Major Concepts Kramers Turnover

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1 Major Concepts Kramers Turnover Low/Weak -Friction Limit (Energy Diffusion) Intermediate Regime bounded by TST High/Strong -Friction Limit Smoluchovski (Spatial Diffusion) Fokker-Planck Equation Probability Distribution Function (pdf) An EoM for the distribution, P(v;t) Derivation Smoluchovski Equation Large friction limit of Kramers Equation 6481 Lecture #3 Fokker Planck Equation

2 Transition State Theory Objective: Calculate reaction rates Obtain insight on reaction mechanism Eyring, Wigner, Others.. 1. Existence of Born-Oppenheimer V(x). Classical nuclear motions 3. No dynamical recrossings of TST Keck,Marcus,Miller,Truhlar, Others... Extend to phase space Variational Transition State Theory Formal reaction rate formulas Pechukas, Pollak... PODS -Dimensional non-recrossing DS Full-Dimensional Non-Recrossing Surfaces Miller, Hernandez developed good action-angle variables at the TS using CVPT/Lie PT to construct semiclassical rates Jaffé, Uzer, Wiggins, Berry, Others... extended to NHIM s, etc 6481 Lecture # Chemical Kinetics II (Marcus: Science 56 (199) 153) 3

3 Activated Dynamics: LE Langevin Equation: x = "# th x " $U(x) + % th (t) $x % th (t)% th ( t &) = k B T# th ' t " t & ( ) Identify a Reaction/Dynamic Variable (Order Parameter?) The bath coordinates are subsumed by the Friction and Random Force Kramers Turnover Rates Mel nikov-pollak-grabert-hänggi (PGH) Theory & Rates Shepherd and Hernandez; J. Chem. Phys. 117, (00). (variational MFPT) 6481 Lecture # Chemical Kinetics II 5

4 Kramers Turnover - Summary Three regimes: Low friction Rate proportional to friction 1 V 0 V exp V 0 V Intermediate friction TST applies, rate independent of friction TST rate an upper bound 1 TST = 0 exp V 0 V High friction (Smoluchovski) Rate proportional to inverse friction 1 = 0 exp V 0 V 6481 Lecture #3 Fokker Planck Equation 7

5 Pollak s Solution for the Rate at intermediate friction Let V(x) be determined at a min or saddle, i.e. first derivative is zero Include only nd -order term Rediagonalize the second order terms New effective one-dimensional barrier can be solved exactly by TST Obtains the Grote-Hynes result: k GH = r k 1dTST where r = E. Pollak, J. Chem. Phys. 85, 865 (1986) r +ˆ( r ) and k 1dTST = 0 e V 6481 Lecture # Chemical Kinetics II 8

6 Fokker-Planck Equation, I Recall: Langevin Equations: U(x) ẍ = ẋ x v = v + (t) Let (t, t) be the (stochastic) velocity increment: + (t) (t, t) = v(t) t + 1 M t t+ t (s)dt where (s) (s) M (s) is the (stochastic) force along the interval. Assuming that the stochastic force is Gaussian distributed ( ; v(t)) = M 0 t) 1/ We now construct the pdf at the end of Δt as a convolution over the pdf s at the beginning of the interval exp M ( + v(t) t) 0 t P (v, t + t) = d P (v,t) (,v ) 6481 Lecture #3 Fokker Planck Equation 10

7 Fokker-Planck Equation, II Recalling: We Taylor Expand in time and in the stochastic velocity increment The BC is that there is 0 current at equilibrium & Maxwell distribution is recovered Lecture #3 P (v, t + t) = d P (v,t) (,v ) P (v, t)+ t P t + = P (v, t) d ( ; v) v + 1 v d P (v, t) ( ; v)+ Inserting the Gaussian distribution, integrating over ξ, & cancelling Δt P t = 0 vp(v, t) + v v P (v, t) M Probability Current Drift/Convection Diffusive d P (v, t) ( ; v) Fokker Planck Equation 11

8 Fokker-Planck Equation, III So the Fokker-Planck Equation (for the field-free case) is: P t = What have we gained? An equation which we know how to solve No longer stochastic v vp(v, t) + Intrinsically doesn t make reference to illdefined trajectories. 0 v P (v, t) M 6481 Lecture #3 Fokker Planck Equation 13

9 High friction (Smoluchovski), I Recall: We note: the BC that any trajectories that reach product at say x > is removed, and hence the density there is 0. at steady state, the flux J is independent of position (as according to the continuity equation) Introducing: the EoM for flux implies: 6481 Lecture #3 (x, t) t J(x, t) = 1 F ext (x) M (x, t) x x = + (x) J(x, t) x =0 M J exp(v (x)/k BT ) M (x, t) exp(v (x)/ ) (x) Fokker Planck Equation 14

10 High friction (Smoluchovski), II Recall: x = The solution is the simple quadrature: (x) = M J M J exp(v (x)/k BT ) x x > The rate (or inverse escape time) is then given by the Flux over Population : 1 = J N = J x (x)dx exp(v (x )/ )dx So this gives us the solution of the problem for arbitrary potentials 6481 Lecture #3 Fokker Planck Equation 15

11 High friction (Smoluchovski), III We need to compute the flux integral: (x) = M J For x in the product region. But in this region, the integral is dominated by the maximum of the exponent. Use saddle-point method (real version of Stationary Phase Approximation) V (x) V 1 M (x x ) (x) = M J exp(v / ) x M Note that there is no x-dependence in the final result. x > exp(v (x )/ )dx Lecture #3 Fokker Planck Equation 16

12 High friction (Smoluchovski), IV We now need to compute the population, N: =3 x (x)dx = x exp( V (x)/ ) (x) Note:!(x) is a constant = (x > ) x exp( V (x)/ ) But in this region, the integral is dominated by the minimum of the exponent. Use Laplace s method (real version of Stationary Phase Approximation at a minimum) x (x)dx = M J exp(v / ) V (x) V M 0(x x 0 ) M = J 0 exp( [V 0 V ]/ ) 1 exp( V0 / ) The rate (according to flux-over-population) is: 1 = J N = 0 exp V 0 V M Lecture #3 Fokker Planck Equation 17