Le ture by Professor Tigran T hrakian
Statisti al Me hani s
Tobias Hofbaur
Mar h 26, 2007
Contents 1 Introdu tion and Boltzmann's hypothesis 1.1 Thermodynami probability . . . . . . 1.2 Criteria for equilibrium . . . . . . . . . 1.3 Correlation to Thermodynami s . . . . 1.4 Example: ideal gas equation . . . . . .
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3 3 3 5 6
2 Phase Spa e, Ensemble and Liouville's Theorem 2.1 Phase Spa e and Ensemble . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Liouville's Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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3 Classi al Counting 3.1 Stirling Approximation of N ! for large N . . 3.2 Boltzmann Counting . . . . . . . . . . . . . 3.3 Boltzmann distribution fun tion . . . . . . . 3.3.1 Correlation to known quantities . . . 3.3.2 Cal ulate CV . . . . . . . . . . . . . 3.3.3 Cal ulate average Energy . . . . . . 3.4 Appli ations . . . . . . . . . . . . . . . . . . 3.4.1 Equation of state (gas equation) . . 3.4.2 Equipartition of energy . . . . . . . 3.4.3 Intera tion of Radiation with Matter 3.5 Degenerate levels (states) . . . . . . . . . .
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9 9 11 13 14 16 16 17 17 18 20 22
4 Quantum me hani al ounting 4.1 Pauli ex lusion prin iple . . 4.2 Boltzmann distribution . . . 4.3 Fermi-Dira distribution . . 4.4 Bose-Einstein distribution . 4.5 Summarize FD and BE . . . 4.6 Sign of µ . . . . . . . . . . . 4.7 Radiation Laws . . . . . . .
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1 Introdu tion and Boltzmann's hypothesis Statisti al Me hani s is a mi ros opi theory of Thermodynami s, i.e. Me hani s ( lassi al and qm) of very large (statisti al) number of parti les. 1.1 Thermodynami probability
The state of a system is des ribed by the thermodynami probability Ω i.e. the total number of possible on gurations (mi rostates) the parti les in the system an have to a hieve a given outer state (ma rostate). The more freedom (possibilities) the parti les have, the liklier the state. Equilibrium ist the most probable state i.e. (1.1)
Ωequil = Ωmax
In thermodynami Systems: Ω = Ω(N, E, V ) 1.2 Criteria for equilibrium
Take two systems A1 and A2 with states Ω1 (N1 , E1 , V1 ) and Ω2 (N2 , E2 , V2 ). A1 and A2 intera t and a hieve equilibrium. The ompound system has state Ω(0) = Ω1 Ω2 Energy ex hange
N1 , N2 , V1 , V2 are onstant
E (0) = E1 + E2 = const
As we have equilibrium δΩ(0) = 0. As the state only depends on the energy E1 (or E2 as they are dependent), we get δΩ(0) = ∂Ω(0) ∂E1 δE1 = 0 ∂Ω(0) |¯ ¯ = ∂E1 E1 ,E2
∂Ω1 Ω2 ∂E1
+ ¯ 1 ,E ¯2 E
∂Ω2 Ω1 ∂E1
=0 ¯ 1 ,E ¯2 E
where E¯1 , E¯2 are the energies of A1 and A2 in equilibrium. ∂Ω1 ¯1 ) ∂Ω2 | ¯ = 0 |E¯1 Ω2 (E¯2 ) + Ω1 (E ∂E1 ∂E1 E2
3
We an repla e the partial derivative for E1 as
∂ ∂E1
=
∂E2 ∂ ∂E1 ∂E2
∂ = − ∂E 2
∂Ω1 ¯2 ) − Ω1 (E ¯1 ) ∂Ω2 | ¯ = 0 | ¯ Ω2 (E ∂E1 E1 ∂E2 E2 ∂Ω1 ∂Ω2 1 1 ¯1 ) ∂E1 |E¯1 − Ω2 (E ¯2 ) ∂E2 |E¯2 = 0 Ω1 (E
or
β1 ≡
∂ ∂ ln Ω1 |E¯1 = ln Ω2 |E¯2 ≡ β2 ∂E1 ∂E2
(1.2)
i.e. in equilibrium β1 = β2 ∂ ln Ω is equilibrium parameter (must be related to T by 0th law) β = ∂E Boltzmann's hypothesis
Take a look a the entropy: ∆S (0) = ∆S1 + ∆S2 ∂S2 ∂S1 ∆E1 + ∆E2 ∂E1 ∂E2 ∂S2 ∂S1 (−∆E) + (∆E) = ∂E1 ∂E2 Assuming that heat is owing from A1 to A2 . The se ond Law tells us ∆S > 0, i.e. ∂S1 ∂S2 ∆E − + >0 ∂E1 ∂E2 ∆S (0) =
∂S1 ∂S2 > ∂E2 ∂E1
Re all the Maxwell relation:
∂U ∂S
= T , therefore 1 1 > T2 T1 T1 > T2
Combine
∂ ∂E
ln Ω = β and
∂E ∂S
=T ∂ ∂E ∂ ln Ω ln Ω = βT = ∂E ∂S ∂S
A
ording to Boltzmann this is onstant
∂S 1 = ≡k ∂ ln Ω βT
(1.3)
with a universal onstant k (later Boltzmann's onstant). Integrating yields S + S0 = k ln Ω
The hoi e of S0 = 0 (suggested by Plan k) leads to (1.4) If all "parti les" are on ned to one state then Ω = 1 (impossible), S = k ln 1 = 0. Impossibility of attaining T = 0 from 3rd Law. S = k ln Ω
4
Ex hange of Volume
E1 , E2 , N1 , N2 are onstant. Similar al ulus leads to η≡
∂ ln Ω ∂V
(1.5)
ζ≡
∂ ln Ω ∂N
(1.6)
is xed for A1 , A2 in equilibrium Ex hange of Parti les
E1 , E2 , V1 , V2 are onstant
is xed in equilibrium
1.3 Correlation to Thermodynami s
The quantities β=
∂ |N,V ln Ω ∂E
η=
∂ |E,N ln Ω ∂V
ζ=
∂ |E,V ln Ω ∂N
must be related to thermodynami oordinates (i.e. same values for equilibrium). To see this, we ompare the dierentials for ln Ω. On the one hand we have ∂ ln Ω ∂ ln Ω ∂ ln Ω dE + dV + dN ∂E ∂V ∂N = βdE + ηdV + ζdV
d ln Ω =
On the other hand we have d ln Ω =
1 dS k
The rst Law was T dS = dE + P dV − µdN , so we get P µ 1 dE + − kT kT kT d ln Ω = βdE + ηdV + ζdV d ln Ω =
Therefore β=β
η=
P kT
and in equilibrium: T1 = T2 , P1 = P2 , µ1 = µ2
5
ζ=
−µ kT
(1.7)
1.4 Example: ideal gas equation
As an appli ation we an derive the ideal gas equation 1. The gas onsist of non-intera ting, non-overlapping "parti les" 2. The probability of any parti le to o
upy a ertain state is the same for all parti les The possibilities for one parti le are proportional to V . The possible states for N parti les are proportional to V N , i.e. Ω ∝ V N = cV N η=
as η =
P kT
∂ N ∂ ln Ω = (ln c + ln V N ) = ∂V ∂V V
we get N P = V kT P V = N kT
6
2 Phase Spa e, Ensemble and Liouville's Theorem 2.1 Phase Spa e and Ensemble
Usual on guratoin spa e: (x, y, z) = ~q Usual momentum spa e: (px , py , pz ) = p~ For a N "parti le" system the phase spa e is (~qi , p~i ) with i = 1 . . . 3N Ensemble: olle tion of all states of a system over all time, i.e. a mental opy of all states Time average of all states ≡ Ensemble average Density fun tion ρ(~qi , p~i , t), ρ is the density of points in phase-spa e. The volumeintegral is a norm (not normalized) Z
ρ(~ qi , p~i , t)d3N ~qd3N p~ = norm ≥ 0
The ensemble average of a physi al quantity f is the 1st momentum of f with respe t to the density ρ R hf i =
f (~qi , p~i )ρ(~qi , p~i , t)dτ R ρ(~qi , p~i , t)dτ
with the volume-element in phase-spa e dτ Equilibrium: ∂ρ =0 stationary ensemble ∂t
(2.1)
2.2 Liouville's Theorem
Criterion for equilibrium: ∂ρ ∂t = 0 must be satis ed ∂ρ Continuity equation ∂t + ∇J~ = 0. Here we have ρ(~qi , p~i , t). The velo ity (time-rate hange) of "points" in phase-spa e is ~v = (~qi , p~i ). The urrent at a point in phase-spa e is J~ = ρ~v . So we get the ontinuity equation. ∂ρ + ∇(ρ~v ) = 0 ∂t
7
(2.2)
∇=
∂ ∂ ∂~ qi , ∂~ pi
, therefore N X ∂ ∂ ˙ ˙ ∇(ρ~v ) = (ρ~q) (ρ~p) i + i ∂~ qi ∂~ pi i=1 ! !# " X ∂ ~q˙i ∂ ~p˙ i ∂ρ ˙ ∂ρ ˙ ~q + ρ + ~p + ρ = ∂~qi i ∂~qi ∂~ pi i ∂~ pi i
Now Hamilton's equation of motion are
∂H ~q˙ i = ∂~ pi
∂H ~p˙ i = − ∂~qi
using this we get X ∂ρ ∂H ∂ ∂H ∂ρ ∂H ∂ ∂H ∇(ρ~v ) = − − +ρ ∂~ qi ∂~ pi ∂~ pi ∂~qi ∂~qi ∂~ pi ∂~ pi ∂~qi i X ∂ρ ∂H ∂ρ ∂H − = ∂~ qi ∂~ pi ∂~ pi ∂~qi i
= {ρ, H}P.B.
with the Poisson-Bra kets. Therefore the ontinuity equation gives Liouville's Theorem ∂ρ = − {ρ, H}P.B ∂t
(2.3)
1.
∂ρ ∂t
= 0 if ρ is independent of (~qi , p~i ) i.e. onstant energy, Mi ro anoni al ensemble
2.
∂ρ ∂t
= 0 if ρ = ρ([H]),Canoni al ensemble
8
0.2 3 Classi al Counting 3.1 Stirling Approximation of
N!
for large
N
0.1
Ω = number of mi rostates. e.g. N parti les (identi al), total number of arrangements ! N !. if Ni parti les have energy Ei then Ni ! arrangements are identi al. then Ω = ΠN i Ni ! When N >> 1 then N ! an be expressed analyti ally: Integral representation of n!: Z ∞ Z ∞ f (x, n)dx xn e−x dx ≡ n! = 0
0
0 0
Figure 3.1: Plot of f (x, 2) f (x, n) = xn e−x f ′ (x, n) = nxn−1 e−x − xn e−x = xn e−x nx − 1
f ′ (n, n) = 0
f ′′ (n, n) = −nn−1 e−n < 0
9
2
f is peaked around x = n, i.e. n! re eives most ontribution from small range of x around x = n. Approximate: Z n! ≈
∞
fapprox (x, n)dx
0
fapprox (x, n) = f (n, n) + (x − n)f ′ (n, n) +
(x − n)2 ′′ f (n, n) + . . . 2!
(x − n)2 (−nn−1 e−n ) + . . . = nn e−n + 0 + 2! (x − n)2 n −n =n e 1− + ... 2n ! 2 (x − n)2 1 (x − n)2 n −n 1− ≈n e + + ... − 2n 2! 2n n −n −
fapprox (x, n) = n e
e
“
x−n √ 2n
”2
Therefore n −n
n! ≈ n e
Z
∞
0
√ = nn e−n 2n
−
e Z
“
x−n √ 2n
∞
| −n
√ n! ≈ nn e−n 2nπ
”2
dx
2
e−y dy {z }
√ ∝ π
(3.1)
In the Boltzmann-Hypothesis S = k ln Ω so we need to approximate ln n!: √ ln n! = ln nn e−n 2nπ 1 1 = n ln n − n ln e + ln n + ln 2π 2 2 1 ≈ n+ ln n − n 2 ln n! ≈ n ln n − n
(3.2)
Hooke's Law: F = −kx Model: N links (ea h of unit length), length L of the hain, N ≫ L, i.e. lot of reases.
Example.
N +L 2 N −L # of forward links = 2
# of ba kward links =
10
As Ω is the number of mi rostates possible to a hieve a given outward state, we have Total number of arrangements of Links Ω= (# of arr. of forward l.)(# of arr. of ba kward l.) Ω=
N +L 2
So we get:
N! N −L ! 2 !
N −L S =k ln Ω = k ln N ! − ln N +L 2 ! − ln 2 ! ln N +L + N +L − N −L ln N −L + N −L =k N ln N − N − N +L 2 2 2 2 2 2 S =k N ln N − N +L ln N +L − N −L ln N −L 2 2 2 2 ln N + N −L ln N − N +L ln N +L − N −L ln N −L =k N +L 2 2 2 2 2 2 =k N +L ln N − ln N +L + N −L ln N − ln N −L 2 2 2 2 +L −L + (N − L) ln N2N = − k2 (N + L) ln N2N
De ne κ =
N +L 2N ,
⇒
N −L 2N
=1−κ
S = −kN {κ ln κ + (1 − κ) ln(1 − κ)}
∂F In 1 dimension, For e = Pressure. The Maxwell-relation was P = − ∂V with F = U − T S = −T S as U = const sin e mi ro anoni al ensemble
P =
∂ (kN T {κ ln κ + (1 − κ) ln(1 − κ)}) ∂L
= kN T {ln κ + 1 + (−1) ln(1 − κ) + (−1)} kT {ln κ − ln(1 − κ)} 2 kT N +L = ln 2 N −L ! L 1+ N kT P = ln L 2 1− N =
As
L N
≪ 1 expand around
L N
1+x = 0: ln 1−x = 2x + 23 x3 + . . .
P =
kT L N
3.2 Boltzmann Counting
Phase spa e: (i)
11
1 2N
T
=−
∂F ∂L T
phaseP ell, labeledPby Ei , with Ni parti les in it. The total number of parti les is N , N = i Ni , E = i Ni Ei No. of arrangements of N parti les No. of mi rostates = No. of indistinguishable arrangements Ω= Q
N! i (Ni !)
(3.3)
! S = k ln Q N i (Ni !) "
# Y = k ln N ! − ln( (Ni !) "
= k ln N ! −
i
X
#
ln Ni !
i
But N ≫ 1 and Ni ≫ 1, therefore approximation with Sterling "
X
S = k N ln N − N −
i
#
(Ni ln Ni − Ni )
h i X = k N ln N − (Ni ln Ni )
It is onvenient to introdu e ωi =
Ni N,
i.e.
h
P
wi = 1
i S = k N ln N − (N ωi ln(N ωi ) h i X = kN ln N − (ωi ln N + ωi ln ωi ) i h X X = kN ln N − ln N ωi − ωi ln ωi X S = −kN ωi ln ωi X
For equilibrium we have Seq = Smax , i.e. δS = 0 with respe t to variation δωi δS =
X ∂S δωi ∂ωi i
δωi are arbitrary variations of the fun tion ω(Ei ) subje t to the onstraints: P 1. i Ei ωi = const P 2. i ωi = 1
i.e. set
δS + λ1 δE + λ2 δN = 0
i.e. vary S = S + λ1
X
X Ni Ei +λ2 Ni | {z } | {z } S1
12
S2
(3.4)
•
∂S ∂ωi
∂ = −kN ∂ω i
P
ω ln ω j = −kN (ln ωi + 1) j j δS =
X ∂S X δωi = −kN (ln ωi + 1)δωi ∂ωi i
•
∂S1 ∂ωi
=
∂ ∂ωi N
P
j
i
Ej ωj = N Ei δS1 = N
X
Ei δωi
i
•
∂S2 ∂ωi
=
∂ ∂ωi N
P
j
ωj = N δS2 = N
δS = −kN =N
X
(ln ωi + 1)δωi + λ1 N
i
X
X
δωi
X
Ei δωi + λ2 N
i
[−k(ln ωi + 1) + λ1 Ei + λ2 ] δωi
X
δωi
i
for arbitrary δωi , therefore i.e
−k(ln ωi + 1) + λ1 Ei + λ2 = 0
∀i
ln ωi = −α + βEi
(3.5)
ωi = e−α+βEi P P P We had 1 = i ωi i.e. 1 = i ωi = e−α i e−βEi i.e. X eα = e−βEi ≥ 0 i
(3.6)
eα = Z(β)
Z(β) is alled "partition fun tion" 3.3 Boltzmann distribution fun tion
The Boltzmann distribution fun tion is then given by fB (Ei ) =
e−βEi Z
Z=
X
e−βEi
(3.7)
e−βE dE
(3.8)
i
or for ontinuous energy spe trum e−βE fB (E) = Z
Z=
13
Z
3.3.1 Correlation to known quantities
The entropy was S = −kN
P
i.e.
i ωi ln ωi
S = −kN
X e−βEi
Z
i
(−βEi − ln Z)
X X β ln Z −βE −βE i i = kN Ei e + e Z Z i i | {z } =Z ! βX Ei e−βEi + ln Z = kN Z i
P
i Ei
e−βEi
=
∂ − ∂β
P
i
e−βEi
=
∂ − ∂β Z
i.e.
S = kN
Total energy E=
X
Ni Ei = N
i
X
−
β ∂Z + ln Z Z ∂β
Ei ωi = N
i
X
Ei
i
e−βEi = Z
N X ∂ −βEi N ∂Z NX Ei e−βEi = − e =− = Z Z ∂β Z ∂β i
i
E=−
together we get
N ∂Z Z ∂β
β ∂ + ln Z = S = kN − Z ∂β 1 ∂Z = k −N β + N ln Z Z ∂β S = k (βE + N ln Z)
(3.9)
Relate β Correlate the new β to known quantities: S applied to Thermodynami s: S = kβU + kN ln Z(β) dS(U, β) =
∂S ∂U dU
+
∂S ∂β dβ .
Cal ulate dS :
1 ∂Z dS = kβdU + kN + kU dβ = kβdU + (kU − kU )dβ = Z ∂β = kβdU
14
(3.10)
The relevant Maxwell Relation is "old" β .
∂U ∂S N,V
=
1 T,
i.e. kβ =
1 T
i.e. the "new" β is our
Average Energy Cal ulate the average energy per parti le, i.e. the rst moment of Ei with respe t to
fB (Ei ):
hU i =
X
U = N
Ei
e−βEi 1 ∂Z 1 X ∂ −βEi 1 ∂ X −βEi e =− =− e =− = Z Z ∂β Z ∂β Z ∂β i
i
∂ Re ne notation hU i = − ∂β ln Z Now readjust notation: ZN = Z1N where Z1 is the partition fun tion per "parti le"
∂ ∂ ∂ ln Z1 = − ln Z1N = − ln ZN ∂β ∂β ∂β X X ∂ ∂ U= Ni Ui = N ωi Ui = N − ln Z1 = − ln ZN ∂β ∂β U = N hU i = −N
i
Thermodynami s:
S = kβU + k ln ZN
Relate α Helmholtz: F = U − T S = U − T
1 TU
+ k ln ZN = − β1 ln ZN
F =−
1 ln ZN β
Maxwell relations (involving the potential F ) S=−
∂F ∂T
= N,V
= k ln ZN + kT
∂ (kT ln ZN ) = ∂T
∂ 1 ∂ ln ZN = k ln ZN − ln ZN = ∂T T ∂β
= kβU + k ln ZN nothing new ∂ 1 ∂F = ln ZN useful P =− ∂V N,T β ∂V N,T ∂F 1 α ∂ 1 1 µ= =− ln ZN = − ln Z1 = − ln eα = − ∂N V,T β ∂N V,T β β β α is a hemi al potential.
15
3.3.2 Cal ulate
CV =
∂U ∂T
CV
V
∂β ∂ =− ∂T ∂β
∂ ln ZN ∂β
CV = kβ 2
=
2 1 ∂2 2 ∂ ln Z = kβ ln ZN N kT 2 ∂β 2 ∂β 2
∂2 ln ZN ∂β 2
(3.11)
3.3.3 Cal ulate average Energy
In the Boltzmann distribution, E ranges from 0 to ∞. Consider a system at energy E and al ulate the derivation ∆E from E (at any observation). Estimate ∆E hEi . If this is very small, the mi ro anoni al ensemble is equivalent to the anoni al ensemble. Z e−βE dE Ef (E)dE = E Z Z Z 1 1 ∂ = Ee−βE dE = − e−βE dE Z Z ∂β 1 ∂Z hEi = − Z ∂β hEi =
Z
E
D (∆E)2 = (E − hEi)2 = E 2 − 2E hEi + hEi2 = Z Z Z 2 2 = E f (E)dE − 2 hEi Ef (E)dE + hEi f (E)dE = Z = E 2 f (E)dE − 2 hEi2 + hEi2 = Z 1 = E 2 e−βE dE − hEi2 = Z Z 1 ∂2 e−βE dE − hEi2 = = Z ∂β 2 1 ∂2Z = − hEi2 = Z ∂β 2 1 ∂2Z 1 ∂Z 2 = − = Z ∂β 2 Z ∂β 1 ∂Z 2 1 ∂Z ∂ 1 ∂Z ∂ − = − = ∂β Z ∂β ∂β Z ∂β Z ∂β ∂2 1 ∂Z 2 1 ∂Z 2 = ln Z + 2 − = ∂β 2 Z ∂β Z ∂β ∂2 ln Z = ∂β 2
16
(3.12)
∂ Re all from (3.11) CV = kβ 2 ∂β 2 ln Z ,i.e. 2
(∆E)2 =
i.e.
∆E = hEi
p
CV kβ 2
(3.13)
kT 2 CV hEi
3.4 Appli ations 3.4.1 Equation of state (gas equation)
In anoni al ensemble (Boltzmann distribution): P =−
∂F ∂V
= kT N,T
∂ ∂V
ln ZN N,T
1. Ideal gas ( re all Ω ∝ V N ). Energy spe trum is ontinuous, f (E) = i.e. V (x) = 0, no for es. Z
Z
e−βE dqdp = Z Z p2 p2 −β 2m dqdp = V e−β 2m dp = = e r 2mπ =V β
Z1 =
−βE
e
dτ =
This yields a pressure of P = kT
∂ ∂V
ln ZN = kT N,T
∂ ∂V
r 2mπ ∂ ln V = = N kT ∂V β r ∂ 2mπ = N kT ln V + ln = ∂V β N kT = V
2. non-ideal Gas, e.g. VdW Z1 =
Z
17
e−βE dτ
N ln Z1 = N,T
e−βE Z ,
E=
|~ p|2 2m
E=
p2 2m
+ W (q) Z1 =
Z
e
P =−
∂F ∂V
Z
2
p −β 2m
dp e−βW (q) dq r Z 2mπ = e−βW (q) dq β r Z Z Z 2mπ −βW (q) dq − dq + e dq = β r Z 2mπ −βW (q) = V + e − 1dq β r Z 2mπ 1 = V 1+ e−βW (q) − 1dq β V
= N kT N,T
∂ ∂V
ln Z1 = N,T
r c 2mπ ∂ + ln V + ln 1 + ln = N kT ∂V β V 1 ∂ c = N kT + ln(1 + V ∂V V 1 c c2 = N kT − 2 + 3 + ... V V V 3.4.2 Equipartition of energy
• For a free parti le with mass m in 1 dimension we had Z1 =
energy is
∂ ln Z1 ∂β √ 1 ∂ =− ln 2mπV − ln β ∂β 2 1 1 = = kT 2β 2
=−
• For a 3 dimensional ideal gas the exa t energy is E = Z Z1 = e−βE d3 qd3 p =
2mπ β
18
3/2
V
|~ p|2 2m .
q
2mπ β V
. The average
Therefore the average energy is ∂ ∂ =− ln Z1 = − ∂β ∂β 3 31 = = kT 2β 2
3 3 ln V + ln 2mπ − ln β 2 2
Equipartition: 12 kT per degree of freedom. • Gas of SHO: E =
p2 2m
+ 21 mω 2 q 2 Z1 =
Z
Z
−β
e
„
p2 + 12 mω 2 q 2 2m
p2
Z
«
dpdq
β
e−β 2m dp e− 2 mω r r 2mπ 2π = β βmω 2 2π Z1 = ωβ =
∂ ln Z1 = kT = < E >= − ∂β
2 q2
dq
1 1 + 2 2
kT
Heating bulk system by radiation ≡ system of os illators (ω being the frequen y of the radiation. CV = N
∂ ∂T
< E >= V
∂ N kT = N k = const ∂T
but this is not experimentally observed. p • Relativisti ideal gas: (only kineti energy), E = c m2 c24 + |~ p|2 Z Z √ 2 2 2 −βE 3 3 Z1 = e d pd q = V e−βc m c +p p2 dp sin θp dθp dφp = Z ∞ √ 22 2 p2 e−βc m c +p dp = V 4π 0
19
Now take massless parti les: m = 0 Z
p2 e−βcp dp = Z 1 d = 4πV − p2 e−βcp dp = βc dp Z 4πV −βcp 0 − 2pe dp = =− βc Z 8πV d −βcp 1 = e dp = p − βc βc dp Z 8πV −βcp =− 2 2 0− e dp = β c 8πV = 3 3 β c
Z1 = 4πV
hEi = −
∂ 3 ∂ ln Z1 = − (−3 ln β) = = 3kT ∂β ∂β β
3.4.3 Intera tion of Radiation with Matter
Radiation ≡ e.m. waves with frequen y ω Matter ≡ atoms, mole ules, os illator model with Hooke's onstants given by ω of the radiation Classi ally: ω has ontinuous spe trum i.e. from SHO Z1 =
Z
−βE 3
3
d pd q =
e
2π ωβ
1 1 ⇒ hEi = kT + kT 2 2
CV =
∂U ∂T
U = N hEi = N kT
not observed
= N k = const
V
Plan k's Hypothesis: Energy has dis rete spe trum (in the ontext of radiation intera ting with mi ros opi matter). Energy is absorbed in dis rete pa kets ("quanta") En = n~ω
So we get Z1 =
X n
e−β~ω =
∞ X
e−β~ω
n=0
20
n
=
1 1 − e−β~ω
i.e. ∂ ∂ ln Z1 = ln(1 − e−β~ω ) = ∂β ∂β ~ω e−β~ω = = 1 − e−β~ω ~ω = β~ω e −1
hEi = −
1 ∂β ∂ ∂ hEi = N ~ω = β~ω ∂T ∂T ∂β e −1 −1 N ~ω β~ω = =− 2 ~ωe 2 β~ω kT (e − 1)
CV = N
= N ~2 ω 2 kβ 2
eβ~ω
2
(eβ~ω − 1)
T dependent
Examine the limits for T ≫ 1 (i.e. β ≪ 1) and T ≪ 1 (i.e. β ≫ 1) • T ≫ 1, β ≪ 1, lassi al limit
hEi =
~ω 1 ~ω =≈ = = kT eβ~ω − 1 β~ω β
As for the lassi al SHO (equipartition) • T ≪ 1, β ≫ 1, qm limit hEi =
~ω ~ω −β~ω ≈ eβ~ω − 1 e
∂β ∂ −β~ω ∂ hEi = N ~ω e = ∂T ∂T ∂β = −N ~ωkβ 2 (−~ω)e−β~ω =
CV = N
CV = N ~2 ω 2 kβ 2 e−β~ω
i.e. CV −T−→0 −→ 0 as 3rd Law and agrees with experimental result Note.
Cal ulate with orre t qm energy: E = (n + 21 )~ω 1 hEi 1 − e−β~ω ~ω ~ω = + β~ω 2 e −1
Z1 = e−β
~ω 2
with the zero point enery
=−
~ω 2
21
~ω ∂ ∂ ln e−β 2 + ln 1 − e−β~ω = ∂β ∂β
Einstein spe i heat
~ω = eβ~ω − 1 1 ∂ = kθ = ∂T e Tθ − 1 θ= ~ω k
∂ CV = ∂T
θ
eT kθ 2 = 2 2 θ T eT − 1
For system with a range of frequen ies sum up all ontribution, given the density fun tion ρ = ρ(θ)
Einstein: ρE (θ) = δD (θ − θE ) CVtot
=
Z
CV (θ, T ), ρE (θ)dθ =
= CV (θE , T ) ( 1 θ ≤ θD Debye: ρD = 0 θ > θD CVtot
=k
θ
Z
θD
Z
xD
0
=k
θ2 eT θ 2 dθ = 2 T eT − 1 x2 ex
(ex − 1)2 Z xD x2 ex
T dx =
0
= kT
dx (ex − 1)2 | {z } xD →∞ π2 −−−−→ 3
0
CVDebye = kT · const 3.5 Degenerate levels (states)
e.g. more than one state with energy E . Boltzmann distribution Z1 =
Z
e−βE dµ(E)
dµ(E) = dE maybe, but not ne essary. Typi ally dµ(E) = g(E)dE where g(E) =no. of states/unit energy range. Re all equipartition: Z1 =
Z
−βE 3
e
3
d pd q = V
Z
2
p −β 2m 3
e
22
d p=
2mπ β
3/2
V
But Z1 =
So
Z
−βE
e
Z1 =
Invert to nd g(E) 1 g(E) = 2πi
Use the result
1 2πi
R s′ +i∞ s′ −i∞
3
dEd q = V
Z
2mπ β
3/2
β ′ +i∞
Z
βE Z1 (β)
e
ds =
e−βE dE =
V =
V
β ′ −i∞
esx sn+1
Z
(
xn n!
0
V β
in orre t
g(E)e−βE dEdq 3
(2mπ)3/2 dβ = 2πi
Z
eβE dβ β 3/2
x>0 . As E > 0 x≤0
g(E) = (2mπ)3/2
E 1/2 1 2 !
Alternative: Count no. of states of energies up to E : M (E). The number of states is proportional to the volume of a ball in state spa e. M (E) ∝
√
with p = 2mE 1/2 we get M (E) ∝
4π 3 p 3
4π (2m)3/2 E 3/2 3
The number of states per unit energy range is
dM ∝ E 1/2 dE
same result Example.
1. Arti ial example: En = n~ω (QM SHO) with degenera y g(En ) = n Z1 =
X
ne−βn~ω =
n
1 ∂ −βn~ω e = ~ω ∂β 1 ∂ X −β~ω n = e =− ~ω ∂β 1 ∂ 1 =− = ~ω ∂β 1 − e−β~ω e−β~ω = 2 (1 − e−β~ω )
=
X
−
23
∂ ∂ ∂ ln Z1 = − (−β~ω) + 2 ln e−β~ω − 1 = ∂β ∂β ∂β 2 (−~ω)e−β~ω = = ~ω + −β~ω e −1 2e−β~ω = = ~ω 1 − −β~ω e −1 2 = ~ω 1 − = 1 − eβ~ω 1 1 = 2~ω + β~ω 2 e −1
hEi = −
In the limit for T ≫ 1, β ≪ 1
1 2 1 + = ~ω + hEi ≈ 2~ω 2 β~ω beta ∂ ∂ hEi = 2N kT = 2N k = const CV = N ∂T ∂T
24
lassi al limit
2. SHO with degenera y g(n) = n2 Z1 =
X
g(n)e−βn~ω =
n
X
X
n2 e−βn~ω =
n
∂2
1
e−βn~ω = 2 ∂β n 1 ∂ 2 X −β~ω n e = = 2 2 2 ~ ω ∂β n
=
~2 ω 2
1 ∂2 1 = 2 2 2 ~ ω ∂β 1 − e−β~ω 1 ∂ e−β~ω = =− ~ω ∂β (1 − e−β~ω )2 =
1 =− ~ω
= = = =
−~ωe−β~ω
2
(1 − e−β~ω )
e−β~ω
+2
−2
e−β~ω
−β~ω
3 (−1)(−~ω)e
(1 − e−β~ω )
!
=
e−β~ω e−β~ω
2 3 = (1 − e−β~ω ) (1 − e−β~ω ) e−β~ω 1 − e−β~ω + 2e−β~ω e−β~ω
= 3 (1 − e−β~ω ) e−β~ω − e−β~ω e−β~ω + 2e−β~ω e−β~ω (1 − e−β~ω )
3
=
e−β~ω + e−β~ω
3 = (1 − e−β~ω ) 1 + e−β~ω = e−β~ω 3 (1 − e−β~ω )
i ∂ h ∂ −β~ω + ln 1 + e−β~ω − 3 ln 1 − e−β~ω = ln Z1 = − ∂β ∂β 1 1 = ~ω − e−β~ω (−~ω) + 3 (−1)e−β~ω (−~ω) = −β~ω 1+e 1 − e−β~ω e−β~ω e−β~ω +3 = = ~ω 1 + 1 + e−β~ω 1 − e−β~ω 1 3 = ~ω 1 + β~ω + β~ω e e −1
hEi = −
25
Classi al Limit: T ≫ 1, β ≪ 1
3 1 + hEi −−−→ ~ω 1 + 2 + β~ω β~ω 1 3 ≈ ~ω 1 + + kT = 2 ~ω 1 kT = 3~ω + 2 ~ω ∂ hEi = 3k = const CV = N ∂T T ≫1
~
2
=
2
3. QM rotator: El = hHi = l(l+1)~ , (H = |L| 2I 2I ). There are 2l + 1 values of m for ea h l i.e. ea h El is (2l + 1) fold degenerate Z1 =
X
g(l)e−βEl =
i
X ~2 (2l + 1)e−βl(l+1) 2I = l
X = (2l + 1)e−θl(+1)β l
a) T ≪ 1, β ≫ 1 extreme QM limit, i.e. we an trun ate to l = 0, 1, 2 Z1 = 1 + 3e−2θβ + 5e−6θβ
hEi = −
e−2θβ + 5e−6θβ ∂ ln Z1 = 6θ ∂β 1 + 3e−2θβ + 5e−6θβ
i.e. hEi is T dependent and CV −T−→0 −→ 0 agrees with 3rd law b) T ≫ 1, β ≪ 1 so we must in lude high l (no trun ation); we an treat l as a
ontinouous variable and repla e the sum by an integration. Z
∞
−l(l+1)θβ
(2l + 1)e dl = 0 1 h −l(l+1)βθ i∞ = e =− θβ 0 1 kT = = θβ θ
Z1 =
hEi = −
Z
0
∞
1 ∂ −l(l+1)θβ e dl = −θβ ∂l
1 ∂ 1 ∂ ln = ln(θβ) = θ= ∂β θβ ∂β θβ
1 = kT β CV = k = const =
26
4 Quantum me hani al ounting 4.1 Pauli ex lusion prin iple
No two Fermions an o
upy the same state. If parti les are identi al then the state fun tion ψ(x1 , x2 ) also des ribes the system with
x1 and x2 inter hanged. i.e.
|ψ(x1 , x2 , . . . , xN )|2 = |ψ(x2 , x1 , . . . , xN )|2
or
ψ(x1 , x2 , . . . , xN ) = ±ψ(x2 , x1 , . . . , xN )
i.e. there are two types of (identi al ) parti les in QM • ψ is symmetri under inter hange, BOSONS, integer spin, e.g. γ, g, π
• ψ is antisymmetri under inter hange, FERMIONS, half-integer spin, e.g. e− , p, n
Then if we put two (identi al) Fermions in the same state
ψ(x1 , x2 , . . . , xN ) = −ψ(x1 , x1 , . . . , xN )1 ψ(x1 , x1 , . . . , xN ) = 0
i.e. no two Fermions an o
upy the same state. Now ounting di tated by QM. 1. No two Fermions in a state, i.e. 0 or 1 parti le per state ( annot use Stirling approximation) 2. Fermions & Bosons annot be labeled (indeterministi ) In phase spa e, a phase ell ontains Ni ≫ 1 parti les of energy Ei . Divide ea h ell into G > N elementary ells, ni = NGi an be small. For the Boltzmann hypothesis, we need the thermodynami probability Ω. De ne: Ωα = thermodynami al probability of ea h elementary phase- ell ˜ i = thermodynami al probability of i-th phase ell Ω Ωα =
q G
˜i Ω
sin e ea h one of the G elementary phase ells is equaly likely to be o
upied by a parti le 0. Re over Boltzmann distribution (Classi al ounting, deterministi ) 1. Fermi-Dira distribution (antisymmetri under inter hange) 2. Bose-Einstein distribution (symmetri under inter hange)
27
4.2 Boltzmann distribution Ni ˜i = G Ω Ni !
, as Ni ≫ 1 approximate with Stirling. ln Ni ! = Ni ln Ni − Ni Ni ! = eNi ln Ni −Ni = NiNi e−Ni
Use Ni = Gni Gni Gni Gni eGni e Ge G ˜i = = = Ω Gn (Gni ) i Gni ni q G ˜ i = n−ni eni ⇒ Ωi = Ω i Y Y −n ⇒Ω= Ωi = ni i eni i
i
And so the entropy is S = k ln Ω = k ln
Y
Ωi = k
i
Previous version S = −k
X
ln Ωi = k
i
P
X X (−ni ln ni + ni ) = −k (ni ln ni − 1) i
i ωi ln ωi
i
with ωi =
Ni N
4.3 Fermi-Dira distribution
˜i = Ω
G! (G − Ni )!Ni !
As G ≫ 1, Ni ≫ 1 approximate with Sterling ˜i = Ω = = = ⇒ Ωi =
e−G GG eNi −G (G − Ni )G−Ni e−Ni NiNi GG
(G − Gni )(G−Gni ) (Gni )Gni GG
i − ni )G−Gni nGn i −ni G ni (1 − ni )ni −1
GG (1 q G
=
=
=
GG (G − Ni )G−Ni NiNi GG
i GG−Gni (1 − ni )G−Gni GGni nGn i
1
(1 −
˜ i = n−ni (1 − ni )ni −1 Ω i
28
=
i ni )G−Gni nGn i
=
=
So the entropy is S = k ln
Y
Ωi = k
i
= −k
X i
X
ln Ωi = k
i
X i
[−ni ln ni + (ni − 1) ln(1 − ni )] =
[ni ln ni + (1 − ni ) ln(1 − ni )]
4.4 Bose-Einstein distribution
In Bose statisti s any number of parti les (A) in given state B . States B and parti les A are not individually labeled. Assign ea h parti le A to the state on their left BAABAAABABAABB...
How many possibilities are there? We have G + Ni "parti les" to distribute, the leftmost parti le has to be a B , so G possibilities for this, the rest an hoose freely, i.e. (G + Ni − 1)! possible arrangements. As the parti les are indistinguishable Ni ! are the same, as the states are indistinguishable, G! are the same. So we get ˜ i = G(G + Ni − 1) = (G + Ni − 1)! Ω G!Ni ! (G − 1)!Ni ! −G−N i (G + N )G+Ni (G + Ni )! e i ≈ = = −G G −N i G!Ni ! e G e NiNi = =
GG GGni (1 + ni )G+Gni (G + Gni )G+Gn i = = i GG (Gni )Gni GG GGni nGn i (1 + ni )G+Gni
=
i nGn i
G = (1 + ni )1+ni ni−ni
⇒ Ωi = (1 + ni )1+ni ni−ni
i.e. the entropy is
S = k ln
Y
Ωi = k
i
=k
X i
= −k
X i
ln (1 + ni )1+ni ni−ni =
[(1 + ni ) ln(1 + ni ) − ni ln ni ] =
X i
[ni ln ni − (1 + ni ) ln(1 + ni )]
4.5 Summarize FD and BE (BE) (−) (±) = Ωi . Ωi = ni−ni (1 ∓ ni )ni ∓1 De ne Ωi(F D) = Ω(+) i , Ωi
S (± = k ln
Y i
(±)
Ωi
=k
X i
[−ni ln ni + (ni ∓ 1) ln(1 ∓ ni )]
29
Use al ulus of variation: δS =
X ∂S δni ∂ni
subje t to the onstraints of xed energy and xed number of parti les. δS + λ1 δE + λ2 δ
X
Ni
i
δE =
P
i
Ei δni , δN =
P
!
=0
i δni
( ) X ∂S ∂ =k [−ni ln ni + (ni ∓ 1) ln(1 ∓ ni )] = ∂ni ∂ni i (ni ∓ 1) = = k − ln ni − 1 + ln(1 ∓ ni ) ∓ 1 ∓ ni 1 ∓ ni ∓(1 ∓ ni ) = k ln −1∓ = ni 1 ∓ ni 1 ∓ ni = = k ln ni X ni δS = −k ln 1 ∓ ni i
i.e.
X
δni
i
ni + λ1 Ei + λ2 = 0 −k ln 1 ∓ ni
δni being arbitrary (subje t to the onstraints imposed) −k ln
i.e.
ni + λ1 Ei + λ2 = 0 1 ∓ ni ln
ni = −(α + βEi ) 1 ∓ ni ni = (1 ∓ ni )e−(α+βEi )
ni (1 ± e−(α+βEi ) ) = e−(α+βEi )
e−(α+βEi ) 1 ± e−(α+βEi ) 1 ni = α+βE i ± 1 e ni =
or with ontinuous spe trum
∀i
f (E) =
1 eα+βE ± 1
30
Evaluate S : S=k
X i
= −k
[−ni ln ni + (ni ∓ 1) ln(1 ∓ ni )] =
X i
[ni (ln ni − ln(1 ∓ ni )) ± ln(1 ∓ ni )] =
X = −k ni ln i
= −k "
X
=k α "
i
ni ± ln(1 ∓ ni ) = 1 ∓ ni
[−(α + βEi )ni ± ln(1 ∓ ni )] =
X
ni + β
i
= k αN + βE ∓
X
Ei ni ∓
X
ln(1 ∓ ni )
i
X i
#
ln(1 ∓ ni ) =
#
If we now pro eeded like before to employ the Maxwell Realtion ∂U ∂S N,V = T ⇒ P ∂S 1 i ln(1 ∓ ni ) is not independent of ∂U N,V = T , we meet an obsta le. The term E , so β annot be identi ed as before. %beginalign* Here, we pro eed with the alternative path of subje ting the Helmholz free energy F = U − T S to the variational prin iple. Note that now T appears expli itly so we do not need the Lagrange multiplier to x the energy. Using only the onstraint xing the number of parti les, we arrive at exa tly the same result, without having to identify β. To identify the orresponding Lagrange ∂F multiplier α, re all the Maxwell Relation ∂N = µ, T,V "
F = U − T S = U − T k αN + βU ∓ = −kT αN ± kT
i.e. µ=
X i
ln(1 ∓ ni )
∂F ∂N
T,V
X i
#
ln(1 ∓ ni ) =
= −kT α
α = −βµ
i.e. the FD/BE distribution is: 1 eβ(E−µ) ± 1 1 f± (E) = β(E−E ) 0 ± 1 e f± (E) =
31
(4.1) (4.2)
E0 = µ = µ(T ). F = N E0 ±
X
1
ln 1 ∓
eβ(E−E0 ) ± 1 X = N E0 ∓ ln 1 ± eβ(E0 −E) i
= N E0 ±
X i
ln
eβ(E−E0 ) eβ(E−E0 ) ± 1
!
=
i
If you don't take the Gth root you get a better result for S and do not have to
heat on E and N . For more realisti onsiderations you an even take Gi being the number of states for ea h energy level. The result is the same Note.
4.6 Sign of
µ
1. Bose-Einstein: fBE = has to be > 0
1 . eβ(E−µ) −1
N=
Suppose we al ulate the number of parti les, it Z
g(E) >0 −1
eβ(E−µ)
i.e. eβ(E − µ) > 1 for arbitrary E i.e. e−βµ > 1
(4.3)
µ < 0 ∀T
See Figure 4.1 for plot. 2. Fermi-Dira fF D =
1 eβ(E−µ) +1
Z
g(E) dE = +1 Z g(E) dE = eβµ βE e + eβµ
N=
If µ < 0 (µ = −a2 )
eβ(E−µ)
Z
g(E) dE eβE + e−βa2 R R Consider now T → 0, i.e. β → ∞. Then βEg(E)−βa2 dE → 2
N = e−βa
e
whi h is onvergent, i.e. whi h is absurd, i.e.
2
+e
g(E) dE eβE
=I <∞
T →0
N = e−βa I −−−→ 0 µ(T = 0) > 0
See Figure 4.2 for plot. For T = 0 the distribution fun tion is the Heaviside step-fun tion (Figure 4.3)
32
(4.4)
10
f(x)
9 8 7
fBE
6 5 4 3 2 1
Boltzmann tail
0 E
Figure 4.1: Bose-Einstein distribution
1
f(x)
0.8
fFD
0.6
0.4
0.2 Boltzmann tail 0
E0 E
Figure 4.2: Fermi-Dira distribution for T > 0
f(x)
1
fFD
0.8 0.6 0.4 0.2 0
E0 E
Figure 4.3: Fermi-Dira distribution for T = 0
33
4.7 Radiation Laws 1 . Photons Photon gas (E = ~ω (QM)) obeys Bose-Einstein statisti s f (E) = eβ(E−E 0 ) −1 are superrelativisti , i.e. their number is not xed, i.e. the Lagrangian multiplier α xing N is zero. α = 0 ⇒ E0 = µ = 0. The total energy is
U=
Z
Ef (E)dτ
dτ is the volume-element in phase spa e, i.e. dτ = d3 x4πp2 dp. Be ause of the un ertainty relation we annot make both dx and dp in nitesimaly small, so we have to divide by ~3 Z 4πV Ef (E)p2 dp U= 3 ~
As for radiation holds p =
E c
=
~ω c
we hange variables to get
4πV ~3 U= 3 3 ~ c
Z
~ωf (~ω)ω 2 dω
In order to a
ount for the two possible polarisations we have to multiply by 2. The energy/unit volume is Z 3 8π~ U = 3 V c
ω dω eβ~ω − 1
1. Rayleigh -Jeans: energy/unit volum/frequen y for long wavelength ǫ=
8π~ ω 3 8π 8π~ ω 3 ≈ = 3 ω2 3 βω~ 3 c e −1 c β~ω βc
2. Wien's law: energy/unit volume/frequen y for small wavelength ǫ=
8π~ ω 3 8π~ 8π~ ω 3 ≈ = 3 ω 3 e−β~ω c3 eβω~ − 1 c3 eβ~ω c
3. Stefan's Law: energy dependen e on T Z U ω3 8π~ = 3 dω = V c eβ~ω − 1 Z x3 8π~ 1 dx = = 3 c (β~)4 ex − 1 Z 8πk4 x3 = 3 3 T4 dx = c ~ ex − 1 8πk4 π 4 4 U = 3 3 T = σS−B T 4 V c ~ 15
34
Example.
Consider a superrelativisti radiation gas of Bosons/Fermions and give ( 1 0 ≤ E ≤ ED g(E) = 0 E > ED
Cal ulate CV and dis uss the limits. U=
Z
ED
0
• QM Limit, β ≫ 1 U=
Z
ED
0
=
Z
ED
E dE = ±1
eβE
E dE ≈ βE e ±1
Z
ED 0
E dE = eβE
Ee−βE dE =
0
E D Z E 1 ED −βE = − e−βE + e dE = β β 0 0 i ED eβED 1 h =− + 2 1 − e−βED = β β h i 1 U = − 2 (ED β + 1)e−βED − 1 ≈ k2 T 2 β
i.e.
T →0
CV ∝ T −−−→ 0
as expe ted of QM CV • Classi al Limit, β ≪ 1
1. BE:
U=
Z
0
i.e.
ED
E dE ≈ βE e −1
Z
ED 0
E ED dE = = ED kT βE β
CV = ED k = const
2. FD: Z ED Z E 1 βED x E dE ≈ dE = dx = U= eβE + 1 2 + βE β 0 2+x 0 0 Z 1 2 1 βED D dx = 2 [x − 2 ln(2 + x)]βE = 1− = 0 β 0 2+x β 1 = 2 [βED − 2 ln(2 + βED ) + 2 ln(2)] = β 1 2 + βED = 2 βED − 2 ln = β 2 ED 2 2 U = ED kT − 2k T ln 1 + 2kT Z
ED
35
( ) ∂U E E T D D CV = − = = ED k − 2k2 T 2 ln 1 + + ED ∂T 2kT 2kT 2 1 + 2kT ED ED 2 − = ED k − 2k T 2 ln 1 + 2kT 2kT + ED
i.e. CV is T dependent whi h is absurd.
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