Probability Calculations

Presentation on theme: "Probability Calculations"— Presentation transcript:

1 Probability Calculations

2 The Fundamental Postulate of Statistical Mechanics:
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.

3 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)

4 Statistical Mechanics
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)] 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 summarized in what follows.

5 Ω(E;yk)  ≡ Ω(E = -μH; spin 1 is “up”) ≡ 2
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

6 P(spin 1 is “up”) ≡ (⅔) P(spin 1 is “down”) ≡ (⅓)
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 is “up”) ≡ (⅔) The Probability that spin #1 is “down” is: P(spin 1 is “down”) ≡ (⅓)

7 <μz> = (⅓)μ <μz> ≡ ∑kμzkP(μz= μzk) = (⅔)(μ) + (⅓)(-μ)
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

8 Energy Dependence of the Density of States

9 A goal of the following discussion
We’ve 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 discuss an approximation to this E dependence of Ω(E).

10 P(y = yk) ≡ [Ω(E;yk)/Ω(E)] Discussion of an approximation for the
E dependence of Ω(E). No rigorous math! Physical arguments & “hand waving.” Math details are in many books! 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

11  Ω(E) is an incomprehensibly Rapidly Increasing Function of E!!
To estimate the E dependence of Ω(E), ask how many accessible states are there for a macroscopic system (f ~ 1024) at energy E? An order of magnitude estimate!! The result is abstract, but very significant!! Result is:  Ω(E)  Ef δE (f ~ 1024)  Ω(E) is an incomprehensibly Rapidly Increasing Function of E!!

12 an almost incomprehensibly enormously rapidly varying
The bottom line is that: Ω(E) is an almost incomprehensibly enormously rapidly varying Function of E (!!)

13 Example: A Simple Special Case: An 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.

14 Ω(E) = area between 2 ellipses.
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.

15  Ω(E)  VN ∫(E  E + δE) d3p1d3p2…d3pN
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 ∫(E  E + δE) d3p1d3p2…d3pN

16  [R(E)]3N  (E)(3N/2)  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)

17 This is shown schematically for 2 dimensions in the figure:
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

18 Ω(E) = AEf δE, A = constant
In summary, for the classical 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.