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Unformatted text preview: NOTE: In going from Eq. 2.133 to Eq. 2.134, we integrated over a Gaussian distribution using the socalled Gaussian integral identity: • (no proof) Z ∞ ∞ e a x 2 d x = r π a . (2.140) • This integral is symmetric, so then Z ∞ e a x 2 d x = 1 2 r π a . (2.141) QUESTION: In going from from Eq. 2.106 to Eq. 2.107 we used: ln ( x ! ) = ln ( x ) . Is this really true? ANSWER: Yes, for large x. One way to see this is to apply the Stirling formula: ln ( x ! ) = x ln ( x ) x . (2.142) • Then ln ( x ! ) = ( x ln ( x ) x ) = 1 × ln ( x ) + x × 1 x 1 = ln ( x ) (2.143) 2–70 2.19 Applying the Boltzmann Distribution AIM: Show how we can bridge the gap between the microscopic and macroscopic views of a molecular system using the Boltzmann Distribution. H(q,p) Microscopic World Statistics Boltzmann Macroscopic World REMINDER: When we derived the Boltzmann distribution, we mentioned (Eq. 2.122) that we can calculate a macroscopic property h A i of a system as a weighted average of the microstates, where the weight is the Boltzmann probability: h A i = 1 Z ∑ k A k exp E k k B T (2.144) = ∑ k A k p Boltz k . (2.145) QUESTION: What do we mean by ’weighted average’? 2–71 EXAMPLE: Given N points of mass m i in a plane with coordinates ( x i , y i ) , calculate the centre of geometry : COG ( X , Y ) = 1 N N ∑ i = 1 x i , 1 N N ∑ i = 1 y i ! (2.146) X Y centre of geometry 2 4 6 4 6 2 COG ( X , Y ) = 6 + 2 + 4 3 , 1 + 3 + 5 3 (2.147) 2–72 • The centre of mass is a weighted average : COM ( X , Y ) = 1 M N ∑ i = 1 m i x i , 1 M N ∑ i = 1 m i y i ! , (2.148) where M = ∑ N i = 1 m i . X Y centre of geometry 2 4 6 4 6 2 centre of mass m=0.5 m=1 m=2 M = 2.0 + 1.0 + 0.5 (2.149) COM ( X , Y ) = 2.0 × 6 + 1.0 × 2 + 0.5 × 4 3.5 , 2.0 × 1 + 1.0 × 3 + 0.5 × 5 3.5 = ( 4.57, 2.14 ) (2.150) 2–73 NOTE: A weighted average is also called an expectation value or a mean value . NOTE: When calculating the centre of mass, we are weighting each coordinate with an additional factor : 2 3.5 , 1 3.5 , 0.5 3.5 . NOTE: This is exactly what we have in Eq. 2.145: Centre of Mass Ensemble Average h A i individual mass ⇒ Boltzmann weight m i ⇒ exp E k k B T total mass ⇒ partition function M = ∑ N i = 1 m i ⇒ Z = ∑ k exp E k k B T EXAMPLE: (Two spin system) • Consider a system of two spins s 1 = ± 1 and s 2 = ± 1 ....
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This note was uploaded on 12/05/2010 for the course CHBE 251 taught by Professor Scotty during the Winter '09 term at UBC.
 Winter '09
 scotty

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