Download Phase space in classical physics Quantum mechanically, we can

Phase space in classical physics
Quantum mechanically, we can actually COUNT the number of microstates consistent
with a given macrostate, specified (for example) by the total energy. In general, each
microstate will be specified by one or more quantum numbers. One might, for example,
ask how many microstates are consistent with a given energy (macrostate). This number
of states is what we have been calling the “multiplicity” of a system.
Classically, it’s a bit trickier. A microstate of a gas, for example, is specified by the
positions and momenta of each of the N molecules that make up the gas; that is, molecule
one is specified by p1 and r1 , or if you prefer, by the 6 coordinates p1 x , p1 y , p1z , x1 , y1 , z1 .
The difficulty, of course, is that there are an infinite number of such states, since the
variables are continuous. Let’s finesse this problem for just a moment.
For large N, it will turn out that ONE macroscopic state will be associated with virtually
ALL of the microstates. Here is a place where large numbers lead to intuitively
surprising results. Suppose this room were divided down the middle by a partition, and
all the gas were on one side. Clearly, there are a HUGE number of states corresponding
to the equilibrium state. Now remove the partition. The gas expands and fills the entire
room. As we have seen, the probability that purely by chance, the gas will all come over
to one side again is vanishingly small. Thus, huge as the first number is, it is altogether
negligible compared with the number of states that is available after the partition is
removed. Moreover, for large N, the peak is enormously sharp, so that the likelihood of
even small deviations is tiny.
Let’s return to the initial question: How do we estimate the number of states of a system
for a classical system? Here is one common approach:
•
First, we define the notion of phase space—each momentum and position
coordinate counts as one dimension of phase space. Thus, for a single particle,
there are three degrees of freedom (x, y, and z); and the state of the particle is
defined by those three coordinates, and also by the three coordinates of the
momentum. Thus, one would have a 6 dimensional phase space.
•
Similarly, for a gas of N particles confined to a volume V, one would speak of a
3N degrees of freedom, and a 6 N dimensional phase space.
•
Now, we can calculate the “volume” of the phase space:
Γ = ∫ dp1dp2 dp3 … dp3 N dx1dx2 dx3 … dx3 N
where the integral is over either all states with energy less than E0 , or possibly all
states with the energy in some small range dE , as in E0 < E < E0 + dE — if you
will, the volume of a shell of constant energy. Curiously enough, it won’t matter
which—see below! If the latter, the usual notation is to set
phase space argument
Ω=
page 2 of 5
∫
dp1dp2 dp3 … dp3 N dx1dx2 dx3 … dx3 N
E0 < E < E0 + dE
Before the advent of quantum mechanics, one approach was to look for ways of
categorizing states into “bins,” and then take the limit as the size of each bin → 0.
(Boltzmann). It actually works pretty well, but we won’t do it—discuss reasons. Instead,
we look at G&T’s trick, which you will find in many other books:
•
Consider a quantum mechanical version of a one-dimensional, one particle
system, and count the number of states.
•
The classical version of such a system will have a two-dimensional phase space
(one dimension corresponding to position, the other to momentum). Note that a
two dimensional phase space “volume” has dimensions [ p x ] with units joulesec; note that Planck’s constant has the same units.
•
Calculate the “volume” of phase space (in this case, the area of the twodimensional phase space), and compare this classical system to the corresponding
QM system, in a “semi-classical limit.” One finds that
QM number of states =
classical volume of phase space
h
or in other words, each quantum state corresponds to a classical volume h of
phase space!
Examples:
•
particle in a box: See G&T Sections 4.3.3, 4.3.5, 4.3.6 and class notes.
•
simple harmonic oscillator: See G&T Section 4.3.4 and class notes.
Thus, for a collection of N point masses, free to move in three dimensions, one would
have
QM number of states =
classical volume of phase space
h3N
We refer to this collection of states as an ensemble. And at some point, we need to state
the Postulate of Equal Apriori Probability—all microstates in the ensemble
have equal probabilities. After Gibbs, we call such an ensemble the microcanonical
ensemble.
2
phase space argument
page 3 of 5
Connection to Baierlein
It turns out to be straightforward to calculate the volume of phase space for an ideal gas.
Note that for a single particle, we have
∫ dx dx dx
1
2
3
= V = ∫ dxdydz
Now since it’s an ideal gas, the particles don’t know the others are there, and so we may
treat each one independently. The first step is easy.
Ω=
=
1
h3N
VN
h3N
∫
dp1dp2 dp3 … dp3 N dx1dx2 dx3 … dx3 N
∫
dp1dp2 dp3 … dp3 N
E0 < E < E0 + dE
E0 < E < E0 + dE
2
2
2
p 2 px + p y + pz
=
; thus, for a
For a single particle, the energy E is given by E =
2m
2m
collection of N particles with total energy E in the range E0 < E < E0 + dE , we have
N
3
2mE = ∑∑ pi2α
i =1 α =1
Now this double sum can be thought of as the square of the radius R ( E ) of a 3N
dimensional space:
2mE = R 2 ( E ), or
R = 2mE
The “volume” of an 3N dimensional phase space is just a constant × R 3 N ; hence, we have
the total volume of phase space Γ for all energies E < E0 is
Γ=
B N 3N
B
3N / 2
V R = 3 N V N ( 2mE )
3N
h
h
where B is just a geometry-dependent constant. (If you’re interested, G&T work it out in
an Appendix. But it is never needed in calculations.) Consequently, a surface of constant
energy is given by
Ω=
3N
B N 3 N −1
B
−1
V R
= 3 N V N ( 2mE ) 2
3N
h
h
But of course, for large N, we can neglect the –1, and hence we have
3
phase space argument
page 4 of 5
Ω=
B N
3N / 2
V ( 2mE )
3N
h
Thus, as advertised above, the volume of a shell of constant energy in phase space is
virtually the same as the volume of the entire space for E < E0 . It’s another surprising
consequence of working with large numbers!
For mysterious reasons of my own (it has to do with the nature of identical particles in
quantum mechanics), I will put in an additional factor of N ! in the denominator. Thus,
for an ideal gas, the quantity Ω , which you will see variously described as the volume of
classical phase space, the number of states, or perhaps, the multiplicity, is given by
Ω=
B
3N / 2
V N ( 2mE )
3N
N !h
Note, by the way, that Ω depends on only three macroscopic variables, E , V , and N . At
this point, Baierlein’s mysterious statement becomes obvious. For an ideal gas, the
energy depends only on the temperature. But on pages 31–32, Baierlein is considering a
constant temperature process. Hence the ratio of multiplicities depends only on the ratio
of the volumes raised to the Nth power, as we see from the above equation!
You might be interested to see where we go from here. We will be getting a little ahead
of ourselves, but it’s interesting to see anyway. We shall shortly define the entropy as
S = k ln Ω
where k is Boltzmann’s constant, so-called because it was introduced into physics by
Planck and Einstein.
We shall also define the thermodynamic temperature T as
1  ∂S 
=

T  ∂E V , N
Hence, putting everything together,
1  ∂S 
∂ 
B
3Nk

=
+ Nk ln V +
ln E 
 =
 ln
3N
T  ∂E V , N ∂E  N ! h
2

Note that the first term is just an constant, and hence its derivative is zero. We have
1 3Nk 1
=
2 E
T
3
E = NkT
2
4
or
phase space argument
page 5 of 5
But this result is the same one we got from our simple kinetic theory argument. Here we
have obtained it in a far more general way. It gets even better: It’s easy enough to show
(think about an adiabatic, quasi-static process) that
 ∂E 
p = −

 ∂V  S
But using a theorem that I hope you remember from multivariable calculus,
 ∂E   ∂V   ∂S 

 
 
 = −1 ,
 ∂V  S  ∂S  E  ∂E V
it follows that
p  ∂S 
=

T  ∂V  E
or in other words,
p  ∂S 
∂
=
 =
T  ∂V  E , N ∂V
B
3Nk


+ Nk ln V +
ln E 
 ln
3N
2
 N !h

or
p
1
= Nk , or, written another way,
T
V
pV = N kT
Thus, the defining equations for an ideal gas,
pV = NkT
3
E = NkT
2
follow in a natural way from our phase space argument!!
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