Download Ω (E)

Section 2.4: Probability Calculations
• To calculate the probability of finding a system in
a given state, we use
The Fundamental Postulate
of Statistical Mechanics:
“An isolated system in equilibrium is
equally likely to be found in any
one of it’s accessible states.”
• There will always be an uncertainty in our knowledge
of the system energy ≡ δE. Suppose that we know that
the energy of the system is in the range E to E + δE.
The Fundamental Postulate:
“An isolated system in equilibrium is
equally likely to be found in any
one of it’s accessible states.”
• There is always an uncertainty in our knowledge of the
system energy ≡ δE. Suppose that we know that the
energy of the system is in the range E to E + δE.
• Define: Ω(E) ≡ Total number of accessible states
in this range. y ≡ A macroscopic system parameter
(pressure, magnetic moment, etc.).
• Define: Ω(E;yk) ≡ A subset of Ω(E) for which
y ≡ yk (yk = A particular value of y)
• Let P(y = yk) ≡ probability that y ≡ yk.
By the Fundamental Postulate of
Statistical Mechanics
this is: Pk
 P(y = yk) ≡ [Ω(E;yk)]/[Ω(E)]
• Now, calculate the mean (expected) value of y:
• From probability theory, this is simply:
<y> ≡ ∑kykPk  ∑kyk[Ω(E;yk)]/[Ω(E)]
• Clearly, to calculate this, we need to know
both Ω(E) & Ω(E;yk). This will be discussed
in detail as we go through this chapter!
• In principle (if we know the Ω’s) this calculation is easy.
• Example: Take 3 particles of spin ½ in an
external magnetic field again.
• Suppose that we know from measurement that the
total energy is E = - μH. As we’ve seen, there are
only 3 accessible states for E = - μH. These are:
(+,+,-) (+,-,+) (-,+,+).
Question:
• Probability of finding spin #1 in the “up” position?
• To answer, we need Ω(E) for this problem. We had:
Ω(E)  ≡ Ω(E = -μH) ≡ 3 and
Ω(E;yk)  ≡ Ω(E = -μH; spin 1 is “up”) ≡ 2
• For total the energy E = - μH, the 3 accessible states are:
(+,+,-) (+,-,+) (-,+,+).
• Probability of finding spin #1 in “up” position?
Formally: P(E; y = yk) ≡ [Ω(E;yk)/Ω(E)] or
P(E = -μH; spin 1 is “up”) ≡
[Ω(E = -μH; spin 1 is “up”)/Ω(E = -μH)]
We just saw that Ω(E)  ≡ Ω(E = -μH) ≡ 3 &
Ω(E;yk)  ≡ Ω (E = -μH; spin 1 is “up”) ≡ 2
• So, The Probability that spin #1 is “up” is:
P(spin 1 “up”) ≡ (⅔)
• The Probability that spin #1 is “down” is:
P(spin 1 “down”) ≡ (⅓)
• Since we have the probability, P(E; y = yk) we
can use it to find answers to problems like:
• Calculate the
Mean (average) Magnetic Moment of Spin # 1.
<μz> ≡ ∑kμzkP(μz= μzk)
(Sum goes over k = “up” & k = “down”)
<μz> = P(#1 “up”)(μ) + P(#1 “down”)(-μ)
= (⅔)(μ) + (⅓)(-μ)
so
<μz> = (⅓)μ
Section 2.5: Energy Dependence of
the Density of States
Note!
This section is
one of the most
important
sections in the
chapter!!
• We’ve just seen that the probability that
parameter y has a value yk is:
P(y = yk) ≡ [Ω(E;yk)/Ω(E)]
Ω(E) ≡ # accessible states with energy
between E & E + δE.
 To do probability calculations, we obviously
need to know the E dependence of Ω (E).
A goal of the following discussion
is to find an approximation to the
E dependence of Ω(E).
P(y = yk) ≡ [Ω(E;yk)/Ω(E)]
GOAL: Find an Approximation
for the E dependence of Ω(E).
• No rigorous math! Physical arguments & “hand waving.”
Consider a Macroscopic System
with f degrees of freedom. f is huge! f ~ 1024.
• Energy is in range E to E + δE. Of course, Ω(E)  δE.
Extrapolating earlier discussion of the 1d oscillator:
Ω(E) ~ Phase space volume
in this energy range.
• Define Ω(E) ≡ ω(E)δE. ω(E) ≡ Density of states
Goal: Estimate the E dependence
of Ω(E) (or ω(E)).
So, ~ how many accessible states are
there for a macroscopic system
(f ~ 1024) at energy E?
• Not interested in exact results. We want an
order of magnitude estimate!! The result is
abstract, but very significant!! Let
Φ(E) ≡ Total # of quantum states
for all energies E´ ≤ E.
Φ(E) ≡ Total # of quantum states
for all energies E´ ≤ E.
• Consider 1 “typical” degree of freedom.
ε ≡ energy associated with that degree of
freedom. Let
φ(ε) ≡ total # of quantum states for
this degree of freedom.
• Generally, φ(ε) increases with increasing ε.
So, we can write:
φ(ε)  εα (α ~ 1)
(1)
φ(ε)  ε
(1)
• For the system, replace the energy E by an “average
energy” for a system of f degrees of freedom:
ε  (E/f)
(2)
f degrees of freedom &  φ(ε) states
associated with each.
Total # of states associated with f degrees of
freedom ≡ product of # associated with each:
Φ(E)  [φ(ε)]f (3)
Now, use (1), (2), & (3) together:
• So, we have:
φ(ε)  ε (1),
and ε  (E/f) (2)
Φ(E)  [φ(ε)]f
(3)
• Using (1), (2), & (3) together:
 The total # of states for all energies
E´ ≤ E is roughly: Φ(E)  [φ(ε)]f
So Φ(E)  [E/f]f or Φ(E)  Ef or
So finally: Φ(E) ≡ AEf (4)
A = constant
Ω(E) ≡ # accessible states with
energy between E & E + δE
• So, we can write:
Ω(E) ≡ Φ(E + δE) - Φ(E); δE <<< E
 Expand Φ in a Taylor’s series & keep
only the lowest order term:
 Ω(E)  (Φ/E)δE  [(Ef)/E]δE
Ω(E)  Ef-1δE  Ef δE (f >> 1)
 Ω(E)  Ef δE (f ~ 1024)
 Ω(E) is an incomprehensibly
Rapidly Increasing Function of E!!
•Briefly look at some numbers with
powers of 10 in the exponent to get a
“feel” for the “bigness” of Ω(E)  Ef :
•Briefly look at some numbers with
powers of 10 in the exponent to get a
“feel” for the “bigness” of Ω(E)  Ef :
1. Approximate age of the Universe
in seconds = “only”  1018 s!
This is a 1 with 18 zeros after it!
•Briefly look at some numbers with
powers of 10 in the exponent to get a
“feel” for the “bigness” of Ω(E)  Ef :
1. Approximate age of the Universe
in seconds = “only”  1018 s!
This is a 1 with 18 zeros after it!
2. Universe volume divided by the
volume of a grain of sand = “only”  1090!
This is a 1 with 90 zeros after it!
Ω(E) 
f
E δE,
Consider f ~
24
10 :
• By what factor does Ω (E) change
when E changes by only 1%? Let
r ≡ [Ω (E + 0.01E)/Ω (E)] =
• Evaluate r using logarithms:
f
(1.01)
log10(r) = 1024log10(1.01)  4.32  1021
So, r  10x, where x  4.32  1021, or
r ~ 1 with 4.32  1021 zeros after it!
Ω(E) 
f
E δE,
Consider f ~
24
10 :
• Consider the same problem again, but let E
increase by only 10-6E. By what factor does
Ω (E) change? Let
r ≡ [Ω (E + 10-6E)/Ω (E)] = (1.0 +10-6)f
•Evaluate r using logarithms:
log10(r) = 1024log10(1.0 +10-6)  4.34  1017,
So r  10y, where y = 4.34  1017, or
r ~ 1 with 4.34  1017 zeros after it!
• Finally, consider 2 seemingly similar numbers:
u
10 ,
and
24
10
C=
u=
v
24
D = 10 , v = 2  10
• Are these similar numbers??
• Finally, consider 2 seemingly similar numbers:
u
10 ,
and
24
10
C=
u=
v
24
D = 10 , v = 2  10
• Are these similar numbers?? NO!!!
Their ratio is: (D/C) = 10u
• Finally, consider 2 seemingly similar numbers:
u
10 ,
and
24
10
C=
u=
v
24
D = 10 , v = 2  10
• Are these similar numbers?? NO!!!
Their ratio is: (D/C) = 10u
 That is, D is 10u times (u = 1024!)
larger than C
u = 1 with
24
10
zeros after it!
• The bottom line is that: Ω(E) is
• The bottom line is that: Ω(E) is
An Almost Incomprehensibly
• The bottom line is that: Ω(E) is
An Almost Incomprehensibly
Enormously Rapidly Varying
• The bottom line is that: Ω(E) is
An Almost Incomprehensibly
Enormously Rapidly
Varying
Function of E (!!)
Simple Special Case:
Ideal, Monatomic Gas
• To make this general discussion clearer, as an
example, lets consider a classical ideal,
monatomic gas, with N identical molecules
confined to volume V.
• Lets calculate Ω(E) exactly for this case.
Ideal Gas  NO INTERACTION
between molecules.
~ Valid approximation for real gases
in the low density limit.
• Classical ideal, monatomic gas, N identical
molecules in volume V. Calculate Ω(E) for this case.
Ideal Gas  No Interaction between molecules.
• In this case, the total energy E of the gas is the sum of
the kinetic energies of the N molecules, each of mass m:
E  (2m)-1∑(i = 1,N)(pi)2,
pi = 3d momentum of particle i.
• Ω(E) ≡ # of accessible states in the interval E to E + δE
• Ω(E) ≡ # of cells in phase space between E & E + δE.
• Ω(E)  volume of phase space between E & E + δE.
• In what follows, recall the 1d oscillator where
Ω(E) = area between 2 ellipses.
• Ω(E)  volume of phase space between E & E + δE.
2mE = ∑
(i = 1,N)
∑
(α = x,y,z)
(piα)2
• The energy is independent of the particle positions!
piα = α component of momentum of particle i.
Ω(E)  ∫
(E  E + δE)
d3r1d3r2…d3rNd3p1d3p2…d3pN
A 6N dimensional volume integral!
• The limits E & E + δE are independent of the ri’s
 The position integrals for each ri
can be done immediately:
∫d3ri ≡ V  ∫d3r1d3r2…d3rN ≡ VN
 Ω(E)  VN ∫
d3p1d3p2…d3pN
(E  E + δE)
 Ω(E)  VN ∫(E  E + δE) d3p1d3p2…d3pN
(1)
A 3N dimensional volume integral in p space
• Consider the sum
2mE = ∑(i = 1,N)∑(α = x,y,z) (piα)2 (2)
(2) ≡ “sphere” in 3N dimensional momentum space.
• Briefly consider the case of 1 particle only. (2) is:
2mE = (px)2 + (py)2 + (pz)2
This is a “sphere” in momentum space of “radius”
R(E) = (2mE)½
• For 1 particle, the 3d “sphere volume”  [R(E)]3  (E)(3/2)
• For N particles in 3N dimensional momentum space,
(2) ≡ a “sphere” of “radius” R(E) = (2mE)½
So, the 3N dimensional “sphere volume” is
 [R(E)]3N  (E)(3N/2)
• Lets write:
Ω(E)
 VNG(E)
where:
G(E) ≡ ∫(E  E + δE) d3p1d3p2…d3pN (3)
• G(E) ≡ Volume of “spherical shell” between E & E + δE
This is shown schematically for 2 dimensions in the figure:
• G(E)  [R(E)]3N  (E)(3N/2)
 Ω(E)  VN(E)(3N/2)
• Write: Ω(E) = BVN(E)(3N/2)
• B = constant, Ω(E) = # of accessible states for an ideal
gas in the energy interval E to E + δE
• In summary, for the Ideal Gas, we found:
Ω(E) = BVN(E)(3N/2)δE
= # accessible states of an ideal gas in the energy
interval E to E + δE, B = constant
• In general, we found for f degrees of freedom,
Ω(E) = AEf δE,
A = constant
• For the ideal gas, f = 3N, so we got an E dependence
of E(½)f instead of Ef, but, again, all of this was
Approximate & Order of Magnitude.
• So, no worries about the difference between f & (½)f.