Lecture 2: Intro. Statistical Mechanics

Transcription

1 Lecture 2: Intro. Statistical Mechanics Statistical mechanics: concepts Aims: A microscopic view of entropy: Joule expansion reviewed. Boltzmann s postulate. S k ln g. Methods: Calculating arrangements; Stirling s formula; Fluctuations. Assemblies of quantum oscillators.!/10^6 S k log(g) ln(!) ln! ln February 05 Lecture 2 1

2 Joule expansion Entropy and configuration: Review Joule expansion of ideal gas. 2 degenerate states per molecule: Take an ideal gas and double its volume: For 1 mole: (contains molecules) S R ln(2v/v) R ln 2. Boltzmann: associated the increase in entropy with the increased number of arrangements of the molecules, g. In the present example we have molecules The initial g arrangements become 2 g finally. Entropy for final system (parts 1 and 2) S S 1 + S 2 while the arrangements behave like g g 1.g 2. Guess: S k ln g. k const. February 05 Lecture 2 2

3 Proof of k ln g guess. In the Joule expansion above, S k k [ ( ln 2 g) ln( g) ] ln 2 k R Proof of ln g guess: If, S f(g) S S 1 + S 2 g g 1.g 2 f(g) f(g 1.g 2 ) f(g 1 ) + f(g 2 ). Differentiate w.r.t. g 1, g 2 f (g 1. g 2 ) f (g 1 ) Differentiate w.r.t. g 2, f (g 1 g 2 ) + g 1.g 2 f (g 1. g 2 ) 0 f (g) -f (g)/g f (g) k/g f(g) k ln g + const. Boltzmann s constant February 05 Lecture 2 3

4 Calculating number of arrangements (Maths revision) Factorials: umber of ways of selecting distinguishable objects:.(-1).(-2) 1! 1 less to to choose from, etc. 1st choice from objects!/10^6 ln(!) Combinations: (order unimportant) E.g. distribute distinguishable objects between 2 boxes. n 1 in 1st box, and n 2 (-n 1 ) in second box. Using earlier result. February 05 Lecture 2 4

5 Combinations (Revision).(-1)..(-n 1 +1) (-n 1 ).(-n 1-1)..1 The result from this argument,.(-1)...(-n 1-1).(-n 1 ).(-n 1-1).1! depends on the order (as before) If there are g arrangements independent of order, g(n 1,n 2 ) n 1! n 2!! ways of of arranging n 1 ways irrespective of of order g ( n, n ) xx! ways of of arranging n 2! 1 2 n1! n2! n1!( n1 )! (Same as n C r n!/r!(n-r)!, no. of ways of selecting r from n irrespective of order) Generalisation to particles in r boxes! g, February 05 Lecture 2 5 ( ) n i n1! n2! nr! AA

6 Stirling s approximation (Revision) Dealing with large factorials.!, when is large: ln! ln1+ ln 2 + ln + 1/ 2 1/ 2 ln ln! ln x d x + 1/ 2 [ x ln x x] 1/ 2 Stirling s approximation ((n ln n - n)/n!)/10-6 For our purposes ~ It is an excellent approximation. February 05 Lecture 2 6

7 Distribution function between chambers How many molecules in each chamber? The numbers must fluctuate (while is clearly fixed). Let us calculate the deviation from ideal partitioning, n. Box 1 with (/2+n); Box 2 with (/2-n). Using previous result, A, and Stirling: g ln ( n n ) 1, 2 ( g) ln Using ln(1+x) x-x 2 /2 +.. For small x. ln ( g) g ln 2 2 A sharply-peaked Gaussian. ( / 2 + n)!( / 2 n)! ( / 2 + n) ln( / 2 + n) + / 2 + ( / 2 n) ln( / 2 n) + / ln 2 ( / 2 + n) ln( 1+ 2n / ) ( / 2 n) ln( 1 2n / ) exp 2n 2 ( 2 ) 2n! 6x10 23 n 2 n February 05 Lecture 2 7

8 2 non-degenerate states What if the 2 chambers have different energies? For example molecules in the right chamber have extra potential energy mgh. The entropy comes from S k ln g. S k ln k! ( ( 1 f ))! ( f )! [ f ln f + (1 f )ln(1 f )] Free Energy F U - TS is minimised at equilibrium. (See also Q6, sheet 1.) F F fε + kt f 0 Boltzmann distribution (to be derived more formally later). February 05 Lecture 2 8 [ f ln f + (1 f )ln(1 f )] f 1 f exp ( ε / kt )

9 Quantum oscillators Commence a systematic presentation of statistical mechanics. Counting microstates: (easier in a quantised system) weakly interacting assembly of quantum oscillators. I.e. They interact to exchange energy, BUT Energy levels of each oscillator unaffected by the interaction. For simplicity take energy levels equally spaced mω, Separation Eε. Example: 5 oscillators sharing 4 quanta. Oscillator o o. of states Total no. of microstates 70.B. Even with this small number a Boltzmann distribution emerges. 0 quanta most likely. February 05 Lecture 2 9

10 Counting microstates Counting states: Represent quanta as x and divided into oscillators by. Eg last 2 lines of table; xx x x x x x x Have -1 boundaries separating the quanta (Always have 2 boundaries on the outside). eed to know o. of arrangements of -1 boundaries and m quanta. g(, m) ( 1+ m) ( 1 )! m! ( 8!/4!4! 70, for 5, m 4) Distribution for one-oscillator. oscillator. 0 qu: {(4/5 x 5) + (3/5 x 20) + (3/5 x 10) + (2/5 x 30) + (1/5 x 5)}/ qu: {(0/5 x 5) + (1/5 x 20) + (0/5 x 10) + (2/5 x 30) + (4/5 x 5)}/ qu: {(0/5 x 5) + (0/5 x 20) + (2/5 x 10) + (1/5 x 30) + (0/5 x 5)}/ qu: {(0/5 x 5) + (1/5 x 20) + (0/5 x 10) + (0/5 x 30) + (0/5 x 5)}/ qu: {(1/5 x 5) + (0/5 x 20) + (0/5 x 10) + (0/5 x 30) + (0/5 x 5)}/ ! Boltzmann distribution February 05 Lecture 2 10

11 Concluding remarks Direct calculation of the number of ways of getting, for example, 2 quanta no. of ways of giving remaining 2 quanta (i.e. m-1 quanta) to remaining 4 oscillators (i.e. -1 osc.). (4-1+2)! / (4-1)! (2)! 5! / 3! 2! 10 General calculation umber of ways an oscillator gets n quanta, when m quanta are distributed between oscillators is: 1 + ( m n) 1! + m 2! {( ) } ( ) {( 1) 1 }! ( m n)! ( 2)!( m n)! ote on Q. 3(d), sheet 1 Asks how energy is partitioned between two systems ( 1 and 2 oscillators respectively), when the two are in thermal equilibrium. Evidently, if they have m1 and m2 quanta, we expect m 1 / 1 m 2 / 2. Look at ln(g(m 1 ) and maximise ln g( m1 ) m1 0 Sharpness (curvature) of peak 2 ln 2 g( m1 ) m1 peak February 05 Lecture 2 11