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《Simulations Moléculaires》Cours04IIIa

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Necessary Background of Statistical Physics (2) when the velocity of the center of mass is fixed to zeroKinetic energy.
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Necessary Background of Statistical Physics(2)

Necessary Background of Statistical Physics (2)

Temperature A(3M-2)-K(p n=3 when the velocity of the center of mass is fixed to zero Kinetic energy. Kp 2m

Temperature = = N i N p m K p i 1 2 2 1 ( )   Kinetic energy:   − = ( ) (3 ) 2 p N c K k N N T  Nc = 3 when the velocity of the center of mass is fixed to zero

Pressure Definition aF = oNe( n Working expression 6w

Pressure ( ) 3 1 1 r r N N i i i U N P     = = −     '( ) ( , ) 6 1 1 2 (2) r 1 r 2r1 2 r1 2 r r d d u N         = − ( )          =−  = V Z TV N V F T N T N P k T ln ( , , ) , , Definition: Working expression:

Chemical potential μ=(OF/ON)ry Widom method(test particle method): u=FN+I-FN=-kTIn(ZN+ZN=kTInA3-kTIn(QN+/QN) A=(2B 2/m)2-thermal wave length Q、FxF O n dp 'exp()) 长(exp(-0) where is the interaction potential between the N+l th particle with with all the others X=μ-=-kTln() Lid=kT In(p a)-chemical potential of the ideal gas Drawback: break down at high densities

Chemical potential m = (F/N)T,V Widom method (test particle method): m = FN+1 - FN = -kTln(ZN+1/ZN) = kTln3 - kTln(QN+1/QN)  = (2p 2 /m)1/2 - thermal wave length Drawback: break down at high densities! where is the interaction potential between the N+1 th particle with with all the others, = + = N i r N ri u 1 1 ( , )    exp( ) exp( ( )) exp( ( ( ) )) 1     −   = − − + = + N N N N N N N N N N V d d N V r U r r U r Q Q     m ex = m - m id = -kT ln(N) m id = kT ln( 3 ) - chemical potential of the ideal gas

Properties which can be determined from fluctuations Important remark Fluctuations are ensemble dependent! But averages are not e. g =0 in the microcanonical ensemble but is non zero in the canonical ensemble Heat capacit kT2Cv=nvT NVT NVT -<H(p q) 2 NVT Remark: In general, the numerical precision on fluctuations is poorer than that for the averages <△H(p、qN)N=<△U2N+<△K2Nvr <△K2 NVT 3N(kT)2/2

Properties which can be determined from fluctuations Important remark: Fluctuations are ensemble dependent! But averages are not. e.g., =0 in the microcanonical ensemble but is non zero in the canonical ensemble. Heat capacity kT2CV = NVT = NVT) 2 >NVT = NVT - NVT 2 Remark: In general, the numerical precision on fluctuations is poorer than that for the averages. NVT = NVT + NVT NVT = 3N(kT)2 /2

Isothermal compressibility pkTx=r aP kT(O HVT /OW)VTHVT

Isothermal compressibility kTχT = mVT/mVT ( ) T P T V V   =− 1  kT(mVT /m)V,T = mVT

Transport properties Self-diffusion coefficient Einstein formula 6Dt t→)∞ Expression in terms of velocity auto-correlation function(VACF) D-dt(v(ov(O) Diffusion equation opr d_DVdr) Solution with the initial condition, P(r, t=0=dr-ro (4mD)≈W 4D

Transport properties Self-diffusion coefficient Einstein formula: 6Dt t →  Expression in terms of velocity auto-correlation function (VACF):   =  0 D dt v(t)v(0)   Diffusion equation: ( , ) ( , ) 2 D r t t r t      =   Solution with the initial condition, (r, t=0) = d(r - r0 ): ) 4 exp( 1 ( , ) 2 0 3/2 (4 ) Dt r t r r Dt    − = − p 

Mean square displacement Velocity auto-correlation function <r(t)-r(O)2 ()(O) v(O)v(O)) Remark The method of mean square displacement provides a better numerical precision in the calculation of D

Mean square displacement Velocity auto-correlation function 0 t 1 (0) (0) ( ) (0) v v v t v       t Remark: The method of mean square displacement provides a better numerical precision in the calculation of D

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