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Material Type: Notes; Class: Univ Physics: Thermal Physics; Subject: Physics; University: University of Illinois - Urbana-Champaign; Term: Fall 2008;
Typology: Study notes
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Physics 213: Lecture 7, Pg 1
Physics 213: Lecture 7, Pg 2
Lecture 7 Lecture 7
The Boltzmann Distribution The Boltzmann Distribution
Concept of a thermal reservoir
The Boltzmann distribution
Paramagnetic Spins – MRI
Elasticity of a Polymer
Supplement: Proof ofequipartition, showing its limits
Physics 213: Lecture 7, Pg 4
o
n
1
2
n
1
1
2
2
n
n
o
Physics 213: Lecture 7, Pg 5
Concept of a Thermal Reservoir Concept of a Thermal Reservoir
n
m
q
R
= energy of reservoir
ε
ε
ε
ε
Total energy U = E
n
R
Set
.
R
Key problem
: What is the probability,
n
, that the small system is in
the state n with energy
n
? (Here E
n
= n
ε ε
ε ε
, i.e., n quanta)?
Answer
: The probability of finding the small system in state n is
proportional to the corresponding
R
Small
Reservoir
system
R
n
P
Ω
∝
Physics 213: Lecture 7, Pg 7
n = label of state n (quantum #)E
n
= energy of system in state n
(not necessarily an oscillator)
tot
n
R
E
n
U
R
R
R
Basic Question:
What is the probability that the small system is
in a particular state
(labeled n) with energy E
n
R
n
R
R
i
Ω Ω
Ω Ω
R
=
ΩΩΩΩ
R
(U
tot
n
) = # states of reservoir
when the particle has energy E
n
.
(See previous table.)
The probability of finding thesmall system in a state withenergy E
n
is proportional to the
corresponding number of statesin the
reservoir
P
n
E
n
The entropy of thereservoir decreases whenthe small system extractsE
n
from it. But T
R
is ~ fixed.
Physics 213: Lecture 7, Pg 8
R
n
R
n
R
R
R
o R
R
tot
o
R
R
tot
n
R
n
R
o
n
R
R
R
R
R
E kT
n
σ
σ
σ
σ
σ
σ
σ
σ
σ
σ
−
n
n
kT
kT
n
e
P
Ce
Z
=
≡
Boltzmann
Factor:
n
/
1
Z
n
n
E
kT
n
P
e
−
=
⇒
=
∑
∑
“Partition
function”
Physics 213: Lecture 7, Pg 10
A particular molecule has three states, with energy spacing
ε ε
ε ε
= 10
-
J, as
shown; the molecule is in contact with the environment (reservoir), which hasa temperature of 1000K.
a) What is P
1
, the probability that the molecule is in the middle energy
state?
1
n
E / kT
/ kT
1
E
/ kT
0 / kT
/ kT
2
/ kT
23
3
e
e
e
e
e
e
kT
−
−ε
−
−
−ε
− ε
−
ε
∑
−
−
−
b) What is P
2
, the probability that it is in the highest energy state?
E
2
E
1
E
o
Physics 213: Lecture 7, Pg 11
A particular molecule has three states, with energy spacing
ε ε
ε ε
= 10
-
J, as
shown; the molecule is in contact with the environment (reservoir), which hasa temperature of 1000K.
a) What is P
1
, the probability that the molecule is in the middle energy
state?
b) What is P
2
, the probability that it is in the highest energy state?
kT
/
E
kT
/
E
kT
/
E
kT
/
E
2
2
1
0
2
e
e
e
e
−
−
−
−
E
2
E
1
E
o
1
n
E / kT
/ kT
1
E
/ kT
0 / kT
/ kT
2
/ kT
23
3
e
e
e
e
e
e
kT
−
−ε
−
−
−ε
− ε
−
ε
∑
−
−
−
Physics 213: Lecture 7, Pg 13
Act 1 Act 1
Since P
1
and P
2
both
0 as T
0
must
At zero temperature, the molecule is always
in the ground state.
) What is P
0
at T = 0?
a)
b)
c) 1
A particular molecule has three states, withenergy spacing
εεεε
= 10
-
J, as shown; the
molecule is in contact with the environment(reservoir
),),),),
at temperature T.at temperature T.at temperature T.at temperature T.
E
2
E
1
E
o
) What is P
2
as T
a)
b)
c) 1
Physics 213: Lecture 7, Pg 14
Act 1 Act 1
A particular molecule has three states, withenergy spacing
εεεε
= 10
-
J, as shown; the
molecule is in contact with the environment(reservoir)
at temperature T
.
E
2
E
1
E
o
) What is P
2
as T
a)
b)
c) 1
2
0
1
2
/
0
2
/
/
/
0
0
0
1 3
E
kT
E
kT
E
kT
E
kT
e
e
P
e
e
e
e
e
e
−
−
−
−
−
−
−
−
=
→
=
For very high temperatures, all states are equally likely to be occupied,because the reservoir entropy doesn’t depend much on energy.
a) decreases
b) increases
c) increases, thendecreases
2
as we decrease T?
Physics 213: Lecture 7, Pg 16
The probability of finding the particle with energy E
equals
the number of single-particle states at that energy
(
ΩΩΩΩ
1
) TIMES
the probability that one state is occupied
:
3 states
1 state
E
/
P(E)
3
/
E
kT
e
Z
−
=
0 /
P(0)
/
1/
kT
e
Z
Z
−
=
=
/
1
3
E
kT
Z
e
−
=
P(E) depends
on the # ofstates at E.
/
n
P
/
n
E
kT
e
Z
−
=
Thermal reservoir
temperature T
Small system (single atom, oscillator, spin…)
Probability
one
particular state (n) is occupied:
You may come across a system that is “degenerate”, which means thatmore than one state has the same energy. In this case, you simply needto count the number of degenerate states.
Physics 213: Lecture 7, Pg 17
Paramagnet: Paramagnet:
System of Independent Magnetic Spins System of Independent Magnetic Spins
N spins, each with magnetic
moment m , in contact with athermal bath at temperature T.
Each spin has potential energy:
B
E
n
μ
±
=
n
n
n
Ce
P
=
Physics 213: Lecture 7, Pg 19
n n
-E
n
kT
)
kT
/
B
tanh(
N
M
μ
⋅
μ
=
Let’s plot this function:
down
up
/
/
B kT
B kT
μ
μ
−
kT
kT
kT
kT
down
up
e
e
e e N ) P P ( N M
μ
μ
μ
μ
− +
μ
=
−
μ
=
kT
/
B
down
Ce
P
μ
−
=
B
E
down
μ
=
B
E
up
μ
−
=
kT
/
B
up
Ce
P
μ
=
PartitionFunction
Physics 213: Lecture 7, Pg 20
At sufficiently high temperatures, the ratio
x =
μμμμ
B / kT
is much less than
unity. Using
for small x:
Curie’s Law
Total magnetic moment
of the spin system:
T
2
M
B/T
x
)
x
tanh(
≈
B/kT
)
/
tanh(
kT
B
N
M
μ
μ
⋅
=
High-B, low-T,spins lined up“saturation”
Low-B, high-T,few spins lined upLinear response
