forcing a polymer out of a potential well

Forcing a polymer out of a potential well

Keshav Kumar *, K.L. Sebastian

Department of Inorganic and Physical Chemistry, Indian Institute of Science, Bangalore 560012, India

Received 22 February 2002; in final form 16 April 2002

Abstract

We consider the forced escape of a polymer out of a potential well. The process has two steps – crossing the acti-

vation barrier, and translocation of the chain over the barrier. We suggest that translocation occurs by the motion of a

kink, whose shape and velocity change as more and more beads of the polymer have come out of the potential well. We

find that for short polymers, the translocation time ttrans � lnðn0Þ, where n0 is the number of beads that have crossed the

barrier. For long chains, the translocation time �N, the total number of beads. � 2002 Elsevier Science B.V. All rights

reserved.

1. Introduction

The activated escape of a particle from a meta-stable well is known as the Kramers problem [1,2]and has been studied in a large number of papers.It has been the subject of detailed reviews [3,4].Recently, we [5–7] considered the generalization ofthis problem to the case of polymers. We studiedthe escape of a polymer molecule, modelled as astring of beads, trapped in the less stable well of adouble well potential. The polymer was modelledas a Rouse chain, having N beads. We found thatthe polymer can go over to the more stable well bya kink mechanism. The kink is nothing but thedistortion of the polymer caused by the barrier,and it moves with a constant velocity as it is driven

by the free energy DV that a bead gains in goingfrom the left to the right of the double well (themotion is equivalent to that of a Brownian parti-cle, driven by a constant force). This would lead toa crossing time proportional to N, which is fasterthan was suggested by other mechanisms [8–12](see however, the paper [13]). The kink that wediscuss in this Letter is fundamentally differentfrom the kinks considered by Deutsch [14]. Deu-tsch’s kinks are hairpin structures, formed by partsof a long-chain molecule which are forced to moveinto the gap between two successive fibres of themedium in which the molecule is moving, whileour kink is a distortion in the chain over the bar-rier, which just makes that portion of the chainmore stretched out in comparison with a chainthat is free.

In an interesting paper, Bhattacharjee [15] hasstudied the statistical mechanics of a single chain,trapped in a potential well and subject to a force atone of its ends. He argues that the polymer can bepulled out only if the force exceeds a critical value.

13 June 2002

Chemical Physics Letters 359 (2002) 101–108

www.elsevier.com/locate/cplett

* Corresponding author. Permanent address: Department of

Chemistry, Indian Institute of Technology, Chennai 600 036,

India. Fax: +80-331-6552.

E-mail address: [email protected] (K.L. Sebastian).

0009-2614/02/$ - see front matter � 2002 Elsevier Science B.V. All rights reserved.

PII: S0009-2614 (02 )00652-8

mail to: [email protected]

We have investigated the dynamics [16] of the es-cape of a chain from a potential well as a result ofapplying a force at one of its ends.

In this Letter, we consider a problem that issimilar to the above two problems, but is quiteinteresting in its own right. We imagine that apolymer is initially trapped in a potential well. Aconstant external field is applied, thus forcing thechain to move out of the potential well. The ap-plied field is assumed to be small enough that thepolymer still has a barrier to overcome (see Fig. 1).Unlike the double well case, the free energy gain inescaping from the meta-stable well is not finite, butin principle, very large. So we ask, can one havethe kink mechanism operating in this case? Fur-ther, how does the time of escape depend on thenumber of beads and the external field? An ex-periment to verify the results of the paper shouldbe possible, as is obvious from the recent, veryinteresting experiments of Kasianowicz et al. [17]and Han et al. [18]. Further, the problem has ob-vious relevance to electrophoresis using nano-structures (see [18]).

The first step in the escape is the overcoming ofthe activation barrier. For this, the system has to goover the saddle point on the free energy hyper-surface. For the double well case, Sebastian andPaul [7] have analyzed this problem and arguedthat the escape can occur by two mechanisms: theend-crossing and the hairpin crossing. These cor-respond to overcoming the barrier by a few beads,that are either at an end of the chain, or away from

the two ends. In both cases, activation energy Ea iseasily calculated. For sufficiently long chains, it hasno dependence onN. The second step in the processis the movement of the beads across the barrier. Weshall here refer to this process as translocation andthe time taken for this will be denoted by ttrans.Sebastian and Paul [7] argue that for the doublewell, this time is proportional to N. In the presentcase too, we argue that if the polymer is sufficientlylong, then the time is proportional to N. The keypoint of our argument is that for a very long chainmolecule, the polymer configuration changes withtime, until finally it settles down to a steady state,where the polymer crosses over with a steady ve-locity from one side to the other. If the chain issufficiently long that it settles down to this state bythe time a fraction of its length has crossed over,then the crossing time is proportional to N. Oth-erwise, the crossing time would have a weaker de-pendence on N (see below). In this case too, it ispossible for one to think of end and hairpincrossings, though in the following we consider onlythe end crossing. It is possible to analyze the hair-pin crossing in a similar fashion.

2. The model

The external force converts the potential wellinto a meta-stable well as shown in Fig. 1. Wedescribe the dynamics of the escape process, usingthe one-dimensional Rouse model, governed bythe equation

foRðn; tÞ

ot¼ m

o2Rðn; tÞon2

� V 0ðRðn; tÞÞ þ f ðn; tÞ: ð1Þ

In the above, f is a friction coefficient for the nthsegment. The term mðo2Rðn; tÞ=on2Þ comes fromthe fact that stretching the chain can lower itsentropy and hence increase its free energy. Con-sequently, the parameter m is temperature depen-dent and is equal to 3kBT=l2 (see [19, Eq. (4.5)].They use the symbol k for the quantity that we callm). As the ends of the string are free, the boundaryconditions to be satisfied are

oRðn; tÞon

� �n¼0

¼ oRðn; tÞon

� �n¼N

¼ 0:

Fig. 1. The meta-stable potential well that results on the ap-

plication of an external field. R ¼ 0 is the location of the

maximum of the potential. The beads show the transition state.

102 K. Kumar, K.L. Sebastian / Chemical Physics Letters 359 (2002) 101–108

V ðRÞ is the free energy of a bead located at theposition R. f ðn; tÞ are random forces acting on thenth bead and have the correlation functionhf ðn; tÞf ðn1; t1Þi ¼ 2fkBTdðn� n1Þdðt � t1Þ (see [19,Eq. (4.12)]). The deterministic part of Eq. (1),which will play a key role in our analysis, is ob-tained by neglecting the random noise term. It is

foRðn; tÞ

ot¼ m

o2Rðn; tÞon2

� V 0ðRðn; tÞÞ: ð2Þ

This may also be written as

foRðn; tÞ

ot¼ � dE½Rðn; tÞ�

dRðn; tÞ ; ð3Þ

where E½Rðn; tÞ� is the free energy functional for thechain given by

E½Rðn; tÞ� ¼Z N

0

dnm2

oRðn; tÞon

� �2"

þ V ðRðn; tÞÞ#: ð4Þ

We take the potential to have the formV ðRÞ ¼ UðRÞ � ER, where UðRÞ is assumed to bethe original potential well which traps the polymer.E is the external field, assumed to be independentof position R. We take the minimum of V ðRÞ to beat R ¼ �a0 (see Fig. 1) and the maximum at R ¼ 0.It is convenient to take V ð�a0Þ ¼ 0. The value ofR > 0 such that V ðRÞ ¼ 0 is denoted by Rf . In thefollowing, we shall assume that beyond this point,the potential can be well approximated by just thelinear term, �ER.

2.1. The saddle point and the form of the free energyhyper-surface

The saddle point that is to be crossed to escapefrom the meta-stable well is an extremum of the freeenergy functional in Eq. (4) and may be found bysolving the equation dE½R�=dRðnÞ ¼ 0, which gives

md2Rdn2

¼ V 0ðRÞ: ð5Þ

This has to be solved, subject to the boundaryconditions. It is clear from the analysis of Sebas-tian and Paul [7] that the saddle point (transitionstate) is obtained by solving Newton’s equation (5)for a (fictitious) particle of mass m in the upsidedown potential �V ðRÞ and that the particle has tostart at �a0 and end at Rf , and has to have total

energy equal to zero. Using this solution, one canestimate the activation energy for the crossing. Asthis problem has been discussed in detail [5,7], wedo not discuss it here.

Once the barrier is crossed by the first fewbeads, the remaining beads have to cross it. This iswhat we refer to as translocation and in the case ofthe double well potential, this happens by the kinkmechanism. The kink is the distortion of the chainby the barrier and exists in the segments of thepolymer stretched across the barrier. As beadscross the barrier, the kink moves on the polymer inthe backward direction. As the driving force forthe motion of the kink is finite, being equal to DV ,the kink moves with a steady velocity and for avery long polymer, the translocation occurs in atime ttrans proportional to the number of beads N.Mathematically, this means that Eq. (2) has a so-lution of the form Rsðn� vtÞ. In comparison, in thepresent case, there is only one (metastable) welland the beads that move out of the well experiencea potential that keeps on decreasing as they moveforward (V ðRÞ ! �1 as R ! 1). The drivingforce for the motion of the kink would depend onthe number of beads that have crossed and thevelocity with which they move. (In fact one can doa simple estimate and show that steepest descentalong the free energy hyper-surface leads to adecrease in the free energy of the form� �E2n3

0=ð6mÞ, where n0 is the number of beadsthat have crossed the point R ¼ Rf (see Fig. 2).)

Fig. 2. The motion of the chain after the transition state has

been crossed. n0 denotes the number of beads that has passed

the point R ¼ Rf and these exert a force F on the bead at

R ¼ Rf .

K. Kumar, K.L. Sebastian / Chemical Physics Letters 359 (2002) 101–108 103

This means that one does not have a solution ofEq. (2) which is strictly of the form Rsðn� vtÞ andone should be looking for more general solutionsof the partial differential equation (2). As this is anonlinear partial differential equation, general so-lutions are very hard to find and hence we adoptthe following approximate analysis for the prob-lem.

3. The approximate method

3.1. The pulled kink

In this section, we present our analysis of theproblem, which makes use of an adiabatic, steady-state approximation. Our analysis is based on thefollowing physical arguments. The transition stateis shown in Fig. 1. As more and more beads cross,the polymer, on an average, takes up a configu-ration similar to the one shown in Fig. 2. Thenumber of beads that are to the right of the pointR ¼ Rf is denoted by n0. These beads are beingpulled by the linearly decreasing potential thatexists in the region R > Rf . The beads, in turnexert a force on the bead at Rf . As n0 changes, thisforce also changes and therefore, the shape andvelocity of movement of the distortion (the kink)of the polymer in the barrier region also changes.The force that the n0 beads exert on the bead at Rf

is

moRðn; tÞ

on

� �R¼Rf

and we imagine that the kink is able to followadiabatically the change in this force. So we put

F ¼ moRðn; tÞ

on

� �R¼Rf

and assume it to be a constant for the calculation ofthe velocity of the kink. With F constant, Eq. (2)has, in the region R < Rf , a steady state solution ofthe form Rsðn� vtÞ, where v is the velocity withwhich the distortion moves on the chain, and isdetermined by the value of F. We now find thevelocity v approximately, as function of the force F.Substituting Rðn; tÞ ¼ Rsðn� vtÞ into Eq. (2), we get

md2Rs

ds2þ vf

dRs

ds¼ V 0ðRsÞ for Rs < Rf : ð6Þ

In the above we have denoted s ¼ n� vt. If oneimagines s as time, then this is a simple Newtonianequation for the motion of particle of mass m,moving in the upside down potential �V ðRÞ sub-ject to friction. vf=m is the coefficient of friction.The value of s for which RsðsÞ ¼ Rf shall be de-noted by s1. The boundary conditions are

dRsðsÞds

� �s¼�1

¼ 0 anddRsðsÞ

ds

� �s¼s1

¼ F =m:

We can find a solution, satisfying these conditionseasily by first finding a function X ðsÞ, having thefollowing properties: It is the position of a particlewhich follows Eq. (6). As s ! �1, X ðsÞ ! �a0

and the velocity ðdX ðsÞ=dsÞ ! 0. At the time s ¼ 0the particle reaches the point Rf with the velocityðdX ðsÞ=dsÞs¼0 ¼ F =m. It may appear that there aretoo many conditions on X ðsÞ to satisfy. However,v too is an undetermined parameter, and its valueis uniquely fixed by these conditions. RsðsÞ is re-lated to X ðsÞ by RsðsÞ ¼ X ðs � s1Þ. In Fig. 3, weshow this solution X ðsÞ in a plot of the invertedpotential �V ðRÞ in which the particle moves. Thedotted line shows how the total energy of a particlefollowing the path X ðsÞ would change. The parti-cle has an energy zero at s ¼ �1 and an energyF 2=ð2mÞ at s ¼ 0. This means that its total energyincreases, implying that the friction coefficientvf=m for its motion is negative. Thus v is negative –

Fig. 3. The upside down potential �V ðRÞ which determines the

kink. The dotted curve indicates the kink.


that is, the distortion moves backward on thepolymer. Even though it is not possible to solveEq. (6) to find X ðsÞ, it is possible to get an ex-pression for v. On multiplying the equation

md2Xds2

þ vfdXds

¼ V 0ðX Þ

by ðdX=dsÞds and integrating from �1 to 0 andrearranging, we get

v ¼ �F 2

2mfR 0

�1 ds dXds

2: ð7Þ

Now using X0 to denote the solution of the equa-tion mðd2X0=ds2Þ ¼ V 0ðX0Þ, having the energy zeroand satisfying the boundary conditionsX0ð�1Þ ¼ 0 and X0ð0Þ ¼ Rf , we can approximatethe integral in Eq. (7) as follows:Z 0

�1ds

dXds

� �2

¼Z 0

�1ds

dX0

ds

� �2

þZ 0

�1ds

dXds

� �2(

� dX0

ds

� �2):

The integralR 0

�1 dsðdX0=dsÞ2can be evaluated

exactly and is found to be equal toR Rf

�a0dR

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2mV ðRÞ

p. Now,Z 0

�1ds

dXds

� �2(

� dX0

ds

� �2)

¼Z Rf

�a0

dXds

� �dX

�� dX0

ds

� �dX0

�:

In this expression, the integration extends over theinterval ð�a0;Rf Þ and the integrand has the values0 and F =m at the two boundaries. Hence, we canapproximate this integral by ða0 þ Rf ÞF =ð2mÞ.Using this in Eq. (7), we get

v ’ �F 2

2fðmC þ FaÞ ; ð8Þ

where C ¼R Rf

�a0dR

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2mV ðRÞ

pand a ¼ ða0 þ Rf Þ=2.

3.2. The force F

Having found the velocity of movement of thekink, as a function of the force F, we proceed tofind an approximate expression for the force F.

For this, we think of the n0 beads that havecrossed the point Rf . These beads are in the linearpart of the potential (see Fig. 2) and each one ofthem is subject to a force equal to E. So, for theportion of the chain having R > Rf ; Eq. (2) maybe written as

foRðn; tÞ

ot¼ m

o2Rðn; tÞon2

þ E: ð9Þ

As the kink is moving backward with a velocity v,the chain is moving forward with the velocity �v.Hence the beads are coming out from the well witha velocity �v, which means that on an average, then0 beads move forward with the same velocity.Now think of a chain of just n0 units subject to thelinear potential �ER. What is the force that weshould apply at one end so that the beads would,on an average move with a steady velocity �v? Wenow find this force. If they are moving steadily,then for these n0 units, Rðn; tÞ ¼ �vt þ gðnÞ; wheregðnÞ is to be determined. Using this in Eq. (9), wefind that

md2gðnÞ

dn2¼ �fv� E: ð10Þ

As the free end of the chain has no force acting onit, oRðn; tÞ=on ¼ 0 at this end, which meansdgðnÞ=dn ¼ 0 at this end. Using this, and inte-grating Eq. (10), we get

mdgðnÞ

dn

� �n¼n0

¼ �ðfvþ EÞn0;

mdgðnÞ

dn

� �n¼n0

¼ mdRðn; tÞ

dn

� �n¼n0

is the force that has to be applied at one end tokeep the n0 units in a steady state of motion. Now,even though we have found this force, assuming asteady state for a chain of just n0 units, we assumethat these results are applicable to the n0 units thathave come out of the well and crossed the point Rf .Thus, we assume that for the polymer escapingfrom the meta-stable well, the bead at Rf experi-ences a force

F ¼ �ðfvþ EÞn0 ð11Þfrom the n0 units that have crossed this point.


3.3. The time of translocation ttrans

Now we find the velocity (number of beads perunit time) of crossing the barrier, using Eq. (11) inEq. (8). We get

v ¼ �ðfvþ EÞ2n20

2f mC � aðfvþ EÞn0f g : ð12Þ

This can be exactly solved for v. The result is

v ¼�� Cmþ En0ða� n0Þ

þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiC2m2 þ a2E2n2

0 þ 2CEmn0ðn0 � aÞq �

�ðn0 � 2aÞn0f: ð13Þ

There are two solutions, of which we have chosenthe physically acceptable one, for which the mag-nitude of the velocity tends to zero as n0 ! 0. Forsmall n0, �v changes linearly with n0. As n0 in-creases, �v too increases and for large values of n0

it approaches �v ¼ E=f. Hence, for a sufficientlylong polymer, the time required for all the beads totranslocate is proportional to N=E. For not solong a polymer, the regime where velocity changeslinearly with n0 would be applicable and this im-plies that the time of translocation � lnðn0Þ. Onemay also ask: what would be the effect of the noiseterms? The answer is straightforward. The motionof the chain in the forward direction is equivalentto that of a Brownian particle, driven by an ex-ternal field. Such a particle would, on an averagemove with a velocity determined by the externalfield and the friction coefficient. Our conclusionsabove for the behavior of the chain on an average,would not change.

3.4. The exact solution for a Morse type potential

The basic idea of the above analysis is that asmore and more beads cross, the velocity of move-ment of the kink changes and in the limit n0 ! 1,it finally attains a steady-state velocity. Hence, inthis limit, the crossing time is proportional to thenumber of beads in the polymer. The existence ofsuch a steady-state solution can be easily demon-strated for the case where the trapping is due to a

Morse like potential. We take V ðRÞ ¼ �Ae�aRþBe�2aR � ER, where E is the externally applied field,making it possible for the polymer to escape fromthe Morse potential in which it is trapped. Eq. (2)has the following exact solution:

Rðn; tÞ ¼ 1

alnðeanþbt þ cÞ ð14Þ

with

a ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia aA2 � 4BEþ ffiffiffi

ap

AffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiaA2 � 8BE

p� �4Bm

vuut; ð15Þ

b ¼ aE=f ð16Þand

c ¼ 4ffiffiffia

pBffiffiffi

ap

AþffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiaA2 � 8BE

p : ð17Þ

Note that in the limit n ! 1, Rðn; tÞ ! ðanþ btÞ=aand this means that the chain molecule is movingforward with constant velocity b=a ¼ E=f.

3.5. Simulations

To verify the conclusions of the above analysis,we have performed computer simulations for adiscrete bead model. For simulations, it is conve-nient to make use of the reduced variablesR� ¼ R=l, t� ¼ mt=f and V �ðR�Þ ¼ V ðRÞ=ml2 andconvert Eq. (1) to a finite difference scheme as

R�nðt� þ dt�Þ ¼ R�

nðt�Þ � dt 2R�nðt�Þ

��R�

nþ1ðt�Þ �R�n�1ðt�Þ

�� V �0 ðR�

nÞdt� þ ðdrGn Þ: ð18Þ

R�nþ1ðt�Þ denotes the position of the nth bead at the

time t�. drGn is a random noise, having mean zeroand variance hdrGn drG

n0Þi ¼ ð2dt=3Þdnn0 . drGn at dif-

ferent time steps are uncorrelated. V ðR�nÞ was taken

to be a quartic potential joined smoothly with alinear potential so that the potential had the shapein Fig. 1. We took

V �ðR�Þ ¼ k6ðR� þ a0Þ2

�ð3R�2 � 2R�a0 � 4R�a1 þ a20 þ 2a0a1Þ

for R� < R�f . This potential has a maximum at

R� ¼ 0 and a meta-stable minimum at R� ¼ �a0.


R�f is a point to the right of the maximum at R� ¼ 0

such that V ðR�f Þ ¼ 0 and R�

f > 0. For R� > R�f ,

V �ðR�Þ is taken to be linear in R�, with a slopeadjusted such that the potential has continuousfirst derivative.

As simulating the activated crossing of thebarrier needs long simulations, we decided on thefollowing strategy. A given bead, say the mth (saym ¼ 50) from one end (which we shall refer to asA) was kept fixed on top of the barrier. The end Awas put to the right of the barrier and the otherportions of the chain were spread around theminimum of the potential. Then, with the mthbead constrained to be kept fixed at the maximum,the system was evolved in time for 5000 steps (onestep has dt ¼ 0:01) and the average position of theend A is noticed. Then, the bead number m waschanged to a new value such that the bead at theend A has an average potential energy equal tozero. Then we start with this arrangement, andevolve the system for 5000 steps and take that asthe starting point for the simulation of the move-ment of the remaining parts of the chain across thebarrier. Now we relax the constraint on the mthbead and allow all the beads to move according toEq. (18). The time that ðmþ n0Þth (counted fromend A) took to cross the top of the barrier wasaveraged over 100 runs and plotted against n0. Atypical result is shown in Fig. 4.

The simulations have been carried out for avariety of values of the parameters and the resultsshowed the crossing time to be proportional to N,in agreement with the analysis above. As an ex-ample, we give in Fig. 5 the results for a simulationwith k ¼ 0:0031, a0 ¼ 15, a1 ¼ 15:3. In this case,the barrier height is 1.0. Note that for larger valuesof n, the plot is linear, while for lower values of n,it is not. A crossing time verses logðn0Þ plot notgiven here, to save space, demonstrated lnðnÞ de-pendence for initial crossing times.

4. Conclusions

We have analyzed the forced escape of a poly-mer from a potential well, using the Rouse modelfor the dynamics of the polymer. The motion ofthe chain on an average is studied by analyzing thedeterministic part of the equations for the Rousemodel. As it is not possible to solve the equation,we have performed an approximate analysis of theproblem. We suggest that the kink mechanism[5,7], proposed earlier for the polymer in a doublewell potential, can operate in this case too. How-ever, unlike in that case, the velocity and shape ofthe kink would change as more and more beadscome out of the well. The reason for the change isthe pulling force that the beads that have come outwould exert on the kink. We find that because of

Fig. 4. Results of the simulation a0 ¼ 3, a1 ¼ 5, k ¼ 0:05 and

barrier height¼ 2.925. Total number of beads in the chain was

taken to be 10 000.

Fig. 5. Results of the simulation: expanded view, which shows

the crossing of only the beads at the beginning.


this, the kink velocity changes and would tend tothe value of �E=f, as more and more beads comeout. If the chain is very long, the rate of comingout would be determined by this velocity andhence ttrans � N . For shorter chains, the velocitywould be linearly dependent on the number ofbeads that have come out and hence ttrans � lnðn0Þ.These results have been verified by computersimulations.

Acknowledgements

K.K. thanks Alok Paul and S.C. Prasanna fortheir help. The simulations were carried out byAlok Paul.

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forcing a polymer out of a potential well

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