Institute of Natural Sciences Shanghai Jiao Tong University

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1 Institute of Natural Sciences Shanghai Jiao Tong University two 1 May, 24th Joint work with Mathieu Lutz 1 LMBA (UMR 6205), Université de Bretagne-Sud, F-56017, Vannes, France. emmanuel.frenod@univ-ubs.fr

2 Charge particles submitted to Strong Magnetic Field two In Usual : (x, v) = (x 1, x 2, x 3, v 1, v 2, v 3 ) X(t; x, v, s), V(t; x, v, s) X = V V = q (E(X) + V B(X)) m B : Strong Applied piece + Strong Self Induced piece + }{{} 1 B Self } Induced {{ Perturbations } Forgotten E : Self Induced piece (E = Φ, B = A)

3 Helicoidal trajectories - Larmor Radius two Source: S. Jardin s Lectures at Cemracs 10 In Tokamak: Electron Larmor Radius m Ion Larmor Radius 10 2 m

4 Dimensionless Dynamical two Ion Larmor Radius Tokamak size X = V V 10 2 m 10m = E(X) + V B(X) 10 3

5 Simplifications two Skip E Turn to dimension 2: x = (x 1, x 2 ), v = (v 1, v 2 ), with B (x) = (0, 0, B(x 1, x 2 )) B > 1, B(x 1, x 2 ) = A(x 1, x 2 ) = A 2 x 1 (x 1, x 2 ) A 1 x 2 (x 1, x 2 ) X = V, X(0) = x 0, V = 1 B(X) V = 1 ( ) B(X) V2, V(0) = v V 0 1 X 1 X 2 V 1 V 2 V 1 V = 2 1 B(X) V 2 1, B(X) V 1 X 1 X 2 V 1 V 2 (0) = x 01 x 02 v 01 v 02

6 Gyrokinetic model two Z = J B (Z) ( ) Z1 = J Z 2 B (Z) B (Z), Z(0) = z 0 B x 2 (Z) B x 1 (Z), Z(0) = z 0 for magnetic moment J

7 What is hidden two Z = J B (Z), Z(0) = z 0 B (Z) Γ = B (Z) J ( + 2B (Z) 2 B (Z) 2 B (Z) 3 ( B ((Z))) 2), Γ(0) = γ 0 J = 0, J (0) = j 0

8 Key result two IF: In coordinate system r = (r 1, r 2, r 3, r 4 ), a Dynamical writes: with R = P(R) rh(r) P(r) = THEN: (Trajectory R = (R 1, R 2, R 3, R 4 )) H r 3 = 0 M = 0 AND: R 4 = 0 r 4 M(r)

9 Panorama two Usual (x, v) 1:? X = V V = 1 B (X) V 2 Canonical (q, p) H = H (q, p): Q = p H P = q H 3 4: Method (x, θ, v) Almost Canonical (y, θ, v) 5: Method (z, γ, j)

10 Panorama two Usual (x, v) H (x, v), P (x, v) s.t: X V = P x,vh 1:? Canonical (q, p) H (q, p), P (q, p)=s s.t: Q P = S q,p H 2 3 4: Method (x, θ, v) H (v), P (x, θ, v) Almost Canonical (y, θ, k) 5: Method H (y, θ, k), P (y) (z, γ, j) Ĥ (z, j), P (z) = P (z)

11 Canonical two Usual : (x, v) = (x 1, x 2, v 1, v 2 ) Trajectory : (X(t; x, v, s), V(t; x, v, s)) ((X, V) = (X 1, X 2, V 1, V 2)) X = V B(x) = A(x) V = 1 B(X) V Canonical : (q, p) = (q 1, q 2, p 1, p 2 ) Trajectory : (Q(t; q, p, s), P(t; q, p, s)) ((Q, P) = (Q 1, Q 2, P 1, P 2)) q = x x = q p = v + A(x) v = p A(q) Q = X P = V + A(X) X = Q V = P A(Q) Q 1 H P = S q,p H (q, p) = p A(q) 2 2 ( ) 0 I3 S = I 3 0

12 Check of Canonical nature of Canonical two Q P = S q,p H, H (q, p) = 1 p A(q) 2 2 Q = p H (Q, P) = P A(Q) P = q H ( (Q, P) = ( A(Q))T P A(Q) ) ( A) T (p A) = ( A)(p A) + ( A) (p A) Q = P A(Q) P ( A(Q)) ( P A(Q) ) = A(Q) ( P A(Q) )

13 Check of Canonical nature of Canonical Coord. - 2 two Q = P A(Q) P ( A(Q)) ( X = V P ( A(Q)) X = V V = A(X) V P A(Q) ) [ ( Q ) = = A(Q) P A(Q) ] X = Q V = P A(Q) ( P A(Q) ) = A(Q) ( P A(Q) )

14 As by products : Poisson Matrix, Poisson Bracket, Change of Formula two In any coordinate system r = (r 1, r 2, r 3, r 4 ), the Dynamical writes: R = P(R) rh(r) = {I, H}(R) {f, g}(r) = ( r f (r)) (P(r)( r g(r))) (f and g : R 4 R) ({f, g}(r)) i = ( r f i (r)) (P(r)( r g(r))) (f : R 4 R 4 and g : R 4 R) I(r) = r Another coordinate system r = ( r 1, r 2, r 3, r 4 ) with r = ρ(r), r = ρ( r) = ρ 1 ( r) R = P( R) r H( R) H( r) = H( ρ( r)) ( P( r)) ij = { ρ i, ρ j } ( ρ( r))

15 Panorama two Usual (x, v) H (x, v), P (x, v) s.t: X V = P x,vh 1:? Canonical (q, p) H (q, p), P (q, p)=s s.t: Q P = S q,p H 2 3 4: Method (x, θ, v) H (v), P (x, θ, v) Almost Canonical (y, θ, k) 5: Method H (y, θ, k), P (y) (z, γ, j) Ĥ (z, j), P (z) = P (z)

16 Function and Poisson Matrix in Usual two H (x, v) = 1 v 2 ( H (q, p) = 1 2 p A(q) 2 ) 2 ( ) 0 x,v H = v ( ) 0 I2 P (x, v) = Indeed: I 2 ( A(x)) T ( A(x)) v P (x, v) x,v H = ( A(x)) T ( A(x)) v To be compared with: X = V V = A(X) V = 1 B(X) V

17 Formula giving Poisson Matrix two P (x, v) = ( 0 I2 I 2 ( A(x)) T ( A(x)) Change of coordinates: r and r with r = ρ(r) and r = ρ( r) = ρ 1 ( r) ) ( P( r)) ij = { ρ i, ρ j } ( ρ( r)) Here: (x, v) and (q, p) with (x, v) = ξ(q, p) and (q, p) = π(x, v) For instance : ξ 3 (q, p) = p 1 A1(q) and ξ 4 (q, p) = p 2 A2(q) 1 A 1 q 1 (q) ) 1 A 1 q 1 (q) A 1 A 2 ξ 3 (q, p) = 1 ( q 2 (q) 0 1 S = I2 I {ξ 3, ξ 4 }(π(x, v)) = 1 ξ 4 (q, p) = ( A2 (x) A ) 1 (x) q 1 q 2 1 q 2 (q) 0 1

18 (in velocity) two x 2 x 1 ρ θ v (x, θ, v): v = ( ) v, cos θ θ s.t. v = v sin θ

19 Function and Poisson Matrix in H (x, θ, v) = v 2 2 (H (x, v) = 1 v 2 ) 2 two P (x, θ, v) = cos (θ) 0 0 v 0 0 cos (θ) v sin (θ) v sin (θ) v sin (θ) cos (θ) B (x) v 0 sin (θ) cos (θ) B (x) v 0 Singularity in v = 0 and ω = B (x) v > 0

20 Panorama two Usual (x, v) H (x, v), P (x, v) s.t: X V = P x,vh 1:? Canonical (q, p) H (q, p), P (q, p)=s s.t: Q P = S q,p H 2 3 4: Method (x, θ, v) H (v), P (x, θ, v) Almost Canonical (y, θ, k) 5: Method H (y, θ, k), P (y) (z, γ, j) Ĥ (z, j), P (z) = P (z)

21 Method Target two Find a Coordinate (y, θ, k) s.t. Poisson Matrix (P ) shape: M or M (y, θ, k) = Υ(x, θ, v), (x, θ, v) = ξ(y, θ, k), (ξ = Υ 1 ) (P (y, θ, k)) ij = {Υ i, Υ j }(ξ(y, θ, k)), {Υ i, Υ j } = ( Υ i ) ( P ( Υ j )) Needed: {Υ 4,Υ 3 } = 1 (={Υ k,υ θ }={k, θ}) {Υ 1,Υ 3 } = 0 (={Υ y1,υ θ }={y 1, θ}) {Υ 1,Υ 4 } = 0 (={Υ y1,υ k }={y 1, k}) {Υ 2,Υ 3 } = 0 (={Υ y2,υ θ }={y 2, θ}) {Υ 2,Υ 4 } = 0 (={Υ y2,υ k }={y 2, k})

22 First equation processing two {Υ 4,Υ 3 } = 1 (={Υ k,υ θ }={k, θ}) or {Υ 3, Υ 4 } = 1 (={Υ θ Υ k,}={θ, k}) ( ) Υ 3 (= Υ θ ) = (0, 0, 1, 0) T {Υ 3, Υ 4 } = ( Υ 3 ) ( P ( Υ 4 )): penultimate comp. of ( P ( Υ 4 )) ( ) cos(θ) Υ 4 sin(θ) Υ 4 Υ 4 + ω x 1 x 2 v = 1 (ω (x, v) = B(x) v ) (Υ 4 (x, θ, v) = k) Method of Characteristics

23 Method of Characteristics - 1: exact solution two cos(θ) Υ 4 x 1 sin(θ) Υ 4 x 2 + ω Υ 4 v = 1, (ω (x, v) = B(x) v ) Υ 4 cos(θ) Υ 4 + v v sin(θ) Υ 4 = v v B(x) x 1 B(x) x 2 B(x) Υ 4 v=0 = 0 (In fact Υ 4 v=ν = 0 for ν 0) X 1 (θ; v; x, u) s.t. X 1 v X 2 (θ; v; x, u) s.t. X 2 v = v cos(θ) B(X 1, X 2 ), X 1(θ; u; x, u)=x 1 v sin(θ) = B(X 1, X 2 ), X 2(θ; u; x, u)=x 2 v Υ 4 (x, θ, v) = Υ 4 (X (θ; 0; x, v), θ, 0) + 0 = s B(X (θ; s; x, v)) ds v Gives explicit expression of k in terms of (x, θ, v) 0 s B(X (θ; s; x, v)) ds

24 Method of Characteristics - 2: Asymptotic expansion two ( ) X1 (θ; v; x, u) X (θ; v; x, u) = X 2 (θ; v; x, u) X v = v something X close to x X = x+vx v 2 X cos(θ) X = v F(X, v, θ), X (θ; u, x, u) = x, (F(x, θ) = B(x 1, x 2 ) v sin(θ) ) B(x 1, x 2 ) X = x + vf(x, θ) + v 2 L F (F)(x, θ) + v 3 L 2 F(F)(x, θ) +... L(F) linked to derivative Indentifying with: X = x + vx v 2 X gives

25 Method of Characteristics - 3: Asymptotic expansion two Υ 4 (x, θ, v) = X 1 = F(x, θ), X 2 = L F (F)(x, θ),... v 0 v s B(X (θ; v; x, u)) ds = s v B(x) ds + s 2 T 1 ( B(x) ) X 1 ds+ v ( + 2 s 3 T 2 1 ( B(x) ) X 1 + T 1 1 ( B(x) ) X 2) ds = v 2 2B(x) +... (T i linked with the Taylor expansion coefficients) Gives new variable k as an expansion in

26 On other equations - Poisson Matrix in {Υ 4,Υ 3 } = 1 (={Υ k,υ θ }={k, θ}) Processed. Gave k two {Υ 1,Υ 3 } = 0 (={Υ y1,υ θ }={y 1, θ}), {Υ 1,Υ 4 } = 0 (={Υ y1,υ k }={y 1, k}) {Υ 2,Υ 3 } = 0 (={Υ y2,υ θ }={y 2, θ}), {Υ 2,Υ 4 } = 0 (={Υ y2,υ k }={y 2, k}) To be Processed. Will give y and k in terms of (x, θ, v) and expansions in : Υ = Υ 0 + Υ Υ Hence: (y, θ, k) gotten Last term of new Poisson matrix P (y, θ, k): (P ) 12 = (P ) 21 = {Υ 1,Υ 2 }(={Υ y1,υ y2 }={y 1, y 2 }), 0 P (y, θ, k) = B(y) 0 0 B(y)

27 Function in two We know: H (x, θ, v) = v 2 2 H (y, θ, k) = H (ξ(y, θ, k)) with ξ = Υ 1 We do : Υ = Υ 0 + Υ Υ ξ = ξ 0 + ξ ξ H (ξ 0 + ξ ξ ) = H (ξ 0 ) + T 1 ( H )(ξ 0 ) ξ H (y, θ, k) = B(y)k + H 1 (y, θ, k) + 2 H 2 (y, θ, k) First term : Independent of θ

28 Let us take stock two Usual (x, v) H (x, v), P (x, v) s.t: X V = P x,vh 1:? Canonical (q, p) H (q, p), P (q, p)=s s.t: Q P = S q,p H 2 3 4: Method (x, θ, v) H (v), P (x, θ, v) Almost Canonical (y, θ, k) 5: Method H (y, θ, k), P (y) (z, γ, j) Ĥ (z, j), P (z) = P (z)

29 Transform based Method Target - 1 two We have P (y, θ, k): nice shape. But: H (y, θ, k) = H 0 (y, θ, k) + H 1 (y, θ, k) + 2 H 2 (y, θ, k) +... depends on θ. Key result θ independent Function. Target: Change of coordinates (y, θ, k) (z, γ, j) = ζ(y, θ, k) leaving P unchanged, ( P (z, γ, j) = P (z, γ, j), P (y, θ, k) = P (y, θ, k)) parametrized, close to identity, i.e.: ζ(y, θ, k) = [ζ()](y, θ, k) = (y, θ, k) + ζ 1 (y, θ, k) + 2 ζ 2 (y, θ, k) +... Ĥ (z, γ, j) = H (λ(z, γ, j)) = Ĥ0 (z, j) + Ĥ1 (z, j) + 2 Ĥ 2 (z, j) +... (λ = ζ 1 ) + N Ĥ N (z, j) + N+1 Ĥ N+1 (z, θ, j)

30 Transform based Method Target - 2 two We have P (y, u, k, θ): nice shape. But: H (y, θ, k) = H 0 (y, θ, k) + H 1 (y, θ, k) + 2 H 2 (y, θ, k) +... depends on θ. Key result θ independent Function. Target: Change of coordinates (y, θ, k) (z, γ, j) = ζ(y, θ, k) leaving P almost unchanged (up to order N in ) ( P (z, γ, j) = P (z, γ, j) + N Something, P (y, θ, k) = P (y, θ, k) + N Something) parametrized, close to identity, i.e.: ζ(y, θ, k) = [ζ()](y, θ, k) = (y, θ, k) + ζ 1 (y, θ, k) + 2 ζ 2 (y, θ, k) +... Ĥ (z, γ, j) = H (λ(z, γ, j)) = Ĥ0 (z, j) + Ĥ1 (z, j) + 2 Ĥ 2 (z, j) (λ = ζ 1 ) + N Ĥ N (z, j) + N+1 Ĥ N+1 (z, θ, j)

31 A remark two X f = P f ; ϑ i,j, f ij N {( i n=0 ( X f ) n {g, h} = n (( i n=0 ) jn ( X ) n g, n! f ( i (y, θ, k)= n=0 k=0 C k n ) ) jn ( X ) n {g, h} = n! f ( i n=0 ) jn ( X ) n n! f { ( X ) k f g, ( X ) n k } f h ) } jn ( X ) n h + N Something n! f r 1 r 2 r 3 (y, θ, k) r 4 r 1 : (y, θ, k) y 1, r 2 : (y, θ, k) y 2, r 3 : (y, θ, k) θ, r 4 : (y, θ, k) k

32 Consequence of the remark two For g 1,..., g N, α i = min {k N s.t. ki N} (= E( N i ) + 1) Ĥ (z, γ, j) = ζ = ϑ α1,1, ḡ 1 ϑ α2,2, ḡ 2... ϑ α N,N, ḡ N, (λ = ζ 1 ) H (λ(z, γ, j)) = ( αn n=0 ( α1 n=0 Nn n! Since for i, j s.t. ij N ) ( X n ) g, f {( i n=0 jn n! ) ( n ( X ) α2 ) n 2n ḡ n! 1 ( X ) n ḡ n! 2... n=0 ) ( X ) n g N H (z, γ, j) + N+1 Something, (( i n=0 jn n! ( i n=0 jn n! ( X f ) n ) ( X f ) n ) h ) {g, h} } = + N Something ( ˆP (z, θ, j)) k,l = {ζ k, ζ l } (λ(z, θ, j)) = ( P (z, θ, j)) k,l + N 1 Something

33 game to play two Build g 1,..., g N s.t. Ĥ 0 (z, j)+ĥ1 (z, j)+ 2 Ĥ 2 (z, j)+ N Ĥ N (z, j)+ N+1 Ĥ N+1 (z, θ, j) = ( αn H (λ(z, γ, j)) = n=0 Nn n! ( αn = Nn n! n=0 ( α1 n n=0 Ĥ (z, γ, j) = ( α1 ) ( n ( X ) α2 ) n 2n ḡ n! 1 ( X ) n ḡ n! 2... n=0 n=0 ) ( X ) n g N H (z, γ, j) + N+1 Something n! ( X ) n g N ) ( X ḡ 1 ) n ) ( α2 n=0 2n n! ( X ḡ 2 ) n )... ( H 0 (z, γ, j)+h 1 (z, γ, j)+ 2 H 2 (z, γ, j) N+1 Something, )

34 If you play the game... two... with: P (y, θ, k) = T 0 = B(y) 0 0 B(y) ; T 2 (y) = = 1 T 0 + T 2 (y) B(y) 0 0 B(y) Ĥ 0 (z, j) = H 0 (z, γ, j), ( T 0 ḡ 1 ) H0 = H 1 Ĥ1 ( T 0 ḡ 2 ) H0 = V 2 (H 1, H 2, g 1 ) Ĥ2 etc.

35 ... you have two Ĥ 0 (z, j) = H 0 (z, γ, j), Ĥ 1 (z, j) = 1 2π H 1 dγ, 2π 0 ( ) T 0 ḡ 1 H0 = H 1 1 2π H 1 dγ 2π 0 Ĥ 2 (z, j) = 1 2π V 2 (H 1, H 2, g 1 )dγ 2π 0 ( ) T 0 ḡ 2 H0 = V 2 (H 1, H 2, g 1 ) 1 2π etc. 2π 0 V 2 (H 1, H 2, g 1 )dγ

36 At the end of the day two Usual (x, v) H (x, v), P (x, v) s.t: X V = P x,vh 1:? Canonical (q, p) H (q, p), P (q, p)=s s.t: Q P = S q,p H 2 3 4: Method (x, θ, v) H (v), P (x, θ, v) Almost Canonical (y, θ, k) 5: Method H (y, θ, k), P (y) (z, γ, j) Ĥ (z, j), P (z) = P (z)

37 Hence two In : Trajectories (Z, Γ, J ) Z = Something independent of Γ + N+1 Remainder(Z, Γ, J ) Γ = Something complicated J = N 1 Something(Z, Γ, J ) (Z T, Γ T, J T ): Z T = Something independent of Γ Γ T = Something complicated J T = 0 (Z T, Γ T, J T )(t) (Z, Γ, J )(t) C N 1

38 Implementing with N=3 two Z T Γ T J T = J B (Z T B ( Z T ), ) = B ( Z ) T J T ( + 2 (B (Z T )) 2 = 0, B ( Z T ) 2 B ( Z T ) 3 ( B ( Z T )) 2 )

39 two Thank for your attention