1. Maxwell-Boltzmann Statistics

2. Maxwell-Boltzmann Statistics
As we’ve already seen, The Maxwell Boltzmann
Distribution Function at energy ε & temperature
T has the form:
As discussed earlier, Boltzmann
developed statistical mechanics &
was a pioneer in thermodynamics.
His work also contained elements of
relativity & quantum mechanics,
including the existence of atoms
with discrete energy levels.

3. Many believe that thermodynamics
Boltzmann introduced the theory of probability
into a fundamental law of physics, the 2nd
Law of Thermo. In his statistical interpretation of
that law, he broke with the classical view that
fundamental physical laws have to be deterministic.
“With his work, the probabilistic interpretation of
quantum mechanics had already a precedent.” As
already discussed, he constantly battled for
acceptance of his work & struggled with depression
& poor health. He committed suicide in 1906.
Many believe that thermodynamics
was one of the causes.

4. Eherenfest shot himself
Paul Eherenfest was Boltzmann’s
PhD student. He delivered
Boltzmann’s eulogy & carried on
(among other things) the development
of Statistical Thermodynamics for ~30 years.
Similar to 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!!

5. Bridgman committed suicide in 1961!
US physicist Percy Bridgmann (on right in photo
below), the 1946 Nobel Physics Prize winner, was
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!!

6. Maxwell-Boltzmann Distribution Function:
Number of Particles at Energy ε, Temperature T:
A = Normalization Constant (inverse of partition function)
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. This
depends on the problem.

7. g(ε)  “Density of States.”
Assume a continuous distribution of energies &
calculate: g(ε)  Number states with energy
from ε to ε + dε.
g(ε)  “Density of States.”
To do this calculation, we first need to make some
assumptions. 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)

8. 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.
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.

9. ε is a sphere in p space, Every point (px,py,pz) in 3-D momentum (p)
ε = [(p2/(2m)] (1)
Every point (px,py,pz) in 3-D momentum (p)
space corresponds to some energy ε.
(px,py,pz) forms a 3-D grid in p space.
“Free” Particles
From (1),
ε is a sphere
in p space,
as shown in the Figure.

10. g(ε)  Number of states with energy
ε = [(p2/(2m)] (1)
ε is a sphere in p space.
“Free” Particles
To find the density of states g(ε), we’ll count how
many momentum states there are in a region of p
space (density of momentum states) & use (1) to find
the density of energy states g(ε):
g(ε)  Number of states with energy
from ε to ε + dε

11. 4πp2dp = volume of the shell. The number of p states in
“Free” Particles
ε = [(p2/(2m)] (1)
First, rewrite (1) in the form:
The number of p states in
a spherical shell from p to
p + dp is proportional to
4πp2dp = volume
of the shell.

12. Also, ε = [(p2/(2m)], so that
So, the number of states g(p) with momentum
between p & p + dp has the form:
B is a proportionality constant, which we’ll calculate
later. Each p corresponds to a single ε, so
Also, ε = [(p2/(2m)], so that

13. ε = [(p2/(2m)] gives: So,   This gives: We had:
So, the final form is:
The constant C contains B & all the other
proportionality constants together.

14. N  Total number of particles in the system.
To find the constant C, evaluate
N  Total number of particles in the system.
Look the integral up in a table & find:
So that:
n(ε)  Number of molecules with energy between ε
& ε + dε in a gas of N molecules at temperature T.

15. Maxwell-Boltzmann EnergyDistribution
Note: No molecules have E = 0 & only a few have high energy (a few kT or greater),
n(E,T) E

16. Changes in the
distribution with
temperature are shown
in the figure (each
vertical grid line
corresponds to kT).
Note: The distribution for higher T is skewed
towards higher E but all 3 curves have the
same area. Note also: The probability of a
particle having energy greater than 3kT (in this
example) increases as T increases.

17. This is the total energy for the N
The Total Thermal Energy of the System is:
Evaluation of this integral gives:
This is the total energy for the N
molecules, so the average energy per molecule is:
This is exactly the same as the result
obtained from the Equipartition Theorem.

18. This is a very small energy!!!.
Some things to note about this 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?
The average thermal energy at room temperature, kT is about 40 meV, or (1/25) eV.
This is a very small energy!!!.
4. (½)(kT) of energy "goes with" each degree of freedom. (The Equipartition Theorem again!)

19. v Because ε = (½)mv2, the number of molecules having
speeds between v and v + dv can also be calculated.
The result is
n(v) v

20. The speed of a molecule having the
average energy can be found by solving
for v. The result is
This is an rms speed because it is the square root
of the square of an average quantity. Vrms is the
speed of a molecule having the average energy.

21. In contrast, the average speed <v> is calculated from:
The result is:
Compare this with vrms & find:
This makes sense, because the velocity
distribution is skewed towards high energies.

22. Summary of different velocity results:
Now, lets find the most probable speed by
setting (dn(v)/dv) = 0 .
The result is
The subscript “p” means “most probable.”
Summary of different velocity results: