Transcript Spontaneous magnetization for H=0

Ginzburg-Landau theory of second-order
phase transitions
Second order= no latent heat
(ferromagnetism, superfluidity, superconductivity).
Let the transition occur at T=TC
Ordered phase below transition (typically)
Order parameter h (vanishes in disordered phase):
h = macrospopic wave function (superconductivity)
h = magnetization (ferro or antiferro)
Vitaly L. Ginzburg
Example: liquid crystals
Lev Landau
Order parameter
h = excess population at angle Q (or a simple function of Q )
in nematic-isotropic transition in liquid crystals-simplest example (1 real scalar parameter)
One of the most common LC phases is the
nematic. The word nematic comes from
the Greek νημα (nema), which means
"thread”
Alignment in a nematic phase
.
1
(see lectures by A. Salonen)
Tfy-0.3252 Soft Matter Physics, Fall 2009 / E. Salonen
2
Equilibrium Condition
G (h )  U  PV  TS
Gibbs free energy
 G 

 0

h

T
Gibbs free energy must be
minimum at equilibrium at any T
dG  SdT  VdP
  2G 
 2  0
 h T
How G depends on h near the transition Temperature
Order parameter h is 0 or small (since it vanishes in disordered phase)
Near the transition
G  G0   nh n , G0  disordered phase free energy
n
Gibbs free energy must be minimum at
equilibrium in disorderesìd phase
 G 
0
  1  2 2h  ...  1 for h  0.

h

T
Therefore the expansion around critical point starts from second order
G  G0   2h 2   3h 3   4h 4 
At h=0 there is a minimum or a maximum
3
GL Phenomenological model, with terms up to 4° order:
G  G0   2h 2   3h 3   4h 4 
disordered  2 (T )  0, T  TC
G-G0
At the transition T= TC
G must be the same for
nematic and isotropic
(equilibrium).
h
ordered  2 (T )  0
Above TC minimum at h=0 (order parameter vanishes at equilibrium in disordered phase
  2G 
 2   0  2 2  63h   0   2 (T )  0
 h T
  2 (T )  0 taken linear
Below TC maximum at h=0
1
2
 2 (T )  (T  T * ) A0
parameter T * usually different from TC : T *  TC
G  G0  (T  T * )
A0 2
h   3h 3   4h 4 ,  4  0 in order to have overall minimum
2
4
1
G  G0  (T  T * ) A0h 2   3h 3   4h 4 ,
2
1
1
1
often written in the form
G  G0  A0 (T  T * )h 2  Bh 3  Ch 4
2
3
4
Discontinuity of Order Parameter h at Tc
One obtains the discontinuity of order parameter by imposing equilibrium and minimum at Tc
Equilibrium : G  G0 
1
1
1
1
1
1
A0 (TC  T * )h 2  Bh 3  Ch 4  0  A0 (TC  T * )  Bh  Ch 2  0
2
3
4
2
3
4
 G 
1
1
1
*
2
3
Minimum : G  G0  (T  T * ) A0h 2  Bh 3  Ch 4  0  
  A0 (T  T )h  Bh  Ch
2
3
4
 h T
 G 

 0

 h T
 G G
0

 A0 (TC  T * )  Bh  Ch 2  0

1
1
1 2
*
A
(
T

T
)

B
h

Ch  0
 0 C
2
3
4

Divide the first by 2 and subtract:
hC  
2B
3C
B
C
h  h 2  0 and find hc
6
4
Choosing h  0, B  0 since C  0 (minimum condition)
5
hC  
T*
2B
3C
Choosing h  0, B  0
One can write T* in terms of TC and the other parameters. The order
parameter is discontinuous at transition.
1
1
1
G  G0  A0 (T  T * )h 2  Bh 3  Ch 4
Insert hhc back into G-G0 equation
2
3
4
1
1
1
G  G0 at thetransition  h 2 ( A0 (TC  T * )  Bh  Ch 2 )  0
2
3
4
1
1
2B 1
2B 2
A0 (TC  T * )  B (
)  C (
) 0
2
3
3C 4
3C
2 B2
2 B2
*
*
get A0 (TC  T ) 
 TC  T 
9 C
9 CA0
Order parameter versus T
Above, we got the discontinuity
hC  
2B
3C
To obtain h(T) we can start from the equilibrium condition
 G 
*
2
3
0
  A0 (T  T )h  Bh  Ch
 h T
 nematic : h (T ) 
 B  B 2  4 A0C (T  T * )
2C
6
To get rid of the double sign, choose the right h at the transition:
insert
 B  B 2  4 A0C (T  T * )
2 B2
TC  T 
into h (T ) 
9 CA0
2C
*
2 B2
B2
 B  B  4C
B 
9 C 
9 , recalling minimum condition B  0.
 h
2C
2C
|B|
B  2 B  OK with discontinuity h
B 
B
c
 3C
3
3
h


B
2C
2C


 wrong

3C
2
 we choose h (T ) 
 B  B 2  4 A0C (T  T * )
2C
: order parameter versusT .
7
Langevin paramagnetism
Model: "gas" of ions with magnetic moments   g B J
H  g  B J .B magnetization M 
N z
Volume
Quantum partition function
Z
J

m  J
mg  B B
exp[
]
K BT
Paul Langevin
(1872-1946)
 B
1
In the special case J  , with g  2 find Z  2 cosh( B )
2
K BT
 B H 
N
 M  B tanh 

V
K
T
 B 
8
Weiss mean field theory (1907) of Ferromagnetism:
Idea:the effective field gets a contribution
from the magnetization: H eff  H   M ,
 B H 
1
N
λ to be determined J  , g  2 M  B tanh 

2
V
 K BT 
 B ( H   M ) 
N
 M  B tanh 

V
PIERRE-ERNEST WEISS
born March 25, 1865, Mulhouse, France.
died Oct. 24, 1940, Lyon, France.

K BT

The spontaneous magnetization may remain even for H=0
M
  M 
N
B tanh  B

V
K
T
 B 
(Ferromagnetism!)
One can fix the parameter  in terms of Tc as follows:
   M  B M
with H  0 since M small, tanh  B

K BT
 K BT 
2
K BTCV
B M
MV
N


B


 
.
TC 
is very useful.
2
N B
K BTC
N B
VK B
For T  TC
9
  M 
N
B tanh  B
 can be solved graphically.
V
 K BT 
 M
N
x B
 M   B tanh  x  . Graphical solution:
K BT
V
M
Set
tanh  x  
 B   M B 

 
T
K
K
T
M
M  B   B 


N
N
TC




B
B  B 
V
V
 KB 
N  B2 
xT

, TC 
TC
VK B
x=y
1 solution if T>Tc,
2 otherwise
M
  M 
N
B tanh  B
 can be solved graphically.
V
 K BT 
From the graphical solution we gain the trend with T of the M(T) curve.
Increasing T at H=0 M must vanish, but how?
PIERRE-ERNEST WEISS
Spontaneous magnetization for H=0
born March 25, 1865, Mulhouse, France.
died Oct. 24, 1940, Lyon, France.
T Tc from below:
M is very small  x  M
xT
x 3 xT
tanh  x  
. We need the third power: x  
TC
3 TC
M x
K BT
B
 M

B
K BT
x  3(1 
1
T
)
TC
TC  T . Power law!Critical exponent 1/2.
11
A different experiment: fix T =Tc and measure M dependence on H at
for H0 . Since M is small we can expand
 B (H   M ) 
N
M   B tanh 

V
K
T
B C


3
 B (H   M )  1  B (H   M ) 
N
 B [
 
  ...]
V
K BTC
K BT


 3
 
N  B2
N
M
(H   M )  B  B
VK BTC
V
 K BT
N  B2 
Using TC 
, the
VK B
We are left with:
3

3
2
2
2
3
3
 ( H  3H  M  3H  M   M )

N  B2
l.h.s simplifies with
M .
VK BTC
3
 B 
N
N
3
2
2
2
3
3
0
H  B 
(
H

3
H

M

3
H

M


M
).

VK BTC
V
K
T
 B 
2
B
1
Setting M  H 
for
H  0, neglecting H 2 and H 3
3
1
 B  3 3
N
N
3
3
0
H  B 
  M  M  H    3. M  H .
VK BTC
V
 K BT 
Then, neglect of H 2 and H 3 justified a posteriori.
2
B
Power law!Critical exponent 1/3.
12
Behavior for T >Tc , in paramagnetic phase M small
 B ( H   M ) 
N
M   B tanh 

V
K
T

B

N B 2
(H  M )
V K BT
This gives
N
M
V
B 2
B 2
 N  B2
N
H
H , TC 
2
 N B
V K B (T  TC )
VK B
K BT 
V
which is the Curie law. Critical exponent=-1.
Of course this holds in para phase while M is small.
How do these critical exponents arise? Widom in 1965 put forth the scaling
hypothesis for magnetic materials in field B:
G  Gibb's free energy: r , p : G ( r (
T
T
),  p B)  G(( ), B)  G homogeneous.
TC
TC
f homogeneous function  a,b:f( a x,  b y)   f ( x, y)
This produces relations between various critical exponents.
The dependence on approximation scheme and dimensionality is dramatic.
(Weiss theory)
(Weiss theory)
14
Ising Model in 1d , defined on a closed ring
(model invented in 1920 by Wilhelm
Lenz as a thesis for Ernst Ising)
H Ising   Jsn sn 1   H  sn
n
H  H
Simplify notation :
n
H Ising   Jsn sn 1  H  sn
n
n
where the sums over sites and sn is a classical spin taking the values +1,-1.
The Partition function is defined by:
Z  Tre
  H Ising
From Z one derives thermodynamics: F   KT log Z .
Z     e
s1  s2 
  J ( s1s2  s2 s3  sN s1 )   H ( s1  s2 sN )
e
.
sN 
15
Z      e J ( s1s2  s2s3  sN s1 ) e H ( s1  s2 sN ) .
s1  s2 
Rearrange:
sN 
Z      [e Js1s2 e Hs2 ][e Js2 s3 e Hs3 ] [e JsM s1 e Hs1 ].
s1  s2 
sN 
Vs1s2  [e Js1s2 e Hs2 ]
1
2
There is a [ ] factor for each bond; each factor is a matrix product
 Jsn sn1   Hsn1
(VV
)

(
V
)
(
V
)

[
e
]
1 2 sn , sn1
1 sn , s 
2 s , sn1
where V  VV
1 2
called transfer matrix
(V1 ) s , s  e  Jss
(V2 ) s , s  e  Hs ( s, s)
 e J
V    J
e
e  J  e H

e J  0
0   e ( J  H )
   ( J H )
 H 
e
 e
Z      [VV
1 2 ]s1s2 [VV
1 2 ]s2 s3
s1  s2 
sN 
e  ( J  H ) 

e ( J H ) 
[VV
1 2 ]sM s1 .
16
Z      [VV
1 2 ]s1s2 [VV
1 2 ]s2 s3
s1  s2 
sN 
[VV
1 2 ]sM s1 . But pbcimply a Tr :
M
M
Z   [(VV
)
]

TrV
, V  VV
1 2
s1s1
1 2
called transfer
matrix
s1 
V V 
V 

V
V
 
 
recall : Z      [e  Js1s2 e  Hs2 ] [e  Js2 s3 e  Hs3 ] [e  JsM s1 e  Hs1 ].
s1  s2 
 e J
V    J
e
sN 
e  J  e H
 J 
e  0
0   e ( J  H )
   ( J H )
 H 
e  e
e  ( J  H ) 
 ( J H ) 
e

 e ( J  H )  
e  ( J  H ) 
Det[V   I ]    (  J  H )
 0 
 ( J H )
e

 e
 2   e  J (e  H  e   H )  e 2  J  e 2  J  0
 2  2 e  J cosh(  H )  2sinh(2  J )  0
17
 2  2 e  J cosh(  H )  2sinh(2 J )  0
   e  J cosh(  H )  e2  J cosh(  H )2  exp(2 J )  exp(2 J )
   e J cosh(  H )  e J cosh(  H )2  1  exp(4 J )
   e J cosh(  H )  e J sinh(  H )2  exp(4 J )
For H0,
J
  e  e
J
exp(4 J )  e
TrV    
TrV N   N   N
J
 J 2  J
e e
e
J
e
 J
2cosh(  J )

 2sinh(  J )
Z  N  N .
18
We need to note that another reasoning is possible. Instead of
Vs1s2  [e Js1s2 e Hs2 ]
1
2
one can decide to distribute the interactions with H symmetrically among the bonds and
define a different transfer matrix, which is also common in textbooks, with elements:
1
Vsl , sm  e
1
2
Vs2 , s2  [e  Js1s2 e
2
 Jsl sm   H ( sl  sm )
,
 e ( J  H )
V    J
 e
1
 H ( s1  s2 )
2
]
e  J 

e ( J  H ) 
This is transfer matrix has the same trace and the same determinant as the previous one,
hence the eigenvalues are the same.
19
Thermodynamic limit: only large eigenvalue matters
Z  N  N  N
  e  J cosh(  H )  e  J sinh(  H ) 2  exp(4  J )
F   NKT ln    NJ  NK BT ln cosh(  H )  sinh(  H ) 2  e 4  J  .


F
sinh(  H )
Magnetization (after simplification)

.
2

4

J
H
sinh(  H )  e
No spontaneous magnetization.
The peak in specific heat is
due to some short-range
order (increasing T at low
temperatures spins are
reversed and this costs
energy).
20
no ferromagnetism, no phase transition, no long-range order!
1d theory fails to explain magnetism. Why?
Ground
state :

1 broken bond : 
Breaking a single bond breaks long  range order. It costs energy 2J.
S=K B logW.
The free energy change is
F  2 J  K BT ln( N )
For large N there is a gain in F at any T. In 2d is different.
21
Rudolf Peierls Centre for Theoretical
Physics
1 Keble Road
Oxford, OX1 3NP
England
22
Percolation problem: Renormalization group approach to the
onset of conductivity
A.P.Young and R.B. Stinchcombe J. Phys. C 8 L535 (1975)
Consider a regular lattice (sites connected by bonds, with Z bonds per site)
Remove a fraction 1-p of bonds at random. Conductivity is a function of p.
There is a critical value pc such that for p< pc the lattice is not conducting
and consists of clusters of connected sites in a sea of isolated sites. If we
pick two sites far apart the probability of finding a continuous path joining
them gets small. Conversely, for p> pc the film as a whole is conducting ,
even if it contains nonconducting islands. The macroscopic conductivity S
of the lattice is a function SS(p, s, s  bond conductivity .
Let us consider a square lattice:
How to find an approximation to pc ?
By scaling!
For exampe, discard half of the x and
half of the y and consider a new lattice
where the nodes can be connected or
not with some probability. The new
problem is similar. We can exploit that!
23
Decimation
Rescaled length (rescaling factor b )
Rescaled
probability
p  pc  p1 ( p, b)  0
p1  p1 ( p, b)
for b  
p  pc  nonconducting islands disappear
p  pc  system unchanged
for b  
for b    p1  p1 ( pc , b )
24
A similar problem can be posed on the lattice with vertices 1,2, .. or
with the rescaled lattice A,B,…on a larger square mesh.
p1= probability in rescaled network , s1  bond conductivity in
rescaled network
SS(p1, s1
25
in this square lattice case:
rescaling factor b  2,
Renormalization of

bond probability :
 p1
p
at each scaling
p changes by a
factor  ( p).
Rescaled length (rescaling factor b  2 ),
the correlation length index 
b 2   
1
is defined by b  c
ln(b) ln( 2)

 0.8547
ln( ) ln( 3 )
2
By comparison with the exact results one can evaluate approximations.
To proceed find a simple approximation to p1  p1 ( p, b)
Consider only paths involving ,,c,.
What is the probability p1 that nodes A and B of new lattice are connected?
1
B


A
Let


2
D
P( ,  ,  ,  )  probability that  ,  ,  , 
C
are connected,
P( ,  ,  )  probability that α,β,γ are conected but δ is not, and so on.
p1  P( ,  ,  ,  )  P( ,  ,  )  P( ,  ,  )  P( ,  ,  )  P(  ,  ,  )  P( ,  )  P( ,  )
renormalization of p :
p1  p 4  4 p 3 (1  p)  2 p 2 (1  p) 2  2 p 2  p 4
27
Renormalization of
p1  2 p 2  p 4
bond probability :
  p1  4 p (1  p 2 ) p   
 p1
 4 p (1  p 2 )
p
at each scaling p changes by a factor  ( p).
p1  2 p 2  p 4  Fixed point equation:
pc  2 pc 2  pc 4
Fixed points
x  2x2  x4
x  0 or
x 3  2 x  1  0 which yields
Roots : x  0, x  1, x 
x  1 or
x2  x 1  0
5 1
 5 1
 0.618, x 
 1.618
2
2
pc  0 Trivial Fixed point : repeated scaling  no conductance
pc  1 Trivial Fixed point : repeated scaling  perfect conductance
pc  1.6 nonsense fixed point
nontrivial solution :
pc  0.618 (Exact
result
pc  0.5)
28