Lecture 13Lecture 13——Ideas of Statistical Mechanics Ideas of Statistical Mechanics Chapter 4, Chapter 4, Wednesday February 6Wednesday February 6thth•Comments about homework and exam•Finish the spins on a lattice problem•Statistical basis for the 2nd lawReading: Reading: All of chapter 4 (pages 67 All of chapter 4 (pages 67 --88)88)***Homework 4 due Thu. Feb. 7th*******Homework 4 due Thu. Feb. 7th****Assigned problems, Assigned problems, Ch. 4Ch. 4: 2, 8, 10, 12, 14: 2, 8, 10, 12, 14Exam 1: Exam 1: Fri. Feb. 8th (in class), chapters 1Fri. Feb. 8th (in class), chapters 1--44Review:Review:Thu. 7th at 5:30pm, tentatively in NPB1220Thu. 7th at 5:30pm, tentatively in NPB1220A simple model of spins on a latticeA simple model of spins on a latticen1= # of ‘up’ spins; n2= # of ‘down’ spinsN = n1+ n2= total # of spinsEnergy U = −(n1− n2)ε= (N − 2n1)ε12BBεμεμ=−=+Quantum spinsin a magneticfieldB22+−==Magneticmoment μA simple model of spins on a latticeA simple model of spins on a lattice-1 0 10.00.10.20.30.40.50.60.7 S/NkBx1111ln ln2222BxxxxSNk++−−⎧⎫⎛⎞⎛⎞⎛⎞⎛⎞=− +⎨⎬⎜⎟⎜⎟⎜⎟⎜⎟⎝⎠⎝⎠⎝⎠⎝⎠⎩⎭121UnxNNε==−n1> n2T > 0n1< n2T < 0n1= n2T = ∞A simple model of spins on a latticeA simple model of spins on a lattice11ln21BkxTxε−⎛⎞=⎜⎟+⎝⎠-1 0 1-8-6-4-2024682ε/kBTx121UnxNNε==−n1> n2T > 0n1< n2T < 0T = ∞n1= n2A simple model of spins on a latticeA simple model of spins on a latticen1> n2T > 0n1< n2T < 0n1n2n1n201230.00.20.40.60.81.0 MV/μNμB/kBTA simple model of spins on a latticeA simple model of spins on a latticetanhBNBMVkTμμ⎛⎞=⎜⎟⎝⎠The statistical basis for the 2nd lawThe statistical basis for the 2nd lawABUA, SA, TAUB, SB, TBS = SA+ SB()2*AAA22AA1() exp22UUWpUUUπ⎧⎫−⎪⎪∝= −⎨⎬ΔΔ⎪⎪⎩⎭Gaussian distribution(Bell curve)The statistical basis for the 2nd lawThe statistical basis for the 2nd lawThe statistical basis for the 2nd lawThe statistical basis for the 2nd lawA B()()iiiAABBWWUWU=()()AfAABT AUWWUWUU=−∑UT= UA+ UBThe statistical basis for the 2nd lawThe statistical basis for the 2nd
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