STATISTICAL MECHANICS
Thermal Energy Recall that KE can always be separated nto 2 terms: KE system = 1 2 M 2 total v CM KE nternal Rgd-body rotaton and elastc / sound waves Use smplfyng assumptons KE of all partcles n few eqns. In general unrealstc to descrbe KE of every atom Alternatve: descrbe a small sample of atoms nfer KE nternal Ths wll work only f the sample s statstcally sound Goal: fnd how thermal energy s statstcally dstrbuted Elastc collsons KE transfers from faster to slower partcle Equlbrum collsons occur, but dstrbuton of KE s constant
Dstrbuton Functons Let x = result of some measurement Spectrum of values for x can be ether dscrete or contnuous If many measurements are made: Descrbe results usng a probablty dstrbuton functon f(x): Dscrete x g x f x = 1 Normalzaton g x f x Expectaton Value Contnuous x f x dx = 1 g x g x Infnte precson: Prob(exact value) = 0 f x dx f(x ) 2 con flps (# heads) probablty densty p a b b = a f x dx 0 1 2 x
State Space for Ideal Gas Each partcle has 6 degrees of freedom (D.O.F.): r, v Ideal gas of N partcles 6N-dmensonal state space Each pont n state space s called a mcrostate of system A macrostate descrbes possble values for a measurement Example: Many mcrostates produce the same pressure To calculate dstrbuton f(x) for some measurement x: 1) Fnd % of mcrostates whch yeld result x (or x x+dx) densty of states Ω(x) / degeneracy volume n state space 2) Fnd probablty of each mcrostate (ncludng constrants) e.g. (conservaton of energy) and (maxmzaton of entropy )
Densty of States / Degeneracy Goal: count % of mcrostates n a gven macrostate Example: Flp a con N tmes, let x = # of heads Degeneracy (# of mcrostates w/ value x): N x = N! x! N x! Densty of states Ω(E) de for an deal gas Volume of state space whch has energy n range E (E+dE) dv ss = 4 v 2 dv E = 1 2 m v 2 v = 2 E m dv = de 2 E m dv ss = 4 2 E m E de = A 1 E de de m 2 E (more mcrostates at hgher E)
Constrants Wthout constrants all mcrostates equally possble Constrants restrct the state space avalable to system Example: 6-sded de rolled 3 tmes (216 mcrostates) Constrant: <x> = 2 Calculate how many mcrostates Among these mcrostates, fnd dstrbuton of 1's, 2's, etc. Constrants on deal gas N partcles, total energy E 0 Goal: calculate dstrbuton of E for allowable mcrostates only To smplfy calculaton, use dscrete spectrum of energes E Dstrbuton: (n 1 wth energy E 1 ), (n 2 wth energy E 2 ),
Countng Mcrostates Consder measurement wth possble results x Perform measurement N tmes dstrbuton: n Countng the # of mcrostates for a partcular dstrbuton: = N n 1 N n 1 n 2 N n 1 n 2... n 3 p x = n N = N! N n 1! N n 1 n 2!... n 1! n 2! n 3!... N n 1! N n 1 n 2! N n 1 n 2 n 3!... = N! n 1! n 2! n 3!... Strlng's approxmaton (good for large X): ln X! X ln X X ln [ N ln N N ] [n 1 ln n 1 n 1 n 2 ln n 2 n 2 n 3 ln n 3 n 3...] ln n ln n N # of mcrostate arrangements wll be maxmzed for some partcular dstrbuton of n
Entropy ln n ln n N ln N n N ln n N Ths quantty s called entropy S Probablty dstrbuton p for system naturally fnds maxmal S Wthout extra constrant: maxmal S all p are equal (prove!) Wth constrants: non-unform dstrbuton of p p ln p Examples: calculate entropy for 1) con 2) 6-sded de Constrant: based 3-sded de (<x> = 2.5) Calculate p 1, p 2, p 3 such that S s maxmzed Image compresson: Each color pxel has (256) 3 possbltes How much can fle be compressed usng black/whte (256)?
Lagrange Multplers Goal: Maxmze functon S(x 1, x 2, x N ) wth constrants M constrants: C 1 (x 1, x 2, )=0, C 2 (x 1, x 2, )=0, C M (x 1, x 2,...)=0 Usual procedure: solve M constrant equatons for x j Maxmze S wth respect to remanng (N M) x 's Solve (N M) equatons for (N M) unknowns Problem: calculatons are often tedous and/or dffcult Alternatve method: Lagrange multplers Consder the functon Maxmzng S' also maxmzes S Solve (N+M) equatons for (N+M) unknowns S 1 C 1 2 C 2... S ' x 1, x 2,..., 1, 2,... S ' x S max = S ' j S max = 0
Example Mnmze the functon S x, y, z = x 2 y 2 z 2 Subject to the constrant C 1 x, y, z = 6 3 x 2 y z = 0 What does ths look lke n 3-D space? Method 1: Solve constrant for x and plug nto S Method 2: Lagrange multplers Now add another constrant C 2 x, y, z = 5 z = 0 Calculate the new x,y,z whch mnmzes S
Boltzmann Factor Constrants on p for system wth possble energes E : 1) Normalzaton: C 1 p 1, p 2,... = 1 p = 0 2) Total KE of system: C 2 p 1, p 2,... = E p E = 0 Maxmze S p 1, p 2,... = p ln p (usng Lagrange): S p 1 C 1 p 2 C 2 p = 0 ln p 1 1 1 2 E = 0 p = e 1 e 1 e 2 E p = A e E (rename constants) p (E ) Boltzmann Factor descrbes statstcal dstrbutons of energy (gnores densty of states) E Normalze: 1 = p A = = A 1 e E e E normalzaton constant partton functon
Temperature Boltzmann factor contans unknown parameter β When β s large most partcles have low energy Hstorcal dea of temperature : average partcle moton Heat / cold felt by human senses Measured by thermal expanson of substances (arbtrary unts) β and temperature nversely related And measured n dfferent unts (temperature n ºK/ºC/ºF, β n J 1 ) Unts matched up by Boltzmann's constant: 1 k B T k B = 1.38 10 23 J K
Dstrbutons n an Ideal Gas Each possble measurement has ts own dstrbuton: Velocty Components v x v y v z : Energy: Densty of states: v x = A 1 Densty of states: E = A 1 E Boltzmann factor: e m 2 v x 2 k T Boltzmann factor: e E k T Normalzed dstrbuton: f v x = m m 2 v x 2 k T e 2 k T 2 v x = 0 v x = k T m Normalzed dstrbuton: f E 4 = k T 3 E = 3 2 k T E e E k T Maxwell-Boltzmann dstrbuton of energes Note: E = 1 2 m v x 2 v y 2 v z2 = 3 2 m v x2 = 3 2 k T
Standard Devaton Useful statstcal quanttes for a dstrbuton f(x): Expectaton value <x> descrbes center of dstrbuton Standard devaton σ descrbes spread of dstrbuton x x 2 = x 2 x 2 Standard devaton for deal gas velocty component: v x = 0 = v x2 v x, rms v x,rms for deal gas depends on temperature: v x, rms = k T m
Examples Consder a 3-state system: E 1 / 3E 1 / 5E 1 Wth no degeneracy n the states If kt = 2E 1, calculate P(E ) Repeat calculaton f 5E 1 level has 2-fold degeneracy Calculate velocty dstrbuton f(v x, v y, v z ) for deal gas Includng overall normalzaton constant Calculate σ E for Maxwell-Boltzmann dstrbuton
Equpartton of Energy Energy of mcrostate depends of values of DOF: Common stuaton: E = C x 2 (where x = DOF) C x 2 E = C x 2 k T e dx C x 2 e k T dx = C 2 k T 3 C k T C = 1 2 k T Result holds for any value of C (thus any DOF) Examples: translaton / rotaton / vbraton KE of molecules Systems wth energy whch s quadratc n several DOF: Equally dstrbutes the energy among DOF (½ kt each) Explans heat capacty of deal gases
Heat Capacty of an Ideal Gas Heat Capacty heat energy requred for ΔT: Usually whle holdng volume or pressure constant (C V vs. C P ) C P = C V + Nk some heat s used as work Equpartton Theorem: predcts C V = 3 2 N k T T C = Q T = 3 2 N k Matches expermental results for argon / helum (monatomc) Experments on datomc gases (O 2, N 2, etc.): C V = 5 2 N k C V ncreases due to molecular vbraton / rotaton But these DOF requre a mnmum T to turn on C V becomes a functon of T:
Pressure n an Ideal Gas L Usng knowledge of f(v x, v y, v z ) Can relate gas pressure to statstcal quanttes Consder an atom wth a gven velocty v x, v y, v z Makes one collson wth rght-hand wall every t = 2 L v x Each collson delvers mpulse p x = 2 m v x Average force ths atom exerts on rght wall: Total pressure on rght wall: P = N F x A F x = p x t = m v 2 x L = N V m v x2 = N V k T Ideal Gas Law
Thermodynamc Cycles Thermal energy of enclosed gas can do work And have work from external force done on t W = P dv Energy can be added to system by heat source To replensh energy lost due to work Thus creatng a repeatng cycle of work on external object Examples: burnng fuel, solar energy, etc. Gas expands whle hot, compresses whle cold Thermodynamc engnes have theoretcal lmts on effcency Because not all of the thermal energy can be used
PV Dagrams Vsualzaton of thermodynamc cycles Common paths n PV space: Isothermal: temperature = constant P = C V Adabatc: total energy = constant P = C V = C P C V Isobarc: pressure = constant Constant-volume: vertcal lne on PV dagram