Long-Range Correlations in Driven Systems: Understanding Non-Equilibrium Dynamics
Discover how driven systems exhibit long-range correlations unlike equilibrium systems, leading to unique phenomena like long-range order and spontaneous symmetry breaking, as discussed by David Mukamel. Explore examples, consequences, and analogies related to electrostatics, as well as insights into normal and anomalous heat conduction in one dimension.
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Long-range correlations in driven systems David Mukamel Marseille, April 5 2017
Typically, driven systems show long-range correlations in their steady state. This is unlike systems in equilibrium where particles are Correlated only locally (except at the phase transition point). Long-range correlations can even lead to long-range order and spontaneous symmetry breaking in one dimension.
General outline Will first discuss an example where long-rang correlations show up discuss analogy to electrostatics consider some consequences Will present a simple example where long-range order (spontaneous symmetry breaking) takes place in d=1 Normal and anomalous heat conduction in 1d systems
Local drive perturbation T. Sadhu, S. Majumdar, DM, Phys. Rev. E 84, 051136 (2011) Phys. Rev. E 90, 012107 (2014)
Local perturbation in equilibrium Particles diffusing (with exclusion) on a grid occupation number ??= 0,1 N particles V sites ? ? =? Prob. of finding a particle at site k ?
Add a local potential u at site 0 1 1 1 ??? 1 1 0 1 N particles V sites ? = ?/? The density changes only locally. ? 0 ???= ? 1 1 detail balance:
Main Results In d 2 dimensions the density corresponds to a potential of a dipole in d dimensions, decaying as ? ? ~1/?? 1for large r. The current satisfies ?(?)~1/??. The same is true for local arrangements of driven bonds. The power law of the decay depends on the specific configuration. The two-point correlation function ?(?,?) corresponds to a quadrupole In 2d dimensions, decaying as ?(?,?)~1/ ?2+ ?2 ?for ? = 1/2 The same is true at other densities to leading order in ? (order ?2).
Density profile (with exclusion) The density profile ?( ?)~ 1/?2 along the y axis in any other direction 1/?
Evolution equations 1 1 1 ? -2 3 1 2 -1 0 1 1 1 ??? = 0,1 ??? + ?? ??? = ??+11 ?? ??1 ??+1 ?? 11 ?? ??1 ?? 1 ?? ?? ??? = ??(?) ???(?) ?? ? 0,1 = ??+1? + ?? 1? 2??(?)
1 1 1 ? 1 2 -1 0 1 1 1 ?11 ?0 ?01 ?1 ? 11 ?0 ?01 ? 1 ?? (1 ?)?? ?0? + ?? ?0? = ?? ?? ??0(?) ?? = ?1? + ? 1? 2?0? + ? ?0(1 ?1)
1 1 1 ? 1 2 -1 0 1 1 1 ?11 ?0 ?01 ?1 ?11 ?2 ?21 ?1 ?? (1 ?)?? ?1? + ?? ?1? = ?? ?? ??1(?) ?? = ?2? + ?0? 2?1? ? ?0(1 ?1)
The resulting dynamical equations ??? ?,? = ?2? ?,? + ? ? 0 1 ? ?1 [? ?,0 ? ?,?1]
Non-interacting particles Time evolution of density: ??? ?,? = ?2? ?,? + ?? 0,? [? ?,0 ? ?,?1] ?2= ? ? + 1,? + ? ? 1,? + ? ?,? + 1 + ? ?,? 1 4?(?,?) The steady state equation ?2? ? = ??(0)[? ?,0 ? ?,?1] particle density electrostatic potential of an electric dipole
?2? ? = ??(0)[? ?,0 ? ?,?1] ?2? ?, ?? = ? ?, ?? Green s function ? ? = ? + ?? 0 [? ?,0 ?( ?, ?1)] Unlike electrostatic configuration here the strength of the dipole should be determined self consistently.
Greens function ?(?,?;0,0) of the discrete Laplace equation ? 2 1 3 0 ? 2 ? 1 0 0 1/4 1 1/4 1/? 1/4 2 ? 1 1 4 2 2 4/3? ? 3 1 ? ?, ?? 2 ?ln| ? ?0|
determining ?(0) To find ?(0) one uses the values ? 0,0 = 0 ,? 0, ?1 = 1 4 ? ? 0 = 1 ? 4
at large ? ? ? = ? ?? 0 ?1 ? ?2+ ?(1 ?2) density: 2? ? ? =?? 0 1 ?2[ ?1 2 ?1 ? ? + ?(1 current: ?3) ?2 2?
Multiple driven bonds ? ? = ? + ?? ?1 ? ?, ?1 ? ?, ?1+ ?1 + ?? ?2 ? ?, ?2 ? ?, ?2+ ?1 + Using the Green s function one can solve for ? ?1, ? ?2 by solving the set of linear equations for ? = ?11,?2,
Two oppositely directed driven bonds quadrupole field The steady state equation: ?2? ? = ??(0)[2? ?,0 ? ?, ?1 ? ?, ?1] 2 ? ? = ? ?? 0 1 ?2 2 ?1 ? ?2 + ?(1 ?4) 2?
? 2 dimensions ? 2? 0 ???(?) ? = 1 ? ? = ? ? ?? 2 ? ?,?? = 1 ? 2 ? ? ?? 1
The model of local drive with exclusion Here the steady state measure is not known however one can determine the behavior of the density. ??? ?,? = ?2? ?,? + ? ? 0 1 ? ?1 [? ?,0 ? ?,?1] ? = 0,1 is the occupation variable ? ? = ? ? ? 0 1 ? ?1 ?1 ? ?2+ ?(1 ?2) 2?
Local charge: ? ? 0 1 ? ?1 [? ?,0 ? ?,?1] ? ? = ? ? ? 0 1 ? ?1 ?1 ? ?2+ ?(1 ?2) 2? The density profile is that of the dipole potential with a dipole strength which can only be computed numerically.
Simulation results Simulation on a 200 200 lattice with ? = 0.6 For the interacting case the strength of the dipole was measured separately .
Magnetic field analog for ? ? process ? = ln[??,?] ??? ??? for the ?,? bond ? = ln
Zero-change configuration The density is flat however there are currents
Zero magnetic field configuration no currents but inhomogeneous density (equilibrium)
In general non-zero electric field inhomogeneous density non-zero magnetic field currents zero magnetic field equilibrium configuration
( ) ' W C C Given transition rates a necessary and sufficient condition for detailed balance: for every set of microstates C1, ,Ck = 1 ( ) 2 2 ( )... 3 ( ) 1 ( )... 1 2 ( ) 1 1 ( ) W W W k W k k W W k 2 3 1 4 5 Kolmogorov criterion
H=0 in every plaquette is equivalent to satisfying the Kolmogorov criterion ?0= ?ln??,?+1 ???,?+1= ???+1,? ??+1,?= 0
Two-point correlation function ? ?,? =< ? ? ? ? >- ?(r) ?(?) In d=1 dimension, in the hydrodynamic limit ???,? =1 ?g(? ?, ? ?) ? T. Bodineau, B. Derrida, J.L. Lebowitz, JSP, 140 648 (2010).
In higher dimensions local currents do not vanish for large L and the correlation function does not vanish in this limit.
Steady state ? ?,? =< ? ? ? ? >- ?(r) ?(?) ?+ ?? ?,? = ? ?,? ?(?,?) corresponds to an electrostatic potential in 2? induced by ?
In 1d ?,? 0, 1 ??? + ?? ??? + ?? ??? ??(?) = (??? + ?? ?(t))??(t) + ??(?)(??? + ?? ??? ) = ??+1+ ?? 1 2?????? + ??(??+1+ ?? 1 2??) ?? For ? ?,? 1 ??(?,?) ?? = ?? ?,? + ??(?,?)
Symmetry of the correlation function: ? ?,? =< ? ? ? ? >- ?(r) ?(?) inversion particle-hole ??r,s = C ?( ?, ?) ??,?r,s = C ?,1 ?(r,s) at ? = 1/2 ???,? = ? ?(?,?) ??? ? ?,? = 0
Consequences of the symmetry: The net charge =0 (for periodic or open bc) At ? = 0.5 ???,? = ?? ?, ? Thus the charge cannot support a dipole and the leading contribution in multipole expansion is that of a quadrupole (in 2d dimensions). ?(?,?)~1/ ?2+ ?2 ?
For ? 0.5 one can expand ? in powers of ? One finds: The leading contribution to ? is of order ?2. This implies no dipolar contribution in this order due to the symmetry ? ? , (?,?) ( ?, ?) The correlation thus decays at large distances as a quadrupolar potential ?(?,?)~1/ ?2+ ?2 ?
? ?,? = ??1?,? + ?2?2?,? + ?2? 1 ?, ? = ?2? 1?,? ?2? ?, ? = ?2?(?,?) Since ?1= 0 (no dipole) and the net charge is zero the leading contribution is quadrupolar
?+ ?? ?,? = ?1?,? + ?2?,? + ?3(?,?) ?1?,? = ? ? + ??,? 2? ?,? + ? ?,? ?? ??,?+??+ ??,? ? + ? ?,? + ?? 2? ?,? + ? ? ??,? ??,? ??+ ??,? ? ?2?,? = 1 2??,?+?? ??2?2??, ?1??,0+ ? ? + ?? ? ? ? 1 2??,? ?? ??2?2??, ?1??,0+ ? ? + ?? ? ? ? ?3?,? = ? ? ? ? ??,0 ??+?1,0 ??,0 ??+?1,0 1 ??,0 ??+?1,0 1 ??,0 ??+?1,0 + ? ? ? ? ? = ? 0 1 ? ?1 + ?( ?1) 1 ?(0)
Summary I Local drive in ? > 1 dimensions results in: Density profile corresponds to a dipole in d dimensions ? ? ~ 1/?? 1 Two-point correlation function corresponds to a quadrupole in 2d dimensions ?(?,?)~1/ ?2+ ?2 ? At density ? = 0.5 to all orders in ? At other densities to leading (?2) order
The ABC Model B C A dynamics q AB BA 1 q BC CB 1 q CA AC 1 q=1 corresponds to equilibrium and the steady state is homogeneous (fully mixed). question: what is the steady state for q 1? Evans, Kafri, Koduvely, Mukamel PRL 80, 425 (1998) Clincy, Derrida, Evans, PRE 67, 066115 (2003) A. Ayyer, et al JSP , 137(5-6):1166 1204, 2009 O. Cohen, DM RRL 90, 012107 (2014)
Simple argument: q AB BA 1 CCCCA ACCCC q BC CB 1 q BBBBC CBBBB AAAAB BAAAA CA AC 1 AACBBBCCAAACBBBCCC fast rearrangement AABBBCCCAAABBBCCCC slow coarsening AAAAABBBBBCCCCCCAA The model reaches a phase separated steady state
The model exhibits strong phase separation AAAAAAAABBBABBBBBBCCCCCCCCCAA The probability of a particle to be at a distance on the wrong side of the boundary is l lq The width of the boundary layer is -1/lnq
= = N N N Special case A B C The argument presented before is general, independent of densities. For the equal densities case the model has detailed balance for arbitrary q. It turns out that for any microscopic configuration {X} one can define energy E({X}) such that the steady state distribution is ({ }) ln ({ }) E X qE X ({ }) P X q e
AAAAAABBBBBBCCCCC E=0 AB .. BA .. E E+1 BC .. CB .. E E+1 CA .. AC .. E E+1 With this weight one has: = =1 ( ) (... ...) ( ) (... ...) W AB BA P AB W BA AB P BA =q = (... ...) / (... ...) P BA P AB q ? ? ??(?)
= = This definition of energy is possible only for N N N A B C AAAAABBBBBCCCCC AAAABBBBBCCCCCA E E+NB-NC NB= NC Thus such energy can be defined only for NA=NB=NC
= = N N N A B C ( x ) ( x ) = E P q The energy E may be written as 1 N N i ( x ) ( ) = = + + 2 (N ) 3 / E C B A C B A + + + i i k i i k i i k = 1 1 i k 1 2 N (long-range interaction) Alternatively, in a manifestly translational invariant form: 1 1 N N k ( x ) ( ) = i 1 = + + E C B A C B A + + + i i k i i k i i k N = 1 k
