stat mech notes
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8.044 Statistical Physics ISpring 2008
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Language:
Probability 1st Law
Microcanonical ensemble
2nd Law
3rd Law
Canonical ensemble
T=0 Quantum gases
Grand
CanonicalEnsemble
PhaseTransitions
Transport
Processes
8.044 STATISTICAL PHYSICS I
8.044 L1B1
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PROBABILITY
Random variable (ignorance and/or QM)Continuous, discrete, or mixed
Probability density: p(x)px()Histogram
(normalized)p(x)
x
PROB(x < + d) = px()d8.044 L1B2
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px() 0,
px()d=1 ,
bPROB(ax < b) = px()da
Cumulative probability: d
Px() px()d px() = dPx()
Either px() or Px() completely specifies the RV x.
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Example Physical adsorption of a gas
most of the time ( when hot)
x
small fraction of the time (when hot) leaving the surface
y
z
nv
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1
p() p(v) v3 v2
e 22
2
2
3
24
= kTm
vp() p( t )
2sin()cos()
/2 t
e
1
t
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PROB = p()dp()d= 2 sin() cos()d(1/2)d
d = sin()dd
z
x
y
d
d
d
PROB/d = (1/) cos()
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Gaussian density (memorize)
p(x)
(xx0)2
1p(x) =
22e
22
x x + x
h
he = 0.61 h12
0 0
2 parameters
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8.044 Statistical Physics ISpring 2008
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Gaussian density (memorize)
p(x)
(xx0)2
1p(x) =
22e
22
x x + x
h
he = 0.61 h12
0 0
2 parameters
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Example Atom escaping from a cavity
AHole Atom escapes after nth wallencounter
p(n) = (AH)(1 AH)nAT AT
ATotaln
= 0,
1,
2,
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pn(x) =
(AH)(1AH)n(xn)AT AT
n=0
Called a geometric or a Bose-Einstein density
Pn(x)
pn(x)
0 2 4 x 0 2 4 x
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Example Mixed, tdependent RV
Chemical adsorption
Physical adsorption
x
p(x)
x
e- t /
P(x)1
Given: atom on bottom at t= 0e- t /
p(x) = et/ (x)+(1et/) f(x)
8.044 L2B3
x
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Averages
< f(x) > f(x)p(x) dx
p(x)
std.
dev.
x< x > is the mean< x2 > is the mean square< (x< x >)2 > = < (x2 2x < x > + < x >2) >
= < x2 > 2 < x >2 + < x >2
= < x2 > < x >2 is the variance (standard deviation)2
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Example Mean free path n
d L=d/cos
/2
< L > = (d/cos )p() d0/2
= (d/cos ) 2 sin cos d = 2d/2 sin d 0 0
= 2d[/2(cos ) = 2d08.044 L2B6
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Poisson density Events occur randomly along a line
at a rate r per unit lengthL
x
x
p(1)rxas x 0Events are statistically independent
1p(n) = (rL)nerL= 1 < n >n e
n! n!
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Examples of Poisson probability densities0.6
0.25
0.5
= 0.5 0.2 = 2.30.4
0.15
0.3
0.2
0.1
0.10.05
1 2 3 4 5 2 4 6 8 10
0.12
0.1 = 10
0.08
0.06
0.04
0.02
5 10 15 20 25 30 10 20 30 40 50
0.01
0.02
0.03
0.04
0.05
0.06
0.07
= 30
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f(x, y)=
f(, )px,y(, )dd
NEW CONCEPTS:
Reduction to a single variable
px() =
px,y(, )d
Conditional probability density
px( y)dprob.( < x + dgiven that = y)|8.044 L3B3
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p(x, y) =p(x|y)p(y)p(x, y) =p(y|x)p(x)
x and y are statistically independent ifp(x, y) =p(x)p(y) p(x y) = p(x) p(y|x) =p(y)
|
conditioned unconditioned
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1 2 2p(x|y) =2a2 2 |x|
a y
y
2= 0 x >a2 y| |
p(x|y)
a
2
-y
2
a
2
-y
2
x
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Derivation of Poisson density p(n;L)
Given: p(n = 1;x)rx as x 0Start by findingp(n = 0;L)
0 L
x
dL
dL0p(0)p(1)p(n > 1)p(n = 0, dL) 1p(1) = 1r dL
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Poisson Random Variabler(x) S.I. events
p(1) in x rxr is constant x
p(n = 0;L) = erL 1st element ofp(n;L)
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L L+Lx0
p(n;L + L) p(n;L)p(0; L)+p(n1;L)p(1; L) (1rL) (rL)
d p(n;L)p(n;L + L)p(n;L)= rp(n1;L)rp(n;L)
L dL
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d p(n; )Let rL +p(n; ) =p(n1; )
d
1st order, linear DE already known
DE: is the variable and n is an index
Probability: n is the variable and is a parameter
1p(n; ) = n e
n!
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v1 v1 v1
v2v2v21 = 0-1
v2 and v2are S.I.
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p(v2 |v1)
1
0
v2
(v2-v1) when =1
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c
t
v( t )
1
-1
()
c
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Averages< v1v2 > =
v1v2p(v1, v2) dv1 dv2
=
v1v2p(v2|v1)p(v1) dv1 dv2
= v1p(v1)
v2p(v2 v1) dv2 dv1 |
conditional mean = ()v1
2= ()
v1p(v1) dv1 =< v1v2 > /2 =2
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Functions of a random variable
Given: px() and f(x)Find: pf()
f(x)
x
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A. Sketch f(x). Find where f(x) <
B. Integrate to find Pf().
C. Differentiate to find pf().
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Example Intensity of light
I= aE2p(E) =
2
1
2exp[E2/22]
AI
a
a
0
a2
B
/aPI() = /apE() d8.044 L5B3
g
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g
x
a(y) b(y)
da
dy
dy
db
dydy
g
ydy
d b(y)
a(y)g(y,x) dx=
dyg(y,x= b(y))db(y)
dy g(y,x= a(y))da(y)dy +b(y)
a(y)g(y,x)
y dx8.044 L5B4
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C In general
1 1pI() =
2apE(
/a)( 11
2a
) pE(/a)
= 1 12a[pE(
/a) +pE(
/a)]
In our particular case
1 Ip(I) = exp[
2a
2
I 2a2] 1
2
3
4
5
I/ a2
a2 p(I)2
0 1 2 3 48.044 L5B5
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Given = c/ and p()A
Find p()
B
P() =p() d 0 c/
=c/
c/
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C
In general
p() = (c/2) p(c/)
In our casec 1 1 (c/0)2p() =2 2.404 0 exp[(c/0)]1
Let 0 c/0, then1 1
0
4
1p() =2.404 0 exp[(/0)1]1
8.044 L5B8
1As (/ ) 0 e(/0)1
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As (/0)0 exp[(/0)1]1e (/0)
1 1As (/0) exp[(/0)1]1 (1+(/0)11) (/0)
p()(/
0) -1
(/0) -4 e
(/0) -3
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Example Random number generator for programmers
p(x) p(y)
1 1
0 1 x 0 1 y
xand y are statistically independent
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z MAX(x, y) Find p(z)
p(x, y) = p(x)p(y)
Where is MAX(x, y) = ?
Where is MAX(x, y) < ?
A
0 1 x
1
y
p(x,y)=1
p(x,y)=0
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vExample Desorbing atom
p(v,,) = p(v)p()p()p(v) = (1/24) v3 exp[
v2/22]
p() = 2 sin cos p() = 1/2Find p(vz)
z
x
y
n
leaving the surface
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Avz= v cos v
vcos < v < / cos
/2
8.044 L6B5
B
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C
B
/2 /cos Pvz() = pv()p() dd0 0/2 /cos
=
p() pv() dd0 0
dPvz() = /2
p() 1 dpvz() = cos pv(cos )0
d
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0
pvz() =
(2 sin cos )cos
/2 1 124
cos 223exp[ 1 2
cos2 ]d
Let 122
2
cos2 X
dX = 12
2
cos3 (sin )d
= 0 X= 2/22; = /2 X=8.044 L6B7
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pvz() = eXdX=
2 [
2 2/22 2/22 e
X
=
2 exp[
2
/2
2
] >0
p(vz)
0 vz
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Results have a special meaning when
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1) The means are finite (= 0)
2) The variances are finite (=)
3) No subset dominates the sum4) n is large
width
n
1p(s)
mean
n
n s
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Given p(x,y), find p(sx+ y)
A
dx
y = x
x+y =
x
y
BPs() =
d
d
px,y(,
)
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C ps() =
dpx,y(,)
This is a general result; xand y need not be S.I.
Application to the Jointly Gaussian RVs in Section 2
shows that p(s) is a Gaussian with zero mean and a
Variance = 22(1 + ).
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In the special case that xand y are S.I.ps() =
dpx()py() = dpx()py()
The mathematical operation is called convolution.
pq p(z)q(xz)dz= f(x).
8.044 L7B6
Example
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p
Given:
1p(z) = (z/a)nexp(z/a)n!a
1q(z) = (z/a)mexp(z/a)
p(z)
zn
(z/a)n e-z/a
zm!a
0 < z and n,m= 0,1,2, Find: p q
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q(z) q(-z)
0
z 0 z
q(x-z)=q(-(z-x))
z0 x
q(x-z)
z0 x
p(z)
finite product
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x zn m1 1 xzez/ae(xz)/adz=pq
n!m! a2 0 a a1 1=
1
n+m+1ex/a x zn(xz)mdz
n!m! a a 01
=
1 xn+m+1ex/a 1
n(1)
mdn!m! a a 0
n!m!(n+m+1)!
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1
xn+m+1
ex/a1
=pq(n+ m+ 1)! a a
a function of the same class
8.044 L7B10
Example Atomic Hydrogen Maser
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flask
cavity
beam
0
1
RF out
* = 1.4....... GHzH10 about 10 KHz
|n stays) = ?p( twall
8.044 L7B11
ntwall =
tn, Each stay is S.I.
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i=1
p(t|1) = (1/) et/
p(t 2) = p(t 1) p(t 1) = (1/)(t/) et/| | |
p(t 3) = p(t 2) p(t 1) = (1/2)(1/)(t/)2 et/| | |
8.044 L7B12
1 1 t n1et/p(t |n) =
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5 10 15 20 25 30
0.02
0.04
0.06
0.08
0.1
0.12
ep(t|n)(n1)!
0.15
0.10
0.05
p(t | 12)
t /0 10 20 30
8.044 L7B13
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Facts about sums of RVs
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Exact expressions for < s > and Var(s) ifS.I.
p(s) = p(x)p(y) ifS.I.
p(s) slightly more complicated if not S.I.
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usually changes functional form
But not always
Fourier techniques are very useful
8.044 L8B2
Very important special case: Central Limit Theorem
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RVs are S.I. All have identical densities p(xi) Var(x) is finite but < x > could be zero n is large
p(s) Central Limit Theorem:p(s) is Gaussian
n
s n
8.044 L8B3
Ifx is continuous
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1
22e(s)
2/22p(s) =
< s >= n < x >
2
= n
2
x
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Ifxis discrete in equal steps of x
x
22
e(s)2/22 (six)p(s) =
i combenvelope
p(s)
sx
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3 directions, N atoms
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E = total K.E.E = 3NK.E.x = 3N kT2U
Variance(E) = 3N Variance(K.E.x) = 3N(kT)221
exp[(E(3/2)N kT)2p(E) =2{(3/2)N(kT)2 2{(3/2)N(kT)
2 ]} }
3s.d. 2N kT 1
= =3
mean2N kT 32N
8.044 L8B7
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Some terms that must be understood
Microscopic Variable
Macroscopic Variable
8.044 L9B1
Extensive (N) Intensive (= f(N))
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V volumeAareaL lengthP polarizationM magnetization
P pressureS surface tensionF tensionE electric fieldH magnetic field
T temperature
U internal energy
8.044 L9B2
Adiabatic Walls
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Diathermic Walls
Equilibrium
Steady State
Complete Specification:
Independent and Dependent Variables
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Equation of State
P V= N kTV = V0(1 + TKTP)M= cH/(TT0) T > T0
In Equilibrium with Each Other
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OBSERVATIONAL FACTS
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"0 th Law"
equilibriumif A C and B
equilibrium
then A B
Cequilibrium
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" Law 0.5 ? " Many macroscopic states of B can be
in equilibrium with a given state of A
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YA YBA B1 1 YB = f (XB)
alsoXA, YA
XA XB
8.044 L9B6
THEOREM A "predictor" of equilibrium h(X, Y, ...) exists
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only in equilibrium
state variable
many states, same h different systems,
different functional forms
value the same ifsystems in equilibrium
h(X, Y)
Y
X
locus of
constant h
8.044 L9B7
XA
, YA
XC
, YC
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A A C C
XA , YA , XC , YC all free [ PA , VA , PC , VC ]
XA , YA XC , YC
equilibrium
keep same adjust
XC = f 1(YC , XA , YA ) [ PC = PA VA/ V C ]F1(XC , YC , XA , YA ) = 0 [ PC VC - PA VA = 0 ]
8.044 L9B8
XB
, YB
XC
, YC
equilibrium
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B B C C
keep sameadjust
XB = g(YB , XC , YC )
F2(XC , YC , XB , YB ) = 0
solve for XC
XC = f 2(YC , XB , YB )
[ PB = PC VC/ V B ]
[ PC VC - PB VB = 0 ]
[ PC = PB VB/ V C ]
same value as before
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f 1(YC , XA , YA ) = XC = f 2(YC , XB , YB ) 1
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[ PA VA/ V C = PB VB/ V C ]
XA , YA XB , YB
equilibrium
due to 0 th law
F3(XA , YA , XB , YB ) = 0 2
2 F3 factors+YC drops out
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1
For this equilibrium condition
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For this equilibrium conditionh( XA , YA ) = constant = h( XB , YB )
[ PA VA = PB VB ]
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Isotherm at T1
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YB B
1
YA A
1
Isotherm at T1
T2 T2
Isotherm at T1
2
2
XA XB
8.044 L9B12
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THEOREM A "predictor" of equilibrium h(X, Y, ...) exists
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only in equilibrium
state variable
many states, same h different systems,
different functional forms
value the same ifsystems in equilibrium
h(X, Y)
Y
X
locus of
constant h
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Empirical Temperature: t
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P
1
Y
1
low density gas
PV/N = constant2
2
PV/N = constant'
X V
8.044 L10B1
we could possible alternative
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Define t = cg PV/N t' = cg' (PV/N)
Use to find isotherms
in other systems Then in a simple paramagnet
t = cm (M/H)-1 t ' = cm' (M/H) -
Many possible choices for t
8.044 L10B2a
PV = Nkt t = PV/Nk
P
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8.044 L10B2b
0 0.2 0.4 0.6 0.8 1
0
0. 2
0. 4
0. 6
0. 8
1
P
VPV = constant
P
V
t
t ' = (PV/Nk)2
P
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8.044 L10B2c
PV = constant
P
V
t '
0 0.2 0.4 0.6 0.8 1
0
0. 2
0. 4
0. 6
0. 8
1P
V
t ' ' = PV/Nk
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0 0.2 0.4 0.6 0.8 1
0
0. 2
0. 4
0. 6
0. 8
1
8.044 L10B2d
PV = constant
P
V
t ' '
P
V
1
M = cH/t t = cH/M
H
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0.2 0.4 0.6 0.8 1
0
0. 2
0. 4
0. 6
0. 8
1
8.044 L10B2e
H
MH/M = constant
H
M
t
1
V = V0(1+t -P) t = (V-V0) /V0 +P/
P
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0 0.2 0.4 0.6 0.8 1
0
0. 2
0. 4
0. 6
0. 8
1
8.044 L10B2f
(V-V0) / V0 +P = constant
t
(V-V0)(V-V0)
P P
Workthe systemdW = differential of work done on
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= - (work done b y the system)
Hydrostatic system
dW = -PdV F
dx
dW = Fdx = (PA)(-dV/A) = -PdV
8.044 L10B3
dW = FdL
WireFF
P pushes, Fpulls
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2SL F constraint
wire
framefilm
8.044 L10B4
dW = Fdx = (SL)(dA/L) = SdA
dW = SdA
Surface
dW = Fdx = (F)(dL) = FdL
(2 surfaces)
Chemical Cell (battery)dW = EEMF dZCHARGE
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Electric charges Field in absence of matter
dW = EdP as set up by externalsources. Does not include
energy stored in the fielditself in the absence of
Magnetic systems the matter.
dW = HdM
8.044 L10B5
All differentials are extensive
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Only -PdV has a negative sign Good only for quasistatic processes
bW = dW depends on the path
a
W is not a state function
8.044 L10B6
Pisotherm PV= NkTA
low density gas
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V
PY
X
isotherm PV NkT
B
V2
P1
P2
V1
a
a
b
bc
8.044 L10B7
dW = YdX
depends on Y(X)
(a) W1 2 = P1(V2 V1) = P1(V1 V2)
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(b) W1 2 =
(c) W1 2 =
=
P2(V2 V1) = P2(V1 V2)2 2
dV
1P(V) dV = 2NkTdV = NkT
V 1 V1
NkTln V2 = NkTln V1 = P1V1 ln V1V1 V2 V28.044 L10B8
MATH
I) 3 variables, only 2 are independent
-
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F(x,y,z) = 0x= x(y,z), y= y(x,z), z= z(x,y)
x
y ,
x
y1 y
z
z= 1=
yx x x yz zz
8.044 L10B9
xz
wand ifW(x,y,z), then x =
y w yz w
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II) State function of 2 independent variables
S= S(x,y)
S
S
dS= dx+ yx
dy
x y A(x,y) B(x,y)
An exact differential
8.044 L10B10
2S 2S B= = x
Ay = yx xy yx
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yx
necessary condition, but it is also sufficient
Exact differential if and only if
A
y B=
x yx2
dS = S(x2, y2) S(x1, y1) is independent of1Thenthe path.
8.044 L10B11
III) Integrating an exact differential
-
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dS= A(x,y) dx+B(x,y) dy1. Integrate a coefficient with respect to one variable
S
= A(x,y)x y
S(x,y) = A(x,y) dx+f(y)y fixed
8.044 L10B12
2. Differentiate result with respect to other variable
Sy
d f(y)dy
= B(x,y) A(x,y) dx +=y
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y dyyx
3. Integrate again to find f(y)B(x,y)
A(x,y) dx
d f(y) y
=dy
f(y) = {} dy
done
8.044 L10B13
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8.044 Statistical Physics ISpring 2008
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Thermodynamics focuses on state functions: P,V,M,S, . . .
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Nature often gives us response functions (derivatives):
1V 1 V 1 V
V T P T V P T S V P adiabatic
M
T H T
8.044 L11B1
Example Non-ideal gas
Given
-
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Gas ideal gas for large T & VP Nk
=T V VNb P NkT 2aN2= +V T (VNb)2 V3
Find P8.044 L11B2
P PdP
dV + dT
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dP = dV + dT
V T T V P Nk
P =T V dT+ f(V) = VNb dT+ f(V)
NkT=(VNb) + f(V)
8.044 L11B3
P NkT NkT 2aN2=
(V Nb)2+f(V)= (V Nb)2 +V T V 3
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(VNb) (VNb)V T V
2aN
2
aN2
f(V) = dV = + cV3 V2NkT aN2
P= (VNb) V + c2
but c= 0 since PNkT/V as V 8.044 L11B4
Internal Energy U
Observational fact
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final
W
initial
isolated
(adiabatic)
Final state is independent of how W is applied.
Final state is independent of which adiabatic path is
followed.
8.044 L11B5
a state function Usuch that
U W
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U= WadiabaticU = U(independent variables)
= U(T, V) or U(T, P) or U(P,V) for a simple fluid
8.044 L11B6
Heat
If th th i t di b ti dU d/W
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If the path is not adiabatic, dU= d/Wd /W/QdUd
d/Q is the heat added to the system.
It has all the properties expected of heat.
8.044 L11B7
First Law of Thermodynamics
dU = d /W/Q + d
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dU= d /W/Q+ d
U is a state function
Heat is a flow of energy
Energy is conserved
8.044 L11B8
Ordering of temperatures
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T1 T2
dQ
When d/W = 0, heat flows from high T to low T.
8.044 L11B9
Example Hydrostatic System: gas, liquid or simple
solid
Va iables ( ith N fi ed) P V T U
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Variables (with N fixed): P,V,T,U.Only 2 are independent.
/Q dd /Q
CPCV
dT V dT PExamine these heat capacities.
8.044 L11B10
dU= d/Q+ d/W = d/QPdV
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d/Q= dU+ PdVd
We want . We have dV.dT
U U
dU= dT+ dVT V V T
8.044 L11B11
U Ud/Q= dT+ + P dVT V V T
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/Q U
U
dVd = + + P dT T V V T dT
d/Q U =CV dT V T V
8.044 L11B12
CP
d/QdT
P=
UT
V+
UV T + P
VT
PCV V U
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U
= + P VCPCVV
T
The 2nd law will allow us to simplify this further.
UNote that CP = .
T P8.044 L11B13
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Paths Experimental conditions, not just math
floatingfills bath insulation
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~~~~~~~
floating
piston
fills
container
bath insulation
V=0 P=0 T=0 Q=0
8.044L12B1
Q = 0 could come from time considerationsExample Sound Wave
( )
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(x) v
x
too fast for heat to flow out of compressed regions
1v =
S
8.044L12B2
Example Hydrostatic system: an ideal gas, PV=NkTU
New information= 0 ,
V T3 ibl
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3 possible sources
Experiment
freeexpansion
bath initially at T observe Tf
= Ti i
8.044L12B3
No work done so W = 0Tf = TiQ= 0together U = 0 (U/V )T = 0
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together U 0 (U/V)T 0
here quasi-static changes
Physics: no interactions, single particle energiesonly (U/V)T = 0Thermo: 2nd law + (PV= NkT)(U/V)T = 0
8.044 L12B4
Consequences
U UdU = dT+ dVT V V T CV 0
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CV 0
TU = CV(T) dT+ constant0
set=0
3In a monatomic gas one observes CV = 2Nk.3Then the above result gives U= CVT = 2NkT.
8.044 L12B5
U VCPCV = (V T+P) T P
0 ( / ) /
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0T(NkT/P)P=Nk/P
= Nk for any ideal gasApplying this to the monatomic gas one finds
3 5CP = Nk +Nk = Nk
2 25
CP/CV =3
8.044 L12B6
Adiabatic Changes d/Q= 0Find the equation for the path.
Consider a hydrostatic example.
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U Ud/Q= dT+ + P dV = 0T V V T
CV (CPCV)/V
T CPCV) 1 (1) = =V Q=0 CV V V
This constraint defines the path.
8.044 L12B7
Apply this relation to an ideal gas.
1 V
1
N kT
1 N k
1 V
1= = = =V T P V T P P V P V T TPath
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Path
dT T= (1)dV V
dT dV T V= (1) V ln T0 = (1) lnT V0
(1)
T V =T0 V0
8.044 L12B8
Adiabatic Isothermal
TV 1 = c
P adiabatPV = c
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TV 1 = cPV = c
= 5/3 (monatomic)
P V 5/3isotherm
PV= cP V 1dP P
=
dV VdP 5P V
=dV 3V
8.044L12B9
F
insulation
Expansion of an ideal gas
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8.044L12B10
rupture diaphragmadiabatic Q = 0
not quasistatic
W = 0
U = 0
slowly move pistonadiabatic Q = 0
quasistatic
W is negative
U = is negative
T
V
constant U
T
V
adiabat
Starting with a few known facts,
1st /W, and state function math,law, done can find
relations between some thermodynamic quantities,
-
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a general expression for dU,and the adiabatic constraint.
Adding models for the equation of state and the heatcapacity allows one to find
the internal energy Uand the adiabatic path.
8.044 L12B11
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1. The System
Fixed:
E < energy < E+
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8.044 L13B1
Fixed:
EV
N
M or H
P or E
A complete set of
independent thermodynamic
variables is fixed.
Many micro-states
satisfy the conditions.
2. Probability Density
All accessible microscopic states are equally probable.
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Classical
p({p,q}) = 1/ E < H({p,q}) E+ = 0 elsewhere
accessible
{dp,dq}= (E,V,N)
8.044 L13B2
Quantum
p(k) = 1/ E < k|H|k E+
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= 0 elsewhere
(1) = (E,V,N) k,accessible
8.044 L13B3
Let X be a state of the system specified by a subset{p,q }of{p,q}
p(X) =except {p,q}p({p,q}) {dp,dq}
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=
1
except {p,q}{dp,dq} (consistent with X)
=
volume consistent with X
=total volume of accessible phase space
8.044 L13B4
3. Quantities Related to
(E,V,N) H({p,q})
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(E,V,N)(E,V,N) E
= density of states as a function of energy
(E,V,N) = (E,V,N)
8.044 L13B5
Example Ideal Monatomic Gas
qi= x,y,z in a box V = LxLyLz
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y
x,mpi= m y,mz < pi
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y
{ p}
0 0 0
= VNE
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This describes a 3N dimensional spherical surface inthe ppart of phase space with a radius R= 2mE.
8.044 L13B8
Math:
Volume of an dimensional sphere of radius R is/2
R(/2)!
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Sterlings approximation for large Mln(M!) MlnMM
MM! M
e8.044 L13B9
3N/2
(E,N,V) = VN (2mE)3N/2(3N/2)!
4emE 3N/2V N
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VN 3N
1(E,N,V) = 3
2
N
E
{ }
(E,N,V) = 3
2
N
E
{ }
8.044 L13B10
1p(xi) =
=
VN1LyLz= 0 x < Lx
VN Lx1 VN2LyLzLxLz
= = p(xi)p(yj) S.I.p(xi, yj) =
=VN LxLy
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p(pxi) = p({p,q}){
p , dq}=dp
1/ = pxi
Note that differs on each of the three lines, be-ing a generic symbol for the reduced phase volume
consistent with some constraint.8.044 L13B11
2 distributed over other variablespx/2m E(3N1)/23N1
VN4em(E)=
2 E 3N1
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3N1 E 4em1/2=
3N 3
E1 1
2
3N 3N 11 2 +12 2(N3)
(E)
3N 3N
N 2 E 2 A B
8.044 L13B12
3N1 1 1/2A= N 1 1
32N
=
N 1 + 3 3N 3 3N/2
2
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but lim 1 + x = ex
soA
N e
1/2
8.044 L13B13
3N1 21 2 1 2/ < > B=
E1
E
3N
= E 1 3N/2
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where we have used < >E/3N and E= 3N < >.1
so B 3N
e
/2
8.044 L13B14
p(px) =
31/2 1
N e1/2 /24m 3N < > e e
1 /2=
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4m < >
/ =e
2Now use = p2/2mand < >=< px
> /2m.x
x xp(px) =1
e
p2/22
2 < p >x
8.044 L13B15
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= (E,V,N. . .) Volume of the accessible regionof phase space
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of phase space.
8.044 L14B1
4. EntropyS(E,V,N) kln (E,V,N)
kln (E,V,N) Differ only by lnNkln(E,V,N)
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It is a state function.It is extensive.
It is a logarithmic measure of the microscopic de-generacy associated with a macroscopic (that is, ther-
modynamic) state of the system.
k is Boltzmanns constant, units of energy per 0K.8.044 L14B2
5. Statistical Mechanical Definition of Temperature
1Total is microcanonical
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1
2
dQd /W1 2 = 0 interaction between 1 &
Find the most probable E1 2 is so small that can
E be separated18.044 L14B3
p(E1) =
=
1(E1)2(EE1) (E)
lnp(E1) = ln1(E1)+ ln2(EE1)ln (E)
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1= ( S1(E1) + S2(EE1)S(E) )
k
1 S1 S2 lnp(E1) =k E1d E2d = 0E1 /W1=0 /W2=0
8.044 L14B4
The condition for determining E1 is
S1 S2 =/W1=0
E2 dE1 d /W2=0
1 2f of f of
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f f
But this also specifies the equilibrium condition. Thus
S 1= f(T) (in equilibrium)
E d/W=0 T8.044 L14B5
6. Two Fundamental Inequalities What ifE1 = E1?
1 as equilibrium is established.1p(E1) p(E1)
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1(E1)2(EE1) 1(E 1)1)2(EE
1(E
1)1 1) 2(EE
1(E1) 2(EE1)
8.044 L14B6
1) S1(E1)+S2(E E0 S1(E 1) S2(EE1)
S1 S2
S = S1 + S2 increases
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S 0The total entropy of an isolated system always in-
creases or, at equilibrium, remains constant.
8.044 L14B7
Now assume 1 T2 Tbath does not change.2
/Q2 ddE2 d /Q1dS2 = = =T2 T2
Tbath
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d/Q1dS= dS1 + dS2 = dS1 Tbath 0d/Q1
dS1 Tbath
In particular, for systems in equilibrium with a bath
dS= d/Q/T
.
8.044 L14B8
Example Ideal Monatomic Gas
VN 4emE
3N
3N/2=
V
4emE3N
3/2
N
4emE3/2 S(E,N,V) = kNln V
3N
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3N2
EVN 4emE
3N
3N/2
4emE3/2 2 1 E S(E,N,V) = kNln
V
3N ln(
3 N)
8.044 L14B9
The Energy Relation
1 S Nk 3 1 (3/2)Nk =
T E N,V {} 2 E{}= E
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E= (3/2)NkT
Here U= E so CV = U = (3/2)Nk.T V8.044 L14B10
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Microcanonical: E fixed + equal a priori probabilities
microscopic probability densities [S.M.]
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Together with the definition of entropy
temperature scale and 2ND law inequalities [Ther-modynamics]
8.044 L15B1
S 0 (3 4 7) 2ND Lawd dS1 /Q1 (3 6 5)T
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Quasi-static means arbitrarily close to equilibrium.
Necessary for work differentials to applyRequired for = in above 2ND law relations
8.044 L15B2
7. Entropy as a Thermodynamic Variable
S 1/W=0 T gives us TE d
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Other derivatives give other thermodynamic variables.
PdV
d/W = SdA Xidxi +HdM+ EdP+ iFdL
8.044 L15B3
We chose to use the extensive external variables (a
complete set) as the constraints on . Thus
Skln = S( E,V,M, )Now solve for E.
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S(E,V,M, ) E(S,V,M, )We know
dE|d /Q from the 1ST law/W=0 = ddE /W=0 TdS utilizing the 2ND lawd|
8.044 L15B4
Now include the work.
dE = d /W/Q+ d
dE T dS+ d/W
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P dV dE T dS+ SdA + HdM+ EdP+
FdL
The last line expresses the combined
1ST and 2N D laws of thermodynamics.8.044 L15B5
Solve for dS.
1 P HdS= dE+ dV dM
EdP+T T T T
Examine the partial derivatives of S
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Examine the partial derivatives ofS.
S S H1
==E V,M,P T M E,V,P T S P S Xj ==V E,M,P T xj E,xi=xj
T8.044 L15B6
INTERPRETATION
S(E,V)
S SdS =
dE + dV
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V
dS = dE+ dVE V V E
1 P= dE+ dV
T TE
8.044 L15B7
UTILITY
Internal Energy
S(E,V,N)
E
V
=1
T T(E,V,N)E(T, V, N)
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E V T
Equation of State S(E,V,N)
V E=PT P(E,T,V,N)P(T, V, N)
8.044 L15B8
Example Ideal Gas
4 E3/2S(E,N,V) = kln = kNln V 3em N
S kN {} kN P
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SV E,N=
kN{}
{}
V =kNV =
PT
PV= NkT
8.044 L15B9
COMBINATORIAL FACTS
# different orderings (permutations) ofKdistinguish-able objects = K!
# of ways of choosing L from a set of K:
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# ofways ofchoosing L from a set ofK:K!
if order matters(KL)!
K!if order does not matter
L!(KL)!8.044 L15B10
EXAMPLE Dinner Table, 5 Chairs (places)
Seating, 5 people 5 4 3 2 1 = 5! = 120
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Seating, 3 people 5 4 3 = 5! =60 2!
1Place settings, 3 people 5 4 3/6 = 5! = 10 2! 3!
8.044 L15B11
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EXAMPLE 2 Level System
Ensemble of N "independent" systems
ENERGY
| 1 >
N = N0 + N1
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|
N N0 + N1
E = N1
|0 >
0
8.044 L16B1
SURFACE MOLECULES IONS IN A CRYSTAL LOWEST LYING STATES
0 0
ENERGY
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0
E N1
NO WORK POSSIBLE (JUST HEAT FLOW)
8.044 L16B2
1 when N1 = 0 or NN!(E) =N1!(NN1)! Maximum when N1 = N/2
T= (or -S(E)
)
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E = N/2
S(E) = kln (E)E = N
T=0
ET>0 T
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1
S S N1 k= = = [1ln N1 +1+ln(NN1)]T E N N1 E
1/
k NN1 k N = ln = ln N1 N1
1
8.044 L16B4
NN11 = e/kT NN1 =
e/kT + 1
NE= N1 =e/kT + 1
1.0 N1 /N or E/N
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/kT~ e~0.5
1 2 3 kT/4
8.044 L16B5
e/kTE 2
= NkC(e/kT + 1)2T kT
2 Nk 2Nk e/kT low T, high TkT 4 kT
0 4
0.5
C/Nk
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0.1
0.2
0.3
0.4
1 2 3 4kT/
8.044 L16B6
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p(n) =
(N1)! N1! (NN1)!N! (N1 n)! (NN1 1 + n)!
1/N 1 n= 0 NN1 n= 0N1 n= 1 1 n= 1
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N1 p(0) = NN1 = 1
N N
p(0) +p(1) = 1N1 = [e/kT + 1]1 p(1) = N
8.044 L16B8
1
p(n)
0. 5
0 1 n
p(0)
p(1)
1 2 3 4kT/
N
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NE= (0)N p(0)+()N p(1) =
e/kT + 1
But we knew E, so we could have worked backwards
to find p(1).
8.044 L16B9
MICROCANONICAL ENSEMBLEMODEL THE SYSTEM
FIND (E,N,V ....)THERMODYNAMIC RESULTS MICROSCOPIC INFORMATION
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FIND S(E,N,V ....) P(~~~) = '/
1S =TE N,V
PS =V E,N T
etc.
8.044 L16B10
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ENTROPY AND THE 2nd LAW
21 2
1
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Tbath
dS 0 dS1 dQ
1/Tbath
8.044 L17B1
S as a State Function
Note: adiabatic ( d/Q = 0) constant S if thechange is quasistatic. This is the origin of the sub-
script S on the adiabatic compressibility.
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1 V
1 V
T V P T S V P S
8.044 L17B2
Example A Hydrostatic System
S SdS =T V dT+ V TdV by expansion
1 P=
TdU+ TdV from dU= TdSPdV1
U
1
U
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1
U 1 U
= dT+ + P dVT T V T V T
by expansion ofUBut the cross derivatives ofS must be equal.
8.044 L17B3
1 U 1 2U=V T T V T T VT
1
U1
U1
2U1
P+ P =
T2 V T + P + +T T V T V T VT T T Equating these two expressions gives
U P+ P = T
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+V T T VU P
= TVPV T T
New Information! Does not contain S!
8.044 L17B4
CONSEQUENCES a) U Ud/Q= dU+ PdV= dT+ + P dVT V V T
CV T PT V
d/Q P V
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d/Q P V = CV + TdT
PCP
T V T PV
1 V 1 VUse and T .
V TP
V PT
8.044 L17B5
=
P 1 (1) V T P = = T V T V V TV P P T P T
T 2V T 2V
CPCV = TT
V = 1 =T TCV
For an ideal gas P V= N kT = 1/T and T = 1/P.
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Thus
V /T= =N kCPCV 1/P
This holds for polyatomic as well as monatomic gases.
8.044 L17B6
CONSEQUENCES b) Ideal Gas: CVNkT U Nk
P = = T P = PP = 0 V V T VU U
dU= dT
+ dV
= CV dTT V V T
CV 0
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CV 0
TU= CV(T, V) dT+ constant0(U/V)T = 0 for all T CV is not f(V); CV = CV(T).
8.044 L17B7
CONSEQUENCES c) Ideal Gas: SS S
dS= dT+ dVT V V T
d/Q
S
d/Q= T dS = TdT
VCV
T V
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dU PdU= T dSP dV dS= + dV
T TS 1 U P
= + .V T T V T T
0 Nk/V
8.044 L17B8
CV(T) N k dS= dT+ dVT V
T ) VS(T, V) = CV(T dT+ N k ln( ) + S(T0, V0)T0 T V0
For a monatomic gas CV = (3/2)N k.
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T VS(T, V)S(T
0, V
0)= ( 3/2)N k ln( ) + N k ln( )
T0 V0
V T 3/2
= N k ln V0 T0
8.044 L17B9
V T3/2
isentropic (adiabatic)
V2/3
T
are constant
V 5/3P
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V5/3P
8.044 L17B10
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Maxwell Relations
E
E
dE(S, V) =S V
dS+V S
dV expansion
= TdSPdV 1st and 2nd laws
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T P
V S
= S V
8.044 L18B1
TdE(S,L) = TdS+ FdL = F
L S S L
T HdE(S,M) = TdS+ HdM =
M S S M
Observe:
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Observe:
d(TS) = TdS+ SdTd(PV) = PdV+ V dP
8.044 L18B2
Helmholtz Free Energy FETS
S
P
dF= SdTPdV V T
=T V
Enthalpy HE+ PV
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T V
dH= TdS+ V dP P S
=S P
8.044 L18B3
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The Magic Square Mnemonic
-V
S
xE F
dE= TdS+ Xdx-T
S V
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S V(1) = (1)(1)
H P G P T
(+1) T PX
8.044 L18B5
Homework problem 6-2, A Strange Chainl
F
F
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L = Nl tanh(lF/kT)For small extensions
1 =
T8.044 L18B5b
Example Elastic Rod
Given F= (a + bT) (L L0) and CL = T3+ for stability
Find E(T, L) and S(T, L).
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S S dE= TdS
d/Q +
FdL = T T LdT+ T L T +F dL
CL bT(LL0)
8.044 L18B6
E F
GH
L
-T
-F
S S
(1) =
F= b(LL0)L T T L
dE= T3dT+ a(LL0)dL
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E=T4 + f(L)
4
af(L) = a(LL0) f(L) = (LL0)2 + c1
2
8.044 L18B7
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bS(T, L) =
T3
2(LL0)2 + c2
3
A rubber band has a negative thermal expansion coef.dF= (a+ bT)dL+ b(LL0)dT set = 0
+ when extended
L
b (LL0)
< 0 for rubber
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T
F = (a+ bT)
< 0 for rubber
+ for stability
b > 08.044 L18B9
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Heat Engine
Takes a substance around a closed cycle Heat is put into the substance and taken out
Work is taken out
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Efficiency, (work out) / (heat in)
8.044 L19B1
Closed cycle U= Q+W = 0Q= W
Q /QPHEAT IN
d1
(HOT RESERVOIR)
2 1= d /Q/Q+ d
1 2
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HEAT OUT
/Q/Q +
1 22
|QH| |QC|(COLD RESERVOIR)V
8.044 L19B2
Most General Case
Wout = W = Q= |QH| |QC|Wout
= |QH| |QC|= 1 |QC||QH| |QH| |QH|
Very Special Case Example: Carnot Cycle
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Any substance Isothermal and adiabatic changes
8.044 L19B3
PT
1 1'
22 '
Q=0
1'
2
2 '
1
TH
TC
Q=0
S V
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/Q< TdSUse the second law: d
8.044 L19B4a
PT
2' 2
1 1 '
S
1'
22 '
1TH
TC
Q=0
Q=0
V
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DRAWN TO SCALE FOR AN IDEAL GAS: PV=NkT
T H = 1.5 TC S HIGH - SLOW = (3/2) Nk ln2
8.044 L19B4b
1
dS|QH| TH1
2
2 1dS, use dS=
1dS|QC| TC
2 2
1 TC TC1
dS |QC| TC1
1
dS and |QC||QH|
TH
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|QH| TH
TC= 1 |QC|
|QH| 1
TH8.044 L19B5
Arbitrary Engine Cycle
d/QTdSfor each element along the path.
2 2 2d/Q
1TdS Tmax dS
1 1
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/
11 1
positive|QH|
8.044 L19B6
2
1 1d/Q
2TdS, both sides are negative
1 1TdS| Tmin|QC| | 2 | 2 dS|
2Tmin | 1 dS| since dS= 0
T|Q |
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Tmin|QC||QH| Tmax
Tmin= 1 |QC| 1 Tmax|QH|
8.044 L19B7
Carnot cycle in a pure thermodynamic approach
QC TCUsed to define temp. = 1 |QH| | | 1 TH
Used to define the entropy
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/Q dd0 /Q is an exact differential
T T8.044 L19B8
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Refrigerator Run cycle backwards, extract heat at cold
end, dump it at hot end
HEAT EXTRACTED (COLD END)=
|QC|= |QC|WORK DONE ON SUBSTANCE W |QH| |QC|
For the special case of a quasi-static Carnot cycle
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TC=TH TC
8.044 L20B1
As with engine, can show Carnot cycle is optimum.
Practical: increasingly difficult to approach T = 0.
Philosophical: T = 0 is point at which no more
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heat can be extracted.
8.044 L20B2
Heat Pump Run cycle backwards, but use the heat
dumped at hot end.
HEAT DUMPED (HOT END)=
|QH|= |QH|WORK DONE ON SUBSTANCE W |QH| |QC|
For the special case of a quasi-static Carnot cycle
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TH=TH TC
8.044 L20B3
55o F subsurface temp. at 40o latitudeTC
= 286K70o F room temperature
TH = 294K
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294|QH| 37W 8
8.044 L20B4
3rd law lim S= S0T 0
At T = 0 the entropy of a substance approaches aconstant value, independent of the other thermody-
namic variables.
Originally a hypothesisN lt f t h i
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Now seen as a result of quantum mechanicsGround state degeneracy g (usually 1)
S kln g (usually 0) 8.044 L20B5
S
Consequences = 0x T=0
Example: A hydrostatic system
1V
1
S
V T P
= V P T
0 as T0V T 2
CP CV = 0 as T0
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P VKT
S(T)S(0) = TT=0
CV(TT
)dTCV(T)0 as T0
8.044 L20B6
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3 1/21
/2 ep(px) = e
4m 1/2
N e
3N < >
1= e
4m < > /2
2Now use = p2/2mand < >=< px > /2m.x
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p(px) =
1
ep2/2x x
2 < p2 >x
8.044 L13B15
Homework problem on classical harmonic oscillators
1p(pi, qi) =
(2/) < > exp[/< >]1 2=
(2/)kTexp[pi/2mkT] exp[(m2/2kT)qi]2
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=
Ensembles
Microcanonical: E and N fixedStarting point for all of statistical mechanics
Difficult to obtain results for specific systems
Canonical: N fixed, T specified; E variesWorkhorse of statistical mechanics
G d C i l d ifi d d
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Grand Canonical:T
and
specified;E
andN
varyUsed when the the particle number is not fixed
8.044 L21B1
2
1
1 IS THE SUBSYSTEM OF INTEREST.2, MUCH LARGER, IS THE REMAINDER OR THE "BATH".
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ENERGY CAN FLOW BETWEEN 1 AND 2.THE TOTAL, 1+2, IS ISOLATED AND REPRESENTED BY AMICROCANONICAL ENSEMBLE.
8.044 L21B2
For the entire system (microcanonical) one has
volume of accessible phase space consistent with Xp(system in state X) =
(E)
In particular, for our case
p({p1, q1}) p(subsystem at {p1, q1}; remainder undetermined)
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1({p1, q1})2(EE1)=(E)
8.044 L21B3
klnp({p1, q1}) = k
ln 1 + k
ln 2(E
E1)
k
ln (E)
kln 1 = 0 S2(EE1) S(E)
S2(E2)E1S2(EE1) S2(E)
E2
evaluated at E E
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evaluated at E2
= E
8.044 L21B4
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In all cases, including those where the system is too
small for thermodynamics to apply,
p({p1, q1}) exp[H1({p1, q1})]kT
exp[H1({p1, q1})]
kT=
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exp[H1({p1, q1})]{dp1, dq1}
kT
8.044 L21B6
If thermodynamics does apply, one can go further.
S(E) = S1(< E
1>) + S
2(< E
2>)
S2(E)S(E) =S2(E)S2(< E2 >) S1(< E1 >)
(S2(E2)/E2) < E1 >=< E1 > /T < E1 >k lnp({p1, q1}) = H1({p1, q1}) + S1
T T
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T Tp({p1, q1}) = exp[
(< E1 >T S1)] exp[H1({p1, q1})]kT kT
1/Z8.044 L21B7
< E1 >T S1 = U1 T1S1 = F1p({p,q}) = Z
1
exp[
H({p,q})]kT
Z is called the partition function.
ZN(T, V) = exp[H({p,q})]{dp,dq}kT
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= exp[(ET S) F(T, V, N)] = exp[ ]kT kT
8.044 L21B8
In the canonical ensemble, the partition function is
the source of thermodynamic information.
F(T, V, N) = kTlnZN(T, V)
FS(T, V, N) = T V,NF
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FP(T, V, N) = V T,N
8.044 L21B9
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Canonical Ensemble
p(E)
p(E) eE/kT NOT!
E
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p({p,q}) eH({p,q})/kT
8.044 L22B1
ADVANTAGES OF CANONICAL OVER MICROCANONICAL ENSEMBLE
MICROCANONICAL CANONICAL
1) ONE INTEGRATES OVER ALL PHASE SPACE
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8.044 L22B2
SURFACE OFCONSTANT E
2) SEPARATION
eH/kT = eHa/kTeHb/kTlet H= Ha+Hb , then
p({p,q}) = p({p,q}a) p({p,q}b) (a & b are SI)
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Z= ZaZbF = Fa+ FbS= Sa+ Sb etc.
8.044 L22B3
For N similar, non-interacting systems
Z= (Z1)N , F = N F1 , S = N S1
For N indistinguishable particlesN
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(Z1)Z= , correct Boltzmann countingN!
8.044 L22B4
Example Non-interacting classical monatomic gas
N
pipi= i=1Hi
N
Z=(Z1)
NN!H = i=1 2m
2p2 +py+p2zH
1( r) =
x
p,
2m
p, 2+py+pz)/2mkT/Z1xp1( r) = e(p 2 2
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Gaussian px p 2< p >=< p2 +py+p2 >= 3mkTx z= 3/2 kT
8.044 L22B5
Z1 = 2 2 2e(px+py+pz)/2mkT dpxdpydpzdxdydz
h3
3/2= (2mkT)3/2LxLyLz/h3 = V2mkTh2
N3N/2 3/21 V
2mkT
Z(T, V, N) = 1 VN2mkT =N! h2 N! h2
units of cm3
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3/2 N1 V 2mkT
=
N! v0 h2 v08.044 L22B6
F = kTlnZ
= kT
2/303N 2mkTv
NlnN+N+Nln(V/v0)+ l n2 h2
3Nuse N= Nln e= ln(e2/3)2
V 3N 2e2/3mkTv2/3
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kTNln + lnNv0 2 0h2=
8.044 L22B7
1 1F NkT
= (1)(kTN)P = = VNv0
V Nv0 VT,N
3N 1 ()2 () T
F= k[ ] + kTS =
T V,Nk+ k
Nln
ln
2(ev0)
2/3mkTh2
3N2
V 3N+
Nv0 2=
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3E = F+ TS= NkT
2 8.044 L22B8
Find the adiabatic path, S= 0.
V V0 3lnN v0
ln = (ln Tln T0)N v0
2
V 3 T Vln = =
V0
2ln
T0
V0
TT0
3/2
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V0 2 T0 V0
T0
8.044 L22B9
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Canonical Ensemble
CLASSICAL QUANTUM
p({p,q}) = e
H({p,q})/kT
/Z p(state) = eEstate/kT /Z
Z= eH/kT{dp,dq} Z= eEstate/kTstates
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8.044 L23B1
EXAMPLE 2 LEVEL SYSTEM: STATES OF AN IMPURITY IN A SOLID
E = 1
g
g-FOLD
DEGENERACY
* EXCITED
E = 0GROUNDSTATE
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LOCATION INTERNAL
ENERGY LEVELS PHYSICAL DIFFERENCE
8.044 L23B2
STATES: |0>, |1 >, g > | E=0 E=
eEstate/kT = 1 e0 + g e/kT = 1 + ge/kTZ1 = states
p(state) = eEstate/kT
/Z1
1=
1 +/kT for |0 >
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1 + ge
=e/kT
1 + ge/kT for |i > i = 1, g8.044 L23B3
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AssumeN impurities (N1)
0
= 0(V/V0)V0 V
Z= ZN F(T, V, N) = kTlnZ= NkTlnZ11
g(kT2)e
/kT
SF Nk l Z +NkT
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g(kT2)e /kTS= F = Nk lnZ1 +NkT1 + ge/kT
T V
8.044 L23B5
e/kTS= Nk ln(1 + ge/kT) + gNk
kT 1 + ge/kT
g e/kTU= F+ TS= N1 + ge/kT = N p(E= )
F F P = =V T,N T V TV
) /kT
U( g
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kT) e/kT = U= NkT( g
1 + ge/kT V V8.044 L23B6
ALTERNATIVE WAY OF FINDING U
Usually (but not always) U= .
If so, U= H({p,q})p({p,q}) {dp,dq}But Z= c eH({p,q}){dp,dq} 1/kT
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8.044 L23B7
1
1
Z H({p,q})eH({p,q}){dp,dq}= c N,V
eH({p,q})Z= H({p,q}) {dp,dq}
eH({p,q}
){dp, dq}
p({p,q})
Z N,V
Z= U
Z N V
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Z N,V8.044 L23B8
Example Monatomic Gas
1VN
2mkT3N/2
Z= = 3N/2N! h2
1 3N 1
U=
3N/
2 2
3N/2 =
3NkT
2
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N/
2
2
8.044 L23B9
Example 2 Level System
NZ = 1 + ge
U = 1 + ge NN1 + ge N1 ge
gN e/kT
=/kT
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gNe1 + ge/kT
8.044 L23B10
Background Classical Harmonic Oscillator
H(p,x) = p2 + 1Kx22m 2
1 x2
p(p,x) = exp[ p2
] 1 ]2mkT 2mkT 2(kT/K) exp[2(kT/K)
Z = 2m
kT 1 U = 1
K Z
Z
= kT, CV = Nk T
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Z 2m kT U 1K ZZ
kT, CV Nk T
8.044 L23B11
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Polyatomic Gases
1
Non-interacting, identical Z= N! ZN Find Z11Each molecule has # atoms 3# position coordi-nates
3# = 3 + nr + (3#3nr)
C.M. rotation nv, vibration
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nv, vibration
8.044 L24B1
MONATOMIC DIATOMIC LINEAR TRI. NON-LINEAR TRI.Xe HS CO
2H
2O
3
3
3
2
3
2
3
0
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39
49
16
03
8.044 L24B2
C.M. Motion:
Particle in a box Es kT classicalRotation:
(H2 rot = 3.651012 Hz 175 K ) Q.M.Vibration:
(H2 vib = 1.321014 Hz 6,320 K ) Q.M.
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HCM +Hvib +Hrotproblem separatesH=
8.044 L24B3
Vibration
nv
1 2 1 Ki Hvib =
+ 2 i2 aiKiai
2
i=1 2
nv 1 dimensional harmonic oscillators, use Q.M.
2)Hn= nn n= (n+ 1 h n= 0,1,2,
The energy levels are non degenerate
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The energy levels are non-degenerate.
8.044 L24B4
2)h/kT/ en/kTp(n) = e(n+1
n=0
h e(n+
1 h/kT 1h/kT e 2) = e2 h/kT nn=0 n=0
h
h
h
h 1
1
1
1
1 h/kT/ h/kT1e= e2h/kT ep(n) = 1e h/kT n= (1b)bn
72
5
2
32
12
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8.044 L24B5
Geometric or Bose-Einstein
p(n)
n
b 1< n > = =
1b eh/kT1e hh/kT when kT
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e hwhen kT
8.044 L24B6
1For kTh 2 h 1 + h+ 1
kT 2 kT 1kT 1 kT 1 h= 12 h h kTh 1 + 1 2 kTkT 1= 2h
2 ) h (Classical)< >= (< n > + 1 h kT kT
h kT 1 h (Ground state)
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h kT1 h (Ground state)28.044 L24B7
3
2
1
0
-1
1 2 3
kT/h
T
kT
h1
2
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8.044 L24B8
< > d < n > CV = N = Nh
T V
dT
h/kTh2 e
= N k kT eh/kT12
h2 h/kT kTN k e h (energy gap behavior)kT
N k kT h
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N k kT h8.044 L24B9a
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High and low temperature behavior without solving thecomplete problem Consider first the high T limit.
e-/kT
contains stateshkT h
Z1 =
en/kTn=0
1 E/kT kT y kT 1
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1 eE/kTdE kT= eydy= kT
0 h 0 h h1
8.044 L24B10
Zvib = ZNN1
1 Z
= N(N)N1 = NkTUvib = Z
NCvib = Nk
Next, consider the low T limit.e -/kT
kT consider only 2 states
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h12
32h
8.044 L24B11
32 h/kTe 1 h/kTp(n= 1)
e1 h/kT + e3 = h/kT + 1 e
2 h/kT e2h/kTp(n= 0) 1e
< E >= 2N h/kT + 3N h/kT1 h 1e he2= 1N heh+ N h/kT2 < E > h h/kT = N k h2 h/kTCV = = N h e e2
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N k CV Nh e e
T kT2 kT8.044 L24B12
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Angular Momentum in 3 Dimensions
CLASSICAL, 3 numbers: (Lx, Ly, Lz); (|L|, , )
QUANTUM, 2 numbers: magnitude and 1 component
h2 l,m l= 0,1,2L L l,mL2 l,m = l(l+ 1)
Lzl,m = mh l,m m= l,l 1, l2l+1 values
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2l+1 values
8.044 L25B1
Specification: 2 numbers l & m l,m or l,m>|
Molecular rotation
In generalI3
= 01 L2 1 L2 1 L2Hrot =
2I11 +
2I22 +
2I33
L3 = I3 3
= 0 For a linear molecule
1 I1 = I2 I1 + L
2Hrot = (L2
2) =1
L L 2I 2I
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1( 2)2I 2I
8.044 L25B2
1
L2
Hrot = 2IHrot l, m> = l l, m>| |
h2= l(l + 1) l, m>
2I|
l depends on l only;it is 2l + 1 fold degenerate.
l= kR l(l + 1)h2
20R
12R
6R
2R0
/k
9 l = 4
7 l = 3
5 l = 2
3 l = 1l = 01
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h2R (rotational temp.)
2I k8.044 L25B3
< >= =
p(l,m) = 1 el(l+1)R/TZR
ZR= el(l+1)R/T = (2l+ 1)el(l+1)R/Tl,m l
For TR ZR1 + 3e2R/T = 1 + 3e2Rk1 Z 6R k e2Rk
6R k e2R/T
Z 1 + 3e2Rk
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8.044 L25B4
< > CV|rot = N = 6RN k 2R
e2R/T
T T22R2 e2R/T= 3N k (energy gap behavior)
T
For TR, convert the sum to an integral.
ZR (2l+ 1)el(l+1)R/T dl0
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8.044 L25B5
x(l2 + l)R/T dx= (2l+ 1)R/Tdl
1TZRR
exdx= T = 10 R kR
1 Z
(1)(1)Z/
< >= Z= = 1 = kTZ
< > CV|rot = N N k (classical result)T
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8.044 L25B6a
Image removed due to copyright reasons.
Figure 22-1 from
Morse, P. M. Thermal physics. 2nd ed. New York, NY: W. A. Benjamin, 1969. ISBN: 0805372024.
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8.044 L25B6b
H= HCM +Hrot +Hvib
CV(T) = CV|CM + rot vib C V| + C V| all T appears at modest T only at highest T
8 044 L25B7
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8.044 L25B7a
Image removed due to copyright reasons. Image removed due to copyright reasons.
Figure 12-16 from Figure 12-2 from
F. W. Sears, and G. L. Salinger. F. W. Sears,and G. L. Salinger. Thermodynamics, Kinectic Theory, and Statistical Physics. 3rd ed. Thermodynamics, Kinectic Theory, and Statistical Physics. 3rd ed.Reading, MA: Addison-Wesley Pub. Co., 1975
.
ISBN: 020106894X. Reading, MA: Addison-Wesley Pub. Co., 1975. ISBN: 020106894X.
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8.044 L25B7b
Raman ScatteringBEFORE AFTER
i
f
i
f
= f - i = h(i - f )
FREQUENCY CHANGES IN THE SCATTERED LIGHT CORRESPOND TOENERGY LEVEL DIFFERENCES IN THE SCATTERER.
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WHICH ENERGY LEVEL CHANGES OCCUR DEPEND ON SELECTION
RULES GOVERNED BY SYMMETRY AND QUANTUM MECHANICS
8.044 L25B8
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ROTATIONAL RAMAN SPECTRUM OF A DIATOMIC MOLECULE
LEVEL DEGENERACY
BOLTZMANN FACTOR
I()
0
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4(kR /h)
-6(kR /h)
0
8.044 L25B10
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Thermal Radiation Radiation in thermal equilibrium
with its surroundings
k
E0
B0E= E0 ei( rt) k = c|k| k B= B0 ei( rt) B0 = 1kE0 /c
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8.044 L26B1
1 2Time average energy density u= 20|E0|Time average energy flux jE = (cu)1kTime average pressure ( to k) P = uThermal radiation has a continuous distribution of
frequencies.
u(,T)
Peaks near h= 3kBT(h/kB51011 K-sec)
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8.044 L26B2
Spectral Region (Hz) T (K) Thermal Rad.Radio 106 5105
Microwave 1010 0.5 cosmic background
Infrared 1013 5102 room temp.Visible 12 1015 2104 suns surface
Ultraviolet 1016 5105
X ray 10
18
5
10
7
black holes
ray 1021 51010
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8.044 L26B3
ENERGY ABSORBEDABSORPTIVITY (,T)
ENERGY INCIDENT
ISOTROPIC
ENERGY EMITTEDEMISSIVE POWER e(,T)
AREA
ISOTROPIC
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8.044 L26B4
THERMAL RADIATION: PROPERTIES2 ENERGY FLUXES,
IN AND OUT OF CAVITY B
FILTER: FREQUENCY
OR POLARIZATION
CAVITY A
CAVITY B
TA
TB
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OR POLARIZATION
ASSUME TA = TB AND THERMAL EQUILIBRIUM8.044 L26B5
CONCLUSIONS:
u(,T) is independent of shape and wall material
u(,T) is isotropic
u(,T) is unpolarized
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8.044 L26B6
CONSIDER AN OBJECT IN THE CAVITY,IN THERMAL EQUILIBRIUM
dA
COMPUTE THE ENERGY FLUX
T
n
A
ct
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8.044 L26B7
E = (E in cylinder)p(,) dd
sin 1 = (u A cos ct) dd
2 2/2 cos sin 2 1= c u At d d0 2 0 2
1/4 1
1energy flux onto dA= 4c u(,T)
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8.044 L26B8
Momentum Flux
u Plane wave momentum density p=c1k
p = 2|p| since poutpin = | |
8 044 L26B9
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8.044 L26B9
2 cos |p = (E in cylinder)p(,) dd|
c/2= u(,T)At cos2 sin d 2 1 d0 0 2
1/3 1
1
= 3u(,T)At
P(T) = 1 u(,T) d3 0
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8.044 L26B10
Apply detailed balance to the object in the cavity.
Eout = Ein
e dA = (14c u(, T)) dA
e(, T) 1= 4c u(, T) (, T)
This ratio has a universal form for all materials.The result is known as KIRCHOFFS LAW.
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8.044 L26B11
Black Body Radiation
If1 BlackThen e(, T) = 14c u(, T)
OVEN
CAVITY
AT
T
Measure e(, T)
and obtain u(, T)
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8.044 L26B12
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Thermodynamic Approach
u(T)
u(, T) d0
Then
E(T, V) = u(T)V1P(T, V) = 3u(T)
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8.044 L27B1
This is enough to allow us to findu(
T).
dE = T dSP dVE
TS
TP TP
= = V PV T V T1= 1T u(T)3u(T)3
also = u(T)4
u(T) = u(T)
Tu(T) = AT4
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8.044 L27B2
Emissive Power of a Black (= 1) Body
4c u(,T)e(T) = 1e(,T) = 1 4c u(T) = 1AcT44e(T) T4
This is known as the STEFAN-BOLTZMANN LAW.
= 56.7109 watts/m2K4
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8.044 L27B3
Statistical Mechanical Approach
Single normal mode (plane standing wave) in aH?
rectangular conducting cavity.
Ex
z
0 L
( 1xr,t) = E(t) sin(nz/L)E0,0,n,
1x
1y( 1yB0,0,n, r,t) = (nc2/L)1E (t) cos(nz/L)
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8.044 L27B4
Energy density = 10E E+ 1 B B [no 1 r or t average]2 20
V
= 10E2(t) + 11
(nc2/L)2E2(t)2H 2 2 0V 0
= E2(t) + (nc/L)2 E2(t)
2 2
Each mode corresponds to a harmonic oscillator.
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8.044 L27B5
Count the modes.
Enx,ny,nz = |E jsin(nxx/L) sin(nyy/L) sin(nzz/L)eit
|
The unit polarization vector j has 2 possible orthog-onal directions and ni= 1,2,3 .
2 E 2 2 22 2E= 0 2 = c2 (n + ny+ nz)L xt2 c
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8.044 L27B6
nz
nx
ny
GRID SPACING
1 UNIT
R
If the radian frequency <
n2 2 2L
R= x+ ny+ nz = c# modes (freq. < )
= 2 1 483R33= L 33 c
8.044 L27B7
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3d# V 2D() = = L 2 =
2c3d c
D()
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8.044 L27B8
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Normal modes of the radiation field in a rectangularcavity with conducting walls
r,t) =L nx+ ny+ nzEnx,ny,nz,(
c 2 2 2
Harmonic oscillators
VD() = 2c3 2, 0
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8.044 L28B1
Classical Statistical Mechanics
< () >= kBT u(,T) =< () > D() = kBT 2V 2c3
u(T) =
u(,T) d=0
u(, T)CLASSICAL
MEASURED
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8.044 L28B2
Quantum Statistical Mechanics
h< () >= +
e
h/kT1 h/2D() h 3
u(,T) =< () > =2c3 eh/kT1 + z. p. termV
du(,T)To find the location of the maximum, set = 0.
dThe maximum occurs at h/kT2.82.
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8.044 L28B3
h/2kT(1eh/kT)1Z= Zi Zi= estates i
The first factor in the expression for Zi comes fromthe zero-point energy.
F(V,T) = kTlnZ= kT lnZistates i
= kTD() [lnZi] d0
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8.044 L28B4
h/kT)F(V,T) = kT D() ln(1 e d+0
kTV h/kT) d= 2 ln(1 e2c3 0
V
=h3
(kT)4 x2 ln(1 ex) dx2c3 0 4
45
1
2
= h3 (kT)4 V 45 c3
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8.044 L28B5
F 1 2P =
V T = 45 h)3 (kT)4(cF 4 2 k4T3 VS = =
h)3
T V 45 (c1 4 1 2
E = F+ TS= + ) = h)3 (kT)4 V45 45 ( 15 (c
1 1Note: P = 3E/V = 3 u(T) independent ofV.
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8.044 L28B6
NOTE: THE ADIABATIC PATH IS T3V=CONSTANT
T ADIABATIC
T/T0 = (V/V0)-1 / 3
V
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8.044 L28B7
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n( s) often factors into space and spin parts.r, s) = space r)spinr, n ( n (n( s)
space
(x) ex2/2
Hn( x) H.O. in 1 dimensionn
r) e kspace( ir free particle in 3 dimensionsn
8.044 L29B2
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spin
s)n (Spin is an angular momentum so for a given value of
the magnitude Sthere are 2S+ 1 values ofmS.
For the case ofS= 1/2 the eigenfunctions of the z
s are 1/2( s)component of s) and 1/2(
hSz 1/2( s) s) = 1/2(
2
hSz 1/2( s) s) = 1/2(2
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8.044 L29B3
spin s) is not necessarily an eigenfunction ofSz. Forn (example one might have
spin s) = 12
1/2( s)s) +1 1/2(n ( 2
In some cases n( s) may not factor into space andr, spin parts. For example one may find
s) = f(x)1/2( s)n(x, s) + g(x)1/2(
8.044 L29B4
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Many Distinguishable Particles, Same Potential,No Interaction
Lump space and spin variables together
r1, r2, s1 1 s2 2 etc.H0(1) + N) = H0(2) + H0(N)H(1, 2,
In this expression the single particle Hamltonians all
have the same functional form but each has arguments
for a different particle.
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8.044 L29B5
The same set of single particle energy eigenstates isavailable to every particle, but each may be in a dif-
ferent one of them. The energy eigenfunctions of the
system can be represented as products of the single
particle energy eigenfunctions.
{n}(1,2, N) = n1(1)n2(2) nN(N) nN}. There are N #s, but each ni{n} {n1, n2,
could have an infinite range.
H(1,2, N){n}(1,2, N) = E{n}{n}(1,2, N)
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8.044 L29B6
Many Distinguishable Particles, Same Potential,
Pairwise Interaction
N
H(1,2, N) =
H0(i) +1 2
Hint(i, j)
i=1 i=j
The {n}(1,2, N) are no longer energy eigenfunc- tions; however, they could form a very useful basis set
for the expansion of the true energy eigenfunctins.
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8.044 L29B7
Indistinguishable Particles
i iPij f( j ) f( j )Pij)
2 = eigenvalues of( I Pij are + 1,1It is possible to construct many-particle wavefunctions
which are symmetric or anti-symmetric under this in-
terchange of two particles.
Pij (+) = (+) Pij
() = ()
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8.044 L29B8
Identical no physical operation distinguishes be-tween particle i and particle j. Mathematically, this
means that for all physical operators OO, [ Pij] = 0
eigenfunctions ofOmust also be eigenfunctions ofPij.
energy eigenfunctions E must be either (+) orE()E .
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8.044 L29B9
states differing only by the interchange of the spa-tial and spin coordinates of two particles are the same
state.
Relativistic quantum mechanics requiresinteger spin E(+) [Bosons]
half-integer spin E() [Fermions]
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8.044 L29B10
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Composite Particles
Composite Fermions and Composite Bosons
Count the number of sign changes as all the con-stituents are interchanged
Well defined statistics (F-D or B-E) as long as theinternal degrees of freedom are not excited
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8.044 L30B1
The constitutents of nuclei and atoms are e,p& n.Each has S= 1/2.
N even even # of exchanges.(+) B-Ealso N even integer spinN odd odd # of exchanges.() F-Dalso N odd half-integer spin
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Particle Nuclear Spin Electrons Statistics
1H (H1) 2 1 B-E
D (H2) 1 1 F-D1
T (H
3
) 2 1 B-E1He3 2 2 F-D
He4 0 2 B-E
Li6 1 3 F-D
Li7 32 3 B-E
H2 0 or 1 2 B-Ex2 integer ()2 B-E
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Let ( s), ( s), be single particle wavefunctions.r, r, A product many-particle wavefunction, (1)(2), does
not work.
Instead, use a sum of all possible permutations:
(+)2 =
12((1)(2) + (2)(1))
(+) 1 1= N!
n!
permutations
((1)(2)(3)N )
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The antisymmetric version results in a familiar form,
a determinant.
1()
= ((1)(2) (2)(1))2 2
states
(1) (1) (1)
1 (2) (2) (2)()
=
particlesN
N! (3) (3) (3) . . ... ... .
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() = 0 if 2 states are the same since 2 columnsNare equal: Pauli Principle.
()
= 0 if 2 particles have the same ssincerand N2 rows are equal.
Specification: indicate which s.p. s are used.} An # ofentries, each ranging{n, n, n,
from 0 to N but with n= N.
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Fermi-Dirac|1,0,1,1,0,0, Bose-Einstein|2,0,1,3,6,1,
n= E Prime indicates n=NExample Atomic configurations
(1S)2(2S)2(2P)6 Ne(1S)2(2S)2(2P)6(3S)1 Na(1S)1(2S)1 He*
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=
Statistical Mechanics Try Canonical Ensemble
Z(N,V,T) = eE(state)/kTstates
eE({n})/kT{n}
= en/kT
{n}
This can not be carried out. One can not interchange
the over occupation numbers and the over statesbecause the occupation numbers are not independent
(n= N).
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T=0 LOWEST POSSIBLE TOTAL ENERGYBOSE: ALL N PARTICLES IN LOWEST SINGLE PARTICLE STATE
n(
) N()
FERMI: LOWEST N SINGLE PARTICLE STATES EACH USED ONCE
< F, F CALLED THE FERMI ENERGY
n() 1
F
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8.044 L30B9
Free Particle in a Box
( s) = V ei
rspin( h2k2 r, 1 k s) = 2m if B= 0 k
Use periodic boundary conditions.
( x+ nyLy z, r,r+ nxLx y+ nzLz s) = ( s)
Lxx+ m
y22Ly
y+ mz2k= m
x
Lzz
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2/ L x
2/ L y
ky
kx
D(k) = (2Lx)1 (2Ly)1 (2Lz)1
= V(2)3
wave vectors only!
N= 2D(k)43k
3F
kF =
32(N/V)1/3
kz
kx
ky
radius kF
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8.044 Statistical Physics ISpring 2008
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2/ L x
2/ L y
ky
kx
D(k) = (2Lx)
1 (2Ly)
1 (2Lz)
1
= V(2)3
wave vectors only!
N= 2D(k)43k
3F
kF =
32(N/V)1/3
kz
kx
ky
radius kF
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Fermions: Non-interacting, free, spin 1/2, T= 0
kz
kx
ky
kF
Fermi sea
Fermi surface
Fermi wave vector
kF = 32(N/V)1/3Fermi energy
h2k2F = F/2m
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