[PDF] Statistical Mechanics and Thermodynamics of Simple Systems





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  • What is the formula for magnetization in statistics?

    This may be shown as M = Nm/V where M is the magnetization, N is the quantity of the magnetic moment, m is its direction and V is the volume of the sample. Calculate the magnetism in terms of the magnetic fields.
  • What is the formula for magnetization per unit volume?

    For the susceptibility ?, the definition is M =?H, where M is the magnetization (magnetic moment per unit of volume) and H is the magnetic field strength. This ? is dimensionless, but is expressed as emu/cm3.
  • What is the formula for the magnetic moment?

    ?=p×B. The SI unit for magnetic moment is clearly N m T?1. ?=IA×B.
  • The curie law states that in a paramagnetic material, the material's magnetization is directly proportional to an applied magnetic field. But the case is not the same when the material is heated. When it is heated, the relation is reversed i.e. the magnetization becomes inversely proportional to temperature. ? = C/T.
Statistical Mechanics and Thermodynamics of Simple Systems

Handout 6

Partition function

The partition function,Z, is defined by

Z=∑

ieEi (1) where the sum is over all states of the system (each one labelled byi). (a) The two-level system:Let the energy of a system be either-∆=2 or ∆=2. Then

Z=∑

ieEi= e∆=2+ e∆=2= 2cosh(∆ 2 :(2) (b) The simple harmonic oscillator:The energy of the system is (n+1 2 )¯h!wheren=

0;1;2;:::, and hence

Z=∑

ieEi=1 n=0e(n+1 2 )¯h!= e1 2

¯h!1∑

n=0en¯h!=e1 2

¯h!

1-e¯h!;(3)

Using the partition function to obtain functions of state The table below lists the thermodynamic quantities derived from the partition functionZ. Function of state Statistical mechanical expression

U-dlnZ

d

F-kBTlnZ

S=-(@F

@T V=UF T kBlnZ+kBT(@lnZ @T V p=-(@F @V

TkBT(@lnZ

@V T

H=U+pVkBT[

T(@lnZ

@T V +V(@lnZ @V T]

G=F+pV=H-TS kBT[

-lnZ+V(@lnZ @V T] C V=(@U @T VkBT[

2(@lnZ

@T V +T(@2lnZ @T 2) V] You probably only need to remember the rst two; the others can be quickly worked out. 1 kT CV/k B kT S k kT U kTh CV/k B kTh S k kTh U h

The internal energyU, the

entropySand the heat ca- pacityCVfor (a) the two- state system (with energy levels±∆=2) and (b) the simple harmonic oscillator with angular frequency!.

Combining partition functions

Suppose the energy contains two independent contributionsaandbwith energy levelsEaiand E bj, respectively, then

Z=∑

i∑ je(Eai+Ebj) =ZaZb;(4) i.e. the product of the partition functions for theaandbsystems. The generalization to more independent contributions is obvious:Z=ZaZbZc:::. Following from this, ifZ(1) is the partition function for one system, then the partition function for an assembly ofNdistinguishablesystems each having exactly the same set of energy levels (e.g.Nlocalized harmonic oscillators, all with the same frequency) is

Z(N) =ZN(1):(5)

If theNsystems areindistinguishable(e.g. an ideal gas of identical atoms or molecules) then

Z(N) =ZN(1)

N!:(6)

Example: the spin{

1 2 paramagnet In quantum mechanics, a particle with spin angular momentum equal to 1 2 , placed in a magnetic fieldBalong thezdirection, can exist in one of two eigenstates: | ↑⟩, with angular momentum parallel to theBfield, and hence magnetic moment along zequal to-B(costing an energy +BB). | ↓⟩, with angular momentum antiparallel to theBfield, and hence magnetic moment alongzequal to +B(costing an energy-BB). 2 HereB=e¯h=2mis theBohr magnetonand we have used the fact that energy=-·B, and also that for a negatively charged particle (the electron) the angular momentum is antiparallel to the magnetic moment.

Therefore, one spin-

1 2 particle behaves like a two-state system, with the two states having energiesE=±BB, and the single-particle partition function is simply

Z(1) = eBB+ eBB= 2cosh(BB):(7)

A spin-

1 2 paramagnetis an assembly ofNsuch particles which are assumed to benon- interacting, i.e. each particle is independent and "does its own thing". TheN-particle partition function, treating the spin-1 2 particles as distinguishable, is given by

Z(N) =ZN(1) = [2cosh(BB)]N;(8)

and henceFis given by

F=-kBTlnZ(N) =-NkBTln[2cosh(BB)]:(9)

We can work out the total magnetic momentof the paramagnet by computing =-(@F @B T =NBtanh(BB):(10)N N

The behaviour of, given

by eqn (10), is shown in the figure on the left. The magnetizationMis the magnetic moment per unit volume, so M= V =NB V tanh(BB):(11) Themagnetic susceptibilityis defined byM=HwhereHis a small applied field, or more formally=(@M @H T. WhenBB≪1 we can use tanhx≈xforx≪1 to find that

M≈N2BB

V k

BT:(12)

By definition,B=0(H+M) =0(1 +)Hfor a paramagnet. For a weakly magnetic material (like a paramagnet)≪1, and therefore ≈0M B =N02B V k

BT:(13)

This yieldsCurie's law:

∝1 T :(14) ATB

Michaelmas 2011

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