Forced magnetic reconnection and its applications B. H. Oh 1 1 Seoul National University, Department of Nuclear Engineering supervised by Prof. T. S. Hahm 1 KSTAR Conference 2014, 24-26 Feb 2014, Gangwon-do, Korea 1
Introduction Motivation Resonant surface � · � � 0 Magnetic islands? or Surface current? Toroidal magnetostatic Toroidal magnetostatic equilibrium with boundary perturbation axisymmetric equilibrium Two class of equilibrium solutions for perturbing boundary Solution I The equilibrium with the original topology(nested tori) and they have surface currents on this resonant surface The equilibrium with magnetic islands on the resonant surface of the original Solution II equilibrium, but they have no surface currents 2
Introduction In the work by Hahm and Kulsrud, they treated a simple classic model problem suggested by Taylor to determine which one is the correct solution. [Hahm and Kulsrud, Phys. Fluids 28, 2412 (1985)] The geometry of the probelm Sheared magnetic field � �� Perfect conducting wall at x=-a Resonant surface Perfect conducting wall at x=a � � � � �̂ � � � � � , Toroidal magnetic field � � � � � � � . Sheared Introducing the flux function, magnetic field � � � � � 2� � � , � � � � � � cos �� , � � � � �̂ � �̂ � ��. The displacement of The displacement of the boundary is � . the boundary is � . 3
The Time Independent Equilibrium Solutions Review of fundamentals Current sheet (outward direction) Magnetic islands � ~ �� � 0 . � � �� half width of magnetic islands If � � 0 = 0, there are no magnetic islands at x=0. Current sheet leads to discrete jump of the magnetic filed. If � � 0 � 0, there are magnetic islands at x=0. 4
The Time Independent Equilibrium Solutions First, consider the MHD equilibrium equations given by � � � � ��, � � � � 4��, � · � � 0. From the curl of the force balance equation � � � � With boundary condition � � � � � � �� � � � �, � · �� ��� � � � · �� � � � � · �� � � 0, �� � � � � � � � � � � � 0. → it has mirror symmetry � � �� � � � � . Two class of equilibrium solutions are possible. Equilibrium (I) No island formation � � � � � � � sinh �� /sinh �� , � � 0 � 0, � � � �� � � � � �� cosh �� � � � �0 � � � �0 � 2� � �� / sinh �� �� � � � sinh �� . Here � � has a finite jump corresponding to a surface current. Equilibrium (II) � � 0 is finite so the topology � � � � � � � cosh �� / cosh �� . � � 0 � � � �/ cosh �� , has changed and equilibrium (II) has magnetic islands. � � � �� � � � � � �� sinh �� �� � � � cosh �� . � � �0 � � � �0 � 0 There is no surface current since � � is continuous at � � 0 . A general solution is combination of Equilibrium (I) and Equilibrium (II) cosh �� � sinh �� � � � � sinh �� � � � � � � 0 sinh �� . tanh �� 5
The Time Evolution of � Recall the equilibrium solution ���� �� ���� �� ���� �� . � � � � � � 0 cosh �� � � � � � ���� �� time scale for analysis � � � 4�� � Resistive time scale � ≪ � ��/� � ��/� phases A, B � � ~ � ��/� � ��/� phase C � � � �4��� �/� � Hydromagnetic time scale phase D � ≫ � ��/� � ��/� � � The resistive MHD equations which will be used for analysis � where � �� � � · � � � � � · �� � , � � � �̂ · � � � � � � �, � � �̂ � ��, �� � � � · �� � � � � � � � sin �� , � = � � cos �� . 4� � � �. 6
Ideal MHD Solution (A) We linearize the resistive MHD equations and neglect the resistivity because it takes time for the resistivity to play a role in dynamics, then we obtain these equations � � � � � � � � �� � � � � � �� � � �� � � ��� � �� � � ���� � 0, �� � � � � . �� 4� � Combine these equation into � � � �� � � � �� � � � � � � � � � � /� . � � �� � ��, where �� � A similarity solution, � �, � � ����� . It is invariant under transformation � → ��, � → � �� � . ���/� � � � 2 � � �� �� sin � sinh �� � ; � ≪ �, � � � Physical meaning of a similarity solution 4�� �� � � � �� � � � � �� � � � � 2 �� Current distribution get more peaked and confined within the �� � � narrower width at the center, asymptotes to a singular sheet. � � �� � 4�� �� 0 � 4 � , �� ~ � � /� � sinh �� � � 7
Transition from Ideal MHD to Resistive Evolution (B) After current is confined within a layer which is narrow enough at the center, resistivity begins to act. � � � � � ���� � � �� � � ��� � �� � �� � � � ���, 4� But, in the early phase of resistive evolution, � � 0 and � � profiles do not deviate much from the ideal solution. � � � � �� � �� � � 0 � � � �� � � 0 � � � , � (0) from ideal solution in phase A 4� � � sinh �� � � �� � � � /� ). � � �� � � � � � � � 0 � 2 , � ∝ �. (half width of magnetic islands) ∝ t � sinh �� � � � � 8
Resistive Evolution (C) (NONCONSTANT � PHASE) Asymptotic matching consists of: �→� ���� �������� ��������� � lim lim �→� ���� �������� ��������� By using Laplace transformation and introducing dimensionless variables, we obtain � � � � �� � � � 1 Magnetic induction equation Vorticity equation �� � Ψ � �Ω 4Ψ � �� , 4 � � � � �Ψ. � ≡ � � � ��� � � � � � � � � � 4� � �, Combine these equation into where �� � Ψ � �Ω 4Ψ � �� . � �� � � � �Ψ � � � � The interior solution as � → ∞ is lim �� � � �� � Ψ � 2 �/� �� Ψ�0� � �� � �̅, where � Ψ�0��̅. 2 �→� � � ���� �� ���� �� The exterior solution could be obtained from ���� �� . � � � � � � 0 cosh �� � ���� �� � � � � � � � � �Ψ 0 lim �� Ψ � � sinh �� � tanh �� . The exterior solution as � → 0 is �→� � � � � 1 Thus, we obtain a solution where � � � � �� � �̅. Ψ 0 � � �/ 2 �/� �� ���� � , sinh �� �/ tanh �� � To show that phase B is subset of phase C for small t, let � → 0, � → ∞ . � � �� � � � � � � � 0 � 2 � sinh �� � � � � 9
Constant Approximation (D) �Δ � ≪ 1, The constant � approximation. � ≫ � ��/� � ��/� � � �/� � � 0 � � cosh �� � Δ � /2� sinh �� . Interior solution is � �� � � /�� � � Δ � � 1 � 12Ω 3 �/� �/� �/� � � �� � � 2�� �/� � �� � . ��0� � �� � � can be reduced to a real form and evaluated numerically. 8�2� �/� � �/� � � � Γ 3 �/� � � �/� , � � 0, � ~ � , � ≪ 1 � � �/� � cosh �� 5� 4 � � � 2 �/� � Γ 5 � 4 � ��/� , � � 0, � ~ cosh �� 1 � � ≫ 1 An important point is that a time which is long compared to the tearing mode time scale � ��/� � ��/� is required to reach equilibrium (II). 10
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