University of Washington Department of Chemistry Chemistry 553 Spring Quarter 2010 Homework Assignment 3 Due 04/26/10

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1 Universiy of Washingon Deparmen of Chemisry Chemisry 553 Spring Quarer 1 Homework Assignmen 3 Due 4/6/1 v e v e A s ds: a) Show ha for large 1 and, (i.e. 1 >> and >>) he velociy auocorrelaion funcion 1) Saring wih he expression () / s / = + 1 is given by v( ) v( ) e / =. Assume A ( ) A ( ) = αδ ( ) 1 α / / s/ () = + v e v e e A s ds 1 1 ( 1+ ) / ( 1+ ) / ( s1+ s) / v( 1) v e v e ds1dse A s1 A s 1 ( 1+ ) / ( 1+ ) / ( s1+ s) / α 1 δ ( 1 ) ( 1+ ) / ( 1+ ) / s ( 1+ ) / α ( 1+ ) / = + = e v + e ds ds e s s 1 1 / 1 1 ( 1) = e v + αe dse = e v + e e α ( 1+ ) / α ( 1) / ( 1+ ) / 1 / = e v + e e e Where he las resul happens because 1 + is large compared o. The correlaion funcion can only depend upon he absolue difference in ime and i mus decay. So α 1 / α 1 / v ( 1) v e = e b) Saring wih your expression for he velociy correlaion funcion show ha r () = ds1 ds v ( s1) v ( s) = 6 α s1s () / = 1 1 = 1 r ds ds v s v s ds ds e 1 Use he ransformaions:, s + s s = s1 s S =. Now using he echniques from he + + α s / s/ Langevin equaion lecures: r () = ds dse = α dse = α

2 From he equipariion limi: α 6 M =, so r = α = 6 c) Suppose he Langevin force is dela funcion correlaed such ha A A = αδ. y neglecing he derivaive erm in he Langevin equaion 1 1 obain an approximae expression for 1 r ds1 ds v s1 v s () dv v + = A () v v and using his expression evaluae =. How does his expression compare wih ha obained in par b? n wha limi is his approximaion jusified? dv v v Soluion: Sar wih + = A() and assuming a he derivaive is zero: A() =. Then v v = A A ( 1) ( 1) f we assume he velociy correlaion funcion is of he form v( ) v( ) α δ ( ) he mean-squared displacemen is: () = 1 ( 1) = α 1 δ ( 1 ) = α r ds ds v s v s ds ds s s =, 1 1. Same resul as b. This approach is valid in he seady sae limi: >>. ) Consider he Langevin equaion for rownian moion in a resoring poenial: dv M + v+ r = F() is he coefficien of ranslaional fricion, is he resoring force consan, and F() is he Langevin force. a) Show ha his equaion can be wrien in he form: dc () 1 dc () + + ω C () = where C = r r, = M = M / and ω = / M. Soluion:

3 () 1 dr () () () 1 () dr () A() () 1 dc () dr dv M + v+ r = M + + r = F dr + + ω r = d r r dr () () ω () () r + + r r = r A d r r dr r + + ω = d r r 1 d r r + + ω r r = r A = dc + + ω C () = r r r A a b) Assume C() has he form C() = e where a is a consan. Show ha he correlaion a a + funcion C() has he form C() = c+ e + ce where a± = ± 1 8ω and c ± are consans. a a dc () 1 dc () de 1 de a + + ω C () = + + ω e = a a a a a ae + e + ω e = a + + ω = 1/ 1/ ± ( 4ω ) 1± ( 14ω ) a± = = Noe misake above in quesion should be 4 insead of 8. Makes no difference in wha follows. () = + a + a C c+ e ce c) Assume he iniial condiions for he correlaion funcion C() are: C () dc = = C =. Deermine c and c. + = and

4 C = = c+ + c c = c+ d a a a a C + + () = ( c+ e + c e ) = c+ a+ e + cae ca C = = c+ a+ + ca c+ = a+ ca a c+ = = c+ a+ a+ a a c+ 1 = a+ a+ 1 1 a a a a a + a c+ = 1 = = a+ a+ a+ a+ a a+ a a a+ c = c+ = = 1 = a a a a a a a c± =± a a d) The over-damped limi occurs when he resoring force consan is muck less han he coefficien of fricion. Derive an expression for C() in he over-damped limi. n paricular show ha C() canno oscillae in he over-damped limi. 1/ ( ω ) ( ω ) 1± 14 1± 1 a± = where we assume 4ω 1 so ha ( ) 1/ 14ω 1 ω. Then a + and 1 a. a + a a a ae ae 1 / + + ( / / ) C() = c+ e + ce = = e + e a a+ 1 / / e e = e) Anoher approach o obaining C() in he overdamped limi is o assume he dv acceleraion erm is zero such ha M + v+ r = F() reduces o v+ r = F(). Solve for he correlaion funcion C() using his approach.

5 () dr v+ r = F() + r() = F() dc () + C = C = C e = e / / () () dψ 4) Consider he Langevin equaion for rigid rownian roaion : + ψ = F() dϕ where ψ =, is he momen of ineria, is he coefficien of roaional fricion, and F() is he Langevin force for which: F( ) = and F F = αδ () and where α is a consan. a) Fourier ransform he Langevin equaion and hus obain an algebraic equaion for he Fourier ransform of he angular velociy. dψ + ψ = F () dψ iω iω iω e e F e () () + ψ = iω iω iω i e e F e ( ω) ψ () + ψ () = () 1 Ψ = ( iω) Ψ ( ω) + Ψ ( ω) = F ( ω) 1 1 ( ω) iω F ( ω) b) Using he Wiener-Khinchine heorem obain an expression for he auocorrelaion funcion of he angular velociy. ( ω) ( ω) ( ω) iω iω F ( ω) F ( ω) Φ ψ =Ψ Ψ = iω α iω α e F F e δ () ω + β ω + β ω + β = = = According o he Wiener-Khinchine heorem he specral densiy Φ ( ω) ransform of he velociy auocorrelaion funcion. We can obain he velociy auocorrelaion funcion by inverse ransforming he specral densiy: ψ is he Fourier

6 iω 1 iω 1 α e Cω() = ω ω() = ψ ( ω) e dω dω π Φ = π ω + β + 1 α cosω 1 α π 1 α = d ω e e e π = = = ω + β π β where for planar roaion α = + + β / / 5) The Caresian marix represenaion of he roaion operaor R(α,β,γ) is given in Lecure 8. This marix represenaion of a hree dimensional roaion can be easily derived. a) Assume a roaion abou he z axis by an angle α is represened by: cosα sinα R z ( α) = sinα cosα 1 And a roaion around he y axis is given by cos β sin β R y ( β ) = 1 sin β cos β Show ha he produc of he marices Rz ( α ), Ry ( β ), and Rz γ yields he marix represenaion of he hree dimensional roaion operaor R(α,β,γ), given in Lecure 8. cosγ sinγ cos β sin β cosα sinα Rz ( γ) Ry ( β) Rz( α) = sinγ cosγ 1 sinα cosα 1sinβ cosβ 1 cosγ sinγ cosαcos β sinαcos β sin β = sinγ cosγ sinα cosα 1 cosαsin β sinαsin β cos β cosαcos βcosγ sinαsinγ sinαcos β cosγ + cosαsinγ sin β cosγ = cosα cos βsinγ sinαcosγ sinαcos βsinγ + cosαcosγ sin βsin γ = R ( α, β, γ) cosαsin β sinαsin β cos β b) Prove ha he inverse of R(α,β,γ) equals is ranspose. Demonsrae his fac using an π π example: α =, β =, γ = π 4

7 1 π π 1 π π R( 4,, π) = R ( 4,, π) = = 1 1 c) The coefficiens of diffusion associaed wih he roaional moion of a cylinder can be wrien as = and D =, where is he coefficien of fricion associaed wih roaion of he cylinder around i long axis, and is he coefficien of fricion associaed wih moion around an axis perpendicular o he long axis of he cylinder. a) D and D can be hough of as principal values of he diffusion ensor, represened in is principal axis sysem as he 3x3 marix D= D Using Caresian marix mehods, i.e. he roaion marix R and is inverse, calculae he elemens of he diffusion ensor in a frame roaed from he principal axis sysem by π π α =, β =, γ = π. 4 1 D 1 = D 1 D D = D