The Maxwell-Boltzmann Distribution Brennan...
Transcript of The Maxwell-Boltzmann Distribution Brennan...
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The Maxwell-Boltzmann Distribution
Brennan 5.4
Lecture prepared by Melanie Hill
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Maxwell-Boltzmann DistributionScottish physicist James Clerk Maxwell developed his kinetic theory of gases in 1859. Maxwell determined the distribution of velocities among the molecules of a gas. Maxwell's finding was later generalized in 1871 by a German physicist, Ludwig Boltzmann, to express the distribution of energies among the molecules.
Maxwell made four assumptions…
[1] McEnvoy, J.P. and Zarate, Oscar; “Introducing Quantum Theory;” 2004[2] http://www.hk-phy.org/contextual/heat/tep/trans01_e.html
Maxwell pictured the gas to consist of billions of molecules moving rapidly at random, colliding with each other and the walls of the container.
This was qualitatively consistent with the physical properties of gases, if we accept the notion that raising the temperature causes the molecules to move faster and collide with the walls of the container more frequently.[1]
[2]
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The diameter of the molecules are much smaller than the distance
between them
Maxwell-Boltzmann Distribution
Maxwell’s Four Assumptions[1]
xD D
The collisions between molecules
conserve energy
The molecules movebetween collisions withoutinteracting as a constantspeed in a straight line
The positions and velocitiesof the molecules are
INITIALLY AT RANDOMGreat
insight!
[1] McEnvoy, J.P. and Zarate, Oscar; “Introducing Quantum Theory;” 2004
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Maxwell-Boltzmann DistributionBy making these assumptions, Maxwell could compute the probability that a molecule
chosen at random would have a particular velocity.
This set of curves is called the Maxwell Distribution. It provides useful information about the billions and billions of molecules within a system; when the motion of an individual molecule can’t be calculated in practice. [1]
We will derive the Maxwell-Boltzmann Distribution, which will provide useful information about the energy.
[1] McEnvoy, J.P. and Zarate, Oscar; “Introducing Quantum Theory;” 2004
[1]
Raising the temperature causes the curve to skew to the right, increasing the most probable velocity.
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Maxwell-Boltzmann DistributionWhy use statistical mechanics to predict molecule behavior? Why not just
calculate the motion of the molecules exactly?
Even though we are discussing classical physics, there exists a degree of “uncertainty” with respect to the fact that the motion of every single particle at all times cannot be determined in practice in any large system.
Even if we were only dealing with one mole of gas, we would still have to determine characteristics of 6x1023 molecules!!
Maxwell’s theory was based on statistical averages to see if the macrostates, (i.e. measurable, observable) could be predicted from the microstates.
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Maxwell-Boltzmann DistributionIn Section 5.3, it was determined that the thermal equilibrium is established
when the temperatures of the subsystems are equal. So…
What is the nature of the equilibrium distribution for a system of N non-interacting gas particles?
The equilibrium configuration corresponds to the most probable (most likely) configuration of the system.
Consider the simplest case, a system of N non-interacting classical gas particles.
– Classical system:• there are no restrictions on how many particles can be put into any one state
simultaneously• the particles are distinguishable, i.e. each particle is labeled for all time
First, we’ll need to determine the number of microstates within any given configuration, i.e. the number of ways in which N objects can be arranged into n distinct groups, also called the Multiplicity Function.
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Maxwell-Boltzmann DistributionThe Multiplicity Function - determine the number of ways in which N objects
can be arranged into n containers.
Consider the first twelve letters of the alphabet:
a b c d e f g h i j k l
Arrange the letters into 3 containers without replacement. Container 1 holds 3 letters, Container 2 holds 4 letters, and Container 3 holds 5 letters.
| _ _ _ | | _ _ _ _ | | _ _ _ _ _ |
For the 1st slot, there are 12 possibilities. | c _ _ | | _ _ _ _ | | _ _ _ _ _ |
For the 2nd slot, there are 11 possibilities. | c h _ | | _ _ _ _ | | _ _ _ _ _ |
For the 3rd slot, there are 10 possibilities. | c h k | | _ _ _ _ | | _ _ _ _ _ |
etc…
There are 12! possible arrangements if the containers are ignored.
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Maxwell-Boltzmann Distribution(cont.) The Multiplicity Function - determine the number of ways in which N objects can
be arranged into n containers
Since we care about the containers but we don’t care about the order of the letters within each container, we divide out the number of arrangements within each given container, resulting in:
720,27!5!4!3!12
=⋅⋅
There are 27,720 ways of partitioning 12 letters into the 3 containers
The Multiplicity Function:
In general, the number of distinct arrangements of N particles into n groups containing N1, N2, …, Ni, …, Nn objects becomes:
!...!...!!!
21 ni NNNNN
⋅⋅⋅⋅⋅
∏=
=⋅⋅⋅⋅⋅
= n
ii
nini
N
NNNNN
NNNNNQ
1
2121
!
!!...!...!!
!),...,,...,,(
Where Ni is the number of objects in container i
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Maxwell-Boltzmann Distribution
Physically, each container corresponds to a state in which each particle can be put.• In the classical case, there are no restrictions on how many particles that can be
put into any one container or state• See Section 5.7 for the quantum case where there are restrictions for some
particles
The states are classified in terms of their energy, Ei, and degeneracy, gi
Think of the system as n containers with gi subcontainers.
So, how can we arrange Ni particles in these gisubcontainers?
Each level has gi degenerate states into which Ni particles can be arranged
There are nindependent levels
Ei
Ei+1
Ei-1
Degenerate states are different states that have the same energy level. The number of states available is known as the degeneracy of that level.
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Maxwell-Boltzmann Distribution
Since we are dealing with the classical case:• there are no restrictions on how many particles can be put into any one subcontainer
or state simultaneously• the particles are distinguishable, i.e. each particle is labeled for all time
So, how many arrangements of Ni particles in these gi subcontainers are possible?
Consider an example where we have 3 particles and 2 subcontainers:
RG
B
3 particles 2 subcontainers
n containers with gi subcontainers.· · ·
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Maxwell-Boltzmann Distribution(cont.) How many arrangements of Ni particles in these gi subcontainers are possible?
RG
B
R
G B
R
GB R
G
B
R
G
B
R
G B
R
GB
RG
B
There are 8 possible arrangements, or 23
In general, the possible arrangements of Niparticles into gi subcontainers is:
iNig
Therefore, if our system has a particular state that has a particular degeneracy, there is an additional multiplicity of iN
ig for that particular state.
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Maxwell-Boltzmann Distribution(cont.) Therefore, the total Multiplicity Function for a collection of classical particles is:
Constraint 1 implies:
There are two physical constraints on our classical system:1. the total number of particles must be conserved2. the total energy of the system must be conserved
where U is the total energy for the system
∏∏ =
=
=n
i
Nin
ii
niig
N
NNNNNQ1
1
21
!
!),...,,...,,(
NNn
ii == ∑
=1φ
Constraint 2 implies:
UNEn
iii == ∑
=1ψ
The equilibrium configuration corresponds to the most probable (most likely) configuration of the system, i.e. the configuration with the greatest multiplicity! To find the greatest multiplicity, we need to maximize Q subject to constraints.
ConstantConstant
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Maxwell-Boltzmann DistributionFrom previous slide:
∑∑==
−+=n
ii
n
iii NgNNQ
11!lnln!lnln
∏∏ =
=
=n
i
Nin
ii
niig
N
NNNNNQ1
1
21
!
!),...,,...,,(
This equation will be easier to deal with if we take the logarithm of both sides:
xxxx −≈ ln!lnApplying Stirling’s approximation, for large x:
∑∑∑===
+−+−=n
ii
n
iii
n
iii NNNgNNNNQ
111lnlnlnln
0ln =∂∂
−∂∂
+∂∂
jjj NNQ
Nψβφα
In order to maximize, we need to make use of Lagrange multipliers and Constraints 1 and 2:
Almost exact because we’re dealing with verylarge x, 1023 particles!
The derivatives of the 1st and 2nd constraints are zerobecause the constraints are equal to constants
All we are doing is adding and subtractingconstants multiplied by zero!
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Maxwell-Boltzmann DistributionFrom previous slide:
0ln =∂∂
−∂∂
+∂∂
jjj NNQ
Nψβφα
Substituting in ln Q and Constraints 1 and 2:
011ln1ln =−++
⋅+⋅− j
jjjj EN
NNg βα
Taking the derivative, noting that N is constant and the only terms that are nonzero are when i = j:
0lnlnln111 11
=
∂∂
−
∂∂
+
+−+−
∂∂ ∑∑∑ ∑∑
=== ==
n
iii
j
n
ii
j
n
i
n
ii
n
iiiii
j
NEN
NN
NNNgNNNNN
βα
∑∑∑===
+−+−=n
ii
n
iii
n
iii NNNgNNNNQ
111lnlnlnln
0lnln =−+− jjj ENg βα
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Maxwell-Boltzmann DistributionFrom previous slide:
To find β, we need to determine the total number of particles, N, in the system and the total energy, E, of the system.
If the states are closely spaced in energy, they form a quasi-continuum and the total number of particles, N, is given by:
Where D(E) is the Density of States Function found in Section 5.1
∫∞
=0
)()( dEEDEfN
0lnln =−+− jjj ENg βα
jj
j EgN
βα −=ln
jE
j
jj e
gN
Ef βα−==)(
What are α and β?
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Maxwell-Boltzmann DistributionFrom previous slide:
Using the standard integral:
∫∞
+− +Γ
=0
1
)1(n
axn
andxex
23
232)(
πhEmED =E
j
j egN
Ef βα−==)(
∫∞
=0
)()( dEEDEfN
∫∞
−=0
23
232 dEeEemN Eβα
πh
2323
23
23
2
βπα
Γ
= emNh
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Maxwell-Boltzmann DistributionThe total energy, E, of the system can be found by:
Using the standard integral again:
∫∞
+− +Γ
=0
1
)1(n
axn
andxex
23
232)(
πhEmED =E
j
j egN
Ef βα−==)(
∫∞
=0
)()( dEEDEEfE
∫∞
−=0
23
23
232 dEeEemE Eβα
πh
2523
23
25
2
βπα
Γ
= emEh
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Maxwell-Boltzmann DistributionThe average energy per particle is given as:
Where kB is the Boltzmann constant.
Equating the average energy per particle with the ratio of our equations for E and N:
TkNE
B23
=
NETk
em
em
NE
B ==
Γ
Γ
=23
23
2
25
2
2323
23
2523
23
βπ
βπ
α
α
h
h
KeVkB
5106175.8 −×=
TkB23
23
25
23
25 =
Γ
⋅
Γ
β
β
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Maxwell-Boltzmann DistributionFrom previous slide:
TkB23
23
25
23
25 =
Γ
⋅
Γ
β
β
TkB23
2323
23
=
Γ⋅
Γ⋅
β
TkB23
23
=β
TkB
1=β
( ) ( )zzz Γ=+Γ 1
Use the relationship[1]:
[1] http://mathworld.wolfram.com/GammaFunction.html eq. 37
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Maxwell-Boltzmann DistributionNow we have β but we still need α. α will be derived in Section 5.7.
TkB
µα =
Where µ is the chemical potential. In Section 5.8, we will find out that the chemical potential, µ, is exactly equal to the Fermi Energy, EF — Leaving us with an α of:
(Eq. 5.7.26)
TkE
B
F=α
From previous slides:
TkB
1=βjE
j
jj e
gN
Ef βα−==)(
TkEE
j
jj
B
jF
egN
Ef−
==)(
Substituting α and β into f(Ej):
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Maxwell-Boltzmann DistributionFrom previous slide:
Maxwell-Boltzmann Distribution
Reversing terms in the numerator of the exponent:
This distribution gives the number of particles in the jth state, where the jth state has a degeneracy, gj.
If we want to find the probability of finding the particle in the jth state, we need to normalize. We’ll start by dividing the number of particles in the jth state, Nj, by the total number of particles, N.
Using Constraint 1 and f(E):
∑∑=
−
=
==n
j
TkE
j
n
jj
Bj
egeNN11
α
TkEE
j
jj
B
jF
egN
Ef−
==)(
( )TkEE
j
jj
B
Fj
egN
Ef−−
==)(
It’s intuitive! For a given temperature, the chance of higher energy states being occupied decreases exponentially.
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Maxwell-Boltzmann DistributionFrom previous slide:
Solving for eα:
Substituting eα into f(Ej):
∑∑=
−
=
==n
j
TkE
j
n
jj
Bj
egeNN11
α
∑=
−=n
j
TkE
jB
j
eg
Ne
1
α
TkE
jjjB
j
eegNEf−
== α)(
∑=
−
−
=n
j
TkE
j
TkE
jj
Bj
Bj
eg
eNgN
1
Normalizing:
∑=
−
−
=n
j
TkE
j
TkE
jj
Bj
Bj
eg
egNN
1
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Maxwell-Boltzmann DistributionFrom previous slide:
This normalized distribution is the probability, Pj, of finding the particle in the jth state with energy Ej:
Since j is just a dummy index, the probability of finding of a particle having an energy Eris:
∑=
−
−
=n
j
TkE
j
TkE
jj
Bj
Bj
eg
egNN
1
∑=
−
−
==n
j
TkE
j
TkE
jjj
Bj
Bj
eg
egNN
P
1
∑=
−
−
=n
j
TkE
j
TkE
rr
Bj
Br
eg
egP
1
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Maxwell-Boltzmann DistributionIf the degeneracy factor is 1, i.e. no repeatable arrangements, the probability becomes:
If we want to find out the mean value of a physical observable, y, we can make use of the following equation derived in Chapter 1:
For example, the mean energy of a system can be determined by:
∑∑
=
rr
rrr
P
yPy
∑=
−
−
=n
j
TkE
TkE
rB
j
Br
e
eP
1
∑
∑−
−
=
r
TkE
r
TkE
r
Br
Br
e
eEE
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∑
∑−
−
=
r
TkE
r
TkE
r
Br
Br
e
eEE
Maxwell-Boltzmann DistributionFrom previous slide:
The classical partition function:
∑−
=r
TkE
Br
eZ
This denominator occurs very often—every time a mean value is calculated in the Maxwell-Boltzmann Distribution. Since it occurs so often, it is given a special name, the classical Partition Function, Z.
We can use the classical partition function to easily calculate the mean value of the energy.
First, we need to note the following relationships:
∑∑ −−
∂∂
−=r
E
r
Er
rr eeE ββ
β
∑ −=r
EreZ β
TkB
1=β; ⇒
Zeer
E
r
E rr
βββββ
∂∂
−=
∂∂
−=∂∂
− ∑∑ −−
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Z
ZE β∂
∂−
=
Maxwell-Boltzmann DistributionFrom previous slide:
Substituting into the mean energy equation:
ZeEr
Er
r
ββ
∂∂
−=∑ −
∑ −=r
EreZ β
ZZ
Eβ∂∂
−=1
∑
∑−
−
=
r
TkE
r
TkE
r
Br
Br
e
eEE
( )ZE lnβ∂∂
−=
The mean energy equation
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Maxwell-Boltzmann DistributionSummarizing:
∑ −=r
EreZ β
∑
∑−
−
=
r
TkE
r
TkE
r
Br
Br
e
eEE
( )ZE lnβ∂∂
−=
∑=
−
−
=n
j
TkE
j
TkE
rr
Bj
Br
eg
egP
1
TkB
1=β
( )TkEE
j
jj
B
Fj
egN
Ef−−
==)( TkE
B
F=α
( )TkEE
B
F
eEf−−
=)(
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Maxwell-Boltzmann DistributionSummarizing:
( )ZE lnβ∂∂
−=
• For a given temperature, the chance of higher energy states being occupied decreases exponentially
• The area under the curve, i.e. the total number of molecules in the system, does NOT change
• The most probable energy value is at the peak of the curve, when dP/dE = 0
• The average energy value is greater than the most probable energy value
• If the temperature of the system is increased, the most probable energy and the average energy also increase because the distribution skews to the right BUT the area under the curve remains the same
http://www.bustertests.co.uk/studies/kinetics-collision-theory-maxwell-boltzmann-distribution.php
∑=
−
−
=n
j
TkE
j
TkE
rr
Bj
Br
eg
egP
1
( )TkEE
B
F
eEf−−
=)(