B ~ = B~e z,spinsparalleloranti-paralleltomagneticeld

4 Task Theoretical Physics VI - Statistics

4.1 (Spin lattice)

Ateachsiteofaisolatedlatticeanunpairedelectronwithspin

1/2

^islocalized.

Externalmagneticeld

B ~ = B~e z

^{,spinsparallelor}anti-paralleltomagneticeld

s z,i = ± ¹ 2

. Energyofstate

r

^givenby:

E r (B) = − 2µ B B X N

i=1

s z,i

(a)

n

^{spinsparalleltomagneticeld}

⇒ N − n

^spinsanti-parallel.Energyindepan- danceof

n

^:

E n (B) = − 2µ B B 1 2

X n

i=1

1 − 1 2

X N

i=n

1

!

= − 2µ B B 1 2

X n

i=1

1 − 1 2

N − n

X

i=1

1

!

= − 2µ B B

n − N 2

Numberofstatesgiventhroughbinomialdistribution:

Ω n (E n ) = N ! n! (N − n)!

with:

n (E n ) = N 2

1 − E n (B) N µ B B

thisleadsto:

Ω n (E n ) = N !

N 2

1 − ^E N µ ^{n (B)} B B

!

N 2

1 + _{N µ} ^{E n (B)} _{B B}

!

(b)

ApplyingStirling'sformulato

Ω n (E n )

^{, with}

N − n 1 ⇒ N n

^and

n 1

^:

Ω n (E n ) =

N e

^N

n e

n N − n e

N −n = N ^N e ⁿ e ^{− N} e ^{N − n}

n ⁿ (N − n) ^{N −n} = N ^N n ⁿ (N − n) ^{N −n}

Logarithmusnaturalis:

ln Ω n (E n ) = N ln N − n ln n − (N − n) ln (N − n)

inserting

n (E n )

^:

ln Ω n (E n ) = N ln N − N 2

1 − E n (B) N µ B B

ln N 2

1 − E n (B) N µ B B

−

N − N 2

1 − E n (B) N µ B B

ln

N − N

2

1 − E n (B) N µ B B

= N ln N − N 2

1 − E N µ B B

ln

N

1

2 − E 2N µ B B

− N 2

1 + E N µ B B

ln

N

1 2 + E

2µ B B

= N ln N − N 2

1 − E

N µ B B ln N + ln 1

2 − E 2N µ B B

− N 2

1 + E

N µ B B ln N + ln 1

2 + E 2µ B B

= N ln N − N

2 − E 2µ B B

ln N − N 2

1 − E N µ B B

ln

1

2 − E 2N µ B B

+ − N

2 + E 2µ B B

ln N − N 2

1 + E N µ B B

ln

1 2 + E

2µ B B

= − N 2

1 − E N µ B B

ln

1

2 − E 2N µ B B

− N 2

1 + E N µ B B

ln

1 2 + E

2µ B B

+

N − N 2 + E

2µ B B − N 2 − E

2µ B B

ln N

= − N 2

1 − E N µ B B

ln

1

2 − E 2N µ B B

− N 2

1 + E N µ B B

ln

1 2 + E

2µ B B

4.2 (Mixture of ideal gases)

Mixture ofidealgaseswith

N i

^{particlesofthekind}

i

⁽

i = 1, . . . , m

^)inavessel

ofvolume

V

^.

Denitionofentropy:

S = − k B ln W (E)

with

W (E) = _N!h ^{Ω(E) 3} N

^,forindistingushableparticleswith

c N = _{N !h} ^{1 3} N

^.

Forthetotalsystemwecan write:

Ω (N 1 , . . . , N m ) = N!

Π ^m _i=1 N i ! Π ^m _i=1 Ω ^N _i ⁱ

Insertingthis intodenitionoftheEntropie:

S = − k B ln Ω (N 1 , . . . , N m ) N!h ^3N

= − k B ln

N !

Π ^m _i=1 N i ! Π ^m _i=1 Ω ^N _i ⁱ N !h ^3N

= − k B ln N!

Π ^m _i=1 N i ! + X m

i=1

N i ln Ω i − ln N !h ^3N

!

ApplyingStirlingformula

N !/Π ^m _i=1 N i ! = Π ^m _i=1 ^N _e N

e N i

N i

≈ Π ^m _i=1

N N i

N i

:

S

ensemble

= − k B ln Π ^m _i=1 N

N i

N i

+ X m

i=1

N i ln Ω i − ln N !h ^3N

!

= − k B ln Π ^m _i=1 N

N i

N i

+ X m

i=1

N i ln Ω i

N!h ^3N

!

= − k B

X m

i=1

N i ln N N i − k B

X m

i=1

N i ln Ω i

N !h ^3N

= − k B

X m

i=1

N i ln N N i

+ S

= S

Mix

+ S

Entropyof anidealgas(Sackur-Tetrode-equation):

S (E, V, N ) = k B N

ln V

N

+ 3 2 ln

4πmE 3N h ²

+ 5

2

leadsto:

S (E, V, N 1 , N 2 , . . . N m ) = X m

i=1

S i + S

Mix

==

X m

i=1

k B N i

ln

V N i

+ 3

2 ln

4πm i E i

3N i h ²

+ 5 2

− k B

X m

i=1

N i ln N N i

= X m

i=1

k B N i

ln V

N i − ln N N i

+ 3 2 ln

4πm i E i

3N i h ²

+ 5 2

= X m

i=1

k B N i

ln

V N i

N i

N

+ 3 2 ln

4πm i E i

3N i h ²

+ 5 2

= X m

i=1

k B N i

ln V

N + 3 2 ln

4πm i E i

3N i h ²

+ 5 2

= k B N ln V N +

X m

i=1

k B N i

3 2 ln

4πm i E i

3N i h ²

+ 5 2 k B N

Denitionofpressure:

p = T ∂S

∂V _E,N

inserting

S

^:

p = k B T 1 V N = 1

β N V = ρ

β ⇔ pV = N k B T

4.3 (Ideal gas in a centrifuge)

Radius

= R

^,height

= L

^{,centrifugefrequency}

= ω

^{andmassofparticle}

= m

^{. The}

z

^{-axisisparallelto theaxisofthe}centrifuge. OneparticleHamiltonian:

H = p ²

2m − ω (xp y − yp x )

(a)

Toshow:

Z 1 = 2πm

h ² β 3/2

2πL mβω ²

e ^{mβω 2 R 2 /2} − 1

partitionfunctionforoneparticleofthegas.

Denitionofpartitionfunction:

Z N = c N

Z

d ^3N q d ^3N p e ^{−βH(~ q N ,~ p N )}

with

c N = _{N !h} ¹ 3 N

^{foridentical particles.}

InsertingHamiltonian:

Z 1 = 1 1!h ³

Z

d ³ q d ³ p e ^{−β p}

2

2m +βω(xp y −yp x )

Integratingovermomentum:

Z 1 = 1 1!h ³

Z d ³ q

Z ∞

−∞

dp x e ^{− 2 β m p 2 x − βωyp x} Z ∞

−∞

dp y e ^{− 2 β m p 2 y +βωxp y} Z ∞

−∞

dp z e ^{− 2 β m p 2 z}

with:

Z ^∞

−∞

dp z e ^{− 2 β m p 2 z} =

s Z ^∞

−∞

dx Z ^∞

−∞

dy exp

− β

2m (x ² + y ² )

=

s Z ^∞

0

dr Z 2π

0

dϕ r exp

− β 2m r ²

= s

2m β π

Z ^∞

0

du exp ( − u)

=

r 2mπ

β

Z ^∞

−∞

dp x e ^{− 2 β m p 2 x − βωyp x} = Z ^∞

−∞

dp x e ⁻ ( 2m ^β p ² _x +βωyp x + ^mβ ₂ ω ² y ² ) ^{+ mβ} 2 ω ² y ²

= e ^{mβ 2 ω 2 y 2} · Z ^∞

−∞

dp x e ⁻

√ β

2 m p x + √ mβ 2 ωy 2

=

r 2m

β e ^{mβ 2 ω 2 y 2} · Z ^∞

−∞

du e ^{− u 2}

=

r 2m

β e ^{mβ 2 ω 2 y 2} · √ π

=

r 2mπ

β e ^{mβ 2 ω 2 y 2} Z ^∞

−∞

dp y e ^{− 2 β m p 2 y +βωxp y} = Z ^∞

−∞

dp y e ⁻ ( 2m ^β p ² _y − βωxp y + ^mβ ₂ ω ² x ² ) ^{+ mβ} 2 ω ² x ²

= e ^{mβ 2 ω 2 x 2} · Z ^∞

−∞

dp y e ⁻

√ β

2 m p y − √ mβ 2 ωx ²

= . . .

^equivalentto

p x . . .

=

r 2mπ

β e ^{mβ 2 ω 2 x 2}

Weget:

Z 1 = 1 h ³

r 2mπ β

³ Z

d ³ q e ^{mβ 2 ω 2 y 2} e ^{mβ 2 ω 2 x 2}

Usingcylindricalcoordinates:

Z 1 =

2mπ h ² β

3/2 Z R

0

dr Z 2π

0

dϕ Z L

0

dz r e ^{mβ 2 ω 2 r 2}

=

2mπ h ² β

3/2

2πL

Z mβω ² R ² /2 0

du

2mβω ² r/2 r e ^u

=

2mπ h ² β

3/2

2πL

mβω ² · [e ^u ] ^mβω

2 R ² /2 0

=

2mπ h ² β

3/2

2πL mβω ² ·

e ^{mβω 2 R 2 /2} − 1

(b)

Using

N

^{identicalparticlesweget:}

Z N = 1 N!h ^3N

Z

d ^3N q d ^3N p e ^{− β}

P N i =1

p 2 i

2 m − ω(x i p y,i − y i p x,i )

Z N = 1 N!h ^3N

Z d ^3N q

· Z ∞

−∞

d ^N p x e ^−β

P N i=1

_p 2 x,i 2 m +ωy i p x,i

· Z ^∞

−∞

d ^N p y e ^{− β}

P N i =1

_p 2 y,i

2 m − ωx i p y,i

· Z ^∞

−∞

d ^N p z e ^{−β P N i=1}

p 2 z,i 2 m

with:

Z ^∞

−∞

dp z e ^{− 2 β m P N i=1 p 2 z,i} = Z ^∞

−∞

dp z e ^{− 2 β m p 2 z} N

=

r 2mπ β

^N

Z ^∞

−∞

d ^N p x e ^{− β}

P N i =1

_p 2 x,i 2 m +ωy i p x,i

= Z ^∞

−∞

dp x e ^{− 2 β m p 2 x − βωyp x} N

=

r 2mπ β

^N

e ^{N mβ 2 ω 2 y 2}

Z ^∞

−∞

d ^N p y e ^{− β}

P N i =1

_p 2 y,i 2 m − ωx i p y,i

=

r 2mπ β

^N

e ^{N mβ 2 ω 2 x 2}

leadsto:

Z N = 1 N !h ^3N

r 2mπ β

^3N Z

d ^3N q e ^{N mβ 2 ω 2 y 2} e ^{N mβ 2 ω 2 x 2}

= 1

N !h ^3N

r 2mπ β

^3N Z

d ³ q e ^{mβ 2 ω 2} ( ^{x 2 +y 2} ) N

Usingresultfrom (a):

Z N = 1 N!

2mπ h ² β

³ 2 ^N 2πL mβω ²

N

·

e ^{mβω 2 R 2 /2} − 1 N

(c)

Denition offreeenergy:

F (T, V, N) = − 1

β ln Z N (T, V )

inserting

Z N

^from(b):

F (T, V, N ) = − 1 β ln

"

1 N !

2mπ h ² β

^{3 N} 2 2πL mβω ²

N

·

e ^{mβω 2 R 2 /2} − 1 N #

= − 1 β

− ln N ! + 3 2 N · ln

2mπ h ² β

+ N ln

2πL mβω ²

+ N ln

e ^{mβω 2 R 2 /2} − 1

(d)

Denition:

p = − ∂F

∂V

Volumeofacylinder:

V = πR ² L

Sincethereisnodirect

V

^{dependance in}

F

^{andpressureisgoingtovaryin}

dependanceof

R

^and

L

^{wecalculatetwodierentpressures}

p R

^and

p L

^{. Therefore}

oneofthemstaysin asavariable.

p R = − ∂F

∂V

L = − ∂F

∂R

∂R

∂V

L = − 1 2 √

πV L

∂F

∂R

L = − 1 2πRL

∂F

∂R L

inserting

F

^from(c)

L = const.

^:

p R = − 1

2πRL

∂

∂R

− 1 β

− ln N! + 3 2 N · ln

2mπ h ² β

+ N ln

2πL mβω ²

+ N ln

e ^{mβω 2 R 2 /2} − 1

= 1

2βπRL

N ∂

∂R ln h

e ^{mβω 2 R 2 /2} − 1 i

= N

2βπRL

e ^{mβω 2 R 2 /2}

e ^{mβω 2 R 2 /2} − 1 · mβω ² 2 · 2R

!

= N

2πL

mω ² 1 − e ^{− mβω 2 R 2 /2}

Thisconvergesto

p R V = N k B T

^for

ω → 0

^{,thiscanbeseenusing}L'Hospital.

p L

^:

p R = − ∂F

∂V

R = − ∂F

∂L

∂L

∂V

R = − 1 πR ²

∂F

∂L R

with

F

^,

R = const.

^:

p L = − 1 πR ²

∂

∂L

− 1 β − N

β ln L √

m h ³ ω ² N ·

e ^{mβω 2 R 2 /2} − 1

− 5N 2β ln

2π β

= N

βπR ²

∂

∂L

ln L √ m h ³ ω ² N

= N

βπR ² L

= 1 β

N V

Thisistheidealgasrelation

p L V = N k B T

^{,whichseemslikelyforthe}

z

^-axis,

sincethere is noreasonforanyother dependance there (especially thereis no

dependanceof

ω

^).

Totalforceatouter walls:

F = p · A

with

F R = p R A R

= N

2πL

mω ² 1 − e ^{−mβω 2 R 2 /2}

πR ² F R = N

2L

mω ² R ² 1 − e ^{− mβω 2 R 2 /2}

and

F L = p L A L

= 1

β N

πR ² L 2πRL F L = 2N

Rβ

4.4 (Doppler broadening)

Atomswith mass

m

^, temperature

T

^{emit light in}

x

-direction,movingwith

v x

leadstodopplereect:

ν = ν 0

1 + v x

c

Using

v x = ^p _m ^x

^andHamiltonianforfreeparticle

H = P N i=1

p ² _i 2m

^:

ν = ν 0

1 + p x

mc

(a)

Denition:

h A i =

R d ^3N q R

d ^3N p A (~ q N , ~ p N ) e ^{− βH(~ q N ,~ p N )} R d ^3N q R

d ^3N p e ^{− βH(~ q N ,~ p N )}

h ν i = ν 0

R d ^3N q R

d ^3N p 1 + _mc ^{p x}

e ^{− β P N i =1 p}

2 i 2 m

R d ^3N q R

d ^3N p e ^{− β P N i=1 p}

2 i 2m

= ν 0

R d ^3N q R

d ^3N p e ^{− β P N i =1 p}

2 i 2 m

R d ^3N q R

d ^3N p e ^{−β P N i=1 p}

2 i 2 m

+ ν 0

mc

R d ^3N q R

d ^3N p p x e ^{− β P N i =1 p}

2 i 2 m

R d ^3N q R

d ^3N p e ^{−β P N i=1 p}

2 i 2 m

= ν 0

second term vanishes becauseof symmetry reasons(integrating over sym-

metricintervall,but

p x

^{isodd,whileexponent(}

p ² = p ² _x + p ² _y + p ² _z

⁾

p ² _x

^iseven).

(b)

Denition:

∆ν = r D

(ν − h ν i ) ² E

= r D

ν ² − 2ν h ν i + h ν i ² E

= q

h ν ² i − h ν i ² ν ²

calculation:

ν ²

= ν ₀ ²

R d ^3N q R

d ^3N p 1 + _mc ^{p x} 2

e ^{− β P N i =1 p}

2 i 2 m

R d ^3N q R

d ^3N p e ^{−β P N i=1 p}

2 i 2m

= ν ₀ ² + ν ₀ ² (mc) ²

R d ^3N q R

d ^3N p p ² _x e ^{− β P N i =1 p}

2 i 2 m

R d ^3N q R

d ^3N p e ^{− β P N i =1 p}

2 i 2 m

1 + _mc ^{p x} 2

=

1 + 2 _mc ^{p x} + _mc ^{p x} 2

,rsttermleadsto

ν ₀ ²

^{,whilenumeratorand}

denominator integralsarethesame,secondtermcancelsbecauseofsymmetry.

Integralsoflasttermhaveto becalculated:

denominator:

Z d ^3N q

Z

d ^3N p e ^{− β P N i=1 p}

2 i 2m =

Z d ^3N q

Z ^∞

−∞

dp e ^{− β p}

2 2m

3N

=

r 2mπ β

^3N Z d ^3N q

numerator:

Z d ^3N q

Z

d ^3N p p ² _x e ^{− β P N i =1 p}

2 i

2m =

Z d ^3N q

Z ^∞

−∞

dp x p ² _x e ^{− 2m β P N i =1 p 2 x,i} Z ^∞

−∞

dp e ^{− 2m β p 2} 2N

=

r 2mπ β

^2N Z d ^3N q

Z ^∞

−∞

dp x p ² _x e ^{− 2 β m p 2 x} N

solvingonedimensional integral:

Z ^∞

−∞

dp x p ² _x e ^{− 2m β p 2 x} = Z ^∞

−∞

dp x p x · p x e ^{− 2m β p 2 x}

=

− p x · m β e ^{− 2 β m p 2 x}

^∞

−∞

+ m β

Z ^∞

−∞

dp x e ^{− 2 β m p 2 x}

= m

β

r 2mπ β

insertingin numeratorintegral:

Z d ^3N q

Z

d ^3N p p ² _x e ^{− β P N i =1 p}

2 i

2 m =

r 2mπ β

^2N m

β

r 2mπ β

^N Z d ^3N q

=

r 2mπ β

^3N m β

N Z d ^3N q

insertingnumeratoranddenominatorintegral:

ν ²

= ν ₀ ² + ν ₀ ² (mc) ²

hq 2mπ β

i 3N

m β

N R d ^3N q hq 2mπ

β

i 3N R d ^3N q

= ν ₀ ² · 1 + m ^N (mc) ² β ^N

!

Thisleadsto:

∆ν = q

h ν ² i − h ν i ² = s

ν ₀ ² + m ^N

(mc) ² β ^N · ν ₀ ² − ν ₀ ² = m ^{N 2} mcβ ^{N 2} · ν 0

with

N = 1

^{,thisiswhatweneed,sinceweonlyhavealookatoneaverage-}

particle:

∆ν = r k B T

mc ² · ν 0

(c)

UsingMaxwellvelocitydistribution:

w (~v) d ³ v = mβ

2π ³ 2

exp

− βmv ² 2

d ³ v

wegetthevelocitydistributionfor

v x

^:

w (v x ) dv x = mβ

2π ³ 2

exp

− βmv ² _x 2

Z ^∞

−∞

dv y exp − βmv _y ² 2

! Z ^∞

−∞

dv z exp

− βmv _z ² 2

= mβ

2π ³ 2

exp

− βmv ² _x 2

Z ^∞

−∞

dv exp

− βm 2 v ²

2

= mβ

2π ³ 2

exp

− βmv ² _x 2

r 2π βm

²

= mβ

2π ¹ 2

exp

− βmv ² _x 2

whilewegotthedirect correlation

I (ν) dν = w (v x ) dv x

^with

v x = c ν 0

(ν − ν 0 )

^and

dv x = c ν 0

dν

weget:

I (ν ) dν = 1

√ π

mc ² β 2ν ₀ ²

¹ 2

exp − βmc ² (ν − ν 0 ) ² 2ν ² ₀

! dν

Forthenonrelativisticcase

k B T mc ²

^,with

β = _k ¹

B T

^{,wehavetoconsider,}

that

βmc ² → ∞

^{. Whilethesquareroottermwilldivergethe}exponentialterm willconvergeto zero. Substituting

x = q

mc ² β 2ν 0 ²

weget:

I (ν ) = x

√ π exp

− x ² (ν − ν 0 ) ²

whichisarepresentationof thedelta functionfor

x → ∞

^:

x lim →∞ I (ν) = δ (ν − ν 0 )

This means, we only get an intensity at

ν 0 = ν

^{, while}

ν 0

^{is the emitted}

frequencyoftheatoms. Thismeans,thevelocityoftheatomsdoesn'thaveany

eect andthedopplereect doesn'toccur. Meaning

v x c

^:

ν = ν 0



 

 1 + v x

|{z} c

→ 0



 

 ≈ ν 0