Announcements 9/26/12
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Prayer
Exam 1 starts Saturday morning, goes until Thursday
evening
On Friday at the start of class I will talk a bit about what
to expect for the exam
Exam review session: Today, 5 pm, room C295
 Come with probems for me to discuss! (HW,
optional HW, old exams, etc.)
Pearls
Before
Swine
From warmup
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Extra time on?
a. “What is entropy” handout
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Other comments?
a. What is the best thing for us to do to prepare for the test
we have next week?
b. Will there be another review by a TA or is today's the only
one?
From warmup
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Why does heat flow from hot to cold when the process of
energy exchange between two objects is "random". (How
can you get directed motion of heat, when energy is being
exchanged both ways?!)
a. There is more energy on the hot side than the cold side
which means that although it is random, there are more
possibilities for heat to flow from the hot to cold than
from cold to hot. There is also heat being exchanged
from cold to hot but there is a lot less of it so the heat
generally flows from hot to cold.
Microstates vs Macrostates
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Reminder
a. Left microstate: part of the “royal flush” macrostate
b. Right microstate: part of the “garbage” macrostate
c. The most common macrostates are those with the
most microstates
p.413a
Marble Example
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50 red, 50 green in a bag. Draw 4.
a. Microstates?
b. Macrostates?
“most disordered”
 most entropy
About HW 13
Hint: same as DS for a
“free expansion”
HW 13, cont.
(Various questions about micro- and macrostates follow.)
HW 13, cont.
HW 13-4: my plot
From warmup
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When two systems A and B can exchange
energy, the entropy of system A *always*
decreases when system A gives energy to
system B. If that's so, why would energy ever
spontaneously flow from system A to system
B? (It often will. When? Why?)
a. Because system B is at a lower temperature
and the increase in entropy of system B will
be greater than the decrease in entropy of
system A.
Clicker question:
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Which of the following is the equation that
relates the # microstates (W) to entropy (S)?
a. S = CV W
b. S = kB eW
c. S = kB lnW
d. S = tan(W)
e. S = tan-1(W)
Dice
You roll two dice. What are the microstates?
(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),…
 How many microstates are there?
 How does that compare to the number of
microstates for rolling one die?
 How many microstates if we roll 3 dice?
 What are the macrostates for 2 dice? (sum of
numbers)
 What is the most likely macrostate?
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MANY Dice
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You roll 1023 dice with your left hand.
a.How many microstates are there?
You roll 1023 dice with your right hand.
a.How many microstates are there?
How many microstates are there in the
COMBINED system?
Isn’t this ridiculous?
Solution: Use logarithms
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S = some constant  ln(#microstates)
[units of J/K]
a. Much more manageable numbers.
b. Combining two systems:
Stot = C  ln(#microstates1  #microstates2)
= C  ln(#microstates1) + C  ln(#microstates2)
= S1 + S2
c. 2nd Law: System in macrostate with most microstates
 System in macrostate with largest S
System and Reservoir
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System: E1
Large reservoir: E2
Etot = E1 + E2 (const. volume so no work)
0 = dE1 + dE2
Want to maximize S: take dS/dE1, set = 0
dS
d

 S1  S2 
dE1 dE1
dS1 dS 2


dE1 dE1
dS1 dS 2


dE1 dE2
dS1 dS 2


dE1 dE2
From warmup
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In the "What is entropy?" handout, what was
significant about the equation dS1/dE1 = dS2/dE2?
a. It provide a value that is the same for both
systems and relates to the inverse of the
temperature: (J/K)/J = 1/K.
Temperature
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dS/dE is the same for two systems in thermal contact!
Temperature is also the same for two systems in
thermal contact!
a. dS/dE has units of 1/K, so…
Compare to
dS 1

dE T
dQ
dS 
T
We are assuming no
work, so dE=dQ (First
Law)
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This “works” if the constant is chosen properly:
S = kB ln(#microstates)
Small system with 2 possible energies:
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E1A vs E1B (1 microstate each)
Probability of system 1 being in state A vs state B?
P1A ~ (#microstates of system 1 having energy E1A) 
(#microstates of system 2 having energy E2A = E – E1A)
Let #microstates of E1A = 1 for now.
Same thing for state 1B…
P1A # microstates of 2 for case A

P1B # microstates of 2 for case B
e S2 A
 S
e 2B
kB
kB
Math…
P1A # microstates of 2 for case A e S2 A

 S
P1B # microstates of 2 for case B e 2 B
kB
kB
S 2 A  S 2 ( E  E1 A )
“of”, not
“times”
dS 2
 S2 ( E ) 
 E1 A
dE
E1 A
 S2 ( E ) 
T
(same with S 2 B )
 E1 A / k BT
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P1A e
Result:
 E /k T
P1B e 1B B
The Boltzmann Factor
Pbeing in state with energy E ~ e
 E / k BT
“Boltzmann Factor”
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Prob is proportional to BF, but not equal
 Must normalize: Prob = BF/(sum of all BFs)
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Worked Problem: Suppose an atom has only
two available energy levels, which are
separated by 210-23 J. If the temperature is
1.5 K, what is the probability the atom is in the
lower state?
Maxwell-Boltzmann Velocity Distribution
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E = ½mv2
What’s probability of having speed 5 vs speed
10?
P1A e E1 A / kBT
 E /k T
P1B e 1B B
v
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2
1
 mv 2 / k BT
e 2
1
 mv A2 / k BT
2 2
e
P1 A v A

1
P1B
 mvB 2 / k BT
2 2
vB e
Multiplicities
 (Number of states with speed v) ~ v2
Maxwell-Boltzmann Velocity Distribution
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The result:
P(speed v) 
v
2
1
 mv 2 / k BT
e 2
1 2
 mv / k BT
2 2
v e


1 2
 mv / k BT
2 2
v e
dv
0
 Exactly the equation given for the
velocity distribution in your textbook! (after
you do the integral, e.g. with Mathematica)