Transcript Part VI

Maxwell-Boltzmann Statistics
• The Maxwell-Boltzmann Distribution Function at energy ε
& temperature T: f ε = A e-ε/kT .
 
• Boltzmann developed statistical mechanics
& was a pioneer of quantum mechanics.
• His work contained elements of relativity &
quantum mechanics, including discrete atomic energy levels.
• He “introduced the theory of probability into a fundamental law of
physics & thus in his statistical interpretation of the 2nd Law of
Thermodynamics he broke with the classical prejudice, that
fundamental laws have to be strictly deterministic.” “With his
pioneering work, the probabilistic interpretation of quantum
mechanics had already a precedent.”
• He constantly battled for acceptance of his work. He also struggled with
depression & poor health. He committed suicide in 1906.
Many believe that thermodynamics was the cause.
• Paul Eherenfest was Boltzmann’s
student. He delivered Boltzmann’s
eulogy & carried on (among other
things) the development of statistical
thermodynamics for nearly 3 decades.
• Not unlike his mentor Boltzmann,
he was filled with self-doubt & was
deeply troubled by the disagreements
between his friends (Bohr, Einstein, etc.)
which arose during the development of
quantum mechanics.
Eherenfest shot himself to death in 1933!!
• US physicist Percy
Bridgmann (man on right
in photo), the 1946 Nobel
Physics Prize winner, was
also a thermal & statistical
physics pioneer. He studied
the physics of matter under
high pressure.
Bridgman committed suicide in 1961.
There’s no need to worry! I’ve never known a student who
didn’t survive a course in thermal & statistical physics!!
Maxwell-Boltzmann Distribution Function:
f  ε  = A e-ε/kT .
Number of Particles at Energy ε, Temperature T:
nε = A gε  e
-ε/kT
.
A is a Normalization Constant
Integrate n(ε) over all ε to get N = Total Number of Particles
ε = Particle Energy, k = Boltzmann's constant
T = Temperature in Kelvin.
g(ε) = Number of States With Energy ε.
g(ε) depends on the problem.
n  ε  dε = A g  ε  e-ε/kT dε .
Assume a continuous distribution of energies &
calculate g(ε)  Number states with energy ε in the
range ε to ε + dε.
g(ε)  The “Density of States.”
To do this calculation, we first need to make some assumptions.
Specifically, we need to assume that we’ve calculated the energies ε
of the particles. Consistent with the assumption of an ideal gas, we
assume that each particle is “free” so that
ε = [(p2/(2m)] = (½)mv2 (1)
To calculate g(ε), it turns out to be easier to find the number of
momentum states corresponding to a momentum p, & then use (1) to
change variables from p to ε. Corresponding to every value of
momentum p is a value of energy ε.
“Free” Particles
ε=
n  ε  dε = A g  ε  e-ε/kT dε .
[(p2/(2m)]
=
(½)mv2
(1)
ε = a sphere in p space.
Momentum is a 3-dimensional vector & every
point (px,py,pz) in 3-D momentum (p) space
corresponds to some energy ε. Think of
(px,py,pz) as forming a 3-D grid in p space.
Now, count how many momentum states there
are in a region of space (density of momentum states) & use (1) to
find the density of energy states:
g(ε)  Number states with energy ε in the range ε to ε + dε
First, rewrite (1) in the form:
p =
2mε =
p2x + p y2 + p 2z .
The number of p states in a spherical shell from p to p + dp is
proportional to 4πp2dp = volume of the shell.
So, the number of states g(p) with momentum between p & p + dp
has the form:
g p  dp = B p2dp
B is a proportionality constant, which we’ll calculate later.
Each p corresponds to a single ε, so
g  ε  dε = g p  dp = B p2dp .
Also, ε = [(p2/(2m)] = (½)mv2 so that
p2 = 2mε
p = 2mε
So, p2 dp  ε ε-1/2 dε ,
We had
dp = 2m
1 -1/2
ε dε ,
2
This gives g  ε  dε  ε ε-1/2 dε .
n  ε  dε = A g  ε  e-ε/kT dε .
Constant C contains B & all
the other proportionality
constants together.
So, this now has the form:
n  ε  dε = C ε e-ε/kT dε .
n  ε  dε = C ε e-ε/kT dε .
To find the constant C, evaluate


0
0
N =  n  ε  dε = C 
ε e-ε/kT dε
N  Total number of particles in the system.
Look the integral up in a table & find:
The result is
N =
So that n  ε  dε =
C
2
  kT 
2N
 kT 
3/2
3/2
,
ε e-ε/kT dε .
n(ε)  Number of molecules with energy between ε & ε + dε in a
sample containing N molecules at temperature T.
n  ε  dε =
2N
 kT 
3/2
ε e-ε/kT dε .
Plot of the Distribution:
Notice that “no” molecules have E = 0, few molecules
have high energy (a few kT or greater), & there is no
maximum of molecular energy.
This is how the distribution changes with
temperature (each vertical grid line corresponds to kT).
Notice that the distribution
for higher temperatures is
skewed towards higher
energies but all three curves
have the same total area
(makes sense).
Notice how the probability of a particle having
energy greater than 3kT (in this example) increases
as T increases.
The Total Energy of the System is

E=
 ε n  ε  dε =
0
2πN
 kT 
3/2

3/2
-ε/kT
ε
e
dε .

0
Evaluation of the Integral Gives
3NkT
E=
.
2
This is the total energy for the N molecules, so the average
energy per molecule is
3
ε=
2
kT ,
This is exactly the result obtained from the elementary
kinetic theory of gases.
3NkT
E=
.
2
3
ε = kT ,
2
Some things to note about the ideal gas energy:
1. The energy is independent of the molecular mass.
2. Which gas molecules will move faster at a given
temperature: lighter or heavier ones? Why?
3. The average energy at room temperature, kT is about
40 meV, or (1/25) eV. This is not a large energy.
4. (½)(kT) of energy "goes with" each degree of freedom.
Because ε = (½)mv2, the number of molecules having
speeds between v and v + dv can also be calculated.
n  v  dv =
2  N m3/2
 kT 
3/2
2
v e
-mv 2 /2kT
dv .
n(v)
The result is
v
The speed of a molecule having the average energy comes
from solving
2
mv
3
ε=
= kT
2
2
for v. The result is
2
v rms = v =
3kT
.
m
vrms is the speed of a molecule having the average energy ε
It is an rms speed because we took the square root of the
square of an average quantity.
The average speed v is calculated from:

v=
 v n(v) dv
 n(v) dV
0

.
0
The result is
v=
Comparing this with vrms:
8kT
.
m
v rms = 1.09 v .
Because the velocity distribution curve is skewed towards
high energies.
Find the most probable speed by setting (dn(v)/dv) = 0 .
The result is
vp =
2kT
.
m
The subscript “p” means “most probable.”
Summary of the different velocity results:
2
v rms = v =
3kT
.
m
v=
8kT
. vp =
m
v rms = 1.09 v
2kT
.
m