Discussion: Einstein solid
Clicker A question #1
A hot piece of metal sits on block of ice: why can’t there be thermal transfer of energy
from ice to metal, making the ice colder and the metal hotter? A pen lies on a table:
why can’t the pen jump up into the air, leaving the table colder? Statistical issues
arise. The governing principle is “the second law of thermodynamics” (explain briefly
that the first law is essentially just the energy principle, that energy inputs minus
outputs equal change in energy of the system).
May sound fuzzy, but can make quantitative! Easiest system is the balland-spring model of a solid, because atoms don’t move far from their
equilibrium positions, so don’t have to worry about probabilities
associated with spatial distributions. Einstein solid: special case of the
ball-and-spring model where we ignore collective motions and for purposes of
counting probabilities pretend that each atom is an independent oscillator. Actually
THREE independent oscillators: x, y, and z, thanks to the separation of K and U terms.
VPython Demo: Einstein solid
wells_oscillator.py
Show program wells_oscillator.py to emphasize that
one atom is modeled as three oscillators.
Reminder of quantized energy for harmonic oscillator. We’ll measure from the ground
state, often in units (“quanta”) of w . We’re interested in the probabilities for
finding energy distributed in certain ways: why doesn’t the pen jump off the table,
with some energy from the table moving into the pen?
Ponderable: Size of Quanta
Relevant data:
Atomic Mass (kg/mol)
Pb 207 E -3
Al 27 E -3
Fe 56 E -3
Density (kg/m3)
11.4 E 3
2.7 E 3
7.9 E 3
Young’s Modulus (N/m2)
13.8 E 9
69 E 9
198 E 9
Calculate matom, datom, and q
Chapter 12
1
A Groups:
Pb matom = 207 ´10
-3 kg
mol
6 ´10
23 atoms
mol
r =11.4 cmg =11.4 ´10 3 mkg =
3
3
(
)(
= 3.45 ´10 -25 kg
matom
3
datom
d=
)
3
matom
r
= 3.12 A
kbond = Ydatom = 13.8 ´10 9 mN2 3.12 ´10 -10 m = 4.31 mN
kbond
= 1.05 ´10 -34 J ×s
matom
(
Spacing =
)
4.31 mN
= 3.7 ´10 -22 J
-25
3.45 ´10 kg
3.7 ´10 -22 J
= 2.3 meV
1.6 ´10 -19 eVJ
B Groups:
Al matom = 27 ´10
-3 kg
mol
6 ´10 23 atoms
mol
r =2.7 cmg =2.7 ´10 3 mkg =
3
3
(
)(
= 0.449 ´10 -25 kg
matom
3
datom
d=
3
matom
)
r
= 2.55A
kbond = Ydatom = 69 ´10 9 mN2 2.55 ´10 -10 m = 17.6 mN
kbond
= 1.05 ´10 -34 J ×s
matom
(
Spacing =
)
17.6 mN
= 20.7 ´10 -22 J
0.449 ´10 -25 kg
20.7 ´10 -22 J
= 13 meV
1.6 ´10 -19 eVJ
C Groups:
Fe matom = 56 ´10
-3 kg
mol
6 ´10 23 atoms
mol
r =7.9 cmg =7.9 ´10 3 mkg =
3
3
(
= 0.930 ´10 -25 kg
matom
3
datom
)(
d=
matom
3
)
r
= 2.27A
kbond = Ydatom = 197 ´10 9 mN2 2.27 ´10 -10 m = 44.8 mN
Spacing =
kbond
= 1.05 ´10 -34 J ×s
matom
(
)
44.8 mN
= 23 ´10 -22 J
0.930 ´10 -25 kg
23 ´10 -22 J
= 14 meV
1.6 ´10 -19 eVJ
In the Einstein model the x, y, and z oscillations each involve 2 half-length
springs, so the effective stiffness is actually 4 times the individual k and so
the quanta are twice these values.
Chapter 12
2
Tangible: Activity - Filling states
WID 1173500
counting
How many different ways can 1 quantum of energy go into a system of three
oscillators (one atom)? 100, 010, 001 so three different ways.
A groups: How many different ways can 2 quanta of energy go into a system of three
oscillators (one atom)? 200, 020, 002, 110, 101, 011 so six different ways.
Don’t show them this yet:
( q + N - 1)! = ( 2 + 3 - 1)! = 4! = 4 ´ 3 ´ 2 ´ 1 = 6
q!( N - 1)!
2!( 3 - 1)! 2!2! ( 2 ´ 1) ( 2 ´ 1)
B groups: List microstates for 2 quanta of energy shared among 4 one-dimensional
oscillators: 2000, 0200, 0020, 0002, 1100, 1010, 1001, 0110, 0101, 0011. Note that
this is not really physical: corresponds to 1 and 1/3 atoms.
Don’t show them this yet:
Chapter 12
3
( q + N - 1)! = ( 2 + 4 - 1)! = 5! = 5 ´ 4 ´ 3 ´ 2 ´ 1 = 10
q!( N - 1)!
2!( 4 - 1)! 2!3! ( 2 ´ 1) ( 3 ´ 2 ´ 1)
C groups: List microstates for 4 quanta of energy shared among 2 one-dimensional
oscillators: 40, 04, 31, 13, 22. Note that this is not really physical: corresponds to 2/3
of an atom.
Don’t show them this yet:
( q + N - 1)! = ( 4 + 2 - 1)! = 5! = 5 ´ 4 ´ 3 ´ 2 ´ 1 = 5
q!( N - 1)!
4!( 2 - 1)! 4!1! ( 4 ´ 3 ´ 2 ´ 1)
Once drawings are done, have the groups use W =
( q + N - 1)!
q!( N - 1)!
q is the number of quanta and N is the number of places to put them. Calculate for 1
quanta in a single atom (3 oscillators):
q + N - 1 ! 1+ 3 - 1 ! 3! 3 ´ 2 ´ 1
=
=
=
=3
2 ´1
q! N - 1 !
1! 3 - 1 ! 1!2!
(
(
) (
)
(
)
)
Then have a,b,c groups redo their work to verify the formula works.
Consider three oscillators -- one atom: how many ways can we arrange 4 quanta of
energy? Show 400, 040, 004, then have students find the remaining arrangements for
parceling out 4 quanta into three oscillators. These are different “microstates” of one
“macrostate” -- total energy is 4 quanta. We find 15 different ways: 400, 040, 004,
310, 301, 130, 031, 013, 103, 220, 202, 022, 112, 121, 211. (N = 15)
( q + N - 1)! = ( 4 + 3 - 1)! = 6! = 6 ´ 5 ´ 4 ´ 3 ´ 2 ´ 1 = 15
q!( N - 1)!
4!( 3 - 1)! 4!2! ( 4 ´ 3 ´ 2 ´ 1) ( 2 ´ 1)
Chapter 12
4
Discussion: Fundamental assumption of statistical mechanics
Fundamental assumption of statistical mechanics: all of these “microstates” are
equally likely. Sounds plausible; real justification is that starting from this assumption
we reach conclusions that are consistent with the way the world actually works.
Clicker A questions: section 2, 12.2a is just the previous calculation.
Ponderable: Most probable sharing
Quantum Spreadsheet.xls
Next consider two atoms, N = 6 oscillators, that are
sharing q = 4 quanta. Find number of ways to divide
energy between the two atoms: 0+4, 1+3, 2+2, 3+1,
4+0. Students do part of these calculations, table by
table. It was very helpful to explicitly list possibilities,
first for the 0+4, which they just did, (400, 040, 004,
301, 310, 202, 220, 211, 103, 130, 112, 121, 013, 031,
022 = 15 microstates) and then for the 1+3 quanta
arrangement
Do a,b,c groups:
Atom1 with 001 and Atom2 having 003, 030, 300, 012, 102, 021, 121, 210, 201, 111.
Atom1 with 010 and Atom2 having 003, 030, 300, 012, 102, 021, 121, 210, 201, 111.
Atom 1 with 100 and Atom2 having 003, 030, 300, 012, 102, 021, 121, 210, 201, 111.
Gives a total of 30 microstates.
Now calculate the 2+2 for the first atom as q=2, N=3 gives
q + N - 1 ! 2 + 3 - 1 ! 4!
4 ´ 3´ 2 ´1
=
=
=
=6
q! N - 1 !
2! 3 - 1 ! 2!2! 2 ´ 1 2 ´ 1
(
(
) (
)
(
)
)
(
)(
)
Or 200, 020, 002, 110, 101, 011
Second atom is the same, so 6 times 6 is 36.
Then all plot “histogram” (explain what such a bar chart is,) using Excel. (Quantum
Spreadsheet.xls) What sharing is most probable? Equally shared energy.
Two ways of thinking about this result: probability of finding a system in one of these
arrangements, or consider 100 such systems and look at this instant to see how many
of these 100 systems have a particular arrangement.
Chapter 12
5
VPython: Statistical Mechanics I
WID 2465378 charts
Shell given to students.
12.P.69a: use VPython to compute the same thing
as from the previous activities and plot a histogram
of the number of microstates for two atoms (6
oscillators) sharing 4 quanta of energy; should get
the results we obtained by hand and with Excel. Horizontal axis is the number of
quanta in the first atom (reading left to right) or, equivalently, the number of quanta in
the second atom (reading right to left)
12.P.69b: Next consider two small blocks (actually nanoparticles) containing 300
oscillators (100 atoms) and 200 oscillators (about 67 atoms), sharing 100 quanta of
energy. (Too big for Excel.) The blocks are made of the same material, so 1 quantum
of energy is the same for oscillators in both blocks. Just adjust the previous program.
Note that they can mouse down on the graph to read values.
N1 = 300, N2 = 200
N1=250=N2
N1=100, N2=400
N1=400, N2=100
The numbers get REALLY big!
12.P.70 has you plot the natural log of the results, to make it
possible to see the wings of the distribution. It is this
logarithm that is related to the “entropy” of an object. Note
that in VPython ln(x) is written log(x).
Plot ln(#ways), which has the practical advantage of being able to see the wings of the
distribution, and being able to see the contributions of each block separately. Note that
the position of the maximum is still 60/40; taking the logarithm doesn’t change that.
Talk about the meaning of the individual contributions (block 1, and block 2); as you
put more energy into the system, the number of ways to arrange the energy grows,
which is not a big surprise. Note that the highest value for the larger block is larger
than the highest value for the smaller block. You can think of the #ways vs. E for
block 2 as going right to left on the graph. Emphasize that where the two individual
curves cross has no physical significance; what matters is where the total #ways is
maximum (60/40).
Chapter 12
6
maxq = 60, log(maxways) = 264
4.5e114 (can’t use calculator unless it has
probability functions built in, but Excel or Wolfram Alpha will work)
Off-peak doesn’t mean less total energy; same total energy, but fewer ways to arrange
it. As you move off the 60/40 peak in the direction of say 58/42, the number of ways
to arrange energy in the larger block (red curve) decreases faster than the number of
ways to arrange energy in the smaller block increases. (Red curve for object #1 has a
steeper slope than blue line for object #2. Also note the direction of the blue line from
right to left.) Vice versa when moving to 62/38 sharing.
Remind students that when they made the blocks be say 100 and 400 oscillators, the
peak shifted to 20/80 sharing, etc. Summary: Most probable distribution is to share
energy according to size of objects. Nevertheless inside one block, equal energy on
each oscillator (each atom) is just ONE of a gigantic number of ways of arranging that
block’s share of the total energy, and so this is highly unlikely. In fact, ANY one
particular microstate is highly unlikely. In the 300-oscillator block, there are 1.7e96
microstates!
Peak gets narrower as you increase the number of oscillators (atoms); see Figure
12.21 on page 481 (above). What if you go to 1e23 atoms, with lots of energy? The
distribution is so incredibly narrow that the probability of ever seeing two
macroscopic objects very far from the most probable energy sharing is zero for all
practical purposes. Even with our nanosystem of 300 and 200 oscillators, how likely
would it be to find all 100 quanta in block 1 and none in block 2? (From Table on pg
481 and Figure 12.21, get omega = 2.772e81 so can divide (Excel) by 7e114 then
times 100 to get 4e-32) Show the Excel spreadsheet (Table pg 481.xls). and 30/20
oscillators
Discussion: Entropy
Define entropy as S = kln(#ways), where the Boltzmann constant k = 1.4e-23 J/K.
Show that it is an additive property of objects.
State the second law of thermodynamics: If a closed system is not in equilibrium, the
most probable consequence is that the entropy will increase. Note that this doesn’t
Chapter 12
7
mean that the entropy of a (sub)system can’t decrease: it could be in contact with
surroundings whose entropy increases even more, so that the total entropy does
increase.
We begin to see why the pen doesn't leap off the table: it can happen but is wildly
improbable.
Discussion: Temperature
Key concept: Give the argument leading to an entropy-based definition of
temperature. Two blocks are in equilibrium:
S = S1 +
S2
dS
=0
dq1
dS1
dq1
S2
dS 2
dq1
S1
Peak is only really possible state and there the slope is zero. (Don’t forget that these
are really strings of quantized dots, not a real line.) The total entropy is S = S1 + S2,
and with slope zero the following is true:
dS dS1 dS2
=
+
=0
dq1 dq1 dq1
Since q2 = 100 - q1 then dq2 = -dq1 and we get (at equilibrium)
dS1
dS2
=0
dq1 dq2
A physical property of equilibrium that is shared by the two blocks is temperature, so
these ratios are related to temperature. To see the relationship, note that as you move
from left to right on the plot, we are moving quanta from the first block to the second
block approaching equilibrium. The red slope is steeper than the blue slope and the
first object is warmer than the second. So the slope is inversely related to
temperature. (Can’t use quanta of energy since it varies with system. So instead we
use Einternal, the internal energy above the ground state. This corresponds to
temperature on the Kelvin scale.
1 DS
=
T DE
Chapter 12
-
8
Discussion: Review
Table pg 481.xls
Review. Show them Table pg 481.xls spreadsheet and show that numbers get too big
for Excel. Show how more oscillators in first object shifts the most probable number
of quanta (the peak) into the first object also.
Discuss question on pg 482. What happens with real numbers of atoms?
Discuss Fig 12.28. Show what happens when start with q1 = 90 then go to q1 = 60.
Same final slope, and since S/E = 1/T, same final temp.
Discuss Fig. 12.29.
Chapter 12
9
Blue slope greater than orange
slope, so for same E blue gives
greater S, means lower
temperature for second object.
Final slopes the same, so same
final temperature.
Ponderable: Activity - What is this thing called “quanta”?
WID 1172916 blocky
Consider a 3 kg block of aluminum. One mole of aluminum has a mass of 27 grams
(0.027 kg). From Young’s modulus we determined that the stiffness of the interatomic
bond is 16 N/m, but in the Einstein model the x, y, and z oscillations each involve 2
half-length springs, so the effective stiffness is 64 N/m. What is the energy in joules of
one quantum of energy?
kg
0.027 mol
kg
= 4.5 ´ 10 -26 atom
23 atoms
6 ´ 10 mol
w=
ks
= 1.05 ´ 10 -34 J × s
m
(
)
64 mN
4.5 ´ 10 -26
kg
atom
= 3.96 ´ 10 -21 J
Ponderable: Activity - Heat Capacity
WID 1172920 XL
Metal Calc.xls has answers
Aluminum calculation
and Metal Calc Shell (on Resources page) shows
They will do (Ch 12 HW 3) on WebAssign a calculation of the following kind for q =
20, 21, 22, but without the calculation of the heat capacity per atom. They just
Chapter 12
10
reviewed with Aluminum in “What is this thing called quanta”.
Show them 100 atoms of Aluminum calculations in the spreadsheet:
k = 1.38E-23
Aluminum
one quantum = 3.98446E-21
q
omega
E
S
20
4.9100E+26
7.9689E-20
8.4856E-22
21
4.4400E+27
8.3674E-20
8.7896E-22
22
3.8500E+28
8.7658E-20
9.0878E-22
heat cap/atom
1.5650E-23
from
midpoint
to
mid-point
ks = 64
hbar = 1.05E-34
Mmole = 0.027


T
3.9845E-21
3.0403E-23
131.06
3.9845E-21
2.9823E-23
133.60
Show that adding one quantum of energy raises temp (from 131.06 to 133.60). That
tells us per atom heat capacity. C = E/T, divide by 100. So for 20-21-22 (middle of
energy jump to middle of next energy jump is just one quantum of energy) get
DE E21®22 - E20®21
3.9845 ´ 10-21
J/K
C=
=
=
¸ 100 atoms = 1.565 ´ 10-23 atom
DT T21®22 - T20®21 133.60 - 131.06
Have them build their own spreadsheet and analysis for 100 atoms of different metals
(gold, tungsten, and copper: a,b,c groups).
They will need the q and omega values, ks, and Mmole.
Chapter 12
11
 values are given, E = q*1 quantum, S = kln(), differences are by subtraction,
T = 1/(dS/dE).
Comment that the heat capacity rises with temperature. Make the comment that at
high temperatures (which may be room temperature!) the heat capacity reaches the
classical value of 3kB per atom (4.2e-23 J/K/atom).
Reaches a plateau of 3k =
4.2 e-23 per atom at high
temp (like room).
C
Would have won Nobel prize 100 years ago.
Chapter 12
12
0
T
12
Ponderable: Activity - Phase changes
WID 1173554
phaser
SKIPPED
1 g of ice at 0C + 335 J yields 1 g of liquid water at same temperature. The entropy
increases. Latent heat of fusion Lf = 335 J/g. What is the change in entropy?
1 DS
DE 335 J
=
so DS =
=
= 1.2 KJ
T DE
T
273K
Atoms freer to move around so can arrange things even more ways. We don’t know
anything about counting microstates in water.
1 g of liquid water at room temp 20C + energy yields 1 g of liquid water at 100C.
What is the change in entropy? Since temperature changes, have to integrate
373 cdT
373 dT
æ 373 ö
dE
=ò
= cò
= 4.2 KJ ln ç
= 1.0 KJ
÷
T =293 T
293 T
293 T
è 293 ø
So just melting with no T raises entropy by 1.23 J/K, then raising temp from 20 to
100 only raises by 1 J/K. This is a major difference between solids and liquids.
DS = S373 - S293 = ò
T =373
(
)
Some abstractions:
A certain object (for which the Einstein solid is not a good model) has entropy
1 dS
. Find the temperature T as a
S = bE1/ 2 , where b is a constant. We know that =
T dE
function of energy E.
Chapter 12
13
1 dS
=
=
T dE
( )= bE
d bE
1
2
dE
- 12
2
so T =
2 12
E
b
Assuming temp didn’t rise by much because large object. Notice signs. Adding
energy made entropy go up (because more ways to arrange energy) and temp went up.
Suppose the object has a mass of 1 gram. Calculate the heat capacity on a per-gram
2 1
basis as a function of the temperature T, starting from T = E 2 .
b
dE
C=
=
dT
æ b2 2 ö
dç T ÷
è 4
ø
dT
( )
2
b2 d T
b2
=
= T
4 dT
2
Ponderable: Boltzmann Distribution
WID 1173700 energy (have each group do their part and share with others at table.
SKIPPED
A: Problem 12.X.15, pg 498. One air molecule gets bumped off desk at room temp.
How high does it go if it doesn’t hit any other molecules? i.e. “typical height” when
kT = Mgy
1.38 ´ 10 -23 KJ 300K
kT
y=
=
= 8.7 km
kg
Mg æ 0.029 atom
ö
N
çè 6 ´ 10 23 atoms ÷ø 9.8 kg
mol
(
)
B: Problem 11.X.13 Marbles on desk- Marbles of mass M = 10 g are lying on the
floor. They are in thermal equilibrium with their surroundings. What is the typical
height above the floor for one of these marbles? That is, for what value of y is Mgy =
kT ?
1.38 ´ 10 -23 KJ 300K
kT
y=
=
= 4 ´ 10 -20 m
N
Mg
( 0.010 kg ) 9.8 kg
(
)
C: Problem 12.X.16 (also Clicker question #12.7d)
Chapter 12
14
r = rsea levele r
r sea level
=e
Mgy
kT
- Mgy
kT
=e
-
æ 0.029 kg ö
atom 9.8 N 8848 m
)
ç
kg (
23 atoms ÷
è 6´10 mol ø
(1.38´10
-23 J
K
)300 K
= 0.36
VPython Lab: Statistical Mechanics II
WID 2465342
lastone
Students finish 12.P.71 and 12.P.72, starting from a shell and their own program for
12.P.70 (log of #ways).
Unlike 12.P.69 and 12.P.70, where the emphasis
was on two blocks and their equilibrium, in 12.P.71
the emphasis is on a single block with its
dependence of entropy on energy, from which can
be calculated the temperature as a function of
energy. A secondary consideration is doing the
calculation for a second block and noting that at the
60/40 split in energy the temperatures are the same.
A possible source of confusion is that we plot q2 to the left, so occasionally a student
thinks that the temperature is negative due to what appears as a negative slope. But in
fact S2 rises with increasing E2, and the temperature in block 2 is positive.
12.P.72
Start from shell and previous programs. Shell gives
the students the experimental data in the form of a
VPython program for the heat capacity as a
function of temperature for aluminum and lead, on
a per-atom basis. (These data are given in the
textbook on a per-mole basis.)
A good technique is to define additional variables, q1a = q1+1 and q1b = q1+2. Then
use q1, q1a, and q1b to calculate three different values of the entropy. One can then
calculate the two values of entropy difference needed to determine two temperatures
T-lower and T-upper, from which the heat capacity can be calculated as C =
dE/dT/Na, where Na = 35 atoms. Note that dE here corresponds to one quantum of
Chapter 12
15
energy, hbar*sqrt(4*ks/m). It makes sense to calculate dEAl and dEPb for aluminum
and lead, and use these to calculate and plot the heat capacities for the two metals.
A quick check on whether a program is doing the right thing is to see whether the
high-temperature plateau in the plot is approximately 4.2e-23 J/K/atom.
Bugs: Not dividing by 35 to put theory on a per-atom basis to compare with
experiment. E = hbar*(4*ks/m)**1/2 instead of E = hbar*(4*ks/m)**(1/2), so there
was no square root. And there is a problem with 1/2 evaluating to 0 if one does not
invoke "from __future__ import division". Better to use sqrt(…).
* Writing q in the combinatoric formula when q+1 or q+2 was needed.
* Incrementing q an extra time so that T-lower involved q and q+1, whereas T-upper
involved q+2 and q+3.
* Using aluminum mass for lead calculation, or using grams instead of kilograms, or
not dividing by Avogadro’s number for the mass of one atom.
Discussion: Boltzmann Distribution
Probability of finding energy E in a small system in contact with a large reservoir is
E
-E
proportional to W(E)e kT , where e- kT is the “Boltzmann factor” and W(E) is the
number of microstates corresponding to energy E.
SKIPPED
THIS
K Trans
Mgy
é - kT ù é - kT ù é - EkTvib
êe
ú êe
ú êe
ûë
ë
ûë

In
Erot
ù é - kT ù
ú êe
ú
ûë
û
an
ideal
gas,
Boltzmann
factor
is
Ponderable: Boltzmann Distribution
Do problem 12.X.14 on page 496
A microscopic oscillator has its first and second excited states 0.05 eV and
0.10 eV above the ground state energy. Calculate the Boltzmann factor for
the ground state, first excited state, and second excited state, at room
temperature.
e
e
-
-
E
kBT
E
kBT
=e
=e
-
-
0
(1.38e-23)(300)
(0.05)(1.6e-19)
(1.38e-23)(300)
Chapter 12
= e0 = 1
= e-1.9 = 0.14
16
e
-
E
kBT
=e
-
(0.05)(1.6e-19)
(1.38e-23)(300)
= e-1.9 = 0.14
for 0.10 eV get 0.021.
What if 6000K? 0.82
Clickers for section 12.5-7
Chapter 12
17