we start consid ering phase transitions. van der waals...

Lecturer: Anna Lipniacka Date: 6&13/05/2003

File: /home/lipniack/statistic_physics/lecture6/lecture6.sdd

lecture 6

The Outline, lecture 6

− Van der Waals Equation of State

− Vapor pressure equation

− Critical points of the Van der Waals gas − exercise, nucleation

− ferromagnetism

− Landau theory of phase transitions

Phase transitions

next time (Tuesday next week)

We start consid ering phase transitions. Van der Waals equation if state is what lets us to consider interactions between "gas" molecules, we can hope to use it to describe phase transitions.



lecture 6

The Summary,lecture 5

The Gibbs free energy defined as : has a minimum in thermalequilibrium at constant temperature and pressure

G�

U � � � � pV

�G�N p, � � � ,

�G� � p,N

� � � ,�

G�p � ,N

� V

G N, p, N � p, The law of mass action states that ratios of concentration of the reactants in chemical equilibrium are a function of the temperature alone�

j

nj

j � K �

dG � � dN � d � � � Vdp and dG � �G�N p, � dN � �

G� � p,N

d � � �G�p � ,N

dp

we got it considering the minimum of G, and ν are reaction coefficients, n are reactants concentrations

we considered the following form of chemical potential: � � � � ln Zint � � ln

n

nQ�

j

nQj

�j exp

� �jF

jint

� � K � where F = −τ ln Z



lecture 6

Van der Waals gas

Distance between molecules, R

Pot e

ntia

l ene

rgy

~ φ(

r)

0

Repulsive hard interaction

attractiveinteraction

Van der Waals gas has interactions between between gas particles, which is strongly repulsivewhen particles approach, and weakly attractive at long distances. Van der Waals interactionsare a model of interactions of neutral molecules with electric (or magnetic) dipole moments.Repulsive interactions arises on distances when the electron clouds from different particles start to "feel" each other.

We have to modify the equation of state of the non−interacting ideal gas, to take into accountVan der Waals forces

pV � N � � p � � p V � Nb � N �pressurecontribution fromattractive forces

Effective volume decreasedue to hard core repulsion

F � U � �We will consider the formof free energy

including the interaction, toderive ∆p

VR V � Nb



lecture 6

The mean field method

F � U � � Uideal gas ! " U � � and p � # F

# V $ ,N

We will calculate ∆ U frominteractions with"mean field" approach

b

each particle takes an effective volume band feels mean interaction of other particles r

The energy of a given particle (1) in the field produced byother particles will be :%

U1 & '

i ( 2

N )r

1,i & 'i ( 2

N )r

i * +V

dN , dV)

r dV

where we have placed our "test" particle in the center of coordinate system. We willnow assume that the concentration of particles does not depend on r, so dN/dV =n. We have defined here :

= +V

n)

r dV & n +b

Vbox )

r dV & n - 2a

+b

Vbox )

r dV . - 2a

We have now " U 1

2N " U

1 � a

N2

V

F Fideal gas

� 2aN2

V N ln

N

VRn

Q

� 1 � aN2

V

p � # F

# V $ ,N

N

VR

� aN

V

2

We want to calculate pressure modification due anergy component. To calculate this mean interaction energy for "gas" in the box we use mean field approach − all particles "bath" in mean interaction field. We calculate energy change of a test particle, than to calculate total energy change of the "gas" we have to multiply by 1/2N where N is the number of particles. Why 1/2 N ? To not to count energy changes from two−particle interactions twice− this is 1/2 factor is typical in mean field approach.

Note that parameter "a" has dimensions of energy*volume, while parameter "b" has dimensions of volume



lecture 6

Van der Waals equation of state

p N

V � Nb� a

N

V

2

with we get p ! aN

V

2

V � Nb N

aN =3,Nb=1/3 ,(N=3)2

p

V

τ=0.85τ=0.90

τ=1.0

τ=1.05

p

V

ideal gas, N=3, a=0, b=0

τ=1.05

τ=0.85

arbitrary units of p and V

pV=Nτ

We obtained Van der Waals equation of state. Note that b and a parameters will depend on the type of "substance", we would like to get the equation in such a form that is "substance" independent. We hope that the equation will describe both gas and liquid (solid?)



lecture 6

Critical points of Van der Waals Gas

To study the interesting behavior of the Van der Waals equation one has to expressp,V,τ in their natural units for this type of gas. Parameters a and b will in general dependon the type of gas (they depend on interaction potential), lets see if we can write Van der Waalsequation in some generic form.

p / aN

V

2

V 0 Nb 1 2 N

Lets define: pc 3 a

27b2; V

c 3 3Nb ; 4c 3 8a

27b

and re−write the equation usingthese new coefficients:

p

pc

5 3

V 6 Vc

2

V

Vc 7

1

3 8 86 3 99 c

= : p 5 3:V 2:V 7

1

3 8 8

3:9;

p 3 p

pc

;;V 3 V

Vc

;;4 3 44

c

τ=0.85 τ_cτ=0.90τ_c

τ=1.0 τ_c

<p

<V

in terms of thesevariables all VdWgases look alike

1

1point of inflection:where local maximum and minimum coincide

;p 3 ;

V 3 ;4 3 1

Defining so called critical p,T,V lets us define "reduced"− dimensionless parameters of Van der Walls equation , and we can write it in a form which does not depend explicitly on a,b any more. Written in reduced parameters Van der Waals equation should be good for many substances.

We hope that the p−V isotherms as on the picture will describe liquid−gas isotherms . We leave VdW equation for a while and discuss real p−V isotherms.



lecture 6

Isotherm definition

The pressure− volume dependence for a given temperature and quantity of matter (eq. p= N τ/ V ) is called ISOTHERM. It is determined by the Helmholtz free energy of the substance:

p � # F V, ,N

# V $ ,N

We will consider isotherms of a real gas, which molecules can associate themselves into liquid or solid PHASE. Phase is a portion of the system uniform in composition.Liquid and vapour (gas) may coexist on a section of isotherm, if it is for a temperaturebelow critical temperature, above which only one phase can exist.Vapour =gas in equilibrium with liquid

p

V

liquid + gas

liquid

gas

Some critical temperatures in KHe −> 5.2 , H_2 −> 33Ne 44.4 O_2 −> 154Xe 289.7 H_2 0−> 647.1

for T < criticaltemperature

To get pressure at constant temperature we must use derivative of free energy F , as it is relevant to systems at constant temperature .

Below critical temperature, part of the isotherm will show liquid−gas coexistence, above critical temperature there is no difference between gas and liquid phase.



lecture 6

Phase coexistence curve.

For certain conditions of pressure and temperature various phases might coexist− stay inequilibrium. The equilibrium conditions for phase coexistence are equilibrium conditionsfor two systems in thermal, diffusive and mechanical contact: −> τ, µ and p must equalize

Example for gas<−> liquid coexistence condition: =g > =

l;p

g > pl; ?

gp, = > ?

lp, =

Lets assume two phases can coexist for certain p and τ, lets try to find if they can coexist for p+∆p and τ+ ∆τ. Our purpose is to find p(τ) along which two phases can coexist.

@ p A %p, B A % B & @ p, B A C @

C p

%p A C @

C B % B

?g

p D E p, = D E = > ?l

p D E p,= D E =if so:

@g

p, B A C @g

C p

%p A C @

g

C B % B & @l

p, B A C @l

C p

%p A C @

l

C B % BC p

C B &C @

l

C B - C @g

C B p

C @g

C p- C @

l

C p F

We will try to find now the phase coexistence curve− such a pressure for a given temperature that phases can coexist− flat part of p−V diagram for a given temperature.

We specify the conditions for phase coexistence, thermal, mechanical (pressure) and diffusive (chemical potential) equilibrium.



lecture 6

Clausius Clapeyron equation

Lets use G to express the derivatives: # G

# N p, $ � , # G

# p,N

� � , # G

# p $ ,N

VG N, p, N � p, C @

C p & 1

NC G

C p & V

N. v ; C @

C B & 1

NC G

C B & - GN

. - s

C p

C B &C @

l

C B - C @g

C B p

C @g

C p- C @

l

C p F& s

g- s

l

vg

- vl

entropy and volume per particle

Gas−liquid coexistence curve. The number of particlesin each phase will vary along the curve, only N_g+N_l=const=N. Remembering that (τ∗σ)=Heat in a reversible process at constant temperature, we define the latent heat −> heat that must be supplied per particle to move it from liquid to gas phase at the constant pressure and temperature −> "adjust" its entropy reversibly.

L H I sg J s

l latent heat of vaporization

C p

C B & LB %v

Clausius−Clapeyron equation for vapor pressure−>verified experimentally to high accuracy



lecture 6

Side remark, enthalpy and latent heat

Heat transfer at constant pressure is described by the change of enthalpy :

H � U ! pV K dH dU ! pdV ! Vdp d � � pdV ! pdV ! Vdp d � ! Vdp

Thus at constant pressure :

dH d � ! Vdp K " Hp

" �p

L " � Hgas

� Hliquid

Thus latent heat :

Values of enthalpy can be found from integrating heat capacity at constant pressure:

" Hp

" �p

Cp

� # �# p

# H

# p

K H L Cpd



lecture 6

approximate phase diagram for vapor −liquid (solid)

C p

C B & LB %v

with and assuming vg M v

l N O v P vg

P I Q p

we get C p

C B & LpB 2 R C ln p

C B & LB 2

if we further assume that the latent heat of vaporization does not depend on temperature we get

ln p = L0S T U const ; V p T W p

0exp X L

0S Tin standard units, with latent heatper mole we get: p T W p

0exp

X L0

N0k

bT

; N0 W 6.02Y 1023 particles S mol

p [mm Hg]

10 /T [1/K]

10

1

10

6

−2

1.5 5.0

3

critical point

liquid water

ice

Although for water the latent heat has substantial variationwith temperature , the vapor pressure in equilibrium withliquid and solid is quite well approximated by the straightline (1/T vs ln p)

0 C

Triple point: in triple point all three phases can coexistat the same pressure and temperature so ?

gp, = > ?

lp, = > ?

sp, =

gas



lecture 6

Phase diagram for water.

Lets consider solid−liquid, and liquid−gas phase diagrams. From Clausius−Clapeyron equationwe have :

C p

C B l Z solid& s

l- s

s

vl

- vs

& LmeltingB vl

- vs

C p

C B gas Z liquid& L

vaporizationB %v * L

vaporizationB vg

& Lvap

pB 2

C p

C B gas Z solid& L

vaporizationB %v * L

vaporizationB vg

& Lvap

pB 2

in case of water v_l < v_s , lets assume it does notdepend on pressure or temperature

lets assume also that all latent heats aretemperature−independent, and takeexperimental values for them

pl [ solid \ Lmelting

vl] v

s

ln ^ _ ^0 \ L 1_ gram ds dl

ds] d

l

ln T _ T0

=

=] 9̀ 334 J _ cm3 ` ln T _ T0

= ] 3̀ 104atm ` ln T _ T0

p a gas b liquid c p0exp

b L0

N0k

bT d p

0exp

b 4910K

T

p0 e 3.01 f 105 atm ;T

0 e 273.15 K

p [a

tm]

T [K]In the "log" scale triple water point is visible, next page

Here are my atempts to find phase diagram for water− gas −liquid −solid and solid−liquid coexistence, and to find out semi−theoretically triple point of water. Note that I had to take the latent heat and the p_0 and T_0 from experiment

Note that the diagrams describe as well hot the boiling and melting temperature changes with pressure. Note that because the volume per particle in ice is bigger than in water the melting temperature of water decreases a bit with pressure. This is "water anomaly".



lecture 6

Phase diagram for water−plots

From the formulas calculated before, in linear and log scale. Latent heats are taken fromthe experiment, also the p_0 and T_0 parameters, which are integration constants haveto be taken from the experiment.

p0 g 3.01 h 105 atm ;T

0 g 273.15 K

T [K]

p [a

tm]

p [a

tm]

T [K]

ice liquid water gas ice liquid water

gas

triple pointof water−vapour pressureat 273.15 K is around0.01 atm



lecture 6

Example model system for gas−solid equilibrium

From Clausius Clayperon equation we can calculate only the shape of phase diagram, as a function of measurable parameters, but we cannot calculate this parameters. For this we need a model of a solid (or liquid) in which we will be able to calculate µ, in analogy with what we do for the ideal gas model.

Below, we develop an oscillator model for "ideal solid", roughly applicable to "ideal liquid"too.

idealgas

Each atom in the structure is a bound harmonic oscillator with energy states

ideal solid

nhi � j0� j

0is the energy of bounded atom referred to a free atom at rest, so is the binding energy

j0

1.2"

� j0

0nh k l m

0

Zs

� nn

exp� nho � p

0� � expp

0� nn

exp� nho� � exp p

0q �1� exp � ho q �

partition function in the solid, per atom, Helmholzt and Gibbs free energy per atom of solidF = −τ ln Z , G=F+pv=µ

Chemical potential, temperature, pressure in thegas and in the solid have to be equal. As volumeper atom (v) in the solid is much smaller thanin gas, we can neglect pv term for the solid and write and:F = −τ ln Z ~G ~ µ r

s � exp�� s exp � ln Z

s� 1

Zs

We build here a model for the solid, which we can use for example to calculate chemical potential for the solid, and from there using the equality of chemical potentials for the gas and solid phase in equilibrium, to calculate the vapor pressure equation.

To calculate chemical potential we should calculate G per atom, but we will make an approximation for the solid that G/atom=F/atom, thus neglect p*volume_per_atom term.



lecture 6

Oscillator model, gas−solid equlibrium

m t2 u h2

3v 2 wn

Q

xig y exp z { y N

V nQ

n

nQ

y p{n

Q

rs � exp

� � s exp � ln Zs

� 1

Zs

� pg� nQ

and pg

� ps

From gas−solid equilibrium condition we have λ_| } λ_~ :

so : p � � nQ

Zs

� � nQexp

� p0� 1� exp

� ho�p � � 5� 2 m

2� h2

3� 2

exp� p

0� 1� exp� ho�

Coexistence for gas−solid equilibrium,or vapor pressure in equilibrium with solid. Notice that we assume that there are only vapor + solid in the bottle, no other gas exercising the pressure on the solid. What do we have to modify, if there would be ?

What can we learn from our formula ? Example 1) How many atoms of the solid evaporated?

pV W N T , V N W V T 3� 2 m

2� h2

3� 2exp

X �0T 1X exp

X h �T

Example 2) connection with Clausius−Clapeyron:

C p

C B & Lvap

pB 2 R C ln p

C B & LvapB 2

if h � � � � 1� exp � h � � � � h �� ln p e 3� 2ln � � �0

� � � const

Lvap & 3

2B A �

0

energy to get theatom from the solid+kinetic energy in thegas

We assume here that the chemical potential of vapor will be that of the ideal gas, so "absolute activity" = n/n_q



lecture 6

Van der Waals phase transition

Van der Waals equation models interaction between molecules, so it could serve as a simple model of phase transition between gas and liquid.

<p

<V

liquid

gas

� � 0.82�c

<V

1

<V

2

Lets try to find points on the isotherm in whichchemical potential of the liquid phase is the sameas chemical potential of the gas phase

Lets assume that

Gl

N, p V1 W N �

lp V

1 =

N �g

p V2 W G

gN, p V

2

dG � dN � d � ! Vdp Vdp

Lets calculate the difference of G between points 1 and 2, to verify our conjecture

Gg

� Gl

�V

1

V2

V dp 0

all atoms inliquidall atoms in gas

if hatched areas belowand above the line are the same

If hatched areas are the same , indeed G in point 1 and 2 are equal, and gas and liquid can coexist between V_1 and V_2. The proportion of gas and liquid phases must be such thatthey fill all available volume

coexistence line

We are trying to find now the gas−liquid coexistence line on the isothert (p−V relation) of the Van der Waals equation of state. The liquid vs gas can coexist if they have the same G. We have to find an "iso−G line" on our plot



lecture 6

Gibbs energy of Van der Waals gas.

As we might have judged from previous example , G ( τ,p,N) for Van de Waals "gas" might have a non−unique behavior as a function of p for a given N and τ . This is becauseVand der Waals "gas" equation describes in fact both liquid and gas phase, so for a givenp there might be several values of G allowed, corresponding to different phases.

Lets calculate G = F+pV

F W T N lnN

V X Nb nQ

X 1 X aN2

V

pV W T NV

V X NbX a

N2

V

=T N lnN

V X Nb nQ

X 1 X 2aN2

V U T NV

V X NbAlas, G cannot be analytically written as a function of p (instead of V). So we have to study it numerically. Example behavior:

p � pc

G�cN

0.7 0.9

−0.40

−0.48liquid

gas

vaporpressure

� g 0.9 �c

p� pc

G�cN

0.8 1.0 1.2

−0.30

−0.40liquid

gas

� g �c



lecture 6

Gibbs energy of VdW

G=T N lnN

V X Nb nQ

X 1 X 2aN2

V U T NV

V X NbpV W T N

V

V X NbX a

N2

V

� � �c

G�cN

0.986 0.988

−1.452

−1.454

liquid

gas

equilibriumbetween gas andliquid

p g 0.95 pc

� � �c

G�cN

0.9 1.0 1.1

−1.40

−1.50

liquid

gas

p g pc

The phase with the smaller G or (µ) is the stable phase in given conditions. If:

?g

p, = > ?l

p, =there is equilibrium between gasand liquid and phase transition can occur

For VdW (first order phase transition):� ?g� =

p

� � ?l� =

p � � g

�� l �

�G

g� =p

� �G

l� =p

Second order phase transition:� ?g� =

p>

� ?l� =

p

but

� 2?g� = 2

p

� � 2?l� = 2

p

# G

# p,N

� � ,

�G� �

p � �



lecture 6

Nucleation

Another interesting effect connected with gas −> liquid transition is nucleation −> condensationof vapor into droplets: If a transition from gas to liquid could begin. However, this transition has to go via droplets of liquid, which have higher energy− thus higher chemical potential than the bulk of the liquid. Higher energy is due to surface tension effects.

vapor

liquid

@l

p, B ¡ @g

p, B

Lets calculate the chemical potential of the liquid in the droplet

�l

droplet W �l

bulk U surface energy

N

We assume : surface En 8 4¢ R2£ andN 8 4¢ 6 3 R3nl¤

ldroplet 3 ¤

lbulk ¥ 3¦

R nl

so even if �l

bulk p, § �g

p,there is a potential difference to overcome to start to build small droplets¨

gp, © ª ¨

ldroplet « ¨

gp, © ª ¨

lbulk ª 3¬

R nl

« ¨ ª 3¬R n

l ® 0 if R ¯ 3 °± � nl

From size of R droplets grow spontanously, as their chemical potential is smaller than theseof the vapor− condensation centers "important" to help overcoming the small R barrier.



lecture 6

Nucleation, Gibbs energy.

R² 3 ³´ µnlFor there is spontanous condensation on the droplet, however droplets

of slightly smaller size do not evaporate spontaneusly, as this would increase their Gibbsenergy. Consider the Gibbs energy difference between the droplet and the correspondingvolume of the vapor:� G � N � ¶

gp, � � ¶

ldroplet � � 4

�3

R3� ¶ � 4� R2 ·∆ G

R of adroplet

R\ 3 ³´ µn

l

R\ 2 ³´ µn

l

spontaneus growth ofdroplets

evapora−tion ofdroplets

0

droplets do notevaporate

The R value at which the G barrier is the highest is called R critical :

Rc

� 2·� ¶ n

l

� 2·¶

g� ¶

lbulk n

l

we can approximate the vapour chemical potential with the one of ideal gas, and useequilibrium liquid−gas potential for liquidbulk phase: ´ µ \ ^ ln

p

peq

Using example p = 1.1 p_eq (positive ∆µ) we get for · � 72erg q cm2,� � 300K � R

c� 10̧ 6cm



lecture 6

Summary of last time

We introduced Van der Waals equation of state, which can describe both liquidand gas phase (to some extent)

p ! aN

V

2

V � Nb N

The coexistence curve between two phases must satisfy Clausius Clapeyron equation:

C p

C B & LB %v

where L is the latent heat and ∆¹ is the volume differenceper particle of two phases.

� ?g� =

p

� � ?l� =

p � � g

�� l �

�G

g� =p

� �G

l� =p

L � " � �1

� �2

and " v v1

� v2

Van der Waals "gas" undergoes first order phase transition with non−zero latent heat.

Latent heat is defined as L=∆σ∗τ where the change of entropy is (for example) per particle−> between two states. If the two states, liquid and solid for example have different entropy per particle, than the latent heat is non−zero.

Van der Waals (VdW) equation describes "gas" with interactions , thus it can describe both liquid and gas state and can be used to describe phase transition,between liquid and gas state. Phase transition can occur when both liquid and gas state have the same chemical potential− the same G per particle. From the shape of G as a function of pressure and temperature we see that its derivative changes at phase transition−> we have first order phase transition here.



lecture 6

The Outline, lecture 6

− Van der Waals Equation of State

− Vapor pressure equation

− Critical points of the Van der Waals gas − exercise, nucleation

− ferromagnetism

− Landau theory of phase transitions

Phase transitions

last week

today

Today we consider ferromagnetism and Landaustheory for phase transitions. We will use the

model of paramagnet ( spin systems ) to describe a ferromagnet. We will see that the ferromagnet udergoes a transition from ordered state ( ferromagnet) to disordered non−magnetic state a a certain critical temperature.



lecture 6

Mini−repetition of thermodynamic relations

º » ln g U,V,N ,14 » ¼ º

¼ U|V,N

, p 3 ½ ¼ U¼ V ¾ ¿ dUN À const 3 4 d º ½ pdV

dU 3 4 d º ½ pdV ¥ ¤ dN

F » U ½ 4 º ¿ dF 3 dU ½ º d 4 ½ 4 d º ¿ dFN À const 3 ½ º d 4 ½ pdV

º 3 ½ ¼ F¼ 4 V

, p 3 ½ ¼ F¼ V Á , ¤ » ¼ F¼ N|V, Á , ¿ dF 3 ½ º d 4 ½ pdV ¥ ¤ dN

heat work

G » F ¥ pV ¿ dG 3 dF ¥ pdV ¥ V dp ¿ dG 3 ½ º d 4 ¥ Vdp ¥ ¤ dN

º 3 ½ ¼ G¼ 4 p,N

, V 3 ¼ G¼ p Á ,N

, ¤ 3 ¼ G¼ N|

p, Á G 3 ¤ 4 , p N

ºN 3 ½ ¼ ¤

¼ 4 p,N

,V

N 3 ¼ ¤¼ p Á ,N

F 3 ¤ N ½ pV ¿ U º ,V,N 3 ¤ N ½ pV ¥ º 4



lecture 6

Ferromagnetism

Ferromagnet is a substance which has a non−zero magnetic moment even in the absenceof the external magnetic field. We developed a model for a paramagnet some time ago−lets try to reuse it now.

Reservoir, τ B

M Â nmexp 2mB Ã Ä Å 1

exp 2mB Ã Ä Æ 1

M Ç <2s>m / V

2s È N É Ê N Ë ,nÈ N Ì V

We have calculated before the magnetization (density of magnetic moment ) as a functionof the external magnetic field B. Assume now that there is no external magnetic field,but the the magnetic moments itself produce effective field proportional to magnetization.This is the CENRAL assumption for ferromagnets.

BE

Â Í M Î M Â nmexp 2m Í M Ã Ä Å 1

exp 2mB Í M Ã Ä Æ 1

introduce : ÏM Ð M

nm and ÏÑ Ð ÑÒ m2nÓ

M Â exp 2ÓM Ã ÓÄ Å 1

exp 2ÓM Ã ÓÄ Æ 1

so we get

can we solve this ? Note that thereis always M=0 solution, is thereanother one?

ÔM

Ô© « 1Ô© « 0.7

Ô© « 0.5

Ô© « 0.3if or there is no non−zero solution formagnetization. Critical temperature:

Ô© Õ 1 © Õ Ö m2n ©c

« Ö m2n

The spin systems we used before are our models of paramagnets. You have calculated in homework problems the density of magnetic moment as a function of magnetic field ( external) and temperature. We will use now this result.

We assume that there is no external magnetic field, but the magnetic moment s itself produce the magnetic field. We will see what is the solution for magnetization in this condition.

We see that for the temperatures less than critical temperature defined here there exists two solutions for magnetization. One of them is zero, and the other one non−zero. For temperatures above critical temperature only zero solution exists in absence of external magnetic field



lecture 6

Ferromagnetism, magnetization

ÓM Â exp 2

ÓM Ã ÓÄ Å 1

exp 2ÓM Ã ÓÄ Æ 1

Â tanh

ÓMÓÄ Î ÓÄ Â Ó

M

arctanhÓM

ÔM

for the non−zero solutionÔ© × ©©c

no magnetization

maximalmagnetization

Critical temperature above which there is no more spontaneousmagnetization is called Curie temperatureFor higher temperatures, there willbe still magnetization in a response to theexternal magnetic field−paramagneticmaterial. Critical temperature for iron= 1043 K.

©c

« Ö m2n

Critical temperature depends on the coefficient α between magnetization and effective magnetic field, on concentration of magnetic moments (spins) , and on the value of the elementary magnetic moment ( m) .

Our system will undergo ferromagnet−paramagnet phase for Curie temperature



lecture 6

Helmholtz energy of ferromagnetic system

F 2s,B, t wU Ø Ù t w Ø <2s>mB Ú t 1

2N 1 Ú 2s

Nln

1

21Ú 2s

NÚ 1

2N 1 Ø 2s

Nln

1

21Ø 2s

N

M Û <2s>m

VB

e W Ü M , U W X 12

V Ü M2

Paramagnet:

<M Ý M

nm� 2s

N,

<� Ý ��

c

Ý �Þ m2n

F ßM , tN t

c

w Ø 0.5 ßM2+

+ 0.5 ßt 1Ú ßM ln 1Ú ßM Ú 1 Ø ßM ln 1 Ø ßM Ø 2ln2

F ßM , tN t

c à Ø 0.5 ßM2 + 0.5 ßt ßM2 Ú 2

3ßM4 Ø 2ln2 Ú .. =

= Ø ßt ln2 Ú 0.5 ßM2 ßt Ø 1 Ú ßM4

3

Ferromagnet

ÔM

F

N ©c Ô© « 0.3

Ô© « 0.5Ô© « 0.7

Ô© « 1.0

stablephase

stable phase

Minimum at 0 for τ>τ_c á Fp

á Ä V

Â á Ff

á Ä V

We will write down now the Helmholtz free energy for ferromagnet. We will use the result for paramagnet at the top of the page. The result for paramagnet can be obtained directly from the definition of the free energy.

We note that the energy for paramagnet is U=−2<s>B, where 2<s> is the spin excess . The entropy for a given spin excess can be obtained as a logarithm of the number of states with a given spin excess which we calculated long time ago. We express temperature and spin excess in terms of reduced temperature and reduced magnetization, which is directly related to spin excess.

To get the formula which is valid for ferromagnet we assume that magnetic field is not external but internal effective field proportional to magnetization. As this is an effective field this introduces additional factor 1/2 in energy calculations− we cannot count twice the interaction of the same magnets.

We get the free energy formula for ferromagnets . Magnetization will be found from the minimum condition for F. (why are we using F not G).

The real value of F the system will take will be that of the minimum.



lecture 6

Helmholtz free energy of ferromagnet, continued

ßF â F ßM , tN t

c à Ø 0.5 ßM2 + 0.5 ßt ßM2 Ú 2

3ßM4 Ø 2ln2 Ú .. = Ø ßt ln2 Ú 0.5 ßM2 ßt Ø 1 Ú ßM4

3

What values magnetization will take ? We can obtain it from minimum of F ãFã äM Á W ä

M äT X 1 U 43

äM3 W 0V ä

M W 0 å 43

äM2 W 1X äT

We see that for temperature greater than critical temperature,<� Ý �

�c

æ1

the only solution for magnetization is zero. For

there exist also another solution:

1ç <� è 0äM2 W 3

41 X äT

What is F then ? if éê ë êê

c

ì1í éM î 0 í éF î ï éê ln2

ð� ��c ñ 1� ð

M2 � 3

41� ð� � ð

F � � ð� ln2� 3

81� ð� 2 � 3

161� ð� 2 � � ð� ln2 � 3

161� ð� 2

We calculate now the magnetization and the real F the system will take. We see that real F has a smooth transition at temperatures around critical temperature, there is no jump in the value of F. Verify the derivative of F over the reduced as a function of reduced temperature , you will see that it has no jump around critical temperature.



lecture 6

Landau theory of phase transitions

Description of phase transitions in systems self−interaction can be described with meanfield theory. Consider the system at constant volume and temperature. In this casethe stable phase should be described by the minimum of F=U−στ

Assumption F= F(ξ,τ) where ξ is "order parameter" (like magnetization) which canbe independently specified for the system. In equilibrium : ò W ò

0 TConsider F_L (ξ,τ) ó U(ξ,τ) − τσ(ξ,τ) where U and σ are taken for any ξ

Equilibrium value of the order parameter should minimize F_L, in this case F_L is equal to theactual Helmholtz energy of the system at a given temperature

F T W FL

ò0, T ô F

Lò , T for ò õ ò

0

We will consider only the systems, in which F_L can be expanded in EVEN powers of ξ in the absence of external fields (energy and entropy are invariant under exchangeξ <−> −ξ )

FL

ò , T W g0 T U 1

2g

2 T ò 2 U 14

g4 T ò 4 U 1

6g

6 T ò 6 U ..

When thinking of Landau theory of phase transitions it is good to think of systems like ferromagnets. Here the magnetization is the order parameter, and the system can be described as such at the constant volume. Thus the state of the system should be given by the minimum of F, when system is in the contact with reservoir of temperature. Magnetization can be specified independently, but in fact the system will take such a magnetization that the F is at the minimum. We can calculate this magnetization and then put it "back" to F−> this will be the real F the system takes. For the ferromagnet we just calculated that g0=−τln2, g2=(1−τ), g4=4/3



lecture 6

Second order phase transitions

FL

ò , T W g0 T U 1

2g

2 T ò 2 U 14

g4 T ò 4 W g

0 T U 12 Ü T X T 0

ò 2 U 14

g4

ò 4consider :

constvalid in some temperature rangeãF

Lã ò Á W Ü T X T 0ò U g

4ò 3

has to be equal to 0 in equilibrium. It has roots :

ò 2 W Üg

4

T 0X T and ò W 0

ξ

© « 1.5©0

© « ©0

© « 0.7 ©0© « 0.5 ©0© « 0.3©

0

Ö « 1,g4

« 2.0

For same sign α and g_4 we have minimum at 0for τ>τ_0 , and for τ< τ_0 ò 2 W Ü

g4

T 0X T

What is the Helmholtz free energy :

F T W g0 T for T ¯ T 0

F T W g0 T X Ü 2

4g4

T X T 0

2 for T ô T 0

disorderedphase

ordered phase

FL

ª g0

©

We calculate the order parameter for minimal Landau free energy, and then Helmholtz free energy of the system



lecture 6

Helmholtz free energy for second order phase transition

What is the Helmholtz free energy :

F T W g0 T for T ¯ T 0

F T W g0 T X Ü

4g4

T X T 0

2 for T ô T 0

disorderedphase

ordered phase

öö0

FL

ª g0

©

ordered phase

disorderedphase

There is smooth transition of F between disordered andordered phase: derivative does not change= no latent heat=second order phase transitionöö

0

"magnetization"=order parameter÷ ö÷ ö ø 0

ù0

^ \ 0

ù0

^ ú 0

We see that the F behaves smoothly on the transition between ordered and disordered phase, there derivative of F is smooth, thus there is no difference in entropy between ordered and disordered phase at critical temperature. This is the second order phase transition



lecture 6

First order phase transitions

Are characterized by non−smooth behavior of F at phase transitions, for this we needtwo distinct minima of F , the deeper minimum will correspond to stable phase. Thistype of transitions are characteristic for liquid−gas, gas−solid transitions, and there existlatent heat. We postulate :

FL

ò , T W g0 T U 1

2 Ü T X T 0ò 2 X 1

4g

4 T ò 4 U 16

g6 T ò 6 U ..ã

FLã ò û W Ü T X T 0

ò X g4 T ò 3 U g

6ò 5

Ü T X T 0X g

4ò 2 U g

6ò 4 W 0 å ò W 0

c ü

0

© « ©0

ü c

At critical temperature, different from τ_0 bothminima are of the same depth, the system can "flip" from ordered phase to disordered phase.As both minima never "merge" derivative of Fchanges at phase transition. Order parameterξ "jumps" to another minimum for temperaturesgreater than critical temperature



lecture 6

Summary

F_L (ξ,τ) ó U(ξ,τ) − τσ(ξ,τ) where U and σ are taken for any ξIn the Landau theory, the free energy function is taken at any value of order parameter:

It is in minimum with respect to the order parameter when system is in thermal equilibrium

first order phase transition is characterized by latent heat. Landau function has two distinctminima, they become equally deep at critical temperature� ?

1� =p

� � ?2� =

p � � 1

�� 2 �

�F

1� =V

� �F

2� =V

Second order phase transition:�F

1� =V

>�

F2� =

V

but

� 2?g� = 2

p

� � 2?l� = 2

p

Landau function has one minimum, which moves smoothly to the zero value of theorder parameter with the increase of the temperature

we start consid ering phase transitions. van der waals...

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