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9.2 Maxwell Velocity Distribution

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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 9

Statistical 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)

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9.1: Historical Overview

Statistics and probability

 New mathematical methods developed to understand the Newtonian physics through the eighteenth and nineteenth centuries.

Lagrange around 1790 and Hamilton around 1840.

Added significantly to the computational power of Newtonian mechanics second-order differential equations

Pierre-Simon de Laplace (1749-1827)

Major contributions to the theory of probability  possible in principle to have perfect knowledge of the physical universe

Benjamin Thompson (Count Rumford) 1798

Put forward the idea of heat as merely the motion of individual particles in a substance

James Prescott Joule 1843

Demonstrated the mechanical equivalent of heat  the falling energy was transferred to internal energy in the water.

James Clark Maxwell 1850

Brought the mathematical theories of probability and statistics to bear on the physical thermodynamics problems

Showed that distributions of an ideal gas can be used to derive the observed macroscopic phenomena

His electromagnetic theory succeeded to the statistical view of thermodynamics

Einstein 1905

Published a theory of Brownian (random) motion, a theory that supported the view that atoms are real.

 Statistical physics is necessary regardless of the ultimate nature of physical reality.

 When the number of particles is large, it is highly impractical to study individual particles if one is more interested in the overall behavior of a system of particles—for example, the pressure, temperature, or specific heat of an ideal gas.

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9.2: Maxwell Velocity Distribution

The velocity components of the molecules are more important than positions, because the energy of a gas should depend only on the velocities.

Let’s define a velocity distribution function

As Laplace pointed out, we could, in principle, know everything about an ideal gas by knowing the position (x, y, z), and velocity (vx, vy, vz) of every molecule.

 6 parameters  6-dimensional Phase Space Pierre-Simon de Laplace (1749-1827)

possible in principle to have perfect knowledge of the physical universe

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Maxwell Velocity Distribution

Maxwell proved the probability distribution function is proportional to k

is Boltzmann’s constant

Rewrite this as the product of three factors with

T is the absolute temperature

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g(v

x

) dv

x

is the probability that the x component of a gas molecule’s velocity lies between v

x

and v

x

+ dv

x

.

 If we integrate g(v

x

) dv

x

over all of v

x

, it must be 1.

Maxwell Velocity Distribution

Maxwell Velocity Distribution function

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Maxwell Velocity Distribution

The mean value of v

x2

The mean value of v

x

 The results for the x, y, and z velocity components are identical.

 The mean translational kinetic energy of a molecule:

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9.3: Equipartition Theorem

Equipartition Theorem:

In equilibrium a mean energy of ½ kT per molecule is associated with each independent quadratic term in the molecules energy.

 The independent quadratic terms may be quadratic in coordinate, velocity component, angular velocity component, or anything else when squared is proportional to energy.

 Each independent phase-space coordinate is called a degree of freedom for the system.

For example, in a monatomic ideal gas, each molecule has a kinetic energy

There are three degrees of freedom 

 Mean kinetic energy is

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Let’s check our calculation, , by measuring helium’s (He, monoatomic gas) heat capacity at constant volume.

 In a gas of N helium molecules, the total internal energy is

 The heat capacity at constant volume is

 For the heat capacity for 1 mole, N = NA

The measured molar heat capacity of He is very close to this value!

Monoatomic gas

Diatomic molecule

Equipartition Theorem for Monoatomic Molecules

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Equipartition Theorem for Diatomic Molecules with a rigid axis

Consider the rigid rotator model for a oxygen molecule.

The molecule rotates about either the x or y axis.

The corresponding rotational energies are

There are five degrees of freedom (3 translational and 2 rotational).

 the energy per molecule is

Why not 3 degrees in rotational?

In the quantum theory of the rigid rotator, the allowed energy levels are

Because the mass of an atom is confined to a small nucleus,

Iz is orders of magnitude smaller than Ix and Iy.

 A small value of Iz leads to a high energy, relative to that obtained with Ix or Iy and comparable quantum numbers.

 Because the rotational energy is relatively low and small quantum numbers are required, thus only rotations about the x and y axes are allowed!

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Consider diatoms connected to each other by a massless spring.

 How many degrees of freedom does this add?

 7 degrees of freedom (3 translational, 2 rotational, and 2 vibrational). 

Spring connector

 The heat capacities of diatomic gases are temperature dependent, indicating that the different degrees of freedom are “turned on” at different temperatures.

Equipartition Theorem for Diatomic Molecules with a massless spring

7 2 kT

Molar Heat Capacity of H

2

One from the separation: because the potential energy is

One from the velocity: because the vibrational kinetic energy is

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9.4: Maxwell Speed Distribution

Let us return to the Maxwell velocity distribution:

It is also useful to turn this into a speed distribution:

It is not possible simply to assume that  Why?

 Consider some distribution of particles in 3-dim space:

 Similarly, the spherical shell in velocity space is 

 Therefore, the desired speed distribution is

: Maxwell Speed Distribution

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Maxwell Speed Distribution

: Most probable speed (peak speed)

: Mean speed (average speed)

: root-mean-square (rms) speed (associated with K)

*

rms

: the standard deviation of the molecular speeds

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Maxwell Speed Distribution

Compute the mean molecular speed in the light gas hydrogen (H2) and the heavy gas radon (86Rn), both at room temperature 293 K.

The mass of the hydrogen molecule is twice that of a hydrogen atom (neglecting the small binding energy), or 2 x 1.008 u = 2.02 u.

The mass of radon is 222 u.

 Most other gases have molecular masses that fall between these two extremes

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9.5: Classical and Quantum Statistics

There is no restriction on particle energies in classical physics.

 Particles are distinguishable.

There are only certain energy values allowed in quantum systems.

 Particles are not distinguishable.

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Classical Distributions

Rewrite the Maxwell speed distribution in terms of energy:

 Maxwell-Boltzmann energy distribution

 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

Relative probability that an energy state is occupied at T

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Quantum Distributions

In quantum theory, particles are described by wave functions.

Identical particles cannot be distinguished from one another if there is a significant overlap of their wave functions.

Characteristic of indistinguishability that makes quantum statistics different from classical statistics.

 If the particles are distinguishable, the possible configurations for distinguishable particles in either of two energy states, A and B:

 the probability of each is one fourth (0.25).

 If the particles are indistinguishable, the possible configurations are

 the probability of each is one third (~ 0.33).

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Quantum Distributions

Because some particles do not obey the Pauli exclusion principle, two kinds of quantum distributions are needed.

 Fermions:

Particles with half-spins that obey the Pauli principle.

 Bosons:

Particles with zero or integer spins that do not obey the Pauli principle.

For Fermions:  Fermi-Dirac distribution

For Bosons:  Bose-Einstein distribution

 They differ only by the normalization constant and by the sign attached to the 1 in the denominator.

This sign difference causes a significant difference in the properties of bosons and fermions.

 Compare the Maxwell-Boltzmann distribution:

the F-D and B-E distributions reduce to the classical M-B when Bi exp(E) is much greater than 1.

This means that the Ml-B factor A exp(-E) is much less than 1.

(that is, the probability that a particular energy state will be occupied is much less than 1).

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Distribution functions

The normalization constants for the distributions depend on the physical system being considered.

Because bosons do not obey the Pauli exclusion principle, more bosons can fill lower energy states.

Three graphs coincide at high energies – the classical limit.

 Maxwell-Boltzmann statistics may be used in the classical limit.

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Classical and Quantum Distributions

Assume that the Maxwell-Boltzmann distribution is valid in a gas of atomic hydrogen.

What is the relative number of atoms in the ground state and first excited state at room temperature?

In the ground state (n =1) of hydrogen there are two possible configurations for the electron,  g(E1) = 2.

There are eight possible configurations in the first excited state (n=2),  g(E2) = 8.

for atomic hydrogen

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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

2

1 2

2 2

E mv p

m p 2mE

px2p2y pz2

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 Eg pp dpg pBp dp

2 2

2 2 2 2

pmEpdpmdEp dpmE mdE

3

( ) 2

g EB m E dE n E( )

B 2m EdE A3

exp(E)C EdE

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

 

3/ 2

0 0

E 2N

N n E dE C Ee dE C

  n E dE( ) 2N

3Eexp(

E dE)

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