Dynamics of interacting vortices on trapped Bose-Einstein condensates. Pedro J. Torres University of Granada

Dynamics of interacting vortices on trapped Bose-Einstein condensates Pedro J. Torres University of Granada

Joint work with: P.G. Kevrekidis (University of Massachusetts, USA) Ricardo Carretero-González (San Diego State University, USA) D.J. Frantzeskakis (University of Athens, Greece) S. Middelkamp and P. Schmelcher (Universität Hamburg, Germany) David S. Hall and D.V. Freilich (University of Massachusetts, USA)

Publications: P.J. Torres, R. Carretero-González, S. Middelkamp, P. Schmelcher, D.J. Frantzeskakis and P.G. Kevrekidis, Vortex Interaction Dynamics in Trapped Bose-Einstein Condensates, Communications on Pure and Applied Analysis 1 (11) 1589-1615. S. Middelkamp, P. J. Torres, P. G. Kevrekidis, D.J. Frantzeskakis, R. Carretero-González, P. Schmelcher, D.V. Freilich, and D.S. Hall, Guiding-center dynamics of vortex dipoles in Bose-Einstein condensates., Phys. Rev. A, Vol. 84, 1165(R) (11). P. J. Torres, P. G. Kevrekidis, D.J. Frantzeskakis, R. Carretero-González, P. Schmelcher, and D.S. Hall, Dynamics of Vortex Dipoles in Confined Bose-Einstein Condensates., Phys. Lett. A, Volume 375, Issue 33 (11), pp. 344-35

Physical background A Bose-Einstein condensate (BEC) is a state of matter of a dilute gas of weakly interacting bosons confined in an external potential and cooled to temperatures very near absolute zero ( K or -73.15 -C). Under such conditions, a large fraction of the bosons occupy the lowest quantum state of the external potential, at which point quantum effects become apparent on a macroscopic scale.

Physical background Vortices can be created on a BEC. a) 5 m 6 1 18 4 3 36 4 445 ms b).65 c) 8 1 6 Separation (units of ) d).6.55.5.45.4.35 1 3 4 5 time (ms) Coordinates in the trap ( m) 6 4 4 6 8 4 3 6 1 18 4 3 36 4 445 ms

Physical background The dynamical evolution of such vortices is a natural question

The model for n vortices ẋ k = c(r k,t)s k y k b j k ẏ k = +c(r k,t)s k x k + b j k y k y j S j rjk, x k x j S j rjk, k = 1...n, where (x k,y k ) coordinate of vortex k S k = ±1 charge of vortex k r jk = (x j x k ) + (y j y k ) separation between vortex j and vortex k r k = x k + y k distance of vortex k to the center. c(r k,t) trap coefficient, positive and T- periodic in time.

The case c(r i, t) c(t). Two co-rotating vortices. Fix n =, S 1 = S = 1. ẋ 1 = c(t)y 1 b y 1 y r, ẋ = c(t)y b y y 1 r ẏ 1 = +c(t)x 1 + b x 1 x r, ẏ = +c(t)x + b x x 1 r,

The case c(r i, t) c(t). Two co-rotating vortices. Fix n =, S 1 = S = 1. Change: ẋ 1 = c(t)y 1 b y 1 y r, ẋ = c(t)y b y y 1 r ẏ 1 = +c(t)x 1 + b x 1 x r, ẏ = +c(t)x + b x x 1 r, s 1 = x 1 + x d 1 = x 1 x s = y 1 + y d = y 1 y.

The case c(r i, t) c(t). Two co-rotating vortices. Fix n =, S 1 = S = 1. ẋ 1 = c(t)y 1 b y 1 y r, ẋ = c(t)y b y y 1 r Gives: ẏ 1 = +c(t)x 1 + b x 1 x r, ẏ = +c(t)x + b x x 1 r, s 1 = c(t)s s = c(t)s 1

The case c(r i, t) c(t). Two co-rotating vortices. Fix n =, S 1 = S = 1. Solution: ẋ 1 = c(t)y 1 b y 1 y r, ẋ = c(t)y b y y 1 r ẏ 1 = +c(t)x 1 + b x 1 x r, ẏ = +c(t)x + b x x 1 r, with C(t) = t c(s)ds. s 1 = Acos (C(t) + B) s = Asin (C(t) + B),

The case c(r i, t) c(t). Two co-rotating vortices. d 1 = c(t)d bd (d 1 + d ), d = +c(t)d 1 + bd 1 (d 1 + d ).

The case c(r i, t) c(t). Two co-rotating vortices. Change to polar coordinates: From r = d 1 + d, we get d 1 = c(t)d bd (d 1 + d ), d = +c(t)d 1 + bd 1 (d 1 + d ). d 1 = r cos ϕ, d = r sin ϕ, rṙ = d 1 d 1 + d d =.

The case c(r i, t) c(t). Two co-rotating vortices. Change to polar coordinates: From r = d 1 + d, we get d 1 = c(t)d bd (d 1 + d ), d = +c(t)d 1 + bd 1 (d 1 + d ). d 1 = r cos ϕ, d = r sin ϕ, rṙ = d 1 d 1 + d d =. Hence, r is a constant of motion.

The case c(r i, t) c(t). Two co-rotating vortices. The angular component ϕ = arctan (d /d 1 ) yields d 1 ϕ = d 1d d r = b r + c(t). Since r is constant, ϕ(t) = b t + C(t) + D, r where D is an arbitrary constant. The general solution reads ( ) b d 1 = R cos R t + C(t) + D, ( ) b d = R sin R t + C(t) + D, where D and R are arbitrary constants. Hence, the case of two corotating vortices is explicitly solvable. Generically, one finds quasi-periodic solutions and there is a sequence of periodic solutions that can be obtained by fine-tuning R.

The case c(r i, t) c(t). Two co-rotating vortices. y i 4 4 4 4 x i 4 s, d 5 5 5 5 s 1, d 1 x i, y i 4 5 s i, d i 5 1 3 4 5 6 t Figure: Periodic orbit corresponding to two vortices with constant trapping coefficient c(t) =.1, vortex-vortex interaction coefficient b = 1. The vortices are initially placed at (x 1(), y 1()) = (, ) (see [blue] square) and (x (), y ()) = (4, ) (see [green] circle).

The case c(r i, t) c(t). Two co-rotating vortices. y i 4 4 4 4 x i 4 s, d 5 5 5 5 s, d 1 1 x i, y i 4 5 s i, d i 5 5 1 15 5 3 t Figure: Same as before but for a quasi-periodic orbit for a constant trapping coefficient c(t) = π/.

The case c(r i, t) c(t). Two co-rotating vortices. 4 5 y i s, d 4 4 4 x i 5 5 5 s 1, d 1 4 x i, y i 4 5 s i, d i 5 4 6 8 1 t Figure: A quasi-periodic orbit for c(t) =.1(1 + ǫ sin(ωt)) and b = 1 with ǫ =.5 and ω = π/3.

The case c(r i, t) c(t). Two co-rotating vortices. y i 4 4 4 4 x i 4 s, d 5 5 5 5 s, d 1 1 x i, y i 4 5 s i, d i 5 4 6 8 1 t Figure: Same as in Fig. 3 but for a larger perturbation strength of ǫ =.5. The apparent complex motion displayed in the top-left panel (in original coordinates) is nothing but a quasi-periodic orbit that is better elucidated in transformed coordinates in the bottom panel.

The case c(r i, t) c(t). Two counter-rotating vortices. Fix n =, S 1 = 1,S = 1. ẋ 1 = c(t)y 1 + b y 1 y r, ẋ = c(t)y b y y 1 r ẏ 1 = +c(t)x 1 b x 1 x r, ẏ = c(t)x + b x x 1 r,

The case c(r i, t) c(t). Two counter-rotating vortices. Fix n =, S 1 = 1,S = 1. ẋ 1 = c(t)y 1 + b y 1 y r, ẋ = c(t)y b y y 1 r ẏ 1 = +c(t)x 1 b x 1 x r, ẏ = c(t)x + b x x 1 r, ṡ 1 = +b d r c(t)d, and ṡ = b d 1 r + c(t)d 1, d 1 = c(t)s, d = +c(t)s 1.

The case c(r i, t) c(t). Two counter-rotating vortices. Changing time τ = C(t) t c(s)ds yields the equivalent system ṡ 1 = +f(τ) d r d, ṡ = f(τ) d 1 r + d 1, d 1 = s, d = +s 1, (1) where f(τ) b/c(t(τ)).

The case c(r i, t) c(t). Two counter-rotating vortices. d 1 + d 1 = f(τ) d 1 r, d + d = f(τ) d r. Recalling that r = d 1 + d, one notes that this Newtonian system has a radial symmetry. This type of systems plays a central role in Celestial Mechanics. Change to polar coordinates: d 1 = r cos(ϕ), d = r sin(ϕ), leads to () r + r = l r 3 + f(τ) r, (3) ϕ = l r, (4) where l r ϕ = d 1 d d d 1 = r1 r = const. is the angular momentum of the solution (d 1,d ).

The case c(r i, t) c(t). Two counter-rotating vortices. Equation (3) is decoupled and can be studied separately. If r(t + T) = r(t): ϕ(τ) = τ l r ds and r(t + T) = r(t) ϕ(t + T) = ϕ(t) + θ, where θ ϕ(t) = T l/r ds is the rotation number. Coming back to the Cartesian coordinates d = (d 1,d ) and using the more convenient complex notation, d(t + T) = e iθ d(t). Therefore, from a T-periodic solution of (3) we get a quasi-periodic solution of the original system. If θ = (stationary case) the solution d is T-periodic, whereas if θ = π/k then d is kt-periodic (subharmonic of order k).

The case c(r i, t) c(t). Two counter-rotating vortices. r + r = l r 3 + f(τ) r For simplicity, take c(t) = 1 + ǫ sin(ωt). Theorem Assume that ω >. Then, for any l, (3) has a T-periodic solution such that b ( π ) r(t) > r = 1 + ǫ cos. ω

The case c(r i, t) c(t). Two counter-rotating vortices. Sketch of the proof. We summarize the technique employed in [Torres, JDE, 3]. ω > implies that the linear operator Lr := r + r has a positive Green s function G(t,s) such that ( π ( π m := cot G(t,s) M := 1/sin. ω) ω) A T-periodic solution of Eq. (3) is a fixed point of the operator Ar = T [ l G(τ, s) r 3 + f(τ) ] ds. r Apply Krasnoselskii s fixed point Theorem for cone compression-expansion with the cone { P = u X : min u m } x M u, where X the Banach space of the continuous and T-periodic functions with the norm of the supremum.

The case c(r i, t) c(t). Two counter-rotating vortices. In consequence, there is a continuous branch of T-periodic solutions (l,r l ) of Eq. (3). Now we analyze the stability of such solutions. Theorem Take l. Then, there exists an explicitly computable ω l such that if ω > ω l then r l is linearly stable as a T-periodic solution of Eq. (3). Proof. The linearized equation around r l is ÿ + a(τ)y = with a(τ) = 1 + 3l r 4 l + f(τ) rl. Now, use r(t) > r and apply Lyapunov s stability criterion.

The case c(r i, t) c(t). Two counter-rotating vortices. It is possible to prove analytically the presence of KAM dynamics in Eq. (3), by using a Ortega s third order approximation method as in [Torres,Adv.Nonlinear Stud.,]. Theorem Let us assume l =, ω >, ω {3,4 }. Then the T-periodic solution r of Eq. (3) is of twist type except possibly for a finite number of values of ǫ.

The case c(r i, t) c(t). Two counter-rotating vortices. 5 1 y i s, d 5 5 5 5 5 x i 1 1 1 s, d 1 1 5 x i, y i 5 1 5 s i, d i 5 1 4 6 8 1 t Figure: Opposite charge vortices S 1 = 1 (see solid [blue] line) and S = 1 (see dashed [green] line) with constant trapping coefficient c(t) =.1, vortex-vortex interaction coefficient b = 1. The vortices are initially placed at (x 1(), y 1()) = (, ) (see [blue] square) and (x (), y ()) = (4, ) (see [green] circle).

The case c(r i, t) c(t). Two counter-rotating vortices. 1 5 1 y i s, d 5 1 1 1 1 x i 1 1 s, d 1 1 1 5 x i, y i 5 1 1 s i, d i 1 4 6 8 1 t Figure: An irregular orbit corresponding to c(t) =.1(1 + ǫ sin(ωt)) and b = 1 with ǫ = 1 and ω =.1.

The Fetter case. In a more realistic situation, the precession frequency depends also on r. Fetter [1] proposed ω ( ) ( ) r c(r,t) Ω(r) = 8µ(1 r /R ) 3 + ω r µ 5ωz ln, ω r where ω r (ω z ) is the radial (axial) trap frequency, R is the Thomas-Fermi radial extent of the condensate (can be normalized to 1), and µ is the chemical potential.

The Fetter case. The model is or in complex variables ẋ k = S k Ω(r k )y k b ẏ k = S k Ω(r k )x k + b n j k n j k iz k = S k Ω(r k )z k + b y k y j S j rjk, x k x j S j rjk. n j k S j z k z j r jk

Conserved quantities and integrabitlity. The system is hamiltonian. Define H(z 1,...,z n ) = Ω where Ω Ω(). Then, n ln(1 rk) + b 4 k=1 n k=1 j k n S j S k ln(rjk) S k ẋ k = H y k, S k ẏ k = H x k for every k = 1,...,n. H is a the first integral, i.e., a first conserved quantity along the orbits of the system. A second integral of motion is V = which is a form of inertial momentum. n S k rk, k=1

Conserved quantities and integrabitlity. The presence of two conserved quantities or dynamical invariants guarantees integrability in the classical Liouville sense for the case n =. This implies that the energy level sets are compact and the phase space is foliated by invariant tori. On each of these, the motion is quasi-periodic with two frequencies.

The Fetter case with n =, S 1 = 1, S = 1. Theorem The two vortices never collide, i.e., they are separated by a computable minimal distance.

The Fetter case with n =, S 1 = 1, S = 1. Theorem The two vortices never collide, i.e., they are separated by a computable minimal distance. Taking exponentials on the hamiltonian, we have (1 r 1) Ω (1 r ) Ω r b/ 1 = C > t, where C = (1 r 1 () ) Ω (1 r () ) Ω r 1 () b/. Note that < C < b/ because r i () < 1 (i = 1,) and < r 1 () <. Then, r 1 (t) > C 4/b t,

The Fetter case with n =, S 1 = 1, S = 1. Theorem There is an equilibrium, which is unique up to rotations, given by b b (x 1,y1) = ( 4Ω + b,), (x,y) = ( 4Ω + b,)

The Fetter case with n =, S 1 = 1, S = 1. To emphasize the rotational invariance of our system, it is convenient to work on the polar coordinates (r 1,r,θ). The (unique) equilibrium is (r1,r,θ b ) = ( 4Ω, b +b 4Ω,π). +b Theorem The equilibrium (r1,r,θ ) is stable (in the sense of Lyapunov).

The Fetter case with n =, S 1 = 1, S = 1. The hamiltonian H is still a conserved quantity which in the new variables reads H(r 1,r,θ) = 1 [ Ω ln(1 r 1) + Ω ln(1 r) + b ] ln(r 1 + r r 1 r cos θ). The equilibrium (r1,r,θ ) is a critical point of H. The hessian matrix evaluated on (r1,r,θ ) is 7b + 1Ω 1 + b Ω b + 4Ω b + 4Ω 8 7b + 1Ω + b Ω. b One can prove easily that this matrix is positive-definite by Sylvester s criterion. Hence H attains a minimum at (r 1,r,θ ), H is a Lyapunov function and the equilibrium is, in fact, stable.

The Fetter case with n =, S 1 = 1, S = 1. a) 5 m 6 1 18 4 3 36 4 445 ms b).65 c) 8 1 6 Separation (units of ) d).6.55.5.45.4.35 1 3 4 5 time (ms) Coordinates in the trap ( m) 6 4 4 6 8 4 3 6 1 18 4 3 36 4 445 ms

The Fetter case with n =, S 1 = 1, S = 1. a) 5 m b) c) 6 1 18 4 3 36 4 445 ms d) 6 1 6 e) 8 1 6 f) 1 1 6 6 8 4 6 4 4 4 4 4 6 6 8 6 4 3 8 4 3 1 4 3 g) 6 1 6 h) 8 1 6 i) 1 1 6 6 8 4.45 Hz 1.5 Hz 6 -.5 Hz 4 4 4 4 4 6 6 8 6 4 3 8 4 3 1 4 3 in-trap radius ( m) in-trap radius ( m)

Guiding Center Equilibria and their Stability We are looking for guiding centers about which the vortices oscillate. These guiding centers might themselves be precessing around at some frequency ω. For the case of symmetric guiding centers this precession frequency will be zero. We adopt a trial solution of the form z 1 (t) = r 1 exp(iωt) and z (t) = r exp(iωt + iπ) = r exp(iωt). (5) where r 1 and r are now constant, as is ω, the precession frequency of the guiding center.

Guiding Center Equilibria and their Stability Fix α = 1 1 r1, β = 1 1 r, and γ = 1 r1 = 1 s, (6) where s = r 1 = r 1 + r is the separation distance between the two guiding centers.

Guiding Center Equilibria and their Stability Fix α = 1 1 r1, β = 1 1 r, and γ = 1 r1 = 1 s, (6) where s = r 1 = r 1 + r is the separation distance between the two guiding centers. After some algebra, one finds that the precession frequency should be ω = 1 [ ( r1 Ω (α β) + γb r )]. (7) r r 1 Note that if r = r 1, then α = β and ω =, as we expect. On the other hand, once the position of r is fixed, the location of the first vortex r r 1 can be found from the third order polynomial Ω βr 3 + b r r Ω (1 + β) r b r =. (8)

Open problems. The Fetter case with two corotating vortices (n =,S 1 = S = 1) can be handled in a similar way. Dynamics of many vortices, both in the Fetter and non-fetter case. Collisions. Effect of a more complicated precession frequency (for instance non symmetric).

Thank you for your attention!!!