VPython: Statistical Mechanics I
WID 628603 (Problems #’s are 12.P.69-12.P.70 in the 3rd ed.)
Shell given to students.
11.P.38a: 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.
11.P.38b: 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.
N1 = 300, N2 = 200
N1=250=N2
N1=100, N2=400
N1=400, N2=100
The numbers get REALLY big!
11.P.39 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).
maxq = 60, log(maxways) = 264 4.5e114 (can’t use calc, Excel will work)
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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 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
11.23 on page 383. 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? (Use Figure 11.23 and spreadsheet to show approximately
2.7e81÷7e114 *100% = 4e-32 % of the time. Show the Excel spreadsheet and 30/20
oscillators (Table pg 383.xls).
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