Stat Mech Notes

8/6/2019 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.

8.044 L1B3


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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

8.044 L1B4


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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

8.044 L 1B5a


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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()

8.044 L1B6


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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

8.044 L1B7


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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

8.044 L1B7


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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,

8.044 L2B1


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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

8.044 L2B2


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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

8.044 L2B4


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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!

8.044 L2B7


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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

8.044 L2B8


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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

8.044 L3B6


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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

8.044 L3B8


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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

8.044 L3B9


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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)

8.044 L4B1


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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

8.044 L4B2


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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!

8.044 L4B3


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v1 v1 v1

v2v2v21 = 0-1

v2 and v2are S.I.

8.044 L4B6


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p(v2 |v1)

1

0

v2

(v2-v1) when =1

8.044 L4B8


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c

t

v( t )

1

-1

()

c

8.044 L4B10


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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

8.044 L4B11


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Functions of a random variable

Given: px() and f(x)Find: pf()

f(x)

x

8.044 L5B1


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A. Sketch f(x). Find where f(x) <

B. Integrate to find Pf().

C. Differentiate to find pf().

8.044 L5B2


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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/

8.044 L5B7


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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

8.044 L5B9


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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

8.044 L6B1


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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

8.044 L6B2


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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

8.044 L6B4


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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

8.044 L6B6


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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

8.044 L6B8


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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

8.044 L7B3


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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(,

)

8.044 L7B4


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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 + ).

8.044 L7B5


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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

8.044 L7B7


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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

8.044 L7B8


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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)!

8.044 L7B9


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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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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

http://ocw.mit.edu/http://ocw.mit.edu/


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Facts about sums of RVs
http://ocw.mit.edu/termshttp://ocw.mit.edu/terms


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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.

8.044 L8B1


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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

8.044 L8B4


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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

8.044 L8B5


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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

8.044 L9B3


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Equation of State

P V= N kTV = V0(1 + TKTP)M= cH/(TT0) T > T0

In Equilibrium with Each Other

8.044 L9B4

OBSERVATIONAL FACTS


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"0 th Law"

equilibriumif A C and B

equilibrium

then A B

Cequilibrium

8.044 L9B5

" 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

8.044 L9B9

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

8.044 L9B10

1

For this equilibrium condition


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For this equilibrium conditionh( XA , YA ) = constant = h( XB , YB )

[ PA VA = PB VB ]

8.044 L9B11

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

8.044 L9B7

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


8 044 Statistical Physics I


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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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F i f ti b t iti th t i l T f U i it htt // it d /t


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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





283/417

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


285/417

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





296/417

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


299/417

8.044 L26B3

ENERGY ABSORBEDABSORPTIVITY (,T)

ENERGY INCIDENT

ISOTROPIC

ENERGY EMITTEDEMISSIVE POWER e(,T)

AREA

ISOTROPIC


300/417

8.044 L26B4

THERMAL RADIATION: PROPERTIES2 ENERGY FLUXES,

IN AND OUT OF CAVITY B

FILTER: FREQUENCY

OR POLARIZATION

CAVITY A

CAVITY B

TA

TB


301/417

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


303/417

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


305/417

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


306/417

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.


307/417

8.044 L26B11

Black Body Radiation

If1 BlackThen e(, T) = 14c u(, T)

OVEN

CAVITY

AT

T

Measure e(, T)

and obtain u(, T)


308/417

8.044 L26B12


8.044 Statistical Physics I

Spring 2008



309/417

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


311/417

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.


314/417

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


315/417

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


316/417

3d# V 2D() = = L 2 =

2c3d c

D()


317/417

8.044 L27B8



Spring 2008



318/417

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


319/417

8.044 L28B1

Classical Statistical Mechanics

< () >= kBT u(,T) =< () > D() = kBT 2V 2c3

u(T) =

u(,T) d=0

u(, T)CLASSICAL

MEASURED


320/417

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.


321/417

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


322/417

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


323/417

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



Spring 2008



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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

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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(

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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 eigenfunctions; 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).

8.044 L30B8


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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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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

8.044 L

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