MAXWELL - BOLTZMANN STATISTICS

Harvinder Kaur,

Associate Professor,

Head of Department, Physics

GCG 11, Chandigarh

INTRODUCTION The Maxwell-Boltzmann distribution is

an important relationship that finds many applications in Physics and Chemistry. 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.

The Maxwell-Boltzmann distribution can be derived using Statistical Mechanics. It corresponds to the most probable energy distribution, in a collisionally-dominated system consisting of a large number of non-interacting particles. Since interactions between the molecules in a gas are generally quite small, the Maxwell-Boltzmann distribution provides a very good approximation of the conditions in a gas. This probability distribution is named after James Clerk Maxwell and Ludwig Boltzmann.

PHASE SPACE

In Mathematics and Physics, phase space, introduced by Willard Gibbs in 1901, is a space in which all possible states of a system are represented, with each possible state of the system corresponding to one unique point in the phase space. For mechanical systems, the phase space usually consists of all possible values of position and momentum variables

A combination of position and momentum space is known as phase space which is a six dimensional space. So, if a system consists of n particles then we need 6n co-ordinates to describe the behaviour of the system in the phase space.

At any instant of time, suppose one of the particles have its position co-ordinates lying between x and x+x,y and y+y, z and z+z and let its momentum coordinates lie between px and px+ px, , py and py+ py ,pz and pz+ pz. Then, the particle is said to lie in a phase space compartment having a volume,

= xyzpxpypz.

PHASE SPACE DIVISION Let the compartments in phase space be further divided into a very large number of elementary cells of equal

volume, d.

d = dx dy dz dpx dpy dpz = h3o (say)

where ho has the dimensions of length * momentum.

In Classical Physics the choice of phase space cell size (ho) is entirely in our hands. But in Quantum Mechanics we are not free to make ho as small as we like. The minimum volume of an elementary cell in phase space in Quantum Mechanical system is h3, where h is Planck’s constant

Total number of elementary cells in phase space = Total volume in phase space ------------------------------------- Volume of one elementary cell = dxdydz dpx dpy dpz

-----------------------------------------------

d where d = h3

o (which may approach zero) in Classical system h3

in Quantum Mechanical system

TYPES OF STATISTICS

STATISTICS

CLASSICAL STATISTICS(MAXWELL – BOLTZMANN)

QUANTUM STATISTICS

FERMI-DIRAC STATISTICS

BOSE - EINSTEIN STATISTICS

MAIN DIFFERENCES IN THREE STATISTICS

PARAMETER CLASSICAL OR M-B STATISTICS

BOSE-EINSTEIN STATISTICS

FERMI-DIRAC STATISTICS

ParticlesParticles are distinguishable

Particles are indistinguishable

Particles are indistinguishable

Size of phase space cell

The size of phase space cell can be as small as we require

The size of the phase space cell cannot be less than h3

The size cannot be less than h3

Number of cells

If ni be the number of particles and gi the number of cells then gi>>ni so ni/gi<<1. Thus,number of cells can be made as large as possible

The number of cells is less than or comparable to the number of particles gi < ni

ni/gi 1

The number of cells has to be greater or equal to the number of particles so ni /gi 1

Restriction on particles No restrictionNo restriction Restriction due to Pauli

Exclusion Principle

Two particle distribution in two cells

The particles are distinguishable and can be arranged in four ways

The particles are indistinguishable and can be arranged in three ways

No two particles can occupy the same cell. Hence, only one arrangement

A B

B A

ABAB

BASIC APPROACH IN THE THREE STATISTICS

The basic approach in three statistics is essentially the same but we get the different results due to different assumptions. The common approach is as follows :

In any dynamic isolated system, the total number of particles (n) and the total energy (U) has to remain constant

n = ni = constant

dn = dni = 0 ----(1)

dU = ui dni = 0 ----(2)

When the system is in equilibrium then it exists only in the most probable state. From the assumptions of a given kind of statistics, we calculate thermodynamical probability (W) for any given macrostate. In the most probable state of the system W must be maximum. For all natural systems W is a very large number. We, therefore, deal with ln W. For most probable state W is maximum

d(ln W) = 0 -------(3)

The three equations (1), (2) and (3) must be simultaneously satisfied by the system irrespective of the kind of statistics to be applied. These three conditions can be incorporated into a single eq. by the method of Lagrange’s undetermined multipliers. We multiply eq. (1) with - , eq. (2) by - and add to eq. (3)

d(ln W) - dni - ui dni = 0 d(ln W) - ( + ui ) dni = 0 ----- (4)

MAXWELL – BOLTZMANN STATISTICS APPLIED TO AN IDEAL GAS IN EQUILIBRIUM

Consider n molecules of an ideal gas enclosed in a vessel of volume V, moving about in all directions, colliding against one another and against walls of the enclosure with energies ranging from 0 to u k . Let U represents the total energy of all the molecules in the gas which remains constant for an isolated system . In the equilibrium position , the system is in the most probable state.

Suppose n1,n2,---,nk be the number of molecules, g1,g2---,gk be the number cells in the phase space corresponding to the energy intervals 1,2 ,---,k respectively. Then,

kThermodynamic Probability , W(n1 , n2-----------nk) = n! (gi)ni/ni! i=1

k k ln W = ln (n !)+ ni ln gi - ln (ni!) i=1 i=1

Taking Natural Logarithms

Apply Stirling formula, ln n! = n ln n –n

k kln W = n ln n – n + ni ln gi - (ni ln ni - ni)

i=1 i=1

u1 u2 uk

MAXWELL – BOLTZMANN STATISTICS CONTINUED

ui dni = 0 …….(7)

k ( ln gi –ln ni - - ui) dni = 0 ………(8)

i=1

ln gi –ln ni - - ui =0

In the equilibrium state, W is maximum, so that

(d ln W) = 0

kd ln W = ( ln gi –ln ni) dni

i=1

k ( ln gi –ln ni) dni = 0 ………(5)

i=1

i.e.,

dni = 0 …..(6)

Multiplying eq. 6 with - and eq 7 with - and adding to eq. 5 we get,

ni = gi/(e eui) ni e-u

i

This law is known as Maxwell-Boltzmann law of energy distribution and e-ui is

Boltzmann constant

i.e.,

Hence,

NUMBER OF PHASE SPACE CELLS

g(p)dp = (4V / ho

3) 0 p2 dp

Number of molecules in a unit interval around u are

U = p2/2m

n(u) = g(u) e- e-u

g(p)dp = ( V4p2 dp) / ho3

Considering the spherical volume between the momentum value p and p+dp

So, in momentum coordinates,

where, = Total volume in momentum space & V = Total volume in position space

Total number of available phase space cells,

n(p) = g(p) e- e -p2 /2m

Number of cells in phase space corresponding to the momentum interval p to p+dp

g(p) dp = dxdydz dpx dpy dpz

---------------------------------

ho3

n(p) dp = 4Vp2 dp e- e -p2 /2m ---------------

ho3

or = (V/ ho

3)(4/3) p3 max= V / ho

3

VALUES OF AND

n = n(p) dp = 4V e- p2 e -p2 /2m dp ---------------

ho3

n(p) dp = 4 n (/2m) 3/2 p2 dp e -p2

/2m

As stated earlier, total number of molecules, n,

Using standard integral, x2 e-ax2 dx = /(4 a3/2 )

n = n(p) dp = 4V e- ( / (/2m) 3/2 )---------------

4ho3

Therefore, e- = (n ho3 /V) (/2m) 3/2

U = p2/2mUsing

n(u) du = (2n/ ) () 3/2 ue -u du ……..(9)

VALUES OF AND CONTINUEDTotal energy for the system of n molecules is given by,

THIS VALUE OF IS APPLICABLE TO ALL THREE STATISTICS

i.e.,

U = u n(u) du = 3/2 nkT

3/2 nkT = (2n/ ) () 3/2 u3/2e -u du

Using standard integral, x3/2 e-ax dx = (3/4 a2) (/a)

we get,

= 1/KT

Substituting the value of in eq. 9,

n(u) du = (2n/ (kT)3/2 ) u. e –u/kt du ………..(10)

n(p) dp = (2n/ (mkT)3/2 ) p2e –p2/2mkt dp

Using p = m v

n(v) dv= (2nm3/2/ (kT)3/2 ) v2e –mv2/2kt dv ………...(11)

Eq. 10 and 11 are known as Maxwell-Boltzmann law of distribution of energy and law of distribution of velocity respectively

GRAPHICAL DISTRIBUTION OF MAXWELL-BOLTZMANN SPEED DISTRIBUTION

Let f = n(v)/n denote the fraction of molecules which have speeds lying in a unit velocity interval around the value v.

If we plot f(v) against v, using equation (11), we get this graph for a particular value of temperature, T. The value corresponding to the most probable speed (vmp), df/dv = 0f = Cv2 e –mv2/2kT

C = 2 (m/kT)3/2

vmp = f/v = (2kT/m)

The average speed of molecules = v is defined as :

v n(v) dvv = ----------- n(v) dv

vrms = ( v 2) = (3kT/m)

The root mean square speed (vrms) is given by the expression :

v = (8kT/m)

MAXWELL –BOLTZMANN DISTRIBUTION : AN EXPERIMENTAL VERIFICATIONI.F Zartman and C.C Ko performed an experiment to study the distribution of velocities in a molecular beam and hence to verify the theoretical results obtained from Maxwell Boltzmann velocity distribution.

Apparatus An oven (in which bismuth was heated at about 800oC Slits S1, S2 and S3 along one side of the drum A drum (D) rotating about Z axis (~ 100 rps) A glass plate P, fixed at the face opposite to S3. The whole system is enclosed in a vessel which is highly evacuated to ensure that Bi molecules can go into the drum from S1 to drum without suffering any collision with air molecules

To Vacuum Pump

WORKING OF ZARTMAN & C. C. KO EXPERIMENT

When the drum is rotating the slit S3 comes in line with the S1 and S2 once and a bit of Bi vapours enter into the drum. These vapours will be deposited on the different points on the glass plate depending upon their velocities and speed of the rotation of the drum.

The oven is maintained at fixed temperature and the speed of rotation of drum is also constant. The slower the molecule, more is the time taken by it to reach the glass plate. So, slower molecules get deposited at larger distance from the point O.

The distance from point O is thus a measure of molecular speed. The density of Bi atom at various points is determined. This gives the relative number of Bi molecules at various speeds and this distribution is found to agree with the theoretical predictions of Maxwell-Boltzmann Statistical Law Of Molecular Speeds

SUMMARY

A knowledge of 6n coordinates in the phase space for all the n particles of the system completely specifies the position and momentum of the particles.

Maxwell – Boltzmann distribution treats the particle distinguishable since number of cells is much greater than the number of particles.

The M–B distribution describes particle speeds in gases, where the particles do not constantly interact with each other but move freely between short collisions. It describes the probability of a particle's speed as a function of the temperature of the system, the mass of the particle.

The most probable, average and root mean square speeds of particles are (2kT/m) , (8kT/m) and (3kT/m) respectively.

Thanks