, the mean speed was given by According to the Drude model, the conductivity should be proportional to T 1/2 1 1 v T But for most conductors is very nearly proportional to T 1 Also, classically the heat capacity of the electron gas may be given by 3 x (1/2 R). This is not consistent with experimental results: Only about 0.02 R per mole at room T. Clearly a different theory is needed to account for the observed values of electrical conductivity and heat capacity as well as the temperature dependence of the conductivity.

7 Quantum Theory of Electrical Conduction Electrons are fermions, and therefore we need to use the Fermi-Dirac distribution. The problem is to find g(e ), the density of state (the number of allowed states per unit energy) Use the allowed energy values of the free electron to be quantized in a three-dimensional infinite square-well potential. The number of allowed states up to radius r (or up to energy E = r 2 E 1 ) is directly related to the spherical volume (4/3)r 3. The exact number of states up to radius r is 2 due to spin degeneracy 1/8 because the quantum numbers be positive. 3-dimensional number space

If the total number of electron at T =0 is N, then Calculate the Fermi energy and Fermi temperature for copper.

8 Quantum Theory of Electrical Conduction : the density of states (the number of allowed states per unit energy) Note, at T = 0 the Fermi energy is the energy of the highest occupied energy level. If the total number of electron at T =0 is N, then Calculate the Fermi energy and Fermi temperature for copper. Fermi energy of conductors N N n E L V 3 3 F Number density of conductors The number density of conduction electrons in copper is given by Table 9.3 as Note that E F changes little between T = 0 and room temperature.

9 Quantum Theory of Electrical Conduction At T = 0: F FD = 1 the mean electronic energy: the internal energy of a system with N electrons: At T > 0, let s find the electronic contribution in a conductor to the heat capacity:

$near E F will be able to absorb thermal energy. Therefore the fraction of electrons capable of participating in this thermal process is on the order of kt/e F.$

10 Quantum Theory of Electrical Conduction At T > 0, let s find the electronic contribution in a conductor to the heat capacity: At T > 0, Because the energy are filled up to E F, only those electrons near E F will be able to absorb thermal energy. Therefore the fraction of electrons capable of participating in this thermal process is on the order of kt/e F. The exact number of electrons depends on temperature, because n(e) changes with temperature. In general, T = 0 T > 0 For 1 mole; N = N A, N A k=r, E F = kt F :

11 Quantum Theory of the heat capacity and conductivity Heat capacity: Arnold Sommerfield found a value for α = π 2 / 4 at room temperature With the value T F = 80,000 K for copper, we obtain c V 0.02R, which is consistent with the experimental value! Quantum theory has proved to be a success in a case failed classically! Electrical conductivity: Conducting electrons are loosely bound to their atoms. these electrons must be at the highest energy level at room T the highest energy level is close to the Fermi energy We should use (for copper)

12 9.7 Bose-Einstein Statistics Let s derive the Planck formula for blackbody radiation and investigate the properties of liquid helium. Blackbody Radiation Blackbody: a nearly perfectly absorbing cavity that emits a spectrum of EM radiation. [Derivation of Planck formula] In quantum theory we must use the Bose-Einstein distribution because photons are bosons with spin 1. For a photon (= free (massless) particle) in a box, the momentum is also quantized: The energy of a photon is pc, so E pc The number of allowed states up to radius r defined by 2 due to two possible polarization 1/8 because the quantum numbers be positive. hc E pc r 2L (B BE = 1)

13 Derivation of Planck formula Convert the number distribution to an energy density distribution u(e). E ue ( ) ne ( ) 3 L for all photons in the range E to E de Energy density (energy per unit volume) per unit wavelength inside the cavity To convert the spectral energy density to a spectral power density distribution, in the SI system, multiplying by a constant factor c/4 is required from dimension analysis and geometry analysis. Power per unit area per unit wavelength for radiation Planck formula of blackbody radiation Planck did not use the Bose-Einstein distribution to derive his radiation law. Nevertheless, it is an excellent example of the power of the statistical approach to see that such a fundamental law can be derived with relative ease. This problem was first solved in 1924 by the young Indian physicist Satyendra Nath Bose (for whom the boson is named). Einstein s name was added to the distribution because he helped Bose publish his work

14 Liquid Helium Helium is an element with a number of remarkable properties. It has the lowest boiling point of any element (4.2 K at 1 atmosphere pressure), thus liquid helium can be used to cool materials to 4.2 K and lower. It has no solid phase at normal pressures. Liquid helium is also quite interesting in its own right In 1924 Kamerlingh Onnes measured the density of liquid helium as a function of temperature. In 1932 W. Keesom and K. Clausius measured the specific heat of liquid helium. Something extraordinary is happening near 2.17 K. That temperature is commonly referred to as the critical temperature (Tc), transition temperature, or simply the lambda point, a name derived from the shape of the curve in the figure.

Superfluid below the lambda point The rate of flow increases dramatically as the temperature is reduced because the superfluid has a low viscosity.

15 Liquid Helium As the temperature is reduced from 4.2 K toward the lambda point, the liquid boils vigorously. At 2.17 K the boiling suddenly stops. What happens at 2.17 K is a transition from normal phase to superfluid phase. Superfluid below the lambda point The rate of flow increases dramatically as the temperature is reduced because the superfluid has a low viscosity. The flow rate of liquid helium through a capillary tube as a function of temperature Liquid helium forms a Creeping film when the viscosity is very low. the creeping film flows up and over the (vertical) walls of its container.

16 Liquid Helium Fritz London claimed (1938) that liquid helium below the lambda point is part superfluid and part normal (two-fluid model). As the temperature approaches absolute zero, it approaches 100% superfluid. The fraction of helium atoms in the superfluid state: With two protons, two neutrons, and two electrons, the helium atom 4 He is a boson and therefore subject to Bose-Einstein statistics. Superfluid liquid helium is referred to as a Bose-Einstein condensation. not subject to the Pauli exclusion principle all particles are in the same quantum state Such a condensation process is not possible with fermions because fermions must stack up into their energy states, no more than two per energy state. 3 He isotope (one less neutron) is a fermion and superfluid mechanism is radically different than the Bose-Einstein condensation. (Tc = 2.7 mk)

17 Liquid Helium To estimate Tc for the superfluid phase of 4 He, Bose-Einstein statistics may be used. BUT, for simplicity, let s take the fermions density of states function we have already developed and adjust it for the superfluid case. The only difference for bosons is that they do not obey the Pauli principle, and therefore the density of states for spin-zero bosons should be less by a factor of 2. The number distribution n(e ) is now In a collection of N helium atoms the normalization condition is

18 Liquid Helium Use minimum value of B BE = 1; this result corresponds to the maximum value of N. Rearrange this, The result is T 3.06 K. The value 3.06 K is an estimate of T c. The result is a bit off, because we used a density of states derived for a noninteracting gas rather than a liquid. Still, we have come within 1 kelvin of the correct value of Tc. Bose-Einstein Condensation in other Gases By the strong Coulomb interactions among gas particles it was difficult to obtain the low temperatures and high densities needed to produce the condensate. Finally success was achieved in 1995, by using 87 Rb atom gas. First, they used laser cooling to cool their gas of 87 Rb atoms to about 1 mk. Then they used a magnetic trap to cool the gas to about 20 nk. In their magnetic trap they drove away atoms with higher speeds and further from the center. What remained was an extremely cold, dense cloud at about 170 nk.

19 Symmetry in Wave Functions Definition of Symmetric and Antisymmetric Wave Functions Consider a system of two identical (indistinguishable) particles labeled 1 and 2. Put the wave function that describes the system of two particles (1,2). Because of indistinguishability, interchanging them cannot change the probability density, In the case (1,2) = (2,1), the wave functions are called symmetric In the case (1,2) = - (2,1), the wave functions are antisymmetric Symmetric and antisymmetric Wave Function: Consider the wave functions of the two individual particles, (1) and (2). In a system of two noninteracting particles the overall wave function can be expressed as a product of individual wave functions: That is, (1,2) = (1) (2). Now suppose that the two particles are in different states, labeled a and b. The net wave function for this system is a linear combination of the two possible combinations of the particles in states a and b: : Symmetric wave function : Antisymmetric wave function

20 Wave Functions for Bosons and Fermions Bosons have symmetric wave functions, and fermions have antisymmetric ones. Fermions Antisymmetric wave function: Suppose particles 1 and 2 are fermions in the same state: a = b The antisymmetric wave function gives A = 0! Thus, two fermions can t occupy the same quantum state. Bosons Symmetric wave function: Suppose particles 1 and 2 are fermions in the same state: a = b The symmetric wave function becomes the probability density for this state is nonzero! Thus, two bosons can occupy the same state.

CHAPTER 9 Statistical Physics

CHAPTER 9 Statistical Physics 9.1 9.2 9.3 9.4 9.5 9.6 9.7 Historical Overview Maxwell Velocity Distribution Equipartition Theorem Maxwell Speed Distribution Classical and Quantum Statistics Fermi-Dirac