Announcements 2/4/11
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Exam starts Tuesday, goes until next Tuesday
(late fee on last day after 5 pm)
Exam review session: today, 3-4:30 pm,
room C261
 Come with questions!
On Monday I will talk a little about what’s on the
exam.
Reading quiz
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Which of the following was a “macrostate”
listed in the marble example?
a. 1 green, 3 red
b. 50 red, 50 green
c. red, red, green, red
d. the number 6
Reading quiz
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Which of the following is the equation that
relates the # microstates (W) to entropy (S)?
a. S = kB eW
b. S = kB lnW
c. S = tan(W)
d. S = tan-1(W)
e. S = W
Reading quiz
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Which of the following is the factor that tells
you how likely states are to be occupied at a
given temperature?
a. Boltzmann factor
b. Einstein factor
c. Fermi factor
d. Maxwell factor
e. Plank factor
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
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.
How many microstates are there?
You roll 1023 dice with your right hand.
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
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

P1A e
Result:
 E /k T
P1B e 1B B
The Boltzmann Factor
P(beinginstate with energy E) ~ e
 E / kBT
“Boltzmann Factor”
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Prob ~ BF; Prob = BF/(sum of all BFs)
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

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 with, e.g. Mathematica)