Physical Science in 13 problems or Shut up and calculate ...witek/Classes/PHY321/Refs/...Contents...
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Physical Science in 13 problems or Shut up and calculate - its not as hard as you think. Chemistry & Physics 400 L. G. Sobotka Departments of Chemistry and Physics Washington University, St Louis, MO, 63130 August 27, 2012
Transcript of Physical Science in 13 problems or Shut up and calculate ...witek/Classes/PHY321/Refs/...Contents...
Physical Science in 13 problems
or
Shut up and calculate -
its not as hard as you think.
Chemistry & Physics 400
L. G. Sobotka
Departments of Chemistry and Physics
Washington University, St Louis, MO, 63130
August 27, 2012
Contents
I Introduction 4
1 Physical Chemistry 0 - Chem. 400 : Physical Science in 12 problems 5
1.1 Who, When and Where . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.2 Idea, content and grading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.3 Syllabus: The 12 problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
2 The Structure of Physical Science 6
2.1 Classical Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
2.2 Quantum Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
2.3 Statistical Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2.4 Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2.5 Kinetics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
II Lecture Notes 10
1 Acid-Base Titration 11
2 Classical Mechanics 12
2.1 Object, DoF and Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2.2 Templates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2.2.1 Stokes’ law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
3 Quantum Mechanics 14
3.1 Object, DoF and Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
3.2 History [1] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
3.3 Bound states in a finite well . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
3.3.1 Bound states in a 3D “rounded” well. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
3.3.2 Bound states in single (SW) and double well (DW) potentials [2] . . . . . . . . . . . . . . . . 16
4 Time Evolving states 17
4.0.3 Time evolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
4.0.4 A superposition of energy eigenstates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
5 Black-body radiation and PS 18
5.1 Phase space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
5.2 Relativity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
5.3 Black-body radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
5.4 An ideal classical (Non-relativistic) gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
1
6 Heat Capacities 20
6.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
6.2 Heat Capacity Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
6.3 ∆( ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
7 Phase changes 22
7.1 Geometric interpretation of and . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
7.2 Example - immiscible liquids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
8 Equilibrium Constants 24
8.1 Partition functions for a simple systems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
8.1.1 Logic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
8.1.2 Translation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
8.1.3 Rotation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
8.1.4 Vibration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
8.1.5 Electronic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
8.1.6 Equal Partition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
8.1.7 Total partition function for a unit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
8.2 Chemical Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
9 Solutions 28
9.1 Multicomponent systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
9.1.1 The variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
9.1.2 The physical meaning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
9.1.3 Back to the ideal gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
9.2 Gibbs-Duhem Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
9.3 Ideal Solutions (IS) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
9.4 Real Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
10 Kinetics-MB 31
10.1 Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
10.2 Boltzmann → . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
11 Kinetics-Master 33
11.1 Objective & Statement of Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
11.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
12 The rise and fall of Easter Islanders. 34
12.1 Predator-prey problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
12.2 Easter-Island problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
13 Causality and Kramers-Kronig relations. 36
13.1 Complex Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
13.2 Dielectric response. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
13.3 Kramers-Kronig Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
14 Trigonometric 38
15 Constants 39
2
III MathCad handouts 40
3
Part I
Introduction
4
Chapter 1
Physical Chemistry 0 - Chem. 400 :
Physical Science in 12 problems
1.1 Who, When and Where
Lecturer: Professor Lee G. Sobotka
Lectures: Tu 2:30-4:00, LS 301 or by appointment,
see below.
Office: 259 Radiochemistry (RC)
Texts: None and All (i.e. the library.)
Office hrs and Help Session: Th. 12:00 (bring
a sack lunch) - 3:00, LS 460/RC 259
I am happy to teach this class independent study
(including in the evenings) so that it can fit into the
schedule of any student.
1.2 Idea, content and grading
Exercises related to general chemistry, classical mechan-
ics, quantum mechanics, statistical mechanics, thermo-
dynamics, and kinetics, will be solved with numerical
software. Each exercise will be accompanied by a lec-
ture, a software template solving a problem and an ex-
ercise that can be solved with the template. The soft-
ware will allow us to focus on, and treat in a transparent
fashion, physical problems without the unworldly ideal-
izations and contrivances found in textbooks.
Each exercise is accompanied with 1-3 pages of
notes. These notes should be read before class, reread af-
ter class and again when the material is discussed in the
core (required) course (if you have not taken it already.)
Each time you will understand a little more. (Learning
is usually like this - each time through a subject you
understand more than the last time.)
The software the instructor will use is MATHCAD.
This package is available on the computers in the LS
computer, in the student area outside my office and can
be purchased for about 100$. If you desire to use another
package (MATLAB, MAPLE or MATHEMATICA), feel
free to do so, the computing logic is directly transferable.
The exercises are designed to build up an arsenal of
tools for solving physical problems. Most will be VERY
EASY and take only a few minutes, others will take sev-
eral hours. These exercises will be graded on a 0, 1, 2,
or 3 point basis. These numbers correspond to: 0 = not
turned in (or turned in without name) or unreadable, 1
= kind of got it, 2 = perfect, 3 = I’m impressed, send
me your soft-copy file because I want to keep it!
1.3 Syllabus: The 12 problems
1. Titrations, equilibrium constants and fitting.
2. CM: HO and terminal velocity
3. QM: Stationary states
4. QM: Single and double well with dynamics
5. PS: Phase Space and Black-body radiation
6. Heat capacities: C( ) ⇒ U(T), H(T), S(T), A(T),
G(T)
7. EOS: Equations of State and the topology of (1 order)
Phase Transitions (PT)
8. Partition Functions and Equilibrium Constants
9. Activity coefficients and the Gibbs — Duhem relation
10. Maxwell-Boltzmann & Transition State logic
11. Coupled DEQs: PS (e+— e−) decay
12. Projects
(a) The rise and fall of the Easter Islanders (kinetics),
(b) Kramers-Kronig relations (causality), or
(c) The TOV equation (the nuclear equation of state
and neutron-star profiles)
5
Chapter 2
The Structure of Physical Science
2.1 Classical Mechanics
Object
To propagate all 6N coordinates in time.
{rp}=0 =⇒ {rp}(Bold denotes vectors).
Degrees of freedom
All 6N positions and momenta.
Rules
The rules are time reversal invariant. With q and p
generalized coordinates in physical space (answers the
“where” question) and momenta (answers the ”where to”
question) respectively, the rules can be stated 3 ways.
1. F = a =•p
2. (•)−
= 0 with = −
3.• = −
and
• =
with = +
Stationary states
The 6N coordinates evolve in time toward stationary
states. There are stable equilibrium configurations
(like the bottom of a potential valley), unstable equi-
librium configurations (pendulum “stuck” on top) and
metastable configurations (local but not absolute min-
ima.)
Why things happen
Application of (classical) forces under constraints such
as E conservation.
Comments
1. We often deal with conservative systems for which
the forces are derivable from a scalar potential .
F = −∇ or in one dimension, = −
2. The sympletic structure of the conjugate variables
representing “where is the system” and “where is
the system going” defines the mechanics.
3. The 6N coordinates evolve in time.
2.2 Quantum Mechanics
Object
The same as CM.
Desire to: “{rp}=0” =⇒ “{rp}”However the best that can be done is propagate
probability amplitudes for each r and p
Degrees of freedom
All 6N positions and momenta.
Rules
The rules are time reversal invariant. The probability
distributions for individual coordinates evolve in time
according to a wave equation.
HΨ = }Ψwhere
Ψ( ) = ()()
If H 6= () then
()H()=}()
H= }
=
H = and () = −} Therefore,
Ψ() = ()−}
6
Stationary states
We often focus on stationary states. These states (so-
lutions of the time independent SE) have well defined
stationary probability distributions. However the sub-
ject is “mechanical” and the rules generate trajectories
for the distributions of all 6N variables.
Why things happen
The same as CM (with some conditions which nature
has seen fit to apply.)
Comments.
QM is a refinement of CM. While is may seem completely
different, it has the same purpose. QM reduces to CM if
the deBroglie wavelengths ( = ) are small compared
to the scale of the objects in the problem.
2.3 Statistical Mechanics
Object
To propagate distributions in coordinate and momentum
space for an ensemble of particles. The 6 coordi-
nates are not followed. While CM and QM are most
often employed when is small, SM is most often em-
ployed when is large. The trajectories of the r0and p0 are not followed, but the distribution (on) saythe “coordinate” () is. Usually this is still too de-
manding so just a few moments of this distribution are
subjected to analysis.
Degrees of freedom
Selected distributions are usually characterized by the
lowest few moments. Higher moments become more and
more important as the system gets smaller.
averages:
=
P
=
() ()
2nd moment:
2 =
P2
=
2 () ()
2 = 2 − 2
Rules
Laws of probability. One must know if; a) the units are
labelled (distinguishable) or not (indistinguishable) and
b) if the laws of QM allow for multiple occupancy of
quantum states (eigen solutions).
Systems evolve toward maximum probability and
the evolution is not time reversal invariant.
This mechanics does not “know” the individual p’s
and r’s.
Stationary states
The stationary states are stationary “distributions”.
Individual p’s and r’s are evolving but the distributions
reach stationary forms defined by statistical arguments.
These distributions are not the same as the QMical dis-
tributions. The fomer are for the ensemble, the latter
for EACH of the 6N variables.
Why things happen
Increase in (QMical) options.
Comments
1. The distributions are defined by the interactions,
constraints and statistical theory.
2. Allows for the generation of macroscopic variables
(P,V, T.....) and their interrelationships.
Examples:
No interactions =⇒ =
consideration of
a) attractive interaction
b) excluded volume
c) uniform density
leads to =⇒ ( + 2)( − ) =
2.4 Thermodynamics
Object
Evolution of selected macroscopic variables.
{ }
Degrees of freedom
Newtonian sector:
Thermal sector:
Chemical sector:
These are conjugate pairs, “force-displacement” in
the first two sectors, “potential-amount” in the third. In
all cases the product has the units of energy. Generally
we will choose one from each sector to be independent
and the other will be dependent. Note that one coordi-
nate from each sector is extensive (proportional to the
amount of “stuff”) while the other is intensive (unival-
ued throughout matter in equilibrium). This means that
7
any function constructed from: sums of products of the
elements in a sector (e.g. = + ) will be
extensive with the units of energy. One often chooses
to divide through by the number of moles to get mo-
lar quantities. Texts often denote the molar quantities
with an overbar or (my preferance) a lower case, ie
= − or = −
Rules
The rules are NOT time reversal invariant. State func-
tions (of the macroscopic variables) evolve to station-
ary states. The characteristics (topology) of the state
functions w.r.t. variations in the selected variables are
defined by the rules of differential calculus.
Stationary states
Thermodynamic stationary states are dynamical states
in terms of the individual p0 and r0 , but the macrovari-ables are time stationary.
Why things happen
Minimization of energy variables or maximization of the
QMical options. The entropy = ln(QMical op-
tions).
Comments
The “Thermo” mindset: “Dear father Newton, we have
come to you on bended knee to present a slight extension
of your science. We hope you find it to your liking.”
1. Newtonian: = F • r = () =
where ≡
∴ = −
+ − compress− + expand
2. Add thermal sector:
= − +
We will use for the thermodynamic internal en-
ergy of a system.
3. Add chemical (association) sector:
= − + +P
This sector can be viewed as a type of “chemical”
generalized force-response. The work is done when
atom “associations” are destroyed or created. How-
ever this sector is so important for the world around
us that it deserves its own separate sector. (The
chemical potential is a “potential”, an imbalance
of which produces a force to relocate or reassociate
atoms.) Other sectors must be added if work in done
by the application of electric or magnetic fields. (If
one is concerned with atmospheric chemistry, work
done against a gravitational field must be added.)
Note that this is a sum of quantities that are exten-
sive, as one (and only one) term in each product is
extensive.
4. , (and the 0), are generalized forces (or po-tentials) corresponding to,
= −( ) = +(
) =
+( )
and the 0 are the generalized responsesto these “forces”.
5. Thermo. laws
(a) Energy is conserved: = d+ d
(b) Every isolated system that is left to itself will
evolve toward a condition of maximum QMical
options.
∆ = ≥ 0
(c) lim→0 = 0 for perfect (crystalline) state. = ln(QMical.options) thus at = 0 there
is but one option in a “perfect” (crystalline)
state.
The famous epitaph on Boltzmann’s tomb-
stone is.
= ln( ). The “” stands for
“Wahrscheinlichkeit”. Look it up. Note that
Boltzmann’s work proceeded the development
of QM.
2.5 Kinetics
Object
To follow “trajectories” in the macroscopic variable
space of thermodynamics. One is most often interested
in the time development of the amount of every species.
Rules
Differential equations (usually coupled) provide a set of
“stepping rules”. These Master equations take the initial
conditions [concentrations] down stream in time to define
conditions later in time. The master equations are of the
form
[]= gain terms - loss terms.
8
Stationary states
If the master equations provide a means for reac-
tions in both directions (¿) a stationary condition interms of the amounts of each material will untimately
be reached. When this “chemical” equilibrium condi-
tion is achieved, the forward and backward rates will be
equal as the gain and loss terms in the Master equations
will be of equal magnitude and opposite in sign.
A pseudo stationary state can be created when the
system is not closed and an influx of new materal (reac-
tants) offset the steady loss from other terms in the Mas-
ter equation. This common situation is called “steady
state”. Most manufacturing (including biological) is
of this sort. (The exception is “batch-mode” process-
ing.) A steady-state condition for intermediates can be
achieved in a closed system for a rather broad window
in time, under conditions that are not uncommon.
Velocity and Energy distributions
The rates of reactions usually depend on the amount of
energy brought into the reaction collision. Therefore the
velocity and energy distributions of the potential reac-
tants are central issues.
Why things happen
Nothing is really at equilibrium. Kinetics is the study of
nature’s response to an “out of equilibrium” condition.
We either force a response by taking a system close to
equilibrium and nudging it away from equilibrium (e.g.
excitation with a laser pulse) or mixing substances for
which a different atomic level association has a lower
(free) energy than the associations in the reactants. The
approach to equilibrium can be “unimolecular” (isolated
decay) or the result of reactions. In either case the indi-
vidual “atomic level” processes are time reveral invari-
ent and must follow all the rules of meachanics (energy
and momentum conservation.) However as the flow to-
ward equilibrium is always towards the condition with
the most microscopic states, the “arrow of time” is im-
pressed on kinetics. Movies of the approach to equilib-
rium are valid, in our universe, only when run forward.
One more point on the ultimate final state - the
equilibrium state - is in order. Essentially all the kinet-
ics we observe is from one unstable or metastable con-
dition to a lower (free) energy condition which is itself
metastable. The utilitarian use of the idea of an equi-
librium state is that the distribution on the variables we
choose to study are stationary. However “stationary”
only means unchanging on a time scale which has tested
the limits of our patience. I will reenforce this point by
two examples.
1. Consider the diamond in my wife’s wedding ring.
At = 1 atm graphite is more stable than dia-
mond. The diamond is UNSTABLE wrt conversion
to graphite. We say the diamond is “metastable”
wrt graphite. Its lifetime will certainly exceed my
wife lifetime (nature’s patience with her) and cer-
tainly her patience with me.
2. The most stable nuclei are 56Fe and 56Ni. Given
enough time, nuclei would interact with one another
and all matter would convert to these nuclei. How-
ever nuclei are essentially isolated from one another
by 1) repelling e− clouds at low energy and 2) thelike-charged nuclei at high energy. Thus once made
(and removed from a nucleosynthesis source - star
cores or SuperNova) the mass number of each nu-
cleus is fixed. On the other hand, “unimolecular”
decays do occur. The primary example here is “”
decay, e.g. 14C →14N++ e− + We can then say
the elements we have on earth are the result of equi-
libration along an “isobar” (constant A=N+Z) but
not along the mass degree of freedom. The mass dis-
tribution is not the absolute equilibrated one, but
as far as we are concerned Prob() is fixed. The
element distribution is (almost always) fixed be-
cause isobaric equilibration has been achieved prior
to running the chemical reaction.
9
Part II
Lecture Notes
10
Chapter 1
Acid-Base Titration
We will consider an “exact” treatment of a titration
of a monoprotic acid with a strong base. In doing so we
will explicitly consider material and charge conserva-
tion. The expression derived is appropriate to fit data
and extract the initial concentration of the acid, its
value and even . All equilibrium constants are func-
tions of temperature. (We will see why in a subsequent
chapter.) To execute this exercise we will introduce the
degree of titration T, a quantity which is initially 0
and goes to 1 at the equivalence point. We will also end
up inverting our independent and dependent variables
to perform the fit. Our logic follows the prescription of-
fered in [1], therein one can also find the generalization
to treat polyprotic acids.
1. The rxn of interest is:
+2¿ 3+ + −; ≡ [3
+][−][]
As usual the ¿ is meant to signify equilibrium.
2. We define ≡ [][]
where,
[] = the initial concentration of the acid and
[]= is the increasing concentration of the base,
assuming that no reaction occurs in the titration
beaker. Note that [] is complicated by the chang-
ing dilution problem as the titration proceeds.
3. Charge conservation leads to
[3+] + [+]
= [−] + [−]
[+]= [+] = [], if the base is strong.
4. Material Conservation lead to
[]= [] + [−]
5. Returning to the degree of titration
=[][]
=
[−]+[−]−[3+]
[]
= [[−]−[3
+]
[]] + { [−]
[]}
= [
[−]−[3+]
[]] + { [−]
[]+[−]}
6. Multiplying the last term by (1[−]1[−] ) we get
= [[−]−[3
+]
[]] + { 1
[]
[−]+1}
7. Switching the order in the denominator of the sec-
ond term, using the -log function “”, and recalling
the definition of ≡ [3+][−]
[] we find
= [10− − 10−
[]] + { 1
1 + 10− }
There is one line of algebra I have left out. The
space below is for you to fill it in. (Write neatly.)
8. I switched the order of the two terms in the denomi-
nator of the {} term because 10− is usually
1 through most of the titration region. (Con-
sider values of this term for a strong acid)
In this derivation, I have left as a variable as
it is only = 14 at T = 298 K.
Exercise 1.0.1 Look up and plot ( ). Determine
the temperature of the titration near the equivalence point
by comparing () to ( )
References
[1] G. Heil and H. Schafer, Chem. Edu. 7, 339 (2002).
11
Chapter 2
Classical Mechanics
2.1 Object, DoF and Rules
Object
The object of the CMics exercise is to propagate all 6N
coordinates (3N space and 3N momenta ) in time.
{qp}= =⇒ {qp}
Degrees of freedom
The DoF’s are all 6N positions and momenta.
Rules
The rules are time reversal invariant and commonly for-
mulated in one of three ways. Newton’s stepping rules
focused on the forces which can generally be derived from
a potential. The forces and be dispensed with if either
the “Lagrangian” ( ≡ kinetic energy - potential energy)or “Hamiltonian” ( ≡ kinetic energy + potential en-
ergy) are known. What you know about the physical
system dictates which approach leads most directly to
the time evolution of the system.
1. F = a =•p
2. (•)−
= 0 with
≡ (•)− ()
3.• = −
and
• =
with ≡ +
Sympletic structure
For conservative systems there is a clear sympletic
structure between the position and momentum.
The position coordinates (r) tell the momentum coordi-
nates how to change (•p) and the momentum coordinates
(p) tell the position coordinates how to change (•r) The
“stepping rules” (sometimes called ”equations of mo-
tion”) for the change in the mated conjugate variables
are defined by 0, , or and the appropriate equa-tions.
Stationary states
The 6N coordinates evolve in time toward stationary
states. There are stable equilibrium configurations (like
the bottom of a potential valley), unstable equilibrium
configurations (pendulum stuck on top) and metastable
configurations (local but not absolute minima.)
Why things happen
Application of (classical) forces under constraints such
as E conservation.
Comments
1. We will deal with conservative systems. In these
cases the forces are derivable from a scalar potential
( ).
F = −∇ or in one dimension, = −
2. The 6N coordinates evolve in time.
3. The dynamics (and the equations that do the imple-
mentation) are (should be) time reversal invariant.
2.2 Templates
1. Focus points
(a) Coupled “conjugate variables” and the
symplectic structure
(b) Trajectories Phase space {rp}
2. HO example - see template and note
(a) Kinetic (T) and potential (V) terms
(b) Stepping rules from
(c) Reversibility (for standard HO)
(d) Classical HO, damped classical HO, and the
Quantum HO.
12
3. Sphere falling in a viscous medium.
The tempate does this problem numerically. Lets
do it analytically and you can compare. Consider a
the forces on the sphere, gravity down, friction up.
(Is it clear why the latter is up?) Keeping in mind
the scaler will be less than 0
= • = − −
• = − −
with =
= − − = −( + )
(+)
= −Which integrates toR ()(=0)
(+)
=R 0−0
1ln( + )|()
(=0)= −
ln(+)
+= −
=(+)
−−
The = ∞ limit, i.e. the terminal velocity, is .....
well lets move on to getting realistic frictional forces
from Stokes law.
2.2.1 Stokes’ law
The frictional force on spherical objects with radius
moving with very small velocity through a medium
with (dynamic) viscosity can be calculated with
Stokes’ law
= (6)
in the limit of smooth “Laminar” (or Poiseville)
flow. The units of “dynamic viscosity” are
[Pa·s=N·=kg/m·s] while the “kinematic viscosity” ≡
has units of [m2s]. The latter is useful when
the ratio of viscous to inertial forces is relevant.
substance Dynamic Viscosity
[Pa·s=kg/m·s]dry air (25) 184 · 0−6water (25) 098 · 10−3Blood (37) 35 · 10−3Ethylene glycol 161 · 10−3Glycerol (25) 1420 · 10−3
Exercise 2.2.1 Both analytically and numberically de-
termine (and plot) v(t) and v(∞) (the terminal velocity)of a Cu ( = 896 g/cm3) “bb” of 1 mm diameter falling
through glycerol. Calculate the terminal Reynolds num-
ber.
The condition for Laminar flow is a small “Reynolds
number” (Re). The Reynolds numbers is a measure of
the relative significance inertial to viscous “forces”.
Re ≡
=
=
where is the velocity, the kinematic viscosity, and L
is the lengh scale of relevance.
13
Chapter 3
Quantum Mechanics
3.1 Object, DoF and Rules
Object
The same as CM.
Desire to: “{qp}=” =⇒ “{qp}”However the best that can be done is propagate proba-
bility amplitudes.
Degrees of freedom
All 6N positions and momenta.
Rules
The rules are time reversal invariant. The probability
distributions for the individual coordinates evolve in time
according to a wave equation.
HΨ = }Ψwhere
Ψ( ) = ()()
If H 6= () then
()H()=}()
H= }
=
H = and () = −} Therefore,
Ψ( ) = ()−}
Stationary states
We often focus on stationary states. These states (so-
lutions of the time independent SE) have well defined
stationary probability distributions. However the sub-
ject is “mechanical” and the rules generate trajectories
for the distributions of all 6N variables.
Why things happen
Basically the same as CM.
Comments.
QM is a refinement of CM. While is may seem completely
different, it has the same purpose. QM reduces to CM if
the deBroglie wavelengths ( = ) are small compared
to the scale of the objects in the problem.
3.2 History [1]
Energy quantization
1. In order to explain the spectrum of blackbody radi-
ation (BBR), M. Planck (1901) added two rules to
classical physics.
(a) The energy of an electromagnetic (EM) field
must be an integral multiple of a fundamental
unit.
(b) The fundamental unit is proportional to the
frequency with a proportionality constant
that bears his name. Hence we speak of a pho-
ton ≡ 1 quantum of EM energy.
2. Einstein (1904) showed that this energy quantiza-
tion could be used at the elementary particle level
to explain the photoelectric effect.
∆ =
3. A.H. Compton (1923) showed that in X-ray scatter-
ing experiments, the photon carries a momentum
=∆
=
=
That is, unlike classical (nonrelativistic) particles,
the massless photon energy is linear in its momen-
tum rather than quadratic.
4. To explain the spectrum of the hydrogen atom, N.
Bohr (1913) gave a “quantization rule” for the elec-
tron motion which identified “energy levels” in the
14
Hy atom. Hence the picture: the “allowed” ener-
gies of the EM field and the matter can be shown
in a level diagram and spectroscopy becomes the
study of spectroscopic transitions. Note the follow-
ing features.
(a) Every system has a ground state. A vacuum is
the ground state of the EM field.
(b) For matter:
i. level spacing generally gets smaller as the
energy increases.
ii. there is a continuum limit. Above a
threshold energy, an electron becomes free
and its energy essentially (for all practical
purposes) becomes unquantized.
iii. Spectroscopy is energy exchange,
EM field ¿ matter.
Wave Mechanics
1. The classical EM theory of Maxwell treated light
like a wave. Planck and Einstein added a particle
character. L. deBroglie (1923) reversed the argu-
ment.
=
2. Davisson and Germer (1927) and G.P. Thomson
(1928) showed that a beam of electrons scattered
by a thin gold foil showed the sort of interference
rings expected for waves of this wavelength.
3. in 1926 P. Debye, on hearing a student seminar
on deBroglie’s wave hypothesis, asked “if there are
waves, what is the wave equation”? A young E.
Schrodinger, who was in attendance, went off and
constructed one. If (in 1 D)
Ψ( ) = 2(−)
(a) then from deBroglie’s hypothesis
−[ ~2
2]2Ψ( )
2=
2
2Ψ( )
= |− ()|Ψ( )
(b) and from Einstein’s hypothesis
[~]Ψ( )
= Ψ( ) = Ψ( )
(3.1)
(c) Eliminating the gives,
~Ψ()
= − ~2
2
2Ψ()
2+ ()Ψ( )
(d) Note that this is a dynamical equation, i.e.
it tells us how Ψ( ) evolves in time: a new
mechanics - QM - was born.
(e) However it is NOT obvious what Ψ( ) rep-
resents. The simplest rule isR |Ψ( )|2 = Prob( ≤ ≤ ) at time t.
Stationary states
A solution to eq. 3.1 with = has the form
Ψ( ) = −~() (3.2)
with
− ~2
200() + ()() = () (3.3)
If we add the (stationary) boundary condition,
()→±∞→ 0
we find that there are solutions to equations 3.2 and
3.3 ONLY for some values of E () At these energy
“eigen values” we have prescribed “eigen” wave functions
(). This process has selected the stationary states by
insisting that there is no probability at large distances.
3.3 Bound states in a finite well
We will do two exercises related to the stationary states.
The examples are constructed to force you to triangulate
(get a “fix”) on the essence of QM. The third exercise is
extended into the dynamical realm.
15
3.3.1 Bound states in a 3D “rounded”
well.
Exercise 3.3.1 The template does not explain how I got
the kinetic ∇2 form. The radial part of ∇2, in D dimen-sions (all that is relevant for “sherical” symmetry) is
∇2 = −(−1) (−1)
Using the chain rule, show
the the equation solved in the template is equivalent to
the Schrodinger equation in 3D.
Exercise 3.3.2 Calculate the eigen values and func-
tions for the hydrogenic wave functions for Z=1 and X.
Also find the most probable r for n = 1, 2 and 3 for
Hydrogen.
3.3.2 Bound states in single (SW) and
double well (DW) potentials [2]
Consider the sharp finite walled potential shown in the
template. This potential is qualitatively appropriate for
a () electron in a polydiene. An electron with energy
is unbound. We will consider the region 0 ≤
Equation (3) becomes
00() +2
~2[ − ()]() = 0
with
2[− ()]~2
0 0
0 0 ≤ ≤
0
Exercise 3.3.3 The number of bound states of a single
square well is = 1 +√2· First do a dimensional
analysis and convince yourself that the solution must be
this combination of parameters, “Experimentally” verify
this result. (The limit as V or L → 0 is worth remem-
bering.)
If we define the following quantities
2 ≡ 2~2 and 2 ≡ 2(−)
~2 then we
can write the DEQ’s, general solutions and the subset of
solution space imposed by the condition that the expo-
nentials must decay moving away from the well.
place DEQ general relevant solution
x 0 00()− 2() = 0 +−
0 ≤ x ≤ L 00() + 2 () = 0 + − sin() + cos()
x L 00()− 2() = 0 +− −
There are 5 unknowns {} so we knowwe must have 5 equations. Nature is often difficult to
understand but she is well behaved. In this case this
means that () is continuous and smooth across the
boundaries. IN other words, we can match the values
of and its dervative 0 at each boundary, approach-ing from either side. This leads to 4 equations. [Keep
in mind how the sign changes with derivatives of trig.
functions. Also keep the 0 (‘in’ and ‘out’) straight.]With the 4 matching conditions
(0)→ =
()→ sin() + cos() = −
0(0)→ =
0()→ cos()− sin() = −−
Moving everything over to the LHS, this set of equa-
tions can written in matrix form.⎡⎢⎢⎣1 0 −1 0
0 sin() cos() −− − 0 0
0 cos() −
⎤⎥⎥⎦⎡⎢⎢⎣
⎤⎥⎥⎦ = 0A nontrivial solution to this problem exists when
the determinant is zero.
The roots of this “characteristic” equation give the
eigen values. But we still do not know the full set of 5
coefficients { −} for each eigen value as we are oneequation short. The missing equation comes from the
statement that we require be normalized, that is, the
particle must be somewhere.
1 =R 0−∞22+R 0[ sin + cos ]
2+R∞
2−2
Exercise 3.3.4 Prove that the above leads to the 5th
relation used in the template.
Exercise 3.3.5 Generate the normalization condition
for a DW with 8 coefficients (A, B&C, D&E, F&G and
H, for the 4 regions.)
References
[1] Thirty years that shook Physics: the story of quantum
theory, G. Gamow, Doubleday (1966); QC174.1 G3.
[2] This exercise was taken from the quantum Chemistry
class offered by R. Lovett [WU].
16
Chapter 4
Time Evolving states
4.0.3 Time evolution
If the Hamiltonian (operator) is
H = − ~2
2∇2 + (r) (4.1)
the time evolution of a wave function is determined by
HΨ(r ) = ~Ψ(r )
(4.2)
Any state with a well defined energy has the form
Ψ(r ) = −~(r) (4.3)
However the satisfy
H(r) = (r) (4.4)
Equations 4.2 and 4.3 determine both (r) and
4.0.4 A superposition of energy eigen-
states
Because of the linearity of eq. 4.2, any linear combina-
tion of terms of the form of eq. 4.3 is also a solution to
eq. 4.2. Thus, the initial ( = 0) wave function can be
represented by a linear combination of eigenfunctions
Ψ(r 0) =X
(r)
(with {} any set of constants) and time dependencecan then be written as a sum of separately evolving am-
plitudes
Ψ(r ) =X
−~(r)
Exercise 4.0.6 If a system is prepared in an energy
eigen state, prove that the distribution function ∗ is
stationary in time.
Exercise 4.0.7 If Ψ(r 0) = 1√2|0() + 1()| (where
() are the first two eigen states of the double square-
well problem) give an expression for |Ψ(r )|2 which in-volves only real quantities. In the process show that the
time evolution is periodic with a period
P = 1−0
= ∆= 1
”the Ei nstein frequency”
This result is used to create an animation in the
MATHCAD template. Note
cos(~ ) =~
+− ~
2and sin(~ ) =
~ −− ~
2
References
[1] The notes for this chapter follow those from R.
Lovett’s Quantum Chemistry course.
ENTER final “pencil and paper work” here
for YOUR records.
17
Chapter 5
Black-body radiation and PS
5.1 Phase space
Each “particle” exists somewhere in a (phase) space of
2D dimensions. That is, to specify the coordinates of any
“particle” you need to specify the position and (conju-
gate) momenta in each dimension - ( ) In standard
3 dimensional space, this amounts to 6 coordinates. As
there is a position for each momentum and the product
of each pair has units of the units of the volume of PS
is It is therefore convenient to measure the volume
of PS in units of There is another important reason
to use this as a measure of PS volume. Each quantum
state occupies exactly (I will resist proving this to
you - although it is rather easy once you have learned
the quantum HO.) The number of distinguishable states
or “cells” in this PS is: the volume of PS/
One is interested in PS volume for the following rea-
sons..
1. For simple quantum transitions the rate is propor-
tional to the number of possible final states. Con-
sider a state-to-state decay which produces a pho-
ton. The rate is determined by the overlap of the ini-
tial and final states (with the appropriate operator)
and the number of photon final states. (The com-
bination of these two factors is codified in “Fermi’s
Golden rule”.)
2. In equilibrium problems, the amount of any partic-
ular “outcome” is directly proportional to the PS
volume allowed by this “outcome”.
3. The rate of complicated processes leading to a par-
ticular “outcome” is proportional to the PS volume
allowed by the critical intermediate leading to this
“outcome”.
Are all parts of the available (single-particle) PS
equally probable? In thermal equilibrium problems
(when quantum statistics is irrelevant) the probability
of occupying a specific (single-particle) cell of PS is ex-
ponentially dependent on the energy of that cell of PS,
∝ − This Boltzmann factor will suppress PS oc-cupation of (single-particle) states with large energy.
However to execute this weighting you need to know the
relationship between energy and momentum. (Hold on,
not so fast - we must use relativity here! I will return to
this shortly.)
In the following we will imagine that position space
is particularly easy. Specifically each and every particle
is in a box of volume and there is no preference for
where. (Consider a particle in a 3-D box, i.e. with no
potential but infinite walls.) The PS volume in some
(spherically symmetric) momentum region p is then
V =RRRR
R
R
=
R
R
R
= 4
R2 or
N() = 43
R2
Where I have invoked the isotropy of space, in the imag-
ined problem. You should have two images in your head:
1. A picture of physical space (x,y,z) in which the par-
ticle must exist with a total volume
2. A spherical shell in the momentum coordinates of
which captures V() = 42 volume.
If we want to determine the total number of states
in volume with total momentum less than some value,
say , we just integrate up to that value.
N = 433
3
If you want to know the number of states with some
momentum we have (for this 3D problem),
N() = 43
2
Note: I left the (the thickness of the spherical shell)
in the equation. It MUST be there! There are no states
with exactly some momentum (area of infinitely thin
shell is 0) This is not a trivial point as an overwhelming
majority of actual and conceptual problems result from
not appreciating how to transform this “shell thickness”.
What do I mean “transform”? Suppose I ask: how many
states with an energy between and + ? You need
18
to transform the 2 AND the to energy. DO NOT
FORGET THIS!
5.2 Relativity
Before I provide a few examples of calculating
volume, I need to remind you of some basic rela-
tivity. The total energy of an body is the root of the
quadratic sum of a rest mass and kinetic energy
=p(2)2 + 22
1. If the object is not moving, the (rest) energy is
= 2
2. If the rest mass is zero (as it is for a photon)
=
This is called the ultra-relativistic (UR) limit.
3. If the object is moving so slowly that the 2 term is
much less than the first, we can expand the square
root in a Taylor series and keep only the first order
correction.
=
2 + 2
2=
2 + 2
2
We call this nonrelativistic (NR).
5.3 Black-body radiation
What is the equilibrium distribution of photon energies
at any ? What is the (equilibrium) distribution of pho-
ton energies on the surface of the sun, earth or in outer
space? Let me rephrase the question in a “leading fash-
ion”. What is the volume of phase space for massless,
spin 1 bosons, as a function of energy?
To answer this question there are two specific things
you need to know in addition the basics above.
1. While objects of spin I, generally have 2I+1 degen-
erate orientations in isotropic space, the spin projec-
tion of zero cannot exist for objects traveling at the
speed of light. (Can you imagine why? If not, ask
for help.) This means that the photon has 2 projec-
tions (helicities, polarizations) for each and every
PS cell.
2. Bosons (force mediating particles) follow an occu-
pation probability (of PS) determined by Bose and
Einstein. (For massless particles the chemical po-
tential = 0)
= ((−) − 1)−1
This occupation probability is derivable from sta-
tistical arguments when one knows that bosons do
not care about the occupancy of a cell. Bosons
can “party-down” in one cell. On the other hand,
Fermions, will not “cohabitate” cells (aside from
the spin degeneracy.)
You now have the background needed to understand
the BB template. (Remember the → Jacobian!)
Exercise 5.3.1 Calculate the photon energy per cubic
cm as a function of T.
5.4 An ideal classical
(Non-relativistic) gas
We now consider a gas with no internal degrees of free-
dom. All it can do is move in a potentailless region
of volume The Boltzmann weighted sum over
phase space is called the partition function. The
partition function is simply a sum over PS where each
element in the sum is weighted by its probability of be-
ing occupied. (From your first days in “calc” you should
have viewed the “R” as what is it - just a big “S” for
sum.)
As we are dealing with only translational motion, we
only need calculate the “translational” partition function
in 3D.
1. = 1
3
R R RR R R
exp[− 12
(2 + 2 + 2)]
= ( 4
3)R2 exp[− 2
2]
2. Now let 2 = 2
2 2 =
= 2( )
= (2 )
(2 )12= (2 )
12
= ( 43)(2 )
32[R2−]
= (43)(2 )
32[√4]
= [ (2
2)32] (Z)
This is simply the volume of weighted by a
Boltzmann factor, the classical factor indicating the
probability of any given (“single-particle”) PS cell be-
ing occupied (as long as the occupation probability is
exceedingly low.)
One important additional point. If the particles are
Fermions, they must occupy different cells. Therefore
the above equation is ONLY valid if the number of cells
greatly exceeds the number of particles.
19
Chapter 6
Heat Capacities
6.1 Overview
The first law of thermodynamics says the change in the
energy of a system is
= heat added TO + work done ON,
= dQ +dW.
The second law of thermodynamics says, the en-
tropy change is equal to the reversible heat transfer di-
vidied by the temperature,
=
or =
The larger and more complicated the system (mole-
cule) the more internal degrees of freedom (DOF) and
the more heat is required to increase the temperature.
The latter is a measure of the energy in any one DOF
≡ ( ) ≈ ( )= (
)
Recall:
=1= − and
=1= +
const v const p
= ( ) = (
)
= () = (
)
∆ =R
∆ =
R
6.2 Heat Capacity Data
1. data for gases are smoothly varying (usually in-
creasing) functions of and are fit by polynomials.
There are two reasons to choose polynomials. First
they can used to describe any smoothly varying
function. Second, they are easily integrated. (Yes
- anticipate that you will have to integrate polyno-
mials.) The difference − is close to for real
gases but the actual difference can be readily calcu-
lated with knowledge of the EoS. [ deg
] = Cost
to heat an IG per mole at constant p. (The con-
stant p condition implies that the volume increases
and thus the gas “system” works against the .)
(a) Simple monatomic gases have: =32
(b) Simple diatomics have =52 at low and
=72 at high .
(c) Linear triatomics have ∼ 52 at low while
non-Linear triatomics have ∼ 62 at low .
The heat capacities of both types of molecules
increases with .
2. ∼ (but ) for liquids. The trends are
complex and no general prescription can be offered.
However a few general comments can be made. One
is that the heat capacity of Hy-bonded liquids can
be very large.
3. ∼ for solids. The form of ( ) is universal if
plotted using a reduced temperature scale (dividing
by a material specific number.) Einstein and De-
bye explained this universal function. The following
comments help to understand the dependence.
(a) The high temperature limit for all solids is
= 3, a universal result known by the
middle of the 1800s and is called the Law of
Dulong and Petit.
(b) At low temperature the heat capacity scales as
∝ 3 This is a result of the “turning on”
of crystal vibrations with increasing tempera-
ture. (All of them are turned on at high
and the 3 result can be understood by just
counting the number of vibrational modes of
the crystal.)
20
(c) The scaling variable for the (to make the
plot universal) scales like Θ ∝q
where
is the mass of the atoms and is the atom-
atom bond strength. Does this give you some
ideas about what is going on?
6.3 ∆( )
We now address the question of how the enthalpy of a
substance or a reaction changes with temperature. Re-
member if you want to have at a temperature different
than the one provided in a tabulation you need to ap-
ply a correction. What correction? What do you need
to know? Well I did not type all this %ˆ&* about heat
capacities to get my jollies!
1. The molar is the rate of increase of with tem-
perature, = () therefore = ( ) . In-
tegrating, ∆ =R( ) Where I have chosen
to remind you of the temperature dependence of the
heat capacity.
2. For a reaction, all we need to appreciate is that in
changing the temperature all the enthalpies, those
of all reactants R and products P, increase in ac-cordance with their own heat capacities.
Consider my favorite reaction: + → +
∆(2) = ∆(1)+R 21[() + ()− ()− ()]
It is convenient (for me) to define,
[] = C = (P)−(R)=PP P −
PR R Thus
∆(2) = ∆(1) +R 21C( )
Note that if, as is generally the case, the individ-
ual heat capacities are represented as polynomials,
C( ) can be represented as a polynomial, de-rived by just summing terms of like order, with +
signs for products and− for reactants. (Don’t for-get the stoichiometric coefficients!) Also note that
while the individual heat capacities increase with T,
C( ) can either increase or decrease with T.
Remark 1 The subscript refers to “one unit of
the reaction” with the stoichiometric coefficients as
written. This means any quantity such as ∆has an implied reaction with specified coefficients.
The change (∆) refers to taking the stoichiomet-
ric coefficient number of moles of R and converting
them into P. The lower case implies that the stoi-chiometric coefficients refer to the number of moles.
Exercise 6.3.1 Extend the template to 500K and in-
clude the → phase transition.
21
Chapter 7
Phase changes
7.1 Geometric interpretation of
and
Consider an generic phase diagram for a pure (one-
component, = 1) system.
S S S
1. Above T
There is no liquid-gas and () and () are
smooth functions with smooth derivatives -p and
respectively.
= − and = − = + − ∴ = +
2. Below T
First-order phase transitions have discontinuities in
A(v) and G(p).
3. The critical point is defined by
() = (
22) = 0.
For a van der Waals (vdw) fluid these conditions fix
the critical point.
= (−) −
2→ =
27
= 3 =827
4. Gallery of figures
(a) Phase boundaries are defined by the equality
of the free energies (or chemical potentials in
a multicomponent system) of the material in
each phase.
(b) Curvatures are sign definite as and are pos-
itive (when in equilibrium)
= − + = − −
(c) Anatomy of first order PT’s.
Mechanical equilibrium requires that the
phases be at the same pressure. If this were not
22
the case, boundaries would move to make it so.
Thus two phases with different molar volumes
in (mechanical) equilibrium must share a com-
mon tangent [ = −( ) ] to the () curve.
The “blister”, between the common tangent,
is part metastable and part unstable. Put an-
other way: imagine you build a box and put
enough material in that box to have a volume
per mole in the “blister” region. You shake
well and peer inside. What will you see? NO
material with the predesignated molar volume
! Phase equilibria has REMOVED an in-
dependent variable. (You can’t beat mother
nature!)
A(v) and P(v) isotherms. The common tangent
construction in A(v) finds the two molar volumes in
mechanical equilibrium (common pressure.)
7.2 Example - immiscible liquids
If the free energy of mixing A and B has the form
of the heavy solid and (if mixed) dashed line in the fig-
ure above, the liquid will phase separate into two liq-
uids of composition and These two compositions
share a slope (common tangent) and these composi-
tions (at this T) have the property () = () and
() = (). That is, both components ( and )
are in chemical equilibrium. Therefore, IF you try to
prepare A solution inside the “blister” (with a value of
inside the blister) - you will find any matter with this
composition.
You can prove that the common tangent has inter-
cepts with the RHS = − • and − • with the
LHS. [• means the quantity x for pure This proof alsoshows that the common tangent imposes the condition
() = () AND () = ()]
The set of figures below examines the cosolution of
simple alcohols in water. These systems have an up-
per “consolution” temperature (). That is, above a
certain temperature the fluids are miscible in all propor-
tions. Below () the liquid separates into two phases
( and ). Panel a) illustrates that the miscibility gap
(in mole fraction) at fixed increases with complexity
of the alcohol. The second b), illustrates how the misci-
bility gap closes with increasing (1 2 3)
closing completely at the (upper) consolution tempera-
ture, ()
Exercise 7.2.1 Define the coexistence region [T()] for
a vdw fluid.
23
Chapter 8
Equilibrium Constants
8.1 Partition functions for a sim-
ple systems.
8.1.1 Logic
1. If the unit is separable (separate QN’s)
= + + + +
the unit partition function is factorable,
=P
− =P
−(++++)
=P
−−−−−
=P−
P−
P−
P−
P−
=
2. If the units are independent and indistinguishable,
the partition function for N units is
= !
8.1.2 Translation
As previously shown (in the absence of a potential)
= [ (2
2)32] (8.1)
8.1.3 Rotation
In this lecture we will only go so far as considering di-
atomics. There are no additional principles in extending
to more complicated molecules.
1. For the diatomic we must consider the rotational
energy and spectrum.
=h2(+1)
2I= h22
2IIt is important to appreciate that this is really a
problem in 2 − Recall for a rigid rotor “dumb-
bell” with the masses separated by , that the mo-
ment of inertia I = 2
2. =P(2 + 1)
−h2(+1)
2I
The degeneracy factor () is the result of collapsing
the 2− problem into one quantum number.
3. For all but 2() at the lowest 0, many rotational
levels are populated. This suggests that we can ap-
proximate thePby an
ROf course, you can always
execute the sum if you want, but I am telling you
the result will be little different from,
=R(2 + 1)
− h2(+1)
2I
4. Let =h2(+1)
2I = [ h2
2I ](2 + 1) Then,
= [2Ih2
]R− = −[]−|∞0 = −[](0− 1)
= [2Ih2
]1
= [
2Ih2
]1
= [
Θ]1
(8.2)
5. The last form uses the characteristic rotational tem-
perature Θ ≡ h22I6. Where did the 1
factor come from? Well in truth
this requires a discussion of the overall symmetry of
the wave function and how this impacts the prod-
uct of the rotational and nuclear partition functions.
This topic is one covered in the main lecture course.
symmetry
heteronuclear 1
homonuclear 2
C symmetry
7. We can now go on to thermodynamics.
(a) = − ln
= −[h22I ](
2Ih2)(−)−2
= = 2(12 )
Why did I write the last version? It is to high-
light the fact that the 1 comes from 2 - 12
contributions from the fact that the problem is
really 2 The rotational Hamiltonian is RE-
ALLY: = 22I + 22I= (2 + 2 )2I
24
The [2 +1] term in the integral is just
like the [2] when determining the area of
a circle by integrating rings.
(b) Contribution to = 2(12) =
(c) There is no contribution to the pressure be-
cause 6= ( )
8.1.4 Vibration
1. =P∞0 −(+
12)
= (−2)[P∞0 − ] = ()[
P∞0 ]
= ()[1 + + 2 + ] where = −
As [] = (1− )−1 the infinite sum of the partition
function has a closed (exact) form
= (−2)[1− − ] (8.3)
2. The contribution to the energy (and thus heat ca-
pacity) varies with Lets consider the high limit.
(a) ≈ ()[1− 1 + (
)− (
)2 12!+ ]
≈ (1)[1− 1 + (
)] = ()−1
(b) = − ln
= − ln()−1
= −(−) 1−2
= = 2(12 )
Note: = 2
2+ 1
22, i.e. has 2 quadratic
terms.
(c) Contribution to = 1
(d) There is no contribution to the pressure be-
cause 6= ( )
(e) Θ ≡
8.1.5 Electronic
1. =P
−
In almost all cases ∆ so the Boltzmann
factor “kills” the contribution from all excited states
to the partition function. As long as this is the
case, all we need to consider is the ground state and
its degeneracy . Defining the zero of energy as
separated atoms (see figure below) we have
= −
2. A general practice is to combine the zero point vi-
brational energy into the ground-state electronic en-
ergy. This makes sense because is not a measur-
able quantity while = +12 is.
0 = − (8.4)
3. The ground state degeneracy is determined by a
classic experiment called by the name of those who
first did it - “Stern-Gerlach”. Usually they are well
understood by calculations. The information
on the ground state structure is encoded in the term
symbol.
molecule H2 O2 NO
Term Symbol 1P
+3P
− 2Q12
g 1 3 2
8.1.6 Equal Partition
1. Consider a Hamiltonian fragment = ++
Our only requirements are that the fragment has
its own QN’s that are not shared with any other
fragment and that is a coordinate.
2. Now let us calculate the classical partition func-
tion fragment corresponding to this Hamiltonian
fragment
=R−∞−∞ −
3. Let =
−1 = −1
=[−1]−1
= [(
)−1 −1]
−1
= −1 ( )
1 = ()−1
4. = ()−1R −∞−∞ −
Defining the numerical value of this integral as
= ()−1 = −1 · −15. = − ln
= −[ ()1
]−1(− 1)−1− = 1
25
6. =1 per unit (at high .)
8.1.7 Total partition function for a unit
if = + + + +
⇒ =
∆ ∆
classical –––––– gd. st. only
1. = [ (2
2)32]
This is the only term for a monatomic gas.
2. = 2
h2
Classical, usually a great approximation.
3. 0 = (1)[1− − ]
Exact sum. The first factor has been removed and
taken into the the electronic partition function.
4. 0 = −
Generally only the ground state contributes for T
1000K.
5. = ()
8.2 Chemical Equilibrium
1. Consider the generalized rxn:
+ ® +
2. The Helmholtz FE is
(a) = − ln(b) = − − +
P
=P
(c) Therefore,
≡ ( )6= = − ( ln
)6=
(d) At equilibrium
= 0 =P
= + − −
3. Logic
The depend on (amounts) while the
(amounts) are such that = 0
4. Example:
(a) let =
! , =
!...
This implies that there are no interactions.
The partition functions are just the ideal unit
partition functions raised to the number of
units / by the indistinguishability factor.
(b) = − ( ln
)
= − [( ln − ln+
]
= − [ln − ln − 1 + 1]= − [ln
]
(c) Equilibrium requires:
+ = +
We can use the form above (in b) and divide
out all the ’s.
ln
+ ln
= ln
+ ln
ln[( )(
)] = ln[(
)(
)]
Dropping the ln0 and collecting like terms,
[
] = [
]
(d) Now we divide through by the appropriate
power of and (re)define []
[] ≡ [] []
[][]= [
( )( )
( )( )]
=( )( )
(e) ≡
= ( )+−−[]
(f) where I have used = assuming we are
dealing with ideal gases.
(g) Activities and fugacities result if the Hamil-
tonian does NOT separate.
Exercise 8.2.1 The MATHCAD template calculates
the equilibrium constant ( ) for 2 + 2 2. It
makes use of the characteristic “rotational” and ”vibra-
tional” temperatures. The values of these for several di-
atomics are listed below. Your excercise is to calculate
( ) for the reaction specified on the template. Assume
monoisotopic species.
26
27
Chapter 9
Solutions
9.1 Multicomponent systems
9.1.1 The variables
For a one component system there are only two inde-
pendent variables. For each additional component we
must add an additional independent variable. These are
the chemical potentials, the 0
1. = + ( ) − ( ) +P
2. = + ( ) + ( ) +P
3. = −( ) − ( ) +P
4. = −( ) + ( ) +P
But wait. If we have a two component system, we
have added 2 chemical potentials. Shouldn’t we have
added just one new variable if we have a two compo-
nent system? YUP! This means there is something these
equations are missing. The two 0 (for a two com-ponent system) must not really be independent. The
Gibbs-Duhem equation is that relation - be patient we
will derive it soon. But what you should expect is ONE
extra relationship between all the extra variables so that
when we have components there are only − 1 trulyindependent extra variables.
9.1.2 The physical meaning
1. = +() = +(
)
2. = −( ) = −( )
3. = +() = +(
)
4. = +() = +(
)
5. = () = (
)
= ( ) = (
)
Note that under “bench-top” conditions, ( ) con-
stant,
= () .
Remark 2 What is for a one component system?
Obviously it is nothing new as we did not need it for
a one component system. Formally,
= () =
That is, how does change upon the addition of
more “stuff”? Well (per mole) of course! Imagine
an ocean of water (at 1 bar and temperature ) and you
add 18 g of water at 1bar and temperature . How much
did the free energy of ocean change? ANS: (2 ).
This is an infinitesimal process and satisfies the logical
definition of a partial.
The best way to think about what the chemical po-
tential is to go back to the partial differential definition.
is the change in the energy function under the appro-
priate conditions. The chemical potential of any species
must be the same in all phases in mechanical and thermal
eq. If this were not the case, material (“that” material)
would transfer to make is so. The chemical potential is
the generalized “force” which is mated to matter trans-
fer, the product of which (chem. pot * amount of matter)
has the units of energy.
9.1.3 Back to the ideal gas
We could ask: what is the chemical potential of an IG
at some arbitrary pressure? But as with any energy,
we really only care about changes as the zero is always
chosen for convenience. So let us recast our question.
What is the change in chemical potential from a reference
state which we choose as the gas at 1 atm? (NOTE
we are changing the pressure but not the temperature!)
First consider the free energy change. As the gas is ideal,
and thus does not interact with anything,
= − +
=
∆ = ln( )
28
Again, using ideality,
− = − and thus
= + ln( ) = + ln( )
if we let = 1 atm and appreciate anything inside a ln
(or exp) is unitless.
9.2 Gibbs-Duhem Equation
You are ready to discover the extra “relation” between
the two added variables (’s) so that there is only one
new truly independent variable in going from 1→ 2 com-
ponents. At equilibrium, the change in the free energy
with some reaction progress variable is zero.
() =P
= 11 + 22 = 0
Thus the extra relation between the new variables is
X = 0 (9.1)
Simple, but this is not the end of this logical devel-
opment. Consider the internal energy with its natural
variables.
1. = ( )
= () + (
) +
P(
) 6=
= − +P
2. The consequence of being a Homo. fxn of degree 1
(see thermo course for proof) is
= () + (
) +
P(
)
= − +P
From which it follows,
= +− − +P+P
3. Equating the two differentials, we get the Gibbs-
Duhem relation:
(a)
− +X
= 0
(9.2)
(b) Which at constant and
X = 0
X = 0
(9.3)
(c) For a C=2 (A and B) system
+=2= 0
=2= −(
) (9.4)
9.3 Ideal Solutions (IS)
An ideal solution is one that obeys Raoult’s Law.
= • =
• (9.5)
Definition 1 () and • (• ) are the vapor pres-
sures (fugacities) of “above” the solution and pure
liquid, respectively. The fugacities are what must take
the place of pressures for non-ideal gases in order to cal-
culate the correct chemical potential. The are readily
calculated from compressibility data.
Definition 2 I will use and as the mole frac-
tions of the liquid phase, vapor phase and total system,
respectively. In almost all cases '
In either version, the statement of is that the
presence of the other component(s) does not influence
the escaping tendency aside from the factor one would
expect based only on fractional occupancy of the liquid
surface regions (scape points.)
Consider the chemical potential of a vapor compo-
nent in phase equilibrium with the same component in
solution.
1. Each component in the solution is in phase (g l)
equilibrium. For the solvent1
1() = 1() = 1() + ln 1
The superscript denotes the reference state which
for gases is 1 atm (or 1 bar.) The second equality
defines the fugacity for which → =
2. On the other hand, for a pure• = 1 system,
•1() = •1() = 1() + ln •1
3. Solving for 1() and inserting result into 1,
1() = 1() = •1() + ln 1•1
= •1() + ln
1•1
•1or
1() = •1() + ln1 (9.6)
29
4. We usually choose the pure• liquid as the liquid ref-erence state, 1() ≡ •1()
= •1(). Therefore,
1() = •1() + ln1 (9.7)
5. The concept behind these equations is simple. The
chemical potential is just equal to that of the chem-
ical potential of a pure liquid scaled DOWN in pro-
portion to the ln of the mole fraction. Why do I
say “scaled down”? What is the sign of ln? Also
appreciate that the ln term is unitless and the
gives the correction term the units of energy.
Remark 3 How should you think about this expres-
sion? As you pollute a liquid with the chemical
potential of drops via ln There is nothing
but statistics at play here. This is the result you
would get if the hetro-interactions (−) were the
same as the homo-interactions (− −)
Remark 4 Using GD it is easy to show that if one
component of a mixture behaves “ideally” (follows
RL) so must the other.
What happens to the thermodynamic quantities
upon mixing? The list below summarizes the results.
∆ = 0
∆ = 0
∆ = 0
∆ = −P ln
∆ = P
ln
9.4 Real Solutions
Real solutions of nonelectrolytes follow some simple
limiting laws. The coefficient () is introduced to quan-
tify the nonideality.
1. The majority species always approaches Raoult’s
Law: 1 = 1•1 in the limit 1 → 1
2. The minority species always approaches Henry’s
Law 2 = 22 in the limit 2 → 0 Note that the
proportionality with is maintained but the slope
constant no longer is that for the pure material.
3. The activity coefficient is defined as the ratio
of the actual partial pressure (fugacity) to what it
would be if the solution were ideal (and followed
RL): = • Thus = + ln
One generally classifies the deviations from ideality
by saying that the system exhibits either + or −deviations from ideality. The figure above shows a −deviation. Such deviations are common when the hetro
interactions are stronger than the homo interactions. As
a result the “escaping” tendency of both species is re-
duced.
The table below summarizes the types of deviations
and the potential types of azeotropes.
type HL Azeo
Ideal = • = 1 No
+ dev = • 1 2 •2 Low T
− dev = • 1 2 •2 High T
We can now rewrite the GD relation in terms of the
(activity) coefficients capturing the nonidealities.P[] = 0P[
+ ln + ln()] = 0P
[ln + ln()]= 0P
+P
[ln()] = 0P +
P[ln()] = 0P
[ln()] = 0 or for = 2
ln() = −
ln() which can be
R arg
Exercise 9.4.1 Prove that the “non-idealities” of A can
be inferred from the “non-idealities” of its “dance part-
ner” from the data provided.
30
Chapter 10
Kinetics-MB
10.1 Integrals
1.R∞−∞ −
2
=p
→p
R∞−∞ −
2
= 1
2.R∞−∞ 2−
2
=√
232
thus 232√
R∞−∞ 2− = 1
3.R∞0
2−2
=1∗3∗5∗(2−1)
2+1
p
4.R∞0
2+1−2
= !2+1
0
10.2 Boltzmann →
1. Velocity.
(a) The velocity distribution in each of the three
coordinates (or however many dimensions one
is working in) is Gaussian.
() = exp{−} = −22
(b) The normalized versionR∞−∞ () = 1 is
() = (
2)12−22
(c) The average “on” any of the velocity compo-
nents is 0 as they are odd functions of the in-
tegration variable. (Plot the argument.) If it
were not zero the gas would have a collective
“flow” velocity and we would have to run to do
any measurements on it.
(d) The average square component (any of the
three) of the velocity is
2 ≡R∞−∞ 2 ()
= ( 2
)12 12
2(2
)32
=
(e) The average square velocity (using the
Pythagorean theorem) is
2 = 2 + 2 + 2
= 3 2 =
3
(f) The “root-mean-square” values of any compo-
nent and of the total velocity are
≡
p 2 =
q
≡√ 2 =
q3
2. We can now go and get the average
=1
2 2 =
3
2
3. ANALYSIS: The 3 is from the fact we worked the
problem in 3 The 2 is from the fact that the is
related to the velocity (or momentum) quadrati-
cally. If we were dealing with a photon gas, the
2→ 1 Also note that,
∝ 2 not 2
4. SPEED distribution.
Let us calculate the probability that a gas molecule
has a certain velocity irrespective of direction. As
you will recall this is called “speed”.
(a) If the probability of having any value of is
independent of the other components then we
have
( )
= () () ()
(If this assumption were not the case, there
would be collective motion in the gas.)
(b) Assuming the gas velocity distribution is iso-
topic and doing the two angular integrals we
can define a probability of a given speed. This
31
is then the probability that the velocity lies in
a spherical shell of radius
() =R R( 2
)32 exp{−22}= 4(
2)322 exp{−22}
(c) From which we can get: i) most probable speed
{[ ()
] = 0} ii) the average speed, and iii) theroot-mean-square speed.
i. = (2
)12
= 1414()12
ii. ≡R∞0
() = (8
)12
= 1596()12
iii. 2 ≡R∞0
2 ()p 2 = (
3)12
= 1732(
)12
iv. For reference the RELATIVE average
speed of two molecules of the same mass
is (using the Pythagorean theorem),
≡√2
As the speed distribution is skewed out to
larger 0 (as there are more PS cells out
there),
(d) The fraction of the speed distribution below,
above and between these benchmark values can
be determined analytically and numerically.
5. Kinetic Energy distribution.
To get the distribution in kinetic energies, we have
to do a little calculus in order to change the differ-
ential elements.
(a) () = 4( 2
)322 exp{−}As =
122 = V =
(b) Therefore we convert to a differential in ,
() = 4(
2)322 exp{−
}
Cancelling the extra 0 0 and 0= (2
)12( 1
)32 exp{−}
= (22
)12( 1
)32 exp{−}
= (4)12( 1
)32 exp{−}
() = 2()12( 1
)32−
6. The fraction of molecules with energies exceeding a
given value , is given by integration.
=R∞
()
= 2( 1)12( 1
)32
R∞
12 exp{−}
= 2(
)12− + erf (
)
Where the correction term is the “error function”
complement defined by
erf() = 2√
R0
−2
& erf () = 1− erf()The complement is the normalized remainder of the
integral. The function goes to 1 (thus erfc goes
to 0), as the argument → ∞. As long as the lowerlimit of integration is well above the peak, the erfc
correction term can be neglected.
erf (solid, with shaded insets) and erfc (dashed) functions.
7. “The kinetics picture”: The hatched area corre-
sponds to the fraction of the kinetic energy distrib-
ution with energy exceeding the activated complex
(†) energy. The side with the smaller fraction ableto traverse the transition region has greater PS vol-
ume - such that the one way rates are equal at equi-
librium. Quantum barrier penetration can lead to
reactions below the barrier, but this contribution in
most chemical reactions of routine interest is neg-
ligible. Quantum does come into play in deterim-
ing the rate of common reactions in that it is the
number of PS cells of intermediate activated com-
plex that deterines the rate of passage through † .This ennumeration is done by SM (as was done in
a preceeding chapter for reactants and products to
determine .)
Exercise 10.2.1 Generate
velocity distributions P(v)with known T, then devise a
fitting procedure to extract it. No binning allowed. See
template.
32
Chapter 11
Kinetics-Master
11.1 Objective & Statement of
Problem
Starting from some complete definition of the status of
a system at some time (t0), determine how it evolves
in time. Generally in chemistry, this means starting
from the initial concentrations and a set of “master equa-
tions”, integrate in time to deduce concentrations at a
later time.
1. Consider: + → +
Note that I did not write an “¿00 as I will not as-sume equilibrium
2. Initial conditions: define []0 []0 []0 []0
3. Master equations:
[]0 ≡ []
= gain of “X” terms - loss of “X” terms.
4. Solve coupled DEQ
11.2 Examples
Chemists and biologists often talk of “integrated rate
laws”. These are the integrated results of particularly
simple Master equations. These “bench-mark” results
are important to help train your scientific intuition.
(Much like the HO and particle-in-a-box problems pro-
vide for QM.)
1. First-order decay results when a master equation
looks like[]
= −[]
Rearrangement and integration lead to the expo-
nential decay of []
2. Reactions resulting from binary collisions are typi-
cally second order which lead to master equations
like[]
= 1[][]− −1[][]
for reactions like + ¿ + If the for-
ward/backward reaction pathways exist, the reac-
tion will come to equilibrium with[]
= 0 thus
1[][] = −1[][] and ( ) =1−1
=[][]
[][]
Some caution must be taken as often binary colli-
sions cannot conserve both energy and momentum
without the presence of a 3 body.
3. A real steady-state condition will occur when ad-
ditional reactants are supplied and products re-
moved (as in most factories, man-made or other-
wise.) A transitory steady-state condition for in-
termediates will occur. Consider the elementary
reactions: +→ 1
1→ 22→
(a)[]
= − [][]
(b)[]
= − [][]
(c)[1]
= [][]− 1[1]
(d)[2]
= 1[1]− 2[2]
(e)[ ]
= 2[2]
Given these master equations, the initial concentra-
tions AND the values for the rate constants, the
time dependence of all species can be calulated at
any time. However IF the decay rate constants 12are large compared to the rate constant controlling
intermediate formation after a short incuba-
tion period, the amounts of the intermediates are
small and rather constant. When this condition is
reached a “stead-state” approximation, i.e.[]
= 0
is valid.
4. Bottom line: If you know your master equations,
rate constants and initial conditions - just set cou-
pled DEQ and solve them (i.e. step through by
integration.)
Exercise 11.2.1 Determine the 3- rate and fraction
in the PS example in the template.
33
Chapter 12
The rise and fall of Easter Islanders.
Steps in solving (simple) kinetics problems.
1. Think about the problem
2. Write coupled DEQ’s
3. Solve em!
4. Plot everything as a function of time.
5. Compare to data, return to 1.
12.1 Predator-prey problem
The example Predator-Prey problem leads to the so
called Lotka-Volterra equations. The prey is assumed
to have unlimited resources (grass) and thus has an ex-
ponential growth term [()
] ∝ . The death
term is assumed to be proportional to the product,
[()
] ∝ ∗ ) as death is when a prey
meets a predator. Conversely, Predator birth is propor-
tional to the product [( )
] ∝ ∗ while
the death term is simply proportional to the number of
predators [( )
] ∝ .
The solutions of these equations generally lead to
two types of “stationary conditions”. These conditions
are mathematical equilibrium points where two orthogo-
nal coordinates, not necessailly the orginal ones ( ),
have zero first dervatives at that point in state ( )
space. The first stationary point is at = = 0. This
point in state space is a saddle, rather than a point of
stable or unstable equilibrium, +ve and -ve second der-
vatives, respectively. The saddle shape tells us that the
dynamics at this point will tend to let one population
return to zero ( )and the other grow without limit (),
if the system is nudged from the stationary point.
The second stationary point is characterized by
complex eigenvalues of the dervative matrix. This point
is an “attractor” in that a system located in the vacin-
ity of this fixed point in state space will orbit around it.
The solutions to the DEQ with initial conditions in the
neighborhood of this fixed point are oscillatory with the
predator’s peaks following the prey’s by 90 Now let us
consider an even more compicated problem.
12.2 Easter-Island problem
In the final exercise we will do the above to construct
a model for the human population of the small Pacific
Ocean Easter island. (This population is known to have
crashed after they denuded the island.) Consider that:
1. Small islands always have a fresh water problem.
Assume a circular island (the problem is worse if
the shape is more complex.) The ratio perime-
ter/area, decreases as −1Thus the drainage, andloss of fresh water, into the ocean becomes a seri-
ous resource problem for small islands. Islands can
only support a fixed upper limit of trees (the island
carrying capacity or CC) or for that matter people!
2. An initially unpopulated island is “found” and pop-
ulated over a short time interval.
3. They start to eat coconuts from the abundant co-
conut palm Trees (T), carve statues (S) from the
inland stone quary and cut down trees to transport
the S’s to the perimeter of the island. They also
need to burn some T’s to keep warm.
4. The S’s do their job and protect the inhabitants
from evil spirits, but not from themselves.
5. The harvesting of T (per person) depends sig-
moidally on the fraction = . If the island
is fully stocked with T, each person could perhaps
cut down 10 trees/year. If there are no T’s, each
person can harvest no T’s per year. Some smooth
curve connects these two limits.
6. New T’s grow, but two factors must be considered.
(a) Only T’s produce seeds for T’s, therefore T
reproduction depends on in some sigmoidal
fashion and
34
(b) The island can only support CC so that if
( − ) = 0, T reproduction goes to 0.
7. People reproduce in proportion to their number and
perhaps live 50 years if they have plentiful resources,
but will die faster if there are not sufficient resources
(such as T’s.)
8. Statues get produced in proportion to the number
of people and their productivity. The latter depends
on tree harvesting. Statues will also fall over at a
slow rate following simple first order kinetics.
Exercise 12.2.1 Create a model for the above by mod-
ification of the preditor-prey model. (See books by Jared
Diamond, Collapse and Guns, Germs and Steel.)
A final comment.
There are many cases where determining the time
dependence of an average quantity misses much of the
science. In such cases, simulations can be done with real
computer codes, often Monte-Carlo, in which many other
factors can be included. Even if one needs such a sim-
ulation, the above DEQ approach is highly informative
as it creates insight and intuition.
35
Chapter 13
Causality and Kramers-Kronig relations.
If the response to a system comes exclusively af-
ter the stimulus (“causal” behavior) then the real and
imaginary parts of the response of related. The relation
is such that if the real part of the response is known over
all frequencies (), the imaginary response is known at
all frequencies. (The inverse is also true.) To make this
concrete, consider the complex index of refraction ,
() = () + ()
where () and () are the real index of refraction and
the extinction coefficient, respectively. If the system is
causal, if either () or () are known, so is the other.
We will actually solve a slightly different problem in this
exercise. That is, if the 90 reflectivity intensity R() is
known, both () and () can be deduced. We will do
this for water. Before we can get to this point, we have
to learn “lesson #1” of complex variables12.
13.1 Complex Variables
1. Let be a complex variable with and being the
real and imaginary parts, respectively.
= ( ) = + = (cos + sin ) =
2. The complex conjugate takes
→ − and thus∗ = (+ )(− ) = 2 + 2 = 2
3. Complex functions of a complex variable, i.e.
() = ( ) + ( )
have real () and imaginary [] functional parts.
For example,
() = 2 = (2 − 2) + [2]
4. A derivative of a complex function exists iff the (like
and cross) Cauchy-Riemann conditions hold.
=
= −
5. If a complex function () is single valued and its
derivatives are continuous (properties that make the
function “analytic”) in a region R, then any closed
integral, fully contained in R, is zero. (See, (b).
Note that this is like the definition of a state func-
tion or a conservative force.)I
() = 0
6. If we contort the contour (C) to form almost com-
pletely concentric contours, the integrals, traversed
in the opposite sense cancel and in the same sense
are equal.I1
() =
I2
()
This is Cauchy’s Integral theorem.
7. Now imagine the argument contains a singular
point, (i.e () =()
− ). As long as the contourC is removed from the point the integralI
()
− = some number.
36
8. Now imagine that the contour is a rubber band, and
stretch it so that it loops and excises the trouble
point and forms two concentric rings (d). Now the
integrand is everywhere analytic on the path and
thus same sense contour integrals C1 and C2 are
equal.I1
()
− =I2
()
−
9. On the RHS, let = + .I1
()
− =I2
(+)
=
I2
(+ )
I1
()
− →0= · ()
I2
= 2()
This, the Cauchy’s Integral formula, provides a
means for calculating the value of complex function
at any point by doing the contour integral.
10. If the integral is only half way around,
+∞Z−∞
()
− = () (13.1)
13.2 Dielectric response.
1. The response of a substance to an electromagnetic
field is described by a complex dielectric function, a
function of the frequency of the stimulus.
() = 0() + 00()
2. The reflectivity amplitude (), reflected intensity
() (of the EM field), real refractive index (),
complex refractive index (), and extinction coef-
ficient (), are related by3:
(a) () ≡ ()
()= ()() = +κ−1
+κ+1
(b) () =()∗()()∗() = ∗ = 2()
(c) () = () + () =p()
(d) 0() = 2 − 2 and 00() = 2
13.3 Kramers-Kronig Relations
1. Finally we get to combine the dielectic response with
Cauchy’s Integral formula. I will do this in a general
sense, considering a general complex function with
the following properties. (The dielectric response
has these properties.)
(a) ()→ 0 as ||→∞(b) 0() is even and 00() is odd w.r.t. (c) Any singularities in () are below the real
axis.
2. Consider the Cauchy integral result,
() = 1
+∞Z−∞
()
−
The “” indicates “principal value” and signifies
the singularity is analytically excised.2
3. Equating the real parts of the LHS and RHS,
0() = { 1
+∞Z−∞
0()+00()
− }
= 1
+∞Z−∞
00()
− = 1 [
0Z−∞
00()
− +
∞Z0
00()
− ]
4. In the 1 integral, let: = −, = −, 00(−) =
− 00() and the limits −∞→ +∞ while 0→ 0
0() = 1 [
0Z∞
(−)00()−− (−)+
∞Z0
00()− ]
= 1 [
∞Z0
00()+
+
∞Z0
00()
− ]
Generating a common denominator yields the first
Kramers-Kronig relation.
0() =2
∞Z0
00()
2 − 2 (13.2)
5. The second KK relations results from equating the
imaginary parts (start back at 3, above).
00() = −2
∞Z0
0()
2 − 2 (13.3)
Exercise 13.3.1 For water generate both () and
() from (). The paper by Berets4, should be used
as a guide.
References
1. Arfkin, Mathematical Methods for Physicists.
2. Boas, Mathematical Methods in the Physical Sciences.
3. Kittel, Introduction to Solid-State Physics.
4. A. Kocak, S. L. Berets, V. Milosevic, and M. Milosevic,
Applied Spec. 60, 1004 (2006).
37
Chapter 14
Trigonometric
sin2 + cos2 = 1
1 + tan2 = sec2
cot2 + 1 = csc2
sin csc = 1
cos sec = 1
tan cot = 1
tan = sin cos
cot = cos sin
tan 6= 1 cotsin(± ) = sin cos ± cos sin cos(± ) = cos cos ∓ sin sin tan(± ) = tan±tan
1∓tan tan sin 2 = 2 sin cos
cos 2 = cos2 − sin2 = 1− 2 sin2 tan 2 = 2 tan
1−tan2
sin2(2) = 1−cos2
cos2(2) = 1+cos 2
tan 2= 1−cos
sin= sin
1+cos
sin sin = 12[cos(− )− cos(+ )]
cos sin = 12[cos(− ) + cos(+ )]
sin cos = 12[sin(+ ) + cos(− )]
sin+ sin = +2 sin(+2) cos(−
2)
cos+ cos = +2cos(+2) cos(−
2)
sin− sin = +2cos(+2) sin(+
2)
cos− cos = −2 sin(+2) cos(−
2)
sin(2− ) = +cos
cos(2− ) = + sin
sin( − ) = + sin
cos( − ) = − cossin( + ) = − sincos( + ) = − cossin(− ) = − sin
cos(− ) = − cos
sin(2 − ) = − sincos(2 − ) = +cos
sin(−) = − sincos(−) = +cos
± = cos ± sin
sin = −−2
cos = +−2
38
Chapter 15
Constants
1 amu = 931494013 MeV/c2
= 1.66053873 ∗10−27 kgm = 9.109382 ∗10−31 kg
= 5.485799 ∗10−4 amu = 511 keV/c2m = 1.6749272∗10−27 kg
= 1.008664926 amu = 939.56533 MeV/c2
m = 1.67262158∗10−27 kg= 1.00727647 amu = 938.271998 MeV/c2
m = 1.67353249∗10−27 kg= 1.007825017 amu = 938.78298 MeV/c2
1= ~
2= 13703608
=2
2= 2817935 · 10−15 m
~ = 10546 · 10−34[ · ] = 10546 · 10−27 [erg·s]= 6582122 · 10−22 [MeV·s]= 6465 [MeV12·amu12·fm]
~ =197.3 [eV·nm]=197.3 [MeV·fm] =
2
= 1973
( )[fm] = 1973
( )[nm]
= 138065 · 10−23[ ] = 0695[ −1]
= 086171 · 10−4[ ]= 086171 ∗ 10−10[
]
= 831[ ·
] = 199[ ·
]
= 008314[ ··
]= 008206[ ··
]
N = 6.022·1023 [#/mole]V(298 1) = 22711[
]
2 = 1.44 [MeV·fm] =1.44 [ev·nm] =1.44·103[ev·pm] ≡ ~
2= 927400899× 10−24 [JT]
≡ ~2
= 505078317× 10−27 [JT] = −19135 n.m. = −96623707× 10−27 J T = +279275 n.m.= 141060761× 10−26 J T = 29.8782458 [cm/ns]
1 [eV] = 160219× 10−12 [erg] =160219× 10−19 [J]= 8066[−1] =26.06 [
]
12398 · 10−4[] = 1[−1] = ˜ ≡
=
1 yr = 365.25 d = 315576× 107 [s]
v( )= 1389 · 10−3
q( )
()= 1389
q( )
()
F = 96485.309 [C/mole e−]1 [
]⇔ 104·10−5 [
] or 1 [
]⇔ 104 [
]
=4~
2
2= 52918 · 10−11
= 885419 · 10−12[ 2
]
= 251532
= 567051 · 10−8[ 24
] or [ 24 ]
39
Part III
MathCad handouts
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