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

CHAPTER 9Statistical Physics

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 1933. 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

: Maxwell-Boltzmann factor for classical system

The number of particles with energies between E and E + dE

Density of state : the number of states available per unit energy range

221

2 2pE mvm

2 2 22 x y zp mE p p p

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( ) ( ) ( )g E g p p dp g p Bp dp 2 22 2 2 2 p mE pdp mdE p dp mE mdE

3( ) 2g E B m E dE 3( ) 2 exp( )n E B m EdE A E C EdE

If the total number of particles is N, the normalization of n(E) gives.

3/2

0 0

2E NN n E dE C Ee dE C

32( ) exp( )Nn E dE E E dE

9.6: Fermi-Dirac Statistics for Fermions BFD: normalization factor g(E): density of state

First we consider the role of the factor BFD:

Because the parameter (= 1/kT ) is contained in FFD, BFD is temperature dependent. Possible to express this temperature dependence as

: EF is called the Fermi energy

FFD = ½ when E = EF Or. Fermi energy can be defined as the energy at which FFD = ½

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 EF.

In the limit as T → 0,

Fermi-Dirac Statistics

Fermi temperature, defined as TF ≡ EF / k When T >> TF, FFD approaches a simple decaying exponential

T > 0

T >> TFT = TF

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 EF. As the temperature increases from 0 (T > 0), the Fermi-Dirac factor “smears out”

In the limit as T 0,

Classical Theory of Electrical ConductionFermi-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.

E J

J E

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.

meE

mFa

meEavd

Acceleration of e by E:

Drift velocity during the average time between electron-ion collisions:

dvneJ )(

meE

neJvd

J

nemE

2

21 nem

(where n is the number density of conduction electrons)

The mean time between collisions: (l : the mean free path, : the mean speed)

2ne

mv

Classical Theory (Drude model) of Electrical Conduction

According to the Drude model, the conductivity should be proportional to T−1/2

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.

2nemv

From the Maxwell speed distribution, the mean speed was given by

1 1v T

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-dimensionalinfinite square-well potential.

3-dimensional number space

The number of allowed states up to “radius” r (or up to energy E = r2E1) 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.

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.

The number density of conduction electrons in copper is given by Table 9.3 as

Note that EF changes littlebetween T = 0 and room temperature.

Fermi energy of conductors

33 F

N N n EL V

Number density of conductors

Quantum Theory of Electrical Conduction

At T = 0:

FFD = 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:

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 EF, only those electrons near EF 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/EF. The exact number of electrons depends on temperature, because n(E) changes with temperature.

In general,

T = 0

T > 0

For 1 mole; N = NA , NAk = R , EF = kTF :

Quantum Theory of the heat capacity and 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

Arnold Sommerfield found a value for α = π2 / 4 at room temperature With the value TF = 80,000 K for copper, we obtain cV ≈ 0.02R,

which is consistent with the experimental value! Quantum theory has proved to be a success in a case failed classically!

Heat capacity:

Electrical conductivity:

(for copper)

9.7 Bose-Einstein Statistics

Blackbody Radiation Blackbody: a nearly perfectly absorbing cavity

that emits a spectrum of EM radiation.

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

Let’s derive the Planck formula for blackbody radiation and investigate the properties of liquid helium.

E pc

[Derivation of Planck formula]

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.

2hcE pc rL

(BBE = 1)

Convert the number distribution to an energy density distribution u(E).

Derivation of Planck formula

3( ) ( )Eu E n EL

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

Liquid Helium

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.

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

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.

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.

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.

“Superfluid” below the lambda point

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 4He 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.

3He isotope (one less neutron) is a fermion and superfluid mechanism is radically different than the Bose-Einstein condensation. (Tc = 2.7 mK)

Liquid Helium To estimate Tc for the superfluid phase of 4He, 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

Liquid Helium Use minimum value of BBE = 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 Tc. 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.

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 87Rb atom gas. First, they used laser cooling to cool their gas of 87Rb 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.

Bose-Einstein Condensation in other Gases

Symmetry in 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

Definition of Symmetric and Antisymmetric Wave Functions

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:

: Antisymmetric wave function

: Symmetric wave function

Wave Functions for Bosons and Fermions Bosons have symmetric wave functions, and fermions have antisymmetric ones.

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.

Fermions Antisymmetric wave function:

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.