Magneto Hydrostatics Ρ

PHYS 777 Plasma Physics and Magnetohydrodynamics 2004 Fall Instructor: Dr. Haimin Wang Lecture 3 Magneto hydrostatics Force Equation (2-21) ∇vv v v ρ = −∇P + vj × B + ρgv ∇t ∇vv magnetohydrostatics means = 0 ∇t v − ∇P + vj × B + ρg = 0 v v ∇× B v mP j = ∇⋅ B = 0 ρ = µ kT define a coordinate along magnetic field lines (s) v dP vj × B = 0 − = ρg cosθ ds dP δs cosθ = δz = −ρg dz z dz P = Po exp Po = P(z = 0) ∫0 Λ(z) kT(z) P Λ(z) = = Scale Height mg ρg P2/18 − z / Λ if Λ is a constant P = Po e r 2 g = g Θ g = 274 m / s 2 Θ r 2 Θ 8 rΘ = 6 .96 × 10 m r Λ = 50 T ( ) 2 m rΘ P3/18 P4/18 4 e.g. T =10 k r = rΘ Λ = 500km r T =106 k Λ = 5×107 ( )2 m rΘ if structure height Scale height ρ term can be ignored 2µP if β = 0 1, pressure term can be dropped as well B0 v v then we have j× B = 0 force- free condition P5/18 pure vertical fields ∂ B2 0 = − (P + ) solutions are same as above ∂x 2µ B2 dP P + = f (z) = −ρg density is not a function of x 2µ dz Horizontal field B = B(z)xr d B2 0 = − (P + ) − ρg dz 2µ 2 z z / Λ d B −z / Λ P = [P0 − e ( )dz]e ∫0 dz 2µ −z / Λ B −z / Λ B −z / 2Λ B P = P0e ρ = ρ0e B = B0e 2 Λ B = (P0 + B0 / 2µ) / ρ0 g modified scale height P6/18 P7/18 Cylindrically Symmetric Fields r B = (0, BΦ (R), BZ (R)) r 1 dB 1 d j = (0,− Z , (RB )) µ dR µR dR Φ ignore g in force equation dP d B2 + B2 B2 + ( Φ Z ) + Φ = 0 (3-16) dR dR 2µ µR Field lines are given by RdΦ dZ = BΦ BZ 2L B twist Φ = ∫∫dΦ = Φ dZ 0 RBZ B 2L 4πL Φ(R) = Φ : pitch of field R B Z (R) Φ Pure Axial Fields µIR R<a 2πa2 BΦ = µI R>a 2πR 1 I 2 2 2 P∞ + µ( ) (a − R ) R<a P = 4 πa2 P∞ R>a P8/18 R<a gas pressure balances with magnetic pressure and tension R>a gas pressure balances with tension P9/18 In lab, this configuration is called linear Pitch Relationship beween current I and number of particles / unit length R0 I = jz 2πRdR ∫0 R N = 0 n2πRdR ∫0 R R 0 2 0 RBΦ R dP = − d(RBΦ ) ∫0 ∫0 µ P T = nK 8π I 2 = KTN —— Bennet’s Relation µ r r Force – Free Field µ j = α B • linear force free field d B2 + B2 B2 ( Φ Z ) + Φ = 0 (3−16, P = 0) dR 2µ µR if α is constant, —— linear force free r r ∇× B dB j = − Z = αB µ dR Φ BΦ = B0 J1(αR) BZ = B0 J 0 (αR) J 0 and J1 are Bessel functions 1 df • Non - linear if B = f(R) B2 = − R Φ 2 dR 1 B2 = B2 − B2 f = gives pure azimuthal fields z Φ R 2 P10/18 Another example: Uniform - twist field B ΦR / 2L B = 0 Φ 1+ Φ2 R2 /(2L)2 B B = 0 Z 1+ Φ2 R2 /(2L)2 Magnetostatic Fields dP d B2 + B2 B2 + ( Φ Z ) + Φ = 0 dR dR 2µ Rµ one simple solution - - - uniform twist B ΦRB B = 0 B = Z Z 1+ R2 / a2 Φ 2L [Φa2 /(2L)2 −1]B2 P(R) = P + 0 ∞ (1+ R2 / a2 )2µ Another simple solution - - - non - uniform twist BZ = B0 2LBΦ Φ0 Φ(R) = = 2 2 RB0 1+ R / a Φ2 B2a2 P(R) = P + 0 ∞ 8µL2 B Effect of Expanding a tube : Φ increases. BZ P11/18 P12/18 Current – Free Fields (potential fields) r r J = 0 = ∇× B / µ r r ∇2 B = 0 B = ∇Ψ Ψ scalar magnetic potential ∇2Ψ = 0 - - - Laplace Equation this is a very likely situation on the Sun general solution ∞ l l −(l+1) m imφ Ψ = ∑∑(almr + blmr )Pl (Cosθ )e l=−0 m= l In term of associated Legendre Polynomial Constants are determined by boundary conditions. Many codes have been developed. e.g. Fig 3.7. Use observed surface magnetic fields as boundary condition. Force-Free Fields r r r r r ∇× B force balance j × B = 0 j = µ r r ∇× B = αB α = 0 , current - free take the divergence r r (B ⋅∇)α = 0 B lies on surface of constant α P13/18 P14/18 when α is constant - - linear force free r r r r r ∇×(∇× B) = α∇× B r − ∇2 B = α 2 (∇2 +α 2 )B = 0 General remarks It can be shown that (Cowling 1976) if fields have minimum energy, must be force free. However, force-free does not mean minimum energy. Virial theorem B2 (xB + yB )B dV = x y z dxdy ∫V 2µ ∫s µ r if one knows B in a surface, magnetic energy can be derived for a 3- D space r r r if j × B = 0 B ≡ 0 force- free field must be anchored down some where in surface An axisymmetric force free poloidal field must be current free P15/18 Simple constant α force free solutions r Simplest solution B = (0, By (x), Bz (x)) 2 2 2 By + Bz = B0 By = B0Sindx Bz = B0Cosαx other more complicated form B = A Coskx e−lz l x 1 A1 = − B0 −lz k By = A2 Coskx e 2 l 1/ 2 −lz A2 = −(1− 2 ) B0 Bz = B0 Sinkx e K in cylindrical polars (R,φ, Z) l B = B J (KR) e−lz R k 0 1 2 l 1/ 2 −lz Bφ = (1− 2 ) B0 J1(KR) e K −lz J: Bessel function Bz = B0 J0 (KR) e Non-constant α force free fields r r r j × B = 0, ∇⋅ B = 0 difficult non - linear equations r one approximation. B independent of y. ∂A ∂A B = B B = − A : flux function x ∂z y z ∂x ∂A ∂B ∂B ∂A ∂B ∂A ∇2 A + B y = 0 y − y = 0 ∂x y ∂x ∂z ∂x ∂x ∂z ∂A ∂B d 1 ∇2 A + B y = 0 ∇2 A + ( B2 ) = 0 P16/18 ∂z y ∂z dA 2 z P17/18 in cylindrical polar system ∂2 1 ∂ ∂2 d 1 ( − + )A + ( b2 ) = 0 ∂R2 R ∂R ∂z 2 dA 2 φ 1 ∂A b 1 ∂A B = − B = φ B = R R ∂z φ R z R ∂R Magnetic Diffusion r ∂B r r = ∇×(V × B) +η∇2 B ∂t r r r ∇× B = αB ∇ ⋅ B = 0 (B ⋅∇)α = 0 if medium is stationary V ≡ 0, & α = const r r then ∇2 B = −α 2 B r ∂B r = −ηα 2 B → diffusion ∂t −ηα 2t −ηα 2t B = B0 e j = j0 e For non - constant α, Sample solution : ∂φ B = B Cosφ B = B Sinφ α = − y 0 z 0 ∂x φ and V are determined by ∂φ ∂ 2φ ∂φ −η +V = 0 ∂t ∂x2 x ∂x ∂φ ∂V η( )2 + x = 0 ∂x ∂x P18/18 Homework 1. Estimate the scale heights in solar photosphere, transition region and corona. 2. A vertical magnetic flux tube should expand from photosphere to corona. If 1% of photospheric area is occupied by the magnetic fields, at what height, 100% of the surface area is occupied by magnetic fields? 3. Derive Bennett’s relation, starting from fundamental magnetohydrostatics equation with cylindrical symmetry. P19/18.

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