Lecture_25

Lecture_25 - PHYS342 Fall2011 Lecture 25: The Partition...

Info iconThis preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
PHYS 342 Fall 2011 Lecture 25: The Partition Function Ron Reifenberger irck anotechnology Center Birck Nanotechnology Center Purdue University Lecture 25 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
In previous lecture, we discussed in detail the statistical predictions for a few particle system with fixed energy. The temperature of the system was ot specified in any of these not specified in any of these discussions. Can we extend the discussion to a large number of particles – this time explicitly include temperature? 2
Background image of page 2
Microscopic Reversibility for Distinguishable Particles consider the interaction between particles in well-defined states E m , v , E Forward process Reverse process m 1 , v 1 , E 1 m 2 , v 2 , E 2 m 1 , v 1 , E 1 2 2 2 n(E 1 ) n(E 2 ) Elastic Interaction Elastic Interaction m 1 , v 3 , E 3 m 2 , v 4 , E 4 m 1 , v 3 , E 3 m 2 , v 4 , E 4 1. In equilibrium, the number of particles undergoing the forward rocess (in a fixed time) should equal the number of atoms n(E 3 ) n(E 4 ) process (in a fixed time) should equal the number of atoms undergoing the reverse process in the same time 2. Let n(E)= number of particles with energy between E and E+dE 3
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
The rate for such an interaction must be proportional to umber of articles each energy state: number of particles in each energy state: R(1+2 3+4) n(E 1 )n(E 2 ) There will also be an inverse interaction rate R(3+4 1+2) n(E 3 )n(E 4 ) n equilibrium you would like to believe that both rates In equilibrium, you would like to believe that both rates are equal, i.e., R(1+2 3+4) = R(3+4 1+2) This implies that n(E )n(E ) = n(E )n(E ) 1 2 3 4 4
Background image of page 4
n[n(E 1 )n(E 2 )] = n[n(E 3 )n(E 4 )] nn(E 1 ) + 2 ) = 3 ) + 4 ) cv We also require energy to be conserved: E + E = E + E 1 2 3 4 One simple way to satisfy both equations at once is to make nn(E) linearly proportional to E. So…. n n(E)= α‐β E minus sign is matter of convenience () : to match equation for ideal gas see Appendix ( 0 ) E E nE ee n e    1 ;( 0 ) B n is constant kT 5 Tells how particles are distributed as a function of E
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
1. If you sum over all possible energy states, you must recover the total number of atoms in the system: / ) ) iB EkT n E e deg . () (0) i all states all states Some energies may have a eneracy gE Nn ne   . . The details size of the degeneracy depends on the exact situation This possibility must be included 2. The probability of a particle having an energy E will be: Boltzmann factor MULTIPLICITY: Number of states at ( 0 )() P(E) B nE n gEe ) B ET k e E energy E The factor Z is the key to calculating ALL macroscopic (0 N n ) all states i e Z Partition function properties of the system, especially when you can write a formula for Z(T)! It’s as important to statistical physics as Ψ is to quantum mechanics.
Background image of page 6
Image of page 7
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 28

Lecture_25 - PHYS342 Fall2011 Lecture 25: The Partition...

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online