PHYS 777 Plasma Physics and Magnetohydrodynamics 2004 Fall Instructor: Dr. Haimin Wang

Lecture 3

Magneto hydrostatics ρ

Force Equation (2-21) ρ ∇vv = −∇Pv + vj × Bv + gv ∇t ∇vv magnetohydρ rostatics means = 0 ∇t v µ − ∇P + vj × B + g = 0 v v ∇× B v mP j = ∇ ⋅ ρB = 0 = kT define a coordinate along magnetic field lines (s) v dP δ vj × B = 0 ρ− = g cos θ δ ds θ dP s cos = z = − g dzρ 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 ifρ structure height Scale height term can be ignored β 2µP if = 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 ∂ µB2 0 = − (P + ) solutions are same as above ∂µx 2 B2 dP P + = f (z) = −ρg density is not a function of x 2 dz Horizontal field B = B(z)xr d µ Bρ 2 0 = − (P + ) − g dz 2 2 z z / Λ d B −z / Λ P = [P0 − e ( )dz]e ∫0 dzµ2 ρ −z / Λ B ρ −z / Λ B −z / 2Λ B P = P0e = 0e B = B0e 2 µ Λ B = (P0 + B0 / 2 ) / ρ0 g modified scale height P6/18 P7/18 Cylindricallyµ Symmetric Fields µ r B = (0, BΦ (R), BZ (R)) r 1 dB 1 d j = (0,− Z , (RB )) dR R dR Φ ignore g inµ force equation µ dP d B2 + B2 B2 + ( Φ Z ) + Φ = 0 (3-16) dR dR 2 R 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 BZ (R) Φ Pureµ Axial Fields π  IR Ra 2 R

 1µ I 2 2 2 P∞ + ( ) (a − R ) Ra P8/18 Rtension 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

π R0 I = j 2 RdR ∫0 z R N =π 0 n2 RdR ∫0

R0 R0 RB R2dP = − Φ d(RB ) ∫0 ∫0 Φ µ P T = nK π8 I 2 = KTN —— Bennet’s Relation µ µ r r Force – Free Field j = α B • linear force free field d B2 + B2 B2 ( Φ Z ) + Φ = 0 (3 −16, P = 0) dR 2 R if α is constant,µ µ —— linear force free r µ r ∇× B dB α j = α − Z = B dR Φ

BΦ = B0 J1( R) 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 = B 2 − B 2 f = gives pure azimuthal fields z Φ R 2 P10/18 Another example: Uniform - twist field B ΦR / 2L B = 0 Φ 1+ Φ 2 R2 /(2L)2 B B = 0 Z 1+ Φ 2 R2 /(2L)2 Magnetostatic Fields µ dP d B 2 + B 2 B 2 + (µ Φ Z ) + Φ = 0 dR dR 2 R one simple solution - - - uniform twist B ΦRB B = 0 B = Z Z 1+ R 2 / a 2 Φ 2L [Φa 2 /(2L)2 −1]B 2 P(R) = P + 0 ∞ (1+ R 2 / a 2 )2 µ Another simple solution - - - non - uniform twist

BZ = B0

2LBΦ Φ 0 Φ(R) = = 2 2 RB0 1+ R / a Φ 2B 2a 2 P(R) = P + 0 ∞ 8µL2 B Effect of Expanding a tube : Φ increases. BZ P11/18 P12/18 µ

Current – Free Fields (potential fields) r r J = 0 = ∇× B / 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. α Force-Free Fields α r r r r r ∇× B force balanceα j × B = 0 j µ= r r ∇× B = B = 0 , current - free take the divergence r r (B ⋅∇) = 0 B lies on surface of constant α P13/18 P14/18 α when α is constant - - linear force free α r r r r r ∇×(∇× B) = ∇× B r − ∇2 B = 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 µ B2 (xB + yB )B dV = x y z dxdy ∫V 2 ∫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 =αB0Cos x other more complicated form −lz l Bx = A1 Coskx e A1 = − B0  −lz k By = A2 Coskx e 2 l 1/ 2  −lz A2 = −(1− ) B0 Bz = B0 Sinkx e K 2 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 2 ∂ 1 ∂ ∂ d 1 2 ( 2 − + 2 )A + ( b ) = 0 ∂R R ∂R ∂z dA 2φ 1 ∂A b 1 ∂A B = − B φ = φ B = R R ∂z R z R ∂R Magnetic Diffusion ηr ∂B r r 2 α = ∇×(V × B) + ∇ B ∂t r r r ∇× B = B ∇ ⋅ B = 0 (B ⋅∇) = 0 if medium is stationary V ≡ 0, & = const r αr then α ∇2 B = − 2 B ηα r α ∂B r = − ηα2 B → diffusion ∂t − 2t −ηα 2t B = B0 e j = j0 e For non - constant , Sample solution : α ∂ By = Bφ0Cos Bz = B0Sin = − φ ∂x and V are determined by φ 2 φ ∂ φ∂ ∂ α η− +Vx = 0 φ ∂t ∂x2 φ∂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