Magnetic knots and groundstate energy spectrum:

Transcription

1 Lecture 3 Magnetic knots and groundstate energy spectrum: - Magnetic relaxation - Topology bounds the energy - Inflexional instability of magnetic knots - Constrained minimization of magnetic energy - Groundstate energy spectra of magnetic knots and links - Bending energy spectra: Magnetic vs. elastic systems Selected references Maggioni, F & Ricca, RL 2009 On the groundstate energy of tight knots. Proc R Soc A 465, Ricca, RL 2008 Topology bounds energy of knots and links. Proc R Soc A 464, 293. Ricca, RL & Maggioni, F 2014 On the groundstate energy spectrum of magnetic knots and links. J Phys A: Math & Theor 47,

2 Magnetic relaxation Magnetic relaxation (Moffatt 1985: consider and perfectly conducting fluid; ϕ : L n L n,ϕ in a viscous dm dt = B B t d 3 X = B u B V( B ( V B ( d 3 X = u ( B B d 3 X = u J B V( B ( V B [ ] d 3 X. Near equilibrium we have: hence Thus: Du Dt dm dt V ( L n u Du Dt u = p + J B u p + J B 2 d 3 X M ( L n,ϕ t t with magnetic energy monotonically decreasing as..

3 Magnetic relaxation to groundstate Let ϕ : L n be La n,ϕ zero-framed magnetic link: N components same flux Φ i = Φ. Lk i = 0 i Magnetic helicity: H = A B d 3 x = Φ 2 Lk V (B ij Magnetic relaxation (Moffatt 1985 under i j { } V,Φ i. -preserving flow: Φ ϕ Φ Φ 1 ϕ Φ 1 knot tightening Φ 2 Φ 2 link tightening Theorem 1 (Freedman 1988; Moffatt l.b.: lim i ; ii, t M t ( = M inf M min = m Φ2 where m is a topological invariant of L N. V

4 Topology bounds the energy Theorem 2 (Arnold 1974; Freedman & He i Mt ( qh ; ii Mt ( 2 Φ 2 C, π V where q > 0 depends on the geometry of supp B. Theorem (Ricca Let L n ( be a zero-framed link. Then, we have: I : M( t q =, π V π V H m = 2 π c min M min = 2 Φ 2 c π min II :. Proof: From and we have from ( Φ i = Φ Lk i = 0 i 1,..., n = 0 H = Φ 2 Lk i + Lk ij = Φ2 i = j i j C = ε r and Lk ij = 1 2 i j r i j i j r ε r i j, we have V Lk ij ;

5 Therefore H = Φ 2 Now, from Theorem 2(ii, we have: also, by using ( C ε r = Lk ij Lk ij M i j Lk ij Φ 2 C or C H Φ 2 i j 2 Φ 2 C π V 2 π above, we have: r i j i j. Φ 2 c min V ;. ( 2 Φ 2 C π V 2 π Φ 2 V H Φ = 2 2 π H V. Hence, from Theorem 2(i follows (I. Moreover, at Theorem 1(ii we have (II. In general, we have: M min c min M min, from Question: do knot types of same c min relax to same groundstate energy?

6 Inflexional magnetic knots Inflexion at I (in isolation: c = 0 inflexion generic behaviour in 3 : X s,t ( = s 2 3 t 2 s 3, ts 2, s 3 I (Ricca & Moffatt 1992 Reidemeister type I move in action: F c ˆn Lorentz force F m (TRACE mission 2002

7 From inflexion-free knots to magnetic braids Definition. A knot in an inflexion-free configuration is a spiral knot. Theorem (Ricca Let inflexional state. Then Corollary. If K t knot, for t >t 0. K t0 K t0 K t0 denote a loose magnetic knot in is in inflexional disequilibrium. is in inflexional disequilibrium, it relaxes to a spiral inflexional instability

8 From minimal knots to spiral knots

9 Knots and links tabulation 10-crossing knot table (Tait component links by KnotPlot (Scharein 2000 up to arbitrarily large # crossings O(10 6

10 Groundstate energy of zero-framed knots and links Under signature-preserving flows, we have: M min c min Relaxation of 0-framed, 5-crossing knots: ϕ( K 5.1 ϕ( K V = 1, Φ = 1, 0-framing c min # knot types

11 Constrained minimization of magnetic energy of knots K magnetic field: ( = πa 2 L ( r,ϑ,s V K tubular knot : ; Mercier (orthogonal system: ( ( B = 0, B ϑ (, r B s r ( B = 0 B = 0, 1 L dφ P dr, 1 2πr dφ T dr + 0, ψ s, ψ ϑ Φ P Φ P Φ T fluxes, : twist parameter: h = Φ P / Φ T Φ T Theorem (Maggioni & Ricca Let us assume that (i {V, Φ} invariant (V = 1, Φ = 1; (ii circular cross-section independent of arc-length; (iii ψ independent of arc-length; (iv L independent of internal twist. Then, we have M λ * ( h = λ 4/3 2π 2/3 + π 4/3 h 2 λ 2/3 = m(λ, h. L * R * ropelength: λ = L * /R *

12 Groundstate energy spectrum: averaging over complexity m min ( h = 3 2 π 2/3 h 4/3 ( h 2 (see also Chui & Moffatt 1992 m(λ, h c min 10 tightening: V = 1, Φ = 1, h = 0 SONO (Pieranski et al RIDGERUNNER (Ashton et al c min h

13 Groundstate energy spectrum of first 250 prime knots V = 1, Φ = 1, h = 0 M * ο = ( 2π 2 tight unknot: m(# K m(# K = m ( λ(# K, 0 = * M ο λ(# K 2π 4/3 7 crossings 8 crossings 9 crossings 10 crossings 6 crossings 5 crossings 4 crossings 3 crossings # K (Rolfsen table

14 Knot energy spectrum by increasing ropelength (Ricca & Maggioni 2014 V = 1, Φ = 1, h = 0 M * ο = ( 2π 2 tight unknot: m(# K m(# K = m ( λ(#, 0 K = * M ο λ(# K 2π 4/3 Tight knots RIDGERUNNER linear fit over c min -families best fit: m(# K = 4.5ln # K # K (increasing ropelength

15 Link energy spectrum by increasing ropelength (Ricca & Maggioni 2014 V = 1, Φ = 1, h = 0 M * ο = ( 2π 2 tight unknot: m(# K m(# K = m ( λ(# K, 0 = * M ο λ(# K 2π 4/3 Tight links RIDGERUNNER linear fit over c min -families best fit: m(# K = 4.5ln # K # K (increasing ropelength

16 Bending energy spectrum of elastic knots V = 1, h = 0 tight torus: E o = π R * 2 π 5/3 e = E b E o = c(s K [ ] 2 ds 2 4/3 π 5/3 e(# K e(# K Tight knots RIDGERUNNER linear fit over c min -families best fit: e(# K = 0.2ln # K # K (increasing ropelength

17 Bending energy spectrum of elastic links V = 1, h = 0 tight torus: E o = π R * 2 π 5/3 e = E b E o = c(s K [ ] 2 ds 2 4/3 π 5/3 e(# K best fit: Tight links RIDGERUNNER linear fit over c min -families e(# K = 0.2 ln # K # K (increasing ropelength

18 Magnetic versus bending energy (Ricca & Maggioni 2017 χ(# K Tight knots: χ(# K = χ(# K 8_4_3 Tight links: χ(# K = # K (increasing ropelength

19 Ropelength versus topological crossing number [ λ(# K ] 4/3 aln # K + b λ(# K in partial agreement with [ ] 4/3 c min ; then, or Oc ( 3/4 min λ(# K Oc ( min ln 5 c min (Buck & Simon 1999; Cantarella et al. 2002; Diao 2003; Diao et al From lower bounds on energy (V = 1, Φ = 1, we have: since then M min = # K ~ A c min 2 π m(# K λ(# K ( M min mc min λ (# K 2π 1/4 c 3/4 3/4 min 2.66 c min m ο = c min π c min and ; c min better than Buck & Simon (1999, where constant (4π/11 3/ , [ ] 4/3 Assumption:, then from we have 3/4 λ(# K c min m (# K mc ( min = c min π,.,