Statistical Mechanics - Physics at Oregon State University

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256 APPENDIX A. SOLUTIONS TO SELECTED PROBLEMS. this can be written in the form G = −kBT o 1 − k log(1 + λe B T ( p2z eB 1 2m + mc (j+ 2 )) ) If we have the free energy, we also have the partition function, of course. or Using N = − G = −kBT 2 eBL2 2πc G = −kBT 2 eBL2 2πc G = −kBT V ∂G ∂µ T,V,B N = V pz,j ∞ L 2π −∞ eB 2π22 ∞ c −∞ eB 2π 2 2 c 1 − k log(1 + λe B T ( p2 z eB 1 2m + mc (j+ 2 )) ) dp j dp and M = − ∂G ∂B j ∞ dp −∞ j 1 p2 eB 1 − k log(1 + λe B T ( 2m + mc (j+ 2 )) ) 1 p2 eB 1 − k log(1 + λe B T ( 2m + mc (j+ 2 )) ) T,µ,V we find 1 λ−1e 1 p2 eB 1 kB T ( 2m + mc (j+ 2 )) + 1 which gives in the limit λ → 0 (for high temperature): eB N ≈ λV 2π22 ∞ dp c 1 p2 − k e B T ( 2m or N ≈ λV N ≈ λV Check: If B goes to zero we get −∞ eB 2π22 ∞ eB p2 2mck e− B T − 2mk dpe B T c −∞ eB 2π22 eB ∞ 2mck e− B T 2mkBT c −∞ N ≈ λV j eB 1 + mc (j+ 2 )) j dxe −x2 1 4π23 3 ∞ 2mkBT dxe −∞ −x2 which is equal to the old result. We can also take the small λ limit in the free energy and get eB G = −kBT V 2π22 ∞ dp c 1 p2 − k λe B T ( 2m −∞ j e −j eB mck B T 1 eB − mck 1 − e B T eB 1 + mc (j+ 2 )) Comparing with the formula for N we see G = −NkBT , which should not come as a surprise, since G = −pV and we are in the limit of an ideal gas!! Therefore:
A.6. SOLUTIONS FOR CHAPTER 6. 257 For small values of B we have N ≈ λV For small values of x we have e−x ≈ 1 − e−2x and hence Hence 1 − x 2x − 2x 2 + 4 3 N ≈ λV M = kBT ∂N ∂B T,µ,V eB 2π22 eB 2mck e− B T 2πmkBT c 1 = x3 2x 1 eB − mck 1 − e B T 1 − x 1 − x + 2 1 ≈ 3x2 2x (1−x)(1+x+1 3 x2 ) ≈ 1 2x (1−1 6 (2x)2 ) eB 2π22 2πmkBT c mckBT 2 1 eB (1 − ) eB 6 mckBT N ≈ λV mkBT 2π23 2πmkBT (1 − 1 2 eB ) 6 mckBT M = kBT λV mkBT 2π23 2πmkBT B 2 e 3 mckBT χ = kBT λV mkBT 2π23 2πmkBT 1 2 e 3 mckBT and using the expression for N at T = 0: χ = kBT N 1 2 e = 3 mckBT N 2 e 3kBT mc A.6 Solutions for chapter 6. Problem 4. A quantum mechanical system is described by a Hamiltonian H = H0 + κV , with [H0, V ] = 0. κ is a small constant. The Helmholtz free energy is Fκ(T ). Calculate the change in Helmholtz free energy, ∆F = Fκ − F0 for this system up to second order in κ kBT . Z = T re −βH = T re −βH0−βκV = T re −βH0 e −βκV
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Statistical Mechanics Henri J.F. Ja
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IV CONTENTS 4.3 Gas of poly-atomic
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VI CONTENTS
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VIII INTRODUCTION Introduction. The
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X INTRODUCTION In chapter nine we d
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2 CHAPTER 1. FOUNDATION OF STATISTI
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Statistical Mechanics - Physics at Oregon State University

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