The Linear Theory of Tearing Modes in periodic, cyindrical plasmas Cary Forest University of Wisconsin 1
Resistive MHD E + v B = ηj (no energy principle) Role of resistivity No frozen flux, B can tear If fluid moves relative to B then ηj = v B J = v B η F = J B = = B2 η v (v B) B η Restoring force B v Vanishingly small eta can allow tearing when B->0! 2
Resistive MHD B t Same energy source as in ideal MHD, but more states are accessible Dimensionless measure of resistivity ( B) = E = v B η µ 0 LHS resistive term B/τ A ηb/a 2 µ 0 = µ 0a 2 /η τ A = τ R τa S τ R = 6.5 10 B θ,teslat 7 τ A n 1/2 20 3/2 e,kev a m, and τ R µ 0a 2 η 50 T 3/2 e,kev Z eff sec S=10 5-10 8 ; although S>>1, its finiteness permits topological changes and new instabilities 3
Experimental Importance Disruptions beta limits Stochastic magnetic fields Realization of reconnection phenomena 4
Consider torus with helical field B = B φ ˆφ + Bθ ˆθ { { toroidal poloidal B Field lines lie on circles 5
Magnetic Surfaces B Ideally, concentric tori In general, there is shear 6
Add a small perturbation in radial magnetic field: b r with b rb 1 Possible sources of perturbations:! -instability! -magnetic field error! -deliberate additional field How can a small perturbation have a large effect on the field structure? 7
If k = 0 then small magnetic fluctuation gives large field line excursion b r B 8
add a perturbation (e.g., an instability) B = B(r) + b r (r) sin(mθ nφ)ˆr Thus, wavenumber consider region near k = m r ˆθ n R ˆφ k = 0 where k B = m r B θ n R B φ = 0 or where q(r) = m n where q(r) = rb φ RB θ 9
Define a coordinate χ, perpendicular to B,ˆr: χ = mθ nφ so that B r = b r (r) sin χ Perturbation is constant along B at one radius χ B φ θ r 10
Field line equation dr = h χdχ B r B dr = h χdχ b r (r) sin χ B r rdr b r = h χ dχ B sin χ r = ± h χ b r cos χ + C 4 B 11
Field lines without perturbations r = constant radius χ 12
Field lines with perturbation r = ± h χ b r 4 B cos χ + C Reconnection has occurred: w = 2 hχ b r B 13
Field lines respond to resonant perturbation in torus q=3 surface n=1, m=3 perturbation: b r = b cos(3θ φ) 14
SXR imaging of m=3 islands in tokamak 15
SXR imaging of flux surfaces With one dominant tearing instability From a reversed field pinch experiment (similar to a tokamak) 16
SXR tomography with 2 tearing modes 17
Disruptions can be due to overlapping islands Asdex-U tokamak 18
References Furth, Killeen and Rosenbluth, Finite Resistivity instabilities of a sheet pinch, Phys. Fluids 6, 459 (1963). Furth, Rutherford and Selberg, Tearing mode in the cylindrical tokamak, Phys. Fluids 16, 1054 (1973). Coppi, Greene and Johnson, Nucl. Fusion 6, 101 (1966). Tokamaks, 3rd ed. Wesson Theory of Tokamak Plasmas, White. 19
Cylindrical Tearing Mode: Linear Theory 1. Solve for magnetic perturbation in outer region: use force balance, ignoring inertia Use ideal MHD, ignoring resistivity 2.Outer solution determines stability, drives reconnection 3.Inner solution solves resistive MHD equations for narrow current sheet Inertia, resistivity both important 4.Matching inner solution to step in B r determines growth rate 20
Outer region! Ion Equation of motion ρ dv dt = J B P in the outer region, inertia is negligible, and plasma is in force equilibrium P = J B and since the curl of a gradient is zero J B = 0 Using B = 0 and J = 0 B J J B = 0 21
Reduced MHD Ordering The large aspect-ratio, reduced MHD ordering will be used. Taking ɛ = a/r 1, the equililibrium quantities are ordered as B θ 1 r db φ dr ɛb φ, and J θ ɛj φ while the perturbed quantities are ordered as b φ ɛb r ɛb θ, and j θ ɛj φ Use the toroidal component. Since J B B J, the equation governing the outer region is B j φ + b J φ = 0 22
Representation of perturbed field Use a vector potential (flux function) to represent the perturbed flux b = a with a = ψ ˆφ : so that b r = 1 r ψ θ, b θ = ψ r which can be related (by Ampere s law) to the perturbed toroidal current density: µ 0 j φ = 2 ψ = 1 r r r ψ r + 1 2 ψ r 2 θ 2 Assuming that the perturbations are double periodic and have the form e i(mθ nφ) b r = i m r ψ 23
Equation governing outer solution Inserting j φ into the governing equation, and noting that J φ only varies in the radial direction so that b J φ = i m dj φ r dr ψ gives F µ 0 2 ψ m r where F (r) = k B = m r B θ(r) nb φ R Equivalently, dj φ dr ψ = 0, = m r B θ(r) ( ) 1 nq(r). m 1 r d dr r dψ dr = m2 r 2 ψ + mµ 0 rf dj φ dr ψ This equation is well-behaved and solvable, except where F = 0 (resonance). 24
Independently solve ψ in two regions: I) 0 < r < r s and II) r s < r < a where r s is the resonant surface where F (r s ) = 0. One way to do this is by by integrating the coupled first order equations from boundaries into resonant surface: d dr ψ = m2 r 2 ψ ψ d dr ψ = ψ r + mµ 0 rf dj φ dr ψ Matching conditions: b r (ψ) continuous, but db r dr (ψ ) is discontinuous. Matching to inner solution will determine growth rate. This discontinuity is quantified by = lim ɛ 0 [ ψ ψ rs +ɛ ψ ψ ] rs ɛ. 25
HOMEWORK: Numerically calculate for tokamak-like profiles of q rb φ RB θ (using Excel, IDL or your favorite programming language). Begin by assuming a current profile of the form ( ) ν j φ = j 0 1 r2 a 2 where ν can be related to the safety factor at the axis q 0 and at the boundary q a : ν = q a q0 1. Typically, q 0 1 and q a 3 5. For this current profile, B θ (r) = µ 0j 0 a 2 2r (ν + 1) ( ) ν 1 ) 1 (1 r2. For these profiles, begin by computing defining a radial grid array r i = i r that covers the range 0 r a with several hundred discrete radii. Then compute the profiles of F (r) = k B for an m = 2, n = 1 tearing mode. Plot it, and then determine the resonant surface r s where F (r s ) = 0. Next, solve for the perturbed flux in the two regions I) 0 < r < r s and II) r s < r < a. Do this is by by integrating the coupled first order equations ψ i+1 = ψ i + r ψ i+1 = ψ i + r ψ i ( m 2 r 2 i a 2 ψ i ψ i r i + mµ 0 r i F (r i ) dj φ dr ) ψ i i from boundaries into the resonant surface. For the inside solution, carry out this integration by starting the couple equations with the boundary conditions ψ(r = 0) = 0 (regularity), ψ (r = 0) = C and integrate outward to r = r s ɛ. Use a similar scheme to integrate the equations backwards from the outer wall at r = a to r = r s +ɛ (assuming that ψ(r = a) = 0 due to a conducting wall). Finally, take your two solutions for the inside and outside regions and scale them to match amplitudes on each side of the resonant surface. Finally, from the scaled solutions, find the derivatives on either side of the resonance to compute. = lim ɛ 0 [ ψ ψ rs +ɛ ψ ψ rs ɛ ] 26
Solution 4 3 2 1 q 0 0.0 0.1 0.2 0.3 0.4 0.5 120 100 80 60 40 j [Amps/cm 2 ] 20 0 0.0 0.1 0.2 0.3 0.4 0.5 0-100 -200-300 djdr -400 0.0 0.1 0.2 0.3 0.4 0.5 B=1 T R=1.5 m a=0.5 m btheta [T] 0.14 0.12 0.10 0.08 0.06 0.04 0.02 0.00 0.0 0.1 0.2 0.3 0.4 0.5 1.5 1.0 0.5 0.0 k. B -0.5 0.0 0.1 0.2 0.3 0.4 0.5 2.0 1.5 1.0 0.5 psi 0.0 0.0 0.1 0.2 0.3 0.4 0.5 m=2,n=1 q a =3.5 q 0 =1.0 r s = 2.8 27
Perturbed flux function 28
Resistive Layer In the resistive layer, ideal MHD breaks down and two key assumptions fail: the neglect of resistivity in Ohm s law and and the neglect of inertia in the force balance equation. Combining Ohm s law, E + v B = µ 0 J with Faraday s law and Ampere s law gives the magnetic diffusion equation: B t = v B η 2 B µ 0 29
Take the radial component of this equation to give b r t = B v r + η 2 b r µ 0 Assuming that the instability time dependence e γt and that the growth rate γ is real, b r = imψ/r and that r is constant in the layer, gives γψ + B θ (1 nq m )v r = η 2 ψ µ 0 The resistive layer is so narrow that radial derivatives in 2 will dominate, so that d 2 ψ dr 2 = µ 0γ η ψ + µ 0B θ η (1 nq m )v r 30
Assuming b r does not vary much (constant ψ), integrating across layer gives jump in : [ 1 dψ inner = ψ(r s ) dr dψ ] rs +ɛ dr rs ɛ = µ ( 0γ 1 + B ) θ nq(r) (1 η γ m ) v r dr ψ(r s ) Furthermore, we can make a linear expansion of F about the resonance surface so that 1 q(r)n m = q q (r r s ) rs so that = µ 0γ η (1 + B θq γq (r r s) ) v r ψ(r s ) dr 31
Equation of motion In the vicinity of the island, inertia must be taken into account to find v r. The φ component of the inertial term is ˆφ ρ dv dt = γρ v θ r = γρ ir m 2 v r r 2 where incompressibility has been used to relate v r and v θ. Including this inertial term in the force balance equation yields γρr 2 m 2 d 2 v r dr 2 = B θ µ 0 ( 1 nq(r) m ) d 2 ψ dr 2 dj φ dr ψ 32
Using the result from Ohm s law that relates d2 ψ dr to v 2 r, d 2 ( v r B 2 dr 2 θ m 2 q 2 ) ργηr 2 q 2 (r r s ) 2 v r = B θm 2 q ρηr 2 q (r r s) ψ m2 dj φ ργr 2 dr ψ Dimensional analysis gives a characteristic scale length δ = ( ρηγr 2 s q 2 B 2 θ m2 q 2 ) 1/4, = (γτ R) 1/4 S 1/2 ( r 2 s L 2 q a 4 ) 1/4 a (γτ R) 1/4 a S 1/2 33
The equation of motion for v r /ψ can be recast as d 2 y dy 2 x2 y = x C, where x = (r r s )/δ, y = ργr2 s q v r m 2 B θ q δ 3 ψ and a inner = γτ R δ a + (1 xy)dx Notes: Only odd part of y contributes to integral; y = x(1 xy) can be solved numerically to find y. Solution for y can be integrated to find a inner = 2.12γτ R δ a 34
Expressing δ in terms of γτ R results in or equivalently a inner = 2.12 (γτ R) 5/4 S 1/2 ( r 2 s L 2 q a 4 ) 1/4, γτ R = 0.55S 2/5 ( a 4 r 2 sl 2 q ) 1/5 (a outer) 4/5 Express δ as δ a = outer) 1/5 0.86(a S 2/5 ( a 4 r 2 sl 2 q ) 1/20 35
Summary Finite resistivity allows topology to change, and islands to form Outer solution determines stability > 0 Unstable Growth rate is a hybrid between resistive and Alfven growth rates: γ (a outer) 4/5 τ 3/5 R τ 2/5 A 36