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**Unformatted text preview: **PHYS 172: Modern Mechanics Lecture 25 – The Boltzmann Distribution Fall 2011 1 Read 12.7 Summary: Foundations
Einstein Model
of Solids (q + N ! 1)!
# microstates
"#=
q !(N ! 1)!
(N oscillators, q quanta)
Fundamental assumption of statistical mechanics
Over time, an isolated system in a given macrostate (total energy) is
equally likely to be found in any of its microstates (microscopic
2
distribution of energy). Summary: Entropy and Temperature S ! k ln "
1
dS
!
T dEint If the initial state is not the most probable, energy is exchanged until the most
probable distribution is reached. 3 Summary: Specific Heat Catom = !Eatom
"
!T !Esystem N atoms
!T
Al Pb 4 Today: The Boltzmann Distribution
Derivation: Choose Your System Wisely
Many Systems or Many Observations?
Simple Applications:
Probabilities for Atomic Excitations
Block of Lead
Biological Physics 5 Boltzmann Distribution
The Boltzmann distribution comes about by cleverly picking our system.
Say that we want to analyze a cup of water.
What system do we pick? A tiny volume of the water! Example: a
macroscopic
volume of water in
a glass of water Ωres, Eres
Ω, E Etot = Eres + E = constant 6 Recall the Pb Nanoparticle (3 atoms) from
Ch 11, p 392-393:
A
q Ω ln Ω S 4 465 6.20 5 1287 6 3003 Temperature: Tq # 1
"S
!
T "Eint $E $q % !& o
=
$S k $(ln ') B C q T(K) 6.20kB 4-5 58.9 7.16 7.16kB 5 62.5 8.01 8.01kB 5-6 66.6 1 !E ( 3 atoms) 1 8 " 10 #22
C=
=
= 3.4 " 10 #23 J/K per atom
3
!T
3 (66.6 # 58.9 ) What if we wanted to know what is happening on one of the three atoms?
For example, with q=6, how often is atom A found with qA=2?
7 What if we wanted to know what is happening on one of the three atoms?
For example, with q=6, how often is atom A found with qA=2?
We can simply count the states with 2 quanta on atom A:
! A (qA = 2 ) = (2 + 3 " 1)!
=6
2!( 3 " 1)! ! BC (qBC = 6 " 2 ) = ( 4 + 6 " 1)!
= 126
4 !(6 " 1)! P= ! A (2 )! BC ( 4 ) 756
=
" 0.25
!total (6 )
3003 ! A (2 )! BC ( 4 ) = 6 " 126 = 756
Pb Nanoparticle (3 atoms)
0.4 Probability 0.3 A B C 0.2 0.1 0
0 1 2 3
quanta on atom A 4 5 8 6 Now, a more interesting question: What is the
probability of finding 2 quanta on atom A if it is in
contact with a large Pb block? (Assume same T as
before.)
!E
!E
#
#
Probability of !E above ground state
~ " A ( !E ) e kT ! e kT
Probability of being in ground state Boltzmann Distribution
1 This is a property of
any small system in
thermal equilibrium
with a large reservoir
at a fixed temperature. Probability ratio 0.8
0.6
0.4
0.2
0
0 1 2 3 4 Quanta on Atom A 5 6 7 9 !E
!E
#
#
Probability of !E above ground state
~ " A ( !E ) e kT ! e kT
Probability of being in ground state Boltzmann Distribution We are now going
to show WHY
this is true 1 Probability ratio 0.8
0.6
0.4
0.2
0
0 1 2 3 4 Quanta on Atom A 5 6 7 10 VIEWER WARNING! The following slides may contain
Derivations or Derivation Byproducts.
Do not operate heavy machinery
while viewing the following slides.
Also, do not go swimming for
at least one hour after viewing 11 Probabilities Reservoir
A ginormous system
with a constant temperature tiny
system
12
WARNING! This slide may contain Derivations or Derivation Byproducts. Probabilities
What is the probability of
finding the system in this state? Reservoir
(Ginormous) Etot Reservoir
(Ginormous) System
(Tiny) 0 (Etot - E) System
(Tiny) E
13
WARNING! This slide may contain Derivations or Derivation Byproducts. Probabilities
Number of ways of
having Esys = E We can find the probability
from the number of ways
of arranging the energy. !res ( Eres )!( E )
P(E ) =
!tot ( Etot )
Reservoir
(Ginormous)
Total number of ways
of arranging energy
in combined system:
Tot = Res + Sys (Etot - E) System
(Tiny) E
14
WARNING! This slide may contain Derivations or Derivation Byproducts. Now Do Some Math...
!res ( Eres )!( E )
P(E ) =
!tot ( Etot ) We can find the probability
from the number of ways
of arranging the energy. 15
WARNING! This slide may contain Derivations or Derivation Byproducts. More Math...
Eres = Etot − E
Use a Taylor Expansion for the Ginormous Reservoir: (E << Eres ) Spot The Temperature! 16
WARNING! This slide may contain Derivations or Derivation Byproducts. More Math...
Eres = Etot − E
Use a Taylor Expansion for the Ginormous Reservoir: (E << Eres ) Spot The Temperature! 17
WARNING! This slide may contain Derivations or Derivation Byproducts. More Math... Exponentiate Everything... BOLTZMANN FACTOR
Very very important! 18
WARNING! This slide may contain Derivations or Derivation Byproducts. VIEWER WARNING! The previous slides did contain
Derivations and Derivation Byproducts.
So, you should not go swimming
for at least one hour after class.
The Derivation is now OVER.
19 Boltzmann Distribution
The probability of finding energy E in a small system in contact with
a large reservoir is
E
!
We just showed
kT
P E " # E $e
WHY this is true () () The exponential part, e-E/kT, is called the Boltzmann factor.
Ω(E) is the number of microstates in the small system at energy E. In many circumstances, Ω(E)
changes so slowly compared to
e-E/kT that it is essentially constant: P (E ) " const # e ! E
kT Ωres, Eres
Ω, E
20 Boltzmann Distribution
The Boltzmann distribution comes about by cleverly picking our system.
Say that we want to analyze a cup of water.
What system do we pick? A tiny volume of the water! Example: a
macroscopic
volume of water in
a glass of water Ωres, Eres
Ω, E Etot = Eres + E = constant 21 Application: Atomic Excitations
How likely is an atom to be in 1st excited,
compared to odds of being in ground state? P ( E1 )
e−E1 /kT
= −E /kT
P ( E0 )
e0 = e−∆E/kT NOTE: kT at room temp = 1/40 eV.
For the above atom, odds of being in first excited state are P ( excited state )=
1 1 st e "E 1 / 40 eV = 1
e 40 "E = 1
e 40#4 = 3 #10!70 Typical atomic energy gaps are big compared to room temp.
A room-temperature box of neon doesn t glow! (Unless you add energy).
22 Application: Block of Lead
ω
8 × 10−22 J
=
= 57K
k
k T=300K at room temperature BOLTZMANN FACTOR
Very very important! Think of adding q quanta of energy to one lead atom...
At room temperature, the Boltzmann factors for exciting vibrations are:
Most probable state
Watch out!
These are
relative
probabilities
23
Application: Block of Lead
T=300K at room temperature
Think of adding q quanta of vibrational energy to one lead atom...
The actual probabilities use a normalization factor Z (partition function). Most likely to find the
vibrations are excited.
On average, q=5. 24
Today: The Boltzmann Distribution
Derivation: Choose Your System Wisely
Many Systems or Many Observations?
Simple Applications:
Probabilities for Atomic Excitations
Block of Lead
Biological Physics 25 Next Time: Boltzmann Applications
Speed Distribution in a Gas
Energy Equipartition and Specific Heat
Pressure and the Ideal Gas Law 26 ...

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