Lecture 21 Counting Statistics. Chapter

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Fall 2010 PHYS 172H: Modern Mechanics

Lecture 21 – Counting Statistics

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Particle spin angular momentum

Electron, muon, neutrino have spin 1/2 :

measurements of any component of their angular momentum yields ±½ħ (along any chosen axis.)

Quarks have spin ½

Protons and neutrons (three quarks) have spin ½

Fermions: spin ½ are ”matter particles” or “building blocks” They obey the Pauli exclusion principle

Two identical Fermions cannot be in the same quantum state in the same place

Two lowest energy electrons in any atom have orbital angular momentum 0

One has spin up, the other has spin down (+- ½ along any chosen axis)

The different spin projections distinguish between the two electrons so that they are not “identical” for Pauli exclusion purposes.

The two spin projections add up to 0, and when added to the above 0 orbital, the TOTAL ANGULAR MOMENTUM is 0

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Particle spin angular momentum

Bosons: integer spin = Force carrying particles

Mesons: (quark+antiquark) have spin 0 or 1: composite, “old” nuclear force

Gluons (elementary, the “new” strong color force) photons,

gravitons,

weak bosons W and Z

All have spin 1. These are the force carriers of Quantum Field Theory

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Particle spin

Rotational angular momentum

Macroscopic objects: quantization of L is too small to notice!

Rotational energies of molecules are quantized

Quantum mechanics: L_x, L_y, L_z can only be integer or half-integer multiple of ħ Quantized values of _L2 = _{l l}

(

+₁

)

_h2 where l is integer or half-integer

These rules are not at all intuitive. Rotation is a richly complicated

phenomenon (as you have been told in lectures), and even more-so in the Quantum world.

But the quantization of the PROJECTION of angular momentum along any chosen axis plays very directly into the structure of the microworld that we hope you get a sense of in this course.

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Reading Question - Chapter 11, sections 1-4: just for discussion

A. The top hand (aces) is more likely to occur.

B. The bottom hand (no aces) is more likely to occur. C. These two hands are equally likely to occur.

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The question was AMBIGUOUS. BUT, if we ask the likelihood of these EXACT two hands with exact cards in exact order shown,

A. The top hand (aces) is more likely to occur.

B. The bottom hand (no aces) is more likely to occur. C. These two hands are equally likely to occur.

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In either hand, the probability P is one out of all possible distinct five-card hands.

Since a given five-card hand can be rearranged in five factorial (5!) ways, the number of ways, N, that five specific cards are dealt from the 52-card deck is 52x51x50x49x48 / 5! --- since it doesn’t matter in what order the 5 cards are dealt.

When dealing, the first card can be any of the 52 cards. There are then only 51 possibilities for the second card, etc.

Clearly if there are N distinct possible five card hands, the probabiliity of being dealt one of them is just P = 1/N

Note that N = 52! / 5!(52-5)! which is a typical binomial

coefficient, in this case for the sixth term in the expansion of (a+b)52 _{[also, by symmetry, for the sixth from last term]}

Simpler example: (a+b)4 _{= a}4 _{+ 4a}3_{b + 6a}2_b2_….

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The Fundamental Assumption of Statistical Mechanics

A fundamental assumption of statistical mechanics is that in our state of macroscopic ignorance, each microstate (microscopic distribution of energy) corresponding to a given macrostate (total energy) is equally probable. A

microstate specifies how each atom or molecule is moving and where it is and its energy level.

This is analogous to saying that all possible distinct 5-card hands are equally likely….

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CLICKER QUESTION #3

Participation, not graded.

You put an ice cube into a styrofoam cup containing hot coffee. You would probably be surprised if the ice cube got colder and the coffee got hotter.

Would this be a violation of the energy principle? A. Yes

B. No

C. The energy principle doesn’t apply in this situation.

The energy principle (cons. of) is the First Law of Thermo. What’s being violated here is what we call the Second Law

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The Need for Statistical Considerations

Many phenomena that WE NEVER OBSERVE are perfectly compatible with our three fundamental principles.

Our considerations thus far are missing something . . .

?

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The Einstein Model of Solids

• assume atoms oscillate independent of each other

• each atom is comprised of 3

independent quantum harmonic

oscillators: N oscillators ↔ N/3 atoms

ΔE = h k

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CLICKER QUESTION #4

A system of 300 oscillators contains 100 quanta of energy. What is the physical meaning of this model?

A) one atom oscillating in 300 dimensions B) 300 atoms, each in the 100th energy level

C) 300 atoms with 100 joules of energy distributed among them D) 100 atoms with 300 joules of energy distributed among them E) 100 atoms with 100 joules of energy among them× h k

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Counting Microstates:

4 quanta, 3 oscillators

All 15 of these microstates

represent the same

macrostate.

Macrostate: given only by macroscopic parameters (energy, volume, etc.)

Microstate: listing energy state of each oscillator

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a

b

c

d

Counting Arrangements of Distinguishable Objects

# arrangements = 4x3 x2 x1 = 4!=24

abcd bacd cabd dabc acdb bcda cbda dbca abdc badc cadb dacb adbc bdac cdab dcab acbd bcad cbadcbad dbac adcb bdca cdba dcba

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Counting Arrangements of Objects

Five distinct objects → 5! = 120 arrangements.

What if the objects aren’t distinct?

Consider the following

arrangement of colored balls:

Interchanging the red balls gives an identical

sequence: thus, the 5! overcounts by a factor of 2 = 2!.

Likewise, there are 3! = 6 ways of interchanging the blue balls. The 5! overcounts by a factor of 3!.

# arrangements =

5!

10

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A Formula for Counting Microstates

N oscillators and q quanta

N-1 bars and q dots

3 oscillators, 4 quanta ↔ 2 bars, 4 dots: # microstates = as before(2 4)! 15

2!4! + =

(

)

(

)

1 !

!

1 !

q

N

q N

+ −

≡ Ω =

−

Generally, # microstates (N oscillators, q quanta)

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Two objects share a total energy E = E₁+E₂. There are 3 ways to arrange an amount of energy E₁ in the first object and 6 ways to arrange an amount of energy E₂ in the second object. How many different ways are there to arrange the total energy E =

E₁+E₂ so that there is E₁ in the first object and E₂ in the other? A) 3 B) 6 C) 9 D) 18 E) 36₌₇₂₉

CLICKER QUESTION #5

object 1 object 2 way 1 way 2 way 3 6 ways 6 ways 6 ways 3 x 6 =18 total ways

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Interacting Objects (2 atoms)

q₁ q₂ Ω₁ Ω₂ Ω₁ Ω₂ 0 4 1 15 15 1 3 3 10 30 2 2 6 6 36 3 1 10 3 30 4 0 15 1 15 atom 1 (3 oscillators) atom 2 (3 oscillators)

4 quanta to distribute over 6 oscillators 9!/(4!5!) = 126 sum (3+2)! 3! 2! (2+2)!/2!2!

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Fundamental Assumption of Statistical Mechanics

The fundamental assumption of statistical mechanics is that, over time, an isolated system in a given macrostate is equally likely to be found in any of it’s microstates.

Thus, our system of 2 atoms is more likely to be in a microstate where energy is split up 50/50.

Of course, there’s 48%

probability of 1,3 or 3.1 share of the quanta if you don’t

distiguish which atom gets more.

Least likely is to have all the energy in a single atom --- n = 0 or 4 atom 1 atom 2 29% 12% 24% 12% 24%

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Fundamental Assumption of Statistical Mechanics

An isolated system in a given macrostate is equally likely to be found in any of it’s microstates.

If there are very many atoms and very many quanta of energy, then the number of states where the energy is fairly uniformly distributed over all the atoms --- vastly outnumbers the number of states where the energy is very unequally distributed.

The probability is for the system to spend most of its time in the likelier configurations --- configurations of greater uniformity or “averageness”.