Vat ly thong ke va nhiet dong luc hoc thong ke DoXuanHoi

8/14/2019 Vat ly thong ke va nhiet dong luc hoc thong ke DoXuanHoi

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LI NOI AU

Cuon sach nay c viet xuat phat t giao trnh vat ly thong ke a giang c lp sinh vie n nam th t khoa Vat ly , trng HSP TP.HCM t mot vai nam qua. Tuy c soan theo tinh than hanh tai khoa Vat ly, trng HSP TP. HCM, nhng noi dung sach cung a c cap t lieu cho sinh vien.

Sach c trnh bay vi no lc ln ve ma t s pham: Ngoai phan bai tap kem theo mo i chng e cung ccung nh e ao sau them nhng kien thc a c phan tch trong phan ly thuyi ln hn c soa n di dang cac van e e sinh vien tap lam quen vi viec nghien cu ti khoa hoc tro n ven va sinh vien thay c cac lnh vc ap dung cua vat ly thong ke, v du nh trong vat ly cothe dung e gi y cho cac sinh vien lam seminar trong nam hoc, luan van tot nang caothem e chuan b cho cac luan van Thac s vat ly. Nhan thc c rang viec nam vng t nhat la moi nge c t nang cao trong quatrnh ao tao la ieu nhat thiet phai co o i vi moi sinh vien nen trong phaluc co kem theo mot danh muc cac t ng oi chieu Viet-Anh-Phap thng thong ke. Hy vo ng rang phan nay se giu p ch cho cac sinh vien khi s dung ngoai ng trong khi

Cung can nhan manh rang theo y kien cua mot so nha vat ly co uy tn tren the gii th phan nhiet on hoc phai c xem nh la he qua cua mon c hoc thong ke, c trnh bay nh

s, co ngha la phat xuat t cac tien e, cung tng t nh mon c hoc lng t nen nh rang mon c hoc thong ke, cung vi c hoc lng t va lythuyet tng o i, hie n ang tao nen motrong cac tru cot cua vat ly hien ai. Cuon sach nay c xay dng tre n tinh than o.

Mot cach tom tat th vat ly thong ke co the c hieu nh la mon hoc khao sat he vat ly xuat phat t cac ac tnh vi mo cua nhng hat ca u tao nen he. Nhng cac ac tnh vi mo naythe c mo ta chnh xac bi c hoc l ng t. V vay, e hieu c c s cua vat ly thong phai nam vng cac tnh chat l ng t cua cac hat vi mo. Tuy nhien, trong cuon sach nay, c hoc lng t c yeu cau mc to i thieu. Nhng ieu g can thiet se c nhac lai trong st giao trnh.

Cu ng nen noi them rang ra t ang tiec la mot so phan quan trong cua vat ly thong ke nh khcua vat chat, hien tng chuyen pha, hie n tng van chuyen,... khong c e cap en trong cuon sacTac gia hy vong rang trong lan tai ban sau se co ieu kien trnh bay cac van e tren.

Do kinh nghiem con t, thi gian lai rat ha n he p nen chac chan cuon sach nay con nhieu tu sot, mongcac ban oc vui long l ng th va ch dan e sach c hoan thien trong lan tai ban sau.Tac gia xin tran trong ngo li cam ta en tha y Hoang Lan, nguyen Trng khoa, va tha y Ly Vnh Be

Trng khoa Vat ly, trng HSP TP. HCM a tao tat ca cac ieu kien thuan lic truyen at en cac sinh vien trong vai nam va qua. ong thi, tac gia cung xin bay to long b PGS-TS Nguyen Khac Nhap va thay ang Quang Phuc a vui long e ra th gi c ban thao sach v go p y cho tac gia.

Ngoai ra, tac gia cung ghi lai ay li cam n en GV Nguyen La m Duy va SV Nguyen Trong Khoalc anh may vi tnh ban thao vi long nhiet tnh va tan tuy nhat.

Cuoi cung, tac gia bay to long cam n en Phong An ban trng HSP T cuon sach nay mau cho ng c in va en tay ba n oc.

TAC GIA


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Chng I

MO TA THONG KE HE V MO

IA Nhng trang thai vi mo kha dIB Phng phap thong ke cho he v mo

IC Tap hp thong ke. Nguyen ly ergodicID Entropi thong ke trong ly thuyet thong tin

Vat ly thong ke co oi tng nghien cu lanhng he v mo , la nhng he cha mot so rahat (nh electron, photon, nguyen t, phan t,); nhng he nay co the ton tvat ly khac nhau : kh, long, ran, plasma va bc xa ien t. Ve phng diennang lng cua mot he v mo c xac nh bi met (va cac boi so va

Trong khi o,he vi mola he co kch thc so sanh c vi kch thc cua ngutc la c o lng bi ( = 10-10m ), va nang lng cua he vi mo se c o bang 1,6.10-19Joule ).

Mot cach n gian nhat e thiet lap moi quan he gia mot he v mo vahang so Avogadro NA6,023.1023 hat.mol-1. o ln cua hang so NA nay cho chung ta thay mphc hp rat ln cua mot he v mo. Chnh v vay ma e khao sat cac phng phap thong ke , e co c nhng ai lng v mo phat xuat t cac tnh c

Trong chng th nhat nay, ta se gap nhng khai niem c ban nhat c ske. ieu au tien la s phan biet giatrang thai v movacac trang thai vi mo kha d at (accessible microstates) cua mot he v mo, ta se thay ro s khac biet gia hai khaminh hoa cua mot he ch co hai hat. Vi th du nay, ta cung se a vao khc va cac hat khong phan biet c; hai khai niem c ban can phai namhe nhieu hat. Sau o, phng phap thong ke se c gii thieu e a ra nthong ke. Trong cac phan tiep theo, nguyen ly ergodic c trnh bay va kha

c a ra da tren ly thuyet thong tin trong trng hp tong quat nhat.I.A Nhng trang thai vi mo kha d I.A.1 Trang thai v mo cua mot he vat lyTrang thai cua mot he vat ly ma ta co the mo ta bi cac ai lng v

con ngi c goi la trang thai v mo cua he. V du nh neu ta xet mot khmo nay co the la the tch, nhiet o, cua khoi kh. Nh vay, mot trang thnh bi cac ieu kien ma he phu thuoc. Chang han oi vi mot he khoben ngoai (he co lap), th nang lng va so hat tao thanh he luon co gia t

I.A.2 Trang thai vi mo lng t cua mot he vat lyTheo quan iem cua c hoc lng t, trang thai vat ly cua mot hat tai m

dien bi mot vect trong khong gian trang thai, o la vect trang thai ket)t( .S tien hoa theo thi gian cua mot trang thai vi mo c mo ta bi phndinger

)t(H)t(dtdi =h , (I.1)

trong oH la toan t Hamilton, toan t lien ket vi nang lng, bang tong T va toan t the nang tng tacU :


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UTH += . (I.2)

Neu goir la vect rieng tng ng vi v trr r cua hat, tch vo hng)t,r ()t(r r= (I.3)

cho ta ham song, ac trng ay u cho trang thai vat ly cua he.Trong trng hp he bao toan (H oc lap oi vi thi gian t), nang lung El cua he trang thal

c xac nh bi phng trnh tr rieng:ii EH lll =

vi i=1, 2, , gl cho biet s suy bien cua he.Tong quat hn, khi oi tng nghien cu la mot he nhieu hat th ham so( q1, q2 , , qf ) theo

cac bien so la toa o qi se ac trng ay u cho he hat. ay, f la so lng t cuChu y rang khi ta noi en trang thai vi mo cua mot he v mo th ta nga

trang thai vi mo lng t. Con neu ta nhan manh en trang thai vi mo co icua he c khao sat thong qua c hoc co ien Newton nh ta se thay. D nhicua chung ta thu c ch la gan ung ma thoi.

Thong thng th mot he v mo luon c at di mot so ieu kien (vhan che (constraint ), chang han nh oi vi mot khoi kh co lap, khong tng tac vi mnang lng va so hat cua he xem nh la nhng ieu kien do moi trng benhien la hai ai lng nay la khong oi. Khi o se ton tai mot so nhngcua he tng ng vi cung mot trang thai v mo nay. So trang thai vi mo, ong vai tro trong yeu trong viec nghien cu vat ly thong ke.

V du: e de hieu van e, ta se xet mot he nhieu hat n gian gom ch hla co the anh dau c la hat A va hat B. Hai hat nay c phan bo treeu nhau la0 =0 , 1 = , va2 =2. Gia s nang lng toan phan cua he c an .Ta hay xet nhng trang thai vi mo kha d cua he tng ng vi trang thai v

H.I.1Ta co the em so trang thai vi mo bang cach dung s o nh hnh tren:

xep tren cac mc nang lng sao cho tong nang lng cua hai hat bang 2. Vay, co tat ca la 3 tthai vi mo kha d: (1), (2), va (3); = 3. V hai hat A va B phan biet c nen hai trang tva (2) phai c xem la khac nhau.

Neu ta gia s hai hat tao thanh he la khong phan biet c th ta se co s

H.I.2Vay khi nay ta co = 2, nho hn so vi trng hp he cac hat phan biet c.

2= 2 B A

1= AB 0= 0 A B

(1) (2) (3)

2= 2

1=

0= 0 (1) (2)


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Bay gi ta gia s rang mc nang lng1 suy bien bac 2 (tc la mc nang lng1, se co hai

trang thai lng t khac nhau). Khi hai hat la phan biet c, ta co = 6 nh c bieu dien trongo sau:

H.I.3( ay, ta gia thiet rang hai hat co the cung mot trang thai lng t)

Con khi hai hat la khong phan biet c, ta se co =4.

H.I.4

I.A.3 Trang thai vi mo co ien mot mc o gan ung nao o, trang thai vi mo cua mot he v mo

co ien. Ta se xet trng hp n gian nhat la trng hp mot hat chuyenrong cho trng hp tong quat hn.

a) Mot hat chuyen ong mot chieuVi khai niembac t dola so toa o can thiet e xac nh v tr cua hat th tr

nay la he co mot bac t do. Ta biet rang trong c hoc co ien, trang thai cta bi toa o suy rong q va ong lng suy rong p, la nghiem cua he phn

=

=

qH p

pHq

&

&

) b5.I(

)a5.I(

vi H la ham Hamilton cua he.Nh vay, ta co the noi rang trang thai c hoc (co ien) cua hat tai moi t

bang mot iem co toa o (q, p) goi laiem phatrong khong gian tao bi hai truc toa oOq vaOp goi

lakhong gian pha , la khong gian hai chieu. V cac ai lng q va p bien thien tiem pha (q, p) vach thanh mot ng trong khong gian pha; o laqu ao pha.

2= 2 A B

1= A B B A AB AB

0= 0 B A(1) (2) (3) (4) (5) (6)

2= 2

1=

0= 0 (1) (2) (3) (4)


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V du: Xet mot dao ong t ieu hoa tuyen tnh co ong nangm2

pT2

= va the nang 22qm21U = ,

vi m va la khoi lng va tan so goc cua dao ong t. Ta co ham Hamilton:

222

qm21

m2 pUTH +=+= ,

m p pHq ==& ,

qmqH p 2=

=& ,

qm pq 2== &&& .

Ta co phng trnh vi phan theo q:0qq 2 =+&& .

)tsin(qq 0 += , vi q0, la hai hang so phu thuoc ieu kien au.),tcos( pqm p 0 +== & 00 qm p = .

e tm qu ao pha, ta thiet lap he thc gia q va p oc lap vi t:

1 p p

qq

20

2

20

2=+ .

Vay qu ao pha la mot ellip co cac ban truc la q0 va 00 qm p = .

H.I.5 H.I.6e em so trang thai vi mo kha d cua hat khi trang thai c hoc cua h

khong gian pha, ta chia eu cac trucOq vaOp thanh nhng lng nhoq va p. Nh vay, khonggian pha trong trng hp nay la mat phang c phan thanh nhng o ch nhat nhbang pq= . Mot trang thai c hoc cua hat tng ng vi mot iem pha nam ta cang chnh xac khi cang nho: trong c hoc co ien, c chon nho tuy y, tc la mo

thanh mot iem chnh la iem pha.Chu y rang theo c hoc lng t,nguyen ly bat nh Heisenbergcho ta he thc: h2 p.q , vi

2h=h (h la hang so Planck). Tc la khong ton tai mot trang thai c hoc vi

c xac nh vi o chnh xac tuy y. Vay moi trang thai vi mo cua hat phco dien tch bang h== 2 pq0 , ch khong phai bi mot iem pha nh trong c hoc

ph= 2

p

O q q

qu ao pha p

p0 (q,p):iem pha

-q0 q0 q

-p0


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b) Trng hp he co f bac t do

Tc la khi nay, he c mo ta bi f toa o suy rong (q1, q2, , qf ) va f ong lng suy rong1,p2, , pf ).

V du:- He gom mot hat chuyen ong trong khong gian ba chieu co v tr xac n1 x ,

q2 y , q3 z ), vay he nay co ba bac t do: f = 3. Khong gian pha tng ng se lchieu: ( q1, q2, q3, p1, p2, p3 ). Moi o ac trng cho mot trang thai vi mo co the tch( )3 pq .

- He co N hat: v moi hat co ba bac t do nen he co so bac t do la: f =khong gian pha 6N chieu.

Vay tap hp cac ai lng (q1, q2, , qf , p1, p2, , pf ) tng ng vi mot iem pha tronggian pha 2f chieu, goi lakhong gian K , e phan biet vi khong gian pha co hai chieu.

Tng t tren, moi trang thai c hoc cua he co f bac t do c bieu diethoa ieu kien: f f 21f 21 p... p p.q...qq = vi nho tuy y theo c hoc co ien.

Nhng theo c hoc lng t, moi trang thai vi mo cua he tren c bieu dtch thoa ieu kien: f f 21f 21 )2( p... p p.q...qq h tuan theo nguyen ly bat nh HeisVay, oi vi he N hat chang han, th moi trang thai tng ng vi mot o trtch( ) N3 N3 h2 =h .

I.A.4 Mat o trang thaiXet trng hp nang lngE cua he v mo co pho lien tuc. Ta chia nang lng

nhoE sao cho E van cha mot so ln nhng trang thai vi mo kha d. Goi)E( la so trang thai vi mkha d co nang lng trong khoangE va EE + . Khi E u nho ma)E( co the c viet:

E).E()E( = , (I.6)(viE u nho, ta ch gi lai so hang au)

trong o )E( oc lap vi o lnE , th )E( c goila mat o trang thai, v thc chat th theo cothc tren,)E( la so trang thai vi mo co c trong mot n v nang lng.

I.A.5 S phu thuoc cua so trang thai vi mo kha d theo nang lngXet trng hp mot khoi kh gom N phan t giong nhau cha trong mot b

lng toan phan cua khoi kh la,EUK E int++=

trong o, K la ong nang cua chuyen ong tnh tien cua cac phan t kh i p cua khoi tam moi phan t; K ch phu thuoc cac ong lng nay:

=

== N

1i

2i N21 pm2

1) p,..., p, p(K K rrrr .

ai lng )r ,...,r ,r (UU N21rrr= bieu th the nang tng tac gia cac phan t, phu th

tng oi gia cac phan t, tc la ch phu thuoc vao v tr khoi tam cua caCuoi cung neu cac phan t khong phai la n nguyen t, cac nguyen

quay hoac dao ong oi vi khoi tam, cac chuyen ong noi tai nay c atai Q1, Q2, , QM va ong lng noi tai P1, P2, , PM. Nh vay, Eint la nang lng cua cac chong noi tai nay va ch phu thuoc vao Qi va Pi (neu la phan t n nguyen t th Eint= 0).

Trng hp ac biet n gian la0U : tng tac gia cac phan t rat nho so vkhac, co the bo qua. Khi o, ta co he kh ly tng. Trng hp nay xay ranho lam cho khoang cach trung bnh gia cac phan t tr nen rat ln.


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Gia s rang ta xet khoi kh ly tng gii han co ien. Khi nay, so)E( co nang lng trong khoang (E , EE + ) se bang so iem pha trong khong gian pha gi

E va EE + :

,dP...dPdP.dQ...dQdQ. pd... pd pd.r d...r dr d...)E( M21M21 N21 N2EE

E1

rrrrrr +

trong o: iiii dzdydxr d =r va

iziyixi dpdpdp pd =r .

V = Vr d ir nen:)E(V)E( 1

N , (I.7a)

vi:

.dP...dPdP.dQ...dQdQ. pd... pd pd...)E(EE

EM21M21 N21i

+

rrr oc lap oi vi V.

Hn na, trong trng hp kh n nguyen t: Eint= 0, va

= =

= N

1i

3

1

2i pm2

1E ,

gom 3N = f so hang toan phng.Vay trong khong gian f-chieu cua ong lng, phng trnh E = const bieu die

knh 2/1)mE2()E(R = .So trang thai nh vay bang so iem pha nam gia hai mat cau co ban kE). Ma

so trang thai cha trongkhoi cauban knh R(E) c tnh:2/f f )mE2(R )E( = ,

nendE

E)E()EE()E(

=+= .

Vay:2 N312 N312f EEE)E( = .

Phoi hp ket qua tren vi (I.7a), ta co:

, (I.7b)

vi N co o ln khoang bang hang so Avogadro. Tc la)E( tang rat nhanh theo N.

Tong quat hn trng hp ac biet tren, ta co the chng minh rang:. (I.7c)

Tc la so trang thai vi mo kha d la ham tang rat nhanh theo nang lntrong cua c hoc thong ke cua he v mo.

Chu y rang trong cong thc (I.7c) tren, ieu ta can chu y la o ln xac cua )E( , do o, ta khong quan tam en so mu cua E la f hay la mot so ha

2 N3 NEAV)E( =

f E)E(


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I.B Phng phap thong ke cho he v mo

I.B.1 Ham phan bo thong keTrc khi a vao nh ngha ham phan bo thong ke, ta nhac lai ngan go

trong ly thuyet xac suat:Mot bien co c goi langau nhienkhi ta khong co u thong tin e biet trc ket

cua mot bien co nh vay c goi la bien ngau nhien.V du:Ket qua cua viec nem mot con xuc sac, hoac:Van toc cua mot phan t kh sau mot lan va cham vi mot phan t khacla cac bien ngau nhien. Goi tap hp cac bien co nay la{em; m = 1, 2, }, va goi Nm la so lan bien co em xuat hien sau N

phep th ong nhat (tc la cac phep th c thc hien trong cung cac ieXac suat cua bien co emc nh ngha la:

, ,

Nmgoi la so bien co thuan li.V Nm, N 0 va Nm N, ta co ngay tnh chat cua Pm:

0 Pm 1.Trong o, Pm= 1 cho ta bien co chac chan va Pm= 0 khi bien co la bat kha (khong the xTrng hp bien ngau nhien co gia tr thc, lien tuc trong khoang (x1, x2) vi x la mot gia tr t

khoang nay: x(x1,x2), vax la gia so tai x, ta goiN(x) la so lan bien co cho ta ket qukhoang (x, x+x), xac suat e ieu nay xay ra la:

N)x( Nlim)x(P

N = , (I.8)

Khi o, neu ton tai mot ham so thc(x) sao cho:

x )x(Plim)x( 0x = , (I.9)th ham(x) c goi lamat o xac suat , hayham phan bo thong ke tnh tai x.

H.I.7Ta co the viet bieu thc cua xac suat nguyen to la:

dx).x()x(dP = . (I.10)

(Ta co the hieu rang ta a khai trien Taylor cua dP(x) theo dx va chi gi lai Trong trng hp ta co ba bien ngau nhien lien tuc, oc lap nhau ( x, y, z ),thong ke la ham theo (x, y, z):(x, y, z). Xac suat nguyen to e x, y, z trong khoangy+dy), (z, z+dz) c viet:

dxdydz).z,y,x()z,y,x(dP = . (I.11a)

Ta co the viet ngan gon hn:r d).r ()r (dP rrr = , (I.11b)

P(x)x x+x+ + + + +O x1 x x2

N N

limP m N

m =


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trong o, k z jyixr

rrrr ++= , dxdydzr d =r la vect toa o va the tch nguyen to trong khchieu qui ve he truc toa o Descartes.

Cong xac suat:Neu hai bien co e1 va e2 la hai bien co xung khac (khong the xay rath xac suat e e1hoac e2 xay ra la

P( e1hoac e2 ) = P( e1) + P( e2), (I.12)vi P( e1) va P( e2) lan lt la xac suat e xay ra e1va xac suat e xay ra e2.

T cong thc (I.12) tren, ta suy raieu kien chuan hoa:1Pm

m = , (I.13)va khi bien ngau nhien la lien tuc, xac suat e x trong khoang (a, b) hoac D c viet:

= b

adx).x() bxa(P (I.14a)

=D

D r d).r ()r (P rrr . (I.14b)

Nhan xac suat:Khi hai bien co e1 va e2 oc lap nhau (tc la viec xay ra bien co nhng en viec xay ra bien co khac), xac xuat e e1 va e2 xay ra ong thi la:

P( e1 va e2) = P( e1).P( e2). (I.15)Khi hai bien lien tuc, oc lap x va y co ham phan bo thong ke lan l(x) va(y), xac suat

nguyen to e ta co ong thi x(x, x+dx) va y(y, y+dy) la,dxdy).y().x(dy)y(.dx)x()y(dP).x(dP)y,x(dP 212121 === trong o dP1(x) va dP2(y) la xac sua

nguyen to e x(x, x+dx) va y(y, y+dy).Vay, ta se nh ngha ham phan bo thong ke cua hai bien (x,y):

)y().x()y,x( 21 = , (I.16)e co

dxdy).y,x()y,x(dP = . (I.17)Mot trng hp quan trong ma ta thng gap la phai tnh(x) khi a biet(x, y). Khi o, ta se s

dung tnh chat sau: ==y

11 dxdy).y,x(dx)x()x(dPD

(I.18a)

=y

dy).y,x()x(D

(I.18b)

V du 1: Ham phan bo thong ke trong toa o cc (r,).

V == d).,r (),r (dPdxdy).y,x()y,x(dP .


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Can chu y rang dien tch nguyen to dxdy khi chuyen sang toa o cc la d th khong phai la t

drd. Ta can phai tnh dien tch nay bang cach cho r bien thien mot lng dr vabien thien mot ld. Dien tch nguyen to trong toa o cc khi nay se la:

d = rdrd.(Chu y rang ta a khong lay hai canh dr va (r+dr).d ma lay dr va rd). Vay:

= rdrd).,r ()r (d e tnh(r) khi biet(r,), ta gia s(r,) = C = const.

Theo (I.18a):

=

==

2

0

2

01 dCrdr rdrd).,r ()r (dP

rC2)r (Crdr 2dr )r ()r (dP1 === Vay(r) c phan bo tuyen tnh theo r. V du 2: Ham phan bo thong ke trong toa o cau (r,,).

Trong toa o cau (r,, ), the tch nguyen to d c tnh bang cach cho r,, bien thien caclng nho dr, d, d. Khi o, theo hnh ve, ta co c:

= ddrdsinr d 2V = ddrdsinr ).,,r (),,r (dP 2 .

e tnh(r) khi a biet(r,,) chang han, ta gia s rang:(r,,) = C = const. Khi o, tng nh cong thc (I.18a), ta co:

===,

2

,

2

,ddrdsinCr ddrdsinr ).,,r (),,r (dP)r (d

=

==

0

2

0

2 ddsindr Cr )r (dP

Tch so cua hai tch phan sau cho ta goc khoi 4 nendr Cr 4)r (dP 2= .

Ma dr )r ()r (dP = 2Cr 4)r ( = . Chu y:e tnh the tch nguyen to khi oi he truc toa o, ta co the dung Jacobien:

= drd.),r ()y,x(d , vi

=

= yr y

xr x

),r ()y,x(I


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Vi x=rcos , y=rsin I = r . d =rdrd .

= ddrd.),,r ()z,y,x(d , vi

==

zzr z

yyr y

xxr x

),,r ()z,y,x(J

Vi x=r.sin .cos , y=r.sin .sin , z=rcos . I=r 2sin . = sin in dr r dV 2 .

I.B.2 Gia tr trung bnh cua mot bien ngau nhienNeu P(ui) la xac suat e bien ngau nhien u co gia tr ui, gia tr trung bnh cua u c tnh:

=i

i

i ii

)u(P

u).u(Pu . (I.19a)

Ta co ngay ket qua khi u la mot hang so = u = const (tr trung bnh cua mot hang so lchnh hang so o).

Trng hp bien ngau nhien u bien thien lien tuc, ta co cong thc:

==du)u(

du).u(.u

)u(dP

)u(dP.uu . (I.19b)

Khi ta co mot ham f(u) theo bien ngau nhien u (v du nh ta muon tnh ong cua mot phan

t kh theo bien ngau nhien la van toc v cua phan t kh chang han, ta se co= f(v) = 21

mv2

,vi m la khoi lng cua phan t kh), ta co cong thc tnh gia tr trung bnh c

=i

i

iii

)u(P

)u(P).u(f )u(f , (I.20a)

va khi bien ngau nhien co gia tr lien tuc trong khoang (a,b):

== b

a

b

a b

a

b

a

du)u(

du)u().u(f

)u(dP

)u(dP).u(f )u(f . (I.20b)

Ta cung co cong thc tng t khi phai tnh tr trung bnh cua ham f(u,vnhien oc lap u va v:

=i j

ji

i ji

j ji

)v,u(P

)v,u(f ).v,u(P)v,u(f , (I.21a)


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trong o: )v(P).u(P)v,u(P jviu ji = vi Pu(ui) va Pv(v j) lan lt la xac suat e cac bien u, vi va v j.

Khi cac bien u, v nhan cac gia tr lien tuc: u(a,b); v(c,d), gia tr trung bnh cua ham f(u,vtnh:

== b

a

d

c

b

a

d

c b

a

d

c

b

a

d

c

dudv).v,u(

dudv).v,u().v,u(f

)v,u(dP

)v,u(dP).v,u(f

)v,u(f . (I.21b)

Chu y rang tat ca cac cong thc (I.19a,b), (I.20a,b), va (I.21a,b) eu tr necac ieu kien chuan hoa:

=i

i 1)u(P , (I.22a)

= b

a1)u(dP , (I.22b)

=i j

ji 1)v,u(P , (I.22c)

va

= b

a

d

c1)v,u(dP , (I.22d)

Khi co hai ham f(u) va g(u) cung bien thien theo bien ngau nhien u, ta co:)u(g)u(f )u(g)u(f +=+ , (I.23a)

Chng minhGia s ta a co cac ieu kien chuan hoa c thoa. Khi nay, theo nh ng

[ ]i i i i ii if(u) g(u) P(u ) f(u ) g(u ) P(u ).f(u )+ = + = )u(g.)u(P ii i+ )u(g)u(f += .

Tong quat, ta co:)u(f .c)u(cf = . (I.23b)

Tng t, ta cung chng minh c rang)u(g.)u(f )u(g).u(f = . (I.23c)

I.B.3 Thang giang cua mot ai lng ngau nhiene anh gia s sai lech trung bnh cua mot bien ngau nhien u oi vi gia cua bien

nay, mot cach t nhien, ta se tnh ai lngu vi uuu=

lao lech khoi gia tr trung b

.Nhng:uuuuu == ,

v uu = , nen0u = , (I.24a)

v ly do la cac thang giang cua bien u quanh gia tr a bu tr lan nhau khi u ln hn hoac n.Nh vay, ta phai tnh gia tr trung bnh cua bnh phng o lech, tc la phng sai 2)u( :


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)uuu2u()uu()u( 2222 ==

Vay, ta co:222 uu)u( = , (I.24b)

o thang giangcua u c nh ngha bi222 uu)u( == , (I.24c)

Trong hai phan tiep theo, ta se khao sat hai phan bo thong ke quan trong, ravan e cua vat ly thong ke.I.B.4 Phan bo nh thcXet phep th gieo ong tien. Moi lan gieo co hai kha nang: mat so hoa

hieu lan lt la (+) va (), va xac suat lan lt la P+ va P_. ieu kien chuan hoa cho ta:1PP =+ + .

Ta tnh xac suat P( N, n ) e co n lan bien co (+) xuat hien sau N lan th ( N ):Gia s sau N lan th, ta co mot chuoi bien co:

4342143421n Nn

1 ))...()(())...()((:S

+++

Xac suat e co chuoi nay la:n Nn1 PP)S(P += Tng t, ta co the co mot chuoi Si nhng bien co vi bien co (+) xuat hien n lan)

xuat hien N-n lan, vi xac suat: n Nni PP)S(P += .V hai chuoi bien co khac nhau la xung khac, P( N, n ) la tong cua tat ca c

(+) va N-n bien co (). o la cach sap xep khac nhau cua n bien co (+) trong N bi

)!n N(!n! NCn N

= .

Vayn Nnn

Nin N

ii PPC)S(PC)S(P)n, N(P

+=== .

Cuoi cung, ta co

. (I.25)

Phan bo nay c goi la phan bo nh thc.Chu y rang ta co cong thc khai trien nh thc:

n Nn N

0n

n N

N baC) ba( ==+ .

T o, ta co the kiem chng ieu kien chuan hoa cua P(N, n):1)PP(PPC)n, N(P Nn Nn N

0n

n N

N

0n=+== ++

== .

n NnPP)!n N(!n

! N)n, N(P +=


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I.B.5 Phan bo Gauss (phan bo chuan)

Phan bo Gaussla phan bo lien tuc, co ham phan bo thong ke cho bi:222)

0xx(

2 e2

1)x(

= . (I.26)Trong o, x0 la v tr cua phan bo; ng bieu dien cua(x) theo x oi xng qua ng tha=

x0, va c goi la be rong cua phan bo. Ta thay ng bieu dien(x) co dang hnh chuong.I.C Tap hp thong ke. Nguyen ly ergodic

I.C.1 S tien hoa theo thi gian cua mot he v moC hoc thong ke co muc ch la mo ta chnh xac nhat co the c nhn

he v mo, xuat phat t nhng ac tnh vi mo cua nhng hat cau tao nen htac, ta phai tnh c bieu thc cua ham Hamilton cua he. Nhng ieu nay k

o vi mo, ham Hamilton ch co the tnh gan ung; he v mo khong bao gdng (la trang thai co nhng ai lng ac trng khong oi theo thi gian), gian.

Mat khac, ta cung khong the hoan toan co lap mot he e khao sat, v moi trng ben ngoai tuy khong ang ke mc o v mo, nhng lai khonxac mc o vi mo.

V nhng ly do tren, ta khong the theo doi chi tiet nhng tnh chat vi mo cudung phng phap thong ke e tnh nhng thang giang gay ra do s khong ohe.

I.C.2 Tr trung bnh theo thi gian

Gia s ta xet mot ai lng co gia tr f(t) bien thien theo thi gian t cuabang, chang han nh so phan t kh n(t) trong mot the tch V nao o cua bnco gia tr thay oi theo thi gian t, v nhng phan t kh chuyen ong hon loc cho tren hnh ve:


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Tr trung bnh theo thi giancua f(t) c nh ngha:

+

=

0t

0tdt)t(f 1limf . (I.27)

Theo nh ngha nay th oi vi he v mo trang thai can bang,f

oc lap vi t0, la thi iem t bat au phep o. Nhng oi vi he co nhng thay oi v mo, vi nhng kh ma ta thchien phep o thf ch mo ta he trang thai can bang mi ma khong gi lai thien. V du nh khi ta rut vach ngan trong bnh cha, so phan t trong the tchat gia tr on nh vi nhng thang giang nho. Tr trung bnh theo thi giann cua so phan t kh trV ch cho ta biet trang thai can bang mi c thiet lap sau o.

I.C.3 Tr trung bnh tren tap hpThay v khao sat mot he v mo duy nhat theo thi gian nh tren, ta co

nhng he giong nhau, at di cung nhng ieu kien v mo. V du nh ta cnhng bnh cha co cung kch thc, cho vao cung mot loai kh, at di cusuat, nhiet o, Khi so he nay la rat ln, ta cotap hp thong ke(hay tap hp Gibbs).


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Tai mot thi iem nhat nh nao o, ta xet tat ca cac he cua tap hp th

trong cung trang thai v mo, nhng co the trong cac trang thai vi mo khach giong nhau mc o v mo, nhng s tien hoa theo thi gian cua chungmo.

Ta goi N la so he cua tap hp thong ke tren, Nl la so he cung trang thai vi mo (l). Ta muon ogia tr cua ai lng f, la f l cua trang thai (l).Tr trung bnh tren tap hpcua f c nh ngha:

ll

l Pf N1

f )( N = , (I.28a)

Neu N rat ln, ta co:l

ll Pf f f

)( N = , (I.28b)

vi N

NlimP

Nl

l = la xac suat e mot he trang thai (l), c goi la xac suat chiem ong trang thai

(l).

I.C.4 Nguyen ly ergodic

Theo tren, ta co hai cach tnh gia tr trung bnh cua mot ai lng nao Nguyen ly sau ay se cho ta biet moi quan he gia hai phng phap tren: Nguyen ly ergodic: Khi he trang thai can bang, gia tr trung bnh tren tap hp

lng vat ly cua mot he tai mot thi iem nao o trung vi gia tr trung bnh theo thi gian cua mot he duy nhat .

Noi khac i, ta co s tng ng gia tr trung bnh theo thi gian va tr trung bnh tren:f f = .

Trong vat ly thong ke, thay v tnh gia tr trung bnh cua mot ai lung theoluon s dung tr trung bnh tren tap hp, co ngha rang ta luon xet mot tap hkhao sat.

I.D Entropi thong keI.D.1 Khai niemTrong lnh vc truyen thong, khi ta khong the biet trc mot cach chac cha

co nao o ma ta can phai dung ly thuyet xac suat, tc la khi o ta khonco nay. e o lng mc o thieu thong tin ve cac bien co, ta a vao khaentropi thong ke.

Xet tap hp M bien co:{em , m = 1, 2, , M}, moi bien co co xac suat tng ng Pm(vay, 0 Pm 1 , va

==

M

1mm 1P ). Entropi thong ke lien ket vi tap hp nay c nh ngha nh s

(k la hang so dng ) (I.29)

vi 0PlnP mm = neu Pm= 0.(Chu y rang phan bo sung cua nh ngha0PlnP mm = khi Pm=0 c a vao e phu hp v

qua 0)xlnx(lim0x

=

).

==M

1mmmM21 PlnPk )P,...,P,P(S


18/163


19/163

BAI TAP

BT I.1 Xet ba electron co nang lng toan phan la E = 2, c phan bo tren ba mc nang lngnhau:0= 0,1=, 2=2. Bac suy bien cua cac mc lan lt la: g0= 3, g1= 2, g2=2.

1/ Ve s o phan bo cac electron tren cac mc nang lng va em so trang cac electron la khong phan biet c va la cac fermion (tc la hai electron khong thai lng t).

2/ Neu goi trang thai v mo la trang thai cua he co nang lng toan phan b va so hat momc nang lng la nh nhau, ta co tat ca bao nhieu trang thai v mo ?BT I.2 Cho he ba mc nang lng1=, 2= 2, 3 = 3, co cac bac suy bien lan lt la g1= 1, g2= 2, g3= 3.

Nhng hat khong phan biet c c phan bo tren ba mc nang lng nayla E = 3, va co so hat khong xac nh. Goi trang thai v mo la trang thai lng E = 3, va so hat tren moi mc nang lng la nh nhau.

Hay ve s o phan bo cac hat tren cac mc nang lung va em so trang thai vi mo kha d tng ng vi so trang thai v mo tren.BT I.3Hay ve qu ao pha trong moi trng hp sau:

1/ Chat iem khoi lng m chuyen ong theo quan tnh.2/ Chat iem khoi lng m ri t do khong van toc au ni co gia toc trong3/ Chat iem M khoi lng m mang ien tch e ( e> 0 ), chuyen ong trong ien trng cua

tch iem +e ng yen. Cho biet v tr va van toc luc au cua M la r0 va v0 = 0.

BT I.4 Xet vect OMv =r co o ln khong oi, quay eu quanh goc O cua trucOx theo chieu dng cuavong tron lng giac.

1/ Tnh xac suat e goc)OM,Ox( co gia tr trong khoang va+d.2/ Tnh mat o xac suat(Vx) e hnh chieu cuaV

rco gia tr la Vx tren trucOx .

Ve ng bieu dien cua(Vx) theo Vx.

BT I.5Hay tnh tr trung bnhn va phng sai 2)n( trong phan bo nh thc.

BT I.6Cho phan bo Gauss: 222)0xx(2

e2

1)x(

= .

1/ Chng minh rang(x) a c chuan hoa.2/ Tnhx va 2)x( .

BT I.7Xet phan bo nh thc: n Nn P.P)!n N(!n

! N)n, N(P +=

trong trng hp N rat ln, n rat ln va c xem nh bien thien lien tuc trong vungn va u xa N (tc laham P(N,n) bien thien cham trong khoang gia n va n-1: )n, N(P)1n, N(P)n, N(P


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2/ Tnhn .3/ Xet phan bo nh thc n Nnn N P.PC)n, N(P += vi P+ > 1 va n 0 , x 0. (Ham phan bo nay ac trqua trnh phan ra phong xa, s bien thien cua so phan t kh theo o cao, )1/ Hay chuan hoa(x).2/ Hay tnh tr trung bnh, phng sai va o thang giang.

BT I.10Theo nh luat Maxwell, so phan t kh co van toc trong khoang [v, v+dv] thc: dN = N(v)dv, vi kT/k E2eAv)v( = , trong o, Ek la ong nang cua moi phan t, T lcua khoi kh, va k la hang so Boltzmann. N la tong so phan t kh.

1/ Xac nh hang so A.2/ Dung cac tch phan Poisson:

1n20

1naxn2

n2 a2!)!1n2(dxexI

2

+

+ == ,

1n2

0

1nax1n2

1n2 aa2!ndxexI

2

+

++

+== ,

e tnh van toc trung bnhv va van toc toan phng trung bnh2v .

Hay so sanhv va 2v vi van toc cai nhien nhat vm.

(vmc nh ngha bi: 0dv)v(d

mvv=

=).

3/ Hay tnhv , 2v va vmcua phan t monooxit cacbon CO 300 K va 1000 K.

BT I.11Xet he vat ly la mot dao ong t ieu hoa tuyen tnh:

x(t) = x0cos(t+).Tr trung bnh theo thi gian cua mot ai lng vat ly nao o f cua dao ong t

= 0 dt)t(f 1f ,

vi la chu k cua dao ong t.

1/ Hay tnhx va2x .

2/ Xet tap hp thong ke cua nhieu dao ong t ieu hoa tuyen tnh co cun va co cungnang lng toan phan. Gia s rang cac gia tr cua pha aueu ong xac suat trong khoang 0 va 2 (gia thietvi chnh tac)

Tr trung bnh tren tap hp cua mot ai lng f c tnh:

=2

0 d)(f 21

f ,Hay tnhx va 2x . Suy ra rang he cac dao ong t nay la tap hp ergodic.3/ Ta co the suy ra rang he gom nhieu dao ong t ieu hoa tuyen tnh la tap

BT I.12Dung phng phap tha so Lagrange e chng minh rang entropi thong ke S cbien co ngau nhien la ong xac suat.


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VAN E I.A

Bai toan bc i ngau nhien (random walks).M rong cho bai toan khuech tan cua mot phan t kh

Mot ngi say ru i ve nha trong tnh trang khong tnh tao, thc hien cac bo la bai toan bc i ngau nhien hai chieu. ay ta xet bai toan mot chieu,ly thong ke. Ta quy c:

Phan t ch co the di chuyen tren mot ng thang. Moi bc i co khoang cach L. Moi bc sang phai hay bc sang trai c thc hien vi xac suat bang n

21 , va moi

bc la oc lap vi cac bc khac.1) Viet bieu thc cua xac suat Pn e co np bc sang phai va nt bc sang trai sau mot so bc

gia tr trung bnh va phng sai2 ng vi np va nt .Tnh xac suat, tr trung bnh va phng sai ng vi quang ng i c :x = (np nt) L = mL.2) Chng minh rang oi vi n rat ln (n >> m), Pn se c tnh theo phan bo Gauss. S dung

Stirling : ln n! n ln n n va cong thc gan ung ln(1 ) .3) Bang nhan xet la xac suat e tm thay phan t trong khoang (x, x +x) laP (n,x) =

x

)m,n(P

trongm catat

, va khoang cach gia 2 o dch chuyen lien tiep la 2L (kiem nghiem la

suy ra so gia tr cua m trong khoangx laL2x . T o suy ra ham phan bo (x, n). Chuan hoa ham nay.

4) Bai toan tren co the ap dung cho ham phan bo (x, t) cua mot phan t kh theo quang x vao thi iem t.

a) Viet ham phan bo nay theo he so khuech tan D =2

L2 vi la thi gian gia hai va cham.

b) Goi Pn(i) la xac suat e phan t vao v tr i sau n bc. Kiem chng rang tPn(i) = q Pn 1(i + 1) + p Pn 1(i - 1)= )1i(P

21)1i(P

21

1n1n ++

c) Viet tai bieu thc tren theo cac bien so mi x = iL va t =L.Dung khai trien Taylor cho cac bieu thc cua P(x,t) va P(x, t - ) e rut ra phng trnh khuech

cho gii han continuum.


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VAN E I.B

Gii thieu phng phap Monte Carlo.Ap dung e tnh dien tch hnh phang, khoi lng

va moment quan tnh mot hnh khoi

2609971874087743729433912

6580161692296899202837996

6987080001422455685078967

8444621430519038338057201

5824802305691790591266916

2128259741966822983073998

5693834262918193761254289

5472915157608121559307147

6775727545476317319884313

1) Bang tren gom nhng so ngau nhien. Chon 35 so trong 7 cot au tien e t ng tron ban knh n v trong th nhat cua mat phang toa o xOy. T co26099, trch ra 4 ch so au: 2609, va tao ra hai so t 26 va 09 theo qui tac r1 = 0.26 va r2 = 0.09, ta se co moiem trong mat phang xOy theo (y = 0.26; x = 0.09). Nh vay, ng vi 7 cot au trong 1 = 35

iem. em so iem m1 nam trong ng tron. Lap t so 111 n

m p = .

Thc hien phep tnh tng t vi 8 cot va 9 cot au tien, ta co cac so 2

22 n

m p = va3

33 n

m p = vi n2 =

40 va n3 = 45.

So sanh p1, p2 va p3 vi gia tr 785.04= . Tnh cac o sai so tng oi. Nhan xet.

Ky thuat tren, s dung cac so ngau nhien, la noi dung chnh cua phng phagoi nay).

2) Phan ln cac ngon ng lap trnh tren may tnh eu co mot ham cho trrandomtrongPascal,rnd trong Basic, cho ta nhng so chuan ngau nhien (pseudo random numbers aleatoires). Viet chng trnh tng ng vi thuat toan (algorithm) sau:

(Dung hamrandomizee co mot day so chuan ngau nhien).Bc 1: Bat au mot day so ngau nhien.Bc 2: Lap lai m lan, n la so nhng so ngau nhien s dung.Bc 3: Chon nhng so ngau nhien x va y sao cho 0 x 1.Bc 4: Neu x2 + y2 1, tang so em: nn + 1.Bc 5: Cham dt bc 2.Bc 6: Cho ra 4 (n/m).Bc 7: Ket thuc.Thuat toan tren cho phep ta tnh c g ? Giai thch.3) Ve hnh cau ban knh n v tam nam tai goc toa o, noi tiep trong mot kho

toa o (1, 1, 1). Gia s ca khoi cau va khoi lap phng eu co mat o khoi banlam N phan bang nhau, vay moi phan co the tch

N1 . Viet thuat toan e tnh the tch khoi cau.

4) Thay oi thuat toan tren e tnh moment quan tnh cua khoi cau oi vi tru


23/163

VAN E I.C

Mat o trang thai cua kh ly tng (KLT)

I/ Trng hp mot phan t kh Xet mot hat phan t KLT c nhot trong mot hop ch nhat co kch thc Lx, Ly, va Lz. Khao sat lng

t cho ta ket qua la khi at nhng ieu kien gii han tuan hoan cho ham song adang song phang:

r .K .iK eV

1 rrr = ,

V: the tch cua hop.

K r

h la ong lng trong trang thai dngK r va nang lng tng ng la:

m2K E

2

K

rhr = ,

m la khoi lng cua hat vah la hang so Planck.Trong trng hp nay, ong lngK

rse b lng t hoa:

zz

zyy

yxx

x eL2e

L2e

L2K ++= lll

r, z,y,xl Z . (*)

Vay nang lngK E r tao thanh pho gian oan.Sau ay, ta gia s kch thc hop rat ln, khi o, khoang cach gia hai vac

nang lng E la bien thien lien tuc. ieu nay cho phep ta tnhmat o trang thai cua hat.1) Chng to rang o bien thien cuaK

rlien he vi E qua bieu thc:

dEEm21dK 212

=h

.

2) em so trang thai cua hat tng ng vi khoang nang lng gia E va E+K r

ma o dai trong khoang K va K+dK; mui nhng vectK

rnay nam gii han gia hai mat cau ba

K+dK. Lp vo cau nay co the tch 4K2dK. Theo cong thc (*), cac vectK r

co nh nam cac nut cmang trong khong gianK

r, tng ng vi cac gia tr khac nhau cua

z,y,xl . Chng minh rang, vi mot phe

x kha chnh xac, ta co so nut nay la

V/8dK K 4

3

2

.

3) T o, hay suy ra bieu thc cua mat o trang thai(E) ng vi nang lng 2123

32 Em.22V)E(h

= bang cach nhan xet rang(E)dE, so trang thai tng ng trong khoang E

cung chnh la so nut ma ta tm thay cau 2).


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II/ Trng hp mot he KLTXet he gom N hat t do, khong ke spin, giong het nhau nhng ban chat khac

he hat ong nhat_ va nh vay, ta khong can e y en tien e oi xng oi voc lap va c nhot vao mot hop ch nhat nh trong phanI/

1/ Viet bieu thc tnh nang lng toan phan E cua he theo cac vectik r

cua hat (i), vi m la khoi

cua moi hat.2/ Ta nh ngha vectK r trong khong gian 3N chieu:}k ,k ,k ,...,k ,k ,k {K Nz Ny Nxz1y1x1=

r

a/ Tnh the tch v3Ncua mot khoi ch nhat nguyen to xac nh bi cac nh cua veK r

trong khong giannay.

b/ Biet rang the tch cua mot khoi cau ban knh K trong khong gian 3N N3

2 N3

N3 K .)1

2 N3(

V+

= , vi )12 N3( + la ham gamma, dung cong thc vi phan e tnh di3N(K)

cua mat cau nay.c/ Theo nh ngha, so trang thaidE)E(

N co nang lng trong khoang E va E+dE c tnh:

N3

N3 N v

dK )K (SdE)E(S = , vi )E( N la mat o trang thai.

T o, hay chng minh:12

N3 N

N ) NE(V). N(C)E(

= , trong o C(N) la he so c tnh:

)12 N3(

1.2mN

23) N(C

2 N3

2+

=

h

d/ Dung cong thc Stirling: x2ex)1x( xx + khi x >> 1 e tnh lai C(N) khi N rat l


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Chng II

PHAN BO VI CHNH TAC.TIEN E C BAN CUA C HOC THONG KE

II.A Trang thai can bang cua mot he v mo

II.B Tien e c ban cua c hoc thong ke. Toan t LiouvilleII.C Cac tham so v moII.D Qua trnh thuan nghch va qua trnh bat thuan nghch

II.A Trang thai can bang cua mot he v moII.A.1 Cac ai lng ac trng cua he v moe ac trng cho trang thai v mo cua mot he vat ly, ta thng dung c

lng, the tch, nhiet o, ap suat, so hat, mat o, Neu nhng tham so nay c xac nh t nhng ieu kien ben ngoai, co

goi latham so ngoai. Chu y rang cac tham so ngoai nay bao gi cung c cho vthc nghiem nao o, bi v ta khong the kiem tra c ay u nhng ieu kMot khi he v mo ang xet co nhng tham so ngoai c an nh roi, th c

cua he se t do bien thien do nhng thang giang vi mo cua he. Cac ai lbien so noi. Vay nhng bien so noi cua mot he c ac trng bi cac phan bo tho

V du: Khi ta an nh gia tr cua nang lng toan phan, the tch, va so hatlng nay c bao toan (khong thay oi). o la cac tham so ngoai. Khi omot thi iem nao o cua he th ai lng nay t do bien thien nen la bie

Neu bay gi ta lai an nh gia tr cua the tch, so hat, va nhiet o cho cua he co nhng thang giang, tc la chu s phan bo thong ke. Vay, the t

tham so ngoai, trong khi nang lng la bien so noi. Nh vay, ta thay rang mot ai lng vat ly (trong v du tren la nang lnghoac la bien so noi tuy trng hp cu the cua bai toan, ta an nh nhkhong oi (vi sai so) va e cho ai lng nao co s bien oi thong ke.

II.A.2 He co lap trang thai can bangMot he v mo c coi lahe co lapkhi he nay khong trao oi nang lng va cung

hat vi moi trng ben ngoai, ngha la nang lng toan phan cua he thai vi mo kha d cua he (co the la vo so) eu phai tng thch vi gia nay.

Nh vay, oi vi mot he co lap, cac ai lng nh nang lng, so hat ttoan nen eu la cac tham so ngoai.Mot he v mo trang thai can bangkhi cac ai lng v mo ac trng cho he khong thi gian. Neu luc au, mot he co lap khong trang thai can bang th sau thi gian hoi phuc), he cung se i ve trang thai can bang.

V du: Xet mot khoi kh luc au c nhot vao mot phan ben trai co the1 cua mot hop. Phacon lai cua hop co the tch V2 trong khong. Hai phan nay c ngan cach bi mot varut C ra, cac phan t kh se dan dan chiem toan bo the tch V1+V2 cua hop. Sau khoang thi gia


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phuc, mat o phan t kh se ong nhat tai moi thi iem trong the tch cuaoi theo thi gian. Trang thai sau cung nay cua khoi kh la trang thai can ba II.B Tien e c ban cua c hoc thong ke. Toan t Liouville

II.B.1 Phan bo vi chnh tacXet mot he S co lap, trang thai can bang, tc la cac ai lng v mo cu

the tch V, so hat N, la oc lap oi vi thi gian. Khi nay co the ton tai thai vi mo kha d tng ng vi trang thai v mo nay. Khong mot nh luattrong nhng trang thai vi mo o, trang thai nao co u tien e xay ra hn nnhan xet tren, ta co the khai quat hoa e phat bieu tien e c ban sau:

oi vi mot he co lap trang thai can bang, tat ca nhng trang thai vi msuat .

Noi khac i, mot he co lap trang thai can bang co the mot trang tbang nhau.

V nang lng E cua he luon c xac nh vi mot sai so E nao o nen ta co cong thcsuat Pl e he trang thai (l) co mc nang lng El:

== 0

constCPl ,

, )EE,E(E

EEEE++

ll (II.1)

Khi nay, ta noi rang he S trang thai phan bo vi chnh tac. Tap hp thong ke gom nhng ht vi he S c goi latap hp vi chnh tac.

Tien e c ban tren a c oi chng vi ly thuyet va thc nghiem. Vcac tnh toan da tren tien e nay eu cho nhng ket qua phu hp vi thc

II.B.2 nh ly LiouvilleTrong c hoc thong ke co ien, mot he co lap co f bac t do c mo ta

ong lng: } p,..., p, p,q,...,q,q{ f 21f 21 . Tai moi thi iem, trang thai cua he c bieiem pha trong khong gian pha.

Xet tap hp thong ke cua he co lap nay. So he cua tap hp co v tr tch pha nguyen to )dp,...,dp,dp,dq,...,dq,dq( f 21f 21 c tnh bi:

f 21f 21f 21f 21 dp...dpdpdq...dqdq) p,..., p, p,q,...,q,q( , (II.2)vi ) p,..., p, p,q,...,q,q( f 21f 21 lamat o so hetrong khong gian pha.

Moi he cua tap hp thong ke chuyen ong theo thi gian, qui nh bi ca,

pHq

ii

=& ,qH p

ii

=& (II.3)

vi ) p,..., p, p,q,...,q,q(HH f 21f 21= la ham Hamilton cua he.V so he trong tap hp thong ke c bao toan nen so iem pha ra khoi

o trong mot n v thi gian phai bang toc o giam cua so iem pha trong tVi ) p,..., p, p,q,...,q,q(v f 21f 21 &&&&&&r = la toc o chuyen ong cua mot iem pha vanr la vect phap

tuyen cua dien tch S bao quanh the tch V tai iem ang xet, so iem pha ri kdS.n.v. rr


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Vay ta phai co

==

S VSdV

dt

ddSvdS.n.v. rrr . (II.4)

Nhng theo nh ly Gauss-Ostrogradski: =S V

dV)A(dSArr

vi la toan t gradient:)

p,...,

p,

p,

q,...,

q,

q(

f 21f 21

,

nen ta co, bang cach ap dung nh ly nay cho vectvA rr

= :

=S V

dV)v(dSv rr .

T (II.4), ta suy ra0dV)v(

tV=

+ r . (II.5)

V he thc tren phai c nghiem ung vi moi the tch V nen ta phai co0)v(

t=+

r (II.6)

Tch vo hng cua cac vect va vr cho ta

0) p( p

)q(qt

f

1ii

ii

i

=

+

+

=&&

0 p p p

pqqq

qtf

1i i

ii

ii

ii

i=

+

+

+

+

=&&&&

0 p p

qq

p p

qqt

f

1i i

i

i

ii

ii

i=

+

+

+

+

=

&&&&

T he phng trnh Hamilton: 0qq

p p

i

i

i

i =+

&& , ta co:


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0 p p

qqt

f

1ii

ii

i=

+

+

=&& . (II.7a)

Cuoi cung, ta co he thc cuanh ly Liouville:

, (II.7b)

va phng trnh tren c goi la phng trnh Liouville.Xet trng hp = const, hoac tong quat hn, trng hp la ham ch cua nang lng E.hang so chuyen ong nen

0qE.

Eq ii=

=

, va 0

pE.

E p ii=

=

.

Vay0

t=

.

He thc tren co y ngha vat ly: S phan bo cac he tren nhng trang thai la khong gian (tng ng vi trang thai can bang).

ac biet, oi vi tap hp vi chnh tac, trang thai can bang tng ng v = constvi E0 < E< E0+E, va = 0 tai cac vung co nang lng khong thoa hai bat ang th

Tom lai, nh ly Liouville cho ta biet rang tap hp thong ke tng ng vtap hp co = const trong khong gian pha, tc la cac trang thai kha d la onghoan toan phu hp vi tien e c ban cua c hoc thong ke.

II.B.3 Toan t LiouvilleT he phng trnh Hamilton:

=

=

.

,

ii

ii

qH

p

pHq

&

&

va vi gia thiet ham H= H( qi, pi ) khong phu thuoc tng minh vao thi gian (0tH = : nang lng E

khong oi theo thi gian), phng trnh Liouville (II.7a) tr thanh

0qH.

p pH.

qtdtd f

1i iiii=

+

=

=. (II.8)

Ta nh ngha dau ngoac Poisson {A,B} cua hai ham A(qi, pi, t) va B(qi, pi, t) nh sau:

{ } =

=

f

1i iiii qB.

pA

pB.

qAB,A . (II.9)

Nh vay, phng trnh (II.8) tr thanh:{ } 0H,

tdtd =+

= . (II.10)

Phng trnh (II.10) la dang thu gon cua phng trnh Liouville. Ta cung co the{ }=

,Ht

. (II.11)

Neu ta nh nghatoan t Liouville:

0dtd =


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)x,E(dy)y;x,E( =

,

1dy)y;x,E(W =

).

Vay

=

lnlnWln ,Ta co)x,E(S

k 1ln = ,

va bang cach at= s

k 1ln ,

trong o, sc goi la entropi vi chnh tac tng phan.).Ss(

k 1expW =

.sk 1

expW

Gia tr cai nhien nhat ym cua bien so noi y la gia tr lam cho W cc ai. Nh vay

0ys

k 1

myy

=

=

.

Khai trien Taylor cua ham squanh iem y=ym:

m

m my y

1 1 1 ss (y) s (y ) .(y y ).k k k y

=

+ +

m

22

m 2y y

1 s(y y ) ...2k y

=

+

Trong tong ben phai tren, so hang th nh triet tieu do ieu kien cc . ong thi, taat

2myy

22 1

ys

k 1

=

=

.

(v sco cc ai tai y=ym nen ta co: 0ys

myy2

2 , nang lng truyen t he T1 ln hn sang he co T1 nhohn.

II.C.2 Ap suat vi chnh tacKhi hai he tng tac nhau dan en s thay oi cua nhng tham so ngoai m

nang lng th o latng tac c . Nh vay, tng tac c cung van lam cho cac mc nathay oi.

Cong c hoc thc hien len hec nh ngha la gia so cua nang lng trung bnh ng vi gia so cua tham so ngoai:EW = .

Cong c hoc ma he thc hien nh vay la:EW'W == .Ta co 0'WW =+ . He thc nay noi len rang nang lng cua toan bo hai he lTuy nhien, e co the khao sat s tng tac c gia hai hay nhieu he, ta ca

cac ai lng lien quan en cac tham so ngoai phai c hoan toan xac nsuat ma ta se a vao sau ay. Vay ta a vao khai niemqua trnh chuan tnh.

Qua trnh chuan tnh cua mot he vat ly la qua trnh trong o he bien othai cuoi bi mot chuoi lien tiep nhng trang thai can bang: tai moi thbang tng ng vi cac gia tr cua cac tham so ngoai an nh cho he. Muontham so ngoai phai u cham sao cho he at c mot trang thai can bang mngoai nay thay oi ang ke. T u cham ay phai hieu la thi gianngoai ln hn nhieu oi vi thi gian hoi phuc cua he.

V du: Khi ta nhot mot khoi kh vao mot xylanh, va c ngan bi mot pistocac tham so ngoai la chieu dai cua xylanh co cha chat kh. Neu khoi kh na

ES

T1

=

Elnk

T1

=


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bang mi sau thi gian la = 10-3 s sau khi khoang cach t ay xylanh en piston b giatrnh co the xem nh chuan tnh khi thi gian chuyen ong cua piston la 0,1s.

Tom lai, khi he thc hien mot qua trnh chuan tnh, cac tham so ngoai thayiem eu co mot gia tr xac nh.

Ta xet he S tng tac vi mot he khac. Goi El la nang lng cua he S trang thai vi mo (l); El laham cua cac tham so ngoai x:

)x,...,x,...,x,x(EE n21 = ll .Khi tham so x bien thien, o bien thien cua nang lng c viet

=

= dxxEdE

n

1

ll .

Khi o, he thc hien mot cong la:

=

== dxxE

dEWn

1

lll . (II.22)

at

. (II.23)

lalc suy rongtng ng vi tham so x, ta co

. (II.24)

Nh vay, trong tap hp thong ke, tr trung bnh suy rong la

=xE

X ,l

l ,

v ta co: = dxXW ,ll .

Sau khi a vao khai niem tong quat cua lc suy rong tren, ta se nh nhng ai lng quan trong ac trng cho he v mo: Ap suat c nh ngha la lc suy rong tng ng vi tham so ngoai la th S trang thai (l) co mc nang lng El:

. (II.25)

nh ngha tren cua ap suat d nhien la ong nhat vi nh ngha cua aphoc co ien. e de thay ieu nay, ta xet lai v du tren cua khoi kh nhot

=xE

X ,l

l

=

=n

1, dxXW ll

VE

P= ll


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khong trao oi hat la thanh phan cau tao nen he, tc la cac he luon lahe kn. Sau ay, ta se xet trhp hai he tng tac nhau va co the trao oi hat, e gii thieu mot khai nie

II.C.3 The hoa hoc vi chnh tacXet he co lap S gom hai he nho S 1 va S 2 tng tac nhau. Hai he S 1 va S 2tng tac nhau co th

trao oi nhiet, trao oi the tch, va ca so hat. Goi E1, V1, N1 va E1, V2, N2 la nang lng, the tchso hat cua moi he S 1 va S 2. V he S co lap nen nang lng E, the tch V, va so hat N c S thoa:

EEE 21 =+ xac nh.VVV 21 =+ xac nh. N N N 21 =+ xac nh.

Neu so trang thai vi mo kha d cua S 1 va S 2 lan lt la1(E1,V1,N1) va2(E2,V2,N2), so trangthai vi mo kha d cua S la

) N,V,E(). N,V,E( 22221111 = .

V12 EEE = , 12 VVV = , va 12 N N N = ,

nen la ham cua E1,V1, va N1: = (E1,V1,N1).

Trang thai cai nhien nhat cua he tng ng vi cc ai tc la ln at cc ai. Khi nay:

0Eln

1=

, 0

Vln

1=

, 0

Nln

1=

.

Ta co lien tiep cac ket qua sau:

=

=

=

+

=

=

=

=

+

=

= = = +

)c29.II(

. Nln

Nln0

Nln

Nln

Nln

Nln

) b29.II( p p

Vln

Vln0

Vln

Vln

Vln

Vln

)a29.II(TT

Eln

Eln0

Eln

Eln

Eln

Eln

2

2

1

1

2

2

1

1

1

2

1

1

21

2

2

1

1

2

2

1

1

1

2

1

1

21

2

2

1

1

2

2

1

1

1

2

1

1

tren, ta a goi1T , 1 p va 2T , 2 p la nhiet o va ap suat vi chnh tac cua S 1 va S 2 . Tuy nhien

ay, ta co them he thc (II.29c), ac trng cho s can bang cua qua trnh tra S 1 va S 2.

Noi chung, ta at:

, hay (II.30) NST

=

Nln1

=


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va goi lathe hoa hoc vi chnh taccua he.Nh vay, khi co s can bang gia hai he S 1 va S 2 , ta co, t (II.29c):

= 21 .

Ket luan: Khi hai he S 1 va S 2 tng tac nhau (nhiet, c, va trao oi hat) va at enbang, ta co nhiet o, ap suat va the hoa hoc cua hai he la bang nhau:

= 21 TT , = 21 p p , va = 21 .

II.C.4 ai lng cng tnh va ai lng cong tnhBa ai lng vat ly ac trng cho trang thai can bang cua hai he tng ta

suat, va the hoa hoc co tnh chat ac biet quan trong ma ta se tm hieu dXet mot tham so v mo y ac trng cho mot he S . Gia s he S c phan lam hai he nho S 1 va S

2 . Goi y1, y2 la cac gia tr cua ai lng nay cua cac he S 1 va S 2. Co hai trng hp co the xa hoac ta co y1+ y2 =y. Khi nay, y c goila ai lng cong tnh.

hoac y1= y2= y. Khi nay, y laai lng cng tnh.Ta co the thay ngay rang cac ai lng sau ay la cong tnh: so hat N, thnang lng E, entropi S (cac ai lng nay lien he trc tiep en so hat cua h

Va cac ai lng sau la cng tnh: mat o khoi lngVM= (that vay, t so cua hai ai

cong tnh tr thanh cng tnh), va theo nh cac phan II.C.1, 2, va 3 ma ta aap suat, the hoa hoc eu la cac ai lng cng tnh.II.D Qua trnh thuan nghch va qua trnh bat thuan nghch

Khi xet mot he v mo th thong thng he c at trong mot so ieu kiehoa bi cac tham so v mo y1, y2,, yn). Nhng trang thai vi mo kha d cua he la tatthai tng thch vi nhnghan che at ra cho he. Nh vay, neu goi la so trang thai vi mo khata co he thc ham:

)y,...,y,y( n21= .Gia s trang thai v mo au tien cua he vi mot so han che nao o

mo kha d lai. Gia s ta lay bt mot so han che i. Khi nay, tat ca nhng trcon la kha d oi vi he, nhng ngoai ra, v so han che tren he b bt i, them nhieu trang thai vi mo na tr thanh kha d oi vi he. Tc la vieket qua la hoac lam tang hoac e khong oi so trang thai vi mo kha d luf :

if Neu if = : he a trang thai can bang va co the mot trong nhng

vi xac suat bang nhau, se van c phan bo vi cung xac suat tren cacthai can bang xem nh khong b nhieu loan. Qua trnh nay laqua trnh thuan nghch. Neu if > : mot khi he a c phan bo mot cach ngau nhien tren cac t

th viec lap lai han che khong the lam cho he t ong ri nhng trang thanhng trang thai vi nhieu han che hn. Khi co mot qua trnh nao o xay ri t trang thai au en trang thai cuoi, ma viec at them hoac lay bt i nay khong the thiet lap lai c trang thai au tien cua he, th qua trnh qua trnh bat thuannghch.


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V du: Xet hai he A va A co nhiet o khac nhau ngan cach bi mot vach

he A va A tng tac nhiet vi nhau khi ta e cho vach ngan truyen nhiet (lathi gian, cac he at en trang thai can bang.

Bay gi ta lai lam cho vach ngan lai cach nhiet (thiet lap lai han che). lai trang thai au tien la cac he A va A co nhiet o khac nhau c. Vayau la mot qua trnh bat thuan nghch.

Chu y rang qua trnh thuan nghch la mot trng hp ac biet, con haunhien eu la qua trnh bat thuan nghch, tc la thong thng ta se co:if > .

Nh vay, ve phng dien v mo, neu goiSla o bien thien cua entropi vi chnh tac cutrnh, th ta co the phat bieu rang: Nhng qua trnh ma ta thng gap la nhng qua trnhtangS > 0. Con qua trnh tng ng viS = 0 la qua trnh thuan nghch.


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BAI TAP

BT II.1Hay chng minh trc tiep nh ly Liouville (khong dung nh ly Gauss-Ostragchuyen ong cua cac iem pha theo thi gian trong khong gian pha nh la chuylong qua mot the tch nguyen to co nh co dang hnh hop.

BT II.2Chng minh rang entropi vi chnh tac la mot ai lng cong tnh.BT II.3Xet mot khoi kh co the tch V, gom N phan t ma the nang tng tac kho

nang, chuyen ong theo qui luat cua c hoc co ien va co nang lng phan 1/ Xet trng hp mot phan t. Tnh so trang thai vi mo kha d cua phan t2/ Suy ra so trang thai vi mo kha d cua N phan t cua khoi kh.3/ Thiet lap cong thc tnh nang lng E cua khoi kh theo N va theo nhiet o .Suy ra dang cua phng trnh trang thai f(p,V,T) = 0, trong o, pla ap suat vi chnh tac cua he

BT II.41/ Chng minh rang nhiet o vi chnh tac cua mot he la mot ai lng dng:> 0.2/ Chng minh rang khi hai he S 1 va S 2 tiep xuc nhiet th nhiet lng luon c hap th

nhiet o vi chnh tac nho hn va c phat ra bi he co nhiet o vi chnh tac cao hBT II.5Xet hai he tng tac nhau trng hp tong quat (trao oi nhiet lng, trao o

hat).Hay xac nh chieu trao oi cua the tch va so hat khi hai he tien en trang th


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VAN E II.A

He co nhiet o tuyet oi am

Biet rang mot electron co trang thai lng t xac nh bi vectz2 S,S,z,y,x

, vi21S = , va

2

1Sz = . Trong trng the xuyen tam, vect trang thai nay laz2z2 S,S,L,L,H

hoac sm,S,m,l,n vi n, l, m,

S va ms lan lt la so lng t chnh, so lng t qu ao, so lng t t, so lng tt spin. Moi mc nang lng En co bac suy bien bang gn = 2n2. Neu ta ch chu y en so bac t do selectron se ac trng bi mot trong hai trang thaism,S bang 2

1,21 hoac

21,

21 , ma ta se k hieu ngan

la+ hoac , tng ng vi cac nang lng xac nhB=+ , va B= neu ta xet he N spin at trt trng eu co cam ng tB

r.

(Cac gia tr+ va tnh t ham Hamilton :=

= N

1iiBHrr , vi ii S

rr = la moment t cua spin Si).

Gia s rang he N spin trang thai phan bo vi chnh tac.1/ Hay tnh nang lng toan phan E cua he. Suy ra so n+ va n cua nhng spin+ va .2/ Tnh so trang thai vi mo kha d cua he e suy ra cac gia tr cua entropi vi cva nhiet o vi

chnh tac Tcua he.3/ Ve ng bieu dien s bien thien cua entropi Scung nh cua

T1 va Ttheo n v nang lng r

gonB N

E

.

Suy ra rang he co the co nhiet o tuyet oi am: T< 0 trong ieu kien xac nh.


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Chng III

PHAN BO CHNH TAC - NG DUNG

III.A He can bang vi he ieu nhietIII.B Gii han nhiet ong lcIII.C ng dung cho he co ienIII.D ng dung cho he lng t

III.A He can bang vi he ieu nhietIII.A.1 Khai niem he ieu nhiet

Xet mot he vat ly S co nang lng E, S c tiep xuc nhiet vi he co nang lng E . Gia she rat ln so vi he S , tc la: E >> E. Do ieu kien nay, khi trang thai can banoi ang ke trong khi trang thai v mo cua hau nh khong oi. He c goi lahe ieu nhiet cua he

S .Khi S trang thai can bang vi , ta noi rang S phan bo chnh tac va cac he tng t S tao

thanh tap hp thong ke goi latap hp chnh tac.Ve mat nh lng, e co ieu kien cho he la he ieu nhiet, ta nhan xet rang he to S

la he co lap, co nang lng tong cong la: Etc= E + E. Theo nh ngha, nhiet o vi chnh tac la:

EEE tcES

T1

=

=

Vay, au tien thT phu thuoc E. Nhng neu la he ieu nhiet th ta phai coT oc lap oi vE.

Ta dung cong thc khai trien Taylor cua ham)E(ES

tai iem tcEE = :

...ES)EE()E(

ES)E(

ES

tcEE2

2

tctc ++

=

=

...ESE)E(

ES

tcEE2

2

tc +

=

=

T o, ta suy ra rang ieu kien nh lng e

=

T1

ES khong phu thuoc nang lng E cu S ,

tc la e he la he ieu nhiet, la:

tctc EE2

2

ES

ESE


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(III.2)

Khi nay, nhiet o T cua S la tham so ngoai (v b ap at t nhiet o vi chnh tacT cua he ), vanang lng E cua heS la bien so noi. (Trong khi phan bo vi chnh tac cho he conhiet o eu la tham so ngoai).

III.A.2 Tha so Boltzmann trong phan bo chnh tacXet he S can bang nhiet vi he ieu nhiet , tc la ieu kien (III.1) c thoa.Goi (l) la mot trang thai vi mo cua S , co nang lng E l . Neu goi Etc la nang lng (vi o

ungE) cua he tong hpS , th nang lung cua he la E phai thoa:tcEEE =+ l , (III.3)

V he tong hp S la he co lap nen ta co the ap dung tien e c ban cuatrang thai vi mo kha d cua S la ong xac suat. Xac suat Pl eS trang thai (l) se bang so tranthai cua S eS trang thai nay (so bien co thuan li) chia cho tong so nhS

(so bien co tong cong) latc. Nhng v khi

S

trang thai (l

) th trang thai co nang l

lEEE tc = , nen

tcP

= l , (III.4)

vi la so trang thai cua (bang so trang thai cua S khiS trang thai (l)).V tc = const nen ta co the viet:

)EEE(CP tc ll == . (III.6)V theo nh ngha, entropi vi chnh tac cua la = lnk S , nen ta co:

)k S

exp(CP

=l .Ta dung cong thc khai trien Taylor cua ham)E(S quanh iem Etc:

...ES.)EE(

21

ES)EE()E(S)E(S

tctc E2

22

tcE

tctc ++

+=

Vi ieu kien (III.1), ta thay ch

gi lai hai so hang au trong he thc tren va vlEEE tc = , T1

T1

ES

tcE

==

, ta co

TE)E(S)E(S tc l= .

Vay)

kTE

k )E(Sexp(CP tc ll =

.

Cuoi cung, ta co cong thc tnh xac suat e he S trang thai (l) co nang lng E l:

(III.7)

= TT

lll

EkTE e1e.1P =

=


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ai lng

= 1kT c goi lanang lng ac trngva Z goi laham tong thong ke, c tnh theo

ieu kien chuan hoa1P)(

=l

l :

. (III.8)

Phan bo thong ke c xac nh bi cong thc (III.7) c goi la phan bo chnh tac.So hang kT/Ee l c goi latha so Boltzmanncua phan bo chnh tac.III.A.3 Phan bo thong ke cua nang lngTrong phan bo chnh tac, nhiet o cua he la tham so ngoai (an nh b

nhiet), trong khi nang lng cua he t do bien thien; o la bien so noi. Sauthong ke cua ai lng quan trong nay.

a) Trng hp tong quat Moi mc nang lng El cua he co the suy bien. Ta goi g(El) la bac suy bien cua El. Vay, trong

trng hp tong quat, xac suat e he co nang lng El phai c nhan len bi tha so g(El):kTEe

Z1)E(g)E(P lll

= , (III.9)va ham tong thong ke c tnh:

=l

ll

E

Ee)E(gZ

Gia tr trung bnh cua nang lng c tnh bi:

==)( )(

EeEZ1P.EE

l llll

l (III.11a)

==l l

lllll

E E

Ee)E(g.EZ1)E(P.E (III.11b)

b) Trng hp xap x lien tuc cua nang lngTa xet trng hp nang lng bien thien lien tuc. ieu kien e phep tnh

hien la: cac mc nang lng phan bo rat sat nhau sao cho trong khoang nanE co mot so raln cac mc, ong thi, ham f(El) e lay tong phai bien thien rat cham sao cho ta co f l)f(E).

Khi nay, neu goi w(E) la mat o xac suat th xac suat e he co nangE+dE la

dE)E(w)E(dP = . (III.12a)Mat khac, neu goi(E) la mat o trang thai th so trang thai co nang lng

E+dE la(E)dE. V xac suat e co mot trang thai nay laE

eZ1

nendE)E(e

Z1)E(dP E= . (III.12b)

So sanh (III.12a) va (III.12b), ta co)E(e

Z1)E(w E= . (III.13)

==)(

E

)(

kT/E eell

ll


43/163

Do ieu kien chuan hoa +

=0E

1)E(dP (E0 la mc nang lng c ban cua he, co ng

nang lng thap nhat), t (III.12b), ta co:

+

=0E

E dE)E(eZ (III.14)

Trong khong gian pha, neu so bac t do cua he la f, th so trang thai co nava E+dE la

f )2( pdqddE)E(h

rr

= ,

vi =

=f

1iidqqd

r , =

=f

1iidp pd

r

Vay

( ) pdqd

21e

Z1)E(dP f

E rr

h= . (III.15a)

Va ta co cong thc tnh ham tong thong ke:

+

=

0E

Ef pdqde)2(

1Z rrh

. (III.15b)

Gia tr trung bnh cua nang lng khi ai lng nay phan bo lien tuc c bo gian oan:

ll

l PEE)(

= dE)E(Ew)E(EdP00 EE

++= .

+

=0E

E dE)E(EeZ1E . (III.16)

Hay

+

=

0E

Ef pdqdEe)2(

1Z1E rr

h. (III.17)

III.A.4 Cac ai lng ac trng cho he chnh tac Ta thng dung he thc toan t sau:

. (III.18)

Nang lung t docua he c nh ngha bi:

(III.19)

Ta co:

EZ

e)E(Z)'Z(ZlnEeZ

E

)(

E ===

=

ll l

l

. (III.20)

TT

=

ZlnkTF =

)Z(lnE=


44/163

Ta nh nghanhiet dung ang tch:

(vi V, N, = const) (III.21a)

vanhiet dung rieng ang tch:, (III.21b)

viA N

Nn = la so mol cua he.

Ta co the chng minh c cong thccua entropi chnh tac:

. (III.22)

Ap suat chnh taccua he c nh ngha bi:

, (III.23)

vathe hoa hoc chnh tac:

. (III.24)

III.B Gii han nhiet ong lcIII.B.1 Khai niem gii han nhiet ong lcTa noi rang mot he v mo at engii han nhiet ong lckhi kch thc he u ln e nh

giang thong ke cua cac bien so noi la khong ang ke. Khi nay, moi bieduy nhat (la gia tr trung bnh va cung la gia tr cai nhien nhat, v o lechnen khong ang ke, nh ta se thay di ay).

He v mo gii han nhiet ong lc c goi lahe nhiet ong lc.III.B.2 Phan bo chnh tac gii han nhiet ong lcTa co cong thc tnh mat o xac suat (III.13):

)E(eZ1)E(w E=

vi(E) la mat o trang thai, nen so trang thai vi mo kha d la:dE)E(= )E(ln)dEln()E(lnln +=

== lnk lnk S . (III.25)

ZlnEZ

)E(SZlnE)E(ln)E(wln ==

. (III.26)

TECV

=

TE

n1CV

=

T

FS

=

VF p

=

NF

+=


45/163

Ta goi Em la gia tr cai nhien nhat:

===

)E(kT1

ES

k 10

E)E(wln

mE tc

T)E(T m = Tc la mot he v mo can bang vi he ieu nhiet th gia tr cai nhien nmcua nang lng la

tr lam cho he nay co nhiet o vi chnh tac bang nhiet o cua he ieu nhieTa dung cong thc khai trien Taylor cua ham lnw(E) quanh gia tr cua Emva ch gi lai ba so h

au:

mm E2

22

mE

mmEw)EE(

21

Ew)EE()E(wln)E(wln

+

+=

V 0Ew

mE=

, va at:

2E

2

2

E2

2

)E(1

ES

k 1

Ew

mm

==

(v lnw co cc ai tai Emnen 0E

w2

2 < ),

ta co

2

2m

m )E(2)EE()E(wln)E(wln

=

= 2

2m

m )E(2)EE(exp)E(w)E(w . (III.27)

T ieu kien chuan hoa

1dE)E(w =

+

,ta co

+

=

1dE

)E(2)EE(exp)E(w 2

2m

m

Do tch phan Poisson

+

=

22

2m )E(2dE)E(2

)EE(exp ,

ta suy ra:

2m )E(2

1)E(w

= . (III.28)

Mat khac, ta co, t cong thc (III.26)

ZlnEk

)E(S)E(wln mmm =

,

tc la


46/163

=

mm

m Ek )E(Sexp

Z1)E(w , (III.29)

So sanh vi (III.28), ta co c:

=

mm2 E

k )E(Sexp.)E(2Z , (III.30)

vi2

2

2 ES

k 1

)E(1

=

. (III.31)

Ket luan: He thc (III.27) chng to rang mat o xac suat bien thien theo nang l

Gauss, tai gia tr Em . V nang lng E va entropi S eu tang ty le vi so hat N cua he nen E)2

bien thien nh so hat N.Vay:

N

1

E

E

m

. (III.32)

Tc la phan bo co sai so tng oi rat nho khi so hat N cua he la rat l

III.B.3 S tng ng gia cac phan bo gii han nhiet ong lcT cong thc tnh nang lng t do:

ZlnkTF = ,ta co

=

2

mm )E(2ln

Z1E

k )E(SkTF .

V Em va S

eu ty le vi N, trong khi so hang cuoi ty le ch vi lnN, nen nhiet ong lc), ta co c

)E(TSEF mm= .

T cong thc tnh entropiTFS

= (III.22), ta cung suy ra c rang:

2m )E(2ln2

k )E(SS += .

Va nh vay, gii han nhiet ong lc:)E(SS m

=

Ta cung a co:)E(TT m= .Va tng t, ta cung chng minh c cho ap suat va the hoa hoc :

)E( p p m= ,

)E( m= .

Ket luan: oi vi he v mo (co so bac t do hay so hat rat ln) trang thcac phan bo vi chnh tac va phan bo chnh tac la tng ng: cac ai lng vi chnh tac ong nhat v


47/163


48/163

Xac suat pdr d) p,r (P rrrr e phan t co v tr trong khoangr r va r dr rr + va co ong lng trong kh

pr va pd p rr + bang so o trong khong gian pha3)2( pdr dh

rr

nhan vi xac suat m2 p 2e e phan t

trong o nay :

3m2 p

)2( pdr de pdr d) p,r (P

2

h

rrrrrr

.

Vay xac suat pd) p(Prr

e phan t co ong lng trong khoang pr va pd p rr + c tnh:m2 p

V

2e pdr d) p,r (P pd) p(P = rrrrrr .

T o, ta co the tnh xac suatvd)v('P rr e van toc cua phan t trong khoangvr va vdv rr + la:vdCevd)v('P 2mv

2 rrr = . (III.35)Ket qua tren c goi la phan bo Maxwellcua van toc phan t kh ly tng n ngu

hang so c tnh bi ieu kien chuan hoa ).

III.C.3 nh l phan bo eu

Trong c hoc thong ke co ien, nang lng cua mot he vat ly la ham c1, q2,, qf va cua f ong lng suy rong p1, p2, , pf :) p,..., p, p,q,...,q,q(EE f 21f 21= (III.36)

Thong thng, ta co the co hai ieu kien sau c thoa:i) Nang lng toan phan c phan lam hai phan:

) p,..., p, p,..., p, p,q,...,q,q('E) p(E f 1i1i21f 21ii ++= ,(III.37a)

trong oi ch phu thuoc pi va phan con lai E khong phu thuoc pi .ii) Hami co dang toan phng theo thanh phan pi cua ong lng:

2iii bp) p( = , vi b = const. (III.37b)

Trng hp ta thng gap lai(pi) la ong nang, ch phu thuoc moi thanh phan ctrong khi the nang khong phu thuoc ong lng.Sau ay, ta se tnh gia tr trung bnh cuai khi he trang thai can bang nhiet.Khi he can bang nhiet o T= (k)-1, he trang thai phan bo chnh tac, gia tr trui

c tnh theo cac tch phan trong toan bo khong gian pha:

+

+

=

f 1E

f 1iE

i

dp...dqe

dp...dqe. (III.38)

Neu ieu kien i) c thoa, ta co:

+

+

+

+

+

+

+

+ =

=

f 1'E

i

f 1'E

ii

f 1)'E(

f 1i)'E(

i

dp...dqe.dpe

dp...dqe.dp.e

dp...dqe

dp...dq.e

i

i

i

i

,

trong o, cac tch phan th nh t so va mau so khong c tnh theo biei, nh vay, cac tch phanay giong nhau va ta co the n gian e co


49/163

=

=

=

+

+

+

+

+

i

i

i

i

ii

i dpelndpe

dpe

dpe

dp.ei

i

i

i

i

Khi ieu kien ii) c thoa:

+

+

+

== dyedpedp.e

22

12ii byi bpii ,

vi bien phu i py 21

= .Vay

=

+

=

+

21dyelnln

21 2 by

i ,

v so hang th nh trong tong e lay ao ham khong phu thuoc.Cuoi cung, ta co

. (III.39)

Tc la gia tr trung bnh cua moi so hang toan phng oc lap cua nang lkT21 .

Ket qua quan trong tren c goi lanh l phan bo eutrong c hoc thong ke co ien.Chu thch:

Neu thay v he thc (III.37a) cho ong lng pi , ta lai co ieu kien cho toa o qi nh sau) p,..., p, p,q,...,q,q,...,q,q('E)q(E f 21f 1i1i21ii ++= . (III.40a)

va2

iiibq)q( = , (III.40b)

th bang cach tnh tng t, ta cung co ket qua sau:kT

21

i = .

Can chu y rang nh l phan bo eu ch ung trong gii han co ien, tkhoang cach gia cac mc nang lng chung quanh gia tr trung bnhE la rat nho oi vi nangac trng : kTE


50/163

0 p.r dtdK 2

N

1iii =

=+

=

rr .

kT23K = . (III.41)

b) Xet mot dao ong t ieu hoa tuyen tnh co ien, c nang toan phan cu22

2

qm21

m p

21UK E +=+= .

Ap dung lan lt nh l phan bo eu cho K va U, ta co

kT21

K = , kT21

U= ,

nen:kTE = . (III.42)

nh l virialXet mot he kh gom N phan t, chiem the tch V. Ta gia s cac phan t

va N, V u ln e he co the c xem la he nhiet ong lc. Xet mot phair r , va

tong hp lc tac dung len phan t laif r

.Virial cua he c nh ngha la:

i

N

1i.i f r rr

== (III.43)Gia s tam quan tnh cua he ng ngh trong he qui chieu quan tnh:0f

N

1ii

rr=

=. Khi nay, virial

oc lap vi goc toa o.Neu goii p

r la ong lng cua phan t (i), cac phng trnh chuyen ong cua

.f dt pd

,m

pdtr d

ii

ii

rr

rr

=

=

T o:( )

i

2i

iiii m p

f .r p.r dtd +=

rrrr

Lay tong tren tap hp N phan t:K 2 p.r

dtd

ii += rr ,vi :

=

= N

1i i

2i

m2 p

K

la tong ong nang cua ca he.Lay tr trung bnh theo thi gian cua he thc tren:

(That vay, tr trung bnh cuadtdg la:


51/163

[ ])0(g)(g1limdtdtdg1lim

dtgd

0

=

=+

+ .

V =

= N

1iii p.r g

rr la ai lng b chan, nen gii han tren bang khong).

Hn na, trang thai can bang, tr trung bnh theo thi gian bang tr tru(nguyen l ergodic), nen:

=21K . (III.44)

He thc tren la noi dung cuanh l virial: trang thai can bang, tr trung bnh cua ocua he bang

21 gia tr trung bnh cua virial cua he.

So sanh vi nh l phan bo eu cho ong nang cua he N phan t (III.41), NkT3= .

III.C.4 Kh l tnga) Ham tong thong ke

e tm lai cac ket qua thc nghiem a co cho kh ly tng, tc la cho (the nang tng tac gia cac hat khong ang ke oi vi ong nang cua mc, au tien, ta xet trng hp mot hat M co toa or r va ong lng pr chuyen ong trong the tV, tc la trong gieng the ba chieu:

=

V)r (MkhiV)r (Mkhi0

)z,y,x(U rr

(III.45)

Vi k z jyixr rrrr ++= la toa o cua M va V la the tch cua mot khoi lap phng

thai chnh tac{ }x,T .Ham tong thong ke Z1 cua mot hat M nay c tnh:

= pdqd)Eexp(

h

1Z31

rr

, (III.46)vi U

m2 pE

2+= la nang lng cua hat, bang tong so cua ong nang

m2 p2 va the nang U.

=

=

+=

.dxdydz)]Uexp(.dp)]m2

pexp(h1

qd)]Uexp( pd)]m2

pexp(h1

pdqd)]Um2

p(exp[h1Z

3

x

2x

2

3

2

31

rr

rr

Ap dung cong thc tch phan Poisson cho tch phan th nhat, ta co

mkT2h1dp)

m2 pexp(

h1I x

2x == . (III.47)

e tnh tch phan th hai, ta nhan xet rang khi MV, U = 0, va khi MV, U , nenVdxdydz)Uexp(J == .

Vay:


52/163

( ) 233

31 mkT2

hVJIZ == . (III.48)

Chu y: e tnh tch phan +

pd)m2

pexp(2 r , ta cung co the dung toa o cau (p, , ):

,)mkT2(dp p)m2

pexp(4

ddpdsin p)m2

pexp( pd)m2

pexp(

23

0

22

222

==

=

+

+

r

bang cach dung cong thc cua tch phan Poisson:

30

at2

a41dtet

2 = +

.

Bay gi, ta xet he hat ong nhat, oc lap, va khong phan biet c trthc (III.45), trang thai chnh tac{ }x,T . Nang lng cua he cho bi:

i N

1i

2i Um2

pE +=

=, (III.49)

va ham tong thong ke cua he c tnh:

= N21 N21 N3 pd... pd pdqd...qdqd)Eexp(h1

! N1Z rrrrrr . (III.50)

(Chu y rang tha so ! N

1 co trong bieu thc cua Z do gia thiet he hat khong phan

V cac hat la ong nhat nen ta co:

)51.III(. pdqd)Um2

p(exph1

! N1

pd... pd pdqd...qdqdUm2

pexph

1! N

1Z

N

iii

2i

3

N21 N21 N

1ii

2i N3

+=

+=

=

rr

rrrrrr

Nhan xet rang bieu thc trong dau moc{ }... chnh la ham tong thong ke cua mot hat1 ma ta atnh c (cong thc (III.48)), ta co:

N1)Z(! N

1Z =

2/ N3 N3

N

)mkT2(hV

! N1

Z = . (III.52)b) Khao sat nhiet ong lc

oi vi he ta xet la he v mo; N u ln e ta co the ap dung cong thln(N!)NlnN ,

e co

N3

2/ N3

h)mkT2(lnVln N Nln NZln ++= .


53/163

Vay, ta co cong thc tnh nang lng t do Helmholtz:

ZlnkTF =

N3

2/ N3

h)mkT2(lnkT]Vln N[lnkTN += . (III.53)

Ap suat chnh tac cua he la

V

NkT

V

F p =

= , (III.54)

va ta co phng trnh trang thai cua kh ly tng:(III.55)

Ta cung tnh c entropi cua he:

TFS

=

, (III.56)

vi S0 la hang so.Cuoi cung, cong thc tnh noi nang cua he kh ly tng c suy ra t:

ZlnEU== ,

va ta co c

(III.57)

Ket qua nay ta co the thu c bang cach ap dung nh l phan bo eu chIII.C.5 Kh van der Waals

Trong chat kh, ta co the nang tng tac gia cac hat (phan t) nho hn s(khi the nang tng tac co the bo qua, ta co chat kh l tng, con neu thec oi vi ong nang cua hat, th he cac hat nay c xem nh la chat ltrang thai chnh tac.

Ham tong thong ke cua he hat phep gan ung co ien la:

= pdqde)2(1

! N1Z E N3

rr

h, (III.58)

trong o:)q(U p

m21E 2i

r+= , (III.59)neu ta gia s he hat la ong nhat, khong phan biet c. Ta co cong thc

= qde. pde

)2(1

! N1Z )q(U

pm2

N3

2i rr

h

r,

Nhng v: N

i2i

m2 p

N

im2

p p

m2 dp pe4 pde pde

2i

2i2

i

=

=

rr

NkT pV =

0STln Nk 23S +=

NkT23EU ==


54/163

( ) 2 N3 p

m2 mkT2 pde2i =

r ,nen

( ) ZumkT2)2(

1! N

1Z 2 N3 N3 =

h , (III.60)

vi

= qdeZ )q(U.urr

(III.61)la ham tong thong ke lien quan en the nang tng tac gia cac hat.

(Trong trng hp 0)q(U r , ta se co: Zu=VN, va ta tm lai c ket qua cho he kh l tTa gia s rang tng tac cua hat th ba vi hai hat ang xet la khong

tng tac cua toan he hat)q(U r la tong cua the nang tng tac cua tng cap hat (i,j)( ) jiij r r u rr (v the nang tng tac gia hai hat ch phu thuoc khoang cach tng oi ji r r

rr gia hai hat (i) va (j) ).Khi nay:

( ) ( )

=i j

jiiji i j

jiij r r u21r r u

21)q(U rrrrr

Thong thng, the nang tng tac cua mot cap hat( ) )(ur r u jiij rr

c cho bi bieu thc:

=

602

120

0u)(LJu, (III.62)

c goi lathe Lennard-Jones, ma ng bieu dien c ve tren hnh III.1. Ta co thrang khi 0 , the tng tac tng ng vhut.

Hoac mot cach s lc hn, the nang tng tac cua tng cap hat cho b(hnh III.2):

=s

00u)(u

0

0

,

,

>


55/163

( )[ ],eVVZ NUxu eff = (III.65)

Bi v co tat ca2

N)1 N( N21 2 cap hat nen neu goiu la the nang tng tac trung bnh c

cap hat, the nang cua toan he c tnh:

u N21Uu

2 NU N eff

2

eff == . (III.66)

V xac suat e moi hat hien dien trong khoang gia va + d la:V

d4 22 , nen:

Vd4)(uu

22

0

=

.

e co gia tr cu the cuau , ta gia s mo hnh khoi cau cng co the ap dung

=

=

00

dVu4.d

Vu4u 2ss00

2s

00 .

Ta lai gia s rang s > 3: ng cong the nang giam u nhanh e tch phantu:

s3Vu4

s3Vu4u s30s00s3s00

0

=

=

+

s3Vu4u

300

= . (III.67)

s3Vu4

N21U

300

eff = .

V NaUeff = , (III.68)

vi

030 us3

33

2a

= , (III.69)Mat khac,the tch loai tr cua he kh la NVx cung c tnh bang so cap hat

21 N2 nhan vi the tc

loai tr cua moi hat la34 o3 nen:

N.3

2V

)34( N

21 NV

3ox

3o

2x

=

=

Vx=b.N (III.70) vi

== 3o3o 3

421

32' b (III.71)

Ta phoi hp cac cong thc (III.60) va (III.65) e co:

( )( ) ( )[ ] NUx2 N3 N3 eff eVVmkT22

1! N

1Z

=h

(III.72)

T o, ta tnh c ap suat trung bnh cua khoi kh:


56/163

( )[ ]eff x U NVVln NV

1V

Zln1 p

=

=

Vi gia tr cuaeff U (III.68) va Vx (III.70), ta co:

2

2

2

2

x V N'a

V' bV NkT

V N'a

VV NkT p

=

=

Cuoi cung ta co c phng trnh trang thai:( ) NkTV' bV

V N'a p 2

2=

+ (III.73)

c goi la phng trnh van der Waalscho kh thc. Neu a= 0, b= 0, ta co phng trnh tkh ly tng.

Ta co nhan xet la trong (III.60), tha so th nhat ve trai cho thay ranhn so vi ap suat cua kh ly tng., bieu th lc tng tac hut tam xa, vatch cua he kh thc giam ty le vi so hat N oi vi the tch cua kh ly tphai c xem nh co kch thc.

T phng trnh (III.73) tren, tng ng vi moi gia tr cua nhiet o T, ta vs bien thien cua ap suat p theo the tch V, goi lang ang nhiet van der Waals. Ta thay rang khi nhiet o T ln hn mot gia tr Tc (goi lanhiet o ti hanTc), moi gia tr cua p

cho duy nhat mot gia tr cua V. ng ang nhiet vi T = Tc goila ng ang nhiet ti han. Cac giatrVc vapc cua the tch va ap suat tng ng vi ng ang nhiet nay tangang c goi lathe tchvaap suat ti han.


57/163

III.D ng dung cho he lng t

III.D.1 He dao ong t ieu hoa lng t tuyen tnhTheo ket qua cua c hoc lng t, cac mc nang lng cua mot dao ong

tan so goc cho bi:+= h)

21n(En , (III.74)

vi cac trang thai kha d cua dao ong t c ac trng bi so lng thay bang khong: n= 0, 1, 2,

Ham tong thong ke cua dao ong t c tnh:

( )

=

=

=

=

+==0n0n0n

E nexp2

exp)21n(expeZ n hhh Nhng ta co

( ) ...ee1nexp 20n

+++=

=

hhh

la tong vo han cac so hang cua cap so nhan co so hang au la 1 va co 1eq >hkT , tc la nang lng ac trng oi vi hai mc nang lng lien tiep nhau cua dao ong t. V khi nay1>>h nen ta co the ducong thc khai trien Taylor cho ham muhe :

+ hh 1e

.kTkT2

E += h (III.77a)

Ta tm c ket qua cua phep tnh co ien. Qua that vay, v>>hkT , nen nh ta a thay, cphep tnh xap x co ien phai em lai ket qua tng oi chnh xac. Xet trng hp nhiet o thap: h . Khi nay, phep tnh gan ung cho:

.e2

E += hhh (III.77b)

Khi T u nho e ta co the bo qua so hang th nh, nang lng trung bnh cmc c ban.

2h


58/163

III.D.2 He Rotato lng t

a) Rotator co ienXet hai hat M1 va M2 co khoi lng m1va m2, cach nhau mot khoang re. Khoi tam G cua he

xac nh trong toa o Descartes:

.mm

r mr mmm

OMmOMmOG21

2211

21

2211++=

++=

rr

Neu chon goc toa o O trung vi khoi tam G:.

mmr

mr

mr

r mr m0r mr m21

e

2

2

1

122112211 +

====+rrr

Moment quan tnh cua he oi vi O la:222

211 r mr mI +=

co the tnh nh sau:,r I 2e=

vi21

21mm

mm+

= , goi la khoi lng rut gon cua he.

Moment ong lng cua he oi vi goc O:.vmr vmr L 222111

rrrrr+=

Neu khong co lc ngoai nao tac dung len he,constL =r

, he quay quanh truc Oz trong mat pvuong goc viL

r vi van toc goc khong oi. Ta co:( ) ( ) ( ) ( )

( )+=+=+=

222

211

222111222111

r mr m

r .r mr .r mv.r mv.r mL

r

.r IL 2e ==r

Khi nay, c nang toan phan cua he la tong ong nang cua hai hat:

.I21

vm21

vm21

E22

22211 =+=

Tc la ham Hamilton lien ket vi nang lng uc tnh theo moment ong lLr

:

.r 2

LI2

LH 2e

22

== (III.78)

(Ta thay bai toan Rotator co ien rut ve bai toan chuyen ong quay cua h , cachgoc O mot khoang re).

b) Rotator lng te khao sat lng t he Rotator, ta chuyen cac ai lng L va H thanh ca

lng va toan t Hamilton theo nguyen l tng ng:

.H

H;L

L Vay, he thc (III.78) cho:

I2LH

2= , (III.79)

vi ket qua co c theo c hoc lng t:,m,)1(m,L 22 lllhl += (III.80)


59/163

trong o vect ketm,l la vect lien ket vi ham ieu hoa cau( ),,Ym l vil nguyen dng hoacnguyen am, va m co the nhan 2l +1 gia tr

m = -l, -l +1,,0,,l +1,l.Nh vay:

,m,Em,I2

)1(m,I2

Lm,H22

llllh

ll l=+==

viEl la gia tr cua cac mc nang lng:

I2)1(E

2 += llhl , (III.81)

cac mc nay co bac suy bien 2l +1.Ham tong thong ke cua phan bo chnh tac

( )( )

( )( )

( ) + +=+==l

llh

lll

l

ll I2/)1(

E

EE 2e12e12eZ (III.82)

Tr trng hp cua phan t H2 co moment quan tnh qua nho, oi vi nhng phan t khac, khoang cach gia hai mc nang lng quay lien tiep tnh t (III.81) v-4eV. Do o nhiet o T u ln, ta co

1I2

)1(E2


60/163

III.D.3 Nhiet dung rieng cua kh ly tng lng nguyen tNang lng E cua kh ly tng lng nguyen t gom nang lng Etr cua chuyen ong tnh tie

khoi tam, nang lng Erotcua chuyen ong quay quanh hai truc vuong goc vi truc nnang lng Evibcua dao ong ieu hoa. Ve phan ong gop cua lp vo electron the nh lai rang cac gia tr cua nang lng kch thch lp vo nay co o lnhiet o vai chuc ngan o Kelvin:

1 eV K 000.12T .Vay, ch vi nhiet o rat cao cac lp vo electron quanh hat nhan mi co

nhiet dung cua chat kh.Tuy nhien, cung can biet rang co nhng biet le cho nhan xet tren, nhat

en cau truc tinh te cua mc nang lng c ban: do lien ket gia momen oS, mc nang lng nay tach ra thanh tng mc di ma khoang cach giatram eV, tc la tng ng vi nhiet o ac trng khoang vai tram o. Tuy nhp, ta co the xem nh mc nang lng c ban la suy bien bac (2L+1)(2S+1)anh hng cua lp vo electron na (ta noi rang cac bac t do electron b o

Tom lai, nhiet o nho hn nhieu oi vi vai chuc ngan o Kelvin, ta ccua chuyen ong tnh tien, chuyen ong dao ong, va chuyen ong quay:

E = Etr + Evib+ Erot oi vi mot mol phan t kh, ta co: RT

23kT N

23E Atr ==

(Vi R = NAk la hang so chat kh)nen nhiet dung rieng tng ng vi chuyen ong tnh tien la

R 23

TC

C tr tr == .

oi vi chuyen ong dao ong va chuyen ong quay, ta a vao cac khaicua dao ongvibT va quayrotT : cac nhiet o rat nho so vivibT (hay so virotT ), cac bac t do cudao ong (hay quay) se ong bang va khong anh hng en gia tr cua nhi

Nen biet rang ta luon corotT < vibT , tc la cac chuyen ong quay de b kch thch hong dao ong. ong thi,vibT cang ln neu lc tng tac lien nguyen t cang lncang nhe (thay oi t vai tram en vai ngan o Kelvin). Trong khi o, nhieong quay thng nho (vai o Kelvin), tr trng hp phan t H2, do co momen quan tnh rat nhnhiet o nay vao khoang 85K. Nh vay, nhiet o T thong thng: T >>rotT va cac bac t do quayong gop vao gia tr cua nhiet dung. Ta co bang sau cho ta cac nhiet o acvibT varotT :

Phan t vibT (K) rotT (K)H2 Cl2 Br2 O2 N2 COHClHBr

621580846322563374310342273787

85,30,350,122,12,92,815,012,0

Vay, oi vi chuyen ong dao ong, khivibTT

8/14/20

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