Lecture 12 Reactions as Rare Events 3/02/2009 Notes by MIT Student (and MZB)
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1 Lecture 1 Reactions as Rare Events 3/0/009 Notes by MIT Student (and MZB) 1. Introduction In this lecture, we ll tackle the general problem of reaction rates involving particles jumping over barriers in energy landscapes. Net time, we will etend our analysis to tackle free energy landscapes. Along the way, we ll learn a useful and neat math tool called asymptotic epansion of integrals. Recall from the last lecture our epression for the mean first passage time, τ, for a particle trapped in an energy well defined by the (symmetric) energy landscape U(): αu ( ) αu ( y ) 1 A A τ = e e dyd D 0 In this epression, we had defined: U( ) E U ( ) =,α = E kt Finally, we note that A was the location of the barrier, and E was the energy barrier at that location (see Figure 1). Figure 1: Symmetric potential well We will now proceed in approimating the integral for the mean first passage time, τ. Note that for α<<1, corresponding to very small energy barriers, we could simply Taylor epand the eponentials in the above epression and integrate. However, we are interested in the case α>>1, corresponding to large energy barriers and rare hopping events. In this scenario, we will find that the epansion we use and the final result won t depend on many of the intermediate details of the U() function, but only on its etreme values. 1
2 Our strategy in this lecture will be as follows: 1. First describe the general technique of asymptotic epansion of integrals.. Then apply asymptotic epansion of integrals to evaluating the integral for mean first passage time.. Asymptotic Epansion of Integrals Note: More about this technique can be found in the notes, in and around Lecture 6, 005. Bender & Orszag also cover this technique. Consider the general integral: I(α ) = f (,α )d As the parameter α gets large or small, it may be the case that f(,α) is negligible ecept in only a certain region of the integral. Then, it would not be necessary to evaluate the entire integral; rather, a very good approimation of the integral could be determined by simply evaluating the important region. To get the flavor of how this could happen, let s draw an arbitrary function ψ(), shown below in Figure : Figure : The arbitrary function ψ(). Now, let s consider the function f(,α)to be e αψ(). We can start to think about what would happen if α became very large. First, any negative component in ψ() would become a very small positive number inside the eponential, and approach zero. Second, the positive components would start to scale, with the largest components stretching more than the smaller components. As a result of this stretching, the function f(,α) will be dominated by the region around its maimum. This is illustrated in Figure 3.
3 Figure 3: The function e αψ() for α=5 (top left), α=5 (top right), α=50 (bottom left), α=50 (bottom right). As we increase α, the function becomes dominated by the region around its maimum value at =8. Thus, taking an integral around this region provides a good approimation of the overall integral. Thus, what we have shown is that we can should be able to approimate the integral of the function f(,α) by taking an integral only around its maimum. More, formally, we epress: ma +ε αψ ( ) α I(α ) = f (,α )d e d + O(e ) ma ε where O(e α ) represents the order of the error in our approimation, and ε is a small distance. Net, we will proceed by approimating the function ψ() at its maimum via a Taylor series. Note that we will be eploiting the fact that at the maimum, the first derivative of ψ() is zero. Thus, as ma, or equivalently as ε 0, we have: C ma ψ ( ) ψ ma ( ) where C is the negative of the second derivative of ψ with respect to. Since we are at a maimum, C will be positive. We note that though this analysis assumes a smooth function, it can be generalized to unsmooth functions as well. In addition, we are free to carry over 3
4 more terms in the Taylor series for ψ() at its maimum. Let s in fact do this as we continue our progress towards evaluating the integral I(α). Using our Taylor epansion for ψ() (with added terms) along with our previous equation, we arrive at: ma +ε C C n n ma ε α[ ( ma ) + ( ma ) ] αψ n! ma n =3 I(α ) e e d Net, we will perform a transformation to make this integral look like a standard Gaussian integral, which we know how to integrate. To do this, we define the variable y and rewrite our equation for I(α) above: y ( ) αc ma I(α ) e e * e +ε αc C n αy y n ( ) ] αψ n! αc dy ma n =3 ac ε αc The first eponential in our integral is a standard Gaussian function. To rid ourselves of the second eponential, we again invoke a Taylor epansion. We can do this because the second eponential goes as α n/+1 for n>3, and we are considering the limit as α goes to infinity and thus the eponential going to zero. This gives us: +ε αc αψ y e ma C n αy n I(α ) e [1+ ( ) ]dy αc n=3 n! αc ε αc As final steps, we do three things: 1) Note that ε(αc ) 1/ goes to infinity for large α, because ε is finite (though small). This allows us to redefine the bounds of our integral as to +. y ) Invoke the standard Gaussian integral, e dy = π. The Appendi describes how to derive this integral. 3) Use only the first term of our Taylor epansion, grouping all the other terms into error bounds. This gives our final epression for I(α)! π αψ ma I(α ) e [1+ O( ) + O( ) + O( ) +...] 3/ αc α α α where the O(α) represent error bounds. Neglecting these errors, we arrive at: I(α ) αψ ma C π e α 4
5 Thus, we have shown what we have set out to prove. As we go to large values of α, the integral I(α) does not depend on the details of ψ(), but rather only its maimum value and the curvature at that maimum. In the Appendi, we show how this technique can be used to derive Stirling s approimation.. Asymptotic Epansion of Integrals Applied to Kramer s Escape Problem Let s use our new approimation technique to evaluate our integral for the mean first passage time: A A 1 αu ( ) αu ( y ) D 0 τ = e e dyd First, we evaluate the integral on the right hand side. The term in the eponential will peak at = A. The maimum of Ũ() is by definition equal to 1. We note that we are only evaluating half a Gaussian since we do not consider the other side of the energy hill, and thus will add a factor of ½ to our integral. This gives us: Rearranging: A 1 τ e D 0 π α U "( α U ( ) π τ D α U"( A A α e e ) α e d ) 1 αu ( ) A 0 All that is left now is to evaluate the remaining integral. We can use the same method as before to evaluate this integral, but since we looking at the negative of U(y) in the eponential, the peak will occur at =0. Again, we need to add a factor of ½ since we are only evaluating half a Gaussian. Putting this all together: Rearranging gives: α e τ D τ αd π α U "( A ) d 1 π ( ) α U "(0) πe α U "( A ) U "(0) Thus, we have finished our derivation for the mean passage time. Given the mean passage time, we can find an epression for the escape rate, R. We define R to be 1/τ. Note that although R needs to be scaled by ½ to account for the fact that a particle reaching the maimum can fall back into the well, it also needs to be scaled by a factor of because a particle in the minimum can escape either to the left or to the right. Thus these scaling factors cancel and R=1/τ, giving us: 5
6 e R α αd U"( ) U"( 0) π Net, let s try to get some sort of an intuition for the second derivative factors that occur in the epression for R above. We can in fact approimate these second derivatives by looking at the width of the energy landscape at a distance kt from the maimum and minimum of our potential, as shown in Figure 4: A Figure 4: The distance kt from the minimum and maimum, and corresponding lengths L 0 and L A. Using these definitions, we can approimate our curvature as: kt U "( A ) EL A U "( 0 ) kt EL 0 which allows us (along with our epression for α=e/kt) rewrite our epression for R as: E D R e kt π L 0 L A Finally, we note that our diffusivity D can be modeled by an atomic length scale times an atomic frequency: L h D τ 0 kt τ 0 where h is Planck s constant, and L represents an atomic spacing. 6
7 L L E R e kt πτ L L 0 0 A Let s see what we can learn from this analysis. First, we note that our prefactor won t affect the final rate very much, but a difference in E/kT will cause huge variations in rates due to its presence in the eponential. Net, we see the effect of a wide or shallow well. For eample, if we had a wide well with a large L 0 and large L A, our overall rate R would go down. This makes intuitive sense, as in this case much time would be wasted flittering around valleys and flat peaks in the energy landscape, rather than climbing the potential well. This lecture concludes by noting that transition state theory is very general, and not restricted to Kramer s escape problem. If we keep higher order terms in our escape rate epression, we d arrive at something of the form: kt kt R R 0[1+ β 1 + β ( ) +...] E E where R 0 is now the escape rate R we calculated previously. back into the Bulter Volmer equation. These corrections could be put Appendi 1. Integration of Gaussian To arrive at the integral I of the Gaussian: I e d We first square the integral: ( + y ) I e ddy And then transform to polar coordinates: π r I e rdrdθ 0 0 This integral is easily evaluated as π, and so we can find I to be π.. Stirling s Approimation Stirling s approimation can also be derived via the asymptotic epansion of integrals technique. We start with the integral definition of the factorial 1 : 1 Weisstein, Eric W. "Stirling's Approimation." From MathWorld A Wolfram Web Resource. 7
8 t n n! I(n) = e t dt 0 which can be rewritten as: n! I(n) = e 0 nψ (,n ) t dt where ψ = ln(t ) n The function ψ reaches its maimum at t=n. Thus: ψ ma = ln(n) 1 As we go to large n, we can use our definition of asymptotic epansion of integrals: I(n) π nc e n ψ ma I(n) π [ n ln( n ) n ] nc e π n I(n) e (n) nc n Net, C is the second derivative of ψ evaluated at n 1, or C =1/n. This gives: 1 n n n I(n) πne n = π e n n+ At this point, we are almost finished. We just need to take the log of both sides of the equation: ln(i(n)) = ln(n! ) (n + 1 )ln(n) n 1 ln(π ) As we go to large n, we recover Stirling s approimation, as we desired: ln(n! ) (n)ln(n ) n 8
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