An autonomous out of equilibrium Maxwell's demon for controlling the energy fluxes produced by thermal fluctuations
AAn autonomous out of equilibrium Maxwell’s demon for controlling the energy fluxesproduced by thermal fluctuations
Sergio Ciliberto ∗ Univ Lyon, Ens de Lyon, Univ Claude Bernard, CNRS,Laboratoire de Physique, UMR 5672, F-69342 Lyon, France (Dated: September 16, 2020)An autonomous out of equilibrium Maxwell’s demon is used to reverse the natural direction ofthe heat flux between two electric circuits kept at different temperatures and coupled by the electricthermal noise. The demon does not process any information, but it achieves its goal by usinga frequency dependent coupling with the two reservoirs of the system. There is no energy fluxbetween the demon and the system, but the total entropy production (system+demon) is positive.The demon can be power supplied by thermocouples. The system and the demon are ruled byequations similar to those of two coupled Brownian particles and of the Brownian gyrator. Thusour results pave the way to the application of autonomous out equilibrium Maxwell demons tocoupled nanosystems at different temperatures.
Nowadays the notion of Maxwell’s demon (MD) isgenerically used to indicate mechanisms that allow a sys-tem to execute tasks in apparent violation of the secondlaw of thermodynamics, such as for example to producework from a single heat bath and to transfer heat fromcold to hot sources. To obtain this result the demon doesnot exchange energy with the system but it has a positiveentropy production rate, which compensate the negativeentropy production of the system. In general the increasein entropy is induced by the fact that the demon needsto analyze the information that it gathers on the sys-tem status[1, 2]. In experiments this apparent violationof the second law is obtained by feedback mechanismswhich often require the use of external devices such asA/D converters, computers etc.[3–6]. Several smart ex-periments [7–9] have implemented these feedback locallyconstructing in this way autonomous Maxwell demons,which do not need the use of external devices as the mea-sure and the feedback are performed in the same place.Several autonomous Maxwell demons have been theoret-ically developed[10–14], but they can be of difficult prac-tical implementation in several devices such as colloidalparticles and mesoscopic electric circuits at room temper-ature. However it has been recently introduced in ref.[15]a new paradigm of MD based on an out of equilibriumdevice, which does not elaborate any information aboutthe system status. It has been shown that the parametersof this device can be suitably tuned in such a way that itdoes not exchange energy (heat or work) with the systembut it has a positive entropy production rate. Thus it hasthe two main requirements of an autonomous MD andit can be more easily experimentally realized because, incontrast to the commonly used definition of MD, it workswithout acquiring and analyzing any information aboutthe system status.We discuss here how to implement an out of equilib-rium MD (OEMD) [15] in electric circuits, which are ∗ E-mail me at: [email protected]
FIG. 1: Diagram of the system (grey box) and of the demon(yellow box). The system is constituted by the two resistances R and R kept respectively at temperature T and T , with T ≥ T . They are coupled via the capacitor C . The ca-pacitors C and C schematize the capacitances of the cablesand of the amplifier inputs. The demon (yellow box), is com-posed by two resistances ( R d and R d ) kept at two differenttemperatures T d and T d .Furthermore the resistance R d isdriven by a voltage generator V s whose out-put is filtered bythe low pas filter composed by the resistance Rs and the ca-pacitance C s . The four voltage generators η k (k=1,2,d1,d2)represent the Nyquist noise voltages of the resistances at thetemperatures of the heat baths. versatile dynamical systems ruled by coupled Langevinequations [16, 17]. Thus our study is quite general be-cause it opens the way to the application of OEMDs tocoupled nanosystems modeled by Langevin equations. Asan example we will show in this article how an OEMD canbe used to reverse the natural direction of the heat flux a r X i v : . [ c ond - m a t . s t a t - m ec h ] S e p between two electric circuits kept at different tempera-tures and coupled by the electric thermal noise [18, 19].In fig.1 we sketch the system (gray box) and the demon(yellow box). We chose for the system this specific circuitbecause the statistical properties of the heat flux havebeen characterized both theoretically and experimentally[18, 19]. Furthermore it is ruled by the same equations ofthe Brownian gyrator [20, 21] and of two Brownian parti-cles coupled by a harmonic potential and kept at differenttemperatures[18], making the result rather general The system (grey box in fig.1) is constituted by tworesistances R and R , which are kept at two differenttemperatures T and T ≥ T . In the figure, the two re-sistances have been drawn with their associated thermalnoise generators η and η , whose power spectral densi-ties are given by the Nyquist formula | ˜ η m | = 4 k B R m T m ,with m = 1 ,
2. The coupling capacitance C controls theelectrical power exchanged between the resistances andas a consequence the energy exchanged between the twobaths. No other coupling exists between the two resis-tances.. The two capacitors C and C represents thesum of the circuit and cable capacitances. All the rel-evant energy exchanges in the system can be derivedby the simultaneous measurements of the voltage V m ( m = 1 ,
2) across the resistance R m and the currents i m flowing through them. When T = T the system is inequilibrium and exhibits no net energy flux between thetwo reservoirs. The circuit equations can be written interms of charges q m flowed through the resistances R m ,so the measured instantaneous currents are i m = ˙ q m .We make the choice of working with charges because theanalogy with a Brownian particle is straightforward as q m is equivalent to the displacement of the particle m [17–19]. A circuit analysis shows that the equations forthe charges are: R ˙ q = V − η , and R ˙q = η − V , (1)with V = − q ( C + C ) + q CX (2) V = − q C + q ( C + C ) X . (3)where X = C C + C ( C + C ) and η m is the Nyquistwhite noise: (cid:104) η i ( t ) η j ( t (cid:48) ) (cid:105) = 2 δ ij k B T i R j δ ( t − t (cid:48) ). Inref.[19] we have shown that eqs.1 fully characterize allthe thermodynamics properties of the system.In this system the work and the heat are defined as˙ W m = CX q m (cid:48) ˙ q m (4)˙ Q m = V m i m = V m V m − η m R m . (5)The quantities ˙ W m is identified as the thermodynamicwork performed by the circuit m (cid:48) on m (cid:54) = m (cid:48) and Q m the heat dissipated by the resistance m [16–19, 22]. Asall the variables are fluctuating, the derived quantities ˙ Q m and ˙ W m fluctuate too. In ref.[18] we computed andmeasured the mean heat flux between the two heat baths,which is given by : (cid:68) ˙ Q (cid:69) = − (cid:68) ˙ Q (cid:69) = C k B ( T − T ) XY . (6)where (cid:104) . (cid:105) stands for mean value and we have introducedthe quantity Y = [( C + C ) R + ( C + C ) R ]. We usethe convention that the heat extracted from a systemreservoir is negative and the heat dissipated is positive. The out of equilibrium demon is sketched in the yel-low box of fig.1 and it is composed by two resistances( R d and R d ) kept at two different temperatures T d and T d (see Appendix A). The two voltage voltage gen-erator η d and η d represent the Nyquist noise voltagesassociated to the two resistances at the heat bath tem-peratures. Furthermore the resistance R d is driven by avoltage generator V s whose out-put is fltered by the thelow pas filter composed by the resistance Rs and the ca-pacitance C s (see Appendix A 1). We notice that demonscheme is similar to that of the system, with a couplingcapacitance C → ∞ , on which the driving V s has beenadded. To design it, we followed the main prescriptionsof ref.[15]: 1) It is out of equilibrium; 2) Either T d or T d has to be smaller than T ; 3) it produces colored noise,obtained in our case by the source V s filtered by R s and C s ;4) It is coupled with the two parts of the system ondifferent frequency ranges, specifically high frequencieswith subsystem 1 and DC coupled with subsystem 2.The choice of V s is very important in order to sim-plify the experimental configuration. Indeed V s can beeither the thermal fluctuations of Rs with a suitablecut-off imposed by the R s C s or an external driving.Many choices are possible and the simplest one is to use V s = V f =constant and Rs = 0. In such a way V f iscoupled with R only and the thermal noises η d and η d are directly coupled with R and high pas filtered for R (see Appendix A 1) The demon is always out of equilib-rium, because, when it is disconnected from the system,the power supplied by V f is entirely dissipated in the de-mon resistances producing a mean heat flux towards thedemon heat baths, even in the case T d = T d . This isa simplified version of the original OEMD of ref.[15] be-cause it requires the use of only one cold source at T d and a DC signal that can be easily generated by ther-mocouples making the demon fully autonomous. We willdemonstrate that this demon can reverse the heat fluxof the system in a wide range of parameters with a zeroenergy flux (heat and work) with the system. The connection of the demon to the system changesthe current distributions and the energy exchanges. Thecircuit analysis shows (see Appendix B), that the currents˙ q k ( k = 1 , , d , d
2) flowing in the resistances R k are nowruled by the following equations : R ˙ q = V − η (7) R ˙ q = η − V (8) R d ˙ q d = η d − V (9) R d ˙ q d = η d + V f − V (10) V = − ( C t + C ) q + C q t X t (11) V = ( C + C ) q t − C q X t (12)where q t = ( q + q d + q d ), C t = C + C d . and X t = C C t + C ( C + C t ).In order to reduce the number of parameters we con-sider the case T d = T d = T d and R d = R d . The heatfluxes in the four reservoirs can be computed using ˙ Q k = v k ˙ q k , where v k is the potential difference on the resistance R k (see Appendix C). Introducing the following param-eters R d = R d R d / ( R d + R d ), R t = R d R / ( R d + R ), Y t = R ( C + C ) + R t ( C + C t ), A = C k B / ( X t Y t ), (cid:104) V (cid:105) = V t = V f R t /R d , B = A R t ( X t R t + R ( C + C ) ) / ( R d R C ),we obtain: (cid:68) ˙ Q (cid:69) = A (cid:18) R t R ( T − T ) + R t R d ( T d − T ) (cid:19) (13) (cid:68) ˙ Q (cid:69) = − A R t R ( T − T ) − B ( T − T d ) + V t R (14) (cid:68) ˙ Q d (cid:69) = − A R t R d ( T d − T ) − B ( T d − T ) ++ V f R t R d ( 1 R d + 1 R ) − V t R (15) (cid:68) ˙ W f (cid:69) = V f R t R d ( 1 R d + 1 R ) (16)where (cid:68) ˙ Q d (cid:69) = (cid:68) ˙ Q d (cid:69) + (cid:68) ˙ Q d (cid:69) is the total heat flux in thedemon reservoirs and (cid:68) ˙ W f (cid:69) is the total power suppliedby the external generator V f . The total energy balancedemon+system is : (cid:68) ˙ Q d (cid:69) − (cid:68) ˙ W f (cid:69) + (cid:68) ˙ Q (cid:69) + (cid:68) ˙ Q (cid:69) = 0 (17)These equations allow us to define the conditions forwhich the demon can reverse the flow without any en-ergy exchange with the system.In absence of the demon the heat flux is given by eqs.6,i.e. (cid:68) ˙ Q (cid:69) = − (cid:68) ˙ Q (cid:69) <
0. Using the demon we want toreverse this flow making (cid:68) ˙ Q (cid:69) > (cid:68) ˙ Q (cid:69) = − (cid:68) ˙ Q (cid:69) because an observer, who measures the heat-fluxof the system, has to establish that heat flows from thecold to the hot reservoir. The condition (cid:68) ˙ Q (cid:69) = − (cid:68) ˙ Q (cid:69) has two important consequences. Firstly it reduces eq.17to (cid:68) ˙ Q d (cid:69) − (cid:68) ˙ W f (cid:69) = 0 , (18)which indicates that all the power supplied by V f is dis-sipated in the demon reservoirs and not in the systemreservoirs. Secondly applying it to eqs.13,14 we find that: V t R = A R t R d ( T − T d ) + B ( T − T d ) (19)Finally using eq.19 and the condition (cid:68) ˙ Q (cid:69) > T d , where the the spontaneousprocess is reversed, finding: T d ≤ T − R d R ( T − T ) (20)The eqs.19 and 20 fix the conditions that allows thedemon to reverse the system heat flux without heat ex-change (eq.18) between the demon and the system. Eq.19indicates that the fraction of the power injected by thedemon and dissipated in R ( V t /R in eq.14) is com-pensated by the heat extracted from the system baths.We can also prove that thanks to eq.19 the demon doesnot perform any work on the system. Indeed the to-tal work performed by the demon on the system is : (cid:68) ˙ W d,s (cid:69) = (cid:68) ˙ W d, (cid:69) + (cid:68) ˙ W d, (cid:69) where (cid:68) ˙ W d, (cid:69) and (cid:68) ˙ W d, (cid:69) are the works performed on subsystems 1 and 2 respec-tively. These can be computed using equations eq.7and eq.8 in which we see that a ”force” proportional to q d + q d is applied on the two subsystems. Thus thework per unit time of these forces are [19]. (cid:68) ˙ W d, (cid:69) = CX t (cid:104) ˙ q ( q d + q d ) (cid:105) (21) (cid:68) ˙ W d, (cid:69) = − C + CX t (cid:104) ˙ q ( q d + q d ) (cid:105) (22)From these two works (computed in Appendix D) weobtain for the total work : (cid:68) ˙ W d,s (cid:69) = − A R t R d ( T − T d ) − B ( T − T d ) + V t R (23)We clearly see that if the condition on V t (eq.19) is veri-fied then (cid:68) ˙ W d,s (cid:69) = 0, i.e. no work is done by the demonon the system. Thus eq.19 and eq.18 insure that totalenergy flux from the demon to the system is zero.However the demon produces entropy and the total en-tropy production rate (cid:68) ˙ S (cid:69) is positive in spite of the factthat the system entropy production rate (cid:68) ˙ S s (cid:69) = (cid:68) ˙ Q (cid:69) (1 /T − /T ) is negative, because (cid:68) ˙ Q (cid:69) > T > T when the demon is ”on”. Thetotal entropy production rate is (cid:68) ˙ S (cid:69) = (cid:68) ˙ Q d (cid:69) T d + (cid:68) ˙ Q (cid:69) (cid:18) T − T (cid:19) (24) FIG. 2: a) Heat fluxes as a function of the demon tempera-ture T d : (cid:68) ˙ Q (cid:69) (blue line) computed when the demon is on,using eqs.13,..,16 and the condition for V t eq.19; (cid:68) ˙ Q (cid:69) (hor-izontal red dashed line) computed (eq.6) when the demon is”off” ; (cid:68) ˙ Q (cid:69) (red circles) and (cid:68) ˙ Q (cid:69) + (cid:68) ˙ Q (cid:69) (black stars) ob-tained from the direct numerical simulation of eqs.7,..,10. b)Demon efficiencies as a function of T d : η s (black line) and η Q (blue line, red circles) computed from eqs.14,16,25 (continuoslines) and obtained from the direct numerical simulations ofeqs.7,..,10 (red circles). The parameters used to compute thecurves in a) and b) are: T = 300K, T = 450K, C = 1nF, C = C = 100pF, R = R = 10MΩ; R d = 3MΩ; Cd = 50pF. To show that (cid:68) ˙ S (cid:69) >
0, we start by taking into accountthat (cid:68) ˙ Q d (cid:69) = (cid:68) ˙ W f (cid:69) (see eq.18) and that (cid:68) ˙ W f (cid:69) > V t R because as we said V t R is a fraction of the total powerinjected into the system by the demon source (computedAppendix E). Furthermore as we want (cid:68) ˙ Q (cid:69) > V t /R > (cid:68) ˙ Q (cid:69) as the otherterms are negative because T > T > T d . As a conse-quence (cid:68) ˙ Q d (cid:69) /T d > (cid:68) ˙ Q d (cid:69) (1 /T − /T ) > (cid:68) ˙ Q (cid:69) (1 /T − /T )and we find (cid:68) ˙ S (cid:69) > < ˙ Q k > =
1% and η Q < Acknowledgments
We acknowledge useful discussion with R.S.Whitney.This work has been supported by the FQXi foundationon grant number FQXi-IAF19-05 ”Information as a fuel in colloids and superconducting quantum circuits ” [1] E. Lutz and S. Ciliberto, Physics Today , 30 (2015).[2] J. M. R. Parrondo, J. M. Horowitz, and T. Sagawa,Nature Phys. , 131 (2015).[3] S. Toyabe, T. Sagawa, M. Ueda, M. Muneyuki, andM. Sano, Nature Phys. , 988 (2010).[4] C. Elouard, D. Herrera-Mart´ı, B. Huard, andA. Auff`eves, Phys. Rev. Lett. , 260603 (2017).[5] T. Admon, S. Rahav, and Y. Roichman, Phys. Rev. Lett. , 180601 (2018).[6] Y. Masuyama, K. Funo, Y. Murashita, A. Noguchi,S. Kono, Y. Tabuchi, R. Yamazaki, M. Ueda, andY. Nakamura, Nature Comm. (2018).[7] G. N. Price, S. T. Bannerman, K. Viering, E. Narevicius,and M. G. Raizen, Phys. Rev. Lett. , 093004 (2008).[8] J. V. Koski, V. F. Maisi, J. P. Pekola, and D. V. Averin,Proceedings of the National Academy of Sciences ,13786 (2014).[9] J. V. Koski, A. Kutvonen, I. M. Khaymovich, T. Ala-Nissila, and J. P. Pekola, Phys. Rev. Lett. , 260602(2015).[10] A. C. Barato and U. Seifert, EPL , 60001 (2013).[11] N. Shiraishi, S. Ito, K. Kawaguchi, and T. Sagawa, NewJournal of Physics , 045012 (2015).[12] A. B. Boyd, D. Mandal, and J. P. Crutchfield, New Jour-nal of Physics , 023049 (2016).[13] M. L. Rosinberg and J. M. Horowitz, EPL , 10007 (2016).[14] Z. Lu and C. A. Jarzynski, Entropy , 65 (2019).[15] R. Sanchez, J. Splettstoesser, and R. S. Whitney, Phys.Rev. Lett. , 216801 (2019).[16] N. Freitas, J.-C. Delvenne, and M. Esposito, Phys. Rev.X , 031005 (2020).[17] R. van Zon, S. Ciliberto, and E. G. D. Cohen, Phys.Rev. Lett. , 130601 (2004).[18] S. Ciliberto, A. Imparato, A. Naert, and M. Tanase,Phys. Rev. Lett. , 180601 (2013).[19] S. Ciliberto, A. Imparato, A. Naert, and M. Tanase, J.Stat. Mech. , P12014 (2013).[20] R. Filliger and P. Reimann, Phys. Rev. Lett. , 230602(2007).[21] S. Cerasoli, V. Dotsenko, G. Oshanin, and L. Rondoni,Phys. Rev. E , 042149 (2018).[22] N. Garnier and S. Ciliberto, Phys. Rev. E , 060101(2005).[23] G. Verley, T. Willaert, C. Van den Broeck, and M. Es-posito, Phys. Rev. E , 052145 (2014).[24] M. Polettini, G. Verley, and M. Esposito, Phys. Rev.Lett. , 050601 (2015).[25] J.-F. Derivaux and Y. De Decker, J. Stat. Mech. , 034002(2019). Appendix A: Several details about the demon
In this section we describe the out of equilibrium demon, composed by two resistances ( Rd and Rd ) kept at twodifferent temperatures T d and T d . The two voltage generators ηd and ηd corresponding to the Nyquist noiseof the two resistances at the temperatures of the heat baths. Furhermore the resistance Rd2 is driven by a voltagegenerator V s whose out-put is fltered by the the low pas filter composed by the resistance Rs and the capacitance C s .
1. The demon as a colored noise generator
The dynamics of the demon can be obtained by writing the Kirchhoff laws for the points V f and V d : V s − V f R s − ˙ V f C f − V f + η d − V d R d = 0 (A1) V f + η d − V d R d − ˙ V d C d − V d − η d R d − i d = 0 (A2)where i d is the current flowing between the system and the demon when the latter is ”on”. From these equations weget: τ f ˙ V f = − V f + R f R s V s + R f R d ( V d − η d ) (A3) τ d ˙ V d = − V d + (cid:18) R d R d V f − i d R d (cid:19) + ξ d (A4)(A5)where we use : R d = R d R d R d + R d , τ d = R d C d , R f = R s R d R d + R s , τ f = R f C s and ξ d = η d1 R d R d1 + η d2 R d R d2 . (A6)As V s is not necessarily the thermal noise of R s , it can be a very large external driving that allows the demon towork, we can simplify this circuit if we assume that R d >> R s τ f ˙ V f = − V f + V s (A7) τ d ˙ V d = − V d − (cid:18) R d R d V f − i d R d (cid:19) + ξ d (A8)These equations show that the demon produces colored noise for the system as it is needed to control the system heatflux.
2. The heat flux in the demon switched off
When V s = 0 and i d = 0 the heat transfers between the two reservoirs of the demon can be obtained from eqs.6 for C → ∞ . Thus the heat transfers in the demon are: (cid:68) ˙ Q d (cid:69) = k B ( T d − T d )( R d + R d ) C d . (A9) (cid:68) ˙ Q d (cid:69) = k B ( T d − T d )( R d + R d ) C d . (A10)When there is an external driving one has also to consider the work performed by the external source V f still in thecase i d = 0,i.e. with the demon is not connected to the system. We can use for V s either thermal fluctuations of Rs with a suitable cut-off imposed by the R s C s or an external driving. In the simplest version we make the choice to usea V f = V s constant and Rs = 0. In such a case the heat dissipated by the two resistances is: (cid:68) ˙ Q d (cid:69) = k B ( T d − T d )( R d + R d ) C d + V f R d ( R d + R d ) (A11) (cid:68) ˙ Q d (cid:69) = k B ( T d − T d )( R d + R d ) C d + V f R d ( R d + R d ) (A12)and doing the energy balance we have (cid:68) ˙ Q d (cid:69) + (cid:68) ˙ Q d (cid:69) = V ( R d + R d ) = (cid:68) ˙ W f (cid:69) (A13)where (cid:68) ˙ W f (cid:69) is the power injected into the demon by the voltage generator. where we use the convention that theheat dissipated in the bath and the work performed on the system are positive. Thus the demon is always out ofequilibrium even at T d = T d Appendix B: The currents in the system+demon circuit
We now connect the demon to the system and this connection changes the current distributions. We write theequations of the circuits in terms of charges. The relationships between currents and charges are0 = − ˙ q − ˙ q c + ˙ q c , ˙ q − ˙ q ct − ˙ qc + ˙ q d + ˙ q d = 0 , ˙ q c = ( V − V ) C, V C = q c and V C t = q ct (B1)where ˙ q k ( k = 1 , , d , d
2) are the currents flowing in the resistances R k . Futhermore ˙ q c , ˙ q c and ˙ q ct are the currentsflowing respectively in the capacitors C, C and C t = C + C d . Solving the system for q c and q ct we find : V = q c C = − ( C t + C ) q + C q t X t (B2) V = q ct C t = ( C + C ) q t − C q X t (B3)where q t = ( q + q d + q d ) and X t = C C t + C ( C + C t ). we can now solve for the four currents flowing in the 2system resistances and the two demon resistances specifically R ˙ q = V − η = − ( C t + C ) q + C ( q + q d + q d ) X t − η (B4) R ˙ q = η − V = − ( C + C ) ( q + q d + q d ) + C q X t + η (B5) R d ˙ q d = η d − V = − ( C + C ) ( q + q d + q d ) + C q X t + η d (B6) R d ˙ q d = η d + V f − V = − ( C + C ) ( q + q d + q d ) + C q X t + η d + V f (B7)We write the equations for the charges because the connection with Brownian particles is straightforward ( q correspondto the dispacement). Furthermore it is more clear to understand the amount of work performed by the demon onthe system and the amount of dissipated heats in the various reservoirs. As the resistances of the demon are just inparallel to R the equations B4,.,B7 can be reduced to: R ˙ q = − ( C t + C ) X t q + CX t q t − η (B8) R t ˙ q t = − ( C + C ) X t q t + CX t q + η t + V t (B9)where we define q t = ( q + q d + q d ) , R t = (1 /R + 1 /R d + 1 /R d ) − , and ,V t = V f R t /R d , η t = ( η /R + η d /R d + η d /R d ) R t . (B10)Using eqs.4,5 we can compute the heat and the work of the different parts of the circuit. We can also use eqs. 6 tocompute the total heat exchanged between the reservoir at T with the reservoirs of resistance R , R d , R d . Defining T t = R t ( T /R + T d /R d + T d /R d ) and Y t = R ( C + C ) + R t ( C + C t ), we get : (cid:68) ˙ Q (cid:69) = C k B ( T t − T ) X t Y t . (B11) (cid:68) ˙ Q t (cid:69) = C k B ( T − T t ) X t Y t (B12)where the contribution of the work performed by V t cancels out in the stationary regime (see appendix E). Eq. B11can be decomposed in the various contributions to the heat fluxes from the three reservoir at ( T , T d and T d ) : (cid:68) ˙ Q (cid:69) = C k B X t Y t (cid:18) R t R ( T − T ) + R t R d ( T d − T ) + R t R d ( T d − T ) (cid:19) (B13) Appendix C: Calculation of the heat fluxes
By the definition eq.5 the heat fluxes in the resistances k = 1 , , d , d (cid:68) ˙ Q k (cid:69) = (cid:28) v k ( v k − η k ) R k (cid:29) . (C1)where v k = V k − V f δ k,d (with V d = V d = V ) is the potential difference on the resistance R k . The mean values canbe evaluated by using the Fourier transforms ˜ v k of v k , which can be written as : (cid:68) ˙ Q k (cid:69) = (cid:90) ∞ (cid:60) (cid:26) ˜ v ∗ k R m (˜ v k − ˜ η k ) (cid:27) dω π . (C2)To evaluate this integral we take into account that the spectral density of ˜ η k is | ˜ η k | = 4 k B T k R k and that (cid:60) { η k η ∗ k (cid:48) } =0 because different noise sources are uncorrelated We have already computed (cid:68) ˙ Q (cid:69) in eq.B11 and eq.B13 using theeqs.6. We need to estimate all the heat fluxes in the demon reservoirs and in the reservoir 2 by computing V and V given by eqs.B2 and B3. Their values depend on q , q t which can be computed solving the system of eqs.B8 and B9in Fourier space. Z ˜ q − a ˜ q t = − ˜ η (C3) − a ˜ q + Z t ˜ q t = ˜ η t + V t √ πδ ( ω ) (C4)where Z = ( C t + C ) /X t + iωR , Z t = ( C + C ) /X t + iω R t and a = C/X t . We find˜ q = − ˜ η Z t + η t a det (C5)˜ q t = ˜ η t Z − η a det (C6)det = Z Z t − a = (1 − ω R R t X t ) + iω Y t X t (C7)Inserting these Fourier transform in eqs.B2 and B3, we get:˜ V = ˜ η [1 + iω R t ( C t + C )] + ˜ η t iωR CX t det (C8)˜ V = ˜ η t [1 + iω R ( C + C )] + ˜ η iωR t CX t det + V t √ πδ ( ω ) (C9)We see that | ˜ V | = | ˜ η | (cid:2) ω R t ( C t + C )) (cid:3) + | ˜ η t | ( ωR C ) (1 − ω R R t X t ) + ( ω Y t ) (C10) | ˜ V | = | ˜ η t | (cid:2) ω R ( C + C )) (cid:3) + | ˜ η | ( ωR t C ) (1 − ω R R t X t ) + ( ω Y t ) + V t πδ ( ω ) (C11) (cid:10) V (cid:11) = (cid:90) ∞ | ˜ V | dω π = | ˜ η | ( X t R R t + R t ( C t + C ) + | ˜ η t | R C Y t X t R R t (C12) (cid:10) V (cid:11) = (cid:90) ∞ | ˜ V | dω π = | ˜ η t | ( X t R R t + R ( C + C ) ) + | ˜ η | R t C Y t X t R R t + V t (C13)where we used the integrals: (cid:90) ∞ Y t (1 − ω R R t X t ) + ( ω Y t ) dω π = 14 and (cid:90) ∞ Y t ω (1 − ω R R t X t ) + ( ω Y t ) d ω π = 14 R R t X t (C14)We can now compute the other heat fluxes :˜˙ Q = (cid:60) (cid:40) ˜ V ∗ ( ˜ V − ˜ η ) R (cid:41) = | ˜ V | R − (cid:60) (cid:26) | ˜ η | R t [1 − iω R ( C + C )]det R X t | det | (cid:27) == | ˜ V | R − | ˜ η | R t (cid:2) (1 − ω R R t X t ) + ω Y t R ( C + C ) (cid:3) R [(1 − ω R R t X t ) + ( ω Y t ) ] (C15) (cid:68) ˙ Q (cid:69) = (cid:90) ∞ ˜˙ Q dω π == | ˜ η t | R ( X t R R t + R ( C + C ) ) + R | ˜ η | R t C − | ˜ η | R t Y t R ( C + C )4 R Y t X t R R t == ( | ˜ η t | R − | ˜ η | R t ) R ( X t R t + R ( C + C ) ) + R t C ( | ˜ η | R − | ˜ η | R )4 R Y t X t R R t + V t R = − k B C X t Y t ( T − T ) R t R − k B X t R t + R ( C + C ) X t Y t R (cid:18) R t R d ( T − T d ) + R t R d ( T − T d ) (cid:19) + V t R (C16)˜˙ Q d = (cid:60) (cid:40) ( ˜ V ∗ − V f √ πδ ( ω )) ( ˜ V − ˜ η d − V f √ πδ ( ω )) R d (cid:41) = | ˜ V | R d − (cid:60) (cid:26) | ˜ η d | R t [1 − iω R ( C + C )]det R d X t | det | (cid:27) + V t ( V t − V f ) R d πδ ( ω ) == | ˜ V | R d − | ˜ η d | R t (cid:2) (1 − ω R R t X t ) + ω Y t R ( C + C ) (cid:3) R d [(1 − ω R R t X t ) + ( ω Y t ) ] + ( V t − V f ) R d πδ ( ω ) (C17) (cid:68) ˙ Q d (cid:69) = (cid:90) ∞ ˜˙ Q d dω π == | ˜ η t | R d ( X t R R t + R ( C + C ) ) + R d | ˜ η | R t C − | ˜ η d | R t Y t R ( C + C )4 R d Y t X t R R t + ( V t − V f ) R d == ( | ˜ η t | R d − | ˜ η d | R t ) R ( X t R t + R ( C + C ) ) + R t C ( | ˜ η | R − | ˜ η d | R )4 R d Y t X t R R t + ( V t − V f ) R d == − A ( T d − T ) R t R d − B R d R d (cid:18) R R d ( T d − T d ) + ( T d − T ) (cid:19) + ( V t − V f ) R d (C18)where A = k B C X t Y t and B = k B R t (X t R t + R (C + C) )X t Y t R R d (C19)Finally using the same method we get: (cid:68) ˙ Q d (cid:69) = − A ( T d − T ) R t R d − B R d R d (cid:18) R R d ( T d − T d ) + ( T d − T ) (cid:19) + V t R d (C20) Appendix D: The work performed by the demon on the system
We follow refs. [18, 19] to compute the power injected by the demon on R and R with R d = R d and T d = T d (cid:68) ˙ W d, (cid:69) = CX t (cid:104) ˙ q ( q d + q d ) (cid:105) = (cid:90) ∞ (cid:60) (cid:40) ˜ V ∗ − ˜ η ∗ R ˜ η d − ˜ V R d iω (cid:41) dω π = (D1)= − CX t (cid:90) ∞ (cid:20) | ˜ η | R t C − | ˜ η t | R C + (1 − ω R R t X t ) R t C ( −| ˜ η | + | ˜ η d | R /R d ) R R d | det | X t (cid:21) dω π = (D2)= − k B C R t X t Y t R d ( T − T t ) = A R t R d R ( T − T ) − A R t R d ( T − T d ) (D3)where we used eqsC8, C9, T t = R t ( T /R + T d /R d ) and eqs.C19. (cid:68) ˙ W d, (cid:69) = − ( C + C ) X t (cid:104) ˙ q ( q d + q d ) (cid:105) = − ( C + C ) X t (cid:90) ∞ (cid:60) (cid:40) ˜ η ∗ − ˜ V ∗ R η d − ˜ V R d iω (cid:41) dω π + V t R = (D4)= − ( C + C ) X t R R d (cid:90) ∞ ( | ˜ η | R t /R − | ˜ η d | R t /R d )[ Y t − R C + C )(1 − ω R R t X t )] X t | det | dω π + V t R (D5)= − k B ( C + C ) R t X t R R d ( T − T d ) + V t R . (D6)Adding a subtracting A/R t T / ( R R d ) from the last expression,taking into account the definition of A and B (eq.C19),that ( C + C )( C t + C ) = X t + C and that k B ( C + C ) R t Y t / ( X t R R d Y t ) = B + A/R t / ( R R d ) we get (cid:68) ˙ W d, (cid:69) = − B ( T − T d ) − A R t R R d ( T − T ) − A R t R R d ( T − T d ) + V t R (D7)0The total work performed by the demon on the system is (cid:68) ˙ W d,s (cid:69) = (cid:68) ˙ W d, (cid:69) + (cid:68) ˙ W d, (cid:69) = (D8)= − A R t R d ( T − T d ) − B ( T − T d ) + V t R (D9)If the condition on the demon potential and temperature eqs.20 and 19 are satisfied then the last equation imposesthat (cid:68) ˙ W d,s (cid:69) = 0, i.e. the demon does not perform any work on the system. Appendix E: The power dissipated by the DC currents only
The power dissipated in the system and demon resistances by the DC external generator V f are: P = 0 , P = V t /R , P d = V t /R d , P d = ( V f − V t ) /R d (E1)with V t = (cid:104) V (cid:105) . Taking into account that V t = V f R t /R d we get the total power supplied by the generator V f (cid:68) ˙ W f (cid:69) = P + P d + P d = V f ( V f − V ) R d (E2)The dissipation of the demon is P d = P d + P d thus P = ˙ W d − P d . This has been used in the text for computingthe total entropyConcerning the work of V t in eq.B9 we clearly see that for the node 2, (cid:104) ˙ q t (cid:105) = 0 if we consider only the DC currentimposed by V f . This means that V t (cid:104) ˙ q t (cid:105) = 0 and there is no contribution to the heat flux ˙ Q t which is the heatexchanged by the equivalent R tt