DILL CH_10 Maxwell Boltzmann law

Transcription

1 DILL CH_1 Maxwell oltzmann law The basc concepts Entropy and temperature The Maxwell oltzmann dstrbuton Speed dstrbuton of the molecules M dstrbuton of harmonc oscllators Ensten's theory of lattce heat Debye sound wave model of the Lattce heat - phonons

2 Statstcal mechancs The theory s the more mpressve the more smple are ts underlyng assumptons and the broader s the scope of phenomena t descrbes. Ths s why classcal thermodynamcs has made a deep mpresson on me. It s, n my vew, the only unversal physcal theory, that n the area where ts assumptons are vald s really permanent. Albert Ensten Ludwg Eduard oltzmann (Feb Sep ) austran physcst and professor Grazssa, München, Lepzg and Wen. orn: 13 June 1831 n Ednburgh, Scotland Ded: 5 Nov 1879 n Cambrdge, Cambrdgeshre, England y treatng gases statstcally n 1866 he formulated, ndependently of Ludwg oltzmann, the Maxwell- oltzmann knetc theory of gases.

3 asc concepts of mcrocanoncal ensamble Assume that partcles do not nteract Each energy level several egenstates. All ways to dstrbute energy level E E are equal. may nclude n partcles to Sngle partcle energy levels: E1, E, E3,.. Occupaton numbers: n; 1,,3,... Macrostate = queue : n1, n, n3,.. Total partcle number s constant N n 1 Total energy s constant: U n E 1

4 Degenerate egenstates? In quantum mechancs states two egenstates can have same energy Lke Hydrogen orbtals and 1sm 1/ 1sm 1/ both have the same energy -13,6 ev but one has spn up and another has spn down. s s The degeneray s then g

5 Identty of partcles In classcal physcs, the partcles can be marked (wthout chasng ther physcal propertes) to to dentfy them ndvdually How many ways two partcles can be selected out of 1? Ths s a standard unordered sample of two out of ten, the number of selectons s equal to the bnomal coeffcent 1!! 1! 45

6 Examples of macrostates boundary condtons N n 1 1 E n E 15e Number of mcrostates P 1! 16 4! 1! 3!!! b P 1! 54 5! 1! 1!! 3! a Macrostate (a) s more than two tmes more lkely than macrostate (b)

7 An example of the mcro-and macrostates Energy n unts of e Equlbrum macrostate Macrostate Partcles Energy N 6 E 6e 11 macrostates n whch 46 mcrostates Average occupaton numbers: n W j k k, j k k k, j W n / 46 = mcrostates n macrostate k n = occupaton number of E n macrostate k j j n j 1,777 1, ,999 4, , , ,1987 6

8 Degenerate states Assume energy level has three partcles and t s twcely degenerate g = Label partcles by a, b and c. Possble dstrbutons: Egenstate 1 Egenstate a,b,c - a,b c a,c b b,c a c a,b b a,c a c,b - a,b,c It can be shown that n the general case there are n g dstrbutons - n our example 3 8 so ths works!! Total number of mcrostates ncludng degeneracy n N! n N n1! n N n1 n! W g 1 g g 3 n1! N n1! n! N n1 n! n3! N n1 n n3! 1 3 g n g n g 1! 1 N... N! n! n! n! n! 1 3 n g n..

9 The most probable macrostate Thermodynamc equlbrum s the macrostate whch has more mcrostates that any other macrostate Optmzaton problem: Usng boundary condton determne W N! g n, n, n n n! 1 3,... obtans maxmum value. N n and U n E such that

10 Maxwell-oltzmann dstrbuton M occupaton numers Partton functon E / k T Here p s a probablty that a sngle partcle resdes on energy level. Q N n Np ge Q E / kt Q ge Dll uses Q for both many partcle partton functon and Sum s over (sometmes) sngle partcle partton functon. egenenergés not egenstates! Total energy The sum over N E / kt d U n E gee knt ln( Q) egenstates s Q dt accounted for by the degeneracy g d N Dll wrtes ths n many partcle case as kt ln( q ) dt where q n notaton g e E / k T there s thus slght dfference

11 Entropy In statstcal mechancs entropy s defned by the number of mcrostates S k lnw Macrostate that has largest entropy corresponds to thermal equlbrum. Energy n unts of e Macrostate Number of mcrostates n each macrostate Whch macrostate has the maxmum entropy? Macrostate 6 because t has the more mcrostates than any other mcrostate. It s at the same tme also the thermodynamc equlbrum!

12 Form dscrete to contnuous M-dstrbuton N E / kt N E/ kt n ge dn E, EdE D Ee de Z Z In dscrete case yhere s n partcles on energy level E In contnuous dstrbuton there are wth ther energy n the band Replace Functon g dn E, E de by D E de e g D E de D E Integrate over all energes! s called densty of states E / k T E/ k T Q g e Q D E e de E, EdE partcles

13 DOS for molecule translaton There s one pont of phase space for each value of speed (vector) : Volume of the phase space = surface area thckness = dv 4 v dv Number of states n a shell dn v,v dv vako 4 v dv DOS D v dn v,v dv / dv constant 4 v Number of states havng absolute value of velocty wthn v, v dv s proportonal to volume of the shell

14 DOS propertes In practce, t s more advantageous to fnd the number of states per unt of energy. DOS per unt energy: (we use standard change of varable technque known from mathematcs) E, EdE / v, vdv / / D E dn de dn dv dv de 1 v E / m E mv dv / de 1 / Em change of varable: vako 4 1 / vako 4 / 1/ D E v Em E m Em C E Constant remans unknown: t s not needed and cancels out n M dstrbuton

15 Partton functon for molecules From dscrete to contnuum dstrbuton Q E / k T Replace sum by ntegral and degeneracy by DOS g D E C E. g e 1 Q CE e de C k T 1/ E / kt 3 ( ) 1/ (From Tables of ntegrals: x expaxdx ) 3/ a

16 Energy dstrbuton of molecules From dccrete to contnuum transformaton we obtan N dn N n ge DEe Q de Q E / k T E/ k T and n the partton functon E / 1/ / 1 kt E kt 3 ( ) Q g e C E e de C k T The unknown constant cancels and we obtan dn N de ( k T ) 3/ E 1/ e E / k T (1) ()

17 Total energy Total energy N E/ kt N U EDEe de cf. U gee Q Q E / k T Ths equals to d(ln Z) U knt dt Usnfg partton functon ln Q ln C k ln T ln C ln T tässä C C k We have total energy 3 U knt 3 Average TRANSLATION energy per molecule s k T!! Note that translaton s actvated at all temperatures.

18 Hpw to measure the speed? The mage shows measurement of the speed of the gas molecules escapng from the oven. On the rght sde the comparson of the dstrbuton of the measured and the M. The data presented on the relatve speed v/v m, where most lkely as a functon of the speed value s v m.

19 Speed and energy dstrbuton O O Energy dstrbuton dn de N ( k T ) 3/ E 1/ e E / k T Speed dstrbuton dn m 4 N dv kt 3/ v e mv /k T

20 Statstcal averages of molecular speed dn dv Average speed v ave 1 N vdn 8k T m 1/ 1,13 v mp Root mean square velocty v mp vmp v ave vrms Most probable value of speed: 3/ dn dv k T m 1 rms ( ) ave N v v v dn 3k T m mv mv /kt 4 N v e vmp kt rms 3 1/ k T v m 1/ 3k T m

21 Thermodynamcal functons Summary of thermodynamcsl functons expressed usng the partton functon lnq N E / k T U kt Here Q q where q ge and E T V, N s sngle partcle energy U S k lnq T F U TS kt lnq F lnq kt N N T V F lnq p kt V V, T, V T, N T, N

22 Calculatng entropy We start from the defnton of entropy S k lnw k p ln p where p n / N j1 Insert M dstrbuton 1 E / kt t j j j j j1 See Dll ch5 p8 j pj Q e t 1 E / kt 1 E j j U S k e ln k lnq k ln Q Q kt Q T lnq kt T

23 Dstngusable partcles - sold We The system s made of two dstngusable partcles A, ther sngle partcle partton functons are for ndependent partcles a A b / kt / kt m 1 m1 q e and q e A The total partton functon s sum over total energes t Generally for ndependent partcles a, b A / a, b E / kt kt A / kt / kt j j m m Q e e e e q q Q q j1 j, m1, m1 N N NONDISTINGUISALE PARTICLES! HOW DO THEY DIFFER? If the partcles are ndstngusable the order n whch they are put to the energy levels s not sgnfcant. Therefore we have to multply by the number of possble permutatons s N!. A

24 N Harmonc oscllators Total energy s constant but they may exchange energy Harmonc oscllators have the same energes 1 E On level E there are n oscllators n N Total energy 1 d U n k NT lnq dt s constant.

25 M-dstrbuton for oscllators M-dstrbuton N n e Qvb Partton funton s vb 1/ / k T 1/ / k T / k T / k T Q e e e Use x / k T e / k T 1 / k T 1 1 e x e 1 x - Note vb x e 1 converges always when k T k T 1 / / Q e e 1

26 Energy Total energy s d U knt lnqvb dt Where 1 ln Qvb ln 1 k T / kt e / kt 1 e vb kt / kt 1 d ln Q / dt k T e Ths gves 1 1 U N N / kt e 1 Ths s zero-pont energy that can be neglected n Thermodynamcs Why?

27 Egenfunctons

28 Hgh temperature lmt T k / Energy os a sngle oscllator U E n N / k T / k T e 1 where average quantum number s n At hgh temperatures T e 11 / k T 1 / k T / k f we also drop the zero-pont energy E k T we have e / k T Ths s called equpartton prncple (for a sngle generalzed coordnate) 1 1 k T knetc energy and k T potental energy for an oscllator 1 1

29 Lattce heat Atoms n a sold are hold together by chemcal bonds Debye theory Ensten ndependent oscllator model

30 Enstenn lattce heat In the Ensten model oscllators are ndependent Each atom oscllates n x,y and z drectons wth the same frequence We can drectly apply M for 3N (N s the number of atoms one 3 1-D oscllators for each atom) dependent harmonc oscllator - neglectng zero pont energy E 3N / k T 1 1 At hgh temperature e. lm E 3Nk T T Ths s n lne wth the equpartton theorem!

31 Debye Lattce heat 1/7 In actual vbratons n solds are not ndependent cf the propagaton of sound n steel! The vbratons of the next neghbor atoms are coupled to form a standng sound waves, ether longtudnal or lateral. One standng wave mples coherent vbraton of all atoms - t can be descrbed by one scalar general coordnate. There are 3N ndependent coordnates and the same number of standng waves. The number of standng waves s equal to the total number of degrees of freedom n the system each atom has 3 degrees of freedom so f there are N atoms n the crystal there are 3N degrees of freedom n total

32 Debyen model /7 Rght panel shows the second longest standng wave. If the length of the crystal s L the wave length of ths wave s L. a Edge of crystal a = lattce constant Shortest possble sound wave. Shortest possble wave (left panel) corresponds to neghbor atoms beng 18 out of phase! There can not be a sound wave wth shorter wave length!

33 Phonon number 3/7 Each standnd wave s a harmonc oscllator f v where max / s v S 1 / 1 / n,1,,3,4.. f v / a. s the velocty of sound. E n hf n There are 3N allowed frequences startng from f v / L up to mn n s s f Red wave has quantum number n= 1. Green wave has n = 3. We say that there s one phonon n the red state and three phonons n the green state! The average energy of a standng wave s E n 1 / hf where n 1 / exp hf / k T 1

34 Energy n Debyes model 4/7 For a dfferental band the energy s du Energy of average quantum number DOS at, oscllator de (1) n 1/ exp hf / kt 1 E hf E E de For total energy ntegrate up to maxmum frequence (cut off frequence) lower lmt s approxmately zero. vs (when max L, hf h ) L 8V DE E where we have ncluded two transverse waves phonos h v 3 3 st

35 3 E, EdE 3 3 E/ kt h vst e Total energy 5/7 Lattce heat U n the band E, E de for a crystal wth volume V L 8V 1 du E de 1 3 (1) Ths s just the transverse sound. It has velocty v st and two dfferent polarzatons. In addton there s longtudal sound (only one component) ts velocty s we ntegreate to cut off E max hf max v sl. When all three waves are n Emax 4V U phonons E de E / kt h vst v sl e 1

36 Above we set max max Cut off frequence 6/7 f v / a where a s lattce constant. If there are N atoms we conclude 1/3 1/3 3 V N Na V a fmax vs / N V E N hvs / V 1/3 s (1) A more exact lmt s obtaned by ntegratng DOS : 3 N D( E) de E de E f we set the transverse and longtudal veloctes equal E max Emax Emax 4V 1 4V 1 3 = max h v st v sl 3h v st v sl 1/3 1/3 3 N hvs 4 V

37 Dulong Pett law 7/7 Defne Debye temperature T D Emax hvs 3 N k k 4 V to get the total energy 1/3 ktd 3 3 E/ kt e 9N 1 U phonons E de k T 1 D One can wrte the ntegral usng x E / k T : T 4 / T 3 D 9kN x U phonons T 3 dx x T e 1 1 / D One can show that at hgh 9RN T mol 3 D T 4 / T 3 x e 1 the equpartton prncple. We have 3N degrees of freedom and for each degree of freedom of potental energy and kt of knetc energy. T D T U 3 knt. We regan x dx

38 Comparng Ensten and Debye models Debye k T D 3 3 E / k T e 9N 1 U phonons E de k T 1 D c V U N R mssä N A T N k V A Energy/atom s at hgh temperatures Ensten wthout zero pont energy U N c T V k R 3k T n both theores U phonons N 1 / kt e 1 c V 1 U T V

39 Energy and heat capacty k T Sngle oscllator average energy U 1 E N / k T e 1 as a functon of kt zero-pont energy dropped away Molar heat capacty c V 1 U T V At hgh temperatures equals 1R.

40 The degrees of freedom Frst atom s n the center of system. The poston of the another atom s Specfed by polar angles and dstance.

41 Rotaton Two atomc molecule has two axs of rotaton. 3D molecules lke methane has three ndependent rotaton axs Klasscal energy of rotaton 1 Erot I L L I L / I Erot I I rot M r Momentum of nerta M M M r M r r r E M1 M l l 1 r

42 Energes of rotaton levels Rotaton energes E l ll l l 1 1 ; l,1,,.. r I Actvaton energy El I Absorpton of lght follows Dpole rules l 1

43 M dstrbuton Defne l l r / Ik / T l M dstrbuton N nl l 1 e Z Partton functon Q l 1 e lm Q le l( l1) / T l( l1) / T Hgh temperature lmt r r rot T rot rot l r le dl T / r / T r r Occupaton probablty Quantum number

44 Rotaton at hgh temperatures l r / T At hgh T Q le dl T / rot (for many partcle d case ths s qrot n Dlls notaton) U knt ln Qrot knt dt d 1 snce ln Qrot. dt T Ths agrees wth equpartton prncple snce for rotaton nthere s only knetc energy. We have 1/ kt for both angles and. r There s a threshold energy for rotaton! It s the dfference between the energy of the frst excted state and the ground state energy. Rotaton begns when average thermal energy per molecule exceeds ths dfference: kt / I T / ki where I s the moment of nerta.

45 Vbraton and heat n a molecule Electronc energy + nuclear repulsn = E s gven by Taylor expanson: E ( r) E ( r r ) r r r r... p p de p 1 d E dr rr dr rr p Ep( r) const (1/ ) k r r const E ( r r ) k p where and d Ep p Energes quantze n 1 / E n const k /, n,1,,3,... dr rr M1M = relatve mass = M M 1 Ths s the equlbrum dstance of atoms n a molecule

46 Occupaton number of vbraton levels Occupaton numbers v N 1/ v / T n e Q vb Parttofuncton Q e vb / k 1/ / T / k,1,,3,.. v / T v / T e e Ths s exactly the same as n the of harmonc oscllator. We denote v v Qvb e 1 e v T 1 / T / v Occupaton probablty Quantum number

47 Ocllator energy Vbraton heat at hgh T d 1 1 U knt ln Qvb Nkv Nk dt v / T e 1 settng / k : 1 E N NkT Heat energy U Nk T or for each molecule k T 1 1 of whch kt knetc energy and k agrees wth ekvpartton prncple / T v At hgh T / T 1 e 1 / T v v v T potental energy, ths

48 Prncple of Ekvpartton Each actve degree of freedom gans ½ k T of both knetc and potental energy (potental energy does not exst always) A degree of freedom s actve f ts threshold energy s << k T

49 Expermental lattce heats Methane CH 4 has 5 atoms and 15 degrees of freedom. 3 degrees are related to translaton and 3 to rotaton other 9 are nactve at room temperature Average heat at room temperature: 3 x (1/)kT + 3 x (1/)kT = 3kT. Expermental value 3,5 kt. U C V R kt

50 Heat n a datomc translaton U kn 1 3 kt Rotaton 1 Urot kt Vbraton U U U vb vb, kn vb, pot kt 1 kt kt Energy per molecule [ k T ] 7 / 5/ 3/ A schematc presentaton

51 Combned rotaton-vbraton transton Vbraton levels Rotaton levels Molecules have both rotaton and vbraton energy There are hundreds of rotaton levels between two vbraton levels When a photon s absorbed both rotaton and vbraton quantum numbers change. Vbraton quantum number decreases or ncreases by one the same apples for the rotatonal quantum number f ncreases P-branch R-branch

52 Characterstc temperatures

53 Electronc partton functon For noble gases and atoms wth closed subshell fnfguratons the electronc exctaton energes are so large that the excted states are not occuped. The sngle partcle electronc partton functon s gven by qel g g1 e... all states 1 / kt sum over

54 Electrons n 3D box Electron states n a cube,, / sn n x sn n y sn n z a a a 3/ 1 3 x y z a oundary condtons y z a y z x z x a z x y x y a,,,,,,,,,,,, wave vectors k x n1; k y n; k z n3 where n1, n, n3 1,,3,.. a a a and energes E n n n ma 1 3 Edge of the cube = a. Outsde the cube potental energy s nfnte - nsde t s zero volume of the cube V = a 3.

55 DOS for electrons 1/4 There s one egenstate for each set n 1, n, n 3 of postve ntegers Calculate the number of states that have energy smaller than some orbtrary chosen value E. Choose n, n, n 1 3 ma such that ma n n n E n n n E 1 3 or 1 3 We splt the space of ntegers nto unt cubes each unt cube corresponds to one egenstate n 1, n, n 3.

56 DOS for electrons /4 Calculate the number of states n1 n 3 n 3 such that R R E ma Now set R E ma E ma dn1dn dn R 8 3

57 DOS for electrons 3/4 Pcture shows a cube correspondng to one egenstate R n 1 n3 n3 Those unt cubes for whch ma n n n E (1) 1 3 have dstance from orgn smaller than R R ma ma E Note f R E ma n n n E 1 3 whch contradcts Eq. (1)

58 DOS for electrons 4/4 Insert R Ema to obtan 3 / 1 4 Ema 8 3 Ths s the total numner of states wth energy smaller than E. If D( E) s the densty of states / unt energy ths means 3 / E 1 4 E ma DEdE 8 3 Take dfferental of both sdes wth respect to E to obtan(note that V a ) D E 3 / m VE 1 / 4 3 Multply by because of the electron spn

59 DOS n bg box DE Dscrete set of energes becomes a contnuum

60 Note DE DE foton d / de d / de DOS of phonons Densty of standng waves / unt wave length s same for electrons and phonons D phonon Change varable D E electron foton D d D de Ths follows from E E elektron phonon elektron elektron hv because sound / whle for electrons p h h snce de rogle gves m m p

61 Change varable DOS for phonons d de d DE D phonon DE de d de elektron phonon elektron phonon Use wavelength to energy rato for electrons and phonons d de phonon hv s Note hv / E ; E hv / de h h 1/ h Note ; E 3 d elektron m me m Insert n Eq. (1) and express the densty as a functon of energy 3 / m DE VE phonon 8 V 1 8 V () 3 3 E hv h v s s s 1/ s 3 / 1/ h m h h V 3 3 m hv s m m hv s For transverse phonons spn degeneracy of electrons compensates the polarzaton of sound for longtudnal sound dvde () by! (1)