How a Nonequilibrium Bath and a Potential Well Lead to Broken Time-Reversal Symmetry - First Order Corrections on Fluctuation-Dissipation Relations

: The noise that is associated with nonequilibrium processes commonly features more 1 outliers and is therefore often taken to be Lévy noise. For a Langevin particle that is subjected to 2 Lévy noise, the kicksizes are drawn not from a Gaussian distribution, but from an α -stable distribu-3 tion. For a Gaussian-noise-subjected particle in a potential well, microscopic reversibility applies. 4 But it appears that the time-reversal-symmetry is broken for a Lévy-noise-subjected particle in a 5 potential well. Major obstacles in the analysis of Langevin equations with Lévy noise are the lack 6 of simple analytic formulae and the inﬁnite variance of the α -stable distribution. We propose a 7 measure for the violation of time-reversal-symmetry and we present a procedure in which this 8 measure is central to a controlled imposing of time-reversal-asymmetry. The procedure leads to 9 behavior that mimics much of the effects of Lévy noise. Our imposing of such nonequilibrium 10 leads to concise analytic formulae and does not yield any divergent variances. Most importantly, 11 the theory leads to simple corrections on the Fluctuation-Dissipation Relation. 12


Introduction
The Fluctuation-Dissipation Relation (FDR) is as simple as it is profound [1].For a particle in a fluid we have D/(k B T) = µ. ( On the left hand side of this equation, D is the diffusion coefficient of the particle, k B 16 is Boltzmann's constant, and T is the absolute temperature.The product k B T has the 17 dimension of energy and it is the characteristic "quantum" of Brownian motion.On the 18 right hand side µ denotes the mobility, i.e., µ = v/F where v is the average speed of the 19 particle in the fluid when it is subjected to a force F.

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On the molecular level there are other manifestations of the FDR.Take a resistor with leads to an infinite standard deviation and is also behind the more frequent occurrence 77 of outliers [18,19].Consider a particle in a potential V(x) that is subjected to α-stable noise.We have for the Langevin Equation for the position x(t) of an overdamped particle: In the course of a simulation with timesteps of length ∆t, a noise contribution ξ(t i ) at the i-th step is taken as ξ α (t i ) = θ α,i ∆t 1/α−1 , where θ α,i is the i-th random number in the sequence of steps.The random numbers θ i,α are drawn from an α-stable distribution with a scale factor of one.They are independent and identically distributed (IID).The 1/α in the exponent of ∆t guarantees that the diffused distance scales correctly if different ∆t's are taken.Equation ( 3) furthermore shows how the scale factor is effectively an amplitude.For Gaussian noise, i.e. α = 2, the scale factor c is the usual √ 2D, where D is the diffusion coefficient.The variable β denotes the coefficient of friction.The coefficient of friction is the inverse of the aforementioned mobility µ, i.e. β = 1/µ.The left-and right-hand-side of Eq. ( 3) have the dimension of force.We consider a small segment of a trajectory going from (t i , x i ) to (t i + ∆t, x i + ∆x i ).Multiplying with ∆x i , we obtain the involved energies: Here the term on the left-hand-side indicates the amount of energy that is "dissipated out" in time ∆t.The term in square brackets on the right-hand-side denotes the net force on the particle.The net force is made up of the force due to the potential, F(x i ) = −dV(x i )/dx, and the force due to agitation by the bath.Multiplying the net force by the distance ∆x i over which the force is applied, we obtain the work done on the particle over the segment x i , substituting ∆x i = [F(x i )/β + cξ α (t i )]∆t on the right-hand-side, and using again ξ α (t i ) = θ α,i ∆t 1/α−1 , we find: This equation again describes the energy traffic for the particle in a time interval ∆t.

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On the left-hand-side the first term is the dissipated energy and the second term is the 81 energy associated with going up or down in the potential.The twofold expression on the 82 right-hand-side describes the energy that is "fluctuated into the particle."The second 83 term would be the only contribution in case of a flat potential.The first term accounts 84 for the interplay between the random kicks, ξ α (t i ), and the deterministic F(x i ).If the 85 Brownian kick and the deterministic force are in the opposite direction and such that 86 they balance each other out, then the right-hand-side terms in Eq. (4) will add up to zero 87 and the particle will not move.

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For the case of α = 2, Eq. ( 4) readily reduces to something more familiar.On the left-hand-side, the existence of a basin of attraction implies that the changes V(x)| x i+1 x i will ultimately average to zero.Furthermore, we have θ 2,i = 0 and F(x i ) = 0 in this case.As the kicksizes θ 2,i are not correlated to F(x i ), we also have F(x i )θ 2,i = 0 on the right-hand-side.The Gaussian case, α = 2, also leads to θ 2 2,i = 1.With c 2 = 2D and invoking the FDR, βD = k B T, we then obtain: This equation tells us that the long time average of the energy that is fluctuated into 89 the particle is 2k B T per timestep.Note that in the continuum limit, ∆t → 0, an infinite 90 amount of energy flows through each particle in the system in any finite amount of time.

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The Langevin Equation, however, is an abstraction.As was mentioned before, in actual reality there is about a picosecond between subsequent collisions of an individual water 93 molecule and this puts a lower limit on the value of ∆t.Equation ( 5) has proven fruitful 94 in the analysis of Brownian ratchets [20,21].

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A lot more infinities start accumulating once we make α < 2. For 0 < α < 2, the 96 variance θ 2 α,i diverges.Effectively this means that the bath has no temperature.No

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FDR can be formulated in this case.For 0 < α ≤ 1, θ α,i is zero-centered, but does 98 not converge.This leads to further problems in working with Eq. ( 4).However, most 99 real-life instances of Lévy noise involve values of α that are between 1 and 2 [12][13][14][15][16][17]. 100 Below we propose a way to "fix" the θ 2 α,i -divergence for small deviations from 101 equilibrium.The idea is inspired by the violation of time-reversal-symmetry that Lévy 102 noise causes for a particle in a potential well.We will derive some simple formulae that 103 we will check against the results of numerical simulation.With Gaussian noise, the most likely upslide and most likely downslide both take n 129 steps.In Ref. [22] it is shown rigorously that in case of Lévy noise, the most likely upslide  For a potential with curvature, there is a problem when simulating the motion of 141 a particle that is subjected to Lévy noise.Imagine that the particle in Fig. 1a is to the 142 left of x = 0 and imagine next that it receives a large Lévy kick to the right.In a simple

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Euler scheme with a timestep ∆t, we would have the deterministic force on the particle, 144 F(x) = −dV(x)/dx, be in the positive direction for the entire interval, even when the 145 particle is actually "climbing" the potential on the part of the potential where x > 0.
146 For a parabolic potential the solution to this "large kick problem" is straightforward. 147 Through scaling of t and x, the Langevin Equation β ẋ = −Ax + βcξ α (t) can be brought 148 into a form ẋ = −λx + ξ α (t) [23].We take λ = 1.Upon discretization of ẋ = −x + ξ α (t), we let the kick at t = t i have a value ξ α (t i ) = K.We then take the curvature of the 153 After taking the data from a computer simulation (with a time interval ∆t) or from a real life system (sampled at a time interval ∆t), we can express the deviation from microscopic reversibility as follows: where ϕ desc and ϕ asc are the fraction of descending steps and the fraction of ascending For Lévy noise there is a power-law-tail and a divergent variance for any α = 2 − ε, where ε is small and positive.For α = 2 the Gaussian is recovered.Nevertheless, the 165 convergence to r = 0 as α approaches 2 appears smooth.

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From Fig. 1c it is also obvious that there is a strong dependence on the timestep ∆t.

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It is for the green curve in Fig. 1c, i.e. ∆t = 0.1, that we get the fastest departure from 168 r = 0.This apparent optimum is not hard to understand.In a parabola V(x) = 1 2 Ax 2 the 169 deterministic downslide starting at In other words, there is a characteristic time t char = β/A for the downslide.If ∆t t char , 171 then the particle will generally be back in the basin of attraction after one timestep.

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This leads to the noise-term being dominant and the contribution due to the slope being 175 negligible.What this means in practice for the downslide is that it will take a very 176 large number of steps, N tot , to get back to the basin of attraction after a Lévy jump.A 177 number N asc ≈ N tot /2 of these steps will be ascending and a number N desc ≈ N tot /2 178 will be descending.The difference N desc − N asc will be small relative to N tot and will 179 lead to r = (N desc − N asc )/N tot → 0. In between ∆t → 0 + and ∆t t char there will be 180 a maximum for the value of r. Figure 1c shows that with a timestep, ∆t, that is about 181 one tenth of the characteristic time, t char , an optimal resolution of the nonequilibrium 182 features is obtained.The parameter r for a Lévy particle in a harmonic potential is further 183 explored in Ref. [23].
"... when all the fast things have happened, but the slow things have not."That is how 185 Richard Feynman once described equilibrium [24].The observations in the previous 186 paragraph on the Lévy particle's relaxation to the basin of attraction put an interesting 187 take on Feynman's premise for the case of our nonequilibrium system.If we take 188 ∆t t char we are indeed not sampling sufficiently fast to see the relaxation happen and 189 we are then looking at the r = 0 that characterizes equilibrium.But if we take ∆t t char , we are sampling too fast and also do not see the relaxation happen as sampling too fast 191 likewise leads to r = 0, i.e. the equilibrium result.So "... when the fast things happen, but we are sampling too fast to see it" would also describe equilibrium.

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Consider a small interval ∆x 0 on the x-axis.With a coefficient of friction β, the energy that is dissipated if ∆x 0 is crossed at a speed v is E = βv(∆x 0 ).Over a long time interval, ∆x 0 is traversed equally often in both directions.Let v 0 be the speed on the downslide and let v 0 (1 + δ) be the speed on the upslide.Close-to-equilibrium means that δ is sufficiently small to justify a first order approximation.With the upslide speed corrected by a multiplicative factor (1 + δ), the number of ascending steps gets a multiplicative factor (1 − δ) relative to the number of descending steps.This leads to: If ∆x 0 is traversed first at an upslide-speed v 0 (1 + δ) and next, on the way back to the basin of attraction, at a speed v 0 , then the dissipated energy is: The energy E eq diss = 2βv 0 (∆x 0 ) is what would be dissipated if, in case of microscopic reversibility, we traverse the two directions with the same speed v 0 .The higher speeds on the upslides, i.e. the violation of microscopic reversibility, lead to the (1 + r) correction factor: For a corrected Energy-FDR we thus find: This equation can be seen as an adjusted form of Eq. ( 5) for the case of a small violation

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In principle, temperature is a characteristic of a system that is at thermodynamic equilibrium.Nevertheless, even in a nonequilibrium setup, we can associate the temperature with the average kinetic energy of the particles, i.e., E kin = 1  2 k B T in case of a 1D system.Let upslide and downslide cover the same distance ∆x 0 .In that case the upslide speed, v up = v 0 (1 + δ), is held for a shorter time than the downslide speed v 0 .This will lead to the average actually being smaller than v 0 (1 + 1 2 δ).However, this is a second order effect in δ (see the short derivation in the Appendix).At first order we thus have: If all of the involved particles have the same mass m and taking the average of the square as the square of the average, we come to a lowest order approximation: E kin = 1 2 mv 2 avg ≈ 1 2 mv 2 0 (1 + δ).With δ = 2r, this leads to The Kubo relation expresses the diffusion coefficient D as the time correlation of the velocity: D = ∞ 0 v(t)v(0) dt [25].With the insights developed in this paragraph we find that this leads to D noneq = D eq (1 + 2r).( 12) Equations ( 11) and ( 12 In this section we view the violation of time-reversal symmetry from a different 211 perspective.We rederive Eqs. ( 11) and ( 12) and check the theoretical results with a 212 stochastic simulation.

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Consider Fig. 2. We take two points, P 1 and P 2 , on the parabolic potential of Fig. 1a.

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The point P 1 is in the basin of attraction and the distance between the points is such that 215 it takes more than one timestep to cover the trajectory.Say it would take n steps to go The basic setup is that of a Euler scheme following a Brownian particle in a parabolic potential V(x) = x 2 /2.We take β = 1 and D = 1 (see text).The timesteps have a length of ∆t = 0.01.The breaking of microscopic reversibility consists in an augmentation of climbing steps that occur at more than 1.87 standard deviations away from x = 0.The parameter δ, which is varied from 0.01 to 0.2 in steps of 0.01, represents the fraction by which a climbing step is made longer.We let σ represent the standard deviation of the augmented Gaussian.The graph shows how the simulations bear out the theoretically derived σ = (1 + δ/2).

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Extension of the FDR to nonequilibrium is a challenge that has been taken up by 269 many and different approaches have been tried [10,30,31].Our starting point is a Lévy-

21 a 26 I 2 ,
conductance G.With no net current flowing, there is still a fluctuating voltage between 22 the two ends of the resistor due to the Brownian motion of the charge carriers (generally 23 electrons) inside the resistor.Conductance can be viewed as the electrical equivalent 24 of mobility, i.e.G = I/V with V being the voltage and I being the current.With this 25 realization it is not hard to understand that the mean square of the fluctuating current, is related to the conductance G through k B T. The relation I 2 = 4k B TG(∆ f ),

78 2 .
Overdamped Particle in a Potential Well Subjected to IID α-stable Noise 79

104 3 . 1 2
Nonequilibrium and Time-Reversal Asymmetry in a Parabolic Well 105 If V(x) is a flat potential, then Eq. (3) describes a simple 1D random walk.Both 106 for α = 2 and α = 2, we then have time-reversal symmetry.If one were to make a 107 movie of the moving particle, it would afterwards not be possible to determine whether 108 the movie is played forward or backward.However, when the Brownian particle is 109 in a 1D potential well, a different situation arises.When subjected to white Gaussian 110 noise, Onsager's microscopic reversibility has to apply, i.e., we still have time-reversal 111 symmetry.But for a Lévy-noise-subjected particle in a 1D potential well, time-reversal 112 symmetry is violated.Below we will explain this violation and elaborate on it.Based 113 on the developed insights, we will formulate a measure for the time-reversal-symmetry 114 breaking and we will derive approximate FDR relations for a nonequilibrium bath.115 Consider the situation with the parabolic well, V(x) = Ax 2 with A > 0, as 116 depicted in Fig. 1a.Let the Brownian fluctuations make the particle go up the parabola to 117 a considerable height x 0 outside the basin of attraction.As the basin of attraction of the 118 potential we take the interval around the minimum where the particle spends around 119 90% of its time.In case of Gaussian noise we have microscopic reversibility.The most 120 likely Brownian kick has magnitude zero.Therefore the most likely downslide is a series 121 of such 'zero' kicks and the most likely upslide is, perhaps unintuitively, the exact reverse.122 The deterministic downslide follows β ẋ = −Ax.This leads to x(t) = x 0 exp[−At/β].123 If we let the downslide run from x 0 all the way back back to x = 0, then we have 124 for the dissipated energy β ẋ(t) dx(t) = β ∞ 0 ẋ2 dt = (β 2 /(2A))x 2 0 .In principle, the 125 downslide to x = 0 takes infinitely long.But if we take as the endpoint a location x * in the 126 basin of attraction, then we have for the time t * to reach x * : t * = β log[x 0 /x * ]/A.With 127 a discretized time, this corresponds to n = t * /(∆t) = (β log[x 0 /x * ])/(A∆t) timesteps.

Figure 1 .
Figure 1.(a) A Brownian particle in a parabolic potential V(x) = x 2 /2.(b) When subjected to Lévy noise (α = 1.5, c = 1), microscopic reversibility, i.e. time-reversal symmetry, no longer appplies.It is obvious from the figure that Lévy jumps lead to the particle "shooting up."After the jump there is a slower relaxation, a "sliding down," back to the basin of attraction.(c) The parameter r (see text) is a measure for the time-reversal asymmetry.The figure shows how r depends on the stability index α of the Lévy noise.There appears to be a smooth approach to r = 0 (time-reversal symmetry) as α approaches 2 (Gaussian noise).∆t represents the time interval taken in a Langevin simulation.
Fig.1b, it is not hard to intuit this apparent violation of time-reversal symmetry.The 132 140

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Figures1b and 1cwere obtained using the latter expression at every timestep.
Figure1cshows r as a function of the stability index α for a Lévy-noise-subjected

193 4 .
Corrections on the FDR for a small Deviation from Microscopic Reversibility 194 As we saw earlier, with Gaussian noise time-reversal symmetry implies that ascent 195 and descent are on average equally fast.With Lévy noise the most likely trajectory from 196 the basin of attraction to a position x 0 high above the minimum takes one timestep.The 197 subsequent most likely descent follows the deterministic pattern that would also ensue 198 if there were equilibrium.Below we analyze such breaking of time-reversal symmetry 199 in a close-to-equilibrium condition.We will come to an intuitive understanding and 200 associated approximate relations.Ultimately, we will derive how the FDR looks for a 201 small deviation from equilibrium.

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of time-reversal-symmetry as quantified by the parameter r.Note that the temperature 204 T eq is still in the formula.On the right-hand-side, the small r leads to a small amount of 205 extra power being "fluctuated in."This power is included in what gets dissipated.
) express, in terms of r, what the effect of the time-reversal-207 symmetry breaking is on the temperature and the diffusion.Equations (10-12) constitute 208 the main result of this work.209 5.A Stochastic Simulation with an Artificial Violation of Microscopic Reversibility 210

216 from P 2 Figure 2 .Figure 3 .
Figure2.The parabolic potential with point P 1 in the basin of attraction and a point P 2 significantly higher.At equilibrium, the upward trajectory ω and its reverse ω are traversed equally often.With a small violation of time-reversal symmetry, i.e. r > 0 and a slightly faster upslide, a mismatch arises.The adjusted upslide ω reaches to P 2 and the Boltzmann distribution in the potential is correspondingly widened.
270noise-subjected particle in a quadratic potential well.The quadratic potential well is 271 generic in the sense that along any potential profile V(x), the first term in the expansion 272 around a minimum at x = x 0 is generally quadratic, i.e.V(x) ≈V(x 0 ) + 1 2 V (x 0 )(x − 273 x 0 ) 2 .274Lévynoise is characterized by the occurrence of frequent outliers.In everyday 275 data-processing practice, however, the reasoning is often in the opposite direction: upon 276 observation of frequent outliers it is inferred that the underlying noise must be Lévy.moments finite.Our scheme does not give rise to divergent integrals and leads to simple 301 expressions for a corrected FDR.The correction only involves the parameter r.302A possible experimental verification of Eqs.(10) and (12) would entail a setup 303 with an energy input; an energy input leading to a bath with particles whose motion 304 violates time-reversal-symmetry.Through following that motion with a probe, the r 305 could possibly be established.The power dissipation or the diffusion coefficient could 306 next be established independently.