lecture 6: paramagnetism and elasticitywa14/camonly/statistical/lecture6.pdf · february 07 lecture...
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February 07 Lecture 6 1
Lecture 6:Lecture 6:ParamagnetismParamagnetism and elasticityand elasticity
Applications of statistical methodsApplications of statistical methods
�� Aims:Aims:� Spin paramagnetism:� Paramagnetic salts� Curie’s Law.
� Entangled polymers� Role of entropy in rubber elasticity.
February 07 Lecture 6 2
ParamagnetismParamagnetism
�� Spin systemsSpin systems� Some atoms in salts have a permanent
magnetic moment.� Example: Gd2(SO4) 3.8H2O, where the Gd3+
ions have a spin moment, S=7/2.
�� General case:General case:� Angular-momentum, quantum number, J, gives
paramagnetic moment
� Component of magnetic moment is quantised
1<Landé factor<21<Landé factor<2
( )[ ] 211+JJg Bµ
Bohr magneton = Bohr magneton = eme 2�
JmJgm JBJ ≤≤−= ,µµ
J = 3
mJ
-3 -2 -1 +1 +2+3
H
February 07 Lecture 6 3
Spin Spin paramagnetismparamagnetism
�� Simplest system: a spinSimplest system: a spin--paramagnet.paramagnet.� In this case there is no orbital, angular
momentum so J=S. Since S=½, there are only 2 values of mJ.� Only two energy levels to consider, with energy
+/-µB.
� Calculate the expectation value of the moment from the weighting given by the BoltzmannDistribution,
( ) ( )�� −−=i
BBi
i kTUkTU expexpµµ
B
m = J -1/2 +1/2
February 07 Lecture 6 4
Curie’s LawCurie’s Law
�� Spin Spin paramagnetismparamagnetism� There are 2 states and, hence, 2 terms in each
summation.� Average moment, at temperature, T.
� In the limit of high-temperature and/orlow field
� The magnetic susceptibility can be measured
�
( ) ( ){ } ( ) ( ){ }( )kTB
kTBkTBkTBkTB
µµµµ µµµµ
tanh
eeee
=+−= −−
Curie’s LawCurie’s Law
kTBkTB
2µµµ <<
→
kT2µχ ∝
( )( )H
MHBχµ
µ+=+=
1o
o
= n <µ>= n <µ>
-2 -1 1 2µB kT/
tanh( )µB/kT
high- ,low- limit
TB
February 07 Lecture 6 5
Paramagnetic saltsParamagnetic salts
�� ExperimentExperiment� Curves at different temperatures and fields
scale to lie on the curve given by Curie’s Law.
� Waldram “Theory of Thermodynamics” Ch 15, p187
Gd3+, S=7/2Gd3+, S=7/2
Cr3+, S=3/2Cr3+, S=3/2
Fe3+, S=5/2Fe3+, S=5/2
<µ>/µB<µ>/µB
B/T (Tesla K-1)B/T (Tesla K-1)
February 07 Lecture 6 6
Pierre CuriePierre Curieand Magnetismand Magnetism
�� Curie’s LawCurie’s Law� The subject of Pierre Curie’s doctoral thesis,
1895, the same year as his marriage to Marie.
� Ferromagnetic to paramagnetic transition at Tc.� Paramagnetism in salts ~ 1/T (Curie’s Law)� Diamagnetism is temperature independent.
� Died 1906, after a street accident.
February 07 Lecture 6 7
22--state system: heat capacitystate system: heat capacity
�� Thermal properties of a 2Thermal properties of a 2--state systemstate system� Thermal Energy
� Heat Capacity
� Note the drop at both high and low temperature.� An exception to the “rule” that systems tend to
the classical, equipartition limit at high T
( ) ( )( ) ( )( )
)/tanh(
ln
2exp1lnlnexpexp
kTBB
Z
BBZ
BBZ
µµβε
βµβµβµβµ
−=∂∂−=
−++=−+=
( )kTBkT
Bk
TC µµε 2
2sech�
�
���
�=∂
∂=
February 07 Lecture 6 8
�� Classical treatment:Classical treatment:� Any stretched string (metal or rubber)
� Length, l; Tension f(T,l).� Tension, f, and other thermodynamic quantities
depend on l and T.
�� Start from the First Law:Start from the First Law:
Entropic contribution to Entropic contribution to elasticityelasticity
lflS
TTTS
T
lfSTU
WQU
Tl
dd
dddddd
�
�� +
∂∂+
∂∂=
+=+=
TTU
ld
∂∂
ll
U
Td
∂∂
AA
February 07 Lecture 6 9
�� Classical analysis continuedClassical analysis continued� Need to relate entropy and tension.� From previous results, differentiating gives
� We have derived a Maxwell relation, which connects the entropy to measurable quantites.
Maxwell relationMaxwell relation
ll TS
TTU
∂∂=
∂∂
flS
Tl
U
TT
+∂∂=
∂∂
lT Tf
lS
lTS
TTlS
T∂∂+
∂∂+
∂∂∂=
∂∂∂ 22
Tl lS
Tf
∂∂−=
∂∂
BB
TlS
TTl
U∂∂
∂=∂∂
∂ 22
lT Tf
lS
lTS
TlT
U∂∂+
∂∂+
∂∂∂=
∂∂∂ 22
yzx
zyx
∂∂∂=
∂∂∂ 22
February 07 Lecture 6 10
Elasticity: general caseElasticity: general caseand metal wireand metal wire
�� What determines the tension?What determines the tension?� f may be a function of l and T. Eq. A gives
� Using B we get
�� Metal wire:Metal wire:� Elastic modulus: µ(T) = µo(1+β(T-To)).� Unstretched length: lo(T) = loo(1+α(T-To)).� β and α are ~10-5.
� Effect is due to the ∂U/∂l term. Entropy is unimportant.
TT lS
Tl
UTlf
∂∂−
∂∂=),(
Direct contribution tointernal energy.
For example throughthe stretching of
intermolecular bonds.
Direct contribution tointernal energy.
For example throughthe stretching of
intermolecular bonds.
“Entropic” contribution.For example through
the ordering ofintermolecular bonds
“Entropic” contribution.For example through
the ordering ofintermolecular bonds
lT Tf
Tl
UTlf
∂∂+
∂∂=),(
1<<∂∂
lTf
fT
February 07 Lecture 6 11
Rubber elasticityRubber elasticity
�� Rubber:Rubber:� Generally have large elastic strain.� In simple cases
� From which,
� Tension in rubber is an effect of entropy.
� Band shortens on heating (at constant f).� Band heats on sudden stretching (constant S)� Entropy decreases on stretching (molecules
unfold).
)(),( ollATTlf −≈
Tl
lT
l
lS
TTf
T
Tf
Tl
UTlf
fTf
T
∂∂≈
∂∂≈
∂∂+
∂∂=�
≈∂∂
),(
( )
( )221
ln
dddddd
oo
l
ol
llATT
CS
lllATTCSTlfUQ
−−���
����
�=
−−=−=
constantsconstantsconstantsconstants
February 07 Lecture 6 12
Rubber elasticityRubber elasticity
�� 11--D statistical modelD statistical model� Take molecules to have 2N links, of length, a.� Each link points right or left.
� N+r point Right; N-r point Left.� Length of the stretched molecule, l = 2ra.� Entropy (from k ln g)
� We know TdS = dU - fdl and U≠U(l).
( ) ( )( ) ( )
( ))ln()()ln()()2ln(2!!
!2ln
rNrNrNrNNnkrNrN
NkrS
−−−++−=
��
���
�
−+=
axmll
akT
NakTr
Nr
Nr
akT
rS
aT
lS
Tf
2
1ln1ln2
dd
21
dd
=≈
��
���
���
���
� −−��
���
� +=
��
���
�−=−=
Expand ln’s for small r/NExpand ln’s for small r/N
Note: T and l dependenceNote: T and l dependence
February 07 Lecture 6 13
Elasticity in rubberElasticity in rubber
�� Molecular modelMolecular model� Without strain With strain
�� ExperimentExperiment� X-ray diffraction from un-strained and strained
samples of rubber.
� Note the diffraction spots showing enhanced order in the strained sample.