Lecture Notes for Phys 641 ”Statistical Mechanics” Vitaly A. Shneidman Department of Physics, New Jersey Institute of Technology, Newark, NJ 07102 (Dated: April 21, 2005)
Abstract These notes are not intended to replace either the textbook or the actual lecture but will contain some of the more cumbersome formulas and graphics, which will save us time in class. I will try to keep most of the notes in a single file; the date and the table of contents will be automatically updated. It will be useful to print out the updated part before the lecture. If you notice errors or typos, please e-mail to
[email protected]
1
Contents
I
Introduction I. Overview
2
II. Math
4
A. Integrals and asymptotic relations
4
1. Gauss integral
4
2. The Stirling formula
4
B. Functions of 2 and more variables
5
1. Complete and incomplete differentials
5
2. Legendre transformations
5
3. Homogeneous functions
6
C. Variational principles
6
1. The Euler equation
6
2. Generalizations: several variables
7
3. ”undetermined” Lagrange multipliers
8
D. Elements of probability theory and stochastic processes
8
1. Probability and probability density
8
2. Several random variables and statistical independence
9
3. Non-trivial example: fluctuation of number of particles in an ideal gas
10
4. Random walk and evolution equations
11
5. Information entropy
15
III. Mechanics
17
A. Newton’s equations
17
B. Lagrange equation
17
1. Vibration of molecules
19
2. Rotation
21
C. Hamilton’s approach
22
1. Liouville’s theorem
23
0
2. Numerical illustrations
23
D. Relativistic
25
E. Limitations of classical mechanics in Statistics: Maxwell’s demon and the Gibbs paradox
26
IV. Quantum Mechanics
27
A. Overview
27
B. Quantum properties of light (and other waves)
27
C. Quasicalssical quantization
28
D. de Broglie wave
29
1. Standing waves
29
2. Tunneling.
30
E. The Shr¨odinger equation (SE)
30
1. What is Ψ?
30
2. How to construct a SE?
31
3. Harmonic oscillator and molecular vibrations
32
F. Schr¨odinger equation in 3D: Quantization of angular momentum
32
1. The SE and separation of variables
32
2. Angular momentum
34
3. Rotation of molecules
35
4. Spherical top
35
5. Symmetric top
35
6. Asymmetric top
36
G. Identical particles
36
1. Implication for rotation
37
2. Density of states
37
H. The volume of phase space and the total number of states V. Thermodynamics
38 39
A. First Law
39
B. 2d Law
39
1. Constant values of T and P in equilibrium
39
2. Specific heat
40 1
C. Thermodynamic potentials (TP)
40
1. Energy
40
2. Enthalpy (heat function)
40
3. Free energy (Helmholtz)
41
4. Gibbs free energy (”thermodynamic potential”)
42
5. Additional independent variables and corrections to TP
42
6. Dependence on the number of particles
42
7. Potential Ω
43
D. Two non-trivial applications
44
1. Expansion of gas in vacuum
44
2. The Joule-Thompson process
44
E. Some general relations and theorems
II
44
1. Minimal values of F (T, V, λ) and Φ(T, P, λ)
44
2. Connection between the change of entropy and work
44
3. Thermodynamic inequalities
45
4. Absence of macroscopic motions inside a body
45
F. Phase Transitions
46
1. Tripple point
46
2. Gibbs phase rule
46
3. Clapeyron equation
47
4. Critical point and corresponding states
48
Foundations of statistical mechanics
VI. Distribution function
49
A. Statistically independent states
49
B. The role of energy
50
VII. Microcanonical ensemble
50
A. Properties of the phase volume
50
B. Equal a priori probability and Entropy
51
C. Relation to ”information” (non-equilibrium) entropy
52
2
1. Temperature
52
D. Quantum
52
E. Example: ideal gas
53
1. Number of states and entropy
53
2. Construction of thermodynamics
54
F. Example: Entropic elasticity
55
1. The model and Γ
55
2. Thermodynamics
55
G. Strong and weak sides of microcanonical ensemble
55
1. Energy dependence of the probability density
56
VIII. Canonical ensemble
57
A. Justification(s) of the Gibbs Distribution (GD)
57
B. Maxwell distribution (MD)
59
C. Equipartition of energy
60
IX. Grand canonical ensemble
60
A. Absence of a phase transition for a finite V
III
62
Ideal systems
X. Classical gas (with quantum molequles)
63
A. Monatomic gas
63
1. Canonical
63
2. Grand canonical
64
3. Electronic partition function
64
B. Diatomic gas
65
1. Vibrational partition function
65
2. Rotational partition function (different nuclei)
67
3. Rotational specific heat
68
XI. Fermi-Dirac (FD) and Bose-Einstein (BE) statistics A. Introduction: the simplest example 3
70 70
B. Grand canonical partition function
70
1. Occupation numbers
70
2. Evaluation of Ξ
71
3. Average occupation numbers
71
C. Non-equilibrium quantum gas
72
D. Role of spin and density of levels
73
E. Evaluation of chemical potential
73
F. Energy
74
G. Equation of state
75
H. Ultra-relativistic gas
76
XII. Bose-Einstein condensation
78
A. Properties of the condensed phase
79
1. Number of particles
79
2. Energy
79
3. Specific heat, T < Tc
80
4. Pressure
80
B. Specific heat at all T
80
1. Pressure at all T and isotherms in parameric representation
81
C. Experimental
82
XIII. Blackbody radiation
82
IV
A. Planck distribution
82
B. Total energy and specific heat
83
C. Radiation pressure
84
Specific heat of solids
XIV. Phonons
85
A. Linear monoatomic chain
85
B. 1D- diatomic; optical branches
87
C. 3D - monoatomic
87
4
A. Notations
88
1. General
88
2. Specific symbols
89
5
Dr. Vitaly A. Shneidman, Phys 641, 1st Lecture
Part I
Introduction I.
OVERVIEW
There are several closely related disciplines which study properties of macroscopic bodies. Each has a somewhat different emphases and its own advantages/disadvantages. Here is a non-rigorous classification. • Thermodynamics. Establishes the most general relations without a detailed study of the molecular structure of the bodies involved. (and in fact, most of the thermodynamic relations were derived in ”pre-molecular” times). In the heart of thermodynamics are the 1st, and especially the 2d Law: dS ≥0 dt
(1)
for a closed system, S being the entropy. Can be traced back to Joule and Clausius, respectively. The thermodynamic relations are valid regardless of the state of the bodies involved, or if their microscopic parts obey classical or quantum laws. However, the thermodynamics cannot predict the equation of state, even the simplest one for an ideal gas p = nkB T
(2)
with p being pressure, and n the density of molecules (although, once such an equation is introduced, thermodynamics can do a lot...). • Statistical Physics. Attempts a better understanding of thermodynamic laws and fluctuations from the fact that bodies are macroscopic and thus can be considered as very large number of smaller parts (each still large on a molecular scale). At the heart of Statistical Physics is the Gibbs distribution for a system in a ”thermostat”: wn =
1 En exp − Z kB T µ
2
¶
(3)
Here wn is the probability to find the system in the nth state, En is the energy and Z follows from the normalization condition,
P
wn = 1. The 2d Law is not questioned,
and it is considered that the correctness of the Gibbs distribution is manifested by the correct prediction of thermodynamic relations. • Statistical Mechanics. Attempts to provide a microscopic justification for thermodynamics, starting from individual molecules. The most significant contribution is by Boltzmann. The problem of a strict microscopic justification of the 2d Law is in no way closed; occasionally (I will give references in class) one can hear from respected member of Stat. Mech. community about possible ”violations of the 2d Law” - people outside of their immediate surrounding usually attribute this to a wrong definition of entropy. There is less reverence towards thermodynamics and one sometimes hears that ”thermodynamics is the limit of Stat. Mech. when N , the number of particles, tends to infinity”. [Wouldn’t that be nice? In reality, however, there are many problems which can be treated thermodynamically, but not starting from statistical mechanics]. Generally, the Stat. Mech. path is hard, and often significant contributions come from mathematical community. Another aspect of Stat. Mech. is the study of model systems for which the above Gibbs distribution is postulated, and the thermodynamic properties are derived from that. For non-trivial models the methods employed are usually hard, but results can be extremely instructive. The best example is the Ising model. Verification of the ergodicity hypothesis (which leads to Gibbs distribution) for simple mechanical systems is also part of the Stat. Mech., very hard. • Kinetic theory. Can be considered as a precursor of Stat. Mech., and can be traced to the times when most people, except for the few best ones, did not believe in molecules. Derived equations for simple gases, as eq. (2), but the real challenge was the explanation of the 2d Law, the main puzzle being the reversibility of the equations of mechanics which seems to contradict the irreversibility of the entropy change. (Boltzmann provided here the most up to date explanation). Today, when talking about kinetic theory one puts a somewhat stronger emphasis on transport properties (heat conductivity, etc.). Boltzmann equation is the central one when describing kinetics, at least for molecules which obey classical mechanics. This equation -we might discuss it closer to the end of the course if there is time- describes the evolution of the 3
distribution of molecules due to binary collisions. From there macroscopic transport coefficient, such as viscosity, coefficient of heat conductivity, etc. can be extracted, although that is not easy. The approach of our course will be the closest to Statistical Physics...
II.
MATH A. 1.
Integrals and asymptotic relations Gauss integral
Z
∞ −∞
2
e−αx dx =
r
π α
(4)
For α → ∞ the integrand → δ-function: δ(x) = α→∞ lim
r
α −αx2 e π
(5)
with Z
² −²
δ(x) dx = 1 ,
Z
∞ −∞
δ(x)f (x) dx = f (0)
for any finite ² > 0 and smooth f (x). HW1: I-61 from McQ. calculate for n = 1, 2, 4, and try general n
2.
The Stirling formula
Consider N! =
Z
∞ 0
N −x
x e dx = N
N +1
Z
∞ 0
e−N φ(z) dz
with φ(z) = z − ln z HW2: check this For N → ∞ very sharp maximum near z = z∗ = 1. Thus, 1 φ(z) ≈ φ (z∗ ) + φ00 (z∗ ) (z − z∗ )2 2 Using Gauss integral, N ! ' N N +1
s
2π e−N φ(z∗ ) N φ00 (z∗ )
4
(6)
Or, N! ' B. 1.
√
N 2πN e µ
¶N
(7)
Functions of 2 and more variables Complete and incomplete differentials
Let us construct dF = X(x, y)dx + Y (x, y)dy
(8)
The integral along a path (I) Z
(I)
dF
can depend on the entire path, or only on the initial and final points. In the latter case, one calls dF a complete differential. For a complete differential, ∂X ∂Y = ∂y ∂x
(9)
Theorem (no proof) In 2D there always exists an integrating factor g(x, y) so that gdF becomes a complete differential, even if dF is not. Examples: Incomplete differentials: dQ (heat), dW (work). Complete, dS (entropy), dE (energy), etc.
2.
Legendre transformations
Given a function of two variables df =
∂f ∂f dx + dy ≡ u dx + v dy ∂x ∂y
(10)
change the differentials from dx and dy to du and dy with the transformation g ≡ f − ux dg = df − u dx − x du = u dx + v dy − u dx − x du = v dy − x du Now, x≡− 5
∂g ∂u
(11)
v≡ 3.
∂g ∂y
Homogeneous functions
Homogeneous function of the order n: f (λx1 , λx2 , . . . , λxN ) = λn f (x1 , x2 , . . . , xN )
N X
xi
i
∂f = nf ∂xi
(12)
(13)
(”Euler theorem”) Preview. Why important in physics? in mechanics: kinetic energy K is a homogeneous function of velocities with n = 2 (even in generalized, non-Cartesian coordinates!); important when conservation of energy is introduced in motion of charged particles (or planets): potential energy U is a homogeneous function ¯ = − 1 U¯ , an important consequence of the of coordinates with n = −1; this leads, e.g., to K 2 virial theorem in thermodynamics: extensive variables n = 1 (energy, etc.) and intensive n = 0 (T or p). Important in many cases, e.g. for solutions one has homogeneous functions of N1 , N2 , . . ., the numbers of molecules, and the Euler theorem leads to famous Gibbs-Duhem equations.
C. 1.
Variational principles The Euler equation
Consider J=
Z
x2 x1
f (y, y 0 , x) dx
y(x) - unknown(!); J = min (or, max). Suppose, we know y(x). Let us perturb it a bit: y(x, α = 0) = y(x) , y(x, α) = y(x, 0) + αη(x)
6
(14)
Now J = J(α) Since η)x) is arbitrary, to ensure a J(α) = min, one needs: d ∂f ∂f − =0 0 dx ∂y ∂y
(15)
These are Euler equations (Lagrange in mechanics - see below). Three major simplifications: • no y-dependence
∂f = const ∂y 0
(16)
(”momentum conservation”) • no explicit x-dependence:
f − y0
∂f = const ∂y 0
(17)
related to energy conservation in mechanics • no y 0 -dependence
∂f =0 ∂y
(18)
(thermodynamics). Examples: minimal surfaces
2.
Generalizations: several variables
several dependent variables y1 , y2 , . . . a separate Euler equation for each variable. several independent variables x1 , x2 , . . . X ∂ j
∂f ∂f − =0 0 ∂xj ∂yxj ∂y
Example: Laplace equation
7
(19)
3.
”undetermined” Lagrange multipliers
Suppose we want to find a maximum of J with an additional constraint Z
φ (y, y 0 , x) dx = 0
Then, construct g = f + λφ and get regular Euler equations for g, which will contain however a yet unknown λ. Those equations can be solved together with the constraint equation. If necessary, λ also can be determined. Example: line with fixed area and min length (semicircle).
D. 1.
Elements of probability theory and stochastic processes Probability and probability density
Let x - discrete random variable, with possible values x1 , x2 , etc. Then X
P (xm ) = 1
m
f¯ ≡ hf i =
X
f (xm ) P (xm )
m
D
E
σ 2 ≡ x2 − x¯2 which is the second central moment. HW: show that D
E
D
E
(x − x¯)2 = x2 − x¯2
Examples: Poisson distribution, 0 ≤ m < ∞: P (m, a) =
1 −a m e a m!
HW: check normalization D
E
m ¯ = a, m2 = a2 + a thus σ 2 = a (HW: check this)
8
(20)
Let x - continuous random variable, Z
f¯ =
p(x)dx = 1 Z
f (x)p(x)dx
Examples: • uniform: p(x) = 1 , 0 < x < 1
(21)
and p(x) = 0 otherwise. (computer random number generators are done like that). • Gaussian
1 (x − a)2 pG (x, a, σ) = √ exp − 2σ 2 σ 2π "
HW: show that for Gaussian x¯ = a and D
E 1 ³ ´n (x − x¯)2n = √ 2σ 2 Γ(n + 1/2) π
Transformation of variables
x → y , p(x) → P (y) = p(x) Z
P (y)dy = 1
HW: write Gauss distribution for y = (x − a)/σ 2.
Several random variables and statistical independence
Consider p (x1 , x2 ) with Z Z
p (x1 , x2 ) dx1 dx2 = 1
Two random variables are statistically independent if p (x1 , x2 ) = p1 (x1 ) p2 (x2 ) Then, n m n hxm 1 x2 i = hx1 i hx2 i
9
dx dy
#
(22)
Central limit theorem Consider N statistically independent random variables x1 , x2 , . . . , xN . For simplicity, x¯1 = x¯2 = . . . = x¯N = 0 Denote D
E
x21 = . . . = σ 2
What is the probability density PN of N 1 X xn ? SN = √ N n=1
First note, hSN i = 0 ,
D
E
2 = σ2 SN
Theorem (no proof): 1 z2 lim PN (z) = √ exp − 2 N →∞ 2σ 2πσ 2 "
#
(23)
Note: individual xn do not have to be Gaussian (p (xn ) even does not enter into formulation) - there just should be many of them!
3.
Non-trivial example: fluctuation of number of particles in an ideal gas
Consider a large number n of non-interacting particles in a large (blue) box, as in Fig. 1. Particles are randomly distributed. The average number of particles, m, ¯ in a selected small (red) counting box is m ¯ = n/k where k is the number of red boxes. Unlike n, the value of m ¯ does not have to be large or integer. However, the actual number of particles can be different from m. ¯ what is the probability to get exactly m particles in a red box? If the total (blue) volume is V and the ”red” volume is v, the robability for a selected particle to get into the box is p = v/V = m/n ¯ . For m particles the probability for each to get inside is pm , and the exact expression is pm = Cnm pm (1 − p)n−m 10
FIG. 1: 500 non-interacting particles randomly distributed in a large (blue) box. Probability to find exactly m particles in a selected red box is given by a binomial distribution.
This depends, however, on the blue box, but intuitively if this box is large, it should not matter. Indeed, for n, V → ∞ (with n/V fixed) one gets a Poisson distribution pm '
1 −m¯ m e m ¯ m!
This is rather accurate - see Figs. 2. Further, for m ¯ À 1 the gaussian is approached with σ2 = m ¯ HW - consider k = 2, the blue box is divided in two red boxes. Consider the difference in the number of particles in the two boxes. The Poisson distribution will not work here, but the Gauss will work if this difference is much smaller than n. Find the Gaussian, compare with exact.
4.
Random walk and evolution equations
Consider a random walk xn+1 = xn + ζ with ζ = ±1 being a random variable. Now consider x2n+1 = x2n + 2xn ζ + ζ 2
11
0.12 0.1 0.08 0.06 0.04 0.02 5
10
15
20
0.04
0.03
0.02
0.01
20
40
60
80
100
120
140
160
FIG. 2: Exact binomial distribution for m ¯ = 10 (left) and m ¯ = 100 (right) shown by symbols. Red line - Poisson distribution (works for any m); ¯ black line - Gauss distribution, which is accurate for large m. ¯ Note that for larger m ¯ the distribution broadens, but becomes narrow on a relative scale m/m, ¯ approaching a δ-function for m ¯ → ∞.
and perform averaging. If ζ is independent of xn , then D
But, hζ 2 i = 1, thus Or,
E
D
E
D
x2n+1 = x2n + ζ 2
D q
E
E
x2n = n
hx2n i =
√ n
Note that! The evolution equation Consider a very large number N of statistically identical ensembles. Then, at a given ”time” t (number of random walk steps) one can talk about the ”distribution function” fk (t), with fk /N corresponding to probability. One has 1 1 fk (t + 1) = fk (t) + fk−1 (t) + fk+1 (t) − fk = 2 2 12
= fk (t) + D [fk−1 (t) + fk+1 (t) − 2fk (t)] , D = 1/2
(24)
Since this equation is linear, an identical equation will hold for the probability p k (t) to occupy a given position k. The equation of this kind is known as the ”evolution” or ’master” equation. It is exact, but we re-arranged it to simplify the following transition. Considering fk (t) as a smooth function of both t and k, one has (with k replaced by x) ∂f ∂2f =D 2 ∂t ∂x
(25)
The solution (”Greens function”) is given by 1 x2 G(x, t) = √ exp − 4Dt 2 πDt (
)
(26)
which is a Gaussian with zero average and with σ=
√
Dt
. HW: consider proper physical units, i.e. random walk with step a and t = τ n. Re-write the results. Exact description For this specific problem an exact description is possible, so that the diffusion approximation can be verified. Consider a total of n steps with equal probabilities of 1/2 to go right or left. If there were m steps to the right (and n − m steps left) there will be a total displacement of m − (n − m) = 2m − n steps. Since the steps are taken in any order, there will be Cnm =
n! m!(n − m)!
possibilities to do that (again a binomial distribution!). All in all, there will be 2 n outcomes of n steps, so that 1 m C 2n n will give the exact probability of going 2m − n to the right from origin. Correspondence of the diffusion approximation with the exact solution is excellent - see Fig. 3 (however, a log plot would reveal that tails are not described that accurately. Why?) Biased diffusion Consider a biased random walk with probability p going right, and q = 1 − p going left on every step. ∂2f ∂f ∂f =D 2 +v , v =p−q ∂t ∂x ∂x 13
(27)
0.1
0.08
0.06
0.04
0.02
-60
-40
-20
20
40
60
FIG. 3: Non-biased random walk (points) and the diffusion approximation 0.1
0.08
0.06
0.04
0.02
-60
-40
-20
20
40
60
FIG. 4: Same as in Fig. 4 for a biased random walk.
HW: derive this, an generalize the exact expression. The Green’s function just drifts with time Gb (x, t) = G(x − vt, t) HW: check this Again, correspondence is excellent - see Fig. 4. Thus, the diffusion equation is a good approximation for random walk. More generally, it is an example of a continuous time evolution equation,
14
5.
Information entropy
Consider a system which can be in one of the N states. The probability to be in state i is pi . Then, S = −k with
N X
pi ln pi
(28)
i=1
k ≡ kB = 1.38 · 10−23 J/o K Note that this is just a formal mathematical definition. It turns out, however, that this information entropy is extremely close to the one used in thermodynamics. Maximization of entropy with constrains We use Lagrange multipliers α, β, etc. • ”microcanonical”
N X
pi = 1
i=1
and − Thus,
N X i=1
[pi ln pi − αpi ] = max
ln pi + 1 − α = 0 or pi = const = 1/N , Smax = k ln N • ”canonical”
N X i=1
Now, − Thus,
N X i=1
pi = 1 , x¯ ≡
N X
xi p i = x 0
i=1
[pi ln pi − αpi + βpi xi ] = max ln pi + 1 − α + βxi = 0
or pi =
1 exp (−βxi ) Z(x)
with Z(x) =
X
exp (−βxi ) , Smax = k ln Z + kβx0
i
15
So far, we leave the Lagrangian multipliers undetermined. In physics, if xi are energy levels, with β = 1/kT we get the Boltzmann distribution. HW: show that x0 = −
∂ ln Z ∂β
Continuous variable: S = −k
Z
p(x) ln p(x) dx
(29)
Note: the units for p(x) in eq. (29) are 1/over units of x. Thus, S will get a constant when we switch to new units. Usually this does not matter since the change of entropy is of interest, but should be kept in mind. Preview: In physics, x is multidimensional, with dx corresponding to the product of d~ri d~pi . To make things dimensionless, h ¯ N has to be introduced, even if otherwise the problem is classical. HW: Calculate S for the Gauss distribution HW: using the constrains Z
∞ −∞
p(x) dx = 1 ,
Z
∞ −∞
x2 p(x) dx = σ 2
find the distribution which leads to maximum of S (it is very similar to the discrete case, but you need to actually evaluate the Lagrange multipliers. Growth of entropy dS ≥0 (30) dt HW: consider the evolution with time of the random walk entropy in the diffusion approximation (with p(x, t) given by the Green’s function) What would it mean if eq.(30) would be violated? Either the definition of entropy or the evolution equations would be bad. Let us show that the diffusion evolution equation is ”good”: Z Z 1 dS d =− (p ln p) dx = − p˙ (ln p + 1) dx = k dt dt !2 Z Ã Z Z 2 ∂ p ∂p 1 ln p dx = D ≥0 − p˙ ln p dx = −D ∂x2 ∂x p
(31)
Note in open space diffusion always leads to increase of entropy. In a closed ”volume” - length L in 1dim., entropy is maximum for p(x) = const = 1/L and does not increase anymore. Dr. Vitaly A. Shneidman, Phys 641, 2d Lecture (cont.)
16
III. A.
MECHANICS Newton’s equations
mi~r¨i = F~i ~r1 , ~r2 , . . . , ~rN , ~r˙ 1 , ~r˙ 2 , . . . , ~r˙ N , t ³
Conservation laws (closed system):
´
(32)
~ = const K + U = const , P~ = const , L
(33)
N N X X ~ = 1 mi mi~ri , M = R M i=1 i=1
(34)
Center of mass
Momentum:
P~ =
N X i=1
Kinetic energy: K= HW: show this
~˙ ≡ M V~ mi~vi = M R
N X 1
N ´2 X 1 1 ³ mi vi2 = M V 2 + mi ~vi − V~ 2 i=1 2 i=1 2
(35)
(36)
Note an important difference: P~ is only the CM motion, while energy also includes motion relative to the CM. Thus, in a collision of solid bodies (N → ∞) P~ is conserved, but energy can be re-distributed over a large amount of microscopic motions, and thus ”lost” from a macroscopic point of view. (i.e., formally energy is there, but it cannot be used to get any work, just leading to heating of the two bodies involved in an inelastic collision).
B.
Lagrange equation
L ~q, ~q˙ , t = K − U ³
´
action =
Z
t2 t1
Ld t
∂L d ∂L = dt ∂ q˙i ∂qi
(37) (38) (39)
Compared to Newton’s equations: • (advantage) deal with scalars (L), not vectors (F~ ), thus easy to change coordinates 17
• (advantage) in problems with constrains forces which do not work also do not enter into the equations (e.g., pendulum) • (advantage) conservation laws - ”automatic”. If qi does not enter L then the component of a ”generalized momentum” ∂L = const ∂ q˙i
(40)
∂L − L = const ∂ q˙i
(41)
pi ≡ If no explicit time in L, then X i
q˙i
(if quadratic dependence of K on q˙i and U does not depend on q˙i then regular energy conservation K + U = const) • (advantage) can be generalized beyond classical mechanics (relativistic and quantum); the variational formulation especially convenient • (disadvantage) hard to deal with empirical forces (e.g. friction) which do not comply with energy conservation First application of Euler’s homogeneous function: show that eq.(41) indeed corresponds to regular energy conservation if K is a quadratic function of q˙i and U is independent of q˙i Example: Physical pendulum 1st method: 1 L = ml2 θ˙2 + mgl cos θ 2 Thus, ml2 θ¨ + mgl sin θ = 0 2d method (”undetermined” Lagrange multipliers): ´ 1 ³ L = m r˙ 2 + r2 θ˙2 + mgr cos θ + λ(l − r) 2
Thus (3 equations), r2 θ¨ + mgr sin θ = 0 r¨ + λ − mg cos θ − mr θ˙2 l−r =0 Thus, same equation for θ and λ = mg cos θ + mr θ˙2 (force!) 18
1.
Vibration of molecules
Index-free notations Familiar: ~a · ~b =
X
ai b i
(42)
i
Thus can construct more complex expressions with every one · corresponding to summation ˆ is a matrix (”tensor of the 2d (convolution) with respect to one dummy index. E.g., if M ˆ · ~a is a vector (”tensor of the 1st rank”) with components rank”), then M ˆ · ~a M
´
ˆ ~a · M
´
³
k
=
X
Mki ai
X
ai Mik
i
or (not the same thing!) ³
k
=
i
I.e. a dot · reduces the rank of a tensor by 1. Double summation is also possible leading to a scalar (”tensor of the 0th rank”): ˆ · ~b = ~a · M
X
ai Mik bk
(43)
i,k
Instead of ”inner product” given by · (which reduces the rank of a tensor) one also can get an ”outer product” which increases the rank. E.g., out of two vectors ~a and ~b can construct a matrix Cik = ai bk , or Cˆ = ~a ⊗ ~b ≡ ~a ~b (we will use the latter notation, with no symbol between two side-by side vectors for the outer product; other options, e.g. ∧ are also used in mathematical papers). Note that gradient of a scalar function Φ ∂ Φ ∂~u is also a vector. Thus, ∂ ∂ Φ ∂~u ∂~u will represent a matrix of the second derivatives or ”Hessian”. Molecules Let a set of 3D vectors ~rn (t) (n = 1, . . . , N ) determine the positions of atoms in a molecule, with rn0 being the equilibrium positions. We define a multidimensional vector ³
0 ~u(t) = ~r1 − ~r10 , . . . , ~rN − ~rN
19
´
(44)
With small deviations from equilibrium, both kinetic and potential energies are expected to be quadratic forms of ~u˙ and ~u: 1 ˆ · ~u˙ > 0 K = ~u˙ · M (45) 2 1 U = ~u · kˆ · ~u ≥ 0 2
(46)
ˆ = ∂ ∂ L M ∂ ~u˙ ∂ ~u˙
(47)
∂ ∂ kˆ = − L ∂~u ∂~u
(48)
~u(t) = ~u0 exp (iωt)
(49)
ˆ · ~u0 + kˆ · ~u0 = 0 −ω 2 M
(50)
ˆ −1 · kˆ · ~u0 = ω 2~u0 M
(51)
ˆ · kˆ (”secular matrix”). ω 2 - eigenvalues of a matrix M Normal coordinates Once the secular equation is solved, one can find 3N generalized coordinates Qα which are linear combinations of all 3N initial coordinates, so that L=
´ 1 X³ ˙2 Qα − ωα2 Q2 2 α
(52)
We will not need much explicit Qα , except for in examples - they determine shapes of characteristic vibrations - but rather the very fact that the above representation is possible. (Preview: we will need this in the equipartition theorem and when studying phonons) ¨ α . Show that indeed you get oscillations with frequencies HW: write explicit equations for Q ωα . What happens for ωα = 0? Example: diatomic molecule L=
´ 1 1³ m1 x˙ 21 + m2 x˙ 22 − k1 (x2 − x1 )2 2 2
20
x2 = −
m 1 x1 , X = x 2 − x1 m2
Now 1 1 L = µX˙ 2 − k1 X 2 2 2 with µ=
m1 m2 m1 + m 2
(53)
with ω=
s
k1 m
HW: consider a hypothetical molecule with one atom of mass 1 amu (hydrogen) and the other 20 amu. The spring constant is 1000N/m and the equilibrium distance is 1 ˚ A a) find the frequency b) (optional) estimating the vibrational energy as kB T , with T being the room temperature, find the amplitude of vibrations. Compare this to equilibrium distance - must be small in order to treat vibrations as harmonic. Additional reading: Goldstein. Classical mechanics. A very good description of the matrix approach to molecular vibrations, without computer assistance (for that reason, also a bit too detailed for our purposes).
2.
Rotation
~ × ~r ~v = Ω K=
1Z ρ (~r) v 2 (~r) d V 2
(54)
with ³
~ × ~r Ω
´2
³
~ · ~r = Ω2 r 2 − Ω
´2
one gets 1~ ˆ ~ K= Ω ·I ·Ω 2 21
(55)
with Iik =
Z
³
´
ρ r2 δik − ri rk d V
(56)
Angular momentum: ~ = Iˆ · Ω ~ L
(57)
Iˆ = diag {I1 , I2 , I3 }
(58)
Principal axes of rotation:
1 K= 2
Ã
L2x L2y L2z + + I1 I2 I3
!
(59)
HW. Show that for a diatomic molecule with r being the separation between atoms I = µr2
(60)
with µ being the reduced mass. HW Consider HCl with r ∼ 1˚ A (you can use a more accurate value from McQ). (a) find I; (b) estimating the rotational energy as 1/2kB T , find Ω at room temperature. (c) Alternatively, use L ∼ h ¯ ∼ 10−34 J · s; find Ω. C.
Hamilton’s approach
The first example of Legendre transformations: L ~q, ~q˙ , t → H (~q, p~, t) ³
´
(61)
with p~ =
∂ L ∂ ~q˙
(62)
Then, H = p~ · ~q − L
(63)
∂ ∂ H , p~˙ = − H ~q˙ = ∂~p ∂~q
(64)
Now,
22
1.
Liouville’s theorem
Preview: incompressible liquid Consider a liquid with density ρ and velocity field ~v , then the continuity equation is ∂ ∂ ρ+ · (~v ρ) = 0 ∂t ∂~r
(65)
Now the the full derivative (together with the flow) is given by ∂ ∂ D = + ~v · Dt ∂t ∂~r
(66)
∂ · ~v = 0 ∂~r
(67)
The condition div~v ≡ is equivalent to Dρ =0 Dt
(68)
i.e. incompressible fluid. The Theorem: ”Phase fluid” is incompressible. Indeed, consider a density ρ (~q, p~) Points move in the 6N-dimensional phase space with ”velocities” V~ = (q˙1 , . . . , q˙3N , p˙1 , . . . , p˙ 3N ) ≡ ~q˙ , p~˙ ³
´
(69)
From Hamilton’s equations: divV~ ≡
∂ ˙ ∂ ∂ ∂ ∂ ∂ ˙ · ~q + · p~ = · H− · H=0 ∂~q ∂~p ∂~q ∂~p ∂~p ∂~q
(70)
which is equivalent to incompressible fluid, i.e. ρ (~q(t), p~(t)) = const along the phase trajectory.
2.
Numerical illustrations
See Figs. 5 and 6. 23
(71)
FIG. 5: A one-dimensional particle between two mirrors (”billiard”). Length is 1, mass is 1 and momenta are distributed between 0, 25 and 0.75 (initial coordinates are between 0 and 0.2) (see liouville.nb). The blue line is a phase trajectory with p(0) = 0.5. Black particles are on the boundaries of the phase volume, and red particles are inside. Upper figure - evolution of the phase volume from t = 0 to t = 2.4. Middle figure - a snapshot at t = 4, and lower figure - a snapshot at t = 32. Note: in upper and lower figures (a) preservation of phase volume and (b) particles do not cross boundaries (thus density is const - Liouville’s theorem); in the lower figure there is good mixing. Liouville’s theorem is valid but is hard to track due to very fine boundary. [case (c) is extremely important for statistics]
24
1.5 1 0.5
-1
-0.5
0.5
1
-0.5 -1 -1.5
1.5 1 0.5 -1
-0.5
0.5
1
-0.5 -1 -1.5
FIG. 6: Harmonic oscillator (top) and anharmonic oscillator after about 10 (middle) and about 100 (lower figure) periods; the blue line is a phase trajectory of a selected point. See the liouville.nb with the description of parameters. The Liouville’s theorem is always valid, but the non-spreading of phase points is an artifact of a harmonic oscillator. Otherwise, note good mixing of the phase space at long times. D.
Relativistic
Hrel =
q
p2 c2 + m 2 c4 + U
(72)
Limits : slow Hrel → mc2 +
p2 +U 2m
(73)
fast Hrel → cp + U
25
(74)
E.
Limitations of classical mechanics in Statistics: Maxwell’s demon and the Gibbs
paradox
26
Dr. Vitaly A. Shneidman, Phys 641, 4th Lecture
IV.
QUANTUM MECHANICS
A.
Overview
We consider only quantum properties with immediate relevance to our goals. In most cases the key is the discrete spectrum. Note the following: • discrete spectrum of electromagnetic radiation will be required for a correct description of its thermal equilibrium (Planck’s formula) • discrete spectrum of molecular vibrations and rotations is required when calculating statistical properties of non-monatomic gases • The discrete spectrum of translational motion of molecules in macroscopic bodies is mostly an artificial construction - levels are so dense that there is absolutely no way to resolve them. Nevertheless even if only conceptual, consideration of such levels is crucial for a proper definition of entropy Another essentially quantum aspect is the indistinguishable nature of particles, leading to Bose-Einstein or Fermi-Dirac statistics, depending on the spin. Myth. The uncertainty principle, the absence of a particle trajectory and probability aspects of quantum mechanics explain statistical nature of matter. In reality, according to the Shr¨odinger equation, evolution of the wave function is fully deterministic; probability appears only when one interprets or predicts the results of measurements. If a state of a system can be described by wave function, it is purely mechanical, with zero entropy which remains as such upon further evolution.
B.
Quantum properties of light (and other waves)
Energy Eph = hν = h ¯ω h ¯ ' 1.055 · 10−34 J · s 27
(75)
p 2
1
-1
-0.5
0.5
1
x
-1
-2
FIG. 7: Quasiclassical quantization. Blue phase trajectories correspond to stationary energy levels. The red area between two close levels corresponds to 2π¯h.
h = 2π¯h This is also valid for other waves (phonons!) Momentum: p~ = h ¯~k or p = h/λ Note: Eph = cp
(76)
as classical. Similar for acoustic wave, with c being speed of sound. Note: in many aspects waves behave like particles (with a different dispersion law), but there is an important difference for statistical mechanics: the number of such ”particles” is not conserved. This will correspond to a zero chemical potential, as we discuss later.
C.
Quasicalssical quantization
I
pi dxi = 2π¯h (n + γi ) , 0 ≤ γi < 1 , n = (0), 1, 2, . . .
Strictly speaking, valid for n À 1, but often works for small n as well. See Fig. 7. HW: use the quantization rules to find levels of a harmonic oscillator 28
(77)
D.
de Broglie wave
λ=
h or ~k = p~/¯h p
(78)
HW: find λ for an (a) 1eV electron, (b) a 1eV neutron, (c) a 0.1 mg dust particle with a speed 1 m/s. Note: ANY p, including relativistic (!). Relation to energy: from Erel = Non-relativistic (E = Erel − mc2 ):
q
c2 p2 + m 2 c4
h 2mE
(79)
hc E
(80)
E = hν
(81)
λ = c/ν
(82)
λ= √
Relativistic: m → 0, p → E/c and λ= This is exactly as for a photon:
and
Myth: for a good understanding of QM you always need the Shr¨odinger equation. No! In very many cases the de Broglie picture is sufficient, and allows for a deep analogy with optics.
1.
Standing waves
Example (non-relativistic) Particle in a cubic box in every direction nλ/2 = a , n = 1, 2, . . . Thus, E=
´ π2h ¯2 ³ 2 2 2 , nx , ny nz = 1, 2, . . . + n + n n z y x 2ma2
(83)
(84)
Example (ultra-relativistic, or any wave). Wave in a cubic box Same eq. (83), which is now just a classical wave equation (we consider a single polarization). 29
Thus, E = cp =
πc¯h 2πc¯h = · (nx + ny + nz ) , nx , ny , nz = 1, 2, . . . λ a
(85)
Note degeneration of levels for D > 1.
2.
Tunneling.
p2 + U (x) = E 2m p=
q
2m (E − U )
classical: E < U - forbidden. QM: µ Z
Ψ ∼ exp i
k(x)d x
¶
with k = p/¯h. If E > U - oscillates, if E < U exponential decay. (
R
|p|d x Ψ ∼ exp − h ¯
)
(86)
Transmission probability ¾ 2Z q D ∼ |Ψ| = exp − 2m(U − E)d x h ¯ ½
2
(87)
with integration over classically forbidden region E < U .
E. 1.
The Shr¨ odinger equation (SE) What is Ψ?
M. Born: |Ψ|2 proportional to probability density. Two situations: • infinite motion. In classics - E > U (x) for all accessible x. Experimental - scattering problem. Z
|Ψ(x)|2 dx = ∞ 30
• finite motion. In classics - E ≥ U (x) only for a finite interval of x. Experimental discrete energy levels. Z
|Ψ(x)|2 dx = 1
(88)
Note: in Statistics we will mostly discuss the 2d, discrete option. (Scattering is important, e.g. in non-ideal bose gas, but will not be considered in this part of the course.) Even for practically unlimited motion we will consider an artificial bounding ”box” to make the spectrum discrete, and only then tend the size of the box to infinity, if necessary.
2.
How to construct a SE?
The Hamilton operator H=
¯2 ∂2 p2 ˆ =−h + U (x) → H + U (x) 2m 2m ∂x2
(89)
ˆ = i¯h ∂Ψ HΨ ∂t
(90)
Ψ(x, t) = ψ(x)e−iEt/¯h
(91)
h ¯ 2 ∂2ψ + [E − U (x)] ψ = 0 2m ∂x2
(92)
Steady-state: fixed E
Free particle: U = 0.
with k =
√
h ¯ 2 ∂2ψ + Eψ = 0 2m ∂x2 ψ(x) ∝ e±ikx
(93)
2mE. And Ψ(x, t) ∝ e±ikx−iωt
with ω = E/¯h. Note: k, ω - very large(!) Example. Discrete spectrum: U (x) = 0 for 0 < x < a and U (x) = +∞ otherwise. (same as de Broglie waves).
31
3.
Harmonic oscillator and molecular vibrations
1 U (x) = mω 2 x2 2 SE:
(94)
d2 ψ 2m 1 + 2 E − mω 2 x2 ψ = 0 2 dx 2 h ¯ ¶
µ
q
(95)
or with ζ = x mω/¯h, 2E − ζ2 ψ = 0 ψ + h ¯ω µ
00
¶
here E must be discrete to satisfy Z
∞ −∞
ψ 2 (x)dx < ∞
. One has 1 h ¯ω En = n + 2 ¶
µ
(96)
Wave functions: ψn ∝ e−ζ
2 /2
Hn (ζ)
(97)
with ψ0 ∝ e−ζ
2 /2
(98)
corresponding to the lowest energy state. Otherwise, Hn (ζ) = (−1)n eζ
2
dn e−ζ dζ n
2
HW: plot several ψn . Molecular vibrations: almost classical - same set of ωα . Some new features - degeneration due to symmetry in polyatomic molecules (will not need much) and tunneling in anharmonic vibrations, e.g. N H3 . [see E. Merzbacher, Quantum Mechanics for additional reading.
F. 1.
Schr¨ odinger equation in 3D: Quantization of angular momentum The SE and separation of variables
The Hamilton operator p2 h ¯2 ˆ ˆ H= + U (~r) → H = − ∆ + U (~r) 2m 2m 32
(99)
2 2 2 2 2 2 ˆ =∂ ψ+∂ ψ+∂ ψ ˆ = ∂ + ∂ + ∂ , i.e. ∆ψ ∆ ∂x2 ∂y 2 ∂z 2 ∂x2 ∂y 2 ∂z 2
(100)
Look for a steady-state solution: ˆ (~r) = Eψ (~r) Hψ
(101)
ˆ =∆ ˆ r − 1 ˆl2 ∆ r2
(102)
In spherical coordinates one has
with ˆl2 given by
Ã
ˆl2 = − 1 ∂ sin θ ∂ sin θ ∂θ ∂θ
!
−
1 ∂2 sin2 θ ∂φ2
ˆ r = 1 ∂ r2 ∂ ∆ r2 ∂r ∂r
(103)
(104)
(although specific form of these operators mostly will not be required...) First we use just spherical symmetry (but not specific U (r)). Look for ψ (r, θ, φ) = R(r)Y (θ, φ)
(105)
From ˆ + 2m [E − U (r)] ψ = 0 ∆ψ h ¯2 ˆ r R − R ˆl2 Y + (E − U )RY 2m = 0 Y∆ r2 h ¯2
(106)
Now divide both sides by RY and multiply by r 2 : 2m(E − U )r 2 1 r2 ˆ ∆r R + = ˆl2 Y 2 R Y h ¯
(107)
Note that the l.h.s depends only on r while the r.h.s only on θ, φ. Thus, both sides should be a const ≡ λ. We thus achieved separation of variables using physical understanding of symmetry. The rest is math. First consider ˆl2 = λY
(108)
Y (θ, φ) = Φ(φ) × Θ(θ)
(109)
now look for
and repeat separation of variables. (HW - do that). Recall that key to discrete eigenvalues - BC. Periodicity in φ leads to Y (θ, φ) = eimφ × Θ(θ) , m = 0, ±1, ±2, . . . 33
(110)
(HW - show that. For θ dependence harder (polynomials in cos θ, but pure math: Ylm - spherical harmonics. ˆl2 Ylm = l(l + 1)Ylm
(111)
Note no m - degeneracy of levels(!). (m - ”magnetic quantum number”, l - ”orbital quantum number”).
2.
Angular momentum
¯ ~ = ~r × p~ → L ~ˆ = h ~r × ∇ L i ~ Relation to rotation by a small angle δ φ:
(112)
~ × ~r δ~r = δ φ and for any function f (~r) one has i ~ ˆ δ φ · Lf h ¯
(113)
ˆ2 = h ˆ2 + L ˆ2 = L ˆ2 + L ¯ 2 ˆl2 L z y x
(114)
¯ ∂ ˆz = h L i ∂φ
(115)
δf = δ~r · ∇f =
ˆ x and L ˆ z - see, e.g. the Merzbacher book. Quantization: We will not need L L2 = h ¯ 2 l(l + 1) , l = 0, 1, ; . . .
(116)
Lz = m¯h , m = 0, ±1, . . . , ±l
(117)
34
3.
Rotation of molecules
Why so well separated? ωel À ωvib À ωrot Large parameter
q
(118)
me /M - will explain in class.
Common to use J for l and K for m. Here we consider heteronuclear molecules. For identical nuclei (will be discussed later) some of the energy levels are forbidden.
4.
Spherical top
L2 2I
(119)
h ¯2 J(J + 1) , J = 0, 1, . . . 2I
(120)
Hrot = ˆ 2 . Thus, and the same in quantum with L2 → L Erot = Degeneracy:
gJ = 2J + 1
5.
(121)
Symmetric top
Let Ix = Iy ≡ I. Then, Hrot =
¶ µ L2 L2 L2z 1 1 L2x L2y + + z = − − 2I 2I 2Iz 2I 2 I Iz
(122)
ˆ 2z are h Since eigenvalues of L ¯ 2 K 2 with K = 0, ±1, . . . J 2 Erot =
h ¯2 h ¯2 J(J + 1) − 2I 2
µ
1 1 K2 − I Iz ¶
(123)
See Fig. 8 where degeneracy of levels is removed for clarity. (see Physical Chemistry with Mathematica, although an old version of Mathematica is used there) Diatomic molecule Iz → 0 35
Energy/meV 16 J J J J
14
= = = =
3 2 1 0
12 10 8 6 4 2 0
K = 0
|K| = 1
|K| = 2
|K| = 3
FIG. 8: Rotational spectrum of N H3 with IA = IB = 2.816 · 10−47 kg · m2 and IC = 4.437 · 10−47 kg · m2 (oblate top).
Thus, K = 0 and Erot
h ¯2 = BJ(J + 1) , B = 2I
B-rotational quantum number. HW - estimate B for HCl; find parameters from McQ
6.
Asymmetric top
NOTHING! (energy levels cannot be determined analytically)
G.
Identical particles
Consider permutation of 2 identical particles Ψ (x1 , x2 ) = eiα Ψ (x2 , x1 ) The next permutation restores the original state: e2iα = 1 36
(124)
Thus, eiα = ±1
(125)
Two possibilities depend on spin: s = 0, 1, . . . - ”+” (symmetric wave function, Bose-Einstein statistics) s=
1.
1 2
,
3 2
, . . . - ”-” (antisymmetric wave function, Fermi-Dirac statistics)
Implication for rotation
Consider a homonuclear diatomic molecule. Let the nuclei be Bosons (e.g. D2 ). Then, for SN = 0 only even J, with same degeneracies. More general: gJ = (SN + 1) (2SN + 1) (2J + 1) , J − even
(126)
gJ = SN (2SN + 1) (2J + 1) , J − odd
(127)
and
Now let the nuclei be fermions (e.g., H2 ) gJ = (SN + 1) (2SN + 1) (2J + 1) , J − odd
(128)
gJ = SN (2SN + 1) (2J + 1) , J − even
(129)
and
Orthohydrogen: parallel nuclear spins Parahydrogen: antiparallel nuclear spins.
2.
Density of states
Will be discussed in class. Introduce d Γ = d q1 d p1 d q2 d p2 . . . d q3N d p3N
(130)
which is the classical volume of phase space. If particles are different, from quasicalssical quantization the number of states is dΓ , distinguishable particles (2π¯h)3N 37
(131)
Which is of course unrealistic for large N ∼ NA . If particles are identical, the number of states is much smaller 1 dΓ , identical particles N ! (2π¯h)3N H.
(132)
The volume of phase space and the total number of states
will be done in class. Message - the number of states increases extremely rapidly with energy, and the distance between levels practically goes to zero. This will be the key to introducing statistical entropy.
38
Dr. Vitaly A. Shneidman, Phys 641, 5th Lecture
V.
THERMODYNAMICS
Recommended reading: Thermodynamics by Enrico Fermi or LL, Ch.2 (at this point you can ignore harder passages and those explicitly related to mechanics or statistics); Ch. 1 & 2 by K. Huang are also very goog.
A.
First Law
dE = dR + dQ
(133)
d R = −pd V
(134)
R - work, Q - heat. (Joule experiment). Note: R, Q depend on path; E -depends only on the state (thus d E is complete or ”perfect differential”).
B.
2d Law
dS ≥
dQ T
(135)
Equality is achieved for reversible processes. Then one can write dE = TdS − PdV
(136)
Note: unlike d Q the differential of entropy d S is a perfect differential. In a general case there are irreversible processes with a ”>” sign in the above. Entropy thus grows in thermally isolated systems with d Q = 0, reaching maximum in complete equilibrium. various formulation of the 2d law are equivalent (will be discussed in class).
1.
Constant values of T and P in equilibrium
Will be discussed in class. 39
2.
Specific heat
General definition C=
dQ dT
for a given reversible process. What is constant while T is changing need to be specified. E.g.
¯
(137)
¯
(138)
∂S ¯¯ ¯ Cv = T ∂T ¯V
∂S ¯¯ Cp = T ¯ ∂T ¯p
C. 1.
Thermodynamic potentials (TP) Energy
”Natural variables” E(S, V ) Simple derivatives T =
∂E ∂E , P =− ∂S ∂V
(139)
(often, do not have to indicate what is constant). Also, clear physics: d Q = d E , V = const
(140)
d R = −d E , S = const
(141)
Also, if there are other variables λ which can change with constrains S = const, V = const, then E(S, V, λ) = min
2.
Enthalpy (heat function)
Want to go from S, V (natural variables for E) to S, P . Use Legendre transformation d W = d E + d (P V )
40
or W = E + PV
(142)
d W (S, P ) = T d S + V d P
(143)
d Q = d W , P = const
(144)
with
Now, for an isobaric process
(similar to energy for V = const). Similarly, ¯
∂W ¯¯ Cp = ¯ ∂T ¯p
(145)
(note, need to indicate T as const, since T is not a ”natural variable” for W .) For thermally isolated isobaric processes, dQ = 0, P = const, one has W = const This is interesting, since V can change and thus work is made by the system(!)
3.
Free energy (Helmholtz)
Want to use T, V as ”natural variables”. Again, Legendre transformation d F = d E − d (T S) or F = E − TS
(146)
d F (T, V ) = −Sd T − P V
(147)
with
Note, work d R = d F , T = const HW: show that E = −T
2
Ã
41
∂ F ∂T T
!
(148) V
4.
Gibbs free energy (”thermodynamic potential”)
Natural variables - P, T . d Φ(P, T ) = d F + d (V P ) = −Sd T + V d P
(149)
Φ = F + PV = W − TS = E − TS + PV
(150)
or
HW: show that W = −T 5.
2
Ã
∂ Φ ∂T T
!
P
Additional independent variables and corrections to TP
d E = T d S − P d V + Λd λ If λ is an independent variable, same corrections appear in d F , d Φ and d W . For small changes (only!) (δE)S,V = (δF )T,V = (δΦ)P,T = (δW )S,P
(151)
If λ is allowed to change when either T, V or T, P are const, then the corresponding derivatives are negative, and when equilibrium with respect to λ is reached F = min , T, V = const
(152)
Φ = min , T, P = const
(153)
or
6.
Dependence on the number of particles
The first, and most important example of an extra independent variable is the number of particles, N . Then, all thermodynamic potentials get a correction µd N with µ being the ”chemical potential”. In addition, if s, v,... are entropy, volume, etc. per particle, we expect that E = N e(s, v) , F = N f (v, T ) , W = N w(s, P ) , Φ = N φ(P, T ) 42
Thus, immediately µ = Φ/N
(154)
d µ = −sd T + vd P
(155)
and
Condition of equilibrium Consider a system of two parts with not necessarily same pressure. Then, will show µ1 = µ 2 or µ = const
(156)
in the body. Equilibrium in external field Now, correction to energy is µ0 d N + u (~r) d N so that µ = µ0 + u is the new chemical potential. Or, µ0 + u (~r) = const
7.
(157)
Potential Ω
One has d F = −Sd T − P d V + µd N Now want to have µ as independent variable. Legendre transformation gives d Ω(T, V, µ) = d F − d (µN ) = −Sd T − P d V − N d µ
(158)
Ω = F − µN = F − Φ = −P V
(159)
with
Looks like the simplest potential, but watch the independent variables! the rest will be discussed in class
43
Dr. Vitaly A. Shneidman, Phys 641, 6th Lecture
D. 1.
Two non-trivial applications Expansion of gas in vacuum
will be discussed in class
2.
The Joule-Thompson process
will be discussed in class
E.
Some general relations and theorems
1.
Minimal values of F (T, V, λ) and Φ(T, P, λ)
For T, V = const dE dS dQ dS dF = −T = −T ≤0 dt dt dt dt dt
(160)
dΦ dW dS dQ dS = −T = −T ≤0 dt dt dt dt dt
(161)
For T, P = const
2.
Connection between the change of entropy and work
Suppose we want to bring a small part of a system out of equilibrium. This will reduce the entropy and will require some work R. The work is minimum for a quasistatic process. One has ∆S = −
Rmin T0
(162)
We indicate T0 , the original temperature, since it can become different in the subsystem during the process. We will not prove this relation (it is not hard, but requires some patience and time - see LL), but it will be important when describing fluctuations. For T = T0 = const and V = const Rmin = ∆F , T, V = const 44
(163)
or Rmin = ∆Φ , T, P = const
3.
(164)
Thermodynamic inequalities
Follow from consideration of a small subsystem in a ”thermostat” and requirement that the total entropy can only be reduced upon infinitesimal changes. First derivatives give T, P > 0. Second derivatives give - see LL
and
Ã
CV > 0
(165)
!
(166)
∂P ∂V
<0 T
Metastable states: above holds, but for a finite deviation the total entropy is increased.
4.
Absence of macroscopic motions inside a body
Consider i subsystems. Then S=
X
Si
i
Ã
Mi vi2 Ei − 2
!
with ~ = P
X
Mi~vi = const
i
. Then, ~ = max S + ~a · P From Mi~vi ∂ Si = − ∂~vi T get vi = const (i.e. motion as a ”rigid” body - Galileo). Rotation - harder.
45
(167)
F.
Phase Transitions
µ1 (P, T ) = µ2 (P, T )
(168)
Peq = P (T )
(169)
q ≡ w2 − w1 = T (s2 − s1 )
(170)
Solution:
Latent heat (per molecule):
Positive q for transition to a high-temperature stable phase (in class). HW. Find q/kB T B , the boiling temperature for steam and any other saturated vapor. Use CRC Handbook (better), or R. Reid et al. Properties of Gasses and Liquids.
1.
Tripple point
in class Note: for 2 phase equilibrium everything is determined by T - 1 degree of freedom, a line in (P, T ) diagram. For 3 phase equilibrium - one fixed point, zero degrees of freedom. (Alternatively, for just one phase both P and T are free to choose, i.e. two degrees of freedom). All these are examples of the Gibbs rule of phases - will discuss more with multicomponent systems:
2.
Gibbs phase rule
Consider L phases in contact and n components, with ³
ΦM P, T, N1M , . . . NnM
´
being the Gibbs free energy of the M -th phase. The chemical potentials for each phase and component are µM i =
∂ ΦM , 1 ≤ i ≤ n; , 1 ≤ M ≤ L ∂NiM
(171)
Then, L µM i = µi , for any i ≤ n , M ≤ L − 1
46
(172)
(note: i is same, phases can be different. This gives a total of n × (L − 1) equations for 2 + L(n − 1) independent variables M P, T and cM 2 , . . . cn M M in each phase, with cM k ≡ Nk /N1 (e.g.)
One thus has the difference between the number of unknowns and the number of equations: 2 + L(n − 1) − n (L − 1) = 2 + n − L
(173)
which is the number of ”degrees of freedom”.
3.
Clapeyron equation
(return to one-component). Take a derivative of eq. (168) for P = Peq (T ) to get q s2 − s 1 d Peq = = dT T (v2 − v1 ) v2 − v 1
(174)
Similarly, d Teq /d P . HW. Use CRC Handbook for Chemistry and Physics. Find q, v1 , v2 for ice-water. Which pressure is required for ice to melt at −20o C?. How realistic is it for skates? Liquid-gas transition: Here v 2 À v1 , P v 2 ' k B T and q d ln Peq = dT kB T 2 If q ≈ const
µ
Peq ∝ exp −
47
q kB T
(175)
¶
(176)
p
2 1.5 1 0.5
1
2
3
4
5
6
v
FIG. 9: Van der Waals isotherms for reduced pressure p = P/Pc and volume v = V /Vc . The reduced temperature t = T /Tc (from top to bottom) is 1.15, 1.05, 0.95 and 0.85. Note that for t < 1 part of the blue isotherm is not stable and will be replaced by a horizontal red line. The latter follows from Maxwell construction with the areas below and above the line being equal. 4.
Critical point and corresponding states
in class + handout HW. Consider Problem 1-36 from McQ. Make a good plot of Cp (T ), S(T ), W (T ), Φ(T ) for the entire process (from low-T solid to vapour at 1 atm. (Alternatively, can use some element, or water from CRC Handbook, but need to cover all three phases.). Optional (will contribute to mid-term): also find E, F assuming a linear thermal expansion coefficient for both liquid and solid phases.
48
Dr. Vitaly A. Shneidman, Phys 641, 7th Lecture
Part II
Foundations of statistical mechanics Reading: LL, Ch.1 & 3 (skip harder parts), McQ, Ch.2 & 3, K. Huang, Ch.6
VI.
DISTRIBUTION FUNCTION
We again consider the probability density ρ (~q, p~) similarly to the one we discussed in connection with the Liouville’s theorem. Only now this function is static, i.e. has no explicit t-dependence [although points keep moving in the phase space!]. The best thing to imagine would be one of the lower Figs. 5 or 6, after a good mixing. We also do not keep track of normalization so far (having in mind later quantum analogies). The average of any function f (~q, p~) is given by hf i =
R
f (~q, p~) ρ (~q, p~) d ~qd p~ R ρ (~q, p~) d ~qd p~
(177)
Alternatively, one could follow a single phase point to find a time average 1ZT f (~q(t), p~(t)) d t f¯ = lim T →∞ T 0
(178)
And they are expected to be identical
A.
hf i = f¯
(179)
ρ12 (~q1 , p~1 , ~q2 , p~2 ) = ρ1 (~q1 , p~1 ) ρ2 (~q2 , p~2 )
(180)
Statistically independent states
49
B.
The role of energy
The Liuoville’s theorem shows that ρ is an integral of motion. Thus, one expects it to be a function of the integrals of motion. There can be many of them, but only additive can matter. Indeed, for two non-interacting parts of a body ln ρ12 = ln ρ1 + ln ρ2
(181)
~ L. ~ In the absence of translational motion and rotation Thus, ρ can depend only on E, P, of a system (”in a box”) only E remains, i.e. for a given subsystem i ln ρi = αi − βEi (~qi , p~i )
(182)
The constant αi can depend on subsystem, β is the same to ensure additivity. Let us show that dependence of energy (Hamiltonian) indeed leads to time-independent ρ. From Liuoville’s theorem ∂ ∂ ∂ dρ ≡ ρ + ~q˙ · ρ + p~˙ · ρ = 0 dt ∂t ∂~q ∂~p Recall ∂ ∂ ~q˙ = H , p~˙ = − H ∂~p ∂~q and for ρ = ρ(H) ∂ dρ ∂ ∂ dρ ∂ ρ= H, ρ= H ∂~q d H ∂~q ∂~p d H ∂~p Thus indeed ∂ ρ=0 ∂t VII. A.
(183)
MICROCANONICAL ENSEMBLE Properties of the phase volume
Having in mind the subsequent quantum discussion, introduce from the start dΓ =
1 d p~d ~q N ! (2π¯h)3N
(184)
Total number of (quantum) states in the elementary phase ”volume” d p~d ~q will coinside with d Γ. 50
In a finite volume Γ(E) =
Z
H(~ p,~ q )
dΓ À 1
(185)
The density of states is Ω(E) =
dΓ dE
(186)
Thus, for a small ∆ ¿ E ∆Γ ≡
Z
E
d Γ ' Ω(E)∆
(187)
Due to a rapid increase of Γ with energy (recall our example with ideal gas), one has Γ(E) '
Ω(E) d ln Ω/d E
(188)
Since d ln Ω/d E is a ”modest” function (compared to Ω), Γ and Ω are ”almost the same”, and so is ∆Γ for any reasonable ∆. This will be used below. HW: Examine 3-15 in McQ. Try to generalize for non-monotonic Ω.
B.
Equal a priori probability and Entropy
Consider an ensemble of phase points evenly distributed between E and E + ∆. Again,
the best thing to imagine would be one of the lower Figs. 5 or 6. Define
S = kB ln ∆Γ
(189)
(Boltzmann). Alternatively, could use S = kB ln Γ
(190)
S = kB ln Ω
(191)
or
(although in the latter case S will slightly depend, up to an insignificant constant, on the energy units). Additivity and equivalence will be discussed in class. HW: Consider one classical particle in a cubic box. Find Γ(E) and plot S(E) in all three definitions (they do not have to be too close for just one particle). Use ∆ = 0.1E
51
C.
Relation to ”information” (non-equilibrium) entropy
Let us consider a finite phase volume ∆Γ and not assume a constant ρ(E) (as in the microcanonical distribution) and define S = −kB
Z
d Γρ ln ρ ≡ −kB hln ρi
(192)
A proper normalization of ρ is expected Z
∆Γ
ρd Γ = 1
(193)
Then, if one tries to maximize S with the above constrain one gets • ρ(E) = const = 1/∆Γ (microcanonical) • S = kB ln ∆Γ HW. Show this - we did a nearly identical problem in connection with information entropy and Lagrange multipliers
1.
Temperature
Let us consider the distribution (182). Then, when average is taken S = kB β E¯
(194)
(this is not identical to what we did before since now we have an additional constrain of ¯ If S is treated as thermodynamic entropy, T = d E/d ¯ S or fixed E). β=
1 kB T
(195)
This, however, is more related to canonical ensemble, to be discussed separately.
D.
Quantum
Since even in the classical approach we already count the number of states correctly (due to (2π¯h)3N and N !), at this point the main modification is the discrete sum over energy
52
levels, instead of integrals over phase space. Degeneration of levels should be accounted for. Thus, Γ(E) =
X
gi
(196)
Ei ≤E
The rest is the same. The Boltzmann definition of entropy, eq.(189) is the best, provided ∆ exceeds splitting of the energy levels (and if there is no splitting ∆ will not matter at all!). Note that the lowest energy level, E0 is always non-degenerate, g0 = 1. Thus Γ (E0 ) = 1, with S = 0. This corresponds to the 3d Law. HW: Consider one quantum particle in a cubic box. (a) For a few lowest levels find the degeneracies gi . (b) Calculate and plot Γ for the first few levels and entropy in two definitions (via ∆Γ = gi and via Γ; (c) compare with classical; (d) for the few lowest levels remove the degeneracies by making the box slightly non-cubic, a × b × c with, say b = 0.95a and c = 1.05a. Problem 1.19 by McQ can be useful. Examine the structure of energy levels (make a picture). (e -optional) Examine the limit of large E; Fig. 1-1 by McQ can be useful.
E. 1.
Example: ideal gas Number of states and entropy
We discussed Γ in the Quantum Mechanics introduction (in class). Main points: integra√ tion over d q gives V N . Scaling each momentum with 2mE expresses the result through a volume of a unit n-dimensional sphere (with n = 3N ) Cn =
π n/2 (n/2)!
(197)
(I avoid using the gamma-function here not to confuse with the number of states). One has "
V C3N 3/2 Γ= 3 (2mE) N ! (2π¯h)
#N
(198)
Using the Stirling approximations (leading term): N! '
√
N 2πN e µ
¶N
1 2eπ , Cn ' √ πn n µ
One obtains 1 √ Γ(E) = πN 6
"Ã
2πe5/2 2/3 2m² v 3 (2π¯h)2 !
53
¶n/2
#3N/2
(199)
with v = V /N , ² = E/N Or, with
√ λdB = 2π¯h/ 2m²
(200)
the de Broglie wavelenth Ã
1 2πe5/2 √ Γ(E) = 3 πN 6
One requires
!1/2
3N
v 1/3 λdB
(201)
> 0.3λdB v 1/3 ∼ to ensure the [. . .] > 1 (otherwise, quantum statistics is required; strictly speaking a strong inequality, v 1/3 À λdB is needed to justify classical statistics). One has for the density of states 3N Γ(E) 2E
Ω(E) = Entropy:
" µ
¶3/2 #
" µ
¶3/2 #
m² S = N kB ln v h ¯2
(202)
+ N kB × const − ln N − . . .
(203)
or per particle, s = S/N m² s = kB ln v h ¯2 2.
Ã
ln N + kB × const + O N
!
(204)
Construction of thermodynamics
One needs E(S, V ) (or, ²(s, v)). Thus, ²(s, v) ∝ v −2/3 exp Then, T =
!
Ã
∂ ² ∂s
Ã
∂ ² ∂T
(205)
2² 3kB
(206)
3 = kB 2
(207)
= V
2s 3kB
and (per particle) cV = Then, equation of state:
Ã
∂ ² P =− ∂v
!
!
V
= s
54
2² = kB T /v 3v
(208)
F.
Example: Entropic elasticity
This is a simplest model of elasticity of rubber - see R. Kubo and references to texts on polymer therein. For us, the interesting part is that there is no energy in the system, but there is length-dependent degeneration of states which leads to non-zero tension. [onedimensional analog of pressure]. The microcanonical ensemble is at its hight.
1.
The model and Γ
A chain consists of n links with length a each. Let is calculate Γ as a function of the length of the chain, x = a (n+ − n− ) with n+,− being right- and left-oriented links, respectively. The problem is identical to the one we had in one-dimensional random walk: Γ=
Cnn+
nn ∼ n+ n− n+ n−
Thus, S/nkB = ln n − 2.
n+ n− ln n+ − ln n− n n
Thermodynamics
F = 0 − T S = const + kB T (n+ ln n+ + n− ln n− ) With
Ã
∂ f= − F ∂x
!
= T
d n+ n+ kB T ln dx n−
The first term leads to Hook’s law (will be completed in class).
G.
Strong and weak sides of microcanonical ensemble
to be discussed in class
55
(209)
1.
Energy dependence of the probability density
Here we justify in more detail eq. (182). Using the additivity of energy, on the one hand, and multiplicativity of the probability density for independent subsystems on the other, one has ρ(E)ρ(²) = ρ(E + ²) Here ² can be anything, but let it be small. Then ρ(E) ρ(0) + ²ρ0 (0) = ρ(E) + ²ρ0 (E) ¡
¢
From here, ρ(0) = 1 , ρ0 (E) = ρ(E)ρ0 (0) and ρ(E) ∝ exp (−βE) , β = −ρ0 (0) This is equivalent to what we had in logarithmic form
56
(210)
Dr. Vitaly A. Shneidman, Phys 641, 8th Lecture Extra reading:
Ch. 7 by McQ.
VIII.
CANONICAL ENSEMBLE
Consider a subsystem which is part of a large system (a ”thermostat”). Then the prob-
ability wn to find this subsystem in a state with energy En is given by
1 −En/kB T X , Q = n e−En/kB T (211) wn = e Q with Q known as the partition function. Similarly, in classical statistics (with corrected d Γ) ρ(p, q) =
Z 1 −E(p,q)/kB T e , Q = e−E(p,q)/kB T d Γ Q
(212)
or in terms of energy and density of states d Γ/d E (for which we also used Ω in the previous lecture) Z dΓ d Γ 1 −E/kB T e , Q = e−E/kB T dE ρ(E) = dE Q dE
(213)
In a sense, the last classical expression is the most instructive since it shows the presence of dΓ dE
and a rapidly decaying exponential e−E/kB T . The competition ¯ which corresponds to the actual average of those leads to a sharp maximum near some E, a rapidly growing term,
energy of a subsystem (see below) -se Fig. 10.
A.
Justification(s) of the Gibbs Distribution (GD)
There are several ways to introduce the GD. For example, one could theat the entire large system as a microcanonical ensemble with a large energy E and then consider the probability to get a certain energy En ¿ E in the subsystem. This is proportional to d Γ/d E and the latter is proportional to the exponential of the entropy over k B . When expanding S (E − En ) in powers of En one gets the inverse temperature and the GD. This is done in LL or in Huang, and we will use something similar later to study fluctuations. 57
12 10 8 6 4 2 0.5
1
1.5
2
2.5
FIG. 10: Distribution of energies, d Γ/d E exp (−E/kB T ) in the Gibbs ensemble as a function of reduced energy E/(N kB T ) for an ideal gas for different values of the number of particles, N . From down up (blue lines): N = 4, 16, 64, 256, 1024. A sharp maximum appears due to competition of an increasing density of energy levels and a decaying exponential. [Imagine what happens for N ∼ 1023 (!)]. The red box indicates the corresponding microcanonical distribution for the largest N. HW: Read the corresponding description in LL or in Huang
The other way we will describe does not rely on the microcanonical distribution, but will allow to highlight once again the condition of maximum entropy (and is closer to McQ). Consider µZ
¶
−ρ ln ρ + αρ − βEρ d Γ = max
(214)
with α, β the Lagrange multipliers. From here ρ=
1 exp (−βE) Q
with a constant Q given by normalization. For entropy one has S = −kB hln ρi = kB β E¯ + kB ln Q Compare this to S=
(215)
E¯ − F T
Thus, β= 58
1 kB T
(216)
and
F = −kB T ln Q
(217)
and the canonical distribution can be written as
F − En wn = exp kB T
(218)
The last form is the most useful in practice and is also valid for classical statistics with wn → ρ(p, q), En → E(p, q). HW: (a) Show that ¯ = − ∂ ln Q = −T 2 ∂ F E ∂β ∂T T
(219)
¯ (b) Calculate Q for classical ideal gas; find F (V, T ), E(V, T ) and the equation of state
B.
Maxwell distribution (MD)
For classical energy X1
mvi2 + U (~q)
(220)
1 −U/kB T Y −p2i /2mkB T e e Q i
(221)
E=
i
one has ρ (~p, ~q) =
2
Thus, can write a probability distribution for an individual momentum p2 d wp~ = a exp − 2mkB T Ã
!
d px d py d pz
(222)
with the constant a given by the normalization condition Z
d wp~ = 1
HW: Find a; note that p2 = p2x + p2y + p2z , leading to a product of three integrals
Similarly, for the velocity distribution m d w~v = 2πkB T µ
¶3/2
mv 2 exp − 2kB T Ã
59
!
d vx d vy d vz
(223)
Note: MD is applicable to very non-ideal systems (!), as long as motion is classical. In practice, MD mostly describes translational motion of molecules (since motion of atoms in a molecule, as well as rotations are often quantum). Since the exponential depends only on v 2 , convenient to re-write in spherical coordinates m d w~v = 2πkB T µ
¶3/2
mv 2 exp − 2kB T Ã
!
v 2 sin θd θd φd v
(224)
Or, distribution in v m d wv = 4π 2πkB T µ
¶3/2
mv 2 exp − 2kB T Ã
!
v2d v
(225)
HW: Check normalization HW: (a) Plot the MD for neutrons at 300o K and 3000o K. (b) write the normalized distribution in energies (c) Find the mean energy and the mean square energy
C.
Equipartition of energy
in class
IX.
GRAND CANONICAL ENSEMBLE
Let us relax the condition N = const and require only a fixed ¯≡ N
X
N
N
Z
ρN d Γ N
The earlier conditions remain the same, only now summation in N is added: XZ N
ρN d Γ N = 1 ,
XZ
EN (p, q)ρN d ΓN = E¯
N
Look for (omit sub-index N in ρ, E, d Γ) XZ N
ρN d Γ (−ρ ln ρ + αρ − βEρ + γN ρ) d Γ = max
with α, β, γ being Lagrange multipliers. Thus, − ln ρ − 1 + α − βE + γN = 0 60
(226)
or ρ ∝ exp (−βE + γN ) With normalization, 1 exp (−βE + γN ) Ξ
ρN (p, q) = Ξ=
X
eγN
N
Z
(227)
e−βE d ΓN
(228)
Consider the entropy ¯ + kB ln Ξ S = −kB hln ρi = kB β E¯ − kB γ N and compare this with ¯ −V P (V, T, µ) = E¯ − T S − µN (at this point I avoid using Ω = −P V for the potemtial on the left). Thus, the same β = 1/kB T and γ= with P V = kB T ln Ξ = kB T ln
µ kB T
Ã
X
e
(229)
µN/kB T
N
X
e
n
−EnN /kB T
!
(230)
for quantum (and sum over n replaced by integral over d Γ for classical). HW: Calculate Ξ for a classical gas
With z = eµ/kB T
(231)
one has Ξ=
X
z N QN (V, T )
(232)
N
Next, ¯= N
∂ ∂ ln Ξ ≡ kB T ln Ξ ∂ ln z ∂µ
(233)
and ∂ ln Ξ E¯ = − ∂β
61
(234)
distribution 0.6 0.5 0.4 0.3 0.2 0.1 -4 -2 distribution
2
4
dimensionless vx, absV
0.4 0.3 0.2 0.1 1 distribution 0.4
2
3
4
5
6
reduced energy
0.3 0.2 0.1 5
10
15
20 velocities, km per s
p
FIG. 11: Maxwell distribution in various coordinates. Upper figure: distribution of v x / kB T /m p
(blue) and v/ kB T /m (red). Middle figure - distribution of dimensionless energies, E/kB T . Lower figure: distribution of thermal neutrons over absolute velocities for T = 300 K (blue) and T = 3000 K (red). A.
Absence of a phase transition for a finite V
in class HW: MD (solutions). For normalization and constants - see maxwell.nb which also derives different versions of MD and averages, and has graphics. Distribution of energies: wE (E) = √
√ 2 E π (kB T )
3/2
e−E/kB T
∞
3 kB T 2 0 D E Z ∞ 15 2 2 E 2 wE (E)d E = k T E2 = 4 B 0 D E ¡ ¢ 2 2 ¯ 2 = 3 kB E2 − E T 2 ¯= E
Z
(235)
EwE (E)d E =
62
(236) (237)
Dr. Vitaly A. Shneidman, Phys 641, 9th Lecture
Part III
Ideal systems X.
CLASSICAL GAS (WITH QUANTUM MOLEQULES) A.
Monatomic gas
Reading: Ch. 5-1 in McQ. Notations: kB → k and often β = 1/kT will be used. No bar is used over E or N - their thrmodynamic values are always implied. We considered before almost all relation as examples. Here is a summary.
1.
Canonical
For Boltzmann statistics: QN (V, T ) = q=
Ã
2πmkT h2
1 N q N! !3/2
V
(238)
Free energy (A in McQ): Ã
qe 2πmkT F = −kT ln Q = −N kT ln = −N kT ln N h2 P =−
!3/2
V e N
(239)
∂F = N kT /V ∂V
(equation of state) E=−
∂ 3 ln Q = N kT ∂β 2
Entropy: S = (E − F )/T = . . . HW: make sure you can reproduce all the above relations in detail
63
(240)
2.
Grand canonical
This is an overkill for ideal classical gas since everything can be obtained from a canonical distribution. Nevertheless Ξ(V, T, µ) =
QN eβµN = eqz , z ≡ eµβ
(241)
∂ ln Ξ = qz ∂ ln z
(242)
X N
N= Thus,
µ = kT ln (of course, could get as
∂ F ∂N
N q
(243)
- HW: show that ).
HW: derive an analog of all above relations for a two-dimensional gas, with N particles over an area σ
3.
Electronic partition function
2 ˆ = pn + H ˆ el H 2m
q = qtrans qel , qel =
X
gi e−β²i
(244)
i
with gi being the degeneration (ωi in McQ). Meaning - only a few lowest levels - otherwise ionization(!). One has (with subindex ”0” corresponding to above translational part): F = F0 − N kT ln qel h
E = E0 + (N/qel ) g1 (² − ²0 ) e−β(²−²0 ) + . . . HW: derive this
P = N kT /V (same, since qel does not depend on V ) 64
(245) i
(246)
µ = µ0 − kT ln qel
(247)
S = S0 − . . .
(248)
HW: Caculate corrections to E/N , F/N , S/N , µ for F from Table 5-1 in McQ at T=300K. (use only one lower level). (optional) plot those as a function of T
B.
Diatomic gas
Reading: Ch. 6 in McQ. Notations: we use h ¯ ω instead of hν in McQ. We will still use E0 , F0 , etc. for the translational part (same formulas with m → M ). 2 ˆ int ˆ = pCM + H H 2M
(249)
where M = m1 + m2 is the mass of molecule. Thus, q = qtrans qint
(250)
ˆ int ' H ˆ el + H ˆ rot + H ˆ vib H
(251)
qint = qel qvib qrot
(252)
Next,
and
Ignore the electronic excitations for simple molecules - they are of the order of eV À kT room , so that higher leves are not excited. Thus, E ' E0 + Evib + Erot
(253)
q
(254)
and similarly for F , etc.
1.
Vibrational partition function
Classical: ω=
65
k 0 /m0
with k 0 being the force constant and m0 = m1 m2 / (m1 + m2 )
(255)
qvib,clas = kT /¯hω
(256)
the reduced mass. Then,
HW: show this
Thus, Evib,clas = N kT (”equipartition theorem”) and Fvib,clas = N kT ln (¯hω/kT ) HW: show this
Quantum ²n − ²0 = n¯hω , ²0 = h ¯ ω/2
(257)
Thus, qvib = with Θvib =
1 1−
(258)
e−Θvib /T
h ¯ω , [Θvib ] =o K k
(259)
see Table 6-1 for different molecules. Then, ³
Fvib = N kT ln 1 − e−Θvib /T ³
´
Evib = N kΘvib / eΘvib /T − 1 and Cvib = N k (Θvib /T )2 HW: Check the above 2 relations
´
eΘvib /T 2 (eΘvib /T − 1)
(260) (261)
(262)
HW: Explore the classical limit for T À Θvib ; there will be a
constant difference with the strictly classical expression since we calculate the energy from ² 0 in the quantum case.
66
cvib 1 0.8 0.6 0.4 0.2 0.5
1
1.5
2
2.5
3
reduced temperature
FIG. 12: Vibrational contribution to specific heat, Cvib /N k as a function of T /Θvib 2.
Rotational partition function (different nuclei)
Classical: ²r =
qr,clas
(pθ )2 (pφ )2 + 2I 2I sin2 θ
T 1 Z Z exp {−β²r (pθ , pφ , θ)} d pθ d pφ d θd φ = = 2 (2π¯h) Θr
with
h ¯2 Θr ≡ 2Ik
(263)
(264)
HW: Show this
See Table 6-1 for Θr . The rest is the same as for vibrations, with Θvib → Θr ; in particular the equipartition theorem holds, and Cr,clas = N kT
(265)
²r,J = kΘr J(J + 1) , J = 0, 1, . . .
(266)
zr = e−Θr /T
(267)
Quantum:
Notations:
Then, qr (z) =
∞ X
(2J + 1)z J(J+1)
(268)
J=0
Limits: T → 0, z → 0 (much faster !).
qr = 1 + 3z 2 + 5z 6 + . . . 67
(269)
4
10 8
3
6 2 4 1 2
2
4
6
8
10
12
5
10
15
20
25
30
FIG. 13: Relative populations of rotational levels at lower (left) and higher (right) temperatures [in both cases, however, temperature is noticeably larger than Θr - otherwise only a handfull of levels would be visible] . On the x-axes is the number J.
T → ∞, z → 1, series almost diverges.
P
→
J
R
d J. Goes to classical limit.
HW: Check this. Hint: replace summation by integration.
3.
Rotational specific heat
Generally, needs to be considered numerically (see the Mathematica notebook vibrot.nb for details). The major contribution to the sum for qr (z) (see also Fig. 13) comes from Jmax
1 T '√ − 1/2 = 2Θr −2 ln z µ
¶1/2
− 1/2
The number of contributin terms around Jmax is also of the order of Jmax [this can be found from the second derivative, or from the integral representation which is just an exponential of −J(J + 1)Θr /T ]. About 5Jmax will be sufficient for an accurate qr with an error of about 1%. To get the specific heat we need a symbolic derivative. Note that: ∂ ∂ Θr ∂ ∂ = −kΘr z , = 2z ∂β ∂z ∂T T ∂z (which is useful when calculating the energy and the specific heat, respectively). When taking a second derivative of qr , a typical term is J 2 (1 + J)2 (1 + 2J)z J(J+1) which has a slightly larger maximum at Jc,max
3T ≈ 2Θ µ
68
¶1/2
< 1.5Θr ), then Jc,max is not If the numerical goal is to reprodice Fig. 3 in LL (with T ∼
large, and taking 15 terms will be more then enough. The expected error, of the order of
the first neglected term at the highest T , is virtually zero (about 10−60 ); on the other hand, > (2/3)Θr × (15)2 ∼ 150Θr . the 15-term approximation will become inapplicable for T ∼ The specific heat is shown by a blue line in Fig. 14. To get the asymptotes (red lines) we do the following. At high T the Euler-MacLaurin formula is used which approximates the infinite sum by an integral, and provides further corrections. One has (see McQ): qr '
1 Θr T + + + . . . , T À Θr Θr 3 T
The first term is classical; two extra terms are needed for the second derivative in C r . Thus, in addition to McQ we get for energy: Θ2 E = N k T − Θr /3 − r + . . . 45T Ã
and
Θ2r + ... Cr = N k 1 + 45T 2 Ã
!
!
, T À Θr
(270)
(271)
This indicates an overshoot over the classical value, but otherwise is not an accurate approximation, as seen from Fig. 14. For T ¿ Θr we use eq.(269). Then, ³
E = 6N kΘr e−2Θr /T 1 − 3e−2Θr /T and Cr = 12N k
µ
Θr T
¶2
³
e−2Θr /T 1 − 6e−2Θr /T
´
´
, T ¿ Θr
When applicable, this approximation is very accurate - see Fig. 14.
69
(272)
specific heat 1.2 1 0.8 0.6 0.4 0.2
0.2
0.4
0.6
0.8
1
1.2
1.4
t
FIG. 14: Reduced rotational specific heat Cr /N k as a function of reduced temperature t = T /Θr (blue line). Red lines are the low- and the high-temperature asymptotes, eqs. (272) and (271), respectively. Dr. Vitaly A. Shneidman, Phys 641, 10th Lecture
XI. A.
FERMI-DIRAC (FD) AND BOSE-EINSTEIN (BE) STATISTICS Introduction: the simplest example
in class
B.
Grand canonical partition function
Reading: McQ, Ch.4
1.
Occupation numbers
E=
X
²k nk , N =
k
QN =
X
nk
k
X0 Y³
{nk }
e−β²k
k
(where prime means that the sum is restricted by 70
P
´n k
nk = N ).
(273)
FD: nk = 0, 1
(274)
nk = 0, 1, 2, . . .
(275)
BE:
2.
Evaluation of Ξ
Ξ=
eβµN QN =
X
X X0 Y³ N {nk }
N
0r, Ξ=
max nk ³ Y X k
eβµ e−β²k
k
eβ(µ−²k )
´n k
(276)
´n k
(277)
nk =0
one has, respectively With the FD and BE restrictions on nmax K nmax k
X ³
eβ(µ−²k )
nk =0
´n k
³
= 1 ± eβ(µ−²k )
´±1
Thus, with upper sign for FD and lower for BE Ξ≡ 3.
Y
Ξk =
Y³ k
k
1 ± eβ(µ−²k )
´±1
(278)
Average occupation numbers
n ¯ k = kB T
Ã
∂ ln Ξk ∂µ
!
= V,T
1 eβ(²k −µ)
(279)
±1
Then, N=
X
n ¯k
(280)
n ¯ k ²k
(281)
k
E=
X k
Ω=
X k
³
Ωk , Ωk = −kT ln Ξk = ∓kT ln 1 ± eβ(µ−²k )
(and pV = −Ω). Boltzmann limit: for ²k − µ À kT n ¯ k = eβ(µ−²k ) ¿ 1 71
´
(282)
C.
Non-equilibrium quantum gas
Reading: class notes or LL, Ch.IV-55. Fermi: for each level j with degeneracy Gj there are N
∆Γj = CGjj ways to distribute Nj particles. Using the Stirling formula for large Gj , nj one obtains for the entropy S=k
X j
ln ∆Γj ' k
X j
{Gj ln Gj − Nj ln Nj − (Gj − Nj ) ln (Gj − Nj )}
HW: check this
With n ¯ j ≡ Nj /Gj ≤ 1 (forFermi) one has S = −k
X j
Gj {¯ nj ln n ¯ j + (1 − n ¯ j ) ln (1 − n ¯ j )}
(283)
Using method Lagrange multipliers, find maximum of S with constrains X
Gj n ¯j = N ,
X
²j G j n ¯j = E
j
j
this gives ³
n ¯ j = 1/ eα+β²j + 1
´
HW: show that
For Bose N
∆Γj = CGjj+Nj −1 (check with the elementary example above). The rest is similar, giving S = −k
X j
Gj {¯ nj ln n ¯ j − (1 + n ¯ j ) ln (1 + n ¯ j )}
HW: show this
Maximization with the same constrains gives ³
n ¯ j = 1/ eα+β²j − 1 HW: show that
72
´
(284)
D.
Role of spin and density of levels
dΓ = gdV
dpx dpy dpz h3
(285)
with g = 2s + 1
(286)
s-spin dN (~p) = n ¯ (²(p)) dΓ dNp =
with
4πgV p2 dp n ¯ (²) h3
(287)
√ ²d² dNp m3/2 √ n(²) ≡ V R β(²−µ) dN² = d² = gV √ 2 3 ²d²¯ d² e ±1 2π h ¯
(288)
m3/2 R ≡ g√ 2 3 2π h ¯
(289)
HW: Show that this tends to Maxwell distribution in the classical limit. Hint: consider large negative µ and use the classical expression for µ = µBol - see below.
E.
Evaluation of chemical potential
The total number of particles is √ gm3/2 Z ∞ ²d² N =√ 2 3 β(²−µ) V ±1 2π h ¯ 0 e
(290)
This is an implicit equation for the chemical potential as a function of density. With
one can write
z = ∓eβµ
(291)
√ N g(mkT )3/2 Z ∞ xdx =∓ √ 2 3 1 x V e −1 0 2π h ¯ z
(292)
The integral can be expressed through a special function, Polylogarithm : ¯√ ¯Z ¯ ¯ ¯ π ¯ ∞ √xdx ¯ ¯ ¯ ¯ ¯ ¯ ¯ = ¯ ¯ ν(z) ≡ ¯¯ Li (z) 3/2 1 x ¯ ¯ ¯ 2 e −1 0 z
73
(293)
reduced density
2.5 2 FD
1.5 1 0.5
-10
-8
-6
-4
BE
-2
z
FIG. 15: Reduced density ν(z) (see caption to next figure) as a function of z = ∓e βµ , as follows from eq. (294). Negative z correspond to Fermi-Dirac (FD) and positive z to Bose-Einstein (BE) statistics, respectively. For a given density (horizontal red line) intersection with the curve determines the value of z and thus the chemical potential. Note that the BE curve stops at z = 1 at the value of 2.31516, and for higher density there are no solutions for the BE statistics (upper red line). This corresponds to the Bose-Einstein condensation.
(see http://mathworld.wolfram.com/Polylogarithm.html or the function ”PolyLog[n, z]” in Mathematica). The equation for the chemical potential can be written as √ g(mkT )3/2 π N =∓ √ 2 3 Li3/2 (z) ≡ R (kT )3/2 ν(z) V 2 2π h ¯
(294)
and is shown in Fig. 15. Results of the actual (numerical) solution of this equation using Mathematica are shown in Fig. 16.
F.
Energy
E = RV with
Z
∞ 0
²3/2 d² = RV (kT )5/2 ν² (z) β(²−µ) e ±1
¯Z ¯ ¯ ¯ ¯ ∞ dx x3/2 ¯ ¯3√ ¯ ¯ ¯ ¯ ν² (z) ≡ ¯¯ πLi5/2 (z)¯¯ ¯=¯ 1 x ¯ 4 e −1 0 z
HW: check this - Mathematica can be helpful
74
(295)
(296)
reduced mu 2 FD Bolt
1 0.5 -1
1
1.5
2
scaled density
2.5
BE
-2 -3 FIG. 16: Reduced chemical potential µ/kT as a function of the scaled density (N/V ) · n√ h io 2π 2 ¯h3 / g(mkT )3/2 (which is of the order of (N/V ) · Λ3 /g with the thermal deBroglie wavelength Λ and g = 2s + 1 the degeneration associated with the spin s). In the BE case the chemical
potential is strictly zero if density exceeds the critical value. The dashed line corresponds to Boltzmann statistics and coinsides with both quantum statistics for large negative µ; extension of the Boltzmann curve into the region of large positive µ is formal since any real particles will act as quantum here. G.
Equation of state
Ω ≡ −P V = ∓kT RV
Z
∞ 0
√
³
²d² ln 1 ± eβ(µ−²)
´
(297)
Integrating by parts, get 2 PV = E 3
(298)
HW: show that
This is just like classical(!). Correction to classical ideal gas equation P V /N kT = 2ν² (z)/3ν(z)
(299)
z P V /N kT ≈ 1 − √ 4 2
(300)
Expand for small z
HW: check this - Mathematica can be useful
For the first quantum correction the Boltzmann approximation can be sufficient when evaluating z. 75
With
P µBol (P, T ) = kT ln g(kT )5/2
one has
Ã
2π¯h2 m
!3/2
N = kT ln gV
Ã
2π¯h2 mkT
!3/2
(301)
z ≈ zBol = ∓eµBol /kT
(302)
π 3/2 Nh ¯3 PV =1± + ... N kT 2g V (mkT )3/2
(303)
Note that pressure is increased for Fermi and reduced for Bose. Also, correction is of the order N Λ3 /V g, and should be small. Note: McQ defines 3
Λ = V /qtrans =
µ
h √ 2πmkT
¶3
This is fine, but then he writes (p. 83) ”the condition of applicability of ... Boltzmann statistics is ... Λ3 /V ¿ 1”. This is wrong: the condition is much more strict, Λ3 /V ¿ 1/N . Next, he does not include the g-factor when evaluating corrections to the classical ideal gas law. For bosons with zero spin this is fine, and eq. (10-48) is equivalent to our eq. (303) with g = 1. For fermions, however, the corresponding eq. (10-15) is wrong since there are no fermions with g = 1 (the smallest g = 2 is for electrons). To be fair, McQ does introduce the missing 2 when considering a strongly degenerate electron gas, but he has to start all over to get things right
H.
Ultra-relativistic gas
The average occupation numbers for a given energy level n ¯ (²), as in eq.(279), remain unchanged. The quantum analog of the Maxwell distribution in momentum space, eq.(287) also is the same. [recall that deBroglie wave h/p which is used to count states is also valid in the relativistic case (!)]. The difference comes when one considers energy ² = cp (instead of p2 /2m). This gives 4πg V ²2 d²¯ n(²) (ch)3
(304)
4πg Z ∞ 3 ²dN² = V ² d²¯ n(²) 0 (ch)3
(305)
dN² = The total energy E=
Z
76
Similarly, for Ω one has Z ∞ ´ ³ 1 gV 2 β(µ−²) =− E ² d² ln 1 ± e Ω=∓ 2 3 3 2π (¯hc) 0
(306)
Thus, P =
77
1E 3V
(307)
Dr. Vitaly A. Shneidman, Phys 641, 11th Lecture
XII.
BOSE-EINSTEIN CONDENSATION
What happens if eq.(294) has no solutions, as in the right part of Fig. 15? In other words,
√ π N 1 3/2 (kT ) = ν(1) = ζ(3/2) ≈ 2.31516 V R 2
(308)
determines the highest denstity (or the lowest temperature) when the solution is still possible. [ ζ(x) is the Riemann zeta-function, which appears as the limit of PolyLogarithm as z → 1: ζ(s) = Lis (1) =
∞ X 1 i=1
is
, s>1
(see any good math reference, e.g. Abramowitz and Stegun or the mathworld)]. Since all math was done correctly, the only way one could stray was in replacing the sum over discrete energy levels by an integral. Care should be taken since the average occupation number, n ¯ ² tends to infinity (!) for ² → 0 and µ → 0. Indeed, the lowest level can hold a finite number of particles in order to ensure that eq.(308) (with N reduced by this number) has a solution. [Note: it is possible to isolate the contribution of the lowest level explicitly, obtaining a correction proportional to 1/V - see McQ or Huang. Then all dependences are smooth accross the transition. However, in the limit V → ∞, which is of the main interest, one anyway gets cusped curves, and we shall consider this limit from the start]. If the density v = N/V is fixed, then kTc =
Ã
2 √ πRvζ(3/2)
!2/3
'
0.5714 (Rv)2/3
(309)
HW: Check the dimension HW: Estimate Tc for He - use Table 4-1 to find N/V for indicated T = 4K, and assume that this density is not changing with T
78
reduced No 1 0.8 0.6 0.4 0.2
0.2
0.4
0.6
0.8
1
t
FIG. 17: Reduced number of particles in the ground state N0 /N as a function of reduced temperature t = T /Tc ≤ 1. For higher temperature this number is strictly zero (in the limit of infinite volume). A. 1.
Properties of the condensed phase Number of particles
Note the energy distribution at ² > 0 is still correct at T < Tc . Thus, N²>0 = RV (mkT )
3/2
T ν(1) = N Tc µ
¶3/2
, T ≤ Tc
(310)
The ”missing” particles are in the ground state N0 = N − N²>0
"
T =N 1− Tc µ
¶3/2 #
(311)
see Fig. 17. Remarkably, the ground state which can bearly hold two electrons, can hold an Avogadro number of bosons!
2.
Energy
E = RV (kT )
5/2
ν² (1) T ν² (1) = kT N²>0 ' 0.77kT N ν(1) TC µ
¶3/2
= 0.128
gm3/2 (kT )5/2 V h ¯3
(312)
HW: check the numbers
Note: energy depends only on T, V , but not N - ”extra” particles go to the ground state and do not contribute. HW: show that for µ = 0 one has F = Ω = −2E/3
79
3.
Specific heat, T < Tc
Cv =
4.
Ã
∂ E ∂T
!
5E ∼ T 3/2 , T ≤ Tc 3T
(313)
2E 2 gm3/2 = = 0.128 3 (kT )5/2 3V 3 h ¯
(314)
= V =const
Pressure
!
Ã
∂ Ω P =− ∂V
T,µ=const
Note: no dependence on volume! - horizontal isotherms, similar to two-phase mixture.
B.
Specific heat at all T
Assuming N, V = const (not to write partial derivatives) CV = or Cv =
3kRV
dE d = RV (kT )5/2 ν² [z(T )] dT dT
(315)
" # √ dz π(kT )3/2 5Li5/2 (z)z + 2T Li3/2 (z) 8z dT
In order to eliminate dz/dT we differentiate the equation for the (constant) density n ≡ N/V ³
0 = dn = d R(kT )3/2 ν[z(T )]
´
this gives: dz 3z Li3/2 (z) =− dT 2T Li1/2 (z)
(316)
see bec.nb. (Note that Li1/2 (z) diverges for z → 1) Thus, Cv =
3kRV
√
Li23/2 (z) π(kT )3/2 5Li5/2 (z) − 3 8 Li1/2 (z) "
#
(317)
Or, if we divide by the N k Cv 15 Li5/2 (z) 9 Li3/2 (z) = − Nk 4 Li5/2 (z) 4 Li1/2 (z)
(318)
This is shown in Fig. 18 together with the sub-Tc part. Note that Cv is continuous at the transition, although there is a cusp. At high T it approaches the classical limit of (3/2)N k. 80
c
1.5
1
0.5
1
2
3
4
t
FIG. 18: Reduced specific heat c = Cv /(kN ) as a function of reduced temperature t = T /Tc . For t → 0 it is a power law: c ∼ t3/2 .
Note: both McQ and Huang have the above formula for Cv /N k. LL do not give it, but they derive analytically the change in slope near Tc . McQ suggests to derive Cv as a HW problem, which appears to be too hard
1.
Pressure at all T and isotherms in parameric representation
One has 2 P = R(kT )5/2 ν² (z) 3
(319)
Similarly, V =
N R(kT )3/2 ν(z)
(320)
R is a constant - see eq. (289). In principle, one needs to eliminate z = z(V, T ) (N -fixed) from the second equation, and thus obtain the equation of state P = P (V, T ). Technically, however, this is hard, and it is much better to treat z as a parameter which determines both P and V . The resulting curves are shown in Fig. 19, with z ranging from 0 to 1 (with small z corresponding to large volume and with z = 1 corresponding to the onset of condensation). To obtain the condensation curve we need to get rid of T . Thus Pc3 Vc5
5
=N R
−2
µ
2 ν² (1) 3
which is the black curve in the figure.
81
¶3
ν(1)−5 = const
(321)
P 12 10 8 6 4 2 0.25
0.5
0.75
1
1.25
1.5
1.75
2
v
FIG. 19: Isotherms of ideal Bose-gas in terms of reduced pressure p ∝ P as a function of reduced volume v ∝ V /N . The isotherms correspond to increasing values of T (with equal increments - see pbec.nb for detail). At large v one has the classical ideal gas law pv = const. Transition between the ”gas phase” (blue) and the ”mixed phase” (red) is indicated by the black line with p ∝ v −5/3 (see text). C.
Experimental
Remarkably, could observe BEC in dilute vapors - see http://jilawww.colorado.edu/www/press/images.html for the image, or M.H. Anderson et al., Science, 269, 198 (1995) for the description. They used
87
Rb at a density ∼ 3 · 1012 cm−3 . Which spin to use is an issue, but let us use
s = 0. HW: Estimate Tc . Note: should be VERY small
Note: the actual experiment was performed in a trap, which somewhat changes the conditions of condensation - see www.physics.cornell.edu/sethna/StatMech/ for additional discussion and references
XIII. A.
BLACKBODY RADIATION Planck distribution
One has
82
• ulrarelativistic gas with g = 2 (the latter is a subtle issue, photon has 2 states, not 3 as would a particle with spin s = 1, will discuss in class) • ”particles” are not conserved, thus µ = 0 • energy is h ¯ω The rest is straightforward. n ¯=
1 e¯hω/kT
is the Plank’s formula. Next dNω = h ¯ 3 dN² =
−1
(322)
V ω 2 dω π 2 c3 e¯hω/kT − 1
(323)
Vh ¯ ω 3 dω π 2 c3 e¯hω/kT − 1
(324)
(note that h ¯ enters only in n ¯ , but not in the density of states. Indeed, the latter can be (and was) derived classically at first. HW: Calculate the number of photons,
Density of energy (Plank):
R
dNω
dEω = h ¯ ωdNω =
This is shown in Fig. 20 together with the small- and large-frequency asymptotes: dEω ≈
V kT 2 ω dω , h ¯ ω ¿ kT π 2 c3
(Rayleigh-Jeans, without the restriction - they did not know h ¯) and dEω ≈
Vh ¯ 3 ω dωe−¯hω/kT , h ¯ ω À kT 2 π c3
(Wien, also without the restriction) HW: find dEλ with λ = 2πc/ω; find λmax
B.
Total energy and specific heat
The total energy is given by E=
Z
V (kT )4 Z ∞ w3 dw dEω = 2 π (c¯h)3 0 ew − 1 83
(325)
Ew 1.4 1.2 1 0.8 0.6 0.4 0.2 2
4
6
8
10
w
FIG. 20: The function Ew = w3 / (ew − 1) (blue line) which determines the energy density in the Plank’s formula, with w = ¯hω/kT . Maximum of the blue curve is at ¯hω = 2.822 kT , which determines how the dominant color of radiation changes with temperature (the ”displacement law”). The solid red line is the Rayleigh-Jeans approximation, and the dashed red line (much better) is the Wien’s law.
The integral is π 4 /15. HW: Check this from a handbook or software. Also, calculate the integral approximately, ”by hand”. Hint: see Fig. 20.
Thus, E=4 with σ=
σV 4 T c
(326)
π2k4 60¯h3 c2
(327)
known as the Stefan-Boltzmann constant. HW: Calculate it, with units . Cv =
C.
Ã
∂ E ∂T
!
= 16 V
σV 3 T c
(328)
Radiation pressure
As for any relativistic gas, Ω = −E/3. Thus, P =
84
E 3V
(329)
Dr. Vitaly A. Shneidman, Phys 641, 12th Lecture
Results of the mid-term will be discussed in class
Part IV
Specific heat of solids Reading: McQ, Ch. 11.
XIV. A.
PHONONS Linear monoatomic chain
Let κ be the spring constant; ul - displacement from equilibrium. Force Fl = κ (ul+1 − ul ) − κ (ul − ul−1 ) or m¨ ul = κ (ul+1 + ul−1 − 2ul ) Acoustic wave: a · l → x , (. . .) → a2
∂2u ∂x2
with m/a = ρ, linear density, and aκ = G ρ or
∂2u ∂2u = G ∂t2 ∂x2
2 ∂2u 2∂ u = v ∂t2 ∂x2
with the ”speed of sound” v 2 = G/ρ = a2 κ/m Any u(x − vt) - solution. E.g., exp {i (ωt − kx)} with any ω, k with ω = kv. Return to discreet: look for ul ∝ exp {i (ωt − kx)} , x = la 85
2
1.5
1
0.5
-1.5
-1
-0.5
0.5
1
1.5
2
4
6
2
1.5
1
0.5
-6
-4
-2
FIG. 21:
Then, h
i
h
−mω 2 = κ eika + e−ika − 2 = κ eika/2 − e−ika/2 −
i2
Or, ω(k) = For k → 0, ω ≈ kv. −
s
¯
Ã
!¯
4κ ¯¯ ka ¯¯ ¯ ¯sin m¯ 2 ¯
π π
Brillouin zone. See Fig. 21 (and file phonons.nb). Debye frequency: ωmax ∼
v ∼ 1013 − 1014 Hz a
(infrared)
86
(330)
2.5
2
1.5
1
0.5
-1.5
-1
-0.5
0.5
1
1.5
FIG. 22: D 1 0.8 0.6 0.4 0.2
0.5
1
1.5
FIG. 23: B.
1D- diatomic; optical branches
C.
3D - monoatomic
Three accoustic branches - 1 longitudinal and 2 transverse. the rest will be discussed in class
87
2
t
c
5 4 3 2 1
50
100
150
200
T
FIG. 24: 0.5
1
1.5
2
2.5
3
-2 -4 -6 -8 -10
FIG. 25: APPENDIX A: NOTATIONS 1.
General
Notations are consistent throughout each chapter of the course. Some of the key symbols, e.g. S for entropy are use in the entire text. For dummy variables (e.g., for change of variables in integration) I often use x, y, z, or w, and the variables with only ”local” values will not be listed below. The thermodynamic variables for a large system will be denoted by capital letters, S, E, etc. and used globally. Reduced values per particle will be denoted by a corresponding lower-case letter, s, but will be mostly used locally.
88
2.
Specific symbols
A - not used (used for free energy in McQ, equivalent to our F ) a - lattice constant Cv , Cp - specific heat c - speed of light E - energy F - Helmholtz free energy Fi - forceon a particle (in Chapter on phonons) g - degeneracy associated with the spin h = 2π¯h kB = 1.38 · 10−23 J/o K (occasionally, just k is used when no confusion can happen) ~k - wave vector Li - polylogarithm (special function) m, M - masses of particles N - number of particles n ¯ - occupation number P - pressure p - probability (only in Chapter I) p~ - momentum Q - canonical partition function q - ”partition function” of a single molecule in an ideal gas R - a constant (with dimension!) in quantum gas description, eq. (289) S - entropy T - absolute temperature Tc - temperature of Bose-Einstein condensation u - displacement(in Chapter on phonons) V - volume v - speed of sound (in Chapter on phonons). α - Lagrange multiplier β = 1/kB T Γ - volume of phase space ² - energies of individual particles in ideal gas µ - chemical potential 89
ρ - density; linear density (in Chapter on phonons) κ - elestic constant (in Chapter on phonons) ω - angular frequency
90
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