shihabudheen m m e s kalladi college mannarkkad. statistical distributions- maxwell-boltzmann...

Statistical MechanicsShihabudheen M

M E S Kalladi CollegeMannarkkad

Statistical distributions-Maxwell-Boltzmann statistics –Distribution of molecular energies in an ideal

gas-Average molecular energy- Equipartition theoremMaxwell-Boltzmann speed distribution law-Expressions for

rms speed, most probablespeed and mean speed

The branch of physics called statistical mechanics considers how the overall behavior of a system of many particles is related to the properties of the particles themselves.

The statistics obeyed by any system generally belongs to one of the following two categories:

Classical StatisticsQuantum Statistics.

Classical statistics treats particles as distinguishable whereas quantum statistics treats them as indistinguishable.

Statistical DistributionsStatistical mechanics deals with the behavior of

systems of a large number of particles. We give up trying to keep track of individual

particles.If we can’t solve Schrödinger’s equation in closed

form for helium (4 particles) what hope do we have of solving it for the gas molecules in this room (10really big number particles).

Statistical mechanics handles many particles by calculating the most probable behavior of the system as a whole, rather than by being concerned with the behavior of individual particles.

Four balls and two boxes

1 2

statistical mechanics determines the most probable way in which a certain total amount of energy E is distributed among the N members of a system of particles in thermal equilibrium at the absolute temperature T.

Thus we can establish how many particles are likely to have the energy 1, how many to have the energy 2, and so on.

E N

4 n4

3 n3

2 n2

1 n1

The particles are assumed to interact with one another and with the walls of their container to an extent sufficient to establish thermal equilibrium

More than one particle state may correspond to a certain energy .

If the particles are not subject to the exclusion principle, more than one particle may be in a certain state.

Let W be the number of different ways in which the particles can be arranged among the available states to yield a particular distribution of energies

The greater the number W, the more probable is the distribution.

It is assumed that each state of a certain energy is equally likely to be occupied. (equal a priori probability )

The program of statistical mechanics begins by finding a general formula for W for the kind of particles being considered.

The most probable distribution, which corresponds to the system being in thermal equilibrium, is the one for which W is a maximum, subject to the condition that the system consists of a fixed number N of particles whose total energy is some fixed amount E.

(The No. of particles in a state of energy E) = (The No. of states having energy E) x (probability that a particle occupies the state of energy E).

If we know the distribution function, the (probability that a particle occupies a state of energy E), we can make a number of useful calculations.

Mathematically, the equation is written

g( ) = number of states of energy (statistical weight corresponding to energy )

f( ) = distribution function = average number of particles in each state of energy = probability of occupancy of each state of energy

= n ε g ε f ε

In systems such as atoms, only discrete energy levels are occupied, and the distribution g() of energies is not continuous.

On the other hand, it may be that the distribution of energies is continuous, or at least can be approximated as being continuous. In that case, we replace g(ε) by g(ε)dε, the number of states between ε and ε+dε.

We will find that there are several possible distributions f(ε) which depend on whether particles are distinguishable, and what their spins are.

Three different kinds of particles:

Identical particles that are distinguishable.

Two particles can be considered distinguishable if their separation is large compared to their de Broglie wavelength.

Example: ideal gas molecules. In quantum terms, the wave functions of the

particles overlap to a negligible extent.The Maxwell-Boltzmann distribution

function holds for such particles.

Three different kinds of particles

Identical particles of zero or integral spin that cannot be distinguished one from another because their wave functions overlap.

Such particles, called bosons They do not obey the exclusion principleThe Bose-Einstein distribution function

holds for them. Photons are bosonsWe shall use Bose-Einstein statistics to account

for the spectrum of radiation from a blackbody.

Three different kinds of particles

Identical particles with odd half-integral spin that also cannot be distinguished one from another.

Such particles, called fermions,They obey the exclusion principle, The Fermi-Dirac distribution function holds

for them. Electrons are fermions We shall use Fermi-Dirac statistics to study the

behavior of the free electrons in a metal that are responsible for its ability to conduct electric current.

Statistics

Classical statistics

Maxwell-Boltzmann Statistics

Quantum statistics

Bose-Einstein statistics

Fermi-Dirac

statistics

The Maxwell-Boltzmann distribution:We know that classical particles which are

identical but far enough apart to be distinguishable obey Maxwell-Boltzmann statistics.

Or Two particles can be considered distinguishable if their separation is large compared to their de Broglie wavelength.

The Maxwell-Boltzmann distribution function states that the average number of particles fMB(ε) in a state of energy ε in a system of particles at the absolute temperature T is

-ε/kTf ε = A e .

The is the distribution function for ‘n’ no. of particles in ‘g’ states.

The number of particles having energy ε at temperature T is

A is like a normalization constant; we integrate n(ε) over all energies to get N, the total number of particles.

ε is the particle energy, k is Boltzmann's constant

-ε/kTn ε = A g ε e .

Molecular Energies in an ideal Gas

We assume a continuous distribution of energies so that, the no. of particles having energies between ϵ and ϵ+dϵ is given by

where g(ϵ) dϵ is the no of states in the energy range ϵ and ϵ+dϵ. g(ϵ) is called density of states.

It turns out to be easier to find the number of momentum states corresponding to a momentum p, and transform back to energy states.

-ε/kTn ε dε = A g ε e dε .

Corresponding to every value of momentum there is a value of energy.

Momentum is a 3-dimensional vector quantity. Every (px,py,pz) corresponds to some energy.

We need to find how many momentum states are in this spherical shell.

We count how many momentum states are there in a region of space and then transform to the density of energy states.

2 2 2x y zp = 2 m ε = p + p + p .

The Maxwell-Boltzmann distribution is for classical particles, so we write

The number of momentum states in a spherical shell from p to p+dp is proportional to 4πp2dp (the volume of the shell).

Thus, we can write the number of states having momentum between p and p+dp as

2g p dp = B p dp

where B is a proportionality constant, which we will worry about later.

Because each p corresponds to a single ε,

2g ε dε = g p dp = B p dp .Now,

2 -1/21p = 2mε p = 2mε dp= 2m ε dε ,

2

so that 2 -1/2p dp ε ε dε , -1/2g ε dε ε ε dε .

and -ε/kTn ε dε = C ε e dε .

The constant C contains B and all the other proportionality constants lumped together.

To find the constant C, we evaluate

-ε/kTn ε dε = C ε e dε

-ε/kT

0 0N = n ε dε = C ε e dε

where N is the total number of particles in the system.Now

The result is 3/2C

N = kT , 2

kTaaa

dxex ax \12

1

0

2

1

so that

-ε/kT3/2

2 Nn ε dε = ε e dε .

kT

This is the number of molecules having energy between ε and ε+dε in a sample containing N molecules at temperature T.

“It forms the basis of the kinetic theory of gases, which accurately explains many fundamental gas properties, including pressure and diffusion.”“The Maxwell-Boltzmann distribution also finds important applications in electron transport and other phenomena.”

Here’s a plot of the distribution:

Continuing with the article, the total energy of the system is

3/2 -ε/kT3/2

0 0

2πNE = ε n ε dε = ε e dε .

kT

Evaluation of the integral gives

This is the total energy for the N molecules, so the average energy per molecule is

3NkTE = .

2

3ε = kT ,

2

exactly the result you get from elementary kinetic theory of gases.

Calculation:

aadxex ax

20

2

3

4

3

kTkT

kT

NE

2

2

3 4

3

}{

2

3NkTE = .

2

Results of Kinetic TheoryKE of individual particles is related to the

temperature of the gas:

½ mv2 = 3/2 kT

Where v is the average velocity.

Boltzmann DistributionDemonstrated

that there is a wide range of speeds that varies with temperature.

Equipartition of EnergyEach degree of translational freedom takes ½

kT. KEx +KEy+KEz = ½ kT + ½ kT + ½ kT

KEtotal = 3/2 kT

This is true for single point masses that possess no structure.

Each new DOF requires ½ kT of energy.Each new DOF contributes ½ kT to the total

internal energy of the gas. This is the Equipartition Theorem.

Equipartition of EnergyFor molecules, i.e. multi-atom particles, there

are added degrees of freedom.

Internal Energy of Di-AtomThree translational DOF + 2 rotational DOF

= 5 DOF.Each DOF contributes ½ kt, so the internal

energy of a diatomic gas is,U = 5/2 NkT,

For a gas of N molecules.

Because ε = mv2/2, we can also calculate the number of molecules having speeds between v and v + dv. The result is

23/2

2 -mv /2kT3/2

2 N mn v dv = v e dv .

kT

Here’s a plot (number having a given speed vs. speed):

“We” (Beiser) call this n(v).

The speed of a molecule having the average energy comes from solving

for v. The result is

2mv 3ε = = kT

2 2

2rms

3kTv = v = .

m

vrms is the speed of a molecule having the averageenergy .

ε

It is an rms speed because we took the square root of the square of an average quantity.

The average speed can be calculated fromv

0

0

v n(v) dvv = .

n(v) dV

The result is

8kT

v = .m

Comparing this with vrms, we find that

rmsv = 1.09 v .

You can also set dn(v) / dv = 0 to find the most probable speed. The result is

p

2kTv = .

m

The subscript “p” means “most probable.”

Summarizing the different velocity results:

rms

3kTv =

m 8kT

v = m

rmsv = 1.09 v

p

2kTv =

m

Plot of velocity distribution again:

n(v

)

Example 9.4 Find the rms speed of oxygen molecules at 0 ºC.

You need to know that an oxygen molecule is O2. The atomic mass of O is 16 u (1 u = 1 atomic mass unit = 1.66x10-27 kg).

-27-26

2

1.66×10 kgmass of O = 2 16 u = 5.31×10 kg

u

rms

3kTv =

m

0 m/s? 10 m/s? 100 m/s? 1,000 m/s? 10,000 m/s?

-23

rms -26

3 1.38×10 J /K 0+273 Kv =

5.31×10 kg

rmsv = 461 m/s

rmsv = 461 m/s mile / 1610 m 3600 s/h

rmsv = 1031 miles / hour .